Representation Theory: A First Course (Graduate Texts in Mathematics Readings in Mathematics)

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William Fulton Joe Harris

Representation Theory A First Course

With 144 Illustrations

Springer-Verlag New York Berlin Heidelberg London Paris Tokyo Hong Kong Barcelona Budapest

Preface

The primary goal or these lectures is to introduce a beginner to the finitedimensional representations of Lie groups and Lie algebras. Since this goal is shared by quite a few other books, we should explain in this Preface how our approach differs, although the potential reader can probably see this better by a quick browse through the book. Representation theory is simple to define: it is the study of the ways in which a given group may act on vector spaces. It is almost certainly unique, however, among such clearly delineated subjects, in the breadth of its interest to mathematicians. This is not surprising: group actions are ubiquitous in 20th century mathematics, and where the object on which a group acts is not a vector space, we have learned to replace it by one that is (e.g., a cohomology group, tangent space, etc.). As a consequence, many mathematicians other than specialists in the field (or even those who think they might want to be) come in contact with the subject in various ways. It is for such people that this text is designed. To put it another way, we intend this as a book for beginners to learn from and not as a reference. This idea essentially determines the choice of material covered here. As simple as is the definition of representation theory given above, it fragments considerably when we try to get more specific. For a start, what kind of group G are we dealing with-a finite group like the symmetric group G. or the general linear group over a finite field GL.(f4 ), an infinite discrete group like SL.(Z), a Lie group like SL.C, or possibly a Lie group over a local field? Needless to say, each of these settings requires a substantially different approach to its representation theory. Likewise, what sort of vector space is G acting on: is it over C, R, Q, or possibly a field of finite characteristic? Is it finite dimensional or infinite dimensional, and if the latter, what additional structure(such as norm, or inner product) does it carry? Various combinations

vi

Prerace

of answers to these questions lead to areas of intense research activity in representation theory, and it is natural for a text intended to prepare students for a career in the subject to lead up to one or more of these areas. As a corollary, such a book tends to get through the elementary material as quickly as possible: if one has a semester to get up to and through Harish-Chandra modules, there is little time to dawdle over the representations of G 4 and SL3 C. By contrast, the present book focuses exactly on the simplest cases: representations of finite groups and Lie groups on finite-dimensional real and complex vector spaces. This is in some sense the common ground of the subject, the area that is the object of most of the interest in representation theory coming from outside. The intent of this book to serve nonspecialists likewise dictates to some degree our approach to the material we do cover. Probably the main feature of our presentation is that we concentrate on examples, developing the general theory sparingly, and then mainly as a useful and unifying language to describe phenomena already encountered in concrete cases. By the same token, we for the most part introduce theoretical notions when and where they are useful for analyzing concrete situations, postponing as long as possible those notions that are used mainly for proving general theorems. Finally, our goal of making the book accessible to outsiders accounts in part for the style of the writing. These lectures have grown from courses of the second author in 1984 and 1987, and we have attempted to keep the informal style of these lectures. Thus there is almost no attempt at efficiency: where it seems to make sense from a didactic point of view, we work out many special cases of an idea by hand before proving the general case; and we cheerrully give several proors of one fact if we think they are illuminating. Similarly, while it is common to develop the whole semisimple story from one point of view, say that of compact groups, or Lie algebras, or algebraic groups, we have avoided this, as efficient as it may be. It is of course not a strikingly original notion that beginners can best learn about a subject by working through examples, with general machinery only introduced slowly and as the need arises, but it seems particularly appropriate here. In most subjects such an approach means one has a few! out of an unknown infinity of examples which are useful to illuminate fh~ general situation. When the subject is the representation theory of comple~ se'misimple Lie groups and algebras, however, something special happens: once one has worked through all the examples readily at hand-the "classical" cases of the special linear, orthogonal, and symplectic groups-one has not just a few useful examples, one has all but live "exceptional" cases. This is essentially what we do here. We start with a quick tour through representation theory of finite groups, with emphasis determined by what is userul for Lie groups. In this regard, we include more on the symmetric groups than is usual. Then we turn to Lie groups and Lie algebras. After some preliminaries and a look at low-dimensional examples, and one lecture with

Preface

vii

some general notions about semisimplicity, we get to the heart of the course: working out the finite-dimensional representations of the classical groups. For each series of classical Lie algebras we prove the fundamental existence theorem for representations of given highest weight by explicit construction. Our object, however, is not just existence, but to see the representations in action, to see geometric implications of decompositions of naturally occurring representations, and to see the relations among them caused by coincidences between the Lie algebras. The goal of the last six lectures is to make a bridge between the exampleoriented approach of the earlier parts and the general theory. Here we make an attempt to interpret what has gone before in abstract terms, trying to make connections with modern terminology. We develop the general theory enough to see that we have studied all the simple complex Lie algebras with five exceptions. Since these are encountered less frequently than the classical series, it is probably not reasonable in a first course to work out their representations as explicitly, although we do carry this out for one of them. We also prove the general Weyl character formula, which can be used to verify and extend many of the results we worked out by hand earlier in the book. Of course, the point we reach hardly touches the current state of affairs in Lie theory, but we hope it is enough to keep the reader's eyes from glazing over when confronted with a lecture that begins: "Let G be a semisimple Lie group, P a parabolic subgroup, ... " We might also hope that working through this book would prepare some readers to appreciate the elegance (and efficiency) of the abstract approach. In spirit this book is probably closer to Weyl's classic [Wet] than to others written today. Indeed, a secondary goal of our book is to present many of the results of Weyl and his predecessors in a form more accessible to modern readers. In particular, we include Weyl's constructions of the representations of the general and special linear groups by using Young's symmetrizers; and we invoke a little invariant theory to do the corresponding result for the orthogonal and symplectic groups. We also include Weyl's formulas for the characters of these representations in terms of the elementary characters of symmetric powers of the standard representations. (Interestingly, Weyl only gave the corresponding formulas in terms of the exterior powers for the general linear group. The corresponding formulas for the orthogonal and symplectic groups were only given recently by Koike and Terada. We include a simple new proof of these determinantal formulas.) More about individual sections can be found in the introductions to other parts of the book. Needless to say, a price is paid for the inefficiency and restricted focus of these notes. The most obvious is a lot of omitted material: for example, we include little on the basic topological, differentiable, or analytic properties of Lie groups, as this plays a small role in our story and is well covered in dozens of other sources, including many graduate texts on manifolds. Moreover, there are no infinite-dimensional representations, no Harish-Chandra or Verma

viii

Preface

modules, no Steifel diagrams, no Lie algebra cohomology, no analysis on symmetric spaces or groups, no arithmetic groups or automorphic forms, and nothing about representations in characteristic p > 0. There is no consistent attempt to indicate which of our results on Lie groups apply more generally to algebraic groups over fields other than R or C (e.g., local fields). And there is only passing mention of other standard topics, such as universal enveloping algebras or Bruhat decompositions, which have become standard tools of representation theory. (Experts who saw drafts of this book agreed that some topic we omitted must not be left out of a modern book on representation theory-but no two experts suggested the same topic.) We have not tried to trace the history of the subjects treated, or assign credit, or to attribute ideas to original sources-this is far beyond our knowledge. When we give references, we have simply tried to send the reader to sources that are as readable as possible for one knowing what is written here. A good systematic reference for the finite-group material, including proofs of the results we leave out, is Serre [Se2]. For Lie groups and Lie algebras, Serre [Se3], Adams [Ad], Humphreys [Hu I], and Bourbaki [Bour] are recommended references, as are the classics Weyl [Wei] and Littlewood [Lit 1]. We would like to thank the many people who have contributed ideas and suggestions for this manuscript, among themJ-F. Bueno!, R. Bryant, J. Carrell, B. Conrad, P. Diaconis, D. Eisenbud, D. Goldstein, M. Green, P. Griffiths, B. Gross, M. Hildebrand, R. Howe, H. Kraft, A. Landman, B. Mazur, N. Chriss, D. Petersen, G. Schwartz, J. Towber, and L. Tu. In particular, we would like to thank David Mumford, from whom we learned much of what we know about the subject, and whose ideas are very much in evidence in this book. Had this book been written 10 years ago, we would at this point thank the people who typed it. That being no longer applicable, perhaps we should thank instead the National Science Foundation, the University of Chicago, and Harvard University for generously providing the various Macintoshes on which this manuscript was produced. Finally, we thank Chan Fulton for making the drawings. Bill Fulton anp foe Harris

Using This Book

A few words are in order about the practical use of this book. To begin with, prerequisites are minimal: we assume only a basic knowledge of standard first-year graduate material in algebra and topology, including basic notions about manifolds. A good undergraduate background should be more than enough for most of the text; some examples and exercises, and some of the discussion in Part IV may refer to more advanced topics, but these can readily be skipped. Probably the main practical requirement is a good working knowledge of multilinear algebra, including tensor, exterior, and symmetric products of finite dimensional vector spaces, for which Appendix 8 may help. We have indicated, in introductory remarks to each lecture, when any background beyond this is assumed and how essential it is. For a course, this book could be used in two ways. First, there are a number of topics that are not logically essential to the rest of the book and that can be skimmed or skipped entirely. For example, in a minimal reading one could skip §§4, 5, 6, 11.3, 13.4, 15.3-15.5, 17.3, 19.5, 20, 22.1, 22.3, 23.3-23.4, 25.3, and 26.2; this might be suitable for a basic one-semester course. On the other hand, in a year-long course it should be possible to work through as much of the material as background and/or interest suggested. Most of the material in the Appendices is relevant only to such a long course. Again, we have tried to indicate, in the introductory remarks in each lecture, which topics are inessential and may be omitted. Another aspect of the book that readers may want to approach in different ways is the profusion of examples. These are put in largely for didactic reasons: we feel that this is the sort of material that can best be understood by gaining some direct hands-on experience with the objects involved. For the most part, however, they do not actually develop new ideas; the reader whose tastes run more to the abstract and general than the concrete and special may skip many

Using This Book of them without logical consequence. (Of course, such a reader will probably wind up burning this book anyway.) We include hundreds of exercises, of wildly different purposes and difficulties. Some are the usual sorts of variations of the examples in the text or are straightforward verifications of facts needed; a student will probably want to attempt most of these. Sometimes an exercise is inserted whose solution is a special case of something we do in the text later, if we think working on it will be useful motivation (again, there is no attempt at "efficiency," and readers are encouraged to go back to old exercises from time to time). Many exercises are included that indicate some further directions or new topics (or standard topics we have omitted); a beginner may best be advised to skim these for general information, perhaps working out a few simple cases. In exercises, we tried to include topics that may be hard for nonexperts to extract from the literature, especially the older literature. In general, much of the theory is in the exercises-and most of the examples in the text. We have resisted the idea of grading the exercises by (expected) difficulty, although a "problem" is probably harder than an "exercise." Many exercises are starred: the • is not an indication of difficulty, but means that the reader can find some information about it in the section "Hints, Answers, and References" at the back of the book. This may be a hint, a statement of the answer, a complete solution, a reference to where more can be found, or a combination of any of these. We hope these miscellaneous remarks, as haphazard and uneven as they are, will be of some use.

Contents

v

Preface Using This Book

ix

Part 1: Finite Groups l. Representations of Finite Groups §i.t: Dennitions §1.2: Complete Reducibility; Schur's Lemma §1.3: Examples: Abelian Groups; 6 3

2. Characters §2.1: §2.2: §2.3: §2.4:

Characters The First Projection Formula and Its Consequences Examples: 6 4 and 91 4 More Projection Formulas; More Consequences

3. Examples; Induced Representations; Group Algebras; Real Representations §3.1: §3.2: §3.3: §3.4: §3.5:

Examples: 6 5 and 9( 5 Exterior Powers or the Standard Representation or 6 1 Induced Representations The Group Algebra Real Representations and Representations over Subfrelds or C

3 3 5 8

12 12 15 18 21

26 26 31 32 36 39

Contents

xii

4. Representations of 6 4 : Young Diagrams and Frobenius's Character Formula §4.1: Statements of the Results §4.2: Irreducible Representations of 6d §4.3: Proof of Frobenius's Formula

5. Representations of91 4 and GL 2 (f4 ) §5.1: Representations of 9ld §5.2: Representations of GL 2 (F4 ) and SL 2 (F,) 6. Weyl's Construction §6.1: Schur Functors and Their Characters §6.2: The Proofs

Part II: Lie Groups and Lie Algebras 7. Lie Groups

44 44 52 54 63 63 67

75 75 84

89 93

§7.1: Lie Groups: Definitions §7.2: Examples of Lie Groups §7.3: Two Constructions

93 95 101

8. Lie Algebras and Lie Groups

104

§8.1: Lie Algebras: Motivation and Definition §8.2: Examples of Lie Algebras §8.3: The Exponential Map

9. Initial Classification of Lie Algebras §9.1: §9.2: §9.3: §9.4:

104

Ill 114

121

Rough Classification of Lie Algebras Engel's Theorem and Lie's Theorem Semisimple Lie Algebras Simple Lie Algebras

121 125 128 131

tO. Lie Algebras in Dimensions One, Two, and Three

133

§10.1: §10.2: §10.3: §10.4:

Dimensions One and Two Dimension Three, Rank I Dimension Three, Rank 2 Dimension Three, Rank 3

11. Representations of si 2 C §II. I: The Irreducible Representations §11.2: A Little Plethysm §11.3: A Little Geometric Plethysm

133 136 139 141 146 146 151 153

Contents

xiii

12. Representations of si 1 C, Part I

161

13. Representations ofsi 1 C, Part II: Mainly Lots of Examples

175

§13.1: Examples §13.2: Description of the Irreducible Representations §13.3: A Little More Plethysm §13.4: A Little More Geometric Plethysm

Part Ill: The Classical Lie Algebras and Their Representations

175 182

185 189

195

14. The General Set-up: Analyzing the Structure and Representations of an Arbitrary Semisimple Lie Algebra

197

§14.1: Analyzing Simple Lie Algebras in General §14.2: About the Killing Form

197 206

15. si 4 C and st.c

211

§15.1: §15.2: §15.3: §15.4:

211 217 222 227 231

Analyzing sl.C Representations of si4 C and si.C Weyl's Construction and Tensor Products Some More Geometry §15.5: Representations of GL.C

16. Symplectic Lie Algebras

238

§16.1: The Structure of Sp 2 .C and sp 2.C §16.2: Representations ofsp 4 C

238 244

17. sp 6 C and sp 2.C

253

§17.1: Representations of ~;p 6 C §17.2: Representations of sp 2.c in General §17.3: Weyl's Construction for Symplectic Groups

18. Orthogonal Lie Algebras §18.1: SO.,C and t
267 so~C

19. so 6 C, so 7 C, and so.,C §19.1: §19.2: §19.3: §19.4: §19.5:

253 259 262

Representations of so 6 C Representations of the Even Orthogonal Algebras Representations of so 7 C Representations of the Odd Orthogonal Algebras Weyl's Construction for Orthogonal Groups

267 273 282 282 286 292 294 296

Contents

xiv

299

20. Spin Representations of so.,C

299 307 312

§20.1: Clifford Algebras and Spin Representations of so.,C §20.2: The Spin Groups Spin.,C and Spin.,R §20.3: Spin 8 C and Triality

Part IV: Ue Theory

317

319

21. The Classification of Complex Simple Lie Algebras

319 325 330

§21.1: Dynkin Diagrams Associated to Semisimple Lie Algebras §21.2: Classifying Dynkin Diagrams §21.3: Recovering a Ue Algebra from Its Dynkin Diagram

339

22. g 2 and Other Exceptional Lie Algebras §22.1: §22.2: §22.3: §22.4:

339

Construction ofg 2 from Its Dynkin Diagram Verifying That 9 2 is a Lie Algebra Representations of 92 Algebraic Constructions of the Exceptional Lie Algebras

346 350 359

23. Complex Lie Groups; Characters §23.1: §23.2: §23.3: §23.4:

366

366

Representations of Complex Simple Groups Representation Rings and Characters Homogeneous Spaces Bruhat Decompositions

375

382 395

399 399

24. Weyl Character Formula §24.1: The Weyl Character Formula §24.2: Applications to Classical Lie Algebras and Groups

403 415

25. More Character Formulas §25.1: Freudenthal's Multiplicity Formula §25.2: Proof of (WCF); the Kostant M ulliplicity Formula §25.3: Tensor Products and Restrictions to Subgroups 26. Real Lie Algebras and Lie Groups

415 419 424

II

§26.1: Classification of Real Simple Lie Algebras and Groups §26.2: Second Proof of Weyl's Character Formula §26.3: Real, Complex, and Quaternionic Representations

Appendices A. On Symmetric Functions §A. I: Basic Symmetric Polynomials and Relations among Them §A.2: Proofs of the Determlnantal Identities §A.3: Other Detenninantalldenlities

430 430 440 444 45l 453 453 462 465

C-Ontents

xv

B. On Multilinear Algebra §8.l: Tensor Products §8.2: Exterior and Symmetric Powers §8.3: Duals and Contractions C. On Semisimplicily

§C. I: The Killing Form and Carlan's Criterion §C.2: Complete Reducibility and the Jordan Decomposition §C.3: On Derivations D. Cartan Subalgebras §D.l: §D.2: §D.3: §D.4:

The Existence or Carlan Subalgebras On the Structure or Sem!simple Lie Algebras The Conjugacy of Carlan Subalgebras On the Weyl Group

471 471 472 475 478 478 481 483 487 487 489 491 493

E. Ado's and Levi's Theorems

499

§E. I: Levi's Theorem §E.2: Ado's Theorem

499 500

F. Invariant Theory for the Classical Groups §F. I: The Polynomial Invariants §F.2: Applications lo Symplectic and Orthogonal Groups §F.3: Proof or Capelli's Identity

504 504 511 514

Hints, Answers, and References

5 I6

Bibliography

536

Index of Symbols

543

Index

547

PART I

FINITE GROUPS

Given that over three-quarters of this book is devoted to the representation theory of Lie groups and Lie algebras, why have a discussion of the representations of finite groups at all? There are certainly valid reasons from a logical point of view: many of the ideas, concepts, and constructions we will introduce; here will be applied in the study of Lie groups and algebras. The real reason for us, however, is didactic, as we will now try to explain. Representation theory is very much a 20th-century subject, in the following sense. In the 19th century, when groups were dealt with they were generally understood to be subsets of the permutations of a set, or of the automorphisms G L(V) of a vector space V, closed under composition and inverse. Only in the 20th century was the notion of an abstract group given, making it possible to make a distinction between properties of the abstract group and properties of the particular realization as a subgroup of a permutation group or GL(V). To give an analogy, in the 19th century a manifold was always a subset of R"; only in the 20th century did the notion of an abstract Riemannian manifold become common. In both cases, the introduction of the abstract object made a fundamental difference to the subject. In differential geometry, one could make a crucial distinction between the intrinsic and extrinsic geometry of the manifold: which properties were invariants of the metric on the manifold and which were properties of the particular embedding in IR". Questions of existence or nonexistence, for example, could be broken up into two parts: did the abstract manifold exist, and could it be embedded. Similarly, what would have been called in the 19th century simply "group theory" is now factored into two parts. First, there is the study of the structure of abstract groups (e.g., the classification of simple groups). Second is the companion question: given a group G, how can we describe all the ways in which G may be embedded in

2

I. Finite Gmups

(or mapped to) a linear group GL(V}?. This, of course, is the subject matter of representation theory. Given this point of view, it makes sense when first introducing representation theory to do so in a context where the nature of the groups Gin question is itself simple, and relatively well understood. It is largely for this reason that we are starting off with the representation theory of finite groups: for those readers who are not already familiar with the motivations and goals of representation theory, it seemed better to establish those first in a setting where the structure of the groups was not itself an issue. When we analyze, for example, the representations ofthe symmetric and alternating groups on 3, 4, and 5 letters, it can be expected that the reader is already familiar with the groups and can focus on the basic concepts of representation theory being introduced. We will spend the first six lectures on the case of finite groups. Many of the techniques developed for finite groups will carry over to Lie groups; indeed, our choice of topics is in part guided by this. For example, we spend quite a bit of time on the symmetric group; this is partly for its own interest, but also partly because what we learn here gives one way to study representations of the general linear group and its subgroups. There are other topics, such as the alternating group ~14 , and the grou?s SL 2 (1Fq) and GL 2 (Fq) that are studied purely for their own interest and do not appear later. (In general, for those readers primarily concerned with Lie theory, we have tried to indicate in the introductory notes to each lecture which ideas will be useful in the succeeding parts of this book.) Nonetheless, this is by no means a comprehensive treatment of the representation theory of finite groups; many important topics, such as the Artin and Brauer theorems and the whole subject of modular representations, are omitted.

I

I

LECTURE 1

Representations of Finite Groups

In this lecture we give the basic definitions of representation theory, and prove two of the basic results, showing that every representation is a (unique) direct sum or irreducible ones. We work out as examples the case or abelian groups, and the simplest nonabelian group, the symmetric group on 3 letters. In the latter case we give an analysis that will turn out not to be useful for the study or finite groups, but whose main idea is central to the study or the representations or Lie groups. §1.1: Definitions §1.2: Complete reducibility; Schur's lemma §1.3: Examples: Abelian groups; 6 3

§1.1. Definitions A representation of a finite group G on a finite-dimensional complex vector space Vis a homomorphism p: G-. GL(V) of G to the group of automorphisms of V; we say that such a map gives V the structure of a G-module. When there is little ambiguity about the map p (and, we're afraid, even sometimes when there is) we sometimes call V itself a representation of G; in this vein we will often suppress the symbol p and write g · v or gv for p(g)(v). The dimension of V is sometimes called the degree of p. A map q1 between two representations V and W of G is a vector space map q~: V -. W such that

v~w

l. Representations of Finite Groups

4

commutes for every g e G. (We will call this a G-linear map when we want to distinguish it from an arbitrary linear map between the vector spaces V and W) We can then define Ker q1, lm qJ, and Coker qJ, which are also G-modules. A subrepresentation of a representation Vis a vector subspace W of V which is invariant under G. A representation V is called irreducible if there is no proper nonzero invariant subspace W of V. If Vand Ware representations; the direct sum V $ Wand the tensor product V ® Ware also representations, the latter via

g(v® w) = gv®gw. For a representation V, the nth tensor power y®n is again a representation of G by this rule, and the exterior powers t\"(V) and symmetric powers Sym"(V) are subrepresentations 1 of it. The dual v• = Hom(V, C) of Vis also a representation, though not in the most obvious way: we want the two representations of G to respect the natural pairing (denoted ( , )) between v• and V, so that if p: G-+ GL(V) is a representation and p*: G-+ GL(V*) is the dual, we should have for all g E G, v E v, and v• E representation by

v•. This in

turn forces us to define the dual

for all g E G. Exercise l.l. Verify that with this definition of p*, the relation above is satisfied. Having defined the dual of a representation and the tensor product of two representations, it is likewise the case that if V and Ware representations, then Hom(V, W) is also a representation, via the identification Hom(JI, W) = v• ® W Unraveling this, if we view an element of Hom( V, W) as a linear map qJ from V to W, we have (gQJ)(v> "'" gqJ(g-• v>

,

for all v E V. In other words, the definition is such that the diagnJm

v

'1

__!___.

w

1'

v~w

commutes. Note that the dual representation is, in turn, a special case of this: 1 For more on exterior and symmetric powers. including descriplions as quotient spaces of tensor powers, see Appendix R.

§1.2. Complete Reducibility; Schur's Lemma

5

when W = Cis the trivial representation, i.e., gw = w for all we C, this makes v• into a G-module, with grp(v) = rp(g- 1 v), i.e., grp = '(g- 1 )rp. Exercise 1.2. Verify that in general the vector space of G-linear maps between two representations V and W of G is just the subspace Hom(V, W)a of elements of Hom(V, W) fixed under the action of G. This subspace is often denoted Homa(V. W). We have, in effect, taken the identification Hom(V, W) = v• ®Was the definition of the representation Hom(V, W). More generally, the usual identities for vector spaces are also true for representations, e.g.,

V®(U ED W) = (V® U) ED (V® W), t\k(V$ W) =

.mt

NV®NW,

N(v•) = N(v)•, and so on. Exercise 1.3•. Let p: G-+ GL(V) be any representation of the finite group G on an n-dimensional vector space V and suppose that for any g e G, the determinant of p(g) is I. Show that the spaces NV and t\"-t v• are isomorphic as representations of G.

If X is any finite set and G acts on the left on X, i.e., G-+ Aut(X) is a homomorphism to the permutation group of X, there is an associated permutation representation: let V be the vector space with basis {e.. : x eX}, and let G act on V by The regular representation, denoted Ra orR, corresponds to the left action of G on itself. Alternatively, R is the space of complex-valued functions on G, where an element g e G acts on a function oc by (goc)(h) = oc(g- 1/1). Exercise t.4•. (a) Verify that these two descriptions of R agree, by identifying the element e.. with the characteristic function which takes the value l on x, 0 on other elements of G. (b) The space of functions on G can also be made into a G-module by the rule (goc)(IJ) = oc(llg). Show that this is an isomorphic representation.

§1.2. Complete Reducibility; Schur's Lemma As in any study, before we begin our attempt to classify the representations of a finite group G in earnest we should try to simplify life by restricting our search somewhat. Specifically, we have seen that representations of G can be

I. Representations of Finite Groups

6

built up out of other representations by linear algebraic operations, most simply by taking the direct sum. We should focus, then, on representations that are "atomic" with respect to this operation, i.e., that cannot be expressed as a direct sum or others; the usual term for such a representation is Indecomposable. Happily, the situation is as nice as it could possibly be: a representation is atomic in this sense if and only if it is irreducible (i.e., contains no proper subrepresentations); and every representation is the direct sum of irreducibles, in a suitable sense uniquely so. The key to all this is Proposition 1.5. ~f W is a subrepresentation of a representation V of a finite group G, then there is a complementary invariant subspace W' of V, so that V= WE9 W'.

PRooF. There are two ways of doing this. One can introduce a (positive definite) Hermitian inner product H on V which is preserved by each g e G (i.e., such that H(gv, gw) = H(v, w) for all v, we V and g e G). Indeed, if H 0 is any Hermitian product on V, one gets such an H by averaging over G:

L

ll(v, 111) =

ll 0 (gv, gw).

geG

Then the perpendicular subspace W 1 is complementary to W in V. Alternatively (but similarly), we can simply choose an arbitrary subspace U complementary to W. let n 0 : V-> W be the projection given by the direct sum decomposition V = W E9 U, and average the map n 0 over G: that is, take

L

n(v) =

g(n 0 (g- 1 v)).

geG

This will then be a G-linear map from V onto ~V, which is multiplication by IGI on W; its kernel will, therefore, be a subspace of V invariant under G and complementary to W. 0 Corollary 1.6. Any representation is a direct sum of irreducible representations. This property is called complete reducibility, or semisimplicity. We will see that, for continuous representations, the circle S 1, or any compac~ group, has this property; integration over the group (witft respect to an invariant measure on the group) plays the role or averaging in the above proof. The (additive) group IR does not have this property: the representation

a

H(l a) 0

1

leaves the x axis fixed, but there is no complementary subspace. We will see other Lie groups such as SL.(C) that are semisimple in this sense. Note also that this argument would fail if the vector space V was over a field of finite characteristic since it might then be the case that n(v) = 0 for L' e W. The failure

§1.2. Complete Reducibility; Schur's Lemma

7

of complete reducibility is one of the things that makes the subject of modular representations, or representations on vector spaces over finite fields, so tricky. The extent to which the decomposition of an arbitrary representation into a direct sum of irreducible ones is unique is one of the consequences of the following: Schur's Lemma 1.7. If V and W are irreducible rp: V--> W is a G-module homomorphism, then

repre.~entations

of G and

(l) Eitl1er rp is an isomorphism, or rp = 0. (2) If V = W, tlten rp = ..1. ·I for some ..1. E C, I the identity. PROOF. The first claim follows from the fact that Ker rp and Im rp are invariant subspac~s. For the second, since C is algebraically closed, rp must have an eigenvalue ..1., i.e., for some ..1. E C, rp- ..1.1 has a nonzero kernel. By (l), then, we must have rp- ..1.1 = 0, so rp = ..1.1. 0

We can summarize what we have shown so far in Proposition 1.8. For any representation V of a finite group G, tltere is a tlecomposition

wltere tire V, are distinct irreducible representations. The decomposition of V into a direct sum of tire k factors is unique, as are tlte V, tlrat occur and tlteir multiplicities a,. PROOF. It follows from Schur's lemma that if W is another representation of G, with a decomposition W = E9 "'J~bJ, and rp: V--> W is a map of representations, then ({J must map the factor V,EDo• into that factor "'J~b~ for which "'J ~ V;; when applied to the identity map of V to Jl, the stated uniqueness

0

~~

In the next lecture we will give a formula for the projection of V onto V;Ei><>•. The decomposition of the ith summand into a direct sum of a1 copies of is not unique if a1 > I, however. Occasionally the decomposition is written

v;

(1.9)

especially when one is concerned only about the isomorphism classes and multiplicities of the V,. One more fact that will be established in the following lecture is that a finite group G admits only finitely many irreducible representations I~ up to isomorphism (in fact, we will say how many). This, then, is the framework of the classification of all representations of G: by the above, once we have described

I. Representations of Finite Groups

8

the irreducible representations of G, we will be able to describe an arbitrary representation as a linear combination of these. Our first goal, in analyzing the representations of any group, will therefore be: (i) Describe all tile irreducible representations of G.

Once we have done this, there remains the problem of carrying out in practice the description of a given representation in these terms. Thus, our second goal will be: (ii) Find tecltniques for giving tire direct sum decomposition (1.9), and in particular determining tire multiplicities a, of an arbitrary representation V. Finally, it is the case that the representations we will most often be concerned with are those arising from simpler ones by the sort of linear- or multilinearalgebraic operations described above. We would like, therefore, to be able to describe, in the terms above, the representation we get when we perform these operations on a known representation. This is known generally as (iii) Pletllysm: Describe tlte decomposftions, with multiplicities, r~f representations derived from a given representation V, such as V ® V, v•, N(V), Sym~( V), and N(N V). Note that if V decomposes into a sum of two representations, these representations decompose accordingly; e.g., if V = U ED W, then NV=

,f#t Nu ®NW,

so it is enough to work out this plethysm for irreducible representations. Similarly, if V and Ware two irreducible representations, we want to decompose V ® W; this is usually known as the Clebsclr-Gordon problem.

§1.3. Examples: Abelian Groups; 6

3

One obvious place to look for examples is with abelian groups. It does not take long, however, to deal with this case. Basically, we may observe in general that if V is a representation of the finite group G, abelian or not, each g E G gives a map p(g): V-+ V; but this map is not generally a G-modfle homomorphism: for general It E G we will have g(lr(v)) '# h(g(v)).

Indeed, p(g): V ...... V will beG-linear for every p if(and only if) g i.~ in the center Z(G) of G. In particular if G is abelian, and Vis an irreducible representation, then by Schur's lemma every element g E G acts on V by a scalar multiple of the identity. Every subspace of V is thus invariant; so that V must be one dimensional. The irreducible representations of an abelian group G are thus simply elements of the dual group, that is, homomorphisms p: G ....

c•.

§1.3. Examples: Abelian Groups; 6 3

9

We consider next the simplest nonahelian group, G = 6 3 • To begin with, we have (as with any symmetric group) two one-dimensional representations: we have the trivial representation, which we will denote U, and the altematillg representation U', defined by setting gv = sgn(g)v

forge G, v e C. Next, since G comes to us as a permutation group, we have a natural permutation representation, in which G acts on C 3 by permuting the coordinates. Explicitly, if {e 1 , e 2 , e 3 } is the standard basis, then g · e 1 = e,(l), or, equivalently, · g·(z 1 ,

Z2,

z3 )

= (Zg-1(1)• Zg-I(Z)• Zg-1(3)).

This representation, like any permutation representation, is not irreducible: the line spanned by the sum (I, I, I) of the basis vectors is invariant, with complementary subspace V = {(z 1 , z 2 , z3 )

E

C 3 : z1

+ z2 + z3

= 0}.

This two-dimensional representation Vis easily seen to be irreducible; we call it the standard representation of 6 3 • Let us now turn to the problem of describing an arbitrary representation of 6 3 • We will see in the next lecture a wonderful tool for doing this, called character theory; but, as inefficient as this may be, we would like here to adopt a more ad hoc approach. This has some virtues as a didactic technique in the present context (admittedly dubious ones, consisting mainly of making the point that there are other and far worse ways of doing things than character theory). The real reason we are doing it is that it will serve to introduce an idea that, while superfluous for analyzing the representations of finite groups in general, will prove to be the key to understanding representations of Lie groups. The idea is a very simple one: since we have just seen that the representation theory of a finite abelian group is virtually trivial, we will start our analysis of an arbitrary representation W of 6 3 by looking just at the action of the abelian subgroup m- 3 = lL/3 c 6 3 on W. This yields a very simple decomposition: if we take r to be any generator of m- 3 (that is, any three-cycle), the space W is spanned by eigenvectors v1 for the action of r, whose eigenvalues are of course all powers of a cube root of unity w = e 2 ft 1' 3 . Thus,

W= Ef)V,, where

V, = Cv1 and rv1 = w«•v1• Next, we ask how the remaining elements of 6 3 act on W in terms of this decomposition. To see how this goes, let u be any transposition, so that rand u together generate 6 3 , with the relation uru = r 2 • We want to know where u sends an eigenvector v for the action of r, say with eigenvalue cr/; to answer

I. Representations of Finite Groups

10

this, we look at how r acts on u(v). We use the basic relation above to write r(u(v)) = u(r 2 (t')) = u(w 21 ·v) = w 21 ·u(v).

The conclusion, then, is that if vis an eige1111ector for r witlt eigenvalue w 1, tlren u(v) is again an eigenvector for r, witll eigenvalue w 21 • Exercise 1.10. Verify that with 11 = (12), T = (123), the standard representation has a basis a= (w, l, w2 ), p = (l, w, w2 ), with

ra

= woe,

rP

= w 2 p,

lia = (J,

uP

=

IX.

Suppose now that we start with such an eigenvector v for r.lfthe eigenvalue of 11 is w' i= l, then u(v) is an eigenvector with eigenvalue w 21 i= w', and so is independent of v; and v and u(v) together span a two-dimensional subspace V' of W invariant under 6 3 • In fact, V' is isomorphic to the standard representation, which follows from Exercise 1.10. U, on the other hand, the eigenvalue of v is l, then u(v) may or may not be independent of v. If it is not, then v spans a one-dimensional subrepresentation of W, isomorphic to the trivial representation if u(v) = v and to the alternating representation if u(v) = - v. If 11(v) and v are independent, then v + u(v) and v - u(v) span one-dimensional representations of W isomorphic to the trivial and alternating representations, respectively. We have thus accomplished the first two of the goals we have set for ourselves above in the case of the group G = 6 3 • First, we see from the above that t/te only tlrree irreducible representations of 6 3 are tire trivial, alternatirtg, and staudard represelltations U, U' and V. Moreover, for an arbitrary representation W of 6 3 we can write W

=

Ul1ln E9 u•l1lb G) Vl1l';

and we have a way to determine the multiplicities a, b, and c: c, for example, is the number of independent eigenvectors for r with eigenvalue w, whereas a + cis the multiplicity of l as an eigenvalue of 11, and b + cis tlje multiplicity of - I as an eigenvalue of u. ' In fact, this approach gives us as well the answer to our third problem, finding the decomposition of the symmetric, alternating, or tensor powers of a given representation W, since if we know the eigenvalues of r on such a representation, we know the eigenvalues of r on the various tensor powers of W. For example, we can use this method to decompose V ® V, where V is the standard two-dimensional representation. For V ® V is spanned by the vectors IX ®IX, IX® p, P®a, and p ® p; these are eigenvectors for r with eigenvalues 01 2, l, l, and w, respectively, and 11 interchanges IX® a with P® {1, and a® p with p ®IX. Thus a® IX and (J ® (J span a subrepresentation

§1.3. Examples: Abelian Groups; <5 3

II

isomorphic to V, IX® f1 + f1 ® IX spans a trivial representation U, and IX® f1 - f1 ®ex spans U', so V ® V ~ U €9 U' €9 V.

Exercise 1.11. Use this approach to lind the decomposition of the representations Sym 2 V and Sym 3 V. Exercise 1.12. (a) Decompose the regular representation R of 6 3 . (b) Show that SymH 6 V is isomorphic to Symtv ED R, and compute SymkV for all k. Exercise l.l3*. Show that Sym 2 (Sym 3 V) ~ Sym 3 (Sym 2 V). Sym"'(Sym"V) isomorphic to Sym"(Sym"'V)?

Is

As we have indicated, the idea of studying a representation V of a group G by lirst restricting the action to an abelian subgroup, getting a decomposition of V into one-dimensional invariant subspaces, and then asking how the remaining generators of the group act on these subs paces, does not work well for finite G in general; for one thing, there will not in general be a convenient abelian subgroup to use. This idea will turn out, however, to be the key to understanding the representations of Lie groups, with a torus subgroup playing the role of the cyclic subgroup in this example. Exercise 1.14*. Let V be an irreducible representation of the finite group G. Show that, up to scalars, there is a unique Hermitian inner product on V preserved by G.

LECTURE 2

Characters

This lecture contains the heart of our treatment of the representation theory of finite groups: the definition in §2.1 orthe character of a representation, and the main theorem (proved in two steps in §2.2 and §2.4) that the characters of the irreducible representations form an orthonormal basis for the spaee of class functions on G. There will be more examples and more constructions in the following lectures, but this is what you need to know. §2.1: §2.2: §2.3: §2.4:

Characters The nrst projection formula and its consequences Examples: 6 4 and 21 4 More projection formulas; more consequences

§2.1. Characters As we indicated in the preceding section, there is a remarkably effective tool for understanding the representations of a finite group G, called clwracter tl1eory. This is in some ways motivated by the exa,ple worked out in the last section where we saw that a representation of 6 3 was determined by knowing the eigenvalues of the action of the elements t and u e 6 3 • For a general group G, it is not clear what subgroups and/or elements should play the role of ~1 3 , t, and u; but the example certainly suggests that knowing all the eigenvalues of each element of G should suffice to describe the representation. Of cour~e. specifying all the eigenvalues of the action of each element of G is somewhat unwieldy; but fortunately it is redundant as well. For example, if we know the eigenvalues {A1} of an element g e G, then of course we know the eigenvalues {Af} of gt for each k as well. We can thus use this redundancy

§2.1. Characters

13

to simplify the data we have to specify. The key observation here is it is enough to give, for example, just the sum of the eigenvalues of each element or G, since knowing the sums LA.~ of the kth powers or the eigenvalues of a given element g E G is equivalent to knowing the eigenvalues {A.,} of g themselves. This then suggests the following: Definition. If Vis a representation or G, its cltaracter Xv is the complex-valued function on the group defined by

Xv(g) = Tr(glv), the trace of g on V. In particular, we have

xv(llgll- 1 )

= Xv(g),

so that Xv is constant on the conjugacy classes of G; such a function is called a class function. Note that Xv(l) =dim V. Proposition 2.1. Let V and W be representations of G. Tlren Xvew = Xv

+ Xw.

Xv®w = Xv. Xw.

PROOF. We compute the values of these characters on a fixed element g E G. For the action or g, V has eigenvalues {-l 1} and W has eigenvalues {Jt 1}. Then {A.1 + JL1} and {A. 1 • JL1} are eigenvalues for V <:9 Wand V ® W, from which the first two formulas follow. Similarly {A.j" 1 = X,} are the eigenvalues for g on V*, since all eigenvalues are nth roots of unity, with 11 the order of g. Finally, {A.,A.p < j} are the eigenvalues for g on NV, and

:L .t,.t, = ~b.!!f - I f<J

2

.tr;

and since g 2 has eigenvalues {A.f }, the last formula follows.

0

Exercise 2.2. For Sym 2 V, verify that Xsrm>v(O) = Uxv(0) 2

+ Xv(g 2 )].

Note that this is compatible with the decomposition V® V = Sym 2 VE9NV. Exercise 2.3*. Compute the characters of Sym•v and NV. Exercise 2.4*. Show that if we know the character Xv or a representation V, then we know the eigenvalues of each element g of G, in the sense that we

2. Characters

14

know the coefficients of the characteristic polynomial of g: V ~ V. Carry this out explicitly for elements g E G of orders 2, 3, and 4, and for a representation of G on a vector space of dimension 2, 3, or 4. Exercise 2.5. (Tire original fixed-point formula). If V is the permutation representation associated to the action of a group G on a finite set X, show that Xv(g) is the number of elements of X fixed by g. As we have said, the character of a representation of a group G is really a function on the set of conjugacy classes in G. This suggests expressing the basic information about the irreducible representations of a group Gin the form of a clraracter taMe. This is a table with the conjugacy classes [g] of G listed across the top, usually given by a representative g, with (for reasons that will become apparent later) the number of elements in each conjugacy class over it; the irreducible representations V of G listed on the left; and, in the appropriate box, the value of the character Xv on the conjugacy class [g]. Example 2.6. We compute the character table of 6 3 . This is easy: to begin with, the trivial representation takes the values (l, 1, 1) on the three conjugacy classes [I], [(12)], and [(123)], whereas the alternating representation has values (1, -1, 1). To see the character of the standard representation, note that the permutation representation decomposes: C 3 = U EB V; since the character of the permutation representation has, by Exercise 2.5, the values (3, I, 0), we have Xv = xo- xu= (3, 1, 0)- (1, I, 1) = (2, 0, -I). In sum, then, the character table of 6 3 is

3

2

e;J

(12)

(123)

trivial U alternating U' standard V

I -I 0

-I

2

This gives us another solution of the basic problem posed in Lecture 1: if W is any representation of 6 3 and we decompose W into irreducible representations W ~ ue• EB U'l1lb EB Vl1l<, then Xw = axv + bxv· + CXv· In particular, since the functions Xv• Xu· and Xv are independent, we see that W is determilred up to isomorphism by its character Xw· Consider, for example, V® V. Its character is (Xvf, which has values 4, 0, and I on the three conjugacy classes. Since V EB U ED U' has the same character, this implies that V ® V decomposes into V EB U EB U', as we have seen directly. Similarly, V ® U' has values 2, 0, and -1, so V ® U' ~ V.

§2.2. The First Projection Formula and Its Consequenres

15

Exercise 2.7*. Find the decomposition of the representation V 0 " using character theory.

Characters will be similarly useful for larger groups, although it is rare to find simple closed formulas for decomposing tensor products.

§2.2. The First Projection Formula and Its Consequences In the last lecture, we asked (among other things) for a way or locating explicitly the direct sum factors in the decomposition of a representation into irreducible ones. In this section we will start by giving an explicit formula for the projection of an irreducible representation onto the direct sum oft he trivial factors in this decomposition; as it will turn out, this formula alone has tremendous consequences. To start, for any representation V of a group G, we set

V 0 = { v e V: gv = v Vg

G}.

E

We ask for a way or finding V 0 explicitly. The idea behind our solution to this is already implicit in the previous lecture. We observed there that for any representation V of G and any g E G, the endomorphism g: V-> V is, in general, not a G-module homomorphism. On the other hand, if we take the average of all these endomorph isms, that is, we set l

lp =

jGj

Jog

E

End(V),

then the endomorphism fP will beG-linear since have

LY =

L;l1glr- 1. In fact, we

Proposition 2.8. Tlte map fP is a projection of V onto V 6 . PRooF. First, suppose v = fP(w) = (1/IGI) L;gw. Then, for any II e G,

1 lrv = IGI

l

L lrgw = IGI L gw,

so the image of fP is contained in V 0 . Conversely, if v E (l/IGI)Lv = v, so V 0 c Im(rp); and fP o fP = rp.

VG,

then rp(v) = 0

We thus have a way of finding explicitly the direct sum of the trivial subrepresentations of a given representation, although the formula can be hard to use if it does not simplify. If we just want to know the number 111 of copies of the trivial representation appearing in the decomposition of J~ we can do this numerically, since this number will be just the trace of the

2. Characters

16

projection fp. We have

m = dint VG = Trace(rp) 1

I

= iGI JG Trace(g) = IGI JG Xv(g).

(2.9)

In particular, we observe that for an irreducible representation V other than the trivial one, the sum over all g e G of the values of the character Xv is zero. We can do much more with this idea, however. The key is to use Exercise 1.2: if Vand Ware representations of G, then with Hom(V, W), the representation defined in Lecture I, we have Hotn(V, W)G = {G-module homomorphisms from V toW}. If Vis irreducible then by Schur's lemma dim Horn(V, W)G is the multiplicity of V in W; similarly, if W is irreducible, dim Hom(V, W)G is the multiplicity of Win V, and in the case where both V and Ware irreducible, we have I dim HomG(V, W) == { 0

if V;;; W if V j,. W.

But now the character Xnom!V.IVI of the representation Hom(V, W) = V* ® W is given by Xnom(V,W)(g) = Xv(g)· Xw(g).

We can now apply formula (2.9) in this case to obtain the striking

I kG" -

IGI

Xv(g)xw(g)

=

{I

0

if V;;; W if V -:f,. W.

(2.10)

To express this, let Cc~ ... (G) = {class functions on G}

and define an Hermitian inner product on c •••,.( G) by (a,

fJ)

kG

I "a(g)fl(g). = IGI

(2.11)

Formula (2.10) then amounts to Theorem 2.12./rr terms of tltis imler product, the cltaracters of tlte irreducible representations of G al'e orthonormal. For example, the orthonormality of the three irreducible representations of 6 3 can be read from its character table in Example 2.6. The numbers over each conjugacy class tell how many times to count entries in that column. Corollary 2.13. Tl1e number of irreducible representations f!{ G is less titan or equal to tlte 1111ml1er of conj11gacy classes.

§2.2. The First Projection Formula and Its Consequences

17

We will soon show that there are no nonzero class functions orthogonal to the characters, so that equality holds in Corollary 2.13. Corollary 2.14. Any representation is determined by its c/raracter. Indeed if V ~ v,e"• ED··· ED v,.e"•, with the Jt; distinct irreducible characters, then Xv = Ia,xv,• and the Xv, are linearly independent. Corollary 2.15. A representation Vis irreducible

if and only if (Xv. Xv) =

v,.e"•

In fact, if V ~ Vifle, E9 ···ED as above, then (Xn Xv) The multiplicities a, can be calculated via

l.

=La;.

Corollary 2.16. The multiplicity a1 of V, in Vis the inner product of Xv with Xv,. i.e., a, = (Xv• Xv,)· We obtain some further corollaries by applying all this to the regular representation R of G. First, by Exercise 2.5 we know the character of R; it is simply 0

XR(g)

ifg#e if g =e.

= { IGI

Thus, we see lirst of all that R is not irreducible if G # {e}. In fact, if we set R = Jt;fD"•, with V, distinct irreducibles, then

EB

a, = (Xv,• XR)

= I~I Xv,(e) ·I Gl =

dim

J~.

(2.17)

Corollary 2.18. Any irreducible representation V of G appears in the regular representation dim V times. In particular, this proves again that there are only finitely many irreducible representations. As a numerical consequence of this we have the formula

IGI = dim(R) = L dim(Jt;) 2 •

(2.19)

i

Also, applying this to the value of the character of the regular representation on an element g E G other than the identity, we have 0=

L (dim

V.) · Xv,(g)

if g # e.

(2.20)

These two formulas amount to the Fourier inversion formula for finite groups, cf. Example 3.32. For example, if all but one of the characters is known, they give a formula for the unknown character. Exercise 2.21. The orthogonality of the rows of the character table is equivalent to an orthogonality for the columns (assuming the fact that there are as

2. Characters

18

many rows as columns). Written out, this says: (i) ForgE G,

" IGI 7 x(g)x(g) = c(g), where the sum is over all irreducible characters, and c(g) is the number of elements in the conjugacy class of g. (ii) If g and h are elements of G that are not conjugate, then

I

iliJx(h) = o.

r

Note that for g = e these reduce to (2.19) and (2.20).

§2.3. Examples: 6 4 and

~( 4

To see how the analysis of the characters of a group actually goes in practice, we now work out the character table of 6 4 • To start, we list the conjugacy classes in 6 4 and the number of elements of 6 4 in each. As with any symmetric group 6d, the conjugacy classes correspond naturally to the partitions of d, that is, expressions of d as a sum of positive integers a 1 , a2 , ••• ,a., where the correspondence associates to such a partition the conjugacy class of a permutation consisting of disjoint cycles of length a 1 , a 2 , ••• , a•. Thus, in 6 4 we have the classes of the identity element I (4 = l + I + l + 1), a transposition such as (12), corresponding to the partition 4 == 2 + I + I; a threecycle (123) corresponding to 4 = 3 + I; a four-cycle (1234) (4 = 4); and the product of two disjoint transpositions (12)(34) (4 = 2 + 2). Exercise 2.22. Show that the number of elements in each of these conjugacy classes is, respectively, I, 6, 8, 6, and 3. As for the irreducible representations of 6 4 , we start with the same ones that we had in the case of 6 3 : the trivial U, the alterna?ng U', and the standard representation V, i.e., the quotient of the permutation representation associated to the standard action of 6 4 on a set of four elements by the trivial subrepresentation. The character of the trivial representation on the live conjugacy classes is of course (I, I, 1, I, 1), and that of the alternating representation is (1, -I, I, -I, 1). To lind the character of the standard representation, we observe that by Exercise 2.5 the character of the permutation representation on C 4 is xc• = (4, 2, I, 0, 0) and, correspondingly, Xv = Xc• - Xv = (3, I, 0, -l, - 1).

Note that

lxvl = l, so

Jl is irreducible. The character table so far looks like

§2.3. Examples: 6 4 and \!l 4

19

64

trivial U alternating U' standard J!

6

8

6

3

(12)

(123)

(1234)

(12)(34)

I -I

I

I

I

I

-I

3

I

I 0

-I

I -I

Clearly, we are not done yet: since the sum of the squares of the dimensions of these three representations is 1 + I + 9 = 11, by (2.19) there must be additional irreducible representations of 6 4 , the squares of whose dimensions add up to 24 - 11 = 13. Since there are by Corollary 2.13 at most two of them, there must be exactly two, of dimensions 2 and 3. The latter of these is easy to locate: if we just tensor the standard representation V with the alternating one U', we arrive at a representation V' with character Xv· = Xv ·Xu· = (3, -I, 0, I, - 1). We can see that this is irreducible either from its character. (since IXv·l = I) or from the fact that it is the tensor product of an irreducible representation with a one-dimensional one; since its character is not equal to that of any of the first three, this must be one of the two missing ones. As for the remaining representation of degree two, we will for now simply call it W; we can determine its character from the orthogonality relations (2.10). We obtain then the complete character table for 6 4 :

64

trivial U alternating U' standard V V'= V®U'

I 1 3 3

Another W

2

6

8

6

3

(12)

(123)

(1234)

(12)(34)

-I

I 0 0

-I -I I

-I -I

-I

0

2

I I -1 0

I I

Exercise 2.23. Verify the last row of this table from (2.10) or (2.20). We now get a dividend: we can take the character of the mystery representation W, which we have obtained from general character theory alone, and use it to describe the representation W explicitly! The key is the 2 in the last column for xw: this says that the action of (12)(34) on the two-dimensional vector space W is an involution of trace 2, and so must be the identity. Thus, W is really a representation of the quotient group 1 1 If N is a normal subgroup of a group G, a represenlation p: G ... GL(V) is trivial on N if and only if it factors through the quotient

G ... G/N ... GL(V).

Representations of G/N can be identified with representations of G that are trivial on N.

2. Characters

20 6.d{1,(12)(34),(13)(24),(14)(23)}

~

63.

[One may see this isomorphism by letting 6 4 act on the elements of the conjugacy class of ( 12)(34); equivalently, if we realize 6 4 as the group of rigid motions of a cube (see below), by looking at the action of 6 4 on pairs of opposite faces.] W must then be just the standard representation of 6 3 pulled back to 6 4 via this quotient. Example 2.24. As we said above, the group of rigid motions of a cube is the symmetric group on four letters; 6 4 acts on the cube via its action on the four long diagonals. It follows, of course, that 6 4 acts as well on the set of faces, of edges, of vertices, etc.; and to each of these is associated a permutation representation of 6 4 . We may thus ask how these representations decompose; we will do here the case of the faces and leave the others as exercises. We start, of course, by describing the character x of the permutation representation associated to the faces of the cube. Rotation by 180n about a line joining the midpoints of two opposite edges is a transposition in 6 4 and fixes no faces, so x(l2) = 0. Rotation by l20o about a long diagonal shows x( 123) = 0. Rotation by 90° about a line joining the midpoints of two opposite faces shows x(1234) = 2, and rotation by 180° gives x((l2)(34)) = 2. Now (x, x> = 3, so x is the sum of three distinct irreducible representations. From the table, (X, Xu)= (x, Xv·) =(X, Xw) = I, and the inner products with the others are zero, so this representation is U E9 V' E9 W. In fact, the sums of opposite faces span a three-dimensional subrepresentation which contains U (spanned by the sum of all faces), so this representation is U $ W. The differences of opposite faces therefore span V'. Exercise 2.25*. Decompose the permutation representation of 6 vertices and (ii) the edges of the cube.

4

on (i) the

Exercise 2.26. The alternating group IJ( 4 has four conjugacy classes. Three representations U, U', and U" come from the representations of 111 4 /{ I, ( 12)(34), ( 13)(24), ( 14)(23)} ~ Z/3, so there is one more irreducible representation V of dimensi6n 3. Compute the character table, with m = e2 Kfi3:

~~4

4

4

3

(123)

(132)

(12)(34)

u U' U"

v

3

(/)

(/)2

w'

w 0

0

-I

21

§2.4. More Projection Formulas; More Consequences

Exercise 2.27. Consider the representations of e. and their restrictions to IJl 4 • Which are still irreducible when restricted, and which decompose? Which pairs of nonisomorphic representations of e. become isomorphic when restricted? Which representations of IJ( 4 arise as restrictions from e.?

§2.4. More Projection Formulas; More Consequences In this section, we complete the analysis of the characters of the irreducible representations of a generallinite group begun in §2.2 and give a more general formula for the projection of a general representation V onto the direct sum of the factors in V isomorphic to a given irreducible representation W. The main idea for both is a generalization of the "averaging" of the endomorphisms g: V-> V used in §2.2, the point being that instead of simply averaging all the g we can ask the question: what linear combinations of the endomorphisms g: V-> V are G-linear endomorphisms7 The answer is given by

Proposition 2.28. Let a: G-> C be any function on tlte group G, and for tmy represelltatioll V of G set rtJ•• v

= L a(g)·g: V-> V.

Tllen (/J11 ,v is a ltomomorpltism of G-modules for all V function.

if arul only if a is a class

PROOF. We simply write out the condition that ffJ•• v beG-linear, and I he result falls out: we have (/)0 ,

v(lw) = L a(g) · g(lrv) =

I

a(llgll- 1 ) • ltglr- 1 (lw)

(substituting ltglr- 1 for g) = lt(La(lrglt- 1 )·g(v))

= /r(I, a(g) · g( v))

(if a is a class function) = 11((/J•• v(v)).

Exercise 2.29*. Complete this proof by showing that conversely if a is not a class function, then there exists a representation V of G for which ffJ•• v fails to be G-linear. 0 As an immediate consequence of this proposition, we have

22

2. Characters

Proposition 2.30. The number of irreducible representations of G is equal to the number of conjugacy classes of G. Equivalently, their characters {Xv} form an orthonormal basis for C.1••• (G). PROOF. Suppose a: G-> Cis a class function and (a, xv) = 0 for all irreducible representations V; we must show that a == 0. Consider the endomorphism

f/J«,Y

L a(g)·g: V-> V

=

as defined above. By Schur's lemma, f/J«,Y ..1.

= ..1.· Id; and if 11 =dim

V, then

l == -· n trace(f/J« . v) 1 = -· II

L a(g)xv(g)

IGI--

= -(a, Xv•)

n

=0.

v

La(

Thus, rp., = 0, or g)· g = 0 on any representation V of G; in particular, this will be true for the regular representation V = R. But in R the elements {g e G}, thought of as elements of End(R), are linearly independent. For example, the elements {g(e)} are all independent. Thus a(g) = 0 for all g, as 0 required. This proposition completes the description of the characters of a finite group in general. We will see in more examples below how we can use this information to build up the character table of a given group. For now, we mention another way of expressing this proposition, via the representation rir•g of the group G. The representation ring R(G) of a group G is easy to define. First, as a group we just take R(G) to be the free abelian group generated by all (isomorphism classes of) representations of G, and mod out by the subgroup generated by elements of the form V + W- (V $ W). Equivalently, given the statement of complete reducibility, we can just take all integral linear combina,tions V, oft he irreducible representations V, of G; elements of R(G) are correspondingly called virtual representations. The ring structure is then given simply by tensor product, defined on the generators of R(G) and extended by linearity. We can express most of what we have learned so far about representations of a finite group Gin these terms. To begin, the character defines a map

La,·

x: R(G)-> c •••••(G) from R(G) to the ring of complex-valued functions on G; by the basic formulas of Proposition 2.1, this map is in fact a ring homomorphism. The statement that a representation in determined by its character then says that xis injective;

§2.4. More Projection Formulas; More Consequences

23

the images of x are called virtual cl1aracters and correspond thereby to virtual representations. Finally, our last proposition amounts to the statement that

x induces an isomol"plrism The virtual characters of G form a lattice A~ 7/.c in Co~ ••• (G), in which the actual characters sit as a cone A 0 ~ N< c 7L<. We can thus think of the problem of describing the characters of G as having two parts: first, we have to lind A, and then the cone A0 c A (once we know /\ 0 , the characters of the irreducible representations will be determined). In the following lecture we will state theorems of Artin and Brauer characterizing A® Q and /\.. The argument for Proposition 2.30 also suggests how to obtain a more general projection formula. Explicitly, if W is a fixed irreducible representation, then for any representation V, look at the weighted sum l .~ "Xw(g)·g 1/1 = IGI

E

End(V).

By Proposition 2.28, 1/J is a G-module homomorphism. Hence, if Vis irreducible, we have 1/1 = A.· Id, and l ..1.=--Tracei/J dim V

l . - - •fV= W = { dim V

0

ifV#W.

For arbitrary V,

1/'v =dim

kG

l "Xw(g)·g: W·IGI V--> V

(2.31)

is the projection of V onto the factor consisting of the sum of all copies of W appearing in V. In other words, if V = ~e·•, then

ffi

n1 = dim

V. ·-1!._

l:

GlgeG

Xv,(gj · g

(2.32)

is the projection of V onto ~Ell••. Exercise 2.33*. (a) In terms of representations V and W in R(G), the inner product on c ..... (G) takes the simple form (V, IV)= dim HomG(V, IV).

2. Characters

24

(b) tr X e Cci•••(G) is a virtual character, and (X. x) = I, then either X or -X is the character of an irreducible representation, the plus sign occurring when X( I)> 0. U (X, X)= 2, and x(l) > 0, then X is either the sum or the difference of two irreducible characters. (c) If U, V, and Ware irreducible representations, show that U appears in V ® W if and only if W occurs in V* ® U. Deduce that this cannot occur unless dim U 2 dim Wfdim V. We conclude this lecture with some exercises that use characters to work out some standard facts about representations. Exercise 2.34*. Let V and W be irreducible representations of G, and L 0 : I'-> W any linear mapping. Deline L: V-> W by I L(v) = -

I

IGI geG

g- 1 • L 0 (g · v).

Show that L = 0 if V and Ware not isomorphic, and that Lis multiplication by trace(L 0 )/dim(V) if V = W. Exercise 2.35*. Show that, if the irreducible representations of G are represented by unitary matrices [cf. Exercise 1.14], the matrix entries ofthese representations form an orthogonal basis for the space o[all functions on G [with inner product given by (2.11)]. Exercise 2.36*. U G1 and G2 are groups, and 1'1 and 1'2 are representations of G1 and G2 , then the tensor product V1 ® V2 is a representation of G1 x G2 , by(g 1 x g 2 )·(v 1 ® v2 ) = g 1 ·v 1 ®g 2 • v2 • To distinguish this "external" tensor product from the internal tensor product-when G1 = G2 -this exter11al tenso,. product is sometimes denoted V1 1EJ V2 • If Xr is the character of V,, then the value of the character x of 1'1 lEI V2 is given by the product: x(g, x 02) = x.x2(g2).

If V1 and 1'2 are irreducible, show that V1 1EJ V2 is also irredu~ible and show that every irreducible representation of G1 x G2 arises this way. In terms of representation rings,

In these lectures we will often be given a subgroup G of a general linear group GL(V), and we will look for other representations inside tensor powers of V. The following problem, which is a theorem of Burnside and Molien, shows that for a finite group G, all irreducible representations can be found this way.

§2.4. More Projection Formulas; More Consequences

25

Problem 2.37*. Show that if Vis a faithful representation of G, i.e., p: G ~ GL(V) is injective, then any irreducible representation of G is contained in some tensor power ve" of V. Problem 2.38*. Show that the dimension of an irreducible representation of G divides the order of G. Another challenge: Problem 2.39*. Show that the character of any irreducible representation of dimension greater than 1 assumes the value 0 on some conjugacy class of the group.

LECTURE 3

Examples; Induced Representations; Group Algebras; Real Representations

This lecture is something of a grabbag. We start in §3.1 with examples illustrating the use of the techniques of the preceding lecture. Section 3.2 is also by way of an example. We will see quite a bit more about the representations of the symmetric groups in general later; §4 is devoted to this and will certainly subsume this discussion, but this should provide at least a sense of how we can go about analyzing representations of a class of groups, as opposed to individual groups. In §§3.3 and 3.4 we introduce two basic notions in representation theory, induced representations and the group algebra. Finally, in §3.5 we show how to classify representations or a finite group on a real vector space, given the answer to the corresponding question over C, and say a few words about the analogous question for subfields ofC other than R. Everything in this lecture is elementary except Exercises 3.9 and 3.32, which involve the notions of Clifford algebras and the Fourier transform, respectively (both exercises, of course, can be skipped). §3.1: §3.2: §3.3: §3.4: §3.5:

Examples: 6 5 and "1( 5 Exterior powers of the standard representation of 6 1 Induced representations The group algebra Real representations and representations over subfields of C

§3.1. Examples: 6 5 and

~( 5

We have found the representations of the symmetric and alternating groups for n ~ 4. Before turning to a more systematic study of symmetric and alternating groups, we will work out the next couple of cases.

§3.1. Examples: 6 5 and

27

~( 5

Representations of the Symmetric Group 6

5

As before, we start by listing the conjugacy classes of 6 5 and giving the number of elements of each: we have 10 transpositions, 20 three-cycles, 30 four-cycles and 24 five-cycles; in addition, we have 15 elements conjugate to (12)(34) and 10 elements conjugate to (12)(345). As for the irreducible representations, we have, of course, the trivial representation U, the alternating representation U', and the standard representation V; also, as in the case of 6 4 we can tensor the standard representation V with the alternating one to obtain another irreducible representation V' with character Xv· = Xv · Xv·· Exercise 3.1. Find the characters of the representations V and V'; deduce in particular that V and V' are distinct irreducible representations. The first four rows of the character table are thus 10

20

30

24

15

20

6,

(12)

(123)

(1234)

(12345)

(12)(34)

(12)(345)

u

1 -1 2 -2

-1 -I

1 I 0 0

-1 -I

U'

v V'

4 4

1 -1 0 0

Clearly, we need three more irreducible representations. Where should we look for these? On the basis of our previous experience (and Problem 2.37), a natural place would be in the tensor products/powers of the irreducible representations we have found so far, in particular in V ® V (the other two possible products will yield nothing new: we have V' ® V = V ® V ® U' and V' ® V' = V ® V). Of course, V ® V breaks up into N Vand Sym 2 V, so we look at these separately. To start with, by the formula X!l'v(g)

= f(xv(l7) 2 -

Xv(l1 2 ))

we calculate the character of NV: X11•v

= (6, 0, 0, 0,

I, -2, 0);

we see from this that it is indeed a fifth irreducible representation (and that

NV® U' = NV, so we get nothing new that way). We can now find the remaining two representations in either of two ways. First, if n1 and n2 are their dimensions, we have

5! = 120 = 12 + 12 + 42 + 42 + 62 + nf + n~, so n~ + n~ = 50. There are no more one-dimensional representations, since these are trivial on normal subgroups whose quotient group is cyclic, and IJ£ 5

28

3. Examples; Induced Representations; Group Algebras; Real Representations

is the only such subgroup. So the only possibility is 11 1 = n2 = 5. Let W denote one of these live-dimensional representations, and set W' = W ® U'. In the table, if the row giving the character of W is a 1 a2

(5

a3

a4

a~

a6 ~

that of W' is (5 -IX 1 IX 2 -a 3 a 4 a 5 -1X 6 ). Using the orthogonality relations or (2.20), one sees that W' 1: W; and with a little calculation, up to interchanging W and W', the last two rows are as given:

6,

u U'

v

V' Nv

w W'

I I 4 4 6 5 5

10

20

30

24

15

20

(12)

(123)

(1234)

(12345)

(12)(34)

(12)(345)

I -I 2 -2 0 I -1

1 I I I 0 -I -1

I -I 0 0 0 -I

I I -I -I I 0 0

From the decomposition V G) U and V* = V. The perfect pairing 1

vX

= c~,

1 1 0 0

-2

we have also /\4 V

I -I -I I 0 I -I

= NC 5 =

U',

I\ 3 V--+ /\4 v = U',

taking v x (v 1 " v2 " v3 ) to 11" v1 " v2 " v3 shows that NV is isomorphic to v• ® U' = v·. Another way to lind the representations Wand W' would be to proceed with our original plan, and look at the representation Sym 2 JI. We will leave this in the form of an exercise: Exercise 3.2. (i) Find the character of the representation Sym 2 V. (ii) Without using any knowledge of the character table of 6~, use this to show that Sym 2 Jl is the direct sum of three distinct irreducible representations. (iii) Using our knowledge of the first live rows of the character table, show that Sym 2 Vis the direct sum of the representations U, V. and a third irreducible representation W Complete the character table for 6~. Exercise 3.3. Find the decomposition into irreducibles of the representations Nw, Sym 2 W, and V ® W 1 H V and Ware 11-dimensional vector spaces, and U is one dimensional, a perfect pairing is a bilinear map (1: V x W-+ U such that no nonzero vector v in V has fl(v, W) = 0. Equivalently, the map V ... Hom(W, U),. ® U, VH(WH(I(v, w)), is an isomorphism.

w•

§3.1. Examples:

6~

and lll 5

29

Representations of the Alternating Group

~1 5

What happens to the conjugacy classes above if we replace 6d by 111 4 ? Obviously, all the odd conjugacy classes disappear; but at the same time, since conjugation by a transposition is now an outer, rather than inner, automorphism, some conjugacy classes may break into two. Exercise 3.4. Show that the conjugacy class in 6d of permutations consisting of products of disjoint cycles oflengths b 1 , b2 , ••. will break up into the union of two conjugacy classes in 111 4 if all the b,. are odd and distinct: if any bt are even or repeated, it remains a single conjugacy class in 111 4 • (We consider a fixed point as a cycle of length l.) In the case of 111~, this means we have the conjugacy class of three-cycles (as before, 20 elements), and of products of two disjoint transpositions (15 elements); the conjugacy class of live-cycles, however, breaks up into the conjugacy classes of(l2345) and (21345), each having 12 elements. As for the representations, the obvious first place to look is at restrictions to 111 5 of the irreducible representations of 6~ found above. An irreducible representation of 6s may become reducible when restricted to Ills; or two distinct representations may become isomorphic. as will be the case with U and U', V and V', or W and W'. In fact, U, V, and W stay irreducible since their characters satisfy (x. x) = 1. But the character of Nv has values (6, 0, -2, 1, 1) on the conjugacy classes listed above, so (x, x) = 2, and NV is the sum of two irreducible representations, which we denote by Yand Z. Since the sums of the squares of all the dimensions is 60, (dim Y) 2 + (dim Z) 2 = 18, so each must be three dimensional. Exercise 3.5. Use the orthogonality relations to complete the character table of Ills:

Ill,

u v

20

15

12

12

(123)

(12)(34)

(12345)

(21345)

I -I

I -I

I 4

I

0

w

5

-I

y

3

0

z

3

0

0

-I -I

~+JS

0

~j5

2

2

l-j5

~-~

2

2

The representations Y and Z may in fact be familiar: Ills can be realized as the group of motions of an icosahedron (or, equivalently, of a dodecahedron)

30

3. Examples; Induced Representations; Group Algebras; Real Representations

and Y is the corresponding representation. Note that the two representations 111 5 ..... GL 3 (R) corresponding to Y and Z have the same image, but (as you can see from the fact that their characters differ only on the conjugacy classes of (12345) and (21345)) differ by an outer automorphism ofiJI 5 • Note also that NV does not decompose over 0; we could see this directly from the fact that the vertices of a dodecahedron cannot all have rational coordinates, which follows from the analogous fact for a regular pentagon in the plane. Exercise 3.6. Find the decomposition of the permutation representation ofiJI 5 corresponding to the (i) vertices, (ii) faces, and (iii) edges of the icosahedron. Exercise 3.7. Consider the dihedral group D2., defined to be the group of isometries of a regular n-gon in the plane. Let f ~ lfn c: D2• be the subgroup of rotations. Use the methods of Lecture 1 (applied there to the case 6 3 ~ D6 ) to analyze the representations of D2.: that is, restrict an arbitrary representation of D2 • to f, break it up into eigenspaces for the action off, and ask how the remaining generator of D2 • acts of these eigenspaces. Exercise 3.8. Analyze the representations of the dihedral group D2 • using the character theory developed in Lecture 2. Exercise 3.9. (a) Find the character table of the group of order 8 consisting of the quaternions { ± l, ±i, ±j, ±k} under multiplication. This is the case m = 3 of a collection of groups of order 2m, which we denote Hm. To describe them, let C,. denote the complex Clifford algebra generated by v 1 , • •• , vm with relations vf = - l and v,· vi = -vi· V~o so C,. has a basis v 1 = v1, • ••• • v1., as I = { i 1 < · · · < i,} varies over subsets of { l, ... , m }. (See §20.1 for notation and basic facts about Oifford algebras). Set H,. = { ± v1 :

Ill is even} c: (c:,••")*.

This group is a 2-to-1 covering of the abelian 2-group of m x m diagonal matrices with ± I diagonal entries and determinant I. The center of Hm is {±I} if m is odd and is { ± l, ± v1l. ...• m)} if m is even. The other conjugacy classes consist of pairs of elements { ± v1 }. The isomorphisms1of c:,••n with a matrix algebra or a product of two matrix algebras give a 2"-dimensional "spin" representationS of H 2.+t, and two 2"- 1-dimensional "spin" or "halfspin" representations s+ and s- of H 2 •• (b) Compute the characters of these spin representations and verify that they are irreducible. (c) Deduce that the spin representations, together with the 2m-t onedimensional representations coming from the abelian group Hm/{ ± l} give a complete set of irreducible representations, and compute the character table for Hm.

§3.2. Exterior Powers of the Standard Representation of s.

31

For odd m the groups H,. are examples of extra-SfJecia/2-groups, cf. [Grie], [Qu]. Exercise 3.10. Find the character table of the group SL 2 (Z/3). Exercise 3.ll. Let H(Z/3) be the Heisenber(J group of order 27:

H(Z/3)

~ {( ~ ~

; ). n, b,' e Z/3} c SL,(Z/3).

Analyze the representations of H(l./3), first by the methods of Lecture (restricting in this case to the center

z~

m!

H··Z/3}

~ Z/3

of H(Z/3)), and then by character theory.

§3.2. Exterior Powers of the Standard Representation of 6d How should we go about constructing representations of the symmetric groups in general? The answer to this is not immediate; it is a subject that will occupy most of the next lecture (where we will produce all the irreducible representations of 6~). For now, as an example of the elementary techniques developed so far we will analyze directly one of the obvious candidates: Proposition 3.12. Each exterior power NV of tl1e standard representation V of is irreducible, 0 ~ k ~ d - l.

6~

PROOF. From the decomposition c~ = VEB and only if(Xc•· Xc•) = 2. Similarly, since

u, we see that

Nc~ =(NV® /\0 U)EB(N-• V® NU)

vis irreducible if

= NVEB (\H V,

it suffices to show that (x, x) = 2, where xis the character of the representation Nc~. Let A = {I, 2, ... , d }. For a subset B of A with k elements, and (J e G = 6~, let if g(B) #- B

{g}s

= {

~

-I

if g(B)

=

Band gl 8 is an even permutation

if g(B) = Band gl 8 is odd.

32

3. Examples; Induced Representations; Group Algebras; Real Representations

Here, if g(B) Then x(g)

= B, gl 8 denotes the permutation of the set B determined by g.

=I {g}B, and

I (I {g}B) 2

<x. x> = -d'•f&G . I

= -d1

II

I II

1•G B

=

C

{oh{ll}c

l -d1 L: > I;(sgn gl 8 )·(sgn g!c), • B

~I

where the sums are over subsets Band C of A with k elements, and in the last equation, the sum is over those g with g(B) = Band g(C) = C. Such g is given by four permutations: one of B n C, one of Bl B n C, one of C1 B n C, and one of A 1B v C. Letting I be the cardinality of B n C, this last sum can be written l

d1 ~ ~ .~. bo ~•-• <• ~•-• A< 6~•••• (sgn a) 2 (sgn b)(sgn c) =

-d'. L >"C l!(d- 2k +I)! ( • B

>

lte ~k-r

sgn

b)( }~t-r sgn c). tE

These last sums are zero unless k - I = 0 or 1. The case k = I gives

}r ~ k!(d- k)! = ~(~)k!(n- k)! = t. Similarly, the terins with k- I= t also add up to I, so (x, x) = 2, as required.

0 Note by way of contrast that the symmetric powers of the standard representation of 6~ are almost never irreducible. For example, we already know that the representation Sym 2 V contains one copy of the trivial representation: this is just the statement that every irreducible real representation (such as V) admits an inner product (unique, up to scalars) invariant under the group action; nor is the quotient of Sym 2 V by this trivial subrepresentation necessarily irreducible, as witness the case of 6 5 .

§3.3. Induced Representations If H c G is a subgroup, any representation V of G restricts to a representation of H, denoted ResZ V or simple Res V. In this section, we describe an important construction which produces representations of G from representations of H. Suppose V is a representation of G, and W c V is a subspace which is H-invariant. For any gin G, the subspace g · W = {g · w: we W} depends only on the left coset gH of g modulo H,sincegh· W = g·(la · W) = g· W; for a coset

33

§3.3. Induced Representations

u in G/H, we write u· W for this subspace of V. We say that Vis induced by W if every element in V can be written uniquely as a sum of elements in such translates of W, i.e.,

I'=

EB

u·W

<JeG/H

In this case we write Jl = Indf, W

= Ind W

Example 3.13. The permutation representation associated to the left action of G on G/H is induced from the trivial one-dimensional representation W of H. Here V has basis {e.,: u e G/H}, and W = C · e11 , with H the trivial coset. Example 3.14. The regular representation of G is induced from the regular representation of H. Here V has basis {e.: g e G}, whereas W has basis {e,: h eH}. We claim that, given a representation W of II, such I' exists and is unique up to isomorphism. Although we will later give several fancier ways to see this, it is not hard to do it by hand. Choose a representative g., e G for each coset u e GfH, withe representing the trivial coset H. To see the uniqueness, note that each element of V has a unique expression v = L g., w.,, for elements w., in W Given gin G, write g ·g., = g, · h for some t e G/H and It e H. Then we must have g·(g.,w.,)

= (g·g.,)w., = (g,"IJ)w., =

g,(hw.,).

This proves the uniqueness and tells us how to construct V = lnd(W) from W Take a copy W" of W for each left coset u e G/Il; for we W, let g., IV denote W", so every the element of W" corresponding to IV in W Let V =

EB

<JeG/If

element of V has a unique expression v = l:g.,w., for elements w., in W Given g e G, define

g·(g.,w.,)=g,(IJw.,)

ifg·g.,=g,·ll.

To show that this defines as action of G on V, we must verify that g' · (g ·(g., w.,)) = (g' ·g)· (g., IV.,) for another element g' in G. Now if g' · g, = g, · h', then

g' · (g ·(g., IV.,))= g' · (g,(hw.,)) = g,.(h'(hw.,)). Since (g' ·g)· g., = g' · (g ·g.,) = g' · (Jr' h = g, · h' ·I•, we have (g'·g)·(g.,w.,)

= g,((lr'·lt)w.,) =

g,{ll'·(hw.,)),

as required. Example 3.15. If W =

EB W,, then Ind IV= ffilnd W,.

The existence of the induced representation follows from Examples 3.14 and 3.15 since any W is a direct sum of summands of the regular representation.

34

3. Examples; Induced Representations; Group Algebras; Real Representations

Exercise 3.16. (a) If U is a representation of G and W a representation of H, show that (with all tensor products over C)

U ® Ind W = Ind(Res(U) ® W). In particular, Ind(Res(U)) = U ® P, where Pis the permutation representation of G on G/11. For a formula for Res(lnd(W)), for W a representation of II, see [Se2, p. 58]. (b) Like restriction, induction is transitive: if H c K c G are subgroups, show that Indf,(W) = Ind~(Ind~ W). Note that Example 3.15 says that the map lnd gives a group homomorphism between the representation rings R(/J) and R(G), in the opposite direction from the ring homomorphism Res: R(G) -• R(H) given by restriction; Exercise 3.l6(a) says that this map satisfies a "push-pull" formula a· Ind(P) = lnd(Res(a) · p) with respect to the restriction map. Proposition 3.17. Let W be a representation of 11, U a representation of G, and suppose V = Ind W. Tlten any H-module l10momorpltism U. i.e., Hom 8 (W, Res U)

= HomG(Ind W,

U).

/n particular, tltis universal property determines lnd W up to canmtical

isomorpllism. PROOF.

With V =

EB...

G!H

u ·Was before, define

q; on u · W by

-1

u·W~W~U~U,

which is independent of the representative g" for u since
0

To compute the character of V = Ind W, note that g e G maps uW to guH~ so the trace is calculated from those cosets u with gu = u, i.e., s- 1gs e II for s e u. Therefore, Xlnd

L

w(g) =

Xw(s- 1gs)

(s e u arbitrary).

(3.18)

ga=t1

Exercise 3.19. (a) If Cis a conjugacy class of G, and C n H decomposes into conjugacy classes D1 , ••. , D, of H, (3.18) can be rewritten as: the value of the character of Ind W on C is

IGI ' ID,I

Xtnd

w(C) = IHI,~

TCf Xw(D,).

(b) If W is the trivial representation of H, then

§3.3. Induced Representations

35

Corollary 3.20 (Frobenius Reciprocity). If W is a representation of H, and U a representation of G, then

PROOF. It suffices by linearity to prove this when W and U are irreducible. The left-hand side is the number of times U appears in Ind W, which is the dimension of HomG(Ind W, U). The right-hand side is the dimension of Hom,(W, Res U). These dimensions are equal by the proposition. 0

If Wand U are irreducible, Frobenius reciprocity says: tiJe numher of times U appears in lnd W is the same as the number of times W appears in Res U. Frobenius reciprocity can be used to find characters of G if characters of Hare known. Example 3.21. We compute lnd~ W, when H = 6 2 c G = 6 3 , W = V2 (the standard representation) = U~ (the alternating representation). We know the irreducible represenatations of 6 3 : U3 , U3, V3 , which restrict to U2 , Ui = V2 , U2 EB Ui, respectively. Thus, by Frobenius, lnd V2 = U] EB 1'3 . Example 3.22. Consider next H = 6 3 c G = 6 4 , W = 1'3 . Again we know the irreducible representations, and Res U4 = U3 , Res U~ = Ui, Res 1'4 = U3 €B V3 [the vector

v4 = {(x,, Xz, X], x4): I xi= 0} Ui EB v;, with V3 = V3 , and Res W4 = J.-; (as one may

(1,1,1, -3)e

is fixed by II], Res V~ = see directly). Hence, Ind V3 = V4 €B V~ ffi W4 • (Note that the isomorphism Res W4 = V3 actually follows, since one W4 is all that could be added to 1'4 ffi ~~ to get Ind V3 .)

Exercise 3.23. Determine the isomorphism classes of the representations of 6 4 induced by (i) the one-dimensional representation of the group generated by (1234) in which (l234)·v = iv, i = (ii) the one-dimensional representation of the group generated by (123) in which (l23)·v = e 2 " 1' 3 v.

.J=l;

Exercise 3.24. Let H = ~~~ c G = 6~. Show that Ind U = U EJj U', Ind V = VEB V', and Ind W = W$ W', whereas lnd Y = Ind Z =NV. Exercise 3.25*. Which irreducible representations of 6 4 remain irreducible when restricted to ~1 4 ? Which are induced from 111 4 ? How much does this tell you about the irreducible representations of 111 4 1 Exercise 3.26*. There is a unique nonabelian group of order 21, which can be realized as the group of affine transformations x 1-4 ax + pof the line over the field with seven elements, with a a cube root of unity in that field. Find the irreducible representations and character table for this group.

36

3. Examples; Induced Representations; Group Algebras; Real Representations

Now that we have introduced the notion of induced representation, we can state two important theorems describing the characters of representations of a finite group. In the preceding lecture we mentioned the notion of virtual character; this is just an element of the image A of the character map

x: R(G)--+ Cc~... (G) from the representation ring R(G) of virtual representations. The following two theorems both state that in order to generate A® Q (resp. A) it is enough to consider the simplest kind of induced representations, namely, those induced from cyclic (respective elementary) subgroups of G. For the proofs of these theorems we refer to [Se2, §9, lO]. We will not need them in these lectures. Arlin's Theorem 3.27. The cltaracters of induced represe1ttations ji·om cyclic subgroups of G generate a lattice of firtite i11dex In A. A subgroup H of G is p-elementary if H prime to p and B a p-group.

=A

x B, with A cyclic of order

Brauer's Theorem 3.28. The characters of induced representations from elementary subgroups of G generate the lattice A.

§3.4. The Group Algebra There is an important notion that we have already dealt with implicitly but not explicitly; this is the group algebra CG associated to a finite group G. This is an object that for all intents and purposes can completely replace the group G itself; any statement about the representations of G has an exact equivalent statement about the group algebra. Indeed, to a large extent the choice of language is a matter of taste. The underlying vector space of the group algebra of G is the vector space with basis {e,} corresponding to elements of the group G, that is, the underlying vector space of the regular representation. We define the algebra structure on this vector space simply by ·

e, · eh = e,h· By a representation of the algebra CG on a vector space V we mean simply an algebra homomorphism CG--+ End(V), so that a representation V of CG is the same thing as a left CG-module. Note that a representation p: G--+ Aut(V) will extend by linearity to a map p: CG--+ End(V), so that representations of CG correspond exactly to representations of G; the left CG-module given by CG itself corresponds to the regular representation.

37

§3.4. The Group Algebra

If { ~} are the irreducible representations of G, then we have seen that the regular representation R decomposes R =

EB

(~)ffidim(ll',)_

We can now refine this statement in terms of the group algebra: we have Proposition 3.29. As algebras,

CG ~

EB End(~).

PRooF. As we have said, for any representation W of G, the map G-+ Aut(JV) extends by linearity to a map CG-+ End(JV); applying this to each of the irreducible representations ~ gives us a canonical map rp: CG-+

EB End(Jf't).

This is injective since the representation on the regular representation is faithfuL Since both have dimension (dim HW, the map is an isomorphism.

L

0 A few remarks are in order about the isomorphism rp of the proposition. First, rp can be interpreted as the Fourier transform, c[ Exercise 3.32. Note also that Proposition 2.28 has a natural interpretation in terms of the group algebra: it says that the center of CG consists of those a(g)e, for which a is a class function. Next, we can thinkof rp as the decomposition of the semisimple algebra CG into a product of matrix algebras. It implies that the matrix entries of the irreducible representations give a ba~is for the space of all functions on G, c[ Exercise 2.35. Note in particular that any irreducible representation is isomorphic to a (minimal) left ideal in CG. These left ideals are generated by idempotents. In fact, we can interpret the projection formulas of the last lecture in the language of the group algebra: the formulas say simply that the elements

L

- · e, E CG dim W · 1 - I'I " L. Xw((J) G

geG

are the idempotents in the group algebra corresponding to the direct sum factors in the decomposition of Proposition 3.29. To locate the irreducible representations W, of a group G [not just a direct sum of dim(W,) copies], we want to find other idempotents of CG. We will see this carried out for the symmetric groups in the following lecture. The group algebra also gives us another description of induced representations: if W is a representation of a subgroup li of G, then the induced representation may be constructed simply by Ind lV= CG®c 11 W,

38

3. Examples; Induced Representations; Group Algebras; Real Representations

where G acts on the first factor: g · (e,. ® w) = e,,. ® w. The isomorphism of the reciprocity theorem is then a special case of a general formula for a change of rings CH--> CG: Homc11 (1.V, U) = HomcG(CG ®cu W, U). Exercise 3.30•. The induced representation Ind(W) can also be realized concretely as a space of W-valued functions on G, which can be useful to produce matrix realizations, or when trying to decompose Ind(W) into irreducible pieces. Show that Ind(W) is isomorphic to

Hom 8 (CG, W) ~ {!: G--> W:f(ltg) = lif(g), VII e H, g E G}, where G acts by (g' · f)(g) = f(gg'). Exercise 3.31. IfCG is identified with the space of functions on G, the function (P corresponding to GII'(g)e,, show that the product in CG corresponds to the convolution • of functions:

:L, ..

(With integration replacing summation, this indicates how one may extend the notion of regular representation to compact groups.) Exercise 3.32•. If p: G--> GL(Vp) is a representation, and
ciJ(rl = I

· p(g).

geG

"

(a) Show that
§3.5. Real Representations and Representations over Subfields of C If a group G acts on a real vector space V0 , then we say the corresponding complex representation of V = V0 ®R C is real. To the extent that we are interested in the action of a group G on real rather than complex vector spaces, the problem we face is to say which of the complex representations of G we have studied are in fact real. Our first guess might be that a representation is real if and only if its character is real-valued. This turns out not to be the case: the character of a real representation is certainly real-valued, but the converse need not be true. To lind an example, suppose G c SU(2) is a finite, non abelian subgroup. Then G acts on C 2 = V with a real-valued character since the trace of any matrix in SU(2) is real. If V were a real representation, however, then G would be a subgroup of 80(2) = S 1, which is abelian. To produce such a group, note that SU(2) can be identified with the unit quaternions. Set G = {±I, ± i, ±1, ±k}. Then G/{ ± 1} is abelian, so has four one-dimensional representations, which give four one-dimensional representations of G. Thus, G has one irreducible two-dimensional representation, whose character is real, but which is not real.

Exercise 3.34*. Compute the character table for this quaternion group G, and compare it with the character table of the dihedral group of order 8.

40

3. Examples; Induced Representations; Group Algebras; Real Representations

A more successful approach is to note that if V is a real representation of G, coming from V0 as above, then one can find a positive definite symmetric bilinear form on V0 which is preserved by G. This gives a symmetric bilinear form on V which is preserved by G. Not every representation will have such a form since degeneracies may arise when one tries to construct one following the construction of Proposition 1.5. In fact, Lemma 3.35. A11 irreducible representation V of G is real if and only if tl1ere is a nondegenerate symmetric bilinear form B 011 V preserved by G. PROOF. If we have such B, and an arbitraf} nondegenerate Hermitian form H, also G-invariant, then

v! v•~ v gives a conjugate linear isomorphism qJ from V to V: given x e V, there is a unique lp(x) e V with B(x, y) = H(
= B(x, y) =

B(y, x) = H(
from which it follows that H(

2nk

p: kf-+

(

cosn

. 2nk

SID-

-sin 2nk) II

2nk

COSII

II

is irreducible over R for 11 > 2 (no line in R is fixed by the action of Z/n), but will be reducible over C. Thus, classifying the irreducible representations of G over C that are real does not mean that we have classified all the irreducible real representations. However, we will see in Exercise 3.39 below how to finish the story once we have found the real representations of G that are irreducible over C. 2

§3.5. Real Representations and Representations over Subfields of C

41

Suppose Vis an irreducible representation of G with x,. real. Then there is a G-equivariant isomorphism V ~ v•, i.e., there is a G-equivariant (nondegenerate) bilinear form B on V; but, in general, B need not be symmetric. Regarding B in

and rioting the uniqueness of B up to multiplication by scalars, we see that B is either symmetric or skew-symmetric. If B is skew-symmetric, proceeding as above one can scale so rp 2 = - Id. This makes V "quaternionic," with rp becoming multiplication 2 by j: Definition 3.36. A quaternionic representation is a (complex) representation V which has a G-invariant homomorphism J: V-> V that is conjugate linear, and satisfies J2 = -I d. Thus, a skew-symmetric non degenerate G-invariant B determines a quaternionic structure on V. Summarizing the preceding discussion we have the Theoem 3.37. Ara irreducible representation V is one and only one of the following:

(I) Complex: Xv is not real-valued; V does not IJave a G-invariant noradegenerate bilinear form. (2) Real: V = V0 ® C, a real representation; V l1as a G-iuvariant symmetric nondegenerate bilinear form. (3) Quaternionic: Xv is real, but V is not real; V l1as a G-invariant skewsymmetric nondegeneate bilinear form. Exercise 3.38. Show that for V irreducible, if Vis complex if Vis real if V is quaternionic. This verifies that the three cases in the theorem are mutually exclusive. It also implies that if the order of G is odd, all nontrivial representations must be complex. Exercise 3.39. Let V0 be a real vector space on which G acts irreducibly, V = V0 ® C the corresponding real representation of G. Show that if Vis not irreducible, then it has exactly two irreducible factors, and they are conjugate complex representations of G. 2

See §7.2 for more on quaternions and quaternonic representatious.

42

3. Examples; Induced Representations; Group Algebras; Real Representations

Exercise 3.40. Classify the real representations of ~( 4 • Exercise 3.41*. The group algebra RG is a product of simpleR-algebras corresponding to the irreducible representations over IR. These simple algebras are matrix. algebras over C, R, or the quaternions 11-B according as the representation is complex, real, or quaternionic. Exercise 3.42*. (a) Show that ati characters of a group are real if and only if every element is conjugate to its inverse. (b) Show that an element a in a split conjugacy class of 11( 1 is conjugate to its inverse if and only if the number of cycles in a whose length is congruent to 3 modulo 4 is even. (c) Show that the only d's for which every character ofiJ( 4 is real-valued are d = l, 2, 5, 6, 10, and 14. Exercise 3.43*. Show that: (i) the tensor product of two real or two quaternionic representations is real; (ii) for any V, V* ® Vis real; (iii) if Vis real, so are all NV; (iv) if Vis quaternionic, NV is real fork even, quaternionic for kodd.

Representations over Sublields of C in General We consider next the generalization of the preceding problem to more general sublields of C. Unfortunately, our results will not be nearly as strong in general, but we can at least express the problem neatly in terms of the representation ring of G. To begin with, our terminology in this general setting is a little different. Let K c C be any sublield. We deline a K-representation of G to be a vector space V0 over K on which G acts; in this case we say that the complex. representation V = V0 ® C is defined over K. One way to measure how many of the representations ofG are defined over a field K is to introduce the representation ring R"(G) of G over K. This is defined just like the ordinary representation ring; that is, it is just the group of formal linear combinations of K-representations of G modulo relations of the form V + W - ( V (9 W), with multiplication given by tensor product. Exercise 3.44*. Describe the representation ring of G over 1R for some of the groups G whose complex representation we have analyzed above. In particular, is the rank of RR(G) always the same as the rank of R(G)? Exercise 3.45*. (a) Show that R"(G) is the subring of the ring of class functions on G generated (as an additive group) by characters of representations defined over K.

§3.5. Real Representations and Representations over Sublields of C

43

(b) Show that the characters of irreducible representations over K form an orthogonal basis for RK(G). (c) Show that a complex representation of G can be defined over K if and only if its character belongs to RK(G). For more on the relation between R 1:(G) and R(G), see [Se2].

LECTURE 4

Representations of Sd: Young Diagratns and Frobenius's Character Formula

Jn this lecture we get to work. Specifically, we give in §4.1 a complete description of the irreducible representations o.f the symmetric group, that is, a construction of the representations (via Young synllnetrizers) and a formula(Frobcnius' formula) for their characters. The proof that the representations constructed in §4.1 are indeed the irreducihle representations of the symmetric group is given in §4.2; the proof of Frobenius' formula, as well as a number of others, in §4.3. Apart from their intrinsic interest (and undeniable beauty), these results turn out to be of substantial interest in Lie theory: analogs of the Young symmetrizers will give a construction of the irreducible representations of SL.C. At the same time, while the techniques of this lecture are completely elementary (we use only a few identities about symmetric polynomials, proved in Appendix A), the level of difficulty is clearly higher than in preceding lectures. The results in the latter half of §4.3 (from Corollary 4.39 on) in particular are quite difficult, and inasmuch as they are not used later in the text may be skipped hy readers who are not symmetric group enthusiasts.

§4.1: Statements of the results §4.2: Irreducible representatimis of 6d §4.3: Proof of Frobenius's formula

§4.1. Statements of the Results The number of irreducible representaton of 6d is the number of conjugacy classes, which is the number p(d) of partitions 1 of d: d = ..1. 1 + · ·· + A.t. A., ~ .. · ~ ll ~ 1. We have 1 It is sometimes convenient, and sometimes a nuisance, to have partitions that end in one or more zeros; ir convenient, we allow some or the J., on the end to be zero. Two sequences define the same partition, or course, ir they di!Ter only by zeros at the end.

45

§4.1. Statements of the Results

2:"'

~ao

"' ( --. I ) p(d)td = I1 n:l 1- t

+t+t 2 +· .. )(1 +t 2 +t 4 + .. ·)(1 +t 3 +"')"" which converges exactly in ltl < l. This partition number is an interesting =(1

arithmetic function, whose congruences and growth behavior as a function of d have been much studied (cf. [Har], [And]). For example, p(d) is asymptotically equal to ( 1/ad)e" .fl, with a = 4j3 and fJ = n.jfji To a partition A.= (A. 1, ••• , A.t) is associated a Young diagmm (sometimes called a Young frame or Ferrers diagram)

with A.1 boxes in the ith row, the rows of boxes lined up on the left. The conjugate partition A.' = (A.'1 , ••• , A.;) to the partition A. is defined by interchanging rows and columns in the Young diagram, i.e., reflecting the diagram in the 45o line. For example, the diagram above is that of the partition (3, 3, 2, l, I), whose conjugate is (5, 3, 2). (Without reference to the diagram, the conjugate partition to A. can be defined by saying A.; is the number of terms in the partition A. that are greater than or equal to i.) · Young diagrams can be used to describe projection operators for the regular representation, which will then give the irreducible representations of 6~. For a given Young diagram, number the boxes, say consecutively as shown:

J 5 7

8

More generally, define a tableau on a given Young diagram to be a numbering of the boxes by the 'integers I, ... , d. Given a tableau, say the canonical one shown, define two subgroups 2 of the symmetric group z If a tableau olher than the canonical one were chosen, one would get different [lroups in place of P and Q, and different elemenls in the group ring, but lhe representations construcled this way will be isomorphic.

46 4. Representations or 6 4 : Young Diagrams and frobenius's Character formula

P = P1 = {g e Sc1: g preserves each row}

and

Q = Q1 = {g e Sc1: g preserves each column}. In the group algebra CS 4 , we introduce two elements corresponding to these subgroups: we set

a1 =

L

e, and bA

goP

=

L

sgn(g) · e,.

(4.1)

geQ

To see what aA and bA do, observe that if Vis any vector space and Sc~ acts on the dth tensor power y®cl by permuting factors, the image of the element a1 e CSc1-+ End(V®cl) is just the subspace lm(aA) = SymA' V®;SymA•V® ... ® Sym 1•V c y®cl, where the inclusion on the right is obtained by grouping the factors of y®cl according to the rows of the Young tableaux. Similarly, the image of bA on this tensor power is Im(bA)

= N'•V® N'•V® .. · ® N'•V c

V®cl,

where JL is the conjugate partition to A.. Finally, we set (4.2)

this is called a Young symmetrizer. For example, when .l =(d), c1c11 = a 1c11 = l:,.e.e,, and the image of c 1c~1 on y®cl is Sym 4 V. When .l=(l, ... , 1), c 1 ~, ... , 11 = b(1, ... , 11 = I,.e.sgn(g)e,,and theimageofc 0 , ... , 11 on y®clisNV. We will eventually see that the image of the symmetrizers c1 in y®cl provide essentially all the finite-dimensional irreducible representations of GL(V). Here we state the corresponding fact for representations of Sc~: Theorem 4.3. Some scalar multiple of cA is idempotent, i.e., d = nAcA, and tlte image of c1 (by right multiplication on CS4 ) i.~ a11 irreducible representation VA of sci. Every irreducible represemation of. sd CQ/1 be obtained in this way for a unique partition. We will prove this theorem in the next section. Note that, as a corollary, each irreducible representation of S 4 can be defined over the rational numbers since c A is in the rational group algebra oscl. Note also that the theorem gives a direct correspondence between conjugacy classes in S 4 and irreducible representations of S 4 , something which has never been achieved for general groups. For example, for ,l =(d), l~c11 = C6c1·

L fEeid

e, = C ·

L ft:Sd'

e,

§4.1. Statements or the Results

47

is the trivial representation U, and when

l'.t .....• , = C6., · .L

..1.

=(I, ... , 1),

sgn(g)e,

=C·

is the alternating representation U'. For c,2,11 =(e.

.L

sgn(g)e,

ge Scr

IE S,

..1.

= (2,

t),

+ e,u,)·(et- eU31) = I+ e[l21- e!t31- e(tJZJ

in C6 3 , and 11( 2 , 11 is spanned by c12 , 11 and (13) · c12 , 11 , so 11( 2 , 11 is the standard representation of 6 3 • Exercise 4.4*. Set A= C64 , so V~ = Ac~ = Aa~bA.

(a) Show that VA~ AbAaA. (b) Show that VA is the image of the map from AaA to AbA given by right multiplication by bA. By (a), this is isomorphic to the image of AbA-+ AaA given by right multiplication by aA. (c) Using (a) and the description of VA in the theorem show that

VA'= VA® U', where ..1.' is the conjugate partition to A. and U' is the alternating representation. Examples 4.5. In earlier lectures we described the irreducible representations of 6., ford ~ 5. From the construction of the representation corresponding to a Young diagram it is not hard to work out which representations come from which diagrams:

e2

rn

e3

[IIJ

u

ITID

fJD

v

alternating

§

trivial

EP e4

8

trivial

v

U' alternating

standard

u

~

~ V'

U'

EB

w

48 4. Representations of 6 4 : Young Diagrams and Frobenius's Character Formula

Ss

EflTI ~

~

u

I I I I I

[f

v w

U'

~

V'

EfP

1\zV

W'

Exercise 4.6*. Show that for general d, the standard representation V corresponds to the partition d = (d- I)+ I. As a challenge, you can try to prove that the exterior powers of the standard representation V are represented by a "hook":

Note that this recovers our theorem that the 1\'V are irreducible. Next we turn to Frobenius's formula for the character XA of V1 , which includes a formula for its dimension. Let C1 denote the conjugacy class in 6 4 determined by a sequence i = (i., i 2 ,

... , i 4 )

with L ai~ = d:

c, consists of those permutations that have i 1 !-cycles, i2 2-cycles, ... , and i4 d-cycles. Introduce independent variables x 1 , ••• , x_, with k at least as large as the number of rows in the Young diagram of A.. Define the power sums lj(x), I S: j S: d, and the discriminallt A(x) by fj(x)= x{ +xi+ .. ·+ xl. A(x) =

n (x, M

(4.7)

x;)·

If f(x) = f(x 1 , ••• , xt) is a formal power series, and (1 1 , of non-negative integers, let

••• ,

ll) is a k-tuple

[/(x)]11 ,, .... 1• 1 =coefficient of x~' · · · · · x!• in f.

(4.8)

Given a partition A.: A.,

~

· · · ~ A.t

~

0 of d, set

12 = A. 2

+ k-

2, ... , It= .t.,

(4.9)

49

§4.1. Statements of the Results

a strictly decreasing sequence of k non-negative integers. The character of VA evaluated on g e C1 is given by the remarkable Frobenlus Formula 4.10

= [t\(x)· n ~(x)'1]

XA(CI)

J

. (1, •...•••1

For eumple, if d = 5, A. = (3, 2), and C1 is the conjugacy class of ( 12)(345), i.e., i1 = 0, i 2 = 1, i 3 = 1, then Xr3.:z1(C,)

= [(xt- X:z)·(x~ + xH(x~ + x~)j 1 4, 21 =I.

Other entries in our character tables for 6 3 , 6 4 , and 6 5 can be verified as easily, verifying the assertions of Examples 4.5. In terms of certain symmetric functions SA called Schur polynomials, Frobenius's formula can be expressed by

n Jj(x)'' = L XA(C,)SA, J

the sum over all partitions A. of d in at most k parts (cf. Proposition 4.37 and (A.27)). Although we do not use Schur polynomials explicitly in this lecture, they piay the central role in the algebraic background developed in Appendix A. Let us use the Frobenius formula to compute the dimension of VA. The conjugacy class of the identity corresponds to i =(d), so dim VA= XA(C1">) = [t\(x) · (x 1 + .. · + x,)d] 11 ,. ... ,1•1. Now t\(x) is the Vandermonde determinant:

XI

The other term is (X

I

" + .. · + x' )d --1...

d!

't·1..... ,.,.I

x'•x'' · 1

2 ··•

· x'•

''

the sum over k-tuples (r 1 , ••• , r,) that sum to d. To lind the coefficient of x:• · ... · in the product, we pair off corresponding terms in these two sums, getting

x!•

L sgn(o} (1

1 -

d! u(k) + 1)1" · (l, - u( I)+ l)!'

the sum over those u in 6, such that 1,_1+ 1 This sum can be written as

-

11(i)

+I

~

0 for all I :5: i :5: k.

50 4. Representations of 6 4 : Young Diagrams and Frobenius's Character Formula

n

d! • . -1· - - ) sgn(u) ip1 - 1)· ... ·(11 - u(k- 1 +I)+ 2) 1! • •• It! wH!• 1~1

cl!

1

r.

l.(r. - I)

11

/1

(1 1

-

I)

By column reduction this determinant reduces to the van der Monde determinant, so dim VA = I

1

,.

~~.II n (I, -

I,),

(4.11)

•. l<J

with 11 = A.1 + k - i. There is another way of expressing the dimensions of the VA. The llook leugt/1 of a box in a Young diagram is the number of squares directly below or directly to the right of the box, including the box once.

In the following diagram, each box is labeled by its hook length:

Hook Length Formula 4.12. d! . d1m VA = = - - - - - (Hook lengths)

n

For the above partition 4 + 3 + l of 8, the dimension of the corresponding representation of 6 8 is therefore 8!/6 · 4 · 4 · 2 · 3 = 70. Exercise 4.13*. Deduce the hook length formula from the Frobenius formula (4.11).

Exercise 4.14*. Use the hook length formula to show that the oniy irreducible representations of 6 4 of dimension less than dare the trivial and alternating representations U and U' of dimension l, the standard representation V and V' = V ® U' of dimension d- I, and three other examples: the two· dimensional representation of 6 4 corresponding to the partition 4 = 2 + 2, and the two live·dimensional representations of 6 6 corresponding to the partitions 6 = 3 + 3 and 6 = 2 + 2 + 2.

§4.1. Statements of the Results

51

Exercise 4.15*. Using Frobenius's formula or otherwise, show that:

= ir- 1; Xrd-2,l,IJ(Cr) = i(ir - l)(ir Xrd-r,r 1(Cr)

- 2)- i2;

Xrd-2,2J(Cr) = Wr - l)(ir - 2)

+ i2- I.

Can you continue this list? Exercise 4.16*. If g is a cycle of length din 6 4 , show that XA(g) is ±I if A. is a hook, and zero if A. is not a hook: ( ={(-1)' XA g) 0

ifA.=(d-s,l, ... ,l),O:$s~d-l otherwise.

Exercise 4.17. Frobenius [Fro I] used his formula to compute the value of XA on a cycle of length m :$ d. (a) Following the procedure that led to (4.11)-which was the case m = 1-show that (4.18) where hm = d!/(d - m)!m is the number of cycles of length 111 (if m > I), and l

= f}

tp(x)

f~r

m

(x - 11),

t/t(x)

= tp(X- m)

fl (X- j + f). }~1

The sum in (4.18) can be realized as the coefficient of x-I in the Laurent expansion of tft(x)ftp(x) at x = oo. Deline the rank r of a partition to be the length of the diagonal of its Young diagram, and let a1 and b1 be the number of boxes below and to the right of the ith box of the diagonal, reading from lower right to upper left. Frobenius . . (Many wnters . ca lled( aral b b ... b t he characteristics o f t he partrt1on. now use

a,)

I

2 "·

r

a reverse notation for the characteristics, writing (b., ... , IJ 1 Ia., For the partition (10, 9, 9, 4, 4, 4, 1):

... , a,) instead.)

I r=4 I

I I

I I

I

. . (20 63 47 69)

c 1aractensltc.~ =

Algebraically, r and the characteristics a 1 < · · ·
/1 1

< · · · < b, are

52 4. Representations of 6 1 : Young Diagrams and Frobenius's Character Formula

{l 1 ,

••• ,1t,k-l-a 1 , ••• ,k-1-a,}

{0, I, ... , k - I, k (b) Show that t/t(x)/rp(x)

and

+ b1 , ••• , k + b,}.

= g(y)//(y), where y = x - d and

n(y- b,) r

/(y) =

r

1~1

n (y +a,+ 1)

n (y- j + 1). m

g(y) = f(y- m)

j=J

1-1

Deduce that the sum in (4.18) is the coefficient of x- 1 in g(x)//(x). (c) When m = 2, use this to prove the formula dim v~ r XA((l2)) = d(d _ I) 1~ (b1(b1 + I)- a1(a 1 + l)). Hurwitz [Hur] used this formula of Frobenius to calculate the number of ways to write a given permutation as a product of transpositions. From this he gave a formula for the number of branched coverings oft he Riemann sphere with a given number of sheets and given simple branch points. Ingram [In] has given other formulas for XA(g), when g is a somewhat more complicated conjugacy class. Exercise 4.19"'. If Vis the standard representation of 6 4 , prove the decompositions into irreducible representations: Sym 2 V ~ U$ V$l-{4 - 2 , 21 ,

V® V = Sym 2 V$NV ~ U $ V$

J-!4-2,2 1 $

J-!4-2,J.t 1·

Exercise 4.20"'. Suppose 1 is symmetric, i.e., 1 = 1', and let q 1 > q 2 > · · · > q, > 0 be the lengths of the symmetric hooks that form the diagram of 1; thus, q 1 = 21 1 - 1, q 2 = 21 2 - 3, .... Show that if g is a product of disjoint cycles of lengths q 1 , q2 , ••• , q, then

§4.2. Irreducible Representations of 6d We show next that the representations VA constructed in the first section are exactly the irreducible representations of 6 4 • This proof appears in many standard texts (e.g. [C-R], [Ja-Ke], [N-S], [Wet]), so we will be a little concise. Let A = C6 4 be the group ring of 6 4 • For a partition 1 of d, let P and Q be the corresponding subgroups preserving the rows and columns of a Young tableau T corresponding to A., let a = aA, b = bA, and let c = cA = ab be

§4.2. Irreducible Representations or 6 4

53

the corresponding Young symmetrizer, so VA= AcA is the corresponding representation. (These groups and elements should really be subscripted by T lo denote dependence on the tableau chosen, but the assertions made depend only on the partition, so we usually omit reference to T.) Note that P n Q = { t }, so an element of 6 4 can be written in almost one way as a product p · q, peP, q e Q. Thus, cis the sum ±e9 , the sum over all g that can be written asp· q, with coefficient ± I being sgn(q); in particular, the coefficient of e 1 in c is t.

2:

Lemma 4.21. (I) For peP, p·a =a· p =a. (2) For q e Q, (sgn(q)q) · b = b · (sgn(q)q) = b. (3) For all peP, q e Q, p·c·(sgn(q)q) = c, and, up to multiplication by a scalar, c is tlte only sucl1 element 111 A.

2:

PROOF. Only the last assertion is not obvious. If n,e, satisfies the condi lion in (3), then "r•9 = sgn(q)n, for all g, p, q; in particular, n,9 = sgn(q)n 1 • Thus, it suffices to verify that n, = 0 if g ¢ PQ. For such g it suffices to find a transposition t such that p = t e P and q = g- 1 tg e Q; for then g = pgq, so n, = -11,. If T' = gT is the tableau obtained by replacing each entry i ofT by g(i), the claim is that there is are two distinct integers that appear in the same row ofT and in the same column of 7"; tis then the transposition of these two integers. We must verify that if there were no such pair of integers, then one could write g = p · q for some p e P, q e Q. To do this, first take p 1 e P and q! e Q' = gQg- 1 so that p 1 T and ql T' have the same first row; repeating on the rest of the tableau, one gets p e P and q' e Q' so that pT = q'T'. Then pT = q'gT, so p = q'g, and therefore g = pq, where q = g- 1 (q')- 1g e Q, as ~~~

0

We order partitions lexicograpllically:

A. > t' if the first non vanishing 11 - Jt 1 is positive.

(4.22)

Lemma 4.23. ( l) If A. > tt. tl1en for all x e A, a A· x · b,, = 0. Ill particular, if A. > tt, then cA · c1, = 0. (2) For all x e A, cA · x · cA =is a scalar multiple of cA. In particular, cA · cA = 11AcA for some IIA e C. PROOF. For (I), we may take x = g e 6 4 • Since g·b,.·g- 1 is the element constructed from gT', where 1" is the tableau used to construct b,., it suffices to show that a A· b,. = 0. One verifies that A.> tt implies that there are two integers in the same row of T and the same column of T'. If t is the transposition of lhe~e integers, then aA·t = aA, t·b,. = -b,., so aA·b,, = aA·t·t·b,, = -aA·b,,, as required. Part (2) follows from Lemma 4.21 (3). 0

Exercise 4.24*. Show that if A. i:

tt. then cA ·A· c1,

= 0; in particular, cA · c,. = 0.

54 4. Representations of 6 4 : Young Diagrams and frobenius's Character fonnula Lemma 4.25. (l) Eacll V4 is an irreducible represer1tation of 6 4 • (2) If A. ¥ JL, then V4 and VP are not isomorpllic. PRooF. For (1) note that c 4 V4 c Cc 4 by Lemma 4.23. If W c V4 is a subrepresentation, then c4 W is either Cc 4 or 0. If the first is true, then V4 = A· cA c W. Otherwise W · W c A· c1 W = 0, but this implies W = 0. Indeed, a projection from A onto W is given by right multiplication by an element rp e A with rp = rp 2 e W · W = 0. This argument also shows that c4 V4 ¥ 0, i.e., that the number n4 of the previous lemma is nonzero. For(2), we may assume A.> Jl· Then cA V1 = CcA ¥0, but cA VP = cA· AcP = 0, so they cannot be isomorphic A-modules. 0

PROOF. Let F be right multiplication by c 4 on A. Since F is multiplication by n 4 on VA, and zero on Ker(cA), the trace ofF is n4 times the dimension of VA. But the coefficient of e, in e, · c 4 is l, so trace(F) = 164 1= d!. 0

Since there are as many irreducible representations V4 as conjugacy classes of 6 4 , these must form a complete set of isomorphism classes of irreducible representations, which completes the proof of Theorem 4.3. In the next section we will prove Frobenius's formula for the character of V4 , and, in a series of exercises, discuss a little of what else is known about them: how to decompose tensor products or induced or restricted representations, how to find a basis for V4 , etc.

§4.3. Proof of Frobenius's Formula For any partition A. of d, we have a subgroup, often called a Young subgroup, 64 = 6A, X . " X 6A, C.. 64.

(4.27)

Let U4 be the representation of 6 4 induced from the trivial representation of 6A. Equivalently, UA = A· a4 , with a 4 as in the preceding' section. Let

t/IA =Xu,= character of U4.

(4.28)

Key to this investigation is the relation between UA and JIA> i.e., between t/JA and the character X.t of VA. Note first that VA appears in UA, since there is a surjection (4.29) Alternatively,

55

§4.3. Proof of Frobenius's Formula

by Exercise 4.4. For example, we have

u,d-1, ., ; ; :; Vrd-•·•' $ Vrd, which expresses the fact that the permutation representation C4 of 6 4 is the sum of the standard representation and the trivial representation. Eventually we will see that every UA contains VA with multiplicity one, and contains only other V,. for 11 > A.. The character of UA is easy to compute directly since UA is an induced representation, and we do this next. Fori = (i I' ... ' id) ad-tuple or non-negative integers with a.i« = d, denote by

I

c,c sd the conjugacy class consisting of elements made up of i 1 1-cycles, 12 2-cycles, ..• , id d-cycles. The number of elements in C1 is easily counted to be d! 11•1 1 !212 i 2 !· ... ·dr•i4 1'

ICI= 1

(4.30)

By the formula for characters of induced representations (Exercise 3.19), ! t/tA(C.) = IC,I [Sd: SA] ·IC1 n SAl

111 r' • ! · .. . ·d 1•id.I •

dl.

1 1 ! · ... · A.t!

d!

•

t

n r=·t l'•'rp I· ... d'••rr4!' k

' I ~ll.p'-·-=------:

1

where the sum is over all collections {r,q: I :s;; p :s;; k, I :s;; q :s;; tl} of nonnegative integers satisfying

+ 'zq + · · · + rkq• r, 1 + 2r, 2 + · · · + drpd·

iq = 'tq

A., =

(To count C,n 6A, write the pth component of an element of6A as a product ofrr 1 !-cycles, rr 2 2-cycles, .... ) Simplifying, t/tA(c,) =

t n ____•q_·- - , d

• '

the sum over the same collections of integers {rrq}· This sum is exactly the coefficient of the monomial XA power sum symmetric polynomial P' 11 =

(4.31)

t=~ '•q!rzq! · · · · · rtq!

=

xt• ·... ·x:• in the

(x 1 + .. · + xA)1 •·(x~ + .. · + x~) 12 • ... ·(xf + .. · + xZ)'•. (4.32)

So we have the formula (4.33)

To prove Frobenius's formula, we need to compare these coefficients with the coefficients w A(I) defined by

56 4. Representations of 6 4 : Young Diagrams and Frobenius's Character Formula

wA(i)

= [.1. · P'11 ].,

l

= (A. 1 + k-

I, 12

+ k- 2, ... , 11).

(4.34)

Our goal, Frobenius's formula, is the assertion that X4 (C1) = w 4 (i). There is a general identity, valid for any symmetric polynomial P, relating such coefficients:

[P]A

= L KpA[A · P]r,., H-r,,.,H-2, ... ,pk1' I'

where the coefficients KpA are certain universally defined integers, called Kostka 1111111bers. For any partitions A. and 11of d, the integer K,.A may be defined combinatorially as the number of ways to fill the boxes of the Young diagram for 11 with 11 l's, A. 2 2's, up to 11 k's, in such a way that the entries in each row are nondecreasing, and those in each column are strictly increasing; such are called semistandard tableaux on 11 of type A.. In particular,

Ku = 1, and K,.A = 0 for 11
The result of Lemma A.28 can be written, using (4.30), in the form

~~ ~ IC11wA(i)wp(i) = oAp·

(4.36)

This indicates that the functions w 4 , regarded as functions on the conjugacy classes of 6 4 , satisfy the same orthogonality relations as the irreducible characters of 6 4 • In fact, one can deduce formally from these equations that the w 4 must be the irreducible characters of 6 4 , which is what Frobenius proved. A little more work is needed to see that wA is actually the character of the representation VA, that is, to prove Proposition 4.37. Let XA = Xv, be tl1e character of VA. Tl1en fo.r any conjugacy . class C1 of 64,

XA(Cr) = wA(i). We have seen in (4.29) that the representation UA, whose character is VA. In fact, this is all that we need to know about the relation between UA and V,. It implies that we have

PROOF.

tJtA, contains the irreducible representation t/1 4 =

L,. 11A,.x,.,

nu ~ I, all nA,. ~ 0.

(4.38)

Consider this equation together with (4.35). We deduce first that each wA is a

57

§4.3. Proof of Frobenius's Formula

virtual character: we can write

=I mApX

wA

1,

mAp E Z.

But the wA,1ike the XA• are orthonormal by (4.36), so

= (wA, wA) = I

l

m~,..

p

and hence w A is ±X for some irreducible character X· (It follows from the hook length formula that the plus sign holds here, but we do not need to assume this.) Fix A., and assume inductively that Xp = w,. for allp. > A., so by (4.35)

+ p>A I K,.AXw

VtA = WA

Comparing this with (4.38), and using the linear independence of characters, D. the only possibility is that wA = XA· Corollary 4.39 (Young's rule). Tl1e integer K ,,A is !11e multiplicil y of the irreducible representation v,. in tl1e induced representation UA: uA ~ vA $

EB K,.A vp.

t/tA = XA

p>A.

+

I

K,,AX,.·

t•> ..1

Note that when A.= (1, ... , 1), UA is just the regular representation, so K,.(l, ... , 11 = dim Vw This shows that the dimension of VA i.v tl1e number of standard tableaux on A., i.e., the number of ways to fill the Young diagram of A. with the numbers from 1 to d, such that all rows and columns are increasing. The hook length formula gives another combinatorial formula for this dimension. Frame, Robinson, and Thrall proved that these two numbers are equal. For a short and purely combinatorial proof, see [G-N-W]. For another proof that the dimension of VA is the number of standard tableaux, see [Jam]. The latter leads to a canonical decomposition of the group ring A = C6d as the direct sum of left ideals Aer. summing over all standard tableaux, with er =(dim VAfd!)·cr, and Cr the Young symmetrizer corresponding toT, cf. Exercises 4.47 and 4.50. This, in turn, leads to explicit calculation of matrices of the representations VA with integer coefficients. For another example of Young's rule, we have a decomposition

•

Urd-•••,

= 1=0 EB "fd-1,1)·

In fact, the only 11 whose diagrams can be filled with d - a l's and a 2's, nondecreasing in rows and strictly increasing in columns, are those with at most two rows, with the second row no longer than a; and such a diagram has only one such tableau, so there are no multiplicities. Exercise 4.40"'. The characters VtA of 6d have been defined only when A. is a partition of d. Extend the definition to any k-tuple a = (a 1 , ••• , at) of integers

58 4. Representations

or 6

4:

Young Diagrams and Frobenius's Character Formula

that add up to d by setting .P. = 0 if any of the a1 are negative, and otherwise ... , a1 in descending order. In this case .Pais the character ofthe representation induced from the trivial representation by the inclusion of 6., x · · · x 6 •• in 6 4 • Use (A.S) and (A.9) of Appendix A to prove the determinantal formula for the irreducible characters XA in terms of the induced characters 1/JP:

1/1. = 1/JA, where A. is the reordering of a 1 ,

XA = )

t:Sk

sgn(rN1rA,+•fiJ-I.A 1 +•f2J-2, ...• A.+•flJ-tJ·

If one writes 1/1. as a formal product 1/1., · 1/1., · ... · 1/1••• the preceding formula can be written

1/JA,+l

1/JA,H-1

VIA,· ..

The formal product of the preceding exercise is the character version of an "outer product" of representations. Given any non-negative integers d 1 , ••• , dt, and representations V, of 6 4 ,, denote by V1 o · · · o 1-) the (isomorphism class of the) representation of 6 4 , d = L d, induced from the tensor product representation V,IBJ·. ·IBJ vk of 6d, X ... X 64k by the inclusion of 6d, X •.• X 64, in 6 4 (see Exercise 2.36). This product is commutative and associative. It will turn out to be useful to have a procedure for decomposing such a representation into its irreducible pieces. For this it is enough to do the case of two factors, and with the individual representations V, irreducible. In this case, one has, for VA the representation of 6 4 corresponding to the partition A. of d and V,. the representation of 6., corresponding to the partition f1. of m, (4.41)

the sum over all partitions v of d + m, with NA,•• the coefficients given by the Littlewood- Richardson rule (A.8) of Appendix A. Indeed, by the exercise, the character of VA o V,. is the product of the corresponding determinants, and, by (A.8), that is the sum of the characters Ni~.x •. When m = 1 and f.l = (m), Vp is trivial; this gives (4.42)

the sum over all v whose Young diagram is obtained from that of A. by adding one box. This formula uses only a simpler form of the Littlewood-Richardson rule known as Pieri's formula, which is proved in (A.7). Exercise 4.43*. Show that the Littlewood-Richardson number NAP• is the multiplicity of the irreducible representation VA IBI VP in the restriction of v. from 6 4 +., to 6 4 x 6.,. In particular, taking 111 = I, f1. =(I), Pieri's formula (A.7) gives

§4.3. Proof of Frobenius's Formula

59

the sum over all A. obtained from v by removing one box. This is known as the "branching theorem," and is useful for inductive proofs and constructions, particularly because the decomposition is multiplicity free. For example, you can use it to reprove the fact that the multiplicity of VA in U,, is the number of semistandard tableaux on p. of type A.. It can also be used to prove the assertion made in Exercise 4.6 that the representations corresponding to hooks are exterior powers of the standard representation. Exercise 4.44* (Pieri's rule). Regard 6d as a subgroup of 6d+m as usual. Let A. be a partition of d and v a partition of d + m. Use Exercise 4.40 to show that the multiplicity of V, in the induced representation lnd(VA) is zero unless the Young diagram of A. is contained in that of v, and then it is the number of ways to number the skew diagram lying between them with the numbers from I to m, increasing in both row and column. By Frobenius reciprocity, this is the same as the multiplicity of VA in Res(V.). When applied to d = 0 (or 1), this implies again that the dimension of v. is the number of standard tableaux on the Young diagram of v. For a sampling of the many applications of these rules, see [Dia §7, §8]. Problem 4.45*. The Murnaglwn-Nakayama rule gives an efficient inductive method for computing character values: If A. is a partition of d, and g e 6d is written as a product of an m-cycle and a disjoint permutation h E 6d-m• then XA(g)

=I: (-t)'(P'x,.(fl),

where the sum is over all partitions Jt of d - m that are obtained from A. by removing a skew hook of length m, and r(p.) is the number of vertical steps in the skew hook, i.e., one less than the number of rows in the hook. A skew hook for A. is a connected region of boundary boxes for its Young diagram such that removing them leaves a smaller Young diagram; there is a one-to-one correspondence between skew hooks and ordinary hooks of the same size, as indicated:

A.= (7,6, 5, 5,4,4, I, I) Jt = (7,4,4,3,3, I, l, I)

hook length = 9, r = 4

For example, if A. has no hooks of length m, then XA(g) = 0. The Murnaghan-Nakayama rule may be written inductively as follows: If g is a written as a product of disjoint cycles of lengths m 1 , m2 , ••• , mr, with the lengths m1 taken in any order, then XA(g) is the sum L (-I )'(·' 1, where the sum is over all ways s to decompose the Young diagram or A. by successively

60 4. Representations of 6 4 : Young Diagrams and Frobenius's Character Formula

removing p skew hooks of lengths m1 , ••• , mP, and r(s) is the total number of vertical steps in the hooks of s. (a) Deduce the Murnaghan-Nakayama rule from (4.41) and Exercise 4.16, using the Littlewood- Richardson rule. Or: (b) With the notation of Exercise 4.40, show that k

"'·· "'·, · •.• · 1/!•.(gJ =

I "'·· "'·, ·... ·1/1.,-.. t/1.,., ·... ·"'·-<'' ·

1=1

Exercise 4.46*. Show that Corollary 4.39 implies the "Snapper conjecture": the irreducible representation V,, occurs in the induced representation VA if and only if 1

I

1=1

..1.1 ~

1

I

Jl 1 for all j ~ I.

1=1

Problem 4.47*. There is a more intrinsic construction of the irreducible representation VA, called a Specht module, which does not involve of the choice of a tableau; it is also useful for studying representations of 6 4 in positive characteristic. Define a tabloid {T} to be an equivalence class of tableaux (numberings by the integers 1 to d) on A., two being equivalent if the rows are the same up to order. Then 6 4 acts by permutations on the tabloids, and the corresponding representation, with basis the tabloids, is isomorphic to VA. For each tableau define an element Erin this representation space, by

1:

the sum over the q that preserve the columns of T. The span of all Er's is isomorphic to VA, and the E/s, where T varies over the standard tableaux, form a basis. Another construction of VA is to take the subspace of the polynomial ring C[x~o ... , x 4 ] spanned by all polynomials Fr, where 1'1· = fl (x 1 - x1), the product over all pairs i < j which occur in the same column in the tableau 1: Exercise 4.48*. Let V). be the representation A· bA, which is the.Tepresentation of 6 4 induced from the tensor product of the alternating representations on the subgroup 6,. = 6 1,, x · · · x 6,.•• where Jl = A.' is the conjugate partition. Show that the decomposition of v; is

- v; =

L: K,.'A' '~·· I'

Deduce that VA is the only irreducible representation that occurs in both VA and u;, and it occurs in each with multiplicity one. Note, however, that in general A · c A 'I= A ·a A f1 A · IJA since A · c A may not be contained in A ·aA.

§4.3. Proof or Frobenius's Formula

61

Exercise 4.49*. With notation as in (4.41), if U' = l-( 1 , ... , 11 is the alternating representation of 6m, show that J.oA o 1~ 1 ..... 11 decomposes into a direct sum @ V,., the sum over all1t whose Young diagram can be obtained from that of A. by adding m boxes, with no two in the same row. Exercise 4.50. We have seen that A = C6d is isomorphic to a direct sum of copies of V~ = Ac 1 , where m~ = dim V~ is the number of standard tableaux on A.. This can be seen explicitly as follows. For each standard tableau Ton each A., let cr be the element of ce;d constructed from T. Then A = @A· cr. Indeed, an argument like that in Lemma 4.23 shows that cr · Cr· = 0 whenever T and T' are tableaux on the same diagram and T > T', i.e., the first entry (reading from left to right, then top to bottom) where the tableaux differ has the entry ofT larger than that ofT'. From this it follows that the sum l:A · cr is direct. A dimension count concludes the proof. (This also gives another proof that the dimension of V~ is the number of standard tableaux on )., provided one verities that the sum of the squares of the latter numbers is d!, cf. [Boe] or [Ke].) 111~

Exercise 4.51 *.There are several methods for decomposing a tensor product of two representations of 6d, which amounts to finding the coefficients C~,•• in the decomposition V~

® V,.

~ l:.C~P• V.,

for A., Jl, and v partitions of d. Since one knows how to express V~ in terms of the induced representations u., it suffices to compute V~ ® u., which is isomorphic to Ind(Res(V1 )), restricting and inducing from the subgroup 6, = 6,, x 6., x · · ·; this restriction and induction can be computed by the Littlewood- Richardson rule. Fords; 5, you can work out these coefficients using only restriction to 6d-J and Pieri's formula. (a) Prove the following closed-form formula for the coefficients, which shows in particular that they are independent of the ordering of the subscripts )., 11, and v:

c~P• =I1 Z(1I') (IJ~(i)ro,.(i)ro.(l), the sum over all i = (i 1, ... , id) with l:aia = d, and with roA(i) = XA(C1) and z(i) = i 1 ! I'• · i 2 ! 21' • ... • idl d'•. (b) Show that CA

d

"1 1

= {

l 0

if 11 = A. otherwise,

c A,,{t, ... ,J) =

{

l if Jl = A.' 0 otherwise.

Exercise 4.52*. Let R 4 = R(Gd) denote the representation ring, and set R = EBT=o Rd. The outer product of(4.41) determines maps

62 4. Representations of 6 4 : Young Diagrams and frobenius's Character formula which makes R into a commutative, graded l. -algebra. Restriction determines maps

Rn+m = R(6n+m)-+ R(6n x 6'") = Rn ® R.,, which defines a co-product o: R-+ R ® R. Together, these make R into a (graded) Hopf algebra. (This assertion implies many of the formulas we have proved in this lecture, as well as some we have not.) (a) Show that, as an algebra,

R

~

l.[Ht. ... , H4 ,

... ],

where Hd is an indeterminate of degree d; H4 corresponds to the trivial representation of 6d. Show that the co-product is determined by

o

c'i(Hn)

= Hn ® 1 + Hn-1 ® H 1 + '' · + I ® Hn.

If we set A= l.[H 1 , ••• , Hd, ... ] = ffiAd, we can identify A4 with the symmetric polynomials of degree d in k ~ d variables. The basic symmetric polynomials in Ad defined in Appendix A therefore correspond to virtual representations of 6 4 • (b) Show that Ed corresponds to the alternating representation U', and

(c) Show that the scalar product ( , ) defined on Ad in (A.l6) corresponds to the scalar product defined on class functions in (2.11 ). (d) Show that the involution .9 of Exercise A.32 corresponds to tensoring a representation with the alternating representation U'. (e) Show that the inverse map from Rd to Ad takes a representation W to )

1

'r' z(i) Xw

(C )1'{1) (I)

•

where z(i) = i1 !1 11 • 12 !21• • ... • i4 !d 1d. The (inner) tensor product of representations of 6d gives a map Rd ® Rd -+ Rd which corresponds to an "inner product" on symmetric functions, sometimes denoted *· (f) Show that prll

* m= p

{0

z(I)P 111

for j of i if j = i.

Since these 1'11 ' form a basis for Ad® 0, this formula determines the inner product.

LECTURE 5

Representations of 21d and GL 2 (1Fq)

In this lecture we analyze the representation of two more types of groups: the alternating groups 111 4 and the linear groups GL 2 (F9 ) and SL 2 (Fq) over finite fields. In the former case, we prove some general results relating the representations of a group to the representations of a subgroup of index two, and use what we know about the symmetric group; this should be completely straightforward given just the basic ideas of the preceding lecture. In the latter case we start es.~ntially from scratch. The two sections can be read (or not) independently; neither is logically necessary for the remainder of the book. §5.1: Representations oC'll4 §5.2: Representations of GL 2 (1F9 ) and SL 2 (Fq)

§5.1. Representations of 2ld The alternating groups ~u.,, d ~ 5, form one of the infinite families of simple groups. In this section, continuing the discussion of §3.1, we describe their irreducible representations. The basic method for analyzing representations of IH 4 is by restricting the representations we know from 6 4 • In general when His a subgroup of index two in a group G, there is a close relationship between their representations. We will see this phenomenon again in Lie theory for the subgroups of the orthogonal groups Let U and U' denote the trivia' and nontrivial representation of G obtained from the two representations of G/H. For any representation V of G, let V' = V ® U'; the character of V' is the same as the character of V on elements of H, but takes opposite values on elements not in H. In particular, Resg V' = Resg V.

so.

o•.

5. Representations

64

or 1!1 4 and GL 2 (f9 )

If W is any representation of H, there is a conjugate representation defined by conjugating by any element t of G that is not in H; if t/1 is the character of W, the character of the conjugate is h1-+ t/J(thC 1 ). Since tis unique up to multiplication by an element of II, the conjugate representation is unique up to isomorphism. Proposition 5.1. Let V be an irreducible representation of G, and let W = Res~ V be the restriction of V to H. Then exactly one of the following holds: (I) Vis not isomorpltlc to V'; W is irreducible and isomorpiJic to its corljugate; Ind~ W ~

V Ef) V'. (2) V ~ V'; W = W' $ W", where W' and W" are irreducible and conjugate but not isomorphic; lnd~ W' ~ Indg W" ~ V. Each irreducible representation of H arises uniquely in tills way, noting that

in case ( 1) V' and V determine tlae same representation. PROOF.

Let

x be the character of V. We have IGI = 21111 =

I

~"H

lx(h)l 2 +

I lx
I~H

Since the first sum is an integral multiple of IH 1. this multiple must be I or 2, which are the two cases of the proposition. This shows that W is either irreducible or the sum of two distinct irreducible representations W' and W". Note that the second case happens when x(t) = 0 for all t ¢II, which is the case when V' is isomorphic to V. In the second case, W' and W" must be conjugate since W is self-conjugate, and if W' and W" were self-conjugate V would not be irreducible. The other assertions in (I) and (2) follow from the isomorphism Jnd(Res V) = V ® (U $ U') of Exercise 3.16. Similarly, for any representation W of H, Res(Ind W) is the direct sum of Wand its conjugateas follows say from Exercise 3.19-from which the last statement follows ~~

0

Most of this discussion extends with little change to the case where H is a normal subgroup of arbitrary prime index in G, cf. [B-tD, pp. 293-296]. Clifford has extended much of this proposition to arbitrary normal subgroups of finite index, c[ [Dor, §14]. There are two types of conjugacy classes c in H: those that are also conjugacy classes in G, and those such that c u c' is a conjugacy class in G, where c' = tct- 1 , t ¢ H; the latter are called split. When W is irreducible, its character assumes the same values-those of the character oft he representation V of G that restricts to W-on pairs of split conjugacy classes, whereas in the other case the characters of W' and W" agree on nonsplit classes, but they must disagree on some split classes. If Xw·(c) = Xw .. (c') = x, and Xw·(c') = Xw .. (c) = y, we know the sum x + y, since it is the value of the character of the representation V that gives rise to W' and W" on c u c'. Often the exact values of x and y can be determined from orthogonality considerations.

§5.!. Representations oPlld

65

Exercise 5.2*. Show that the number of split conjugacy classes is equal to the number of irreducible representations V of G that are isomorphic to V', or to the number of irreducible representations of H that are not isomorphic to their conjugates. Equivalently, the number of nonsplit classes in II is same as the number of conjugacy classes of G that are not in H. We apply these considerations to the alternating subgroup of the symmetric group. Consider restrictions of the representations VA from 6d to ~d· Recall that if X is the conjugate partition to A., then VA'= VA® U',

with U' the alternating representation. The two cases of the proposition correspond to the cases (l) A.' ¥- A. and (2) A.' = A.. If A.' ¥- A., let ») be the restriction of VA to llld. If X = A, let and W;' be the two representations whose sum is the restriction of VA. We have

w;

Res VA = Res VA' = WA

Ind WA = VA$ VA'• lnd

w; =

Ind

w;' = VA,

Res VA=

w; $ Wt

when X
Note that

# {self-conjugate representations of 6d} =

# {symmetric Young diagrams}

= # {split pairs of conjugacy classes in 111 4 } = #{conjugacy classes in 6d breaking into two classes in IJid}·

Now a conjugacy class of an element written as a product of disjoint cycles is split if and only if there is no odd permutation commuting with it, which is equivalent to all the cycles having odd length, and no two cycles having the same length. So the number of self-conjugate representations is the number of partitions of d as a sum of distinct odd numbers. In fact, there is a natural correspondence between these two sets: any such partitioQ corresponds to a symmetric Young diagram, assembling hooks as indicated:

I

-

I

rI '--

-

If A. is the partition, the lengths of the cycles in the corresponding split conjugacy classes are q 1 = 21 1 - 1, q 2 = 21 2 - 3, q 3 = 2).~- 5, . ...

5. Representations of\11 4 and GL 2 (1F,)

66

For a self-conjugate partition A, let XA and XA denote the characters of w; and WI', and let c and c' be a pair of split conjugacy classes, consisting of cycles of odd lengths q 1 > q2 > · · · > q,. The following proposition of Frobenius completes the description of the character table of IJI 4 • Proposition 5.3. (1) If c and c' do not correspond to the partition A, then XA(c)

= XA(c') = x:\(c) = x:l(c') =

h~(c v c').

(2) If c and c' correspond to A, then

XA(c) = x;(c') =X,

xi.(c')

= X~(c) = y,

witlt x and y tlae two numbers

H< -t>"' ± J< -t)"'q, ..... q,). and 111 =

Hfl q

1-

I)= !(d- r).

For example, if d = 4 and A= (2, 2), we haver= 2, q 1 = 3, q 2 = I, and x and y are the cube roots of unity; the representations w; and WI' are the representations labeled U' and U" in the table in §2.3. Ford= 5,). = (3, I, 1), r = 1, q 1 = 5, and we find the representations called Y and Z in §3.1. For d ~ 7, there is at most one split pair, so the character table can be derived from orthogonality alone. Note that since only one pair of character values is not taken care of by the first case of Frobenius's formula, the choice of which representation is WI and which WI' is equivalent to choosing the plus and minus sign in (2). Note also that the integer m occurring in (2) is the number of squares above the diagonal in the Young diagram of A. We outline a proof of the proposition as an exercise: Exercise 5.4*. Step 1. Let q = (q 1 > · .. > q,) be a sequence of positive odd integers adding to d, and let c' = c'(q) and c" = c"(q) be the corresponding conjugacy classes in IJI 4 • Let A be a self-conjugate partition of d, and Jet xJ. and xi be the corresponding characters of IJI4 • Assume that xJ. and x:l take on the same values on each element of IJ( 4 th!lt is not in c' or c". Let u xJ.(c') = x1(c") and v = xJ.(c") = XA(c'). . (i) Show that u and v are real when 111 = !:E(q1 - I) is e~en, and II= v when 111 is odd. (ii) Let 8 = x). - x:\. Deduce from the equation (.9, 8) = 2 that lu - vl 2 = q, · ... ·q,. (iii) Show that A is the partition that corresponds to q and that u + v = (-I)"', and deduce that u and v are the numbers specified in (2) of the proposition. Step 2. Prove the proposition by induction on d, and for fixed d, look at that q which has smallest q 1 , and for which some character has values on the classes c'(q) and c"(q) other than those prescribed by the proposition.

=

67

(i) U r = I, so q 1 = d = 2m + l, the corresponding self-conjugate partition is -l = (m + 1, l, ... , 1). By induction, Step l applies to XA and X1· (ii) If r > 1, consider the imbedding H = IJl,, x IJid-q, c G = IJld, and let X' and X" be the representations or G induced rrom the representations w: IBJ w; and W;' IBJ W:l, where WI and W;' are the representations or IJI,, corresponding to q 1 , i.e., to the self-conjugate partition (!(q 1 - 1), I, ... , 1) of q 1; Wi is one of the representations ofiJld-q, corresponding to (q 2, ..• , q,); and IBJ denotes the external tensor product (see Exercise 2.36). Show that X' and X" are conjugate representations of IJld, and their characters x' and x" take equal values on each pair of split conjugacy classes, with the exception of c'(q) and c"(q), and compute the values of these characters on c'(q) and c"(q). (iii) Let ,9 = x' - x", and show that (.9, .9) = 2. Decomposing X' and X" into their irreducible pieces, deduce that X' = Y $ W{ and X" = Y $ W{ for some self-conjugate representation Yand some self-conjugate partition). of d. (iv) Apply Step l to the characters x~ and x~. and conclude the proof. Exercise 5.5*. Show that if d > 6, the only irreducible representations of illd of dimension less than d are the trivial representation and the (11 - I)dimensional restriction of the standard representation of 6d. Find the exceptions for d ~ 6. We have worked out the character tables for all 6d and illd ford~ 5. With the formulas of Frobenius, an interested reader can construct the tables for a few more d-until the number of partitions of d becomes large.

§5.2. Representations of GL 2 (1Fq) and SL 2 (1Fq) The groups G L 2 (F,) of invertible 2 x 2 matrices with entries in the finite field F9 with q elements, where q is a prime power, form another important series of finite groups, as do their subgroups SL 2 (F9 ) consisting of matrices of determinant one. The quotient PGL 2 (F9 ) = GL 2 (F,)/F; is the automorphism group of the finite projective line IP 1 (1F4 ). The quotients PSL 2 (F,) = SL 2 (F4 )/{ ± 1} are simple groups if q # 2, 3 (Exercise 5.9). In this section we sketch the character theory of these groups. We begin with G = GL 2 (F9 ). There are several key subgroups:

(This "Borel subgroup" B and the group of upper triangular unipotent matrices N will reappear when we look at Lie groups.) Since G acts transitively on the projective line P 1 (F4 ), with 8 the isotropy group of the point (1:0), we have

5. Representations or IJ.ld and GL 2 (1F,)

68 We will also need the diagonal subgroup

D= {

(~ ~)} = f*

x f*,

where we write F ror Ft. Let F' = F,, be the extension or F of degree two, unique up to isomorphism. We can identify GL 2 (1Fq) as the group of all IF-linear invertible endomorphisms of IF'. This makes evident a large cyclic subgroup K = (f')* or G. At least if q is odd, we may make this isomorphism explicit by choosing a generator e for the cyclic group IF* and choosing a square root in F'. Then I and form a basis for F' as a vector space over F, so we can make the identification:

Je

Je

K is a cyclic subgroup or G of order q 2 - I. We often make this identification, leaving it as an exercise to make the necessary modifications in case q is even. The conjugacy classes in G are easily round: Representative

No. Elements in Class

No. Classes

a,= ( x0

ox)

q- I

(~

!)

q-1

b, = Cx,y-(X0

Oy)' X#)'

(q-l)(q-2) 2 q(q - l)

---2

Here cx,,and c1 ,xareconjugate by any

(a -ec)

by(_~ ~).and Jx,y and Jx,-y are conjugate

. To count the number or elements in the conjugacy class c -a or bx, look at the action of G on this class by conjugation; the isotropy group is { (

~ ~)},so the number of elements in the class is the index of this group

in G, which is q 2 - l. Similarly the isotropy group for cx,y is D, and the isotropy group for dx,r is K. To see that the classes are disjoint, consider the eigenvalues and the Jordan canonical forms. Since they account for IGI elements, the list is complete. There are q 2 - I conjugacy classes, so we must find the same number or irreducible representations. Consider first the permutation representation or G on P 1 (F), which has dimension q + I. It contains the trivial representation;

69

§5.2. Representations of GL 2 (1F,) and SL 2 (1F,)

let V be the complementary q-dimensional representation. The values of the character xof Von the four types of conjugacy classes are x(a.. ) = q, x(b.. ) = 0, x(c ... ,) = l, x(d... ,) = -I, which we display as the table: V:

0

q

l

-1

Since (X, x) = l, Vis irreducible. For each of the q- l characters a: f*-+ C* of f*, we have a onedimensional representation U. of G defined by U.(g) = a(det(g)). We also have the representations V. = V ® u•. The values of the characters of these representations are

v.: Note that if we identify

(x

)!

x2

-

a(x)a(y)

a(x 2

a(x)a(y)

-a(x 2

ey 2 )

-

-

ey 2 )

".J!) with' = x+ yJe in f', then X

sy 2 = det(;

6 :)

= Normr.1r(0 =

(·( 9

= (911 .

The next place to look for representations is at those that are induced from large subgroups. For each pair a, Pof characters off*, there is a character of the subgroup B: B-+ B/N = D = IF* x IF*-+

which

c• x c• ..... c•,

takes(~ ~)to a(a)p(d). Let W.., be the representation induced from

B toG by this representation; this is a representation of dimension [G: B] q + l. By Exercise 3.19 its character values are found to be:

w..,:

tX(x)p(x) + 1)tr(x)p(x) tX(x)P(Y) + a(y)p(x) this that W.. 11 ~ w11 •• , that IV. .• ~ u. ~ v•. and that (q

=

0

We see from for a # P the representation is irreducible. This gives !(q - 1)(q - 2) more irreducible representations, of dimension q + l. Comparing with the list of conjugacy classes, we see that there are !q(q - I) irreducible characters left to be found. A natural way to find new characters is to induce characters from the cyclic subgroup K. For a representation q>: K = (F')* -+

c•,

the character values of the induced representation of dimension [G: K] = q 2 - I are lnd(q>):

q(q - I )q>(x)

yJe

0

0

Here again ( = x + E K =(IF')*. Note that lnd(q> 9 ) ~ lnd(q>), so the representations Ind(q>) for q>' # q> give !q(q - I) different representations.

5. Representations of 1114 and GL 2 (1F.)

70

However, these represenations are not irreducible: the character x of Jnd(cp) satisfies (x, x) q- 1 if cpq -1: cp, and otherwise (x, X)= q. We will have to work a little harder to get irreducible representations from these Ind(cp). Another attempt to find more representations is to look inside tensor products of representations we know. We have V. ® U7 = V., 1 , and W... 11 ® U7 ~ W., 7,,7 , so there are no new ones to be found this way. But tensor products of the V..'s and W... 11's are more promising. For example, V ® W.., 1 has character values:

=

q(q

+ I )ex(x)

0

ex(x)

+ ex(y)

0

We can calculate some inner products of these characters with each other to estimate how many irreducible representations each contains, and how many they have in common. For example, (Xv®ll'•·•' Xw,) = 2, (Xtndh•J•Xw,,,) = 1

(Xv®w., 1 • Xv®w,) = q (Xv®w,,,, Xtndh•l) = q

if CfJir• =ex,

+ 3, if CfJir• =ex,

Comparing with the formula (Xtndi•J• Xtndi•J) = q- 1, one deduces that V ® W... 1 and lnd(cp) contain many of the same representations. With any luck, Ind(cp) and W... 1 should both be contained in V ® W... 1 • This guess is easily confirmed; the virtual character X.,= Xv®w,,,- Xw,,,- Xtndi•J

takes values (q - l)ex(x), - ex(x), 0, and -(cp(() + cp(()') on the four types of conjugacy classes. Therefore, (x.,, x.,) = I, and x.,(l) = q - 1 > 0, so x., is, in fact, the character of an irreducible subrepresentation of V ® W...t of dimension q- l. We denote this representation by X.,. These !q(q- I) representations, for rp -1: q>q, and with X.,= X.,., therefore complete the list of irreducible representations for GL 2 (1F). The character table is

qt- I GL 2 (F,)

u.

v.

w.,,

x.,

a• =(x0

~)

a(x 2 ) qa(x 2 ) (q + l)tx(x)fJ(x)

(q-

J)q~(x)

b.=G ~)

ql

c•. , --

+q

(X0 ~)

a(x 2 ) 0 a(x)fJ(x)

a(xy) a(xy) a(x){J(J•) + a(y)fJ(x)

-q~(x)

0

ql- q d•.

,=c

:)~c

tx(C') -a(C')

0 -(rp(C)

+ q~({'))

Exercise 5.6. Find the multiplicity of each irreducible representation in the representations V® W.., 1 and Ind(cp).

71

§5.2. Representations of GL 2 (1f,) and SL 2 (F.)

Exercise 5.7. Find the character table of PGL 2 (F) = GL 2 (F)/F*. Note that its characters are just the characters of GL 2 (F) that take the same values on elements equivalent mod F*. We turn next to the subgroup SL 2 (1F,) of 2 x 2 matrices of determinant one, with q odd. The conjugacy classes, together with the number of elements in each conjugacy class, and the number of conjugacy classes of each type, are Representative

-e=(-1

0) 0 -1

(~ :) G~)

(3) (4)

(-I

(5)

q2- I 2 q2-

q2- I

I)

0 -1

2

(-~ ~.)

(7)

(~ x~}x# ±I

c:

I

2

(6)

(8)

No. Classes

e=(~ ~)

(I) (2)

No. Elements in Class

q2- I 2

~).x #±I

+ I)

q-3 -2-

q(q- I)

2-

q(q

q- l

The verifications are very much as we did for GL 2 (f,). In (7), the classes of

(~ x~t) and ( x~ 1 ~)are the same. In (8), the classes for (x, y) and (x, - y) are the same; as before, a better labeling is by the element (in the cyclic group C=

{C e (IF')*: c•+ 1 = 1};

c-t

the elements ±I are not used, and the classes of (and are the same. The total number of conjugacy classes is q + 4, so we turn to the task of finding q + 4 irreducible representations. We first see what we get by restricting representations from GL2 (Fq). Since we know the characters, there is no problem working this out, and we simply state the results:

u.

(I) The all restrict to the trivial representation U. Hence, ir we restrict any representation, we will get the same for all tensor products by

u;s.

5. Representations of IJid and G L1 (1F,)

72

(2) The restriction V of the v;s is irreducible. (3) The restriction W.. oHV,., 1 is irreducible if lll 2 ~ I, and W. ~ 1-tj, when p = o: or {l = lll- 1• These give !(q- 3) irreducible representations of dimension q + 1. (3') Let y denote the character of IF* with Y 2 = 1, Y ~ I. The ~striction of W., 1 is the sum of two distinct irreducible representations, which we denote W'and W". (4) The restriction of x. depends only on the restriction of q> to the subgroup C, and q> and rp- 1 determine the same representation. The representation is irreducible if q> 2 ~ l. This gives !(q - I) irreducible representations of dimension q - 1. (4') If t/1 denotes the character of C with t/1 2 = I, t/1 f= I, the restriction of X~ is the sum of two distinct irreducible representations, which we denote X' and X".

Altogether this list gives q + 4 distinct irreducible representations, and it is therefore the complete list. To finish the character table, the problem is to describe the four representations W', W", X', and X". Since we know the sum of the squares of the dimensions of all representations, we can deduce that the sum of the squares of these four representations is q 2 + I, which is only possible if the first two have dimension !(q + l) and the other two !(q- l). This is similar to what we saw happens for restrictions of representations to subgroups of index two. Although the index here is larger, we can use what we know about index two subgroups by finding a subgroup H of index two in GL 2 (F9 ) that contains SL 2 (F9 ), and analyzing the restrictions of these four representations to H. For H we take the matrices in GL 2 (F9 ) whose determinant is a square. The representatives of the conjugacy classes are the same as those for GL 2 (F,), including, of course, only those representatives whose determinant is a square, but we must add classes represented by the are conjugate to the elements (

~

!)

elements(~

;). x

E

F*. These

in G L2 (f,), but not in H. These are the

q- l split conjugacy classes. The procedure of the preceding section can be used to work out all the representations of H, but we need only a little of this. Note that the sign representation U' from G/H is U., so that W,, 1 ~ W,, 1 ® U' and X~ ~X~® U'; their restrictions to H split into sums of conjugate irreducible representations of half their dimensions. This shows these representations stay irreducible on restriction from H to SL 2 (f,), so that W' and W" are conjugate representations of dimension !(q + 1), and X' and X" are conjugate representations of dimension !(q - 1). In addition, we know that their character values on all nonsplit conjugacy classes are the same as half the characters of the representations W,, 1 and X~· respectively. This is all the information we need to finish the character table. Indeed, the only values not covered by this discussion are

73

§5.2. Representations of GL 2 (F,) and SL 2 (F,)

G:) G;) (-1 I) (-1 0 -1

W'

s

W"

8)

0 -1

s'

t'

s

t'

.~·

X'

u

v

u'

v'

X"

v

u

v'

u'

The first two rows are determined as follows. We know that s Xw•. ,

(~

(1 1)-t . . = -1) 1 1

II (( I 1)) 1 = . !) 0

· . n a dd"1tton, smce 0

if q is congruent to l modulo 4, and

(I 0

IS

+ t =.

conJugate to

to(~ ~)otherwise, and since

x(g- 1) = x(g) for any character, we conclude that s and t are real if q = I mod(4), and s = Tif q 3 mod(4). In addition, since -e acts as the identity or minus the identity for any irreducible representation (Schur's lemma),

=

x< -g)= x(g) · x(l)/x( -e) for any irreducible character X· This gives the relations s' = t(- l)s and t' = t( -l)t. Finally, applying the equation (X, x) = I to the character of W' gives a formula forst+ tS. Solving these equations gives s, t =! ± !y'roq, where w = t( -l) is 1 or - l according as q = 1 or 3 mod(4). Similarly one computes that u and v are - t ± !Jwq. This concludes the computations needed to write out the character table. ExerCise 5.8. By considering the action of SL 2 (Fq) on the set P 1 (F,), show that SL 2 (F2 ) ~ 6 3 , PSL 2 (1F 3 ) ~ \!1 4 , and SL 2 (F4 ) ~ IJ1 5 • Exercise 5.9"'. Use the character table for SL 2 (F,) to show that PSL 2 (1F,) is a simple group if q is odd and greater than 3. Exercise 5.10. Compute the character table of PSL 2 (1F,), either by regarding it as a quotient of SL 2 (1F,), or as a subgroup of index two in PGL 2 (1F,). Exercise 5.11•. Find the conjugacy classes of GL 3 (F,), and compute the characters of the permutation representations obtained by the action of GL 3 (Fq) on (i) the projective plane P 2 (F,) and (ii) the "flag variety" consisting of a point on a line in P 2 (Fq). Show that the first is irreducible and that the second is a sum of the trivial representation, two copies of the first representation, and an irreducible representation. Although the characters of the above groups were found by the early pioneers in representation theory, actually producing the representations in a natural way is more difficult. There has been a great deal of work extending

74

5. Representations of 111. and G L2 (IF9 )

this story to GL.(F,) and SL.(F,) for n > 2 (cf. [Gr]), and for corresponding groups, called finite Chevalley groups, related to other Lie groups. For some hints in this direction see [Hu3], as well as [Ti2]. Since all but a finite number of finite simple groups are now known to arise this way (or are cyclic or alternating groups, whose characters we already know), such representations play a fundamental role in group theory. In recent work their Lie-theoretic origins have been exploited to produce their representations, but to tell this story would go far beyond the scope of these lecture(r)s.

LECTURE 6

Weyl's Construction

In this lecture we introduce and study an important collection of functors generalizing the symmetric powers and exterior powers. These are defined simply in terms of the Youngsymmetrizers cA introduced in §4: given a representation Vof an arbitrary group G, we consider the dth tensor power of V. on which both G and the symmetric group on d letters act. We then take the image of the action of c 1 on y®d; this is again a representation of G, denoted § 1 (V). This gives us a way of generating new representations, whose main application will be to Lie groups: for example, we will generate all representations of SL.C by applying these to the standard representation C" ofSL.C. While it may be easiest to read this material while the definitions of the Young symmetrizers are still fresh in the mind, the construction will not be used again until §IS, so that this lecture can be deferred until then. §6.1: Schur functors and their characters §6.2: The proofs

§6.1. Schur Functors and Their Characters For any finite-dimensional complex vector space V, we have the canonical decomposition

V® V= Sym 2 V®NV. The group GL(V) acts on V ® V, and this is, as we shall soon see, the decomposition of V ® Vinto a direct sum of irreducible GL(V)-representations. For the next tensor power,

V® V® V= Sym 3 V$NV$ another space. We shall see that this other space is a sum of two copies of an irreducible

6. Weyl's Construction

76

GL(V)-representation. Just as Sym 4 V and NV are images of symmetrizing operators from V® 4 = V ® V ® · · · ® V to itself, so are the other factors. The symmetric group 6 4 acts on V® 4, say on the right, by permuting the factors

(v 1 ® ... ® v4 )·a = v.,111 ® ... ® v,. 141 • This action commutes with the left action of GL(V). For any partition J of d we have from the last lecture a Young symmetrizer c1 in C64 • We denote the image of c1 on V® 4 by § 1 V: § 1 V = lm(c 1 lv@d)

which is again a representation of GL(V). We call the functor' V """"§ 1 V the Schur functor or Weyl module, or simply Weyl's construction, corresponding to k It was Schur who made the correspondence between representations of symmetric groups and representations of general linear groups, and Weyl who made the construction we give here. 2 We will give other descriptions later, cf. Exercise 6.14 and§ 15.5. For example, the partition d = d corresponds to the functor V ,-vv' Sym 4 V, and the partition d = 1 + ··· + I to the functor V ,-vv''NV. We find something new for the partition 3 = 2 + l. The corresponding symmetrizer c 1 is c12 . 11 = I + e021 - e031 - e11321 , so the image of c1 is the subspace of V® 3 spanned by all vectors

v 1 ® t> 2 ® v3

+ v2 ® v1 ® v3 -

If NV® Vis embedded in (v 1

11

V® 3

v3 ® v1 ® u 1

-

u 3 ® v 1 ® v2 .

by mapping

v3 ) ® v2 1-+ v 1 ® v 2 ® v3 - v 3 ® v2 ® t• 1,

then the image of c1 is the subspace of NV® V spanned by all vectors

(v 1

11

v3 ) ® v2

+ (t'z

11

v3 ) ® v 1 •

It is not hard to verify that these vectors span the kernel of the canonical map from NV® V to NV, so we have § 12 ,1) V = Ker(NV® V---+ NV).

(This gives the missing factor in the decomposition of V:® 3 .) Note that some of the § 1 V can be zero if V has small dimension. We will see that this is the case precisely when the number of rows in the Young diagram of J is greater than the dimension of V. 1 The funcloriality means simply that a linear map((!: V-+ IV or vector spaces determines a linear map§,(((!):§, V .... §,IV, with§,(((! o t/1) = §,(((!) o §,(t/1) and §,(Id.) = lds,y 2 The notion goes by a variety of names and notations in the literature, depending on the context. Constructions differ markedly when not over a field or characteristic zero; and many authors now parametrize them by the conjugate partitions. Our choice or notation is guided hy the correspondence between these runclors and Schur polynomials. which we will see are their characters.

77

§6.1. Schur Functors and Their Characters

When G = GL(V), and for important subgroups G c GL(V), these §A V give many of the irreducible representations of G; we will come back to this later in the book. For now we can use our knowledge of symmetric group representations to prove a few facts about them-in particular, we show that they decompose the tensor powers V® 4, and that they are irreducible representations of G L( V). We will also compute their characters; this will eventually be seen to be a special case of the Weyl character formula. Any endomorphism g of V gives rise to an endomorphism of §A V. In order to tell what representations we get, we will need to compute the trace of this endomorphism on §..tV; we denote this trace by Xs,v(g). For the computation, let x., ... , x1 be the eigenvalues of g on V, k =dim V. Two cases are easy. For ). =(d), (6.1)

where H4(x 1 , ••• , x 1 ) is the complete symmetric polynomial of degree d. The definition of these symmetric polynomials is given in (A.l) of Appendix A. The truth of (6.1) is evident when g is a diagonal matrix, and its truth for the dense set of diagonalizable endomorphisrns implies it for all endomorphisms; or one can see it directly by using the Jordan canonical form of g. For ). = (1, ... , I), we have similarly §(J, ... ,IJ

V =NV,

Xsu .......•·(g) = Ed(x 1 ,

••. ,

x1 ),

(6.2)

with E4 (x 1 , ••• , x,) the elementary symmetric polynomial [see (A.3)]. The polynomials H4 and E4 are special cases of the Scllur polynomials, which we denote by SA = SA(x 1 , ••• , x1 ). As). varies over the partitions of d into at most k parts, these polynomials SA form a basis for the symmetric polynomials of degree din these k variables. Schur polynomials are defined and discussed in Appendix A, especially (A.4)-(A.6). The above two formulas can be written Xs,v(g) = SA(x 1 ,

••• ,

x,) for).= (d) and).= (1, ... , l).

We will show that this equation is valid for all ).: Theorem 6.3. (I) Let k =dim V. Tlten 0), then

if

§A V

is zero

n

;., - ).} + j - i

).HI

# 0. If ). =

().t ~ · · · ~ ).l ~

dim§AV=SA(l, ... ,l)=

tsi<J,.;;l

j-i

(2) Let lilA be the dimension of the irreducible representation corresponding to ).. Then

y®d

~

VA

of

sd

(9 §A V®'"•. A

(3) For any g E GL(V), tlte trace of g on § 1 V is tile value of tile Schur polynomial art tlte eigenvalues x 1 , ••• , x, of g on V: Xs,v(g)

= SA(x t• · ·., xk).

6. Weyf's Construction

78

(4) Each §l VIs an irreducible representation of GL(V).

This theorem will be proved in the next section. Other formulas for the dimension of §l V are given in Exercises A.JO and AJI. The following is another: Exercise 6.4*. Show that . ml dtm§lV=dT0(k-i+i>=

.

n (k - h +j) , j

v

where the products are over the d pairs (i,j) that number the row and column of boxes for A, and /1,1 is the hook number of the corresponding box. Exercise 6.5. Show that V@ 4

;;

y®J ~

Sym 3 V E9 NV E9 (§12 , 11 l') 192 , and

Sym 4 V $NV$ (§13 , 11 V) 193 $

(§12, 21 V)e 2 E9 (§12 , 1,1) V) 193 •

Compute the dimensions of each of the irreducible factors. The proof of the theorem actually gives the following corollary:

EBl

Corollary 6.6. If c E C6d, and (C64). c = vler, as representations of 64, then tltere is a corresponding decomposition of GL(V)-spaces: [email protected] = EJ1§lyer,, If x 1, ... , xk are the eigenl!alues of an endomorpllism of V, the trace of tlte induced endomorphism of V@ 4 ·c is I>lSA(x 1 , ... , xt)·

If A and Jt are different partitions, each with at most k =dim V parts, the irreducible GL(V)-spaces §l V and § 11 V are not isomorphic. Indeed, their characters are the Schur polynomials SA and S,.. which are different. More generally, at least for those representations of GL(V) which can be decomposed into a direct sum of copies of the represenations §l V's, the representations are completely determined by their d~aracters. This follows immediately from the fact that the Schur polynomials are linearly independent. Note, however, that we cannot hope to get all finite-dimensional irreducible. representations of GL(V) this way, since the duals of these representations are not included. We will see in Lecture 15 that this is essentially the only omission. Note also that although the operation that takes representations of 6 4 to representations of GL(V) preserves direct sums, the situation with respect to other linear algebra constructions such as tensor products is more complicated. One important application of Corollary 6.6 is to the decomposition of a tensor product § l V ® §P V of two Weyl modules, with, say, A a partition of

79

§6.1. Schur Functors and Their Characters

d and 11 a partition of m. The result is §A v ® §" v;;;

EB. NA,•• §. V;

(6.7)

here the sum is over partitions v of d + m, and N1". are numbers determined by the Littlewood-RichardsotJ rule. This is a rule that gives NA,•• as the number of ways to expand the Young diagram of A, using 11 in an appropriate way, to achieve the Young diagram for v; see (A.8) for the precise formula. Two important special cases are easier to use and prove since they involve only the simpler Pieri formula (A.7). For Jl = (m), we have §A V® Sym'"V;;;

EB §. V,

(6.8)

•

the sum over all v whose Young diagram is obtained by adding 111 boxes to the Young diagram of A, with no two in the same column. Similarly for p = (1, ... , l), (6.9) the sum over all partitions 1t whose Young diagram is obtained from that of

..t by adding m boxes, with no two in the same row. To prove these formulas, we need only observe that §AV®§,.V= y@•.cA® V@'"·c~ = y@•@ y@m. (CA@ ell)=

V@(o+m).

c,

with c = CA ® c,. E C64 ® ce.. = C(64 X 6.,) c ce;d+m· This proves that §A V ® § 11 V has a decomposition as in Corollary 6.6, and the coefficients are given by knowing the decomposition of the corresponding character. The character of a tensor product is the product of the characters of the factors; so this amounts to writing the product SAS,. of Schur polynomials as a linear combination of Schur polynomials. This is done in Appendix A, and formulas (6.7), (6.8), and (6.9) follow from (A.8), (A.7), and Exercise A.32 (v), respectively. For example, from Sym 4 V ® V = Sym4+ 1 V ® § 14 , 11 V, it follows that § 14, 11 V = Ker(Sym4V ® V--+ Sym 4+1 V),

and similarly for the conjugate partition, § 12 ,~, ... ,1) V = Ker(NV ® V--+ N+~ V).

Exercise 6.10*. One can also derive the preceding decompositions of tensor products directly from corresponding decompositions of representations of symmetric groups. Show that, in fact, §A V ® §,. V corresponds to the "inner product" representation v}. 0 v" of e;d+m described in (4.41 ). Exercise 6.11 *.(a) The Littlewood-Richardson rule also comes into the deCflmposition of a Schur functor of a direct sum of vector spaces Vand W. This

6. Weyl's Construction

80

generalizes the well-known identities Sym"(V€9 W) = d~" (SymdV® SymbW),

A"(V® W) = EB (NV®NW). a+,=n

Prove the general decomposition over GL(V) x GL(W): §.(V ® W) = EB N~~~.(§ A v ® §II W), the sum over all partitions )., 11 such that the sum of the numbers partitioned by A and 11 is the number partitioned by 1'. (To be consistent with Exercise 2.36 one should use the notation 1!1 for these "external" tensor products.) (b) Similarly prove the formula for the Schur functor of a tensor product: §.(V® W)

= EBCA,..(§~V®§IIW),

where the coefficients CA,.• are defined in Exercise 4.51. In particular show that

Sym 4 (V® W) = EB§AV®§AW, the sum over all partitions Aof d with at most dim V or dim W rows. Replacing W by W*, this gives the decomposition for the space of polynomial functions of degree d on the space Hom(V, W) over GL(V) x GL(W). For variations on this theme, see [HoJ]. Similarly,

N(V® W) = EB§AV®§A.w, the sum over partitions ). of d with at most dim V rows and at most dim W columns. Exercise 6.12. Regarding GL. c = GL. c

X

{I} c: GL. c

X

GL,.

c c: GL.+ .. c.

the preceding exercise shows how the restriction of a representation decomposes: Res(§.(C"+'"))

= L (N~,..dim §,.(C"'))§A(C").

In particular, form= 1, Pieri's formula gives Res(§,.(C"+t ))

= EB § ~(C"),

the sum over all ). obtained from v by removing any number of boxes from its Young diagram, with no two in any column. Exercise 6.13*. Show that for any partition 11

= (Jit, ... , p,) of d,

N•V®N•V® .. ·®N·V~ ~K~,.§~.v, where

K~,.

is the Kostka number and A' the conjugate of)..

§6.1. Schur Functors and Their Characters

81

Exercise 6.14*. Let 11 = ).' be the conjugate partition. Put the factors of the dth tensor power V 04 in one-to-one correspondence with the squares of the Young diagram of .1.. Show that§~ Vis the image of this composite map:

<8), (N'V)-+ <8), (®"'V)-+ V 04 -+ <8)1 (®~JV)-+ <8)1 (Sym~W), the first map being the tensor product of the obvious inclusions, the second grouping the factors of y®d according to the columns of the Young diagram, the third grouping the factors according to the rows of the Young diagram, and the fourth the obvious quotient map. Alternatively, § 1 Vis the image of a composite map

<8)1 (SymA'V)-+ <8)1 (®A'V)-+

V 04 -+

<8)1 (®"JV)-+ @j (Nil').

In particular, §A V can be realized as a subspace of tensors in V 04 that are invariant by automorphisms that preserve the rows of a Young tableau of )., or a subspace that is anti-invariant under those that preserve the columns, but not both, cf. Exercise 4.48. Problem 6.15*. The preceding exercise can be used to describe a basis for the space§~ V. Let v 1 , ••• , v, be a basis for V. For each semistandard tableau T on)., one can use it to write down an element Vr in @;(N''V); Vr is a tensor product of wedge products of basis elements, lhe ith factor in IV''V being the wedge product (in order) of those basis vectors whose indices occur in the ith column of T. The fact to be proved is that the images of these elements vr under the first composite map of the preceding exercise form a basis for § ~ V. At the end of Lecture 15, using more representation theory than we have at the moment, we will work out a simple variation of the construction of§~ v which will give quick proofs of refinements of the preceding exercise and problem. Exercise 6.16*. The Pieri formula gives a decomposition Sym 4V ® Sym 4 V =

EB §ld+a.4-•l V,

the sum over 0 :S a :S d. The left-hand side decomposes into a direct sum of Sym 2 (Sym 4V) and N(Sym 4 V). Show that, in fact, Sym 2 (Sym 4 V)

= §(24,oJ V $

N(Sym 4 V) =

§(24-1,1)1' $ §(24-3,ll VEe §1u-s.sJ V$

§ 124-2,2! V $

§(24-4,4) V $ · .. , ·· · ·

Similarly using the dual form of Pieri to decompose NV® NV into the sum (B§A V, the sum over all).= (2, ... , 2, l, ... , l) consisting of d- a 2's and 2a l's, 0 :Sa :S d, show that Sym 2 (NV) is the sum of tho~e factors with a even, and N(NV) is the sum of those with a odd.

6. Weyl's Construction

82

Exercise 6.t7•. If Aand Jtare any partitions, we can form the composite functor §,,(§ 1 V). The original "plethysm" problem-which remains very difficult in

general-is to decompose these composites: §,.(§A V) =

EB M1,.§. v, •

the sum over all partitions v of dm, where). is a partition of d and /1 is a partition of m. The preceding exercise carried out four special cases of this. (a) Show that there always exists such a decomposition for some nonnegative integers M 1,.. by constructing an element c in C64,., depending on A and /1, such that §,.(§1 V) is V 04'" ·c. (b) Compute Sym 2 (§12 • 2 J V) and N(§( 2 , 2 J V). Exercise 6.18• "Hermite reciprocity." Show that if dim V = 2 there are isomorphisms SymP(Sym9 V)

~

Symq(SymPV)

of GL(V)-representations, for all p and q. Exercise 6.19•. Much of the story about Young diagrams and representations of symmetric and general linear groups can be generalized to skew Young diagmms, which are the differences of two Young diagrams. If ). and /1 are partitions with Jt 1 ~ ).1 for all i, ).jp denotes the complement of the Young diagram for /1 in that of A. For example, if A= (3, 3, I) and /1 = (2, 1), .l./Jt is the numbered part of

To each .1./Jt we have a skew Schur function S11,., which can be defined by any of several generalizations of constructions of ordinary Schur functions. Using the notation of Appendix A, the following definitions are equivalent: (i)

s1/p

(ii)

s1,,. = IEA;-,.;-~+JI,

(iii)

=

IH1,-pri+JI.

S1/11 -- "L. mG X 0I '

•

0 • • • • Xk' '

where "'• is the number of ways to number the boxes of .1./Jt with a 1 l's, a 2 2's, ... , ai k's, with nondecreasing rows and strictly increasing columns. In terms of ordinary Schur polynomials, we have (iv) where N,,.. 1 is the Littlewood- Richardson number.

§6.1. Schur Functors and Their Characters

83

Each l/p, determines elements a11,., b11,., and Young symmetrizers c11P = a11,.b111, in A = C64 , d = I,l,- p1, exactly as in §4.1, and hence a representation denoted V.~1,, = Ac 11P of 6 4 • Equivalently, V11P is the image of the map Ab11P-+ Aa 11 ,. given by right multiplication by a 111., or the image of the map Aa 11 ,. -+ Ab11,. given by right multiplication by b11 w The decomposition of V11 ,. into irreducible representations is V11 p

(v)

= L Np•l v•.

Similarly there are skew Schur functors § 11P, which take a vector space V to the image of c 11,. on V® 4; equivalently, § 11,. Vis the image of a natural map (generalizing that in the Exercise 6.14) (vi)

<8)1 (Ni-PiV)-+

y®d-+

<8)1 (Sym 1n 1 V),

or (vii) Given a basis v" ... , vk for V and a standard tableau Ton l/p, one can write down an element vT in (8}1(Ni- 11iV); for example, corresponding to the displayed tableau, vT = v4 ® v 2 ® (v 1 A v3 ). A key fact, generalizing the result of Exercise 6.15, is that the images of these elements under the map (vi) form a basis for § 11,.V. The character of § 11P V is given by the Schur function S11P: if g is an endomorphism of V with eigenvalues x 1, ... , xk, then (viii) In terms of basic Schur functors, (ix)

Exercise 6.20*. (a) Show that if).= (p, q), §(p,qJ Vis the kernel of the contraction map cp.q:

SymPV ® Sym9 V-+ Symp+t V ® Symq-t V.

(b) If A. = (p, q, r), show that §(p.q,rJ V is the intersection of the kernels of two contraction maps cp,q ® I, and 1P ® cp,, where 11 denotes the identity map on Sym 1 V. In general, for A.= (1 10 ... , A.k), § 1 V c Sym 1 • V® .. · ® Sym 1•V is the intersection of the kernels of the k - 1 maps

t/11 = 11 , ®"·® 11 ,_, ®c 1 ,. 1,., ® 11,..® '"® 11 .,

I:<;;; i

~

k- I.

(c) For A. = (p, I, .... , I), show that § 1 V is the kernel of the contraction map: §(p,t ..... n V

= Ker(SymPV ® N-pv-+ Symp+t V ® N-p-t V).

In general, for any choice of a between I and k- I, the intersection of

84

6. Weyl's Construction

the kernels of all t/11 except t/1,. is §., V ® §, V, where u = (..1. 1 , ••• , ..l..) and =(..I..,+ a• ... , . 1.1 ); so §A V is the kernel of a contraction map defined on §., V ® §, V. For example, if a is k- 1, and we set r =At, Pieri's formula writes §., V ® Sym'V as a direct sum of §A V and other factors §, V; the general assertion in (b) is equivalent to the claim that §AVis the only factor that is in the kernel of the contraction, ie., t

§A v = Ker(§v ...... A._,) v ® Sym'V-+

v@(d-r+l)

® Sym•-l V).

These results correspond to writing the representations VA c UA of the symmetric group as the intersection of kernels of maps to UA, ..... A,+I.A,.,-I ..... A.·

Exercise 6.21. The functorial nature of Weyl's construction has many consequences, which are not explored in this book. For example, if E* is a complex 4 is also a complex, and the symmetric of vector spaces, the tensor product group 6 4 acts on it; when factors in EP and E, are transposed past each other, the usual sign ( -I)P' is inserted. The image of the Young symmetrizer cA is a complex §A(E,..), sometimes called a Schur complex. Show that if E* is the complex E_ 1 = V-+ E0 = V, with the boundary map the identity map, and ..l. =(d), then §A(E,..) is the Koszul complex

E:

o -+ N-+ N- 1 ® S 1 -+ N- 2 ® S2 -+ · · · -+ N where

N = NV, and S' =

® sd-a

-+

S 4 -+ o,

Sym 1V.

§6.2. The Proofs We need first a small piece of the general story about semisimple algebras, which we work out by hand. For the moment G can be any finite group, although our application is for the symmetric group. If U is a right module over A = CG, let B

= Hom 6 (U,

U)

= {(J): U-+ U: qJ(v·g) = qJ(v)·g, Vv E

U, g

E

G}.

Note that B acts on U on the left, commuting with the right action of A; B is called the commutator algebra. H U = U,GJ•, is an irreducible decomposition with U1 nonisomorphic irreducible right A-modules, then by Schur's Lemma l.7

EB

B

= Efj, Hom 6 (U,GJ••, u,GJ"•) = Efj, M.,(C),

where M.,(C) is the ring of n, x n1 complex matrices. If W is any left A-module, the tensor product U ®A W = U ®c Wjsubspace generated by {va ® w- v ® aw}

is a left B-module by acting on the first factor: b · (v ® w) = (b · v) ® w.

85

§6.2. The Proofs

Lemma 6.22. Let U be a finite-dimensional riglat A-module. (i) For any c e A, t/Je canonical map U ®A Ac ~ Uc is a11 isomorplaism of left 8-modules. (ii) If W = Ac is a11 irreducible left A-module, then U ®A W = Uc is a11 irreducible left B-module. (iii) If W, = Ac1 are the distinct irreducible left A-mod11les, witlt 111 1 tlte dimensio11 of W,, then

U ~ Ef),(U ®A W,)m .., ~ ffir(Uc 1)$m, is the decomposition of U into irreducible left 8-modules. PROOF. Note first that Ac is a direct summand of A as a left A-module; this is a consequence of the semisimplicity of all representations of G (Proposition 1.5). To prove (i), consider the commutative diagram ·c -----+

u

~

c.. U®A A

U·c

l

u

where the vertical maps are the maps v ®a,..._. v ·a; since the left horizontal maps are surjective, the right ones injective, and the outside vertical maps are isomorphisms, the middle vertical map must be an isomorphism. For (ii), consider first the case where U is an irreducible A-module, so 8 =C. It suffices to show that dim U ®;~ W :S I. For this we use Proposition 3.29 to identify A with a direct sum Ef>f~ 1 M.,,C ofr matrix algebras. We can identify W with a minimal left ideal of A. Any minimal ideal in the sum of matrix algebras is isomorphic to one which consists of r-tuples of matrices which are zero except in one factor, and in this factor are all zero except for one column. Similarly, U can be identified with the right ideal of r-tuples which are zero except in one factor, and in that factor all are zero except in one row. Then U ®A W will be zero unless the factor is the same for U and w; in which case U ®A W can be identified with the matrices which are zero except in one row and column of that factor. This completes the proof when U is irreducible. For the general case of(ii), decompose U = Ef>; U{JJ•• into a sum of irreducible right A-modules, so U ®A W = Ef>1(U1®;~ W)mn, = em•• for some k, which is visibly irreducible over B = Ef> M.1(C). Part (iii) follows, since the isomorphism A ~ Ef> W,Eil"'' determines an isomorphism

To prove Theorem 6.3, we will apply Lemma 6.22 to the right CSrmodule U = V® 4 • That lemma shows how to decompose U as a 8-module, where 8

6. Weyl's Construction

86

is the algebra of all endomorphisms of U that commute with all permutations of the factors. The endomorphisms of U induced by endomorphisms of V are certainly in this algebra B. Although B is generally much larger than End(V), we have Lemma 6.23. The algebra B is spanned as a linear subspace of End(V 18") by End(V). A subspace of V 18" is a srtb-B-module if and only if it is invariant by GL(V).

PRooF. Note that if W is any finite-dimensional vector space, then Sym"W is the subspace of W 18" spanned by all w" = d!w ®···®was w runs through W. Applying this to W = End(V) = V* ® V proves the first statement, since End( V18 ") = (V*)18" ® V 18" = W 18 ", with compatible actions of 6.,. The second D follows from the fact that GL( V) is dense in End(V). We turn now to the proof of Theorem 6.3. Note that §AVis UcA, so parts (2) and (4) follow from Lemmas 6.22 and 6.23. We use the same methods to give a rather indirect but short proof of part (3); for a direct approach see Exercise 6.28. From Lemma 6.22 we have an isomorphism ofGL( V)-modules: (6.24) with VA = A· cA. Similarly for UA = A· a A, and since the image of right multiplication by aA on V 18" is the tensor product of symmetric powers, we have SymA• V ® Sym 1 'V ® · · · ® Sym 1•V;;;; V 18"

®A

U1 •

(6.25)

EBP

KpA V,, of A-modules by Young's rule But we have an isomorphism UA;;;; (4.39), so we deduce an isomorphism ofGL(V)-modules

SymA' V ® Sym 1 'V ® · · · ® Sym 1•V;;;;

EB KP 1 §P V.

(6.26)

By what we saw before the statement of the theorem, the trace of g on the left-hand side of(6.26) is the product HA(x 1 , ... , Xt) of the complete symmetric polynomials H 1,(x 1 , ... , Xt)· Let §;.(g) denote the endomorphism of § 1 V determined by an endomorphism g of V. We therefore have H;.(x 1 , ... , xt)

= I:PKP;.

Trace(§11 (g)).

But these are precisely the relations between the functions HA and the Schur polynomials S,. [see formula (A.9)], and these relations are invertible, since the matrix (K,. 1 ) of coefficients is triangular with l's on the diagonal. It follows that Trace(§ 1 (g)) = S1 (x 1 , ••• , xt), which proves part (3). Note that if ,t = (,\ 1, ... , ,\4 ) with d > k and ,tHt 0, this same argument shows that the trace is SA(x 1 , ... , Xt, 0, ... , 0), which is zero, for example by (A.6). For g the identity, this shows that §A V = 0 in this case. From part (3) we also get

*

§6.2. The Proofs

87 dim§AV=SA(I, ... ,I),

(6.27)

and computing SA(I, ... , I) via Exercise A.30(ii) yields part (I).

0

Exercise 6.28. If you have given an independent proof of Problem 6.15, part (3) of Theorem 6.3 can be seen directly. The basis elements vr for§ A V specified in Problem 6.15 are eigenvectors for a diagonal matrix with entries x 1 , .•. , x•• with eigenvalue Xd = xj'' · ... · x:", where the tableau T has a 1 1's, a 2 2's, ... , Ot k's. The trace is therefore IKAdx•, where KAG is the number of ways to number the boxes of the Young diagram of,\ with a 1 1's, a 2 2's, ... , at k's. This is just the expression for SA obtained in Exercise A.3l (a). We conclude this lecture with a few of the standard elaborations of these ideas, in exercise form; they are not needed in these lectures. Exercise 6.29*. Show that, in the context of Lemma 6.22, if U is a faithful A-module, then A is the commutator of its commutator B:

A= {t/1: U-+ U: lfr(bv) = IJlfr(v), Vv

E

U,

bE

B}.

If U is not faithful, the canonical map from A to its bicommutator is surjective. Conclude that, in Theorem 6.3, the algebra of endomorphisms of V®" that commute with GL(V) is spanned by the permutations in $ 4 • Exercise 6.30. Show that, in Lemma 6.22, there is a natural one-to-one correspondence between the irreducible right A-modules U1 that occur in U and the irreducible left B-modules V,. Show that there is a canonical decomposition

u = EJ3 (V, ®c u,) as a left B-module and as a right A-module. This shows again that the number of times V, occurs in U is the dimension of U1, and dually that the number of times U1 occurs is the dimension of V,. Deduce the canonical decomposition

v®" =

EB§AV®c VA,

the sum over partitions,\ of d into at most k = dim V parts; this decomposition is compatible with the actions of GL(V) and 6 4 . In particular, the number of times VA occurs in the representation V®" of 6 4 is the dimension of § 1 V. Exercise 6.31. Let e be an idempotent in the group algebra A = CG, and let U = eA be the corresponding right A-module. Let E = eAe, a subalgebra of A. The algebra structure in A makes eA a left E-module. Show that this delines an isomorphism of C-algebras

E = eAe

~

HomA(eA, eA)

= HomG(U, U) =

B.

88

6. Weyl's Construction

Exercise 6.32. If II is a subgroup of G, and e E C/1 is an idempotent, corresponding to a representation W = CH · e of H, show that CG · e is the induced representation Tnd?,(W). For example, if ,9: II ...... C"' is a one-dimensional representation, then G

Ind 11 (.9) = CG·e,h

1 ~where e 3 = IGI ka .9(g)e,.

PART II

LIE GROUPS AND LIE ALGEBRAS

From a naive point of view, Lie groups seem to stand at the opposite end of the spectrum of groups from finite ones. 1 On the one hand, as abstract groups they seem enormously complicated: for example, being of uncountable order, there is no question of giving generators and relations. On the other hand, they do come with the additional data of a topology and a manifold structure; this makes it possible-and, given the apparent hopelessness of approaching them purely as algebraic objects, necessary-to use geometric concepts to study them. Lie groups thus represent a confluence of algebra, topology, and geometry, which perhaps accounts in part for their ubiquity in modern mathematics. It also makes the subject a potentially intimidating one: to have to understand, both individually and collectively, all these aspects of a single object may be somewhat daunting. Happily,just because the algebra and the geometry/topology of a Lie group are so closely entwined, there is an object we can use to approach the study of Lie groups that extracts much of the structure of a Lie group (primarily its algebraic structure) while seemingly getting rid of the topological complexity. This is, of course, the Lie algebra. The Lie algebra is, at least according to its definition, a purely algebraic object, consisting simply of a vector space with bilinear operation; and so it might appear that in associating to a Lie group its Lie algebra we are necessarily giving up a lot of information about the group. This is, in fact, not the case: as we shall see in many cases (and perhaps all of the most important ones), encoded in the algebraic structure of a Lie alg~bra is almost all of the geometry of the group. In particular, we will 1 In spite of this there are deep, if only partially understood, relations between finite and Lie 8roups, extending even to their simple group classir.cations.

II. Lie Groups and Lie Algebras

90

see by the end of Lecture 8 that there is a very close relationship between representations of the Lie group we start with and representations of the Lie algebra we associate to it; and by the end of the book we will make that correspondence exact. We said that passing from the Lie group to its Lie algebra represents a simplification because it eliminates whatever nontrivial topological structure the group may have had; it "flattens out," or "linearizes," the group. This, in turn, allows for a further simplification: since a Lie algebra is just a vector space with bilinear operation, it makes perfect sense, if we are asked to study a real Lie algebra (or one over any sublield of q to tensor with the complex numbers. Thus, we may investigate first the structure and representations of complex Lie algebras, and then go back to apply this knowledge to the study of real ones. In fact, this turns out to be a feasible approach, in every respect: the structure of complex Lie algebras tends to be substantially simpler than that of real Lie algebras; and knowing the representations of the complex Lie algebra will solve the problem of classifying the representations of the real one. There is one further reduction to be made: some very elementary Lie algebra theory allows us to narrow our focus further to the study of semisimp/e Lie algebra.v. This is a subset of Lie algebras analogous to simple groups in that they are in some sense atomic objects, but better behaved in a number of ways: a semisimple Lie algebra is a direct sum of simple ones; there are easy criteria for the semisimplicity of a given Lie algebra; and, most of all, their representation theory can be approached in a completely uniform manner. Moreover, as in the case of finite groups, there is a complete classification theorem for simple Lie algebras. We may thus describe our approach to the representation theory of Lie groups by the sequence of objects Lie group """ Lie algebra ,..,... complex Lie algebra ~

semisimple complex Lie algebra.

We describe this progression in Lectures 7-9. In Lectures 7 and 8 we introduce the definitions of and some basic facts about Lie groups and Lie algebras. Lecture 8 ends with a description of the exponential map, which allows us to establish the close connection between the first two objects above. We then do, in Lecture 9, the very elementary classification theory of Lie algebras that motivates our focus on semisimple complex Lie algebras, and at least state the classification theorem for these. This establishes the fact that the second, third, and fourth objects above have essentially the same irreducible representations. (This lecture may also serve to give a brief taste of some general theory, which is mostly postponed to later lectures or appendices.) In Lecture 10 we discuss examples of Lie algebras in low dimensions.

II. Lie Groups and Lie Algebras

91

From that point on we will proceed to devote ourselves almost exclusively to the study of semisimple complex Lie algebras and their representations. We do this, we have to say, in an extremely inefficient manner: we start with a couple of very special cases, which occupy us for three lectures (11-13); enunciate the general paradigm in Lecture 14; carry this out for the classical Lie algebras in Lectures 15-20; and (finally) finish ofT the general theory in Lectures 21-26. Thus, it will not be until the end that we go back and use the knowledge we have gained to say something about the original problem. In view of this long interlude, it is perhaps a good idea to enunciate one more time our basic Poi11t of View: The primary objects of interest are Lie groups and their representations; these are what actually occur in real life and these are what we want to understand. The notion of a complex Lie algebras is introduced primarily as a tool in this study; it is an essential tool 2 and we should consider ourselves incredibly lucky to have such a wonderfully effective one; but in the end it is for us a means to an end. The special cases worked out in Lectures 11-13 are the Lie algebras of SL 2 and SL 3 . Remarkably, most of the structure shared by all semisimple Lie algebras can be seen in these examples. We should probably point out that much of what we do by hand in these cases could be deduced from the Weyl construction we saw in Lecture 6 (as we will do generally in Lecture 15), but we mainly ignore this, in order to work from a "Lie algebra'' point of view and motivate the general story. 1

Perhaps not logically so; cf. Adams' book [Ad].

LECTURE 7

Lie Groups

In this lecture we introduce the definitions and basic examples of Lie groups and Lie algebras. We assume here familiarity with the definition of differentiable manifolds and maps between them, but no more; in particular, we do not mention vector fields, differential forms, Riemannian metrics, or any other tensors. Section 7.3, which discusses maps of Lie groups that are covering space maps of the underlying manifolds, may be skimmed and referred back to as needed, though working through it will help promote familiarity with basic examples or Lie groups.

§7.1: Lie groups: definitions §7.2: Examples of Lie groups §7.3: Two constructions

§7.1. Lie Groups: Definitions You probably already know what a Lie group is; it is just a set endowed simultaneously with the compatible structures of a group and a~"' manifold. "Compatible" here means that the multiplication and inverse operations in the group structure

x:GxG .... G and

1: o .... o are actually differentiable maps (logically, this is equivalent to the single requirement that the map G X G .... G sending (x, y) to X. y-l is rc"'). A map, or morphism, between two Lie groups G and H is just a map p: G .... H that is both differentiable and a group homomorphism. In general, qualifiers applied to Lie groups refer to one or another of the two structures,

7. Lie Groups

94

usually without much ambiguity; thus, abelian refers to the group structure, or connected refers to the manifold structure. Sometimes a condition on one structure turns out to be equivalent to a condition on the other; for example, we will see below that to say that a map of connected Lie groups q1: G -> H is a surjective map of groups is equivalent to saying that the differential dqJ is surjective at every point. One area where there is some potential confusion is in the definition of a Lie subgroup. This is essentially a difficulty inherited directly from manifold theory, where we have to make a distinction between a closed subnwnifold of a manifold M, by which we mean a subset X c M that inherits a manifold structure from M (i.e., that may be given, locally in M, by setting a subset of the local coordinates equal to zero), and an immersed submanifold, by which we mean the image of a manifold X under a one-to-one map with injective differential everywhere-that is, a map that is an embedding locally in X. The distinction is necessary simply because the underlying topological space structure of an immersed submanifold may not agree with the topological structure induced by the inclusion of X in M. For example, the map from X to M. could be the immersion of an open interval in IR into the plane R 2 as a figure "6": 11-dimen.~ional

IAnother standard example of this, which is also an example in the category of groups, would be to take M to be the two-dimensional real torus R 2jl. 2 = S 1 X S 1, and X the image in M of a line v c R 2 having irrational slope:

/

v L v

/

/

The upshot of this is that we define a Lie subgroup (or clo.~ed Lie subgroup, if we want to emphasize the point) of a Lie group G to be a subset that is

§7.2. E}lamples of Lie Groups

95

simultaneously a subgroup and a closed submanifold; and we define an immersed subgroup to be the image of a Lie group II under an injective morphism to G. (That a one-to-one morphism of Lie groups has everywhere injective differential will follow from discussions later in this lecture.) The definition of a complex Ue group is euctly analogous, the words "differentiable manifold" being replaced by "comple}l manifold" and all related notions revised accordingly. Similarly, to define an algebraic group one replaces "differentiable manifold" by "algebraic variety" and "differentiable map" by "regular morphism." As we will see, the category of comple}l Lie groups is in many ways markedly different from that of real Lie groups (for e}latnple, there are many fewer complex Lie groups than real ones). Of course, the study of algebraic groups in general is quite different from either of these since an algebraic group comes with a field of definition that may or may not be a subfield of C (it may, for that matter, have positive characteristic). In practice, tho!lgh, while the two are not the same (we will see e}lamples of this in Lecfure 10, for example), the category of algebraic groups over C behaves very much like the category of complex Lie groups.

§7.2. Examples of Lie Groups The basic example of a Lie group is of course the general linear group GL.R of invertible n x 11 real matrices; this is an open subset of the vector space of all n x n matrices, and gets its manifold structure accordingly (so that the entries of the matri}l are coordinates on GL.IR). That the multiplication map GL.R x GL.R-+ GL.IR is differentiable is clear; that the inverse map GL.R-+ GL.R is follows from Cramer's formula for the inverse. Occasionally GL.R will come to us as the group of automorphisms of an 11-dimensional real vector space V; when we want to think of GL.R in this way (e.g., without choosing a basis for V and thereby identifying G with the group of matrices), we will write it as GL(V) or Aut(V). A l"epresentation of a Lie group G, of course, is a morphism from G to GL(V). Most other Lie groups are defined initially as subgroups of GL. (though they may appear in other contexts as subgroups of other general linear groups, which is, of course, the subject matter of these lectures). For the most part, such subgroups may be described either by equations on the entries of an n x n matrix, or as t~e subgroup of automorphisms of V ~ IR" preserving some structure on IR". For example, we have: the special linear group SL.R of automorphisms of IR" preserving the volume element; equivalently, n x n matrices A with determinant I. the group B. of upper-triangular matrices; equivalently, the subgroup of · automorphisms of IR" preserving the flag 1 1 In general, a flag is a sequence or subspaces or a lixed veclor space, each properly conlained in · Jhe nex1; it is a complete nag if each has one dimension larger than the preceding, and partial Olherwise.

7. Lie Groups

96

o = V0 c

V1 c V2 c .. ·

v._, c v. = R",

where l'r is the span of the standard basis vectors e 1 , .•• , e1• Note that choosing a different basis and correspondingly a different flag yields a different subgroup of GL.IR, but one isomorphic to (indeed, conjugate to) B•. Somewhat more generally, for any sequence of positive integers a 1 , ••• , at with sum n we can look at the group of block-upper-triangular matrices; this is the subgroup of automorphisms of R" preserving a partial flag

0 = V0 c V1 c V2 c · · · c

~~-• c J11

= R",

where the dimension of l'r is a 1 + ··· + a1• If the subspace l'r is spanned by the first a 1 + .. · + a 1 basis vectors, the group will be the set of matrices oflhe fonn

"' "' * "' "' "' "' 0 0 "' "' 0 0 0 "'

(

0

The group N. of upper-triangular unipotent matrices (that is, upper triangular with l's on the diagonal); equivalently, the subgroup of automorphisms of IR" preserving the complete flag { V.} where V, is the span of the standard basis vectors e 1 , ••• , e, and acting as the identity on the successive quotients V.+ .fV.. As before, we can, for any sequence of positive integers a 1 , ••• , at with sum 11, look at the group of block-upper-triangular unipotent matrices; this is the subgroup of automorphisms of R" preserving a partial flag and acting as the identity on successive quotients, i.e., matrices of the form I

(

0 0 0

"' "' "' "' "' 0 1 "' 1

0

0

1

Next, there are the subgroups of GL. R defined as the group of transform a· tions of V = R" of determinant 1 preserving some bilinear form Q: V x V-> V. If the bilinear form Q is symmetric and positive definite, the group we get is called the (.~pedal) Ol'thogonal group so.R (sometimes written SO(n); see p. 100). If Q is symmetric and non degenerate but not definite-e.g., if it has k positive eigenvalues and I negative-the group is denoted SOt,rR or SO(k, /); note that SO(k, l) ~ SO(/, k). If Q is skew-symmetric and nondegenerate, the group is called the symplectic group and denoted Sp.R; note that in this case n must be even. The equations that define the subgroup ofGL.R preserving a bilinear form Q are easy to writedown.lfwerepresent Q by a matrix M-that is, we write Q(v, w) = 'v· M: w

§7.2. E){amples of Lie Groups

97

for all v, we lll"-then the condition Q(Av, Aw) = Q(v, w)

translates into the condition that 'v·'A · M·A·w = 'v·M·w

for all v and w; this is equivalent to saying that

'A·M·A

= M.

Thus, for example, if Q is the symmetric form Q(v, w) = 'v ·w given by the identity matri){ M = I., the group so. Ill is just the group of 11 x 11 real matrices A of determinant l such that 'A = A-t. Exercise 7.t•. Show that in the case of Sp 2 • R the requirement that the transformations have determinant l is redundant; whereas in the case of so. R if we do not require the transformations to have determinant I the group we get (denoted O.R, or sometimes O(n)) is disconnected. Exercise 7.2•. Show that SO(k, I) has two connected components if k and I are both positive. The connected component containing the identity is often denoted so+(k, 1). (Composing with a projection onto Ill• or IR1, we may associate to an automorphism A e SO(k, I) automorphisms of R1 and R1; so+(k, I) will consist of those A e SO(k, I) whose associated automorphisms preserve the orientations of R• and R1.) Note that if the form Q is degenerate, a transformation preserving Q will carry its kernel Ker(Q) = {v e V: Q(v, w) = 0 Vw e V} into itself; so that the group we get is simply the group of matrices preserving the subspace Ker(Q) and preserving the induced nondegenerate form Qon the quotient V/Ker(Q). Likewise, if Q is a general bilinear form, that is, neither symmetric nor skew-symmetric, a linear transformation preserving Q will preserve the symmetric and skew-symmetric parts of Q individually, so we just get an intersection of the subgroups encountered already. At any rate, we usually limit our attention to nondegenerate forms that are either symmetric or skew-symmetric. Of course, the group GL.C of complex linear automorphisms of a complex vector space V = C" can be viewed as subgroup of the general linear group GL 2.11l; it is, thus, a real Lie group as well, as is the subgroup SL.C of 11 x 11 complex matrices of determinant 1. Similarly, the subgroups so.c c: SL.C and Sp 2.C c: SL 2 .C of transformations of a complex vector space preserving a symmetric and skew-symmetric nondegenerate bili1zear form, respectively, are real as well as complex Lie subgroups. Note that since all nondegenerate bilinear symmetric forms on a complex vector space are isomorphic (in partie-

7. Lie Groups

98

ular, there is no such thing as a signature), there is only one complex orthogonal subgroup SO.C c SL.C up to conjugation; there are no analogs of the groups S0t,IIR. Another example we can come up with here is the unitary group u. or U(n), defined to be the group of complex linear automorphisms of an It-dimensional complex vector space V preserving a positive definite Hermitian inner product H on V. (A Hermitian form H is required to be conjugate linear in the first 2 factor, and linear in the second: H(J.v, Jtw) = IH(v, w)Jl, and H(w, v) = H(v, w); it is positive definite if H(v, v) > 0 for v ¥ 0.) Just as in the case of the subgroups SO and Sp, it is easy to write down the equations for U(n): for some 11 x n matrix M we can write the form Has

H(v, w) = 'v· M. w, Vv,

wE

c·

(note that for 1-1 to be conjugate symmetric, M must be conjugate symmetric, i.e., 'M = M); then the group U(11) is just the group ofn x n complex matrices A satisfying 'A·M·A=M.

In particular, if II is the "standard" Hermitian inner product H(v, w) = 'v· w given by the identity matrix, U(n) will be the group of n x n complex matrices A such that 'A= A- 1• Exercise 7.3. Show that if H is a Hermitian form on a complex vector space V, then the real part R = Re(H) of H is a symmetric fonn on the underlying real space, and the imaginary part C = lm(H) is a skew-symmetric real form; these are related by C(v, w) = R(iv, w). Both R and C are invariant by multiplication by i: R(iv, iw) = R(v, w). Show conversely that any such real symmetric R is the real part of a unique Hermitian H. Show that if H is standard, so is R, and

Ccorresponds to the matrix J

= ( 0

-I.

I.). Deduce 0

that U(n)

= 0(2n) n Sp 2 .R.

Note that the determinant of a unitary matrix can be any complex number of modulus I; the special unitary group, SU(n), is the sJJbgroup of U(n) of automorphisms with determinant l. The subgroup of GL.C preserving an indefinite Hermitian inner product with k positive eigenvalues and I negative ones is denoted Uu or U(k, 1); the subgroup of those of determinant l is denoted SUu or SU(k, 1). In a similar vein, the group GL.H of quaternionic linear automorphisms of an It-dimensional vector space V over the ring H of quaternions is a real 2 This choice of which factor is linear and which conjugate linear is less common than the other. It makes little difference in what follows, but it does have the small advantage of being compatible with the natural choice for quatcrnions.

§7.2. Ellamples of Lie Groups

99

Lie subgroup of the group GL4 "R, as are the further subgroups of H-linear transformations of V preserving a bilinear form. Since H is not commutative, care must be taken with the conventions here, and it may be worth a little digression to go through this now. We take the vector spaces V to be right H-modules; H" is the space of column vectors with right multiplication by scalars. In this way the n x n matrices with entries in IHI act in the usual way on H" on the left. Scalar multiplication on the left (only) is H-linear. View H = C ED jC ~ C 2 • Then left multiplication by elements of I-ll give C-linear endomorphisms of C 2 , which determines a mapping H-+ M 2 C to the 2 x 2 complex matrices. In particular, H• = GL 1 H Y GL 2 C. Similarly IH" = C" ED jC" = C 2 ", so we have an embedding GL.H c.. GL 2 .C. Note that a C-linear mapping rp: H"-+ H" is D-B-linear exactly when it commutes with j: rp(vj) = rp(v)j. If v = v1 + jv 2 , then v · j = - ii2 + jii 1 , so multiplication by j takes

(vv

1)

to ( 0

2

1

-I) (~1 ). 0

v2

It follows that if J is the matrix of the preceding ·

exercise, then GL.H

={A E GL 2.C: AJ = JA}.

Those matrices with real determinant 1 form a subgroup SL.H. A Hermitian form (or "symplectic scalar product") on a quaternionic vector space Vis an Ill-bilinear form K: V x V-+ H that is conjugate H-linear in the first factor and H-linear in the second: K(v.t, wp) = IK(v, w)11, and satisfies K(w, v) = K(v, w). It is positive definite if K(v, v) > 0 for v # 0. (The conjugate I of a quaternion 1 = a + bi + cj + dk is defined to be a - bi - cj - dk.) The standard Hermitian form on H" is l:ii1w1• The group of automorphisms of an If-dimensional quaternionic space preserving such a form is called the compact symplectic group and denoted Sp(n) or U 11 (n). The standard Hermitian form on H" is 1:v1w1• Exercise 7.4. Regarding Vas a complex vector space, show that every quaternionic Hermitian form K has the form

K(v, w) = 1-l(v, w)

+ jQ(v, w),

where H is a complex Hermitian form and Q is a skew-symmetric complell linear form on V, with Hand Q related by Q(v, w) = 1-l(vj, w), and If satisfying the condition lf(vj, wj) = H(v, w). Conversely, any such Hermitian H is the complex part of a unique K. If K is standard, so isH, and Q is given by the same matrill as in Ellercise 7.3. Deduce that Sp(n) = U(2n) n Sp 2 .C. This shows that the two notions of "symplectic" are compatible. More generally, if K is not positive definite, but has signature (p, q), say the Standard }]=I PJWf - Lf,!:+l VI WI> the automorphisms preserving it form a group Up,,H. Or if the form is a skew Hermitian form (satisfying the same

7. Lie Groups

100

linearity conditions, but with K(w, v) = - K(v~W)), the group is denoted

u:u. Exercise 7.5. Identify, among all the real Lie groups described above, which ones are compact.

Complex Lie Groups So far, all of our examples have been examples of real Lie groups. As for complex Lie groups, these are fewer in number. The general linear group GL.C is one, and again, all the elementary examples come to us as subgroups of the general linear group GL.C. There is, for example, the subgroup SO.C of automorphisms of an n-dimensional complex vector space V having determinant 1 and preserving a nondegenerate symmetric bilinear form Q (note that Q no longer has a signature); and likewise the subgroup Sp.C of transformations of determinant 1 preserving a skew-symmetric bilinear form. Exercise 7.6. Show that the subgroup SU(n) c: SL.C is not a complex Lie subgroup. (It is not enough to observe that the defining equations given above are not holomorphic.) Exercise 7.7. Show that none of the complex Lie groups described above is compact. We should remark here that both of these exercises are immediate consequences of the general fact that any compact complex Lie group is abelian; we will prove this in the next lecture. A representation of a complex Lie group G is a map of complex Lie groups from G to GL(V) = GL.c for an ndimensional complex vector space V; note that such a map is required to be complex analytic.

Remarks on Notation A common convention is to use a notation without subscripts or mention of ground field to denote the real gro11ps: 0(11),

SO(n), SO(p, q),

U(n), SU(n), SU(p, q), Sp(n)

and to use subscripts for the algebraic groups GL., SL., SO., and Sp•. This, of course, introduces some anomalies: for example, so.R is SO(n), but Sp.IR is not Sp(n); but some violation of symmetry seems inevitable in any notation. The notations GL(11, IR) or GL(n, C) are often used in place of our GL.R or GL.c, and similarly for SL, SO, and Sp. Also, where we have written Sp 2 ., some write Sp•. In practice, it seems that those most interested in algebraic groups or Lie algebras use the former notation, and those interested in compact groups the latter. Other common notations are U*(2n) in place of our GL.H, Sp(p, q) for our Up.lU, and 0*(211) for our u:~u.

101

§7.3. Two Comtructions

Exercise 7.8. Find the dimensions of the various real Lie groups GL.R, SL.R, B., N., SO.R, SOt, 1R, Sp 2.R, U(u), SU(n), GL.C, SL.C, GL.n~. and Sp(11) introduced above.

§7.3. Two Constructions There are two constructions, in some sense inverse to one another, that arise frequently in dealing with Lie groups (and that also provide us with further examples of Lie groups). They are expressed in the following two statements. Proposition 7.9. Let G be a Lie group, H a con11ected manifold, a11d (p: H -> G a covering space map. 3 Let e' be a11 element lying over the identity e o.f G. Then there is a 11nique Lie group .~tructure on H such tltat e' is tl1e identity aud ({' is a map of Lie groups; and the kemel of({' i.~ ill the center of H.

r

Proposition 7.10. Let H be a Lie group, and c: Z(H) a discrete suhgroup of its ceuter. Then there is a unique Lie group stmcture on the quotielll group G = H/r st1cl1 that tlte quotient map H-> G is a Lie gr011p map. The proof of the second proposition is straightforward. To prove the first, one shows that the multiplication on G lifts uniquely to a map H x II-> H which takes (e', e') toe', and verifies that this product satisfies the group axioms. In fact, it suffices to do this when H is the universal covering of G, for one can then apply the second proposition to intermediate coverings. D Exercise 7. II"'. (a) Show that any discrete normal subgroup of a connected Lie group G is in the center Z(G). (b) If Z(G) is discrete, show that G/Z(G) has trivial center. These two propositions motivate a definition: we say that a Lie group map between two Lie groups G and His an isogeny if it is a covering·space map of the underlying manifolds; and we say two Lie groups G and If are isogenot1s if there is an isogeny between them (in either direction). Isogeny is not an equivalence relation, but generates one; observe that every isogeny equivalence class has an initial member (that is, one that maps to every other one by an isogeny)-that is, just the univers:~l covering space Gof any one-and, if the center of this universal cover is discrete, as will be the case for all our semisimple groups, a final object G/Z(G) as well. For any group Gin such an equivalence class, we will call Gthe simply connected form of the group G, and G/Z(G) (if it exists) the adjoint form (we will see later a more general definition of adjoint form). This means that rp is a continuous map with the property that every point of G has a neighborhood U such that rp- 1(U) is a dis.ioint union of open sets each mapping homeomorrhically to U.

3

7. Lie Groups

102

Exercise 7.12. If If-+ G is a covering of connected Lie groups, show that Z(G) is discrete if and only if Z(H) is discrete, and then lf/Z(H) ,. GfZ(G). Therefore, if Z(G) is discrete, the adjoint form of G exists and is G/Z(G). To apply these ideas to some of the examples discussed, note that the center of SL. (over IR or C) is just the subgroup of multiples of the identity by an nth root of unity; the quotient may be denoted PSL.IIl or PSL.C. In the complex case, PSL.C is isomorphic to the quotient ofGL.C by its center C* of scalar matrices, and so one often writes PGL.C instead of PSL.C. The center of the group so. is the subgroup {±I} when n is even, and trivial when n is odd; in the former case the quotient will be denoted PSO.R or PSO.C. Finally the center of the group Sp 2 • is similarly the subgroup { ±1}, and the quotient is denoted PSp 2.11l or PSp 2.C. Exercise 7.13*. Realize PGL.C as a matrix group, i.e., find an embedding (faithful representation) PGL.C c:... GLI'IC for some N. Do the same for the other quotients above. In the other direction, whenever we have a Lie group that is not simply connected, we can ask what its universal covering space is. This is, for example, how the famous spin groups arise: as we will see, the orthogonal groups SO. Ill and so.c have fundamental group Z/2, and so by the above there exist connected, two-sheeted covers of these groups. These are denoted Spin. Ill and Spin.C, and will be discussed in Lecture 20; for the time being, the reader may find it worthwhile (if frustrating) to try to realize these as matrix groups. The last exercises of this section sketch a few steps in this direction which can be done now by hand. Exercise 7.14. Show that the universal covering of U(n) can be identified with the subgroup or the product U(n) X R consisting or pairs (g, t) with det(g) ,. ed'. Exercise 7.15. We have seen in Exercise 7.4 that SU(2) = Sp(2) = {qE IH: qq= 1}. Identifying R with the imaginary quaternions (with basis i, j, k), show that, for qq = I, the map v~-+qvqmaps R 3 to itself, and is an isometry. Verify that the resulting map 3

SU(2) = Sp(2)-+ S0(3) is a 2: I covering map. Since the equation qq = l describes a 3-sphere, SU(2) is the universal covering of S0(3~ and S0(3) is the adjoint form of SU(2). Exercise 7.16. Let M2C = C4 be the space of2 X 2 matrices, with symmetric form Q(A, B) = ! Trace(AJJ'l), where 1P is the adjoint of the matrix B; the

§7.3. Two Constructions

103

quadratic form associated to Q is the determinant. For g and h in SL 2 C, the mapping A HgAir-• is in so4c. Show that this gives a 2: I covering SL 2C x SL 2C __.

so4c,

which, since SL 2 C is simply connected, realizes the universal covering of

so4c. ExerCise 7.17. Identify C 3 with the space of traceless matrices in M 2 C, so g e SL 2 C acts by A 1--+ gAy-•. Show that this gives a 2: I covering SL 2 C

--+

S03 C,

which realizes the universal covering or sole.

LECTURE 8

Lie Algebras and Lie Groups

In this crucial lecture we introduce the definition or the Lie algebra associated to a Lie group and its relation to that group. All three sections are logically necessary for what follows; §8. I is essential. We use here a little more manifold theory: specifically, the differential of a map of manifolds is used in a fundamental way in §8.1, the notion of the tangent vector to an arc in a manifold is used in §8.2 and §8.3, and the notion or a vector field is introduced in an auxiliary capacity in §8.3. The Campbell-Hausdorff formula is introduced only to establish the First and Second Principles of§8.1 below; if you are willing to take those on faith the formula (and exercises dealing with it) can be skimmed. Exercises 8.27-8.29 give alternative descriptions of the Lie algebra associated to a Lie group, but can be skipped for now. §8. I: Lie algebras: motivation and definition §8.2: Examples of Lie algebras §8.3: The exponential map

§8.1. Lie Algebras: Motivation and Definition Given that we want to study the representations of a Lie group, how do we go about it? As we have said, the notions of generators and relations is hardly relevant here. The answer, of course, is that we have to use the continuous structure of the group. The first step in doing this is Exercise 8.1. Let G be a connected Lie group, and U c: G any neighborhood of the identity. Show that U generates G. This statement implies that any map p: G-+ H between connected Lie groups will be determined by what it does on any open set containing the

§8.1. Lie Algebras: Motivation and Definition

105

identity in G, i.e., p is determined by its germ at e E G. In fact; we can extend this idea a good bit further: later in this lecture we will establish the

First Priitciple: Let G and H be Lie groups, with G coimected. A map p: G--+ H is uniquely determined by its differential dp.: T,G--+ 1~H at the identity. This is, of course, great news: we can completely describe a homomorphism of Lie groups by giving a linear map between two vector spaces. It is not really worth that much, however, unless we can give at least some answer to the next, obvious question: whiclt maps betwee11 tltese two vector spaces actually arise as differentials of group homomorphisms? The answer to this is expressed in the Second Principle below, but it will take us a few pages to get there. To start, we have to ask ourselves what it means for a map to be a homomorphism, and in what ways this may be reflected in the differential. To begin with, the definition of a homomorphism is simply a~ map p such that p(glt)

= p(g). p(h)

for all g and It in G. To express this in a more confusing way, we can say that a IJOmomorpllism respects the action of a group on itself by l~ft or rigltt muftipllcatioll: that is, for any g E G we denote by m,: G --+ G the differentiable map given by multiplication by g, and observe that a ~ map p: G --+ H of Lie groups will be a homomorphism if it carries m, to m,1,, in the sense that the diagram

G-> H p

commutes. The problem with this characterization is that, since the maps m, have no fixed points, it is hard to associate to them any operation on the tangent space to G at one point. This suggests looking, not at the diffeomorphisms m,, but at the automorphisms of G given by conjugation. Explicitly, for any g E G we define the map

'1',: G--+ G by

('1'9 is actually a Lie group map, but that is not relevant for our present

purposes.) It is now equally the case that a lromomorplrism p rttspect.~ the action of a group G on itself by conjugation: that is, it will carry '1'9 into '1',1,, in the sense that the diagram

8. Lie Algebras and Lie Groups

106

a~n

p

commutes. We have, in other words, a natural map '1': G ~Aut( G). The advantage of working with 'I'~ is that it fixes the identity element e E G; we can therefore extract some of its structure by looking at its differential at e: we set Ad(g) = (d'l',).: T.,G

~

T.,G.

(8.2)

This is a representation Ad: G-.. Aut(T.,G)

(8.3)

of the group G on its own tangent space, called the adjoint representation of the group. This gives a third characterization 1 : a homomorphism p respects the adjoint action o.f a group G on its tangent space T.G at the ider1tity. In other words, for any g E G the actions of Ad(g) on T.G and Ad(p(g)) on T.H must corn mute with the differential (dp).: T.G ~ T.,H, i.e., the diagram

T.,G ~ T.,H Ad(g)

I

IAd(p(~ll

T.,G ~ T.H commutes; equivalently, for any tangent vector v e T.,G, dp(Ad(g)(v)) = Ad(p(g))(dp(v)).

(8.4)

This is nice, but does not yet answer our question, for preservation of the adjoint representation Ad: G-.. Aut(T.G)still involves the map p on the group G itself, and so is not purely a condition on the differential (dp) •. We have instead to go one step further, and take the differential of tile map Ad. The group Aut(T.,G) being just an open subset of the vector space of endomorphisms of T.,G, its tangent space at the identity is naturally identified with End(T.G); taking the differential of the map Ad we arrive at a map ad: T.,G

~

End(7,;G).

(8.5)

This is essentially a trilinear gadget on the tangent space T.,G; that is, we can view the image ad(X)(Y) of a tangent vector Y under the map ad(X) as a 1 "Characterization" is not the right word here (or in the preceding case), since we do not mean an equivalent condition, but rather something implied by the condition that '' be a homomorphism.

107

§8.1. Lie Algebras: Motivation and Definition

function of the two variables X and Y, so that we get a bilinear map T,G

X 1~G--+

T,G.

We use the notation [ , ] for this bilinear map; that is, for a pair of tangent vectors X and Y to G at e, we write [X, Y]~ ad(X)(Y).

(8.6)

As desired, the map ad involves only the tangent space to the group G at e, and so gives us our final characterization: the differential (dp). of a homomorpllism p on a Lie group G respects tlte adjoirtt actiort of tile tartgent space to G on itself. Explicitly, the fact that p and dp. respect the adjoint representation implies in turn that the diagram

T,G~ T,H

l•d(dp(o))

ad(v)l

T,H commutes; i.e., for any pair of tangent vectors X and Y to G at e, dp.(ad(X)(Y))

= ad(dp.(X))(clp.( Y)).

(8.7)

or, equivalently, dp.([X, Y]) = [dp.(X), di'.(Y)].

(8.8)

All this may be fairly confusing (if it is not, you probably do not need to be reading this book). Two things, however, should be borne in mind. They are: (i) It is rwt so bad, in the sense that we can make the bracket operation, as defined above, reasonably explicit. We do this first for the general linear group G = GL"R. Note that in this case conjugation extends to the ambient linear space E = End(R") = M"R ofGL"R by the same formula: Ad(g)(M) = gMg- 1, and this ambient space is identified with the tangent space T,G; differentiation in E is usual differentiation of matrices. For any pair of tangent vectors X and Y to GL"R ate, let y: I--+ G be an arc with y(O) = e and tangent vector y'(O) =X. Then our definition of [X, Y] is that [X, Y] = ad(X)(Y) =

~~r=o (Ad(y(t))(Y)).

Applying the product rule to Ad(y(t))(Y) = y'(O)· Y·y(O)

= y(t) l'y(tf 1, this is

+ y(O)· Y·(-y(Ot 1 ·y'(O)·y(0)- 1 )

=X· Y- Y·X, which, of course, explains the bracket notation. In general, any time a Lie group is given as a subgroup of a general linear group GL"R, we can view its

8. Lie Algebras and Lie Groups

108

tangent space T.G at the identity as a subspace of the space of endomorph isms of R"; and since bracket is preserved by (differentials of) maps of Lie groups, the bracket operation on T.G will coincide with the commutator. (ii) Eve11 if it were that bad, it wo11lcl be worth it. This is because it turns out that the bracket operation is exactly the answer to the question we raised before. Precisely, later in this lecture we will prove the

Second P1·incitJie: Let G and H be Lie groups, with G connected and simply connected. A linear map T.G ~ T.H is the differential of a homomorphism p: G-+ II if and only if it preserves the bracket operation, in the sense of (8.8) above. We are now almost done: maps between Lie groups are classified by maps between vector spaces preserving the structure of a bilinear map from the vector space to itself. We have only one more question to answer: when does a vector space with this additional structure actually arise as the tangent space at the identity to a Lie group, with the adjoint or bracket product? Happily, we have the answer to this as well. First, though it is far from clear from our initial definition, it follows from our description of the bracket as a commutator that tl•e bracket Is skew-symmetric, i.e, [X, Y] = -[Y, X]. Second, it likewise follows from the description of [X, Y] as a commutator that it satisfies the Jacobi identity: for any three tangent vectors X, Y, and Z, [X, [Y, Z]]

+ [Y, [Z, X]] + [Z, [X,

Y]] = 0.

We thus make the Definition 8.9. A Lie algebra 9 is a vector space together with a skew-symmetric bilinear map [,]:gxg-+g satisfying the Jacobi identity. We should take a moment out here to make one important point. Why, you might ask, do we define the bracket operation in terms of the relatively difficult operations Ad and ad, instead of just defining [X, Y] to be the commutator X· Y- Y ·X? The answer is that the "composition" X· Y of elements of a Lie algebra is not well defined. Specifically, any time we embed a Lie group G in a general linear group GL(V), we get a corresponding embedding of its Lie algebra 9 in the space End( V), and can talk about the composition X· Y E End(V) of elements of !1 in this context; but it must be borne in mind that this composition X· Y will depend on the embedding of g, and for that matter need not even be an element of g. Only the commutator X· Y- Y ·X is always an element of g, independent of the representation. The terminology sometimes heightens the confusion: for example, when we speak of embedding a Lie algebra in the algebra End(V) ofendomorphisms of V, the word algebra may mean two very different things. In general, when we want

§8.1. Lie Algebras: Motivation and Delinition

109

to refer to the endomorph isms of a vector space V (resp. IR") as a Lie algebra, we will write gi(V) (resp. gi.R) instead of End(V) (resp. M.R). To return to our discussion of Lie algebras, a map of Lie algebras is a linear map of vector spaces preserving the bracket, in the sense of(8.8); notions like Lie su1Jalgebra are defined accordingly. We note in passing one thing that will turn out to be significant: the definition of Lie algebra does not specify the field. Thus, we have real Lie algebras, complex Lie algebras, etc., all defined in the same way; and in addition, given a real Lie algebra !1 we may associate to it a complex Lie algebra, whose underlying vector space is g ® C and whose bracket operation is just the bracket on g extended by linearity.

Exercise 8.10*. The skew-commutativity and Jacobi identity also follow from the naturality of the bracket (8.8), without using an embedding in gi(V): (a) Deduce the skew-commutativity [X, X] = 0 from that fact that any X can be written the image of a vector by dp. for some homomorphism p: R ~ G. (See §8.3 for the existence of p.) (b) Given that the bracket is skew-commutative, verify that the Jacobi identity is equivalent to the assertion that ad = d(Ad),: g ~ End(g) preserves the bracket. In particular, ad is a map of Lie algebras. To sum up our progress so far: taking for the moment on faith the statements made, we have seen that (i) the tangent space g at the identity to a Lie group G is naturally endowed with the structure of a Lie algebra; (ii) if G and H are Lie groups with G connected and simply connected, the maps from G to II are in one-to-one correspondence with maps of the associated Lie algebras, by associating to p: G -> II its differential (dp),: g ~ ~Of course, we make the Definition 8.11. A representation of a Lie algebra g on a vector space V is simply a map of Lie algrbras

p: g ~ gi(V) = End(V), i.e., a linear map that preserves brackets, or an action of g on V such that [X, Y](v) = X(f(v))- Y(X(v)).

Statement (ii) above implies in particular that representations of a cormected and simply connected Lie gr·oup are in one-to-one correspondence with l'epre-

8. Lie Algebras and Lie Groups

110

sentations of its Lie algebra. This is, then, the first step of the series of reductions outlined in the introduction to Part II. At this point, a few words are in order about the relation between representations of a Lie group and the corresponding representations of its Lie algebra. The first remark to make is about tensors. Recall that if V and Ware representations of a Lie group G, then we define the representation V ® W to be the vector space V ® W with the action of G described by

g(v ® w) = g(v) ® g(w). The definition for representations of a Lie algebra, however, is quite different. For one thing, if g is the Lie algebra of G, so that the representation of G on the vector spaces V and W induces representations of g on these spaces, we want the tensor product of the representations V and W of g to be the representation induced by the action of G on V ® W above. But now suppose that {y,} is an arc in G with y0 = e and tangent vector y~ = X e g. Then by definition the action of X on Vis given by X(v) = _dd

I y,(v)

t r•o

and similarly for we W; it follows that the action of X on the tensor product

v®wis X(11 ® w) =

=

_dd l

I

r:o

(y,(v) ® y,(w))

(;,Lo y,(v))®

w

+

v®(;,Lo y,(w)).

so X(v ® w) = X(v) ® w + v ® X(w).

(8.12)

This, then, is how we define the action of a Lie algebra g on the tensor product of two representations of g. This describes as well other tensors: for example, if V is a representation of the group G, v e V is any vector and v2 e Sym 2 V its square, then for any g e G, g(vl) = g(v)l.

On the other hand, if Vis a representation of the Lie algebra g and X e g is any element, we have

(8.13) One further example: if p: G __. GL(V) is a representation of the group G, the dual representation p': G __. GL(V*) is defined by setting p'(g) = 'p(g- 1 ):

v•- v•.

Dilferentiating this, we lind that if p: g--> gi(V) is a representation of a Lie

Ill

§8.2. Examples of Lie Algebras algebra g, the dual representation of g on V* will be given by p'(X)

= 'p(-X) = -'p(X): V*->

V*.

(8.1:4)

A second and related point to be made concerns terminology. Obviously, when we speak of the action of a group G on a vector space V preserving some extra structure on V, we mean that literally: for example, if we have a quadratic form Q on V, to say that G preserves Q means just that Q(g(v), g(w)) = Q(v, w),

Vg E G and v, wE Jl.

Equivalently, we mean that the associated action of G on the vector space Sym 2 V* fixes the element Q E Sym 2 V*. But by the above calculation, the action of the associated Lie algebra g on V satisfies

+ Q(X(v), w) = 0, VX e !I and ''• wE V (8.15) or, equivalently, Q(v, X(v)) = 0 for all X E g and v E V; in other words, l11e Q(v, X(w))

induced action 011 Sym 2 V* kills the eleme11t Q. By way of terminology, then, we will in general say that the action of a Lie algebra 011 a vector space preserves some st,-uctllre when a correspondi11g Lie group action does. The next section will be spent in giving examples. In §8.3 we will establish the basic relations between Lie groups and their Lie algebras, to the point where we can prove the First and Second Principles above. The further statement that any Lie algebra is the Lie algebra of some Lie group will follow from the statement (see Appendix E) that every Lie algebra may be embedded in gi.R. Exercise 8.16*. Show that if G is connected the image of Ad: G-> GL(g) is the adjoint form of the group G when that exists. Exercise 8.17*. Let V be a representation of a connected Lie group G and p: g-.. End(V) the corresponding map of Lie algebras. Show that a subspace W of Vis invariant by G if and only if it is carried into itself under the action of the Lie algebra g, i.e., p(X)(W) c W for all X in g. Hence, V is irreducible over G if and only if it is irreducible over g.

§8.2. Examples of Lie Algebras We start with the Lie algebras associated to each of the groups mentioned in Lecture 7. Each of these groups is given as a subgroup ofGL(V) = GL.R, so their Lie algebras will be subspaces of End(V) = gi.R. Consider first the special linear group SL.R. If {A,} is an arc in SL.R with A0 = I and tangent vector A~ = X at t = 0, then by definition we have for any basis e 17 ••• , e. of V = R", A,(e 1 )

"

••• "

A,(e.)

= e1

"

·•· "

e•.

8. Lie Algebras and Lie Groups

112

Taking the derivative and evaluating at t = 0 we have by the product rule

~L 0(A,(er) A · • • A A,(e.)) =I e 1 A · • • A X(e A · • · A e.

0=

1)

= Trace(X)·(e 1 A··· A e.).

The tangent vectors to SL.R are thus all endomorph isms or trace 0; comparing dimensions we can see that the Lie algebra si.R is exactly the vector space or traceless 11 x 11 matrices. The orthogonal and symplectic cases are somewhat simpler. For example, the orthogonal group o.R is defined to be the automorphisms A of an n-dimensional vector space V preserving a quadratic form Q, so that if {A,} is an arc in O.R with A0 = I and A~ =X we have for every pair of vectors v, weV

Q(A,(v), A,(w))

= Q(v, w).

Taking derivatives, we see that Q(X(v), w)

+ Q(v, X(w)) = 0

(8.18)

for all v, we V; this is exactly the condition that describes the orthogonal Lie algebra so.R = o.R.In coordinates, if the quadratic form Qis given on V = R" as Q(11, w) = 'v· M·w (8.19) for some symmetric 11 x n matrix M, then as we have seen the condition on A E GL.R to be in o.R is that

(8.20)

'A·M·A = M. Differentiating, the condition on an 11 x so.R of the orthogonal group is that 'X·M

11

matrix X to be in the Lie algebra

+ M·X

{8.21)

=0.

Note that if M is the identity matrix..,-i.e., Q is the "standard" quadratic form Q(v, w) = 'v ·won R"-then this says that so.R is the subspace of skewsymmetric 11 x 11 matrices. To put it intrinsically, in terms of the identification of V with V* given by the quadratic form Q, and the consequent identification End( V) = V ® V* = V ® V, the Lie algebra so.R c End( V) is just the subspace N V c V ® V of skew-symmetric tensors: so.R =NV c End(V)

=

V ® V.

(8.22)

All of the above, with the exception of the last paragraph, works equally well to describe the ~ie algebra sp 2 .R of the Lie group Sp 2.R of transformations preserving a skew-symmetric bilinear form Q; that is, sp 2 .R is the subspace of endomorphisms of V satisfying (8.18) for every pair of vectors v, we V, or, if Q is given by a skew-symmetric 211 x 2rr matrix Mas in (8.19), the

§8.2. Examples of Lie Algebras

113

space of matrices satisfying (8.21). The one statement that has to be substantially modified is the last one of the last paragraph: because Q is skewsymmetric, condition (8.18) is equivalent to saying that Q(X(v), w) = Q(X(w), v)

for all v, wE V; thus, in terms of the identification of V with V* given by Q, the Lie algebra sp 2 .1R c End(V) = V ® V* = V ® V is the subspace Sym 2 V c V® V: (8.23)

Exercise 8.24*. With Q a standard skew form, say of Exercise 7.3, describe Sp 2 .1R and its Lie algebra 1p 2 .R (as subgroup of GL 2 .1ll and subalgebra of gl 2 .1ll). Do corresponding calculation for SOk, 11ll.

a

One more similar example is that of the Lie algebra u. of the unitary group U(n); by a similar calculation we find that the Lie algebra of complex linear endomorphisms ofe• preserving a Hermitian inner product His just the space of matrices X satisfying J:f(X(v), w)

+ H(v, X(w)) = 0,

Vv,

wE

V;

if 1/ is given by H(v, w) = 'v· w, this amounts to saying that X is conjugate skew-symmetric, i.e., that 'X= -X. Exercise 8.25. Find the Lie algebras of the real Lie groups SL.e and the elements in GL.H whose real determinant is 1.

SL.D~

Exercise 8.26. Show that the Lie algebras of the Lie groups B. and N. introduced in §7.2 are the algebra b.R of upper triangular 11 x n matrices and the algebra n.IR of strictly upper triangular n x 11 matrices, respectively. If G is a complex Lie group, its Lie algebra is a complex Lie algebra. Just as in the real case, we have the complex Lie algebras gl.e, sl.e, some, and sp 2.C of the Lie groups GL.C, SL.e, SOme, and Sp 2 .C. Exercise 8.27. Let A be any (real or complex) algebra, not necessarily finile dimensional, or even associative. A derivation is a linear map D: A --> A satisfying the Leibnitz rule D(ab) = aD(b) + D(a)b. (a) Show that the derivations Der(A) form a Lie algebra under the bracket [D, E] = DoE- Eo D. If A is finite dimensional, so is Der(A). (b) The group of automorphisms of A is a closed subgroup G of the group GL(A) of linear automorphisms of A. Show that the Lie algebra of G is Dec( A). (c) If the algebra A is a Lie algebra, the map A -->Dec( A), X H Dx, where Dx(Y) =[X, Y]. is a map of Lie algebras.

8. Lie Algebras and Lie Groups

114

Exercise 8.28*. If o is a Lie algebra, the Lie algebra automorphisms of !J form a Lie subgroup Aut(o) of the general linear group GL(!J). (a) Show that the Lie algebra of Aut(g) is Der(!J). If G is a simply connected Lie group with Lie algebra g, the map Aut(G)-+ Aut(g) by qH~ dqJ is one-to-one and onto, giving Aut( G) the structure of a Lie group with Lie algebra Der(o). (b) Show that the automorphism group of any connected Lie group is a Lie subgroup of the automorphism group of its Lie algebra. Exercise 8.29*. For any manifold M, the C"' vector fields on M form a Lie algebra o(M), as follows: a vector field v can be identified with a derivation of the ring A of C"' functions on M, with v(f} the function whose value at a point x of M is the value of the tangent vector v,. on f at x. Show that the vector fields on M form a Lie algebra, in fact a Lie subalgebra of the Lie algebra Der(A}. If a Lie group G acts on M, the G-invariant vector fields form a Lie subalgebra u0 M of u(M}. If the action is transitive, the invariant vector fields form a finite-dimensional Lie algebra. If G is a Lie group, oo(G) = T.G becomes a Lie algebra by the above process. Show that this bracket agrees with that defined using the adjoint map (8.6). This gives another proof that the bracket is skew-symmetric and satisfies Jacobi's identity.

§8.3. The Exponential Map The essential ingredient in studying the relationship between a Lie group G and its Lie algebra 9 is the exponential map. This may be defined in very straightforward fashion, using the notion of one-parameter subgroups, which we study next. Suppose that X e g is any element, viewed simply as a tangent vector to G at the identity. For any element g E G, denote by m,: G -> G the map of manifolds given by multiplication on the left by g. Then we can define a vector field Vx on all of G simply by setting ''x(g) = (m,)*(X). This vector field is clearly invariant under left translation (i.e., it is carried into itself under the diffeomorphism m, for all g); and it is not hard to see that this gives an identification of g with the space of all left-invariant vector fields on G. Under this identification, the bracket operation on the Lie algebra g corresponds to Lie bracket of vector fields; indeed, this may be adopted as the definition of the Lie algebra associated to a Lie group (cf. Exercise 8.29). For our present purposes, however,'all we need to know is that Vx exists and is left-invariant. Given any vector field v on a manifold M and a point p eM, a basic theorem from differential equations allows us to integrate the vector field. This

§8.3. The Exponential Map

\15

gives a differentiable map cp: I-+ M, defined on some open interval I containing 0, with cp(O) = p, whose tangent vector at any point is the vector assigned to that point by v, i.e., such that cp'(t) = v(cp(t)) for all t in I. The map cp is uniquely characterized by these properties. Now suppose the manifold in question is a Lie group G, the vector field the field vx associated to an element X E g, and p the identity. We arrive then at a map cp: I-+ G; we claim that, at least where cp is defined, it is a homomorpiJism, i.e., cp(s + t) = cp(s)cp(t) whenever s, t, and s + t are in I. To prove this, fix s and lett vary; that is, consider the two arcs a and Pgiven by a(t) = tp(s) · cp(t) and fl(t) = cp(s + t). Of course, a(O) = /1(0); and by the in variance of the vector field Vx, we see that the tangent vectors satisfy a'(t) = Vx(a(t)) and fl'(t) = Vx(fl(t)) for all t. By the uniqueness of the integral curve of a vector field on a manifold, we deduce that a(t) = fl(t) for all t. From the fact that cp(s + t) = cp(s)cp(t) for all sand t near 0, it follows that cp extends uniquely to all of R, defining a homomorphism Cf!x: R-+ G

with cp~(t)

= vx(Cf' )(t) =

(m.,1, 1).(X) for all t.

Exercise 8.30. Establish the product rule for derivatives of arcs in a Lie group G: if a and f1 are arcs in G and y(t) = a(t)· fl(t), then

y'(t) = dm«1,1(P'(t))

+ dn111,1(a'(t)),

where for any g E G, the map m, (resp. n,): G -+ G is given by left (resp. right) multiplication by g. Use this to give another proof that (pis a homomorphism.

Exercise 8.31. Show that Cf'x is uniquely determined by the fact that it is a homomorphism of R to G with tangent vector cp~(O) at the identity equal to X. Deduce that if 1/1: G-+ His a map of Lie groups, then lf'~.x = ~' o Cf'x· The Lie group map lf'x: IR -+ G is called the one-parameter subgroup of G with tangent vector X at tire identity. The construction of these one-parameter subgroups for each X amounts to the verification of the Second Principle of §8.1 for homomorphisms from R to G. The fact that there exists such a one-parameter subgroup of G with any given tangent vector at the identity is crucial. For example, it is not hard to see (we will do this in a moment) that these one-parameter subgroups fill up a neighborhood of the identity in G, which immediately implies the First Principle of §8.1. To carry this out, we define the exporterrtia/ map exp: g-+ G by exp(X) = Cf'x(l).

(8.32)

8. Lie Algebras and Lie Groups

116

Note that by the uniqueness of rpx, we have lPox1(t) = lPx(At);

so that the exponential map restricted to the lines through the origin in g gives the one-parameter subgroups of G. Indeed, Exercise 8.31 implies the characleriza lion: Proposition 8.33. Tire exponential map is tlte wtlque map from g to G taking 0 toe wl10se dijfere11tial at tire origi11 (exp.) 0 : T0 g = g->

~G =

g

is the idelll.ity, artd wltose restrictions to tire lirtes through tire origin in g are olle-parameter su[Jgroups of G. This in particular implies (cf. Exercise 8.31) that the exponential map is natural, in the sense that for any map if!: G-> H of Lie groups the diagram

commutes. Now, since the differential of the exponential map at the origin in g is an isomorphism, the image of exp will contain a neighborhood of the identity in G. If G is connected, this will generate all of G; from this follows the First Principle: if G is con11ected, then the map if! is determined by its differential (dt/1). at tire iderttit y. Using (8.32), we can write down the exponential map very explicitly in the case ofGL.R, and hence for any subgroup ofGL.R We just use the standard power series for the function e"', and set, for X E End(V), exp(X)=

X 2 X3 I +X+ 2 + 6 +···.

(8.34)

Observe that this converges and is invertible, with inverse exp(- X). Clearly, the differential of this map from ·g to G at the origin is the identity; and by the standard power series computation, the restriction of the map to any line through the origin in g is a one-para~eter subgroup of G. Thus, the map coincides with the exponential as defined originally; and by uaturality the same is true for any subgroup of G. (Note that, as we have pointed out, the individual terms in the expression on the right of (8.34) are very much dependent of the particular embedding of Gin a general linear group GL(V) and correspondingly of g in End( V), even though the sum on the right in (8.34) is not.) This explicit form of the exponential map allows us to give substance to

§8.3. The Exponential Map

ll7

the assertion that "the group structure of G is encoded in the Lie algebra." Explicitly, we claim that not only do the exponentials exp(X) generate G, but for X and Y in a sufficiently small neighborhood of the origin in !J, we can write down the product exp(X)·exp(Y) as an exponential. To do this. we introduce first the "inverse" of the exponential map: forgE G c GL.R, we set (g - 1)2 log(g) = (g -I)- - 2-

(g - /)3

+ -·- 3- - · · · e ot.R.

Of course, this will be defined only for g sufficiently close to the identity in G; but where it is defined it will be an inverse to the exponential map. Now, we define a new bilinear operation on oi.R: we set

X* Y = log(exp(X)·exp(Y)). We have to be careful what we mean by this, of course; \9e substitute for g in the expression above for log(g) the quantity · exp(X)·exp(Y) =

(1 +X+~~+ ··}(r + y + ~2 + ···)

= I

+ (X +

Y) +

(Tx2 + X· Y + -y2) 2 + ··· ,

being careful, of course, to preserve the order of the factors in each product. Doing this, we arrive at

X* y =(X

+

Y) +

y2)) + ... (- .(X- 2+ -Y)2- + (X2 T + X. y + -T

=X+ Y +![X, Y]

+ · ··.

Observe in particular that the terms of degree 2 in X and Y do not involve the squares of X and Y or the product X· Y alone, but only the commutator. In fact, this is true of each term in the formula, i.e., the quantity log(exp(X) · exp( Y)) can be expressed purely in terms of X, Y, and the bracket operation; the resulting formula is called the Campbeii-Jiau!idorff formllla (although the actual formula in closed form was given by Dynkin). To degree three, it is

X* Y =X+ Y +![X, Y]

± lz[X, [X, Y]] ± rz[Y, [Y, X]]+···.

Exercise 8.35*. Verify (and lind the correct signs in) the cubic term of the Campbeii-HausdoriT formula. Exercise 8.36. Prove the assertion of the last paragraph thai the power series log(exp(X) · exp(Y)) can be expressed purely in terms of X, Y, and the bracket operation. Exercise 8.37. Show that for X and Y sufficiently small, the power series log(exp(X) · exp( Y)) converges.

8. Lie Algebras and Lie Groups

118

Exercise 8.38*. (a) Show that there is a constant C such that for X, Y E 91., X• Y =X+ Y +[X, Y] + E, where IIEII ~ C(IIXII + IIYII) 3 . (b) Show that exp(X + Y) = lim.~..,(exp(X/n)·exp(Y/n))". (c) Show that exp([X, Y]) =

r

!~~ (exp(~}exr(f}exp( -~}exp( -f)

Exercise 8.39. Show that if G is a subgroup of GL.R, the elements of its Lie algebra are the "infinitesimal transformations" of G in the sense of von Neumann, i.e., they are the matrices in 9l.R which can be realized as limits A,-1 Il. i D - - , A, e G, r., > 0, e, ___, 0. r-o e, Exercise8.40. Show that expis surjective for G = GL.C but not for G = GL; IR ifn >I, or for G = SL 2 C. By the Campbell-Hausdorff formula, we can not only identify all the elements ofG in a neighborhood of the identity, but we can also say what their pairwise products are, thus making preeise the sense in which !land its bracket operation determines G and its group law locally. Of course, we have not written a closed-form expression for the Campbeii-Hausdorlf formula; but, as we witl see shortly, its very existence is significant. (For such a closed form, see [Sel, 1§4.8].) We now consider another very natural question, namely, when a vector subspace I) c !) is the Lie algebra of (i.e., tangent space at the identity to) an immersed subgroup of G. Obviously, a necessary condition is that ~ is closed under the bracket operation; we claim here that this is sufficient as well: Proposition 8.41. Let G be a Lie group, 9 its Lie algebra, and I) c 9 a Lie subalgebra. Tllen the subgroup of tlte group G generated by exp@ is an immersed Sllf,group H witll tangent space T,H = !). PROOF. Note that the subgroup generated by exp(i)} is the same as the subgroup generated by exp(U), where U is any neighborhood of the origin in 1). It will suffice, then (see Exercise 8.42), to show that the image of !) under the exponential map is "locally" closed under multiplication, i.e., that for a sufficiently small disc!!.. c 1), the product exp(ll.) · exp(ll.) (that is, the set of pairwise products exp(X) · exp(Y) for X, YEll.) is contained in the image of!) under the exponential map. We will do this under the hypothesis that G may be realized as a subgroup of a general linear group GL.R, so that we can use the formula (8.34) for the exponential map. This is a harmless assumption, given the statement (to be proved in Appendix E) that any finite-dimensional Lie algebra may be

§8.3. The Exponential Map

119

embedded in the Lie algebra gi.R: the subgroup of GL.R generated by exp(g) will be a group isogenous to G, and, as the reader can easily check, proving the proposition for a group isogenous to G is equivalent to proving it for G. It thus suffices to prove the assertion in case the group G is GL.R. But this is exactly the content of the Campbeii-Hausdorrr formula. 0 When applied to an embedding of a Lie algebra g into gl., we see, in particular, that every finite-dimensiortal Lie algebra is the Lie algebra of a Lie group. From what we have seen, this Lie group is unique if we require it to be simply connected, and then all others are obtained by dividing this simply connected model by a discrete subgroup of its center. Exercise 8.42*. Suppose G0 is an open neighborhood of the identity in a Lie group G sueh that G0 • G0 c G0 and G0 1 = G0 . Suppose H 0 is a closed submanifold of G0 such that H0 • //0 c 1/0 and H01 = H0 • Show that the subgroup IJ of G generated by H0 is an immersed Lie subgroup of G. As a fairly easy consequence of this proposition, we can finally give a proof of the Second Principle stated in §8.1, which we may restate as Second Principle. Let G and H be Lie groups with G simply connected, and let g and I) be their Lie algebras. A lirtear rnap a: g--. ~ is tile differential of a map A: G -> H of Lie groups if and only if a is a map of Lie algebras. PROOF. To see this, consider the product G x H. Its Lie algebra is just g@ 1). Let j c g@ I) be the graph of the map a. Then the hypothesis that a is a map of Lie algebras is equivalent to the statement that j is a Lie subalgebra of g@ I); and given this, by the proposition there exists an immersed Lie subgroup J c G x 11 with tangent space T.J = j. Look now at the map n: J ..... G given by projection on the first factor. By hypothesis, the di[erential of this map dn.: j -. g is an isomorphism, so that the map J -. G is an isogeny; but since G is simply connected it follows t"at n is an isomorpltism. The projection 11: G ~ J--. H on the second factor is then a Lie group map whose di[erential at the identity is a. 0

Exercise 8.43*. If g--. g' is a homomorphism of Lie algebras with kernel I), show that the kernel H of the corresponding map of simply connected Lie groups G -> G' is a closed subgroup of G with Lie group ~- This does not extend to non-normal subgroups, i.e., to the situation when I) is not the kernel of a homomorphism: give an example of an immersed subgroup of a simply connected Lie group G whose image in G is not closed. Exercise 8.44. Use the ideas of this lecture to prove the assertion that a compact complex connected Lie group G must be abelian:

120

8. Lie Algebras and Lie Groups

(a) Verify that the map Ad: G-+ Aut(T.,G) c End(T,G) is holomorphic, and, therefore (by the maximum principle), constant. (b) Deduce that if '1', is conjugation by g, then d'l', is the identity, so 'l',(exp(X)) = exp(d'I',(X)) = exp(X) for all X E 7~G, which implies that G is abelian. (c) Show that the exponential map from T,G toG is surjective, with the kernel a lattice/\, so G = T,G/1\ is a complex torus.

LECTURE 9

Initial Classification of Lie Algebras

In this lecture we define various subclasses of Lie algebras: nilpotent, solvable, semisimple, etc., and prove basic facts about their representations. The discussion is entirely elementary (largely because the hard theorems are stated without proof for now); there are no prerequisites beyond linear algebra. Apart from giving these basic definitions, the purpose of the lecture is largely to motivate the narrowing of our focus to semisimple algebras that will take place in the sequel. In particular, the first part of §9.3 is logically the most important for what follows. §9.1: §9.2: §9.3: §9.4:

Rough classification of Lie algebras Engel's Theorem and Lie's Theorem Semisimple Lie algebras Simple Lie algebras

§9.1. Rough Classification of Lie Algebras We will give, in this section, a preliminary sort of classification of Lie algebras, renecting the degree to which a given Lie algebra g fails to be abelian. As we have indicated, the goal ultimately is to narrow our focus onto semisimple Lie algebras. Before we begin, two definitions, both completely straightforward: First, we define the ce11ter Z(g) of a Lie algebra g to be the subspace of g of elements X e g such that [X, Y] = 0 for all Y e g. Of course, we say g is abelian if all brackets are zero. Exercise 9.1. Let G be a Lie group, g its Lie algebra. Show that the subgroup of G generated by exponentiating the Lie subalgebra Z(g) is the connected component of the identity in the center Z(G) of G.

9. Initial Classification of Lie Algebras

122

Next, we say that a Lie subalgebra IJ c: 9 of a Lie algebra g is an ideal if it satisfies the condition

(X,Y]el)

foraiiXEI),YEg.

Just as connected subgroups of a Lie group correspond to subalgebras of its Lie algebra, the notion of ideal in a Lie algebra corresponds to the notion of normal subgroup, in the following sense: Exercise 9.2. Let G be a connected Lie group, H c: G a connected subgroup and 9 and (J their Lie algebras. Show that His a normal subgroup of G if and only if r, is an ideal of 9· Observe also that the bracket operation on 9 induces a bracket on the quotient space g/IJ if and only if I) is an ideal in g. This, in turns, motivates the next bit of terminology: we say that a Lie algebra 9 is simple if dim g > I and it contains no nontrivial ideals. By the last exercise, this is equivalent to saying that the adjoint form G of the Lie algebra g has no nontrivial normal Lie subgroups. Now, to attempt to classify Lie algebras, we introduce two descending chains of subalgebras. The first is the lower central .~eries of subalgebras g\9, defined inductively by

E!Jto=[o.oJ and @kg=

[g, @k-lg].

Note that the subalgebras .@kg are in fact ideals in g. The other series is called the derived series {E!Jt9}; it is defined by §l• R =

[o, oJ

and

Exercise 9.3. Use the Jacobi identity to show that @ko is also an ideal in g. More generally, if IJ is an ideal in a Lie algebra 9, show that [lJ, IJ] is also an ideal in g; hence all §lk[J are ideals in 9· Observe that we have g,kg c: !Zlk!l for all k, with equality when k = 1; we often write simply !Zl9 for @ 1g = @ 1g and call this the commutator subalgebra. We now make the Definitions (i) We say that 9 is 11ilpoter•t if !Zlk!l = 0 for some k. (ii) We say that !1 is solvable if g,tg = 0 for some k.

§9.1. Rough Classification or Lie Algebras

123

(iii) We say that 9 is [Jerfect if .@g = 9 (this is not a concept we will use much). (iv) We say that g is semisimple if 9 has no nonzero solvable ideals. The standard example of a nilpotent Lie algebra is the algebra n.R of strictly upper-triangular n x 11 matrices; in this case the kth subalgebra .@•9 in the lower central series will be the subspace 11k+I,•IR of matrices A = (a 1•1) such that a1•1 = 0 whenever j ::o::; i + k, i.e., that are zero below the diagonal and within a distance k of it in each column or row. (In terms of a complete nag { V,} as in §7.2, these are just the endomorphisms that carry V, into v,_._,.) It follows also that any subalgebra of the Lie algebra n.R is likewise nilpotent; we will show later that any nilpotent Lie algebra is isomorphic to such a subalgebra. We will also see that if a Lie algebra 9 is represented on a vector space V, such that each element acts as a nilpotent endomorphism, there is a basis for V such that, identifying gi(V) with 9l.R, g maps to the subalgebra n.IR c gi.R. Similarly, a standard example of a solvable Lie algebra is the space b.R-of upper-triangular n x n matrices; in this Lie algebra the commutator [email protected] is the algebra n.IR and the derived series is, thus, .@•b.IR = n 2•-• •• R Again, it follows that any subalgebra of the algebra b.IR is likewise solvable; and we will prove later that, conversely, any representation of a solvable Lie algebra on a vector space V consists, in terms of a suitable basis, entirely of uppertriangular matrices (i.e., given a solvable Lie subalgebra 9 of 9I(V), there exists a basis for V such that under the corresponding identification of gi(V) with gl.R, the subalgebra 9 is contained in b.R c 9l.R). It is clear from the definitions that the properties of being nilpotent or solvable are inherited by subalgebras or homomorphic images. We will see that the same is true for semisimplicity in the case of homomorphic images, though not for subalgebras. Note that 9 is solvable if and only if 9 has a sequence of Lie subalgebras 0 = g0 ::::> 9 1 =? · · · ::::> lh = 0, such that 9l+t is an ideal in g1 and !l,/9Hr is abelian. Indeed, if this is the case, one sees by induction that .@;!l c g1 for all i. (One may also refine such a sequence to one where each quotient 9,/91+ 1 is one dimensional.) It follows from this description that if I) is an ideal in a Lie algebra 9, then 9 is solvable if a11d only if I) and g/l) are solvable Lie algebras. (The analogous assertion for nilpotent Lie algebras is false: the ideal n. is nilpotent in the Lie algebra b. of upper-triangular matrices, and the quotient is the nilpotent algebra b. of diagonal matrices, but b. is not nilpotent.) If 9 is the Lie algebra of a connected Lie group G, then 9 is solvable if and only if there is a sequence of connected subgroups, each normal in G (or in the next in the sequence), such that the quotients are abelian. In particular, the sum of two solvable ideals in a Lie algebra g is again solvable [note that (a + b)/b ~ aj(a n b)]. It follows that the sum of all solvable ideals in g is a maximal solvable ideal, called the radical of g and denoted Rad(9). The quotient 9/Rad(9) is semisimple. Any Lie algebra g thus fits into an exact sequence

9. Initial Classification or Lie Algebras

124

(9.4)

0-.. Rad(g) -4 g-.. nfRad(o) -.. 0

where the lirst algebra is solvable and the last is semisimple. With this somewhat shaky justification (but see Proposition 9.17), we may say that to study the representation theory of an arbitrary Lie algebra, we have to understand individually the representation theories of solvable and semisimple Lie algebras. Of these, the former is relatively easy, at least as regards irreducible representations. The basic fact about them-that any irreducible representation of a solvable Lie algebra is one dimensional-will be proved later in this lecture. The representation theory of semisimple Lie algebras, on the other hand, is extraordinarily rich, and it is this subject that will occupy us for most of the remainder of the book. Another easy consequence of the definitions is the fact that a Lie algebra is semisimple if a11d or~ly if it has 110 11onzero abelian ideals. Indeed, the last nonzero term in the derived sequence of ideals .@kRad(g) would be an abelian ideal in g (cf. Exercise 9.3). A semisimple Lie algebra can have no center, so tile ad}oi"t representation of a semisimple Lie algebra is .faithful. It is a fact that the sequence (9.4) splits, in the sense that there are subalgebras of 9 that map isomorphically onto g/Rad(g). The existence of such a Levi decompositio11 is part of the general theory we are postponing. To show that an arbitrary Lie algebra has a faithful representation (Ado's tf1eorem), one starts with a faithful representation of the center, and then builds a representation of the radical step by step, inserting a string of ideals between the center and the radical. Then one uses a splitting to get from a faithful representation on the radical to some representation on all of g; the sum of this representation and the adjoint representation is then a faithful representation. See Appendix E for details. One reason for the terminology simplefsemisimple will become clear later in this lecture, when we show that a semisimple Lie algebra is a direct sum of simple ones. Exercise 9.5. Every semi simple Lie algebra is perfect. Show that the Lie group of Euclidean motions of R 3 has a Lie algebra g which is perfect, i.e., ~0 = g, but 9 is not semisimple. More generally, if I] is semisimple, and V is an irreducible representation of 1), the twisted product g = { (v, X)lv E V, X e lJ}

with [(v, X), (w, Y)] = (Xw- Yv, [X, Y])

is a Lie algebra with !?tin = !J, Rad(g)

= V abelian, and gfRad(g) = 1).

Exercise 9.6. (a) Show that the following are equivalent for a Lie algebra g: (i) 9 is nilpotent. (ii) There is a chain of ideals 9 =no::::> g 1 => • · · : : > n. = 0 with g,/n,+t contained in the center of g/gi+J· (iii) There is an integer,. such that ad(X 1 ) for all X 1 ,

o

ad(X 2 )

• •• ,

x •.

o ·•· o

Yin g.

ad(X.)(Y) = [X 1 , [X 2 ,

••• ,

[X., Y] . .. ]] = 0

§9.2. Engel's Theorem and Lie's Theorem

125

(b) Conclude that a connected Lie group G is nilpotent if and only if it can be realized as a succession of central extensions of abelian Lie groups. Exercise 9.7*. If G is connected and nilpotent, show that the exponential map exp: g ~ G is surjective, making g the universal covering space of G. Exercise 9.8. Show that the following are equivalent for a Lie algebra g: (i) g is solvable. (ii) There is a chain of ideals g = g 0 ::::1 g 1 ::::1 • • • ::::1 gN = 0 with g1/g 1+ 1 abelian. (iii) There is a chain ofsubalgebras g = !lo ::::1 !l 1 ::::1 • • • ::::1 n. = 0 such that !11+ 1 is an ideal in g1, and g,fg 1+1 is abelian.

§9.2. Engel's Theorem and Lie's Theorem We wili now prove the statement made above about representations of solvable Lie algebras always being upper triangular. This may give the reader an idea of how the general theory proceeds, before we go back to the concrete examples that are our main concern. The starting point is Theorem 9.9 (Engel's Theorem). Let g c gl( V) be any Lie subalgebra .~uclr tlrat et~ery

X E g is a 11ilpotent endomorphism of V. Tlre11 tlrere exists a nonzero vector v E V sucft tftat X(v) = 0 for· all X E g.

Note this implies that there exists a basis for V in terms of which the matrix representative of each X E g is strictly upper triangular: since g kills v, it will act on the quotient V of V by the span of v, and by induction we can find a basis i12 , ••• , vN for V in terms of which this action is strictly upper triangular. Lifting i11 to any v1 E V and setting v1 = v then gives a basis for Vas desired. PROOF OF THEOREM 9.9. One observation before we start is that if X E oi(V) is any nilpotent element, then the adjoint action ad(X): gl(V) ~ gl( V) is nilpotent. This is straightforward: to say that X is nilpotent is to say that there exists a flag of subspaces 0 c V1 c V2 c · · · c J'k c Vu 1 = V such that X("f) c "f_ 1 ; we can then check that for any endomorphism Y of V the endomorphism ad(Xr(Y) carries "f into "fH-m· We now proceed by induction on the dimension of g. The first step is to show that, under the hypotheses of the problem, g contains an ideal l) of codimensiort one. In fact, let l) c g be any maximal proper subalgebra; we claim that l) has codimension one and is an ideal. To see this, we look at the adjoint representation of g; since !) is a subalgebra the adjoint action ad(!)) of I) on g preserves the subspace ~ c g and so acts on g/1). Moreover, by our observation above, for any X E l) ad(X) acts nilpotently on gi(V), hence on g, hence on g/1). Thus, by induction, there exists a nonzero element Y E g/l) killed by ad(X) for all X E !); equivalently, there exists an element Y E 11 not in I) such

9. Initial Classification or Lie Algebras

126

that ad(X)(Y) E I) for all X E 1). But this is to say that the subspace IJ' of !l spanned by I) and Y is a Lie subalgebra of g, in which ~ sits as an ideal of codimension one; by the maximality of~ we have I)' = g and we are done. We return now to the representation of !l on V. We may apply the induction hypothesis to the subalgebra I) of g found in the preceding paragraph to conclude that there exists a nonzero vector v E V such that X(v) = 0 for all X E I); let W c V be the subspace of all such vectors v E V. Let Y be any element of g not in ~; since I) and Y span g, it will suffice to show that there exists a (nonzero) vector v E W such that Y(v) = 0. Now for any .vector wE W and any X E 1), we have

X(Y(w)) = Y(X(w))

+ [X,

Y](w).

The lirst term on the right is zero because by hypothesis wE W, X E I) and so X(w) = 0; likewise, the second term is zero because [X, Y] = ad(X) (Y) E 1). Thus, X(Y(w)) = 0 for all X E ~;we deduce that Y(w) E W. But this means that the action of Yon V carries the subspace W into itself; since Y acts nilpotently on V, it follows that there exists a vector v E W such that Y(v) = 0. 0 Exercise 9.10*. Show that a Lie algebra 9 is nilpotent if and only if ad( X) is a nilpotent endomorphism of g for every X E g. Engel's theorem, in turn, allows us to prove the basic statement made above that every representation of a solvable Lie group can be put in uppertriangular form. This is implied by Theorem 9.11 (Lie's Theorem). Let g c gi(V) be a complex solvable Lie algebra. Tlren there e.xists a nonzero vector v E V that is an eigenvector of X for all X E g. Exercise 9.12. Show that this implies the existence of a basis for V in terms of which the matrix representative of each X E g is upper triangular. PROOF OF THEOREM 9.11. Once more, the lirst step in the argument is to assert that g contains an ideal ~ of codimension one. This time, since g is solvable we know that ~!.l "# g, so that the quotient n = g/~g is a nonzero abelian Lie algebra; the inverse image in !.l of any codimension one subspace of n will then be a codimension one ideal in g. Still following the Jines of the previous argument, we may by induction assume that there is a vector v0 E V that is an eigenvector for all X E 1). Denote the eigenvalue of X corresponding to v0 by A.(X). We then consider the subspace W c V of all vectors satisfying the same relation, i.e., we set W = {v E V: X(v) = A.(X)·v 'v'X el)}.

Let Y now be any element of g not in ~- As before, it will suffice to show that Y carries some vector v E W into a multiple of itself, and for this it is enough

§9.2. Engel's Theorem and Lie's Theorem

127

to show that Y carries W into itself. We prove this in a general context in the following lemma. Lemma 9.13. Let I) be a11 ideal in a Lie algebra g. Let V be a represelltation g, and ,1.: I) ..... C a li11ear functim1. Set

f!f

W ={vi V: X(v) = ,l.(X)·v VX e IJ}. Then Y(W) c W for allY e g. PROOF.

Let w be any nonzero element of W; to test whether Y(w) e W we Jet

X be any element of I) and write

+ [X, Y] (w) ,l.(X)· Y(w) + ,l.([X, Y])· w

X(Y(w)) = Y(X(w)) =

(9.14)

since [X, Y] e 1). This differs from our previous calculation in that the second term on the right is not immediately seen to be zero; indeed, Y(w) will lie h1 W if and only if ,l.([X, Y]) = 0 for all X e 1). To verify this, we introduce another subspace of V, namely, the span U of the images w, Y(w), Y2 (w), ... of w under successive applications of Y. This subspace is clearly preserved by Y; we claim that any X e !) carries U into itself as well. It is certainly the case that f) carries w into a multiple of itself, and hence into U, and (9.14) says that f) carries Y(w) into a linear combination of Y(w) and w, and so into U. In general, we can see that I) carries Yk(w) into U by induction: for any X e l) we write X(Yl(w))

= Y(X(YH(w))) +[X, Y](l'H(w)).

(9.15)

Since e U by induction the first term on the right is in U, and since [X, Y] e I) the second term is in U as well. In fact, we see something more from (9.14) and (9.15): it follows that, in terms of the basis w, Y(w), Y2 (w), ... for U, the action of any X e I) is upper triangular, with diagonal entries all equal to ,l.(X). In particular, for any X E I) the trace of the restriction of X to U is just the dimension of U times ,l.(X). On the other hand, for any element X e l) the commutator [X, Y] acts on U, and being the commutator of two endomorphisms of U the trace of this action is zero. It follows then that ,l.([X, Y]) = 0, and we are done. 0

X(Yk- 1 (w))

Exercise 9.16. Show that any irreducible representation of a solvable Lie algebra g is one dimensional, and .@g acts trivially. At least for irreducible representations, Lie's theorem implies they will all be known for an arbitrary Lie algebra when they are known for the semisimple case. In fact, we have: Propositipn 9.17. Let g be a complex Lie algebra, o.. = g/Rad(g). Every irreducible representation of g is of Ihe form V = V0 ® L, where V0 is an irreducible

9. Initial Classification of Lie Algebras

128

represelltatioll of g.. [i.e., a representation of g tiJat is tri11ial L is a one-dimensional representation. PROOF.

011

Rad(g)], and

By Lie's theorem there is a A. E (Rad(g))* such that W

= {11 E

V: X(v) =A.( X)· v VX

E

Rad(g)}

is not zero. Apply the preceding lemma, with I) = Rad(g). Since Vis irreducible, we must have W = V. Now extend A. in any way to a linear function on g, and Jet L be the one-dimensional representation of g determined by A.; in other words, Y(z) = A.( Y) · z for all Y E g and z E L. Then V ® L"' is a representation that is trivial on Rad(g), so it comes from a representation of g..., a~ required. 0 Exercise 9.18. Show that if g' is a subalgebra of g that maps isomorphically onto g/Rad(o), then any irreducible representation of g restricts to an irreducible representation of g', and any irreducible representation of g' extends to a representation of g.

§9.3. Semisimple Lie Algebras As is clear from the above, many of the aspects of the representation theory of linite groups that were essential to our approach are no longer valid in the context of general Lie algebras and Lie groups. Most obvious of these is complete reducibility, which we have seen fails for Lie groups; another is the fact that not only can the action of elements of a Lie group or algebra on a vector space be nondiagonalizable, the action of some element of a Lie algebra may be diagonalizable under one representation and not under another. That is the bad news. The good news is that, if we just restrict ourselves to semisimple Lie algebras, everything is once more as well behaved as possible. For one thing, we have complete reducibility again: Theorem 9.19 (Complete Reducibility). Let V be a repr·esentation of the semisimple Lie algebra g and W c V a subspace irtvariant undel' the actio11 of g. Tlren there exists a subspace W' c V complementary to Wand invariant umler g. The proof of this basic result will be deferred to Appendix C. The other question, the diagonalizability of elements of a Lie algebra under a representation, requires a little more discussion. Recall lirst the statement of Jordan decomposition: any endomorphism X of a complex vector space V can be uniquely written in the form

X= X,+

x.

§9.3. Semisimple Lie Algebras

129

where X, is diagonalizable, x. is nilpotent, and the two commute. Moreover, X, and Xn may be expressed as polynomials in X. Now, suppose that 9 is an arbitrary Lie algebra, X E 9 any element, and p: 9-+ 9l.C any representation. We have seen that the image p(X) need not be diagonalizable; we may still ask how p(X) behaves with respect to the Jordan decomposition. The answer is that, in general, absolutely nothing need be true. For example, just taking g = C, we see that under the representation

p,: tl-+(t) every element is diagonalizable, i.e., p(X), P2:

= p(X); under the representation

tl-+(~ ~)

every element is nilpotent [i.e., p(X), = 0]; whereas under the representation

p3:

tl-+(~ ~)

not only are the images p(X) neither diagonalizable nor nilpotent, the diagonalizable and nilpotent parts of p(X) are not even in the image p(g) of the representation. If we assume the Lie algebra 9 is semisimple, however, the situation is radically different. Specifically, we have Theorem 9.20 (Preservation of Jordan Decomposition). Let g be a .~emisimple Lie algebra. For any element X E g, there exist X., and X. E g suclt that for any representation p: g-+ gi(V) we have

p(X), = p(X,)

and

p(X). = p(X.).

In other words, if we think of p as injective and g accordingly as a Lie subalgebra of gl( V), tlte diag01ralizable and nilpotent parts of all}' element X of g are again in o and are independellt of the particular representation p. The proofs we will give of the last two theorems both involve introducing objects that are not essential for the rest of this book, and we therefore relegate them to Appendix C. It is worth remarking, however, that another approach was used classically by Hermann Weyl; this is the famous unitary trick, which we will describe briefly.

A Digression on "The Unitary Trick" Basically, the idea is that the statements above (complete reducibility, preservation of Jordan decomposition) can be proved readily for the representations of a compact Lie group. To prove complete reducibility, for example,

130

9. Initial Classification of Lie Algebras

we can proceed more or less just as in the case of a finite group: if the compact group G acts on a vector space, we see that there is a Hermitian metric on V invariant under the action of G by taking an arbitrary metric on V and averaging its images under the action of G. If G fixes a subspace W c V, it will then fix as well its orthogonal complement W1 with respect to this metric. (Alternatively, we can choose an arbitrary complement W' to W, not necessarily fixed by G, and average over G the projection map to g(W') with kernel W; this average will have image invariant under G.) How does this help us analyze the representation of a semisimple Lie algebra? The key fact here (to be proved in Lecture 26) is that ~f g i.~ any complex semisimple Lie algebra, tl1ere exists a (unique) real Lie algebra g0 witlr complexificaticm !Jo ® C = o, sucl1 tlwt the simply connected form of rl1e Lie algehra g0 is a compact Lie group G. Thus, restricting a given representation of g to !) 0 , we can exponentiate to obtain a representation of G, for which complete reducibility holds; and we can deduce from this the complete reducibility of the original representation. For example, while it is certainly not true that any representation p of the Lie group SL.IR on a vector space V admits an invariant Hermitian metric (in fact, it cannot, unless it is the trivial representation), we can (i) let p' be the corresponding (complex) representation of the Lie algebra si.IR; (ii) by linearity extend the representation 11' of si.IR to a representation p" of si.C; (iii) restrict to a representation p"' of the subalgebra su. c si.C; (iv) exponentiate to obtain a representation p"" of the unitary group su •. We can now argue that If a subspace W c Vis invariant under the action of SL.IR, it must be invariant under si.IR; and sincesr.c = si.IR ® C, it follows that it will be invariant under si.C; so of course it will be invariant under su.; and hence it will be invariant under SU •. Now, since SU. is compact, there will exist a complementary subspace W' preserved by su.; we argue that W' will then be invariant under su.; and since si.C = su. ® C, it follows that it wilt be invariant under si.C. Restricting, we see that it will be invariant under sl.IR, and exponentiating, it will be invariant under SL.R.

Similarly, if one wants to know that the diagonal elements of SL.IR act semisimply in any representation, or equivalently that the diagonal elements of si.IR act semisimply, one goes through the same reasoning, coming down to the fact that the group of diagonal elements in su. is abelian and compact.

§9.4. Simple Lie Algebras

131

In general, most of the theorems about the finite-dimensional representation of semisimple Lie algebras admit proofs along two different lines: either algebraically, using just the structure of the Lie algebra; or by the unitary trick, that is, by associating to a representation of such a Lie algebra a representation of a compact Lie group and working with that. Which is preferable depends very much on taste and context; in this book we will for the most part go with the algebraic proofs, though in the case of the Weyl character formula in Part IV the proof via compact groups is so much more appealing it has to be mentioned. The following exercises include a few applications of these two theorems. Exercise 9.21"'. Show that a Lie algebra o is semisimple if and only if every finite-dimensional representation is semisimple, i.e., every invariant subspace has a complement. Exercise 9.22. Use Weyl's unitary trick to show that, for 11 > 2, all representations of SO.C are semisimple, so that, in particular, the Lie algebras so.C are semisimple. Do the same for Sp 2.C and sp 2 .C, n z I. Where does the argument break down for S02 C7 Exercise 9.23. Show that a real Lie algebra o is solvable if and only if the complex Lie algebra 9 ®nC is solvable. Similarly for nilpotent and semisimple. Exercise 9.24"'. If l) is an ideal in a Lie algebra o, show that g is semisimple if and only if~ and g/~ are semisimple. Deduce that every semi simple Lie algebra is a direct sum of simple Lie algebras. Exercise 9.25"'. A Lie algebra is called reductive if its radical is equal to its center. A Lie group is reductive if its Lie algebra is reductive. For example, G L.C is reductive. Show that the following are true for a reductive Lie algebra o: (i) !lilg is semisimple; (ii) the adjoint representation of g is semisimple; (iii) 11 is a product of a semisimple and an abelian Lie algebra; (iv) o has a finitedimensional faithful semisimple representation. In fact, each of these conditions is equivalent to g being reductive.

§9.4. Simple Lie Algebras There is one more basic fact about Lie algebras to be staled here; though its proof will have to be considerably deferred, it informs our whole approach to the subject. This is the complete classification of simple Lie algebras: Theorem 9.26. With five exceptions, every simple complex Lie algebra is isomorphic to eitlter sl.C, so.C, or sp 2 .C for some n.

132

9. Initial Classilication or Lie Algebras

The live exceptions can all be explicitly described, though none is particularly simple except in name; they are denoted g 2 , f4 , e6 , e 7 , and e8 . We will give a construction of each later in the book (§22.3). The algebras si.C (for 11 > l), so.C (for 11 > 2), and sp 2 .C are commonly called the classical Lie algebras (and the corresponding groups the classical Lie groups); the other live algebras are called, naturally enough, the exceptional Lie algebras. The nature of the classification theorem for simple Lie algebras creates a dilemma as to how we approach the subject: many of the theorems about simple Lie algebras can be proved either in the abstract, or by verifying them in turn for each of the particular algebras listed in the classification theorem. Another alternative is to declare that we are concerned with understanding only the representations of the classical algebras si.C, so.C, and sp 2 .C, and verify any relevant theorems just in these cases. or these three approaches, the last is in many ways the least satisfactory; it is, however, the one that we shall for the most part take. Specifically, what we will do, starting in Lecture 11, is the following: (i) Analyze in Lectures ll-13 a couple of examples, namely, sl 2 C and sl 3 C, on what may appear to be an ad hoc basis. (ii) On the basis of these examples, propose in Lecture 14 a general paradigm for the study of representations of a simple (or semisimple) Lie algebra (iii) Proceed in Lectures 15-20 to carry out this analysis for the classical algebras si.C, so.C, and sp 2 .C. (iv) Give in Part IV and the appendices proofs for general simple Lie algebras of the facts discovered in the preceding sections for the classical ones (as well as one further important result, the Weyl character formula). We can at least partially justify this seemingly inefficient approach by saying that even if one makes a beeline for the general theorems about the structure and representation theory of a simple Lie algebra, to apply these results in practice we would still need to carry out the sort of explicit analysis of the individual algebras done in Lectures 11-20. This is, however, a fairly bald rationalization: the fact is, the reason we are doing it this way is that this is the only way we have ever been able to understand any of the general results.

LECTURE 10

Lie Algebras in Dimensions One, Two, and Three

Just to get a sense or what a Lie algebra is and what groups might be associated to it, we will classify here all Lie algebras or dimension three or less. We will work primarily with complex Lie algebras and Lie groups, but will mention the real case as well. Needless to say, this lecture is logically supernuous; but it is easy, fun, and serves a didactic purpose, so why not read it anyway. The analyses of both the Lie algebras and the Lie groups are completely elementary, with one exception: the classification or the complex Lie groups associated to abelian Lie algebras involves the theory or complex tori, and should probably be skipped by anyone not familiar with this subject.

§10.1: §10.2: §10.3: §10.4:

Dimensions one and two Dimension three, rank one Dimension three, rank two Dimension three, rank three

§10.1. Dimensions One and Two To begin with, any one-dimensional Lie algebra

g is clearly abelian, that is,

C with all brackets zero. The simply connected Lie group with this Lie algebra is just the group C under addition; and other connected Lie groups that have g as their Lie algebra must all be quotients of C by discrete subgroups A c C. If A has rank one, then the quotient is just under multiplication. Ir A has rank two, however, G may be any one of a continuously varying family of complex tori of dimension one (or Riemann surfaces of genus one, or elliptic curves over C). The set of isomorphism classes of such tori is parametrized by the complex plane with coordinate j, where the function j on the set of lattices A c C is as described in, e.g., [Ahl]. Over the real numbers, the situation is completely straightforward: the only real Lie algebra of dimension one is again IR with trivial bracket; the simply

c•

10. Lie Algebras in Dimensions One, Two, and Three

134

connected Lie group associated to it is IR under addition; and the only other connected real Lie group with this Lie algebra is IR/Z ~ S 1•

Dimension Two Here we have to consider two cases, depending on whether g is abelian or not.

Case I: g abelian. This is very much like the previous case; the simply connected two-dimensional abelian complex Lie group is just C 2 under addition, and the remaining connected Lie groups with Lie algebra g are just quotients of C 2 by discrete subgroups. Such a subgroup 1\ c C 2 can have rank I, 2, 3, or 4, and we analyze these possibilities in turn (the reader who has seen enough complex tori in the preceding example may wish to skip directly to Case 2 at this point). lfthe rank of 1\ is I, we can complete the generator of 1\ to a basis for C 2 , so that 1\ = le 1 c Ce 1 E9 Ce 2 and G ~ C* x C. If the rank of 1\ is 2, there are two possibilities: either 1\ lies in a one-dimensional complex subspace of C 2 or it does not. If it does not, a pair of generators for 1\ will also be a basis for C 2 over c, so that 1\ = Ze, $l.ez, C 2 = Ce, E9 Cez, and G ~ c• X c•. If on the other hand 1\ does lie in a complex line in C 2, so that we have 1\ = le 1 $lte 1 for some t E C1R, then G = E x C will be the product of the torus C/(Z ~ Zr) and C; the remarks above apply to the classification of these (see Exercise IO.l ). The cases where 1\ has rank 3 or 4 are a little less clear. To begin with, if the rank of 1\ is 3, the main question to ask is whether any rank 2 sublattice I\' of 1\ lies in a complex line. If it does, then we can assume this sublattice is saturated (i.e., a pair of generators for I\' can be completed to a set of generators for /\) and write 1\ = le 1 E9 Zte 1 E9 Ze 2 , so that we will have G = E x C*, where E is a torus as above. Exercise 10.1*. For two one-dimensional complex tori E and E', show that the complex Lie groups G = E x C and G' = E' x C are isomorphic if and only if E ~ E'. Similarly for E x C* and E' x C*. If, on the other hand, no such sublattice of 1\ exists, the .situation is much more mysterious. One way we can try to represent G is by choosing a generator for 1\ and considering the projection of C 2 onto the quotient of C 2 by the line spanned by this generator; thus, if we write 1\ = le 1 $ le 2 $Z(~e, + Pe 2 ) then (assuming pis not real) we have maps

cz

l

C/(Z $ZP}

expressing Gas a bundle over a torus E = C/(Z $ Z{l), with fibers isomorphic

135

§IO.l. Dimensions One and Two

to C*. This expression of G does not, however, help us very much to describe the family of all such groups. For one thing, the elliptic curve E is surely not determined by the data of G: if we just exchange e 1 and e 2 , for example, we replace E by C/(Z E9 Z<X), which, of course, need not even be isogenous to E. Indeed, this yields an example of different algebraic groups isomorphic as complex Lie groups: expressing G as a C* bundle in this way gives it the structure of an algebraic variety, which, in turn, determines the elliptic curve E (for example, the field of rational functions on G will be the field of rational functions onE with one variable adjoined). Thus, different expressions of the complex Lie group Gas a C* bundle yield nonisomorphic algebraic groups. Finally, the case where 1\ has rank 4 remains completely mysterious. Among such two-dimensional complex tori are the a1Jelia11 varieties; these are just the tori that may be embedded in complex projective space (and hence may be realized as algebraic varieties). For polarized abelian varieties (that is, abelian varieties with equivalence class of embedding in projective space) there. exists a reasonable moduli theory; but the set of abelian varieties forms only a countable dense union in the set of all complex tori (indeed, the general complex torus pos..<1esses no nonconstant meromorphic functions whatsoever). No satisfactory theory of moduli is known for these objects. Needless to say, the foregoing discussion of the various abelian complex Lie groups in dimension two is completely orthogonal to our present purposes. We hope to make the point, however, that even in this seemingly trivial case there lurk some fairly mysterious phenomena. Of course, none of this occurs in the real case, where the two-dimensional abelian simply connected real Lie group is just R x Rand any other connected two-dimensional abelian real Lie group is the quotient of this by a sublattice 1\ c IR x IR of rank I or 2, which is to say either IR x S 1 or S1 x S1•

Case 2: g 11ot abelia11. Viewing the Lie bracket as a linear map [ , ]: Ng-+ g, we see that if it is not zero, it must have one-dimensional image. We can thus choose a basis {X, Y} for gas vector space with X spanning the image of [ , ]; after multiplying Y by an appropriate scalar we will have [X, Y] =X, which of course determines g completely. There is thus a unique nonabelian two-dimensional Lie algebra g over either IR or C. What are the complex Lie groups with Lie algebra g? To lind one, we start with the adjoint representation of g, which is faithful: we have

ad(X): XHO,

ad(Y): XH -X,

YHX, or in matrix notation, in terms of the basis {X, Y} for g,

ad(X)

=

(0 I)

0 0 ,

Thesegeneratethealgebrag = at the adjoint form

ad(Y)

(~ ~) c

=

(-1 0)

0 0 ·

gi 2 C;wemayexponentiatetoarrive

I0. Lie Algebras in Dimensions One, Two, and Three

136

Topologically this group is homeomorphic to C x C*. To take its universal cover, we write a general member of G0 as

(e'0 s)1 . The product of two such matrices is given by

(~ ·~)-(~' ~') = (e'~,.

s\e's').

so we may realize the universal cover G of G0 as the group of pairs (t, s) E C x C with group law (t, s) · (t', s') = (t

+ t', .~ + e' s').

The center of G is just the subgroup Z(G) = { (2nin, 0)} ~ l,

so that the connected groups with Lie algebra g form a partially ordered tower G

!

! G.= Gfnl ={(a, b) E C* x C; (a, b)·(a', b') = (aa', b

+ a"b')}.

!

Exercise 10.2*. Show that for n of m the two groups G" and G'" are not isomorphic. Finally, in the real case things are simpler: when we exponentiate the adjoint representation as above, the Lie group we arrive at is already simply connected, and so is the unique connected real Lie group with this Lie algebra.

§ l 0.2. Dimension Three, Rank 1 As in the case of dimension two, we look at the Lie bracket as a linear map from No tog and begin our classification by considering the rank of this map (that is, the dimension of 9'Jo), which may be either 0, I, 2, or 3. For the case

§10.2. Dimension Three, Rank l

137

of rank 0, we refer back to the discussion of abelian Lie groups above. We begin with the case of rank l. Here the kernel of the map [ , ]: Ng- gis two dimensional, which means that for some X e g it consists of all vectors of the form X A Y with Y ranging over all of g (X here will just be the vector corresponding to the hyperplane ker([ , ]) c Ng under the natural (up to scalars) duality between a threedimensional vector space and its exterior square). Completing X to a basis {X, Y, Z} of g, we can write 9 in the form [X, Y] = [X, Z] = 0,

[Y, Z] = o:X

+ py + yZ

for some o:, p, y E C. If either p or y is nonzero, we may now rechoose our basis, replacing Y by a multiple of the linear combination o:X + py + yZ and either leaving Z alone (if p -F 0) or replacing Z by Y (if y 'F 0). We will then have.

[X, Y] = [X, Z] = 0,

[Y, Z] = y from which we see that g is just the product of the one-dimensional abelian Ue algebra CX with the non-abelian two-dimensional Lie algebra C Y ffi CZ described in the preceding discussion. We may thus ignore this case and assume that in fact we have p = y = 0; replacing X by o:X we then have the Lie algebra

[X, Y] =[X, Z]

= 0,

[Y,Z]=X. How do we find the Lie groups with this Lie algebra? As before, we need to start with a faithful representation of g, but here the adjoint representation is useless, since X is in its kernel. We can, however, arrive at a representation of g by considering the equations defining g: we want to find a pair of · endomorphisms Y and Z on some vector space that do not commute, but that do commute with their commutator X= [Y, Z]; thus,

Y(YZ- ZY)- (YZ- ZY)Y = Y 2 Z- 2YZY + ZY 2 = 0 . and similarly for [Z, [Y, Z]]. One simple way to lind such a pair of endof,·. morphisms is make all three terms Y 2 Z, YZY, and Z 2 Yin the above equation :.·.zero, e.g., by making Y and Z both have square zero, and to have YZ = 0 ;, while ZY -F 0. For example, on a three-dimensional vector space with basis ;, '" e2 , and e3 we could take Y to be the map carrying e 3 to e 2 and killing Fe1 and e2 , and Z the map carrying e2 to e 1 and killing e 1 and e3 ; we then have ~t YZ = 0 while Z Y sends e3 to e 1 • We see then that g is just the Lie algebra n 3 ~~~~strictly upper-triangular 3 x 3 matrices. When we exponentiate we arrive ~·, at the group

10. Lie Algebras in Dimensions One, Two, and Three

138

which is simply connected. Now the center of G is the subgroup

Z(G)=

m! ~).bEC} ~c.

so the discrete subgroups of Z(G) are just lattices A of rank I or 2; thus any connected group with Lie algebra g is either G, Gfl, or Gf(Z x Z)-that is, an extension ofC x C by either C, C*, or a torus E. Exercise 10.3. Show that G/A is determined up to isomorphism by the onedimensional Z(G)/A. A similar analysis holds in the real case: just as before, 11 3 is the unique real Lie algebra of dimension three with commutator subalgebra of dimension one; its simply connected form is the group G of unipotent 3 x 3 matrices and (the center of this group being R) the only other group with this Lie algebra is the quotient H = Gfl. Incidentally, the group H represents an interesting example of a group that cannot be realized as a matrix group, i.e., that admits no faithful finite· dimensional representations. One way to see this is to argue that in any irreducible finite-dimensional representation V the center S1 of H, being compact and abelian, must be diagonalizable; and so under the corresponding representation of the Lie algebra g the element X must be carried to a diagonalizable endomorphism of I": But now if v E Vis any eigenvector for X with eigenvalue A., we also have, arguing as in §9.2, X(Y(v))

=

Y(X(v))

=

Y(A.v)

= A.Y(v)

and similarly X(Z(v)) = A.Z(v), i.e., both Y(v) and Z(v) are also eigenvectors for X with eigenvalue A.. Since Y and Z generate g and the representation V is irreducible, it follows that X must act as a scalar multiple A.· I of the identity; but since X = [ Y, Z] is a commutator and so has trace 0, it follows that A. = 0. Exercise 10.4*. Show that if G is a simply connected Lie group, and its Lie algebra is solvable, then G cannot contain any nontrivial compact subgroup (in particular, it contains no elements of finite order). The group H does, however, have an important infinite-dimensional representation. This arises from the representation of the Lie algebra g on the space v orrcoo functions on the real lineR with coordinate X, in which Y, Z, and X are the operators

§10.3. Dimension Three, Rank 2

139 Y:fHnix·f,

df Z:fH dx and X = [ Y, Z] is -ni times the identity. Exponentiating, we see that e'r acts on a function f by multiplying it by the function (cos tx + i ·sin tx); e'z sends f to the function F, where F,(x) = f(t + x), and e'x sends f to the scalar -f. multiple

e_.,,

§10.3. Dimension Three, Rank 2 In this case, write the commutator subalgebra .@9 c g as the span of two elements Y and Z. The commutator of Y and Z can then be written [Y, Z] = 1XY + pz.

But now the endomorphism ad( Y) of 9 carries g into .@9, kills Y, and sends Z to IXY + pz, and so has trace {1; on the other hand, since ad( Y) is a commutator in End(9), it must have trace 0. Thus, p, and similarly IX, must be zero; i.e., the snbalgebra .@9 must be abelian. It follows from this that for any element X E g not in .@9, the map ad(X): .@9 -> .@9 must be an isomorphism. We may now distinguish two possibilities: either ad(X) is diagonalizible or it is not. (Note that for the first time we see a case where the classification of the real Lie algebra will be more complicated than that of the complex: in the real case we will have to deal with the third possibility that ad(X) is diagonalizible over C but not over IR, i.e., that it has two complex conjugate eigenvalues. Though we have not seen it much in these low-dimensional examples, in fact it is generally the case that the real picture is substantially more complicated than the complex one, for essentially just this reason.)

Possibility A: ad(X) i.~ diagonalizable. In this case it is natural to use as a basis for .@g a pair of eigenvectors Y, Z for ad(X); and by multiplying X by a suitable scalar we can assume that one of the eigenvalues (both of which are nonzero) is l. We thus have the equations for 9 [X, Y] = Y,

[X, Z]

= IXZ,

[Y, Z] = 0

(10.5)

for Some IX E C*. Exercise 10.6. Show that two Lie algebras g., g•. corresponding to two different scalars in the structure equations (10.5) are isomorphic if and only if IX= IX' or

140

10. Lie Algebras in Dimensions One, Two, and Three

I'J. = 1/I'J.'. Observe that we have for the first time a continuously varying family of nonisomorphic complex Lie algebras.

To find the groups with these Lie algebras we go to the adjoint representation, which here is faithful. Explicitly, ad(Y) carries X to - Y and kills Y and Z; ad(Z) carries X to -I'J.Z and also kills Y and Z; and ad(X) carries Y to itself, Z to I'J.Z, and kills X. A general member aX - bY- cZ of the Lie algebra is thus represented (with respect to the basis { Y, Z, X} for g) by the matrix a (0 0

0 I'J.a

0

b)

I'J.C •

0

Exponentiating, we lind that a group with Lie algebra g is

G~

m~ J.u,,EC}

cGL,C

Here we run across a very interesting circumstance. If the complex number a. is not rational, then the exponential map from g to G is one-to-one, and hence a homeomorphism; thus, in particular, G is simply connected. If, on the other hand, I'J. is rational, G will have nontrivial fundamental group. To see this, observe that we always have an exact sequence of groups 1-..B-+G-+A-+1,

where

and

Now when I'J. ¢ 0, the group A ;:;;; C is simply connected; but when I'J. E ( ) whatever its denominator-we have A ;:;;; C* and correspondingly 1t 1 (G) = l. Exercise 10.7. Show that G has no center, and hence when I'J. # 0, it is the unique connected group with Lie algebra g. For I'J. E Q, describe the universal. covering of G and classify all groups with Lie algebra g. Observe that in this case, even though we have a continuously varying family of Lie algehras !'! •• we have no corresponding continuously varying

141

§I 0.4. Dintension Three, Rank 3

family of the adjoint (linear) Lie groups; the simply-connected forms do form a family, however. Possibility B: ad(X) is not diagona/izable. In this case the natural thing to do is to choose a basis {Y, Z} of .@g with respect to which ad(X) is in Jordan normal form; replacing X by a multiple, we may assume both its eigenvalues are 1 so that we will have the Lie algebra

[X, Y] = Y,

[X,Z] = Y

+ Z,

[Y,Z] = 0.

(10.8)

With respect to the basis { Y, Z, X} for g, then, the adjoint action of the general element aX - bY- cZ of the Lie algebra is represented by the matrix

(0a aa b +c) c 0

0

0

and exponentiating we lind that the corresponding group is

Exercise 10.9. Show that this group has no center, and hence is the unique connected complex Lie group with its Lie algebra. Note that the real Lie groups obtained by exponentiating the adjoint action of the Lie algebras given by (10.5) and (10.8) are all homeomorphic tQ IR 3 and have no center, and so are the only connected real Lie groups with these Lie algebras. · Exercise 10.10. Complete the analysis of real Lie groups in Case 2 by considering the third possibility mentioned above: that ad(X) acts on .@g with distinct complex conjugate eigenvalues. Observe that in this way we arrive at our first example of two nonisomorphic real Lie algebras whose tensor products with Care isomorphic.

§10.4. Dimension Three, Rank 3 Our analysis of this final case begins, as in the preceding one, by looking for ; eigenvectors of the adjoint action of a suitable element X E g. Specifically, we · claim that we can lind an element HE g such that ad( H): g--+ g has an . eigenvector with nonzero eigenvalue. To see this, observe first that for any nonzero X e !J, the rank of ad( X) must be 2; in particular, we must have Ker(ad(X)) =ex. Now start with any X E g. Either ad(X) has an eigenvector with nonzero eigenvalue or it is nilpotent: if it is nilpotent, then there exists a

10. Lie Algebras in Dimensions One, Two, and Three

142

vector Y e !J, not in the kernel of ad(X) but in the kernel of ad(X) 2 -that is, such that ad(X)(Y) = a.X for some nonzero a. e C. But then of course ad(Y)(X) = -a.X, so that X is an eigenvector for ad(Y) with nonzero eigenvalue. So: choose Hand X e g so that X is an eigenvector with nonzero eigenvalue for ad(H), and write [H, X] = a.X. Since If E .@9, ad(H) is a commutator in End(g), and so has trace 0; it follows that ad(JI) must have a third eigenvector Y with eigenvalue -a.. To describe the structure of g completely it now remains to find the commutator of X and Y; but this follows from the Jacobi identity. We have

[H, [X, Y]] = -[X, [Y, If]]- [Y, [H, X]] =

-[X, a.Y]- [Y, aX]

=0, from which we deduce that [X, Y] must be a multiple of H; since it must be a nonzero multiple, we can multiply X or Y by a scalar to make it l. Similarly multiplying If by a scalar we can assume a is 1 or any other nonzero scalar. Thus, there is only one possible complex Lie algebra g of this type. One could look for endomorphisms H, X, and Y whose commutators satisfy these relations, as we did before. Or we may simply realize that the three-dimensional Lie algebra si2C has not yet been seen, so it must be this last possibility. In fact, a natural basis for si 2 C is

X=(~ ~).

Y=

(~ ~)

whose Lie algebra is given by [H,X] = 2X,

[H, Y] = -2Y,

[X, Y] =H.

(JO.ll)

What groups other than SL 2 C have Lie algebra si 2 C? To begin with, the group SL 2 C is simply connected: for example, the map SL 2 C ..... C 2 - { (0, 0)} sending a matrix to its first row expresses the topological space SL 2 C as a bundle with fiber Cover C 2 - { (0, 0) }. Also, it is not hard to see that the center of SL 2 C is just the subgroup { ± J} of scalar matrices, so that the only other connected group with Lie algebra si 2 C is the quotient PSL 2 C = SI 2 C/{ ±I}. As in the preceding case, the analysis of real three-dimensional Lie algebras g with .@g = g involves one additional possibility. At the outset of the argu· ment above, we started with an arbitrary 11 E g and said that if ad(H) had no eigenvector other than H itself, then it would have to be nilpotent. Of course, in the real case it is also possible that ad( H) has two distinct complex conjugate· eigenvalues). and I. Since ad(H) is a commutator in End(g) and so has trace 0, ). will have to be purely imaginary in this case; and so multiplying If by a real scalar we can assume that its eigenvalues are i and - i. It follows then that we can find X, Y E g with

§10.3. Dimension Three, Rank 2

[H,X] = Y

143

and

[H, Y] =-X.

Using the Jacobi identity as before we may conclude that the commutator of X and Y is a multiple of H; after multiplying each of X and Y by a real scalar we can assume that it is either H or -H. Finally, if [X, Y] = - H, then we observe that we are in the case we considered before: ad(Y) will have X + H as an eigenvector with nonzero eigenvalue, and following our previous analysis we may conclude that g ~ s1 2 1R. Thus, we are left with the sole additional possibility that g has structure equations [H, X]= Y,

[H, Y] =-X,

[X, Y] =H.

{10.12)

This, finally, we may recognize as the Lie algebra sn 2 of the real Lie group SU(2) (as you may recall, the isomorphism su 2 ® C ~ si 2 C was used in the last lecture). What are the real Lie groups with Lie algebras s1 2 R and su 2 ? To start, the center of the group SL 2 R is again just the scalar matrices {±I}, so the only group dominated by SL 2 R is the quotient PSL 2 1R. On the other hand, unlike the complex case SL 2 1R is not simply connected: now the map associating to a 2 x 2 matrix its first row expresses SL 2 1R as a bundle with fiber R over IR 2 - {(0, 0)}, so that n 1(SL 2 R) = l. More precisely PSL 2 R maps to the real projective line 111 1 R, which is homeomorphic to the circle, with fiber homeomorphic to R 2 , so n 1 (PSL 2 R) = l. We thus have a tower of covering spaces of PSL 2 R, consisting of the simply-connected group S with center l and its quotients = S/lll (not all of these are covers of SLziR, despite the diagram below). A rwte: In §10.2 we encountered a real Lie group with no faithful finitedimensional representations; only its universal cover could be represented as a matrix group. Here we find in some sense the opposite phenomenon: the groups Sand S" have no faithful finite-dimensional representations, all finitedimensional representations factoring through SL 2 1R or PSL 2 1R. This fact will be proved as a consequence of our discussion of the representations of the Lie algebra si 2 C in the next lecture. What about groups with Lie algebra su 2 ? To begin with, there is SU(2), which (again via the map sending a matrix to its first row vector) is homeomorphic to S 3 and thus simply connected. The center of this group is again {±I}, so that the quotient PSU(2) is the only other group with Lie algebra su 2 • (Alternatively, we may realize SU(2) as the group of unit quaternions, cf. Exercise 7.15.) · Finally, we remark that there are other representations of the real and complex Lie groups discussed above. As we will see, the Lie algebra so 3 C is isomorphic to sl 2 C, which induces an isomorphism between the corresponding adjoint forms PSL 2 C and S03 C(and between the simply-connected forms SL 2 C and the spin group Spin 3 C). This in turn suggests two more real forms of this group: S03 R and so+(2, 1). In fact, it is not hard to see that S03 1R ~ PSU(2), while so+(2, I)~ PSLziR. Lastly the isomorphism Slit, I® c ~

s"

10. Lie Algebras in Dimensions One, Two, and Three

144

su 1 ® C ~ sl 1 C implies that the real Lie algebra su 1, 1 is isomorphic to either su 1 or si 1 1R; in fact, the latter is the case and this induces an isomorphism of groups SUI. I ~ SL 2 R. We summarize the isomorphisms mentioned in the diagram below:

s

I I I

s.

Spin 3 1R

II

SU(2) = {unit quaternions}

SU(I, 1) = SL 1 1R

~

I

PSL 2 C ~ PSU(2) ={unit quaternions}/± l

I

I

(10.13)

Note also the coincidences:

(10.14) which follow from the fact that Sp refers to preserving a skew-symmetric bilinear form, and for 2 x 2 matrices the determinant is such a form. Exercise 10.15. Identify the Lie algebras so 3 , su 1 , su 1 ,., so 1 , 1 , and verify the assertions made about the corresponding Lie groups in the diagram. Exercise 10.16. For each of the Lie algebras encountered in this lecture, compute the lower central series and the derived series, and say whether the algebra is nilpotent, solvable, simple, or semisimple. Exercise 10.17. The following are Lie groups of dimension two or three, so must appear on our list. Find them: (i) the group of affine transformations of the line (x f-+ ax + b, under composition); (ii) the group of upper-triangular 2 x 2 matrices; (iii) the group of orientation preserving Euclidean transformations of the plane (compositions of translations and rotations). Exercise 10.18. Locate IR 3 with the usual cross-product on our list of Lie algebras. More generally, consider the family of Lie algebras parametrized by real quadruples (a, b, c, d), each with basis X, Y, Z with bracket given by

[X, Y] =aZ +dY,

[Y, Z]

= bX,

[Z, X]= cY- dZ.

§10.4. Dimension Three, Rank 3

145

Classify this Lie algebra as (a, b, c, d) varies in R'\ showing in particular that every three-dimensional Lie algebra can be written in this way. Exercise 10.19. Realize the isomorphism of SU(l, 1) with SL 2 R by identifying them with the groups of complex automorphisms of the unit disk and the upper half-plane, respectively. Exercise 10.20. Classify all Lie algebras of dimension four and rank l; in particular, show that they are all direct sums of Lie algebras described above. Exercise 10.21. Show more generally that there exists a Lie algebra of dimension m and rank 1 that is not a direct sum of smaller Lie algebras if and only if m is odd; in case m is odd show that this Lie algebra is unique and realize it as a Lie subatgebra ofsi.C.

LECTUPE 11

Representations of sl 2 C

This is the first of four lectures-§11--14-that comprise in some sense the heart of the book. In particular, the naive analysis of §I 1.1, together with the analogous pari~ of §12 and §I 3, form the paradigm for the study of finite-dimensional representations of all semisimple Lie algebras and groups. §I 1.2 is less central; in it we show how the analysis carried out in §I 1.1 can be used to explicitly describe the tensor produch of irreducible representations. §11.3 is least important; it indicates how we can interpret geometrically some of the resulls of the preceding section. The discussions in §II .I and §11.2 are completely elementary (we do use the notion of symmetric powers of a vector space, but in a non-threatening way).§ 11.3 involves a fair amount of classical projective geometry, and can be skimmed or skipped by those not already familiar with the relevant basic notions from algebraic geometry. §11.1: The irreducihle representations §11.2: A little plethysm §I 1.3: A little geometric plethysm

§1 LL The Irreducible Representations We start our discussion of representations ofsemisimple Lie algebras with the simplest case, that of si 2 C. As we will see, while this case does not exhibit any of the complexity of the more general case, the basic idea that informs the whole approach is clearly illustrated here. This approach is one already mentioned above, in connection with the representations of the symmetric group on three letters. The idea in that case was that given a representation of our group on a vector space V we first restrict the representation to the abelian subgroup generated by a 3-cycle r. We obtain a decomposition

§II. I. The Irreducible Representations

147

V=EIJV«

of V into eigenspaces for the action oft; the commutation relations satisfied by the remaining elements <1 of the group with respect to t implied that such <1 simply pennuted these subspacec; Jl«, so that the representation was in effect determined by the collection of eigenvalues oft. Of course, circumstances in the case of Lie algebra representations are quite different: to name two, it is no longer the case that the action of an abelian object on any vector space admits such a decomposition; and even if Ruch a decomposition exists we certainly cannot expect that the remaining elements of our Lie algebra will simply permute its summands. Nevertheless, the idea remains essentially a good one, as we shall now see. To begin with, we choose the basis for the Lie algebra s1 2 C from the last lecture:

X=

(0 1) 0

0 '

Y=

(~ ~)

satisfying

[H,X]

= 2X,

[H, Y] = -2Y,

[X, Y] =II.

(11.1)

These seem like a perfectly natural basis to choose, but in fact the choice is dictated by more than aesthetics; there is, as we will see, a nearly canonical way of choosing a basis of a semisimple Lie algebra (up to conjugation), which will yield this basis in the present circumstance and which will share many of the properties we describe below. In any event, let Jl be an irreducible finite-dimensional representation of ~;I 1 C. We star't by trotting out one of the facts that we quoted in Lecture 9, the preservation of Jordan decomposition; in the present circumstances it implies that Tire action of II on V is diagonalizable.

(11.2)

We thus have, as indicated, a decomposition ( 11.3)

where the a nm over a collection of complex numbers, such that for any vector ve V. we have H(v)

= a·v.

The next question is obviously how X and Y act on the various spaces

v.. We claim that X and Y must each carry the subspaces V« into other sub-

spares v•.. In fact, we can be more specific: if we want to know where the image of a given vector v e V. under the action or X sits in relation to the decomposition (11.3), we have to know how H acts on X(v); this is given by the

11. Representations of si 2 C

148

Fundamelllal Calculation (first time): II(X(v)) = X(H(v))

+ [H, X](ll)

= X(a· v) + 2X(v) =(a + 2) · X(v); i.e., if vis an eigenvector for H witl1 eigenvalue a, then X(v) is also an eigenvector for H, witll eigmvalue a + 2. In other words, we have X:~--- ~+2·

The action of Yon each V« is similarly calculated; we have Y(~) c Va_ 2 • Observe that as an immediate consequence of this and the irreducibility of V, all the complex numbers a tllat appear in tl1e decompositioll (11.3) must be co11grtw•t to one anotl1er mod 2: for any a 0 that actually occurs, the subspace

E9

v«n+2n

•e l

would be invariant under s{ 2 C and hence equal to all of V. Moreover, by the same token, the V« that appear must form an unbroken string of numbers of the form {J, p + 2, ... , (J + 2k. We denote by 11 the last element in this sequence; at this point we just know n is a complex number, but we will soon see that it must be an integer. To proceed with our analysis, we have the following picture of the action of si 2 C on the vector space V: X

X

X

~

~

~

y

V n-4

u

II

y

v.. -2

u

f(

y

v.

u

f(

Choose any nonzero vector v e v.; since v.+l = (0), we must have X(v) = 0. We ask now what happens when we apply the map Y to the vector v. To begin · with, we have Claim 11.4. The vectors {v, Y(v), Y2 (v), ... } span V. PRooF. From the irreducibility of V it is enough to show that the subspace W c V spanned by these vectors is carried into itself under the action of s{ 2 C.

Clearly, Y preserves W, since it simply carries the vector Y"'(v) into Y"'+l(v). Likewise, since the vector Y"'(v) is in V..- 2 .. , we have li(Y"'(v)) = (n- 2m)· Y"'(v), soH preserves the subspace W. Thus, it suffices to check that X(W) c W, i.e., that for each m, X carries Y"'(v) into a linear combination of the Y'(v). We check this in turn for m = 0, I, 2, etc. To begin with, we have X(v) = 0 e W. To see what X does to Y(v), we use

149

§11.1. The Irreducible Representations

the commutation relations for si 1 C: we have X(Y(v)) =[X, Y](v) = H(v) =

+

+

Y(X(v))

Y(O)

n·v.

Next, we see that X(Y 2 (v)) =[X, Y](Y(v))

+

Y(X(Y(v)))

+ Y(n-t•) = (n- 2)· Y(v) + n· Y(v). = I:l(Y(v})

The pattern now is clear: X carries each vector in the sequence v, Y(v), Y2 (v), ... into a multiple of the previous vector. Explicitly, we have X(Y"'(v)) = (n + (n- 2) + (n- 4) + ···+(11-2m+ 2))· Y"'- 1 (v), or

X(Y"'(v)) = m(n- m +I)· ym- 1 (v),

(11.5)

as can readily be verified by induction.

0

There are a number of corollaries of the calculation in the above Claim. To begin with, we make the observation that

all tire eigenspaces V,. of l:l are one dimen.~ional.

(11.6)

Second, since we have in the course of the proof written down a basis for V and said exactly where each of H, X, and Y takes each basis vector, the representation Vis completely determined by the one complex number 11 that we started with; in particular, of course, we have that

Vis determined by the col/ectior1 of rJ. occurri1tg in the liecomposition

v = EI1 v,..

(11.7)

To complete our analysis, we have to use one more time the finite dimensionality of V. This tells us that there is a lower bound on the IX for which V. "# (0) as well as an upper one, so that we must have Y~(u) = 0 for sufficiently ~· large k. But now if m is the smallest power of Y annihilating v, then from the : relation ( 11.5), 0 = X(Y"'(u)) = m(n- m +I)· ym-•(v), ;;: and the fact that Y"'- 1(v) 'I' 0, we conclude that n - m + I = 0; in particular, ~' it follows that n is a 110n-negatiue integer. The picture is thus that the eigen~.OValues. a_of_H on V form a string of integers diffe~ng b~ 2 and symmetri~ about ~;.. the ongm m Z. In sum, then, we see that there rs a umque represen tall on v«•J ~i'ror each non-negative integer n; the representation v«•J is (n + I )-dimensional, ~ with

H having eige11ua/ues n, 11

-

2, ... , - n + 2, - n.

11. Representations of si 2 C

150

Note that the existence part of this statement may be deduced by checking that the actions of H, X, and Y as given above in terms of the basis v, Yv, Y2 (v), ... , Y"(v) for V do indeed satisfy all the commutation relations for si 2 C. Alternatively, we will exhibit them in a moment. Note also that by the symmetry of the eigenvalues we may deduce the useful fact that any represen-

tation V of si 2 C suclr that tire eigenvalues of Hall lrave tire same parity and occur witlr multiplicity one is 11ecessarily irreducible; more generally, the /lumber ofirreduci(J{e factors In an arbitrary representation V t!fsi 2 C is exactly the sum of tile multiplicities of 0 and I as eigenvalues of H. We can identify in these terms some of the standard representations ofsi 2 C. To begin with, the trivial one-dimensional representation Cis clearly just Jll 01• As for the standard representation of si 2 C on V = C 2, if x and y are the standard basis for C 2 , then we have H(x) = x and H(y) = - y, so that V = C · Jc ® C · y = v_, ® v, is just the representation Vl'l above. Similarly, a basis for the symmetric square W = Sym 2 V = Sym 2C 2 is given by {x 2 , xy, y 2 },and we have ll(x·x) = x· H(x) + ll(x)·x = 2x·x,

+ H(x)· y = 0, H(y· y) = y· H(y) + H(y)· y = -2y· y,

H(x· Y) = x· H(y)

so the representation W = C · x2 $ C · xy $ C · y 2 = W_ 2 $ W0 ® W2 is the representation Vl 21 above. More generally, the nth symmetric power Sym"V of V has basis {x", x•-• y, ... , y"}, and we have

H(x"-lyt) = (n- k)· H(x)· x•-t-r yk = (n-

+ k· H(y)· x•-kyl-'

2k)·x"-tl

so that the eigenvalues of H on Sym"V are exactly n, n - 2, ... , - n. By the observation above that a representation for which all eigenvalues of II occur with multiplicity 1 must be irreducible, it follows that Sym" V is irreducible, and hence that vl•J = Sym"V.

In sum then, we can say simply that

Any i1·reducible representation of si 2 C is a symmetric power of the (11.8) standard repr-esentation v ~ C 2 . Observe that when we exponentiate the image of si 2 C under the embedding si 2 C-+ sl.+rC corresponding to the representation Sym"V, we arrive at the group SL 2 C when 11 is odd and PGL 2 C when n is even. Thus, the represen· rations of tire group PGL 2 C are exactly the even powers Sym 2 "V. Exercise 11.9. Use the analysis of the representations of si 2 C to prove the statement made in the previous lecture that the universal coverS of SL 2 1R has no finite-dimensional representations.

§11.2. A Little Plethysm

151

§11.2. A Little Plethysm Clearly, knowing the eigenspace decomposition of given representations tells us the eigenspace decomposition of all their tensor, symmetric, and alternating products and powers: for example, if V = v. and W = Wp then V ® W = ® Wp) and v. ® W, is an eigenspace for 1/ with eigenvalue (1. + p. We can use this to describe the decomposition of these products and powers into irreducible representations of the algebra si 2 C. For example, let V ~ C 2 be the standard representation of si 2 C; and suppose we want to study the representation Sym 2 V ® Sym 3 V; we ask in particular whether if it irreducible and, if not, how it decomposes. We have seen that the eigenvalues of Sym 2 V are 2, 0, and -2, and those of Sym 3 V are 3, I, -I, and -3. The l2eigenvalues of the tensor product Sym 2 V® Sym 3 V are thus 5 and -5, 3 and -3 (taken twice), and l and - I (taken three times); we may represent them by the diagram

EB

EBW.

EB

--·--~~~~'~-+~~·»_,~@~~~-~

0

5

The eigenvector with eigenvalue 5 will generate a subrepresentation of the tensor product isomorphic to Sym ~ V, which will account for one occurrence of each of the eigenvalues 5, 3, l, -I, - 3, and -5. Similarly, the complement of Sym~ V in the tensor product will have eigenvalues 3 and -3, and I and -I (taken twice), and so will contain a copy of the representation Sym 3 V, which will account for one occurrence of the eigenvalues 3, I, - I and - 3; and the complement of these two subrepresentations will be simply a copy of V. We have, thus, Sym 2 V®Sym 3 JI~ Sym 5 VE:9Sym 3 VEJ:) V.

Note that the projection map Sym 2 V ® Sym 3 V--> Sym~ V on the first factor is just multiplication of polynomials; the other two projections do not admit such obvious interpretations. Exercise 11.10. Find, in a similar way, the decomposition of the tensor product Sym 2 V®Sym 5 V.

Exercise 11.1 t *.Show, in general, that for a 2 h we have Sym•v ® Sym&v = Symauv EJ:) Sym•+b- 2 V EJ:) · ··

® Sym•-bv.

As indicated, we can also look at symmetric and exterior powers of given representations; in many ways this is more interesting. For examrlc, let

11. Representations of s1 2 C

152

J' ~ C 2 be as above the standard representation of si 2 C, and let W

= Sym 2 V

be its symmetric square; i.e., in the notation introduced above, take W = V121• We ask now whether the symmetric square of Wis irreducible, and if not what its decomposition is. To answer this, observe that W has eigenvalues -2, 0, and 2, each occurring once, so that the symmetric square of W will have eigenvalues the pairwise sums ofthese numbers-that is, -4, -2,0 (occurring twice), 2, and 4. We may represent Sym 2 V by the diagram:

__.--+-~•·-~~@--~~-----4

-3

-2

-1

0

2

3

4

From this, it is clear that the representation Sym 2 W must decompose into one copy of the representation VI">= Sym 4 V, plus one copy of the trivial (one-dimensional) representation:

(11.12)

Sym 2 (Sym 2 V)) = Sym 4 V EB Sym 0 V. Indeed, we can see this directly: we have a natural map Sym 2 (Sym 2 V))-> Sym 4 V

obtained simply by evaluation; this will have a one-dimensional kernel (if x and y are as above the standard basis for V we can write a generator of this kernel as (x 2) · (y 2)- (x · y)l). Exercise t 1.13. Show that the exterior square N W is isomorphic to W itself. Observe that this, together with the above description of Syrri 2 W, agrees with the decomposition of W ® W given in Exercise 11.11 above. We can, in a similar way, describe the decomposition of all the symmetric powers of the representation W = Sym 2 V. For example, the third symmetric power Sym 3 W has eigenvalues given by the triple sums of the set { -2, 0, 2}; these are -6, -4, -2 (twice), 0 (twice), 2 (twice), 4, and 6; diagrammatically,

-----~---6

-4

@

@

·2

0

•

I-

Again, there is no ambiguity about the decomposition; this collection of eigenspaces can only come from the direct sum ofSym 6 V with Sym 2 V, so we must have Sym 3 (Sym 2 V) = Sym 6 V$ Sym 2 V. As before, we can see at least part of this directly: we have a natural evaluation map

§11.3. A Little Geometric Plethysm

153

and the eigenspace decomposition tells us that the kernel is the irreducible representation Sym 2 V. Exercise 11.14. Use the eigenspace decomposition to establish the formula ln/Z)

Sym"(Sym 2 V) =

EB Sym

2"- 4 "V

11=0

for all n.

§11.3. A Little Geometric Plethysm We want to give some geometric interpretations of these and similar decom· positions of higher tensor powers of representations of sl 2 C. One big difference is that instead of looking at the action of either the Lie algebra si 2 C or the groups SL 2 C or PGL 2 C on a representation ~V, we look at the action of the group PGL 2 C on the associated projective space' PW. In this context, it is natural to look at various geometric objects associated to the action: for example, we look at closures of orbits of the action, which all turn out to be algebraic varieties, i.e., definable by polynomial equations. In particular, our goal in the following will be to describe the symmetric and exterior powers of Win terms oft he action of PGL 2 C on the projective spaces PW and various loci in IPW. The main point is that while the action of PGL 2 C on the projective space PV ~ P' associated to the standard representation V is transitive, its action on the spaces IP(Sym"V) ~ IP'" for n > 1 is not. Rather, the action will preserve various orbits whose closures are algebraic subvarieties of IP"-for example, the locus of points

C = {[v· v· ... · v]: v e V} c IP(Sym"V) corresponding to nth powers in Sym"V will be an algebraic curve in P(Sym" V) ~ P", called the rational normal curve; and this curve will be carried into itself by any element of PGL 2 C acting on P" (more about this in a moment). Thus, a knowledge of the geometry of these subvarieties ofPW may illuminate the representation W, and vice versa. This approach is particularly useful in describing the symmetric powers of W, since these powers can be viewed as the vector spaces of homogeneous polynomials on the projective space IP(W*) (or, mod scalars, as hypersurfaces in that projective space). Decomposing these symmetric powers should therefore correspond to some interesting projective geometry. 1 PW here

denotes the projective space or lines through the origin in W. or the quotient space or

JYI{O} by multiplication by nonzero scalars; we wrile [w] ror the point in PlV determined by the llonzero vector w. For W = C'".,, [z0 , ••• , z,.] is the point in P'" = PlV determined by a point (:0 , ... ,

z,.) in C'".,.

II. Representations of ~1 2 C

154

Digression on Projective Geometry First, as we have indicated, we want to describe representations of Lie groups in terms of the corresponding actions on projective spaces. The following fact from algebraic geometry is therefore of some moral (if not logical) importance: Fact 11.15. The group of automorphisms of projective space P"-either as algebraic variety or as complex manifold-is just the group POL. +I C. For a proof, see [Ha]. (For the Riemann sphere P 1 at least, this should be a familiar fact from complex analysis.) For any vector space W of dimension 11 + I, Sym 1 W"' is the space of homogeneous polynomials of degree k on the projective space P" = P W of lines in W; dually, Sym 1 W will be the space of homogeneous polynomials of degree k on the projective space P" = P(W"') oflines in w•, or of hyperplanes in W. Thus, the projective space IP(Sym 1 W) is the space of hypersurfaces of degree kin IP" = IP(W"'). (Because of this duality, we usually work with objects in the projective space IP(W*) rather than the dual space /?Win order to derive results about symmetric powers SymkW; this may seem initially more confusing, but we believe it is ultimately less so.) For any vector space Vand any positive integer 11, we have a natural map, called the Veronese embedding IP'V"' Y P(Sym"V"')

that maps the line spanned by v E v• to the line spanned by v" E Sym" v•. We will encounter the Veronese embedding of higher-dimensional vector spaces in later lectures; here we are concerned just with the case where V is two dimensional, so IP v• = IP 1 • In this case we have a map

t.: IP 1 YIP"= IP'(Sym"V"') whose image is called the rationa/11ormal curve C = c. of clegree 11. Choosing bases {o:,f/} for v• and { ... (n!/k!(n- k)!]o: 1 p•-t ... } for Sym•v• and expanding out (xo: + yp)" we see that in coordinates this map may be given as

[x, y] H [x", x•-• y, x"- 2 y2 ,

.•• ,

xy•-•, y"].

From the delinition, the action of PGL 2 C on IP'" preserves c.; conversely, since any automorphism of IP" lixing c. pointwise is the identity, from Fact 11.15 it follows that the group G of autonwrpllisms of /?" tlrat preserve c. is precisely PGL 2 C. (Note that conversely if W is any (11 + i)-dimensional representation of SL 2 C and IPW ~ IP'" contains a rational normal curve of degree n preserved by the action of PGL 2 C, then we must have W ~ Sym"V; we leave this as an exercise. 2 ) When n = 2, Cis the plmre collie delined by the equation 2 Note that any confu~ion between PW and PW* i~ relatively harmle~s for us here, since the representations Sym"V are iwmorphic to their duals.

§11.3. A Little Geometric Plethysm

155

F(Zo. Z,, Z2) = ZoZ2 -

Zf =

0.

For 11 = 3, Cis the twisted c11bic curve in P 3, and is defined by three quadratic polynomials and More generally, the rational normal curve is the common zero locus of the 2 x 2 minors of the matrix

M = (ZoZt ... Z•-t)

z,z2 ... z. ·

that is, the locus where the rank of M is l.

Back to Plethysm We start with Example ( 11.12). We can interpret the decomposition given !here (or rather the decomposition of the representation of the corresponding Lie group SL 2C) geometrically via the Veronese embedding 12: IP' 1 c..IP' 2. As noted, SL 2C acts on P 2 = JPI(Sym 2V*) as the group of motions ofJPI 2 carrying the conic curve C 2 into itself. Its action on the space Sym 2(Sym 2V)) of quadratic polynomials on IP' 2 thus must preserve the one-dimensional subspace C · F spanned by the polynomial F above that defines the conic C 2 • At the same time, we see that pullback via 12 defines a map from the space of quadratic polynomials on P 2 to the space of quartic polynomials on P 1 , which has kernel C · F; thus, we have an exact sequence 0--+ C = Sym 0 V--+ Sym 2 (Sym 2 V))--+ Sym 4 V--+ 0, which implies the decomposition of Sym 2 (Sym 2 JI)) described above. Note that what comes to us at first glance is not actually the direct sum decomposition (11.12) of Sym 2 (Sym 2 V)), but just the exact sequence above. The splitting of this sequence of SL 2 C-modules, guaranteed by the general theory, is less obvious. For example, we are saying that given a conic curve C in the plane IP' 2 , there is a subspace Uc of the space of all conics in IP' 2 , complementary to the one-dimensional subspace spanned by C itself and invariant under the action of the group of motions of the plane IP' 2 carrying . C into itself. Is there a geometric description of this space? Yes: the following proposition gives one. Proposition 11.16. The subrepresentation Sym 4 V c: Sym 1 (Sym 2 V) is tire space of collies spanned by the family of double lines tangent to tire conic C = C1 • : PROOF.

One way to prove this is to simply write out this subspace in coorJP1 2 as above, the tangent

r· dinates: in terms of homogeneous coordinates Z; on ·.tine to the conic Cat the point [I, a, a 2 ] is the line

II. Representations of 5l 2 C

156

L« = {Z: IX 2Zo- 2aZI

+ z2 = 0}.

The double line 2L« is, thus, the conic with equation IX 4 z~-

4o: 3 ZoZI

+ 2o: 2ZoZ2 + 41X 2 z:

-

4aZ. z2

+ Zi = 0.

The subspace these conics generate is thus spanned by z~. ZoZI, zlz2, z;, and Z 0 Z 2 + 2Zf. By construction, this is invariant under the action of SL2 C, and it is visibly complementary to the trivial subrepresentation C · F = C · (Z0 Z 2 - Zf). For those familiar with some algebraic geometry, it may not be necessary to write all this down in coordinates: we could just observe that the map from the conic curve C to the projective space P(Sym 2 (Sym 2 V)) of conics in P 2 .sending each point p e C to the square of the tangent line to C at p is the restriction to C of the quadratic Veronese map P 2 -+ PS, and so has image a quartic rational normal curve. This spans a four-dimensional projective subspace of P(Sym 2 (Sym 2 V)), which must correspond to a subrepresentation 0 isomorphic to Sym 4 V. We will return to this notion in Exercise 11.26 below. We can, in a similar way, describe the decomposition of all the symmetric powers of the representation W = Sym 2 V; in the general setting, the geometric interpretation becomes quite handy. For example, we have seen that the third symmetric power decomposes Sym 3 (Sym 2 V)

= Sym 6 V$Sym 2 V.

This is immediate from the geometric description: the space of cubics in the plane P 2 naturally decomposes into the space of cubics vanishing on the conic c = c2, plus a complementary space isomorphic (via the pullback map •!> to the space of sex tic polynomials on P 1 ; moreover, since a cubic vanishing on C2 factors into the quadratic polynomial F and a linear factor, the space of cubics vanishing on the conic curve C c P 2 may be identified with the space of lines in IP' 2 • One more special case: from the general formula (I 1.14), we have Sym 4 (Sym 2 V) ~ Sym 8 V E!1 Sym 4 V EJ1 Sym 0 V. Again, this is easy to see from the geometric picture: the space of quartic polynomials on P 2 consists of the one-dimensional space of quartics spanned by the square of the defining equation F of C itself, plus the space of quartics vanishing on C modulo multiples of F 2 , plus the space of quartics modulo those vanishing on C. (We use the word "plus," suggesting a direct sum, but as before only an exact sequence is apparent). Exercise 11.17. Show that, in general, the order of vanishing on C defines a liltration on the space of polynomials of degree n in IP' 2 , whose successive quotients are the direct sum factors on the right hand side of the decomposition of Exercise II. I4.

§I 1.3. A Little Geometric Plethysm

157

We can similarly analyze symmetric powers of the representation U =?< Sym 3 Jl. For example, since U has eigenvalues -3, -1, I, and 3, the symmetric square of U has eigenvalues -6, -4, -2 (twice), O(twice), 2 (twice), 4, and 6; diagrammatically, we have

•

e--f---@--t-@-1-@---j---•-+-•-

·6

This implies that (ll.l8) We can interpret this in terms of the twisted cubic C = C3 c P 3 as follows: the space of quadratic polynomials on P 3 contains, as a subrepresentation, the three-dimensional vector space of quadrics containing the curve C itself; and the quotient is isomorphic, via the pullback map 1;, to the space of sextic polynomials on IP' 1• Exercise 11.19*. By the above, the action of SL 2 C on the space of quadric surfaces containing the twisted cubic curve C is the same as its action on IP'(Sym 2 V*) g; P 2 • Make this explicit by associating to every quadric containing C a polynomial of degree 2 on IP 1, up to scalars. Exercise 11.20*. The direct sum decomposition (11.18) says that there is a linear space of quadric surfaces in IP' 3 preserved under the action of SL 2 C and complementary to the space of quadrics containing C. Describe this space. Exercise 11.21. The projection map from Sym 2 (Sym 3 V) to Sym 2 V given by the decomposition (11.18) above may be viewed as a quadratic map from the vector space Sym 3 V to the vector space Sym 2 V. Show that it may be given in these terms as the Hessian, that is, by associating to a homogeneous cubic polynomial in two variables the determinant of the 2 x 2 matrix of its second partials. Exercise 11.22. The map in the preceding exercise may be viewed as associating to an unordered triple of points {p, q, r} in P 1 an unordered pair of points [s, t} c IP' 1 . Show that this pair of points is the pair of fixed points of the automorphism of IP' 1 permuting the three points p, q, and r cyclically. Exercise 11.23*. Show that Sym 3 (Sym 3 V) = Sym 9 V~Sym 5 V$Sym 3 V, and interpret this in terms of the geometry of the twisted cubic curve. In particular, show that the space of cubic surfaces containing the curve is the direct sum of the last two factors, and identify the subspace of cubics corresponding to the last factor.

158

11. Representations of sl 2 C

Exercise 11.24. Analyze the representation Sym 4 (Sym 3 V) similarly. In particular, show that it contains a trivial one-dimensional subrepresentation. The trivial subrepresentation of Sym 4 (Sym 3 V) found in the last exercise has an interesting interpretation. To say that Sym 4 (Sym 3 V) has such an invariant one-dimensional subspace is to say that tl1ere exists a quartic surface in P 3 preserved under all motions of P 3 carrying tire rational normal curve C = C3 into it.~elf. What is this surface? The answer is simple: it is the tangent developable to the twisted cubic, that is, the surface given as the union of the tangent lines to C. Exercise 11.25*. Show that the representation Sym 3 (Sym 4 V) contains a trivial subrepresentation, and interpret this geometrically. Problem l 1.26. Another way of interpreting the direct sum decomposition of Sym 2 (Sym 2 V) geometrically is to say that given a conic curve C c: P 2 and given four points on C, we can lind a conic C' = C'(C; p 1 , ••• , p4 ) c: P 2 intersecting C in exactly these points, in a way that is preserved by the action of the group PG L 3 C of all motions of P 2 (i.e., for any motion A: IP 2 -> P 2 of the plane, we have A(C'(C; p 1 , • •• , p4 )) = C'(AC; Ap 1 , • •• , Ap4 )). What is a description of this process? In particular, show that the cross-ratio of the four points p1 on the curve C' must be a function of the cross-ratio of the Pi on C, and find this function. Observe also that this process gives an endomorphism of the pencil

{C c p2: Pt> ... , P4 E C} ;;;; P' of conics passing through any four points Pie P 2 • What is the degree of this endomorphism? The above questions have all dealt with the symmetric powers of Sym"V. There are also interesting questions about the exterior powers of Sym"V. To start with, consider the exterior square N(Sym 3 V). The eigenvalues of this representation are just the pairwise sums of distinct elements of {3, I, -1, -3}, that is, 4, 2, 0 (twice), -2, and -4; we deduce that (11.27) Observe in particular that according to this there is a skew-symmetric bilinear form on the space U = Sym 3 V preserved (up to scalars) by the action of SL 2 C. What is this form? One way of describing it would be in terms of the twisted cubic: the map from C to the dual projective space (P 3 )"' sending each point 1' E C to the osculating plane to C at p extends to a skew-symmetric linear isomorphism of IP 3 with (IP 3 )*. Exercise t 1.28. Show that a line in P 3 is isotropic for this form if and only if, viewed as an element of P(/\ 2 V), it lies in the linear span of the locus of tangent lines to the twisted cubic.

159

§11.3. A Little Geometric Plethysm

Exercise 11.29. Show that the projection on the first factor in the decomposition (11.27) is given explicitly by the map F"

G~-->F·dG-

G·dF

and say precisely what this means. Exercise 11.30. Show that, in general, the representation N(Sym"V) has as a direct sum factor the representation Sym 2 "- 2 V, and that the projection on this factor is given as in the preceding exercise. Find the remaining factors of N(Sym"V), and interpret them.

More on Rational Normal Curves Exercise 11.31. Analyze in general the representations Sym 1 (Sym"JI); show, using eigenvalues, that we have Sym 2 (Sym"JI) =

E9 Sym

2 "- 4

"JI.

~~~o

Exercise 11.32*. Interpret the space Sym 2 (Sym"V) of the preceding exercise as the space of quadrics in the projective space P", and use the geometry of the rational normal curve C = c P" to interpret the decomposition of this representation into irreducible factors. In particular, show that direct sum

c.

.E9.,,

Sym 2 •-"•v

is the space of quadratic polynomials vanishing on the rational normal curve; and that the direct sum

is the space of quadrics containing the tangential developable of the rational normal curve, that is, the union of the tangent lines to C. Can you interpret the sums for a ~ k for k > 2? Exercise 11.33. Note that by Exercise t I. II, the tensor power Sym"V ® Sym"V

· always contains a copy of the trivial representation; and that by Exercises I 1.30 and 11.31, this trivial subrepresentation will lie in Sym 1 (Sym"l') if 11 is even and in N(Sym"V) if 11 is odd. Show that in either case, the bilinear form on Sym"V preserved by SL 2 C may be described as the isomorphism of IP'" with (P")* carrying each point p of the rational normal curve C c P" into the osculating hyperplane to Cat p. . Comparing Exercises ll.l4 and 11.31, we see that Sym 2 (Sym"V) ~ 'Sym"(Sym 2 V); apparently coincidentally. This is in fact a special case of a more general theorem (cf. Exercise 6.18):

160

II. Representations of toi2 C

Exercise 11.34. (Hermite Reciprocity). Use the eigenvalues of H to prove the isomorphism Syml(Sym" V) ~ Sym"(Symk V). Can you exhibit explicitly a map between these two? Note that in the examples of Hermite reciprocity we have seen, it seems completely coincidental: for example, the fact that the representations Sym 3 (Sym 4 V) and Sym 4 (Sym 3 V) both contain a trivial representation corresponds to the facts that the tangential developable of the twisted cubic in P 3 has degree 4, while the chordal variety of the rational normal quartic in IP 4 has degree 3. Exercise 11.35*, Show that 1\"'(Sym"V) ~ Sym"'(Sym•+t-"'V). We will see in Lecture 23 that there is a unique clo~ed orbit in P(W) for any irreducible representation W. For now, we can do the following special case. Exercise t 1.36. Show that the unique closed orbit of the action of SL 2 C on the projectivization of any irreducible representation is isomorphic to IP 1 (these are the ralimwlnormal curves introduced above).

LECTURE 12

Representations of sl 3 C, Part I

This lecture develops results for si 3 C analogous to those of§ll.l (though not in exactly the same order). This involves generalizing some of the basic terms of §ll (e.g., the notions of eigenvalue and eigenvector have to be redefined), but the basic ideas are in s<'me sense already in §II. Certainly no techniques are involved beyond those of§ll.l.

We come now to a second important stage in the development of the theory: in the following, we will take our analysis of the representations of si 2 C and see how it goes over in the next case, the algebra si 3 C. As we will see, a number of the basic constructions need to be modified, or at least rethought. There are, however, two pieces of good news that should be borne in mind. First, we will arrive, by the end of the following lecture, at a classification of the representations of sl 3 C that is every bit as detailed and explicit as the classification we arrived at previously for si 2 C. Second, once we have redone our analysis in this context, we willtteed to introduce no further concepts to carry out the c/asslficatio/1 of the finite-dimensional representations of all remaining semisimple Lie algebras. We will proceed by analogy with the previous lecture. To begin with, we started out our analysis of s1 2 C with the basis {H, X, Y} for the Lie algebra; we then proceeded to decompose an arbitrary representation V of si 2 C into a direct sum of eigenspaces for the action of H. What element of t~I 3 C in particular will play the role of H? The answer-and this is the first and perhaps most wrenching change from the previous case-is that no one element really allows us to see what is going on. 1 Instead, we have to replace 1 This is not literally true: as we will see from the following analysis, if II is any diagonal matrix whose entries are independent over 0, then the action or II on any representation V or oiJC determines the repre!<entation (i.e., if we know the eigenvalues of II we know V). Rut (as we will also see) trying to carry this out in practice would be sheer perversity.

12. Representations of si 3 C, Part I

t62

the single element /l e si 2 C with a subspace I) c si 3 C, namely, the twodimensional subspace of all diagonal matrices. The idea is a basic one; it comes down to the observation that commuting diagonalizable matrices are simultaneously diagonalizable. This translates in the present circumstances to the statement that any finite-dimensional representation V of s1 3 C admits a decomposition V = ~ V.,, where every vector 11 e V.. is an eigenvector for every element H e 1). At this point some terminology is clearly in order, since we will be dealing with the action not of a single mat rill II but rather a vector space I) of them. To begin with, by an eigenvector for I) we will mean, reasonably enough, a vector 11 e V that is an eigenvector for every He 1). For such a vector v we can write

H(v)

= IX{H)· v,

(12.1)

where IX( H) is a scalar depending linearly on H, i.e., IX e I>*· This leads to our second notion: by an eigenvalue for the action of I) we will mean an element IX e ~·such that there exists a nonzero element v e V satisfying (12.1); and by the eige11space associated to the eigenvalue IX we will mean the subspace of all vectors 11 e V satisfying ( 12.1 ). Thus we may phrase the statement above as ( 12.2) Any finite-dimensional representation V of sl 3 C l1as a decon.positio11

v= wl1ere

•.

~v

v. is au eigempace for I> ancliX ranges over a finite subset of 1)*.

This is, in fact, a special case of a more general statement: for any semisimple Lie algebra g, we will be able to lind an abelian subalgebra ~ c g, such that the action of I) on any g-module V will be diagonalizable, i.e., we will have a direct sum decomposition of V into eigenspaces V.. for ~· Having decided what the analogue for si 3 C of He si 2 C is, let us now consider what will play the role of X and Y. The key here is to look at the commutation relations [fl, X]= 2X

and

[H, Y] = -2Y

in si 2 C. The correct way to interpret these is as saying that X and Y are eigenvector.~ for the adjoint action of H on si 2 C. In our present circumstances, then, we want to look for eigenvectors (in the new sense) for the adjoint action of I) on si 3 C. In other words, we apply (12.2) to the adjoint representation of si 3 C to obtain a decomposition

(I 2.3) where rx ranges over a finite subset ofl)* and I) acts on each space g. by scalar multiplication, i.e., for any II e I) and Y e g.,

[H, Y] = ad(H)(Y) = 1X(H)· Y. This is probably easier to carry out in practice than it is to say; we are being

12. Representations of si 3 C, Part I

163

longwinded here because once this process is understood it will be straightforward to apply it to the other Lie algebras. In any case, to do it in the present circumstances, we just observe that multiplication of a matrix M on the left by a diagonal matrix D with entries a, multiplies the ith row of M by a1, while multiplication on the right multiplies the ith column by a,; if the entries of M are m1,1, the entries of the commutator [D, M] are tints (a,- a1)m 1, 1. We see then that the commutator [D, M] will be a multiple of M for all D if and only if all but one entry of Mare zero. Thus, if we let E1•1 be the 3 x 3 matrix whose (i, J)th entry is 1 and all of whose other entries are 0, we see that the E1,1 exactly generate the eigenspaces for the adjoint action of I) on g. Explicitly, we have

and so we can write where

L,(~' 0

: 2

0

~)=a,.

a3

The linear functionals IX e ()*appearing in the direct sum decomposition (12.3) are thus the six functionals L 1 - L1; the space 9L,-J.1 will be generated by the element E1,1. To draw a picture

( 12.4)

The virtue of this decomposition and the corresponding picture is that we

can read otT from it pretty much the entire structure of the Lie algebra. Of

12. Representations of sl 3 C, Part I

164

course, the action of !) on g is clear from the picture: I) carries each of the subs paces g. into itself, acting on each g. by scalar multiplication by the linear functional represented by the corresponding dot. Beyond that, though, we can also see, much as in the case of representations of si 2 C, how the rest of the Lie algebra acts. Basically, we let X be any element of g. and ask where ad( X) sends a given vector Y E g11 ; the answer as before comes from knowing how I) acts on ad(X)(Y). Explicitly, we let H be an arbitrary element of IJ and as on page 148 we make the

Fundamet•tal Calculation (second time):

+ [[H, X]. Y] =[X, fl(H)· Y] + [a(H)· X, Y] = (cx(H) + P(H))· [X, Y].

[H, [X, Y]] =[X, [11, Y]]

In other words, [X, Y] = ad(X)(Y) is again an eigenvector for !), with eigen· value a + p. Thus, ad(g.): 9p-> 9o+/l; in particular, the action of ad(g.) preserves the decomposition (12.3) in the sense that it carries each eigenspace g11 into another. We can interpret this in terms of the diagram (12.4) of eigenspaces by saying that each g. acts, so to speak, by "translation"; that is, it carries each space 9p corresponding to a dot in the diagram into the subspace 9o+p corresponding to that dot translated by a. For example, the action of !"lt,-t., may be pictured as

0

(12.5) 0

L1

-

L2

12. Representations of sl 3 C, Part I

165

i.e., it carries gL,-r., into gL,-L,; g1,,-L, into I); I) into gL,-L,• OL,-1., into gL,-Lz• and kills 9L,-L,• 0L,-L,• and gL,-L,· Of COurse, not all the data Can be read Off of the diagram, at least on the basis on what we have said so far. For example, we do not at present see from the diagram the kernel ofad(gL,- 1.,) on 1), though we will see later how to read this off as well. We do, however, have at least a pretty good idea of who is doing what to whom. Pretty much the same picture applies to any representation V of si 3 C: we start from the eigenspace decomposition V = EB v. for the action or I) that we saw in (12.2). Next, the commutation relations for si 3 C tell us exactly how the remaining summands of the decomposition (12.3) of si 3 C act on the space V, and again we will see that each of the spaces g. acts by carrying one eigenspace V11 into another. As usual, for any X e g. and v e V, we can tell where X will send v if we know how an arbitrary element He I) will act on X(v). This we can determine by making the

F11ndamental Calculation (third time):

+ [H, X](v) X(p(H) · v) + (a(H) · X)(t•)

H(X(v)) = X(H(v)) =

= (a(H) + P(JI))· X(v). We see from this that X(v) is again an eigenvector for t/1e action of~. with + fJ; in other words, the action of g. carries V, to V..+IJ· We can thus represent the eigenspaces V.. of V by dots in a plane diagram so that each g. acts again "by translation," as we did for representations of sl 2 C in the preceding lecture and the adjoint representation of si 3 C above. Just as in the case of si 2 C (page 148), we have

eigenvalue a

Observation 12.6. The eigenvalues a occurritlg itl an irreducible representation of si 3 C differ ji·om one other by imegral linear combinations of the t•ecrors L,- L1 e I)'". Note that these vectors L 1 - L1 generate a lattice in I>*, which we will denote by 1\R, and that all the a lie in some translate of this lattice. At this point, we should begin to introduce some of the terminology that appears in this subject. The basic object here, the eigenvalue a e I)* of the a~ti9n of IJ on a representation V of g, is called a weight of the representation; are called, naturally enough, weight the .corresponding eigenvectors in vectors and the spaces v. themselves weight spaces. Clearly, the weights that occur in the adjoint representation are special; these are called the roots of the Lie algebra and the corresponding subspaces g. c g roor spaces; by

v.

12. Representations or s1 3 C, Part I

166

convention, zero is not a root. The lattice Ap., c I>* generated by the roots II is called the root lattice. To see what the next step should be, we go back to the analysis of representations of si 2 C. There, at this stage we continued our analysis by going to an extremal eigenspace V., and taking a vector v E V.,. The point was that since J~ was extremal, the operator X, which would carry V., to V., ... 2 , would have to kill v; so that v would be then both an eigenvector for H and in the kernel of X. We then saw that these two facts allowed us to completely describe the representation V in terms of images of v. What would be the appropriately analogous setup in the case of si 3 C? To start at the beginning, there is the question of what we mean by e.xtremal: in the case of si 2 C, since we knew that all the eigenvalues were scalars differing by integral multiples of2, there was not much ambiguity about what we meant by this. In the present circumstance this does involve a priori a choice (though as we shall see the choice does not affect the outcome): we have to choose a direction, and look for the farthest II in that direction appearing in the decomposition (12.3). What this means is that we should choose a linear functional

extend it by linearity to a linear functional 1: l)* --+ C, and then for any representation V we should go to the eigenspace V., for which the real part of l(ll) is maximal. 2 Of course, to avoid ambiguity we should choose I to be irrational with respect to the lattice Ap.,, that is, to have no kernel. What is the point of this? The answer is that, just as in the case of a representation V of si 2 C we found in this way a vector v e V that was simultaneously in the kernel of the operator X and an eigenvector for ll, in the present case what we will lind is a vector v e V., that is an eigenvector for 1), and at tl1e same time in the kernel of the actiorr of nv for every fJ S!icll that /({J) > 0-that is, that is killed by half tile root spaces g11 (specifically, the root spaces corresponding to dots in the diagram (J 2.4) lying in a half plane). This will likewise give us a nearly complete description of the representation V To carry this out explicitly, choose our functional/ to be given by

l(a 1 L 1

+ a2 L 2 + a 3 L 3 ) =

aa 1

+ ba 2 + ca 3 ,

where a + b + c = 0 and a > b > c, so that the spaces g4 c g for which we have l(ll) > 0 are then exactly 9L,-L,• 9r.,-r.,. and 9t.,-r.,; they correspond to matrices with one nonzero entry above the diagonal.

2 The real-versus-complex business is a red herring since (it will turn out very shortly) all the eigenvalues « actually occurring in any representation will in ract be in the real (in ract, the rational) linear sran or A•.

12. Representations

of~I 3 C,

Part I

167

(12.7)

Thus, for i < j, the matrices E1,1 generate tile positive root spaces, and the Ei.i generate tile negative root spaces. We"set (12.8)

Now let V be any irreducible, finite-dimensional representation of si 3 C. The upshot of all the above is the Lemma 12.9. Tltere is a vector v e V wit/1 tl•e propel"ties that (i) v is an eigenvector for~. i.e. v E V. for some a; a11d (ii) vis killed by E~, 2 , E~, 3 , a11d E2.3.

For any representation V of si 3 C, a vector v e V with these properties is called a l1ighest weigl1t vector. In the case of si 2 C, having found an eigenvector v for H killed by X, we argued that the images of 11 under successive applications of Y generated the representation. The situation here is the same: analogous to Claim 11.4 we have Claim 12.10. Let V be an irreducible representation of si 3 C, and v e V a llig/1est weight vector. Tl1en V is generated by tile images of v undel' successitJe applications of the tllfee operators E2 , 1 , E 3 , 1 , aud E 3 , 2 • Before we check the claim, we note three immediate consequences. First, it says that all the eigenvalues {J e ~· occurring in V lie in a sort of !-plane with corner at a:

12. Representations of sl 3C, Part I

168

Second, we see that the dimension of V,. itself is 1, so that v is the unique eigenvector with this eigenvalue (up to scalars, of course). (We will see below that in fact v is the unique highest weight vector of V up to scalars; see Proposition 12.11.) Lastly, it says that the spaces v,._...1L,-L,, and V.. ....1L,-L,, are all at most one dimensional; since they must be spanned by (E 2 , t)"(v) and (£ 3, 2)"(11), respectively. PRooF OF CLAIM 12.10. This is formally the same as the proof of the corresponding statement for sl 2 C: we argue that the subspace W of V spanned by images ofv under the subalgebra ohi 3 C generated by E 2 , 1 , E 3,., and E 3, 2 is, in fact, preserved by all ofsi 3C and hence must be all of V. To do this we just have to check that E~, 2 , Eu, and E 1, 3 carry W into itself(in fact it is enough to do this for the first two, the third being their commutator), and this is straightforward. To begin with, v itself is in the kernel of E 1 , 2 , E2 , 3 , and E1, 3, so there is no problem there. Next we check that E 2 , 1 (v) is kept in W: we have

E 1, 2 (E2.1(v))

= (E 2, 1(Eu(v)) + [Eu, = a:([EJ,2•

since E 1• 2 (v) = 0 and [E~, 2 , E 2 , 1 ]

E

E 2, 1 ](•')

Ez, 1 ])· v

I); and

E2,3(E2,t(v)) = (Ez.t(Ez,3(v))

+ [E2,3• E2.1](v)

=0 since E 2 , 3 (l') = 0 and [E 2, 3 , Eu] = 0. A similar computation shows that E 3 , 2 (v) is also carried into V by E 1 , 2 and E 2 , 3 • More generally, we may argue the claim by a sort of induction: we let ~ denote any word of length nor less in the letters E 2, 1 and E 3, 2 and take W. to be the vector space spanned by the vectors w.(v) for all such words; note that W is the union of the spaces W,, since E 3 , 1 is the commutator of £ 3 • 2 and E 2, 1 • We claim that Eu and E 2 , 3 carry W, into w._ 1• To see this, we can

12. Representations of si 3 C, Part I

169

write w" as either E 2 , 1 o w._ 1 or E 3 • 2 o W0 _ 1 ; in either case w._ 1(v) will be an eigenvector for~ with eigenvalue p for some p. In the former case we have

E u(w.(v)) = E ~, 2 (El.l (w._ 1(v))) = E2 , 1 (E~, 2 (w._ 1 (v))) E

E 2, 1 (J¥..- 2 )

+ [E~, 2 , E2 , 1 ](w._ 1(v))

+ fl([E~, 2 , El.l])·w.-t(v)

since [E 1 , 2 , E2, 1 ] e I); and E 2 , 3 (w.(v)) = E 2 , 3 (Eu(wd(v))) =

E 2, 1 (E 2 , 3 (w._ 1 (v))) + [E 2 , 3 , E 2 , 1 ](w._ 1(v))

E

Ez. i (W.-z)

c W.-t

since [E 2 , 3 , E 2 , 1 ] = 0. Essentially the same calculation covers the latter case w. = E 3 , 2 o 0 1 , establishing the claim.

w._

This argument shows a little more; in fact, it proves

Proposition 12.11. If Vis any representation of si 3 C and v e V is a higltest weigl1t vector, tl1en tlJe subrepresentation W of V generated by tl1e images of v by successive applications of tl1e three operators E 2 , 1 , E 3 , 1 , and E 3 , 2 is irred11cible. PRooF. Let a: be the weight of v. The above shows that W is a subrepresentation, and it is clear that W,. is one dimensional. if W were not irreducible, we would have W = W' E9 W" for some representations W' and W". But since projection to W' and W" commute with the action of I), we have W,. = E9 w;'. This shows that one of these spaces is zero, which implies that v belongs to W' or W", and hence that W is W' or W". 0

w:

As a corollary ofthis proposition we see that any irreducible representation of sl 3 C has a unique highest weight vector, up to scalars; more generally, the set of highest weight vectors in V fonns a union or linear subspaces 'I' w corresponding to the irreducible subrepresentations W of V, with the dimension of 'l'w equal to the number of times W appears in the direct sum decomposition of V into irreducibles. What do we do next? Well, let us continue to look at the border vectors (E2, 1f(v}. We call these border vectors because they live in (and, as we saw, span) a collection of eigenspaces g., g.+L,-L,• 9o+l(l.,-r., 1•... that correspond to points on the boundary of the diagram above of possible eigenvalues of V. We also know that they span an uninterrupted string of nonzero eigenspaces !laHu.,-r., 1 ~ C, k = 0, 1, ... , until we get to the first rn such that

12. Representations or sl 3 C, Part I

170

(Eu)'"(v) = 0; after that we have is thus:

g«H(L,-I.,)

= (O) for all

k

~ m. The picture

(12.12)

where we have no dots above/to the right of the bold line, and no dots on that line other than the ones marked. The obvious question now is how long the string of dots along this line is. One way to answer this would be to make a calculation analogous to the one in the preceding lecture: use the computation made above to say explicitly for any k what multiple of(E 2 ,d•- 1 (v) the image of(E 2 , 1 t(v) under the map E 1, 2 is, and use the fact that (E2, I r(v) = 0 to determine m. It will be simplerand more useful in general-if instead we just use what we have already learned about representations of si 2 C. The point is, tl1e elements E ~, 2 and E2 • 1, togetlrer wit It tlreir commutator [E 1, 2 , E 2 , .J = H 1 , 2 , span a stlbalgebra of si 3 C isomorpllic to s1 2 C Via a11 isomorpltisrn carrying E~, 2 , E 2 • 1 and H 1 . 2 to tire elements X, Y and H. We will denote this subalgebra by sL,-L, (the notation may appear awkward, but this is a special case of a general construction). By the description we have already given of the action of s1 3 C on the representaV«, we see that the subalgebra tion V in terms of the decomposition V = s 1., _ 1_, will shirt eigenspaces V. only in the direction of L 2 - L 1 ; in particular, the direct sum of the eigenspaces in question, namely the subspace

EB

(12.13) of V will be preserved by the action of s1.,-L,· In other words, W is a representation of sL,-L,;;;: si 2 C and we may deduce from this that tl1e eigenvalues of HI. 2 on W ar·e integral, and symmetric witlr respect to zero. Leaving aside the integrality for the moment, this says that the string of dots in diagram (12.12) must be symmetric with respect to the line
12. Representations or s1 3 C, Part I

171

to si 2 C via an isomorphism carrying Eu, E1,, and H1•1 to the elements X, Y, and H. (Note that Ill,}= - H1,1.) Analyzing the action of the subalgebra sL.-L, in particular then shows that the string of dots corresponding to the eigenspaces g«H(L 3 - L 2) is likewise preserved under reflection in the line (11 2 , 3 , L) = 0 in 1>"'· The picture is thus

Let us now take a look at the last eigenspace in the first string, that is, Vp where m is as before the smallest integer such that (E 2 , 1 r(v) = 0 and p = ex + (m - l)(L 2 - /- 1 ). If v' E V11 is any vector, then, by definition, we have E 2 , t(v') = 0; and since there are no eigenspaces V1 corresponding to y above the bold line in diagram (12.12), we have as well that E2 , 3 (v') = Eu(v') = 0. Thus, v', like v itself, satisfies the statement of Lemma 12.9, except for the exchange of the indices 2 and 1; or in other words, if we had chosen the linear functional I above differently-precisely, with coefficients b > a > c-then the vector whose existence is implied by Lemma 12.9 would have turned out to be v' rather than v. If, indeed, we had carried out the above analysis with respect to the vector ,,, instead of v, we would lind tht all eigenvalues of V occur below or to the right of the lines through {J in the directions of L 1 - L 2 and L 3 - L 1 , and that the strings of eigenvalues occurring on these two lines were symmetric about the lines (H 1 , 2 , L) = 0 and (IIu, L) = 0, respectively. The picture now is (H 1•2 ,L)=0

172

12. Representations of s1 3 C, Part I

Needless to say, we can continue to play the same game all the way around: at the end of the string of eigenvalues {(J + k(L 3 - L 1 )} we will arrive at a vector v" that is an eigenvector for~ and killed by E 3 • 1 and E 2 , 1 , and to which therefore the same analysis applies. In sum, then, we see that the set of eigenvalues in V will be bounded by a hexagon symmetric with respect to the lines (H 1•1, L) = 0 and with one vertex at 11; indeed, this characterizes the hexagon as the convex hull of the union of the images of a: under the group of isometries of the plane generated by reflections in these three lines.

(112.J•L )=0

(12.14)

We will see in a moment that the set of eigenvalues will include all the points congruent to a: modulo the lattice AR generated by the L 1 - L 1 lying on the boundary of this hexagon, and that each of these eigenvalues will occur with multiplicity one. The use of the subalgebras sL,-L, does not stop here. For one thing, observe that as an immediate consequence of our analysis of s! 2 C, all the eigenvalues of the elements H 1•1 must be integers; it is not hard to see that this means that all the eigenvalues occurring iit (12.2) must be integral linear combinations of the L~o i.e., in terms of the diagrams above, all dots must lie in the lattice A,., of interstices (as indeed we have been drawing them). Thus, we have Proposition 12.15. All the eigenvalues of any irreducible finite-dimensional representalion of s1 3 tC must lie in the lattice Aw c I)* generated by tire L 1 and be cmrgruent modulo the lattice AR c ~· generated by tire L 1 - Li. This is exactly analogous to he situation of the previous lecture: there we saw that the eigenvalues of II in any irreducible, finite-dimensional representation of si 2 C lay in the lattice Aw ~ Z of linear forms on CH integral on H, and were congruent toone another modulo the sublatticeAR = 2 · Z generated

12. Representations or!51 3 C, Part l

173

by the eigenvalues of If under the adjoint representation. Note that in the case lL/3; we will see later how this reflects a general pallern. The lattice Aw is called the weight lattice. ofsi 2 CwehaveAw/A~t ~ Z/2,whileinthepresentcasewehaveAw/A~t ~

Exercise 12.16. Show that the two conditions that the eigenvalues of V are congruent to one another modulo AI! and are preserved under reflection in the three lines (111, 1, L) = 0 imply that they all lie in Aw, and that, in fact, this characterizes Aw· To continue, we can go into the interior of the diagram (12.14) of eigenvalues of V by observing that the direct sums ( 12.13) are not the only visible subspaces of V preserved under the action of the subalgebras s 1.,-Lj; more generally, for any (J e ~·appearing in the decomposition (12.2) and any i,j the direct sum

will be a representation of s1.,-I.j (not necessarily irreducible, of course); in particular it follows that the values of k for which VpHcL,-Lj) #- (0) form an unbroken string of integers. Observing that if {J iR any of the "extremal" eigenvalues pictured in diagram ( 12.14), then this string will include another; so that all eigenvalues congruent to the dots pictured in diagram (12.14) and lying in their convex hull must also occur. Thus, the complete diagram of eigenvalues wiJJ look like

(12.17)

~We can summarize this description in

~:·Proposition 12.18. Let V be any in·educible, finite-dimensimral representation '':of si 3C. Tlreu for some rx E Aw c !)"'. tire set of eigemwlues occurrirrg i11 V is

174

12. Representations of si 3 C, Part I

exactly the set of linear jimctionals congruertt to r:1. modulo tire lattice 1\ 11 aud lying i11 tire hexagon with vertices tire images of r:1. 1111der the group gmerclted by reflections irr tire lirres (H1•1, L) = 0. Remark. We did, in the analysis thus far, make one apparently arbitrary choice when we defined the notion of "extremal'' eigenvalue by choosing a linear functional I on~·. We remark here that, in fact, the choice was not as broad as might at first have appeared. Indeed, given the fact that the configuration or eigenvalues occurring in any irreducible finite-dimensional representation of sl 3 C is always either a triangle or a hexagon, the "extremal" eigenvalue picked out by 1 will always turn out to be one of the three or six vertices of this figure; in other words, if we define the linear functional/ to take a 1 L 1 + a2 L 2 + a 3 L 3 to aa 1 + ba 2 + ca 3 , then only the ordering of the three real numbers a, b, and c matters. Indeed, in hindsight this choice was completely analogous to the choice we made (implicitly) in the case of s! 2 C in choosing one of the two directions along the real line. We said at the outset of this lecture that our goal was to arrive at a description of representations of {;[ 3 C as complete as that for si 2 C. We have now, certainly, as complete a description of the possible configurations of eigenvalues; but clearly much more is needed. Specifically, we should have an existence and uniqueness theorem; an explicit construction of each representations, analogous to the statement that every representation of si 2 C is a symmetric power of the standard; and for the purpose of analyzing tensor products of representations of sl 3 C, we need a description not just of the set of eigenvalues, but of the multiplicities with which they occur. (Note that the last question is one that has no analogue in the case of si 2 C: in both cases, any irreducible representation is generated by taking a single eigenvector v e V~ and pushing it around by elements of g~; but whereas in the previous case therewasonlyone way to get from V.. to V11-that is, by applying· Y over and over again-in the present circumstance there will be more than one way of getting, for example, from V.. to V..+r.,-r.,; and these may yield_ independent eigenvectors.) This has been, however, already too long a lecture,· and so we will defer these questions, along with all examples, to the next.

LECTURE 13

Representations of sl 3 C, Part II:

Mainly Lots of Examples

In this lecture we complete the analysis of the irreducible representations of sl 3 C, culminating in §13.2 with the answers to all three of the questions raised at the end of the last lecture: we explicitly construct the unique irreducible representation with given highest weight, and in particular determine its multiplicities. The latter two sections correspond to §11.2 and §11.3 in the lecture on sl 2 C. In particular, §13.4, like §11.3, involves some projective algebraic geometry and may be skipped by those to whom this is unfamiliar. §13.1: §13.2: §13.3: §13.4:

Examples Description of the irreducible representations A little more plethysm A little more geometric plethysm

§13.1. Examples This lecture will be largely concerned with studying examples, giving constructions and analyzing tensor products of representations of sl 3 C. We start, however, by at least stating the basic existence and uniqueness theorem that provides the context for this analysis. To state this, recall from the previous lecture than any irreducible, finitedimensional representation of sl 3 C has a vector, unique up to scalars, that is simultaneously an eigenvector for the subalgebra I) and killed by the three subspaces gL,-L,• 9L, -L,, and flL,- 1.,. We called such a vector n highest weigl1t vector of the representation V; its associated eigenvalue will, of course, be called the hig/1est weight of V. More generally, in any finite-dimensional representation W of s1 3 C, any vector v e W with these properties will be called a highest weight vector; we saw th<~t it will generate an irreducible sub-

176

13. Representations of si 3 C, Part II: Mainly Lots of Examples

representation V of W Finally, from the description given in the last lecture of the possible configurations of eigenvalues for a representation of si 3 C, we see that any highest weight vector must lie in the (!)-plane described by the inequalities (11~, 2 , L) ~ 0 and (11 2 , 3 , L) ~ 0, i.e., it must be of the form (a + b)L 1 + bL 2 == aL 1 - bL 3 for some pair of non-negative integers a and b. We can now state Theorem 13.1. For any IJOir of natural 1111111bers a, b there exists a unique irreducible, .finite-dimen.~ional representatioll ra,b of si 3 C with IJighest weight aL,- bL 3 • We will defer the proof of this theorem until the second section of this lecture, not so much because it is in any way difficult but simply because it is time to get to some examples. We will remark, however, that whereas in the case of si 2 C the analyRis that led to the concept of highest weight vector immediately gave the uniqueness part of the analogous theorem, here to establish uniqueness we will be forced to resort to a more indirect trick. The proof of existence, by contrast, will be very much like that of the corresponding statement for si 2 C: we will construct the representations ra,b out of the standard representation by multilinear algebra. For the time being, though, we would like to apply the analysis of the previous lecture to some of the obvious representations of sl 3 C, partly to gain some familiarity with what goes on and partly in the hopes of seeing a general multilinear-algebraic construction. We begin with the standard representation of sl 3 C on V ~ C 3 • Of course, the eigenvectors for the action of!) are just the standard basis vectors e 1 , e2 , and e 3 ; they have eigenvalues L 1 , L 2 , and L 3 , respectively. The weight diagram for Vis thus

Next, consider the dual representation V*. The eigenvalues of the dual of a representation of a Lie algebra are just the negatives or the eigenvalues of the original, so the diagram of v• is

§13.1. Examples

177

Alternatively, of course, we can just observe that the dual basis vectors et are eigenvectors with eigenvalues - L 1• Note that while in the case of si 2 C the weights of any representation were symmetric about the origin, and correspondingly each representation was isomorphic to its dual, the same is not true here (that the diagrams for V and v• look the same is a reflection of the fact that the two representations are carried into one another by an automorphism of s1 3 C, namely, the automorphism X,__. -'X). Observe also that v• is also isomorphic to the representation NV, whose weights are the pairwise sums of the distinct weights of V; and that likewise V is isomorphic as representation to N v•. Next, consider the degree 2 tensor products of V and v•. Since the weights of the symmetric square of a representation are the pairwise sums of he weights of he original, the weight diagram of Sym 2 V will look like

and likewise the symmetric square Sym 2 v• has weights {- 2L,, - L, -- t {-2L1 - 2L1, L~}:

1}

=

178

13. Representations of si 3 C, Part II: Mainly Lots of Exarnples

We see immediately from these diagrams that Sym 2 V and Sym 2 V* are irreducible, since neither collection of weights is the union of two collections arising from representations of sl 3 C. As for the tensor product V ® V*, its weights are just the sums of the weights {L1} of V with those {- L 1} ofV*, that is, the linear functionals L 1 - L1 (each occurring once, with weight vector e, ®en and 0 (occurring with multiplicity three, with weight vectors e1 ® We can represent these weights by the diagram

er ).

where the triple circle is intended to convey the fact that the weight space V0 is three dimensional. By contrast with the last two examples, this representation is not irreducible: there is a linear map

V® V* -+C given simply by the contraction l'

® u• ~-+ ( v, u*) = u*(t>)

§13.1. Examples

179

(or, in terms of the identification V® V* ~ Hom(V, V), by the tr·ace) that is a map of sl 3 C-modules (with C the trivial representation, of course). The kernel of this map is then the subspace of V ® v• of traceless matrices, which is just the adjoint representation of the Lie algebra sl 3 C and is irreducible (we can see this either from our explicit description of the adjoint representation-for example, E~, 3 is the unique weight vector for [) killed by 9L,-L,• BL,-L,• and 9L,-L1 -or, if we take as known the fact that SL 3C is simple, from the fact that a subrepresentation of the adjoint representation is an ideal in a Lie algebra, and exponentiates to a normal subgroup, cf. Exercise 8.43.) (Physicists call this adjoint representation of £;{ 3 C (or SU(3)) the "eightfold way," and relate its decomposition to mesons and baryons. The standard representation Vis related to "quarks" and v• to "antiquarks." See [S-W], [Mack].) (We note that, in general, if Vis any faithful representation of a Lie algebra, the adjoint representation wilt appear as a subrepresentation of _the tensor V®V•.) Let us continue now with some of the triple tensor products of V and v•, which will be the last specific cases we look at. To begin with, we have the symmetric cubes Sym 3 V and Sym 3 V*, with weight diagrams

and

180

13. Representations of sl 3 C, Part II: Mainly Lots of Examples

respectively. In general it is clear that, in terms of the description given in the preceding lecture of the possible weight diagrams of irreducible representations of si 3 C, the symmetric powers of V and V* will be exactly the representations with triangular, as opposed to hexagonal, diagrams. It also follows from the above description and the fact that the weights of the symmetric powers Sym"V occur with multiplicity 1 that Sym"V and Sym"l'* are all irreducible, i.e., we have, in the notation of Theorem 13.1, Sym"V =

r •. o

and

Sym"V* =

r 0 •••

By way of notation, we will often write Sym"V in place of r •. 0 . Consider now the mixed tensor Sym 2 V ® V*. Its weights are the sums of the weights of Sym 2 V-that is, the pairwise sums of the L1-with the weights of v•; explicitly, these are L 1 + L1 - Lk and 2L 1 - Li (each occurring once) and the L 1 themselves (each occurring three times, as L 1 + L1 - L1). Diagrammatically, the representation looks like

Now, we know right ofT the bat that this is not irreducible: we have a natural map

given again by contraction, that is, by the map

which is a map of s1 3 C-modules. 1 What does the kernel of this map look like? Of course, its weight diagram is 1 Another way to sec that Sym 2 V ® V* is not Irreducible is to observe tha1 if a representation W is generated by a highest weight vector v of weight 2L 1 - L 3 , as Sym 2 V® V* must be if it is irreducible, the eigenvalue L 1 can be taken with multiplicily at most 2, the corresponding eigenspace being generated by E.. 2 o £ 2•3 (11) and f 2 .J o E 1• 2(v).

§13.1. Examples

181

and we know one other thing: certainly any vector in the weight space of 2L 1 - L 3 -that is to say, of course, any multiple of the vector d ® e~-is killed by !lL,-L,• g1_,-L,• and 9L,- 1_3 , so that the kernel or 1 will contain a-n irreducible representation f = f 2 • 1 with 2L 1 - L 3 as its highest weight. Since r must then assume every weight of Ker(r), there are exactly two possibilities: either Ker(1) = r, which assumes the weights Li with multiplicity 2; or all the weights of r occur with multiplicity one and Ker(r) ~ r ffi V. How do we settle this issue? There are at least three ways. To begin with, we can try to analyze directly the structure of the kernel of 1. An alternative approach would be to determine a priori with what multiplicities the weights of r •. b are taken. Certainly it is clear that a formula giving us the latter information will be tremendously valuable-it would for one thing clear up the present confusion instantly-and indeed there exist several such, one of which, the Weyl character formula, we will prove later in the book. (We wiH also prove the Kostant multiplicity formula, which can be applied to deduce directly the independence statement we arrive at below.) /\sa third possibility, we can identify the representations ro,b as Weyl modules and appeal to Lecture 6. Rather than invoke such general formulas at present, however, we will take the first approach here. This is straightforward: in terms of the notation we have been using, the highest weight vector for the representation r c Sym 2v ® v• is the vector ® e1, and so the eigenspace rL, c: r with eigenvalue L 1 will be spanned by the images of this vector under the two compositions E 2 • 1 o E 3, 2 and E 3 , 2 o E 2 , 1 . These are, respectively,

e:

E2 , 1 o E3.2(e: ® ej)

= E 2 , dE 3, 2 (ef) ® eJ + d ® E3, 2 (e;)) = E2.1(-e~®e1) = -2(e 1 ·e 2 )®e!

+ e~ ®ef

and

E3,z o Ez,,(e~ ® eJ) = E3.2(Eu (d)® e1

+ d ® Ez,,(eJ})

= E3.2((2e 1 ·e 2 )® ej) = 2(e 1 • e3 )

® e1 - 2(e 1 • e2 ) ® ef.

182

13. Representations of si 3 C, Part II: Mainly Lots of Examples

r

Since these are independent, we conclude that the weight L 1 does occur in with multiplicity 2, and hence that the kernel of r is irreducible, i.e., Sym 2 V®

v• ~ rz.• EB v.

§13.2. Description of the Irreducible Representations At this point, rather than go on with more examples we should state some of the general principles that have emerged so far. The first and most important (though pretty obvious) is the basic Observation 13.2. If the represelllations V and W have highest weight vectors v and w witiJ weights a. and p, respectively, tiJe11 the vector v ®we V ® W is a highest weight vector of weight a. + p. Of course, there are numerous generalizations of this: the vector v" e Sym• V is a highest weight vector of weight na.. etc. 2 Just the basic statement above, however, enables us to give the PROOF oF THEOREM 13.1. First, the existence statement follows immediately from the observation: the representation Sym 6 V ® Symb v• will contain an irreducible subrepresentation f.,,, with highest weight aL 1 - bL 3 • The uniqueness part is only slightly harder (if less explicit): Given irreducible representations Vand W with highest weight a., let v e Vand we W be highest weight vectors with weight a.. Then (v, w) is again a highest weight vector in the representation V EB W with highest weight a.; let U c V EB W be the irreducible subrepre.~entation generated by (v, w). The projection maps n 1: U- V and n 2 : U- W, being nonzero maps between irreducible representations of sl 3 C, must be isomorphisms, and we deduce that V ~ W.

0 Exercise 13.3•. Let §A be the Schur functor introduced in Lecture 6. What can you say about the highest weight vectors in the representation § A(V) obtained by applying it to a given representation V? To continue our discussion of tensor products like Sym• V ® Sym& v• in general, as we indicated we would like to make more explicit the construction b, which we know to be lying in Sym" V ® Symb of the representation To begin with, we have in general a contraction map ••. &: Sym•v ® Symb v•- Sym•-• V ® Symb-l v•

r...

v•.

analogous to the map 1 introduced above; we can describe this map either (in fancy language) as the dual of the map from Sym•-• V ® Symb-1 v• to Sym• V ® Symb v• given by multiplication by the identity element in 2 One slighlly less obvious statement is this: irthe weigh Is or V are ex 1 , ex 2 , ex, ... with /(ex,) > /{a 2 ) :> ... , then/\"Jipossessesa highest weight vectorwcightex 1 + .... +a•. Notethat~incetheorderin1 or the a, rnay in fact depend on thechoiccort{even wilh the restriction a> h > con thecoefficicniS or I as above), this may in some cases imply the exi~tencc of scverul suhrcprcsent:~lious of 1\'V.

§13.2. Description or the Irreducible Representations V ® V* =

Hom(J~

V); or, concretely, by sending

(v, ..... v.) ® (vf" .... 1-+

183

vn

I: (v~o vJ>(v 1 • ••• ·i1,· ... ·11.)®(r•f · ... A*· ... ·vt).

Clearly this map is surjective, and, since the target docs not have eigenvalue aL 1 - bL 3 , the subrepresentation r •. b c Sym•v ® SymbV* must lie in the kernel. In fact, we have, just as in the case of Sym 2 V ® V* above,

Claim 13.4. Tlte kernel of tfte map '•.b is the irreduclllle representatiou

r.,,.

We will defer the proof of this for a moment and consider some of its consequences. To begin with, we can deduce from this assertion the complete decomposition or Sym• V ® Symh V*: we must have (if, say, h :5: a) b

Sym• V ® Symb V*

=

E9 r.-u-;·

(13.5)

1~o

Since we know, a priori, all the multiplicities of the eigenvalues of the tensor product Sym"V ® SymbJI*, this will, in turn, determine (inductively at least) all the multiplicities of the representations r •. b· In fact, the answer turns out to be very nice. To express it, observe first that if a ~ h, the weight diagram of either r.,b or Sym• V ® Symb V* looks like a sequence of b shrinking concentric (not in general regular) hexagons J-11 with vertices at the points (a- i)L 1 - (b - i)L 3 for i = 0, I, ... , b - I, followed (after the shorter three sides of the hexagon have shrunk to points) by a sequence of [(a- h)/3] + I triangles 1J with vertices at the points (a- b- 3J)L 1 for j = 0, 1, ... , [(a - b)/3] (it will be convenient notationally to refer to T0 as Hb occasionally). Diagram (13.6) shows the picture of the weights of Sym 6 V ® Sym 2 V*:

(13.6)

184

13. Representations of si 3 C, Part II: Mainly Lots of Examples

(Note that by the decoinposition ( 13.5), the weights of the highest weight vectors in Sym•v ® Sym 6 V* will be aL 1 - bL 3 , (a- I)L 1 - (b- l)L 3 , ... , (cr- h)L., as shown in the diagram.) An examination of the representation Sym• V ® Sym 6 v• shows that it has multiplicity (i + l)(i + 2)/2 on the hexagon H, and then a constant multiplicity (b + l)(b + 2)/2 on all the triangles 7); and it follows from the decomposition (13.5), in general, that the representation r•. 6 has multiplicity (i + I) on H, and bon 1). In English, the n111ltiplicitie.~ of r •. h increase by one on each of the concentric hexagons of the eigenvalue diagram and eire constant on the triangles. Note in particular that the description of f 2 , 1 in the preceding section is a special case of this.

PRooF OF CLAIM 13.4. We remark first that the claim will be implied by the Weyl character formula or by the description via Weyl's construction in Lecture 15; so the reader who wishes to can skip the following without dire consequences to the logical structure of the book. Otherwise, observe first that the claim is equivalent to asserting the decomposition (13.5); this, in turn, is equivalent to the statement that the representation W = Sym• V ® Symh v• has exactly b + I irreducible components (still assuming a ~ b). The irreducible factors in a representation correspond to the highest weight vectors in the representation up to scalars; so in sum the claim is equivalent to the assertion that the eigenspace W.. of Sym• V ® Symb Jl* contailts a unique ltigltest weight vector (up to scalars) if« is of tl1e form (a - i)L 1 - (b - i)L 3 fori ~ b, and none otherwise; this is what we shall prove. To begin with, the "none otherwise" part of the statement follows (given the other) just from looking at the diagram: if, for example, any of the eigenspaces W,. corresponding to a point Cl on a hexagon //1 (other than the vertex (a - i)L 1 - (b - i)L 3 of H,) possessed a highest weight vector, the multiplicity of Cl in W would be strictly greater than of( a- i)t 1 - (b- i)L 3 , which we know is not the case; similarly, the fact that the multiplicities of W in the triangular part of the eigenvalue diagram are constant implies that there can be no highest weight vectors with eigenvalue on a 1) for j ~ I. Thus, we just have to check that the weight spaces W... for Cl = (a - i)L 1 - (b - i)L 3 contain only the one highest weight vector we know is there; and we do this by explicit calculation. To start, for any monomial index I= (i 1 , i2 , i3 ) of degree ir = i, we denote by e 1 e Symi V the corresponding monomial (e~·) and define (e*) 1 e Sym'V• similarly. We can then write any element of the weight space "'{.-or.,-(b-f)LJ of Sym• J1 ® Sym 6 V* as

11

L

In these terms, it is easy to write down the action of the two operators E1, 2

§13.3. A Little More Plethysm

185

and E2 • 3 . First, E1. 2 kills both e 1 E V and

eJ E V*, so that we have

E,,2((er+b-l. e 1) ® ((eJ)H · (e*)')) = i 2(er+H.e'')®((eJ)b-l.(e*) 1)

- i 1(ej+H.e 1)®((eJt-'·(e*}"'),

where I'= (i 1 +I, i 2 - t, i3 ) and I"= (i 1 - I, i2 + 1, i 3 ) (and we adopt the convention that e 1 = 0 if i 1 < 0 for any y). It follows that rite vector v above is in the kemel of E1 , 2 if and only if tile coefficients c1 satiify i 2 c1 = (i 1 + l)c,.; and by the analogous calculation that vis in the kernel of E 2, 3 if and only if i3 c 1 = (i 2 + l)cJ whenever the indices I and J are related by j 1 = i 1 , } 2 = i 2 + I, and } 3 = i3 - l. These conditions are equivalent to saying that the numbers i 1 !i2 !i3 !c1 are independent of I. We see, in other words, that vis a highest weight vector if and only if all the coefficients c1 are equal to cji 1!i 2 !i 3 ! for some constant c. 0

§13.3. A Little More Plethysm We would like to consider here, as we did in the case of sl 2 C in Lecture II, how the tensor products and powers of the representations we have described decompose. We start with one general remark: given our knowledge of the eigenvalue diagrams of the irreducible representations of sf 3 C (with multiplicities), there can be no possible ambiguity about the decomposition of any representation U given as the tensor product of representations whose eigenvalue diagrams are known. Indeed, we have an algorithm for determining the components of that decomposition, as follows: l. Write down the eigenvalue decomposition of U. 2. Find the eigenvalue rx = aL 1 - bL 3 appearing in this diagram for which the value of l(rx) is maximal. 3. We now know that U will contain a copy of the irreducible representation r« = r.,b> i.e., U ~ f« $ U' for some U'. Since we also know the eigenvalue diagram off«, we can thus write down theeigenvaluediagram of U' as well. 4. Repeat this process for U'. To see how this goes in practice, consider some examples of tensor products of the basic irreducible representations described so far. We have already seen how the tensor products of the symmetric powers of the standard representation V of sl 3 C and symmetric powers of its dual decompose; Jet us look now at an example of a more general tensor product of irreducible representations: say V itself and the representation f 2 , 1 • We start by writing down the weights of the tensor product: since f 2 • 1 has weights 2L; - L1, L 1 + L1 - Lk, and L 1

186

13. Representations of si 3 C, Part II: Mainly Lots of Examples

(taken twice) and V has weights L1, the tensor product will have weights 3L1 - L1, 2L1 + L1 - Lt (taken twice), 2L1 (taken four times), and L1 + L; (taken live times). The diagram is thus

(One thing we may deduce from this diagram is that we are soon going to need a better system for presenting the data of the weights of a representatiol!. In the future, we may simply draw one sector of the plane, and label weights with numbers to indicate multiplicities.) We know right ofT the bat that the tensor product V ® r 2 , 1 contains a copy of the irreducible representation r 3 , 1 with highest weight 3L 1 - L 3 • By what we have said, the weight diagram of r3 • 1 is

SO the COmplement Of r~.l in the tensor prodUCt V (8) r2, 1Will)ook like

§13.3. A Little More Plethysm

187

One obvious highest weight in this representation is 2L 1 + L 2 - L 3 = L 1 - 2L 3 , so that the tensor product will contain a copy of the irreducible representation rl.l as well; since this has weight diagram

the remaining part of the tensor product will have weight diagram

188

13. Representations

or si 3 C, Part II: Mainly Lots or Examples

which we recognize as the weight diagram of the symmetric square Sym 2 V = r2 •0 of the standard representation. We have, thus, (13.7)

Exercise 13.8*. Find the decomposition into irreducible representations of the tensor products V®r1 , 1 , V®r1 , 2 and V®r3 , 1 • Can you find a general pattern to the outcomes? As in the case of si 2 C, the next thing to look at are the tensor powerssymmetric and exterior-of representations other than the standard; we look first at tensors of the symmetric square W = Sym 2 V. First, consider the symmetric square Sym 2 W = Sym 2 (Sym 2 V)). We know the diagram for Sym 2 W; it is

Now, there is only one possible decomposition of a representation whose eigenv:t.lue diagram looks like this: we must have

§13.4. A Little More Geometric Plethysm

189

Sym 2 (Sym 2 V)) ~ Sym 4 V E9 Sym 2 V*. Indeed, the presence of the Sym 4 V factor is clear: there is an obvious map rp: Sym 2 (Sym 2 V))-+ Sym 4 V

obtained simply by multiplying out. The identification of the kernel of this map with the representation Sym 2 V* is certainly less obvious, but can still be made explicit. We can identify V* with N Vas we saw, and then define a map . t: Sym 2 (N V) -• Sym 2 (Sym 2 V)) by sending the generator (u A !l) · (w " z) e Sym 2 (N J') to the element (u · w) · (v · z)- (11· z) · (v · w) e Sym 2 (Sym 2 V), which is clearly in the kernel of rp. Exercise 13.9. Verify that this map is well defined and that it extends linearly · to an isomorphism of Sym 2 (N V) with Ker(rp). Exercise 13.10. Apply the techniques above to show that the representation N(Sym 2 V) is isomorphic to r2.1• Exercise 13.1 t. Apply the same techniques to determine the irreducible factors of the representation N(Sym 2 V). Note: we will return to this example in Exercise 13.22. Exercise 13.12. Find the decomposition into irreducibles of the representations Sym 2 (Sym 3 V) and Sym 3 (Sym 2 V) (observe in particular that Hermite reciprocity has bitten the dust). Describe the projection maps to the various factors. Note: we will describe these examples further in the following section.

§13.4. A Little More Geometric Plethysm Just as in the case of si 2 C, some of these identifications can also be seen in geometric terms. To do this, recall from §11.3 the definition of the Vermrese embedding: if P 2 = PV* is the projective space of one-dimensional subspaces of V*, there is then a natural embedding of P 2 in the projective space P 3 = . P(Sym 2 V*), obtained simply by sending the point [v*] e P 2 corresponding to the vector v* e V* to the point [ v* 2 ] e P(Sym 2 V*) associated to the vector 11* 2 = v* ·v* e Sym 2 V*. The imageS c P 3 is called the Veronese surface. As in the case of the rational normal curves discussed in Lecture II, it is not hard to see that the group of automorphisms of P 3 carrying S into itself is exactly ·the group PGL 3 C of automorphisms of S = P 2 • . Now, a quadratic polynomial in the homogeneous coordinates of the space · P(Sym 2 V*) ~ ~ 3 will restrict to a quartic polynomial on the Veronese surface ' S = PV*, which corresponds to the natural evaluation map rp of the preceding · section; the kernel of this map is thus the vector space of quadratic poly-

13. Representations of !!I 3 C, Part II: Mainly Lots of Examples

190

nomials in P 5 vanishing on the Veronese surface S, on which the group of automorphisms of P 5 carrying S to itself obviously acts. Now, for any pair of points P = [u*], Q = [v*] e S, it is not hard to see that the cone over the Veronese surface with vertex the line PQ c P 5 (that is, the union of the 2-planes PQR as R varies over the surface S) will be a quadric hypersurface in P 5 containing the Veronese surrace; sending the generator u*. v* E Sym 2 V* to this quadric hypersurface will then define an isomorphism of the space of such quadrics with the projective space associated to Sym 2 V*. Exercise 13.1_1_Y.erify the statements made in the last paragraph: that the union of the PQR is a quadric hypersurface and that this extends to a linear isomorphism P(Sym 2 V*) ~ P(Ker(q7)). Verify also that this isomorphism coincides with the one given in Exercise 13.9. There is another way of representing the Veronese surface that will shed some light on this kernel. If, in terms of some coordinates e1 on V*, we think ofSym 2 V* as the vector space of symmetric 3 x 3 matrices, then the Veronese surface is just the locus, in the associated projective space, of rank I matrices up to scalars, i.e., in terms of homogeneous coordinates Z 1•1 = e1 • e1 on PS, S=

{[Z]:rank(~::: z2.3 ~::: z3,3 ~:::)=

1}.

z~, 3

The vector space of quadratic polynomials vanishing on S is then generated by the 2 x 2 minors of the matrix (Z1,J); in particular, for any pair of linear combinations of the rows and pair of linear combinations of the columns we get a 2 x 2 matrix whose determinant vanishes on S. Exercise 13.14. Show that this is exactly the isomorphism Sym 2 (NV) ~ Ker((p) described above. We note in passing that if indeed the space of quadrics containing the Veronese surface, with the action of the group PGL 3 C of motions of P 5 preserving S, is the projectivization oft he representation Sym 2 V*, then it must contain its own Veronese surface, i.e., there must be a surface T = IP'(V*) c P(Ker(q7)) invariant under this action. This turns out to be just the set of quadrics o.f r-a11k 3 containing the Veronese, that is, the quadrics whose singular locus is a 2-plane. In fact, the 2-plane will be the tangent plane to Sat a point, giving the identification T = S. Let us consider one more example of this type, namely, the symmetric cube Sym 3 (Sym 2 V)). (We promise we will stop after this one.) As before, it is easy to write down the eigenvalues of this representation; they are just the triple sums of the eigenvalues {2L, L 1 + L1 } of Sym 2 V. The diagram (we will draw here only one-sixth of the plane and indicate multiplicities with numbers rather than rings) thus looks like

§I 3.4. A Little More Geometric Plethysm

191

from which we see what the decomposition must be: as representations we have

(13.15) As before, the map to the first factor is just the obvious one; it is the identilicatit-n of the kernel that is intriguing, and especially the identification of the last factor. To see what is going on here, we should look again at the geometry of the Veronese surfaceS c P 5 = P(Sym 2 V"'). The space Sym 3 (Sym 2 V)) is just the space of homogeneous cubic polynomials on the ambient space P 5, and as before the map to the first factor of the right-hand side of (13.15) is just the restriction, so that the last two factors of (13.15) represent the vector space I(Sh of cubic polynomials vanishing on S. Note that we could in fact prove (13.15) without recourse to eigenvalue diagrams from this: since the ideal of the Veronese surface is generated by the vector space I(S)z of quadratic polynomials vanishing on it, we have a surjective map

I(S) 2 ® W

= Sym 2 V"' ® Sym 2 V-+ 1(Sh.

But we already know how the left hand side decomposes: we have Sym 2 V"' ® Sym 2 V = f 2 , 2 ffi r~,, ffi C,

(13.16)

so that I(Sh must be a partial direct sum of these three irreducible representations; by dimension considerations it can only be fu ffi C. This, in turn, tells us how to make the isomorphism ( 13.15) explicit (assuming we want to): we can define a map Sym 2 (N V) ® Sym 2 V--.. Sym 3 (Sym 2 V) by sending (u

1\

v)·(w"

z)®(s·t)~->((ll·w)·(v·z)-(ll·z)·(tJ·w))·(s·c)

192

13. Representations ofsl 3 C, Part II: Mainly Lots of Examples

and then just check that this gives an isomorphism of r 2 • 2 EB C c Sym 2 V* ® Sym 2 V with the kernel of projection on the first factor of the right-hand side of (13.15). What is really most interesting in this whole situation, though, is the trivial summand in the expression (13.15). To say that there is such a summand is to say that there exists a cubic ltypersuiface X in IP' 5 preserved under all alltomorphisms of P' carrying S to itself Of course, we have already run into this one: it is the determinant of the 3 x 3 matrix (Z1) introduced above. To express this more intrinsically, if we think of the Veronese as the set of rank I tensors in Sym 2 v•, it is just the set of tensors of rank 2 or less. This, in turn, yields another description of X: since a rank 2 tensor is just one that can be expressed as a linear combination of two rank I tensors, we see that X is the famous chordal variety of the Veronese surface: it is the union of the chords to S, and at the same time the union of all the tangent planes to S. Exercise 13.17. Show that the only symmetric powers of Sym 2 V that possess trivial summands are the powers Sym 3 l(Sym 2 V)) divisible by 3, and that the unique trivial summand in this is just the kth power of the trivial summand of Sym 3 (Sym 2 V)). Exercise 13.18. Given the isomorphism of the projectivization of the vector space /(S)z-that is, the projective space of quadric hypersurfaces containing the Veronese surface-with P(Sym 2 V*), find the unique cubic hypersurface in /(Sh invariant under the action ofPGL 3 C. Exercise 13.19. Analyze the representation Sym 2 (Sym 3 V)) of si 3 C. Interpret the direct sum factors in terms of the geometry of the Veronese embedding of PV* = P 2 in IP'(Sym 3 V*) = P 9 • Exercise 13.20*. Show that the representations Sym 4 (Sym 3 V)) and Sym 6 (Sym 3 V)) contain trivial summands, and that the representation Sym 12 (Sym 3 V)) contains two. Interpret these. Exercise 13.21. Apply the techniques above to show that the representation N(Sym 2 V) is isomorphic to r2.1· Exercise 13.22*. Apply the techniques above to analyze the representation N(Sym 2 V), and in particular to interpret its decomposition into irreducible · representations. Exercise 13.23. If IP' 5 = P(Sym 2 V*) is the ambient space of the Veronese surface, the Grassmann ian G(2, 5) of 2-planes in P 5 naturally embeds in the projective space P(N(Sym 2 V)). Describe, in terms of the decomposition in the preceding exercise, the span of the locus of tangent 2-planes to the

193

§13.4. A Little More Geometric Plcthysm

Veronese, and the span of the locus of 2-planes in P 5 spanned by the images in s of lines in p v•. Exercise 13.24•. Show that the unique closed orbit of the action of SL 3 1C on the representation r.. b is either isomorphic to P 2 (embedded as the Veronese surface) if either a or b is zero, or to the incidence correspondence

I: = { (p, /): p E I} c P 2 if neither a or b is zero.

X

P 2"

PART III

THE CLASSICAL LIE ALGEBRAS AND THEIR REPRESENTATIONS

As we indicated at the outset, the analysis we have just carried out of the structure of si 2 C and sl 3 C and their representations carries over to other semisimple complex Lie algebras. In Lecture 14 we codify this structure, using the pattern of the examples we have worked out so far to give a model for the analysis of arbitrary semisimple Lie algebras and stating some of the most important facts that are true in general. As usual, we postpone proofs of many of these facts until Part IV and the Appendices, the main point here being to introduce a unifying approach and language. The facts themselves will all be seen explicitly on a case-by-case basis for the classical Lie algebras !>I.e, sp 2 .C, and so.C, which are studied in some detail in Lectures 15-20. Most of the development follows the outline we developed in Lectures 11-13, the main goal being to describe the irreducible representations as explicitly as we can, and to see the decomposition of naturally occurring representations, both algebraically and geometrically. While most of the representations are found inside tensor powers of the standard representations, for the orthogonal Lie algebras this only gives half of them, and one needs new methods to construct the other "spin" representations. This is carried out using Clifford algebras in Lecture 20. We also make the tie with Weyl's construction of representations of G L.C from Lecture 6, which arose from the representation theory of the symmetric groups. We show in Lecture 15 that these are the irreducible representations of si.C; in Lecture 17 we show how to use them to construct the irreducible representations of the symplectic Lie algebras, and in Lecture 19 to give the nonspin representation of the orthogonal Lie algebras. These give useful descriptions of the irreducible representations, and powerful methods for decomposing other representations, but they are not necessary for the logical progression of the book, and many of these decomposilions can also be deduced from the Weyl character for.mula which we will discuss in Part IV.

LECTURE 14

The General Setup: Analyzing the Structure and Representations of an Arbitrary Semisimple Lie Algebra

This is the last of the four central lectures; in the body of it, §14.1, we extract from the examples of§ 11-13 the basic algorithm for analyzing a general semisimple Lie algebra and its representations. It is this algorithm that we will spend the remainder of Part Ill carrying out for the classical algebras, and the reader who finds the general setup confusing may wish to read this lecture in parallel with, for example," Lectures 15 and 16. In particular, §14.2 is less dearly motivated by what we have worked out so far; the reader may wish to skim it for now and defer a more thorough reading until after going through some more of the examples of Lectures 15-20. §14.1: Analyzing simple Lie algebras in general §14.2: About the Killing form

§14.1. Analyzing Simple Lie Algebras in General We said at the outset of Lecture 12 that once the analysis of the representations of sl 3 C was understood, the analysis of the representations of any semisimple Lie algebra would be clear, at least in broad outline. Here we would like to indicate how that analysis will go in general, by providing an essentially algorithmic procedure for describing the representations of an arbitrary complex semisimple Lie algebra g. The process we give here is directly analogous, step for step, to that carried out in Lecture 12 for s1 3 C; the only difference is one change in the order of steps: having seen in the case of s1 3 C the importance of the "distinguished" subalgebras s. ~ sl 2 C c g and the corresponding distinguished elements H. E s. c IJ, we will introduce them earlier here. Step 0. Verify tltat your Lie algebra is semislmple; if not, none of the following will work (but see Remark 14.3). If your Lie algebra is not semisimple, pass as indicated in Lecture 9 to its semisimple part; a knowledge of the representations of this quotient algebra may not tell you everything about

198

14. The General Set-up: Analyzing the Structure

the representations of the original, but it will at least tell you about the irreducible representations. Step I. Find an abelian subalgebra ~ c: 9 acting diagonally. This is of course the analogue of looking at the specific element H in si 2 C and the subalgebra I) of diagonal matrices in the case of sl 3 C; in general, to serve an analogous function it should be an abelian subalgebra that acts diagonally on one faithful (and hence, by Theorem 9.20, on any) representation of g. Moreover, in order that the restriction of a representation V of 9 to I) carry the greatest possible information about V, ~should clearly be maximal among abelian, diagonalizable subalgebras; such a subalgebra is called a CarLan subalgebra. Note that while this step would seem to be somewhat less than algorithmic (in particular, while it is certainly possible to tell when a subalgebra of a given Lie algebra is abelian, and when it is diagonalizable, it is not clear how to tell whether it is maximal with respect to these properties). This defect will, however, be largely cleared up in the next step (see Remark 14.3). Step 2. Let I) act 011 g by the adjoint represematlon, and decompose g accordingly. By the choice of I), its action on any representation of 9 will be diagonalizable; applying this to the adjoint representation we arrive at a direct sum decomposition, called a Cartan decompositio11, 9 = ,, EB <EB g.).

(14.1)

where the action of I) preserves each g. and acts on it by scalar multiplication by the linear functional IX E ~·; that is, for any He I) and X e o. we will have ad(H)(X) = 1X(H)· X. The second direct sum in the expression (14.1) is over a finite set of eigenvalues IX E ~·;these eigenvalues-in the language of Lecture 12, the weights of the adjoint representation-are called the roots of the Lie algebra and the corresponding subs paces g. are called the root spaces. Of course, ~ itself is just the eigenspace for the action of~ corresponding to the eigenvalue 0 (see Remark 14.3 below); so that in some contexts-such as the following paragraph, for example-it will be convenient to adopt the convention that g 0 = IJ; but we do not usually count 0 e ~* as a root. The set of all roots is usually denoted R c: ~·.

As in the previous cases, we can picture the structure of the Lie algebra in terms of the diagram of its roots: by the fundamental calculation of§ 11.1 and Lecture 12 (which we will not reproduce here for the fourth time) we see that the adjoint action of 9. carries the eigenspace 911 into another eigenspace 9•+1!· There are a couple of things we can anticipate about how the configuration of roots (and the corresponding root spaces) will1ook. We will simply state them here as Facts 14.2 (i) each root space g. will be one dimensional. (ii) R will generate a lattice 1\R. c: I)* of rank equal to the dimension o.f ~-

§14.1. Analyzing Simple Lie Algebras in General

(iii) R is symmetric al1out tlte origin, i.e., a root as well.

if a e

199 R is a root, tlren -a e R is

These facts will all be proved in general in due course; for the time being, they are just things we will observe as we do the analysis of each simple Lie algebra in turn. We mention them here simply because some of what follows will make sense only given these facts. Note in particular that by (ii), the roots all lie in (and span) a real subspace of I)*; all our pictures clearly will be of this real subspace. Remark 14.3. If indeed 0 does appear as an eigenvalue of the action of() on g/1), then we may conclude from this that I) was not maximal to begin with: by the above, anything in the 0-eigenspace of the action of() commutes with I) and (given the fact that the g. are one dimensional) acts diagonally on g, so that if it not already in 1), then I) could be enlarged while still retaining the properties of being abelian and diagonalizable. Similarly, the assertion in (ii) that the roots span !)* follows from the fact that an element of I) in the annihilator of all of them would be in the center of g. From what we have done so far, we get our first picture of the structure of an arbitrary irreducible finite-dimensional representation V or g. Specifically, V will admit a direct sum decomposition

v = EB v..

{14.4)

where the direct sum runs over a finite set of a e I)* and () acts diagonally on each V. by multiplication by the eigenvalue a, i.e., for any H e I) and v e V. we will have Jl(v) = a(H) · v.

The eigenvalues a e I)* that appear in this direct sum decomposition are called the weights of V; the v. themselves are called weight .~paces; and the dimension of a weight space v. will be called the multiplicity of the weight a in V. We will often represent V by drawing a picture of the set of its weights and thinking of each dot as representing a subspace; this picture (often with some annotation to denote the multiplicity of each weight) is called the weight diagram of V. The action of the rest of the Lie algebra on V can be described in these terms: for any root p, we have gD: V.-+

V.+P•

so we can think of the action of g6 on Vas a translation in the weight diagram, shifting each. of the dots over by p and mapping the weight spaces correspondingly. Observe next that all the weights of an irreducible representation are congruent to one another modulo the root lattice J\R: otherwise, for any weight a of V the subspace

200

14. The General Set-up: Analyzing the Structure

would be a proper subrepresentation of V. In particular, in view of Fact 14.2(ii), this means that the weights all lie in a translate of the real subspace spanned by the roots, so that it is not so unreasonable to draw a picture of them. Step 3. Find tl1e distinguished subalgebras s. ~ si 2 1C c g. As we saw in the example of si 3 1C, a crucial ingredient in the analysis of an arbitrary irreducible finite-dimensional representation is the restriction of the representation to certain special copies of the algebra si 2 1C contained in g, and the application of what we know from Lecture II about such representations. To generalize this to our arbitrary Lie algebra g, let g. c g be a root space, one dimensional by (i)ofFact 14.2. Then by (iii) of Fact 14.2, there is another roots pace !.J-o c g; and their commutator [g., 9-.J must be a subspace of g0 = ~. of dimension at most one. The adjoint action of the commutator [g., g_.] thus carries each of g. and g_. into itself; so that the direct sum s. =g. E9 g_. E9 [g•• g_.J

(14.5)

Is a subalgebra of g. The structure of s. is not hard to describe, given two further facts that we will state here, verify in cases, and prove in general in Appendix D. Facts 14.6. (i) [g., g_.] -# 0; and (ii) [[g., g_.J, g.] .,. 0.

Given these, it follows that the subalgebra s. is isomorphic to s1 2 1C. In particular, we can pick a basis X. e g., Y, e g_., and H. e [g., g_.J satisfying the standard commutation relations (9.1) for si 2 1C; x. and Y,. are not deter· mined by this, but H. is, being the unique element of [g., g_.] having eigen· values 2 and -2 on g. and g_., respectively [i.e., H. is uniquely characterized by the requirements that H. e [g., g_.] and cx(H.) = 2.] Step 4. Use the integrality of the eigenvalues of the H•. The distinguished elements H. e I) found above are important first of all because, by the analysis of the representations of si 2 1C carried out in Lecture 9, in any representation of s.-and hence in any representation of g-all elgenval~es of the action of H. must be integers. Thus, every eigenvalue Pe fj• of every representation of g must assume integer values on all the H•. We correspondingly let Aw be the set of linear functionals p e ~* that are integer valued on all the H.; Aw will be a lattice, called the weigllt lattice of g, with. the property that

all weights of all representations of g will lie In Aw. Note, in particular, that R c Aw and hence AR c Aw; in fact, the root lattice will in general be a sublattice of finite index in the weight lattice. Step 5. Use the .~ymmetry of tl.e eigenvalues of the H•. The integrality of the

§14.1. Analyzing Simple Lie Algebras in General

201

eigenvalues of the H. under any representation is only half the story; it is also true that they are symmetric about the origin in l. To express this, for any oc we introduce the involution W. on the vector space I)* with + 1-eigenspace the hyperplane

n.. = {P e 1,*: (H.., P> = o} and minus I eigenspace the line s~anned by oc itself.' In English, reflection in the plane n.. with axis the line spanned by oc:

2/J(//,.) W.,(p) = fJ - oc(H.) IX

= fJ -

(14.7)

W.

is the

(14.8)

{J(H.)oc.

Let Ill\ be the group generated by these involutions; Ill\ is called the Weyl group of the Lie algebra g. Now suppose that Vis any representation of g, with eigenspace decomposition V = E9 v,. The weights p appearing in this decomposition can then be broken up into equivalence classes mod oc, and the direct sum (14.9) of the eigenspaces in a given equivalence class will be a subrepresentation of V for 5 4 • It follows then that the set of weights of V congruent to any given Pmod oc will be invariant under the involution W.; in particular,

The set of weights of a11y representatioll of g is i11varia11t under the Weyl group. To make this more explicit, the string of weights that correspond to nonzero summands in (14.9) are, possibly after replacing fJ by a translate by a multiple of oc:

p, p + oc, P+ 2oc, ... , p + moe,

with m = - {J(H.).

(14.10)

(Note that by our analysis of si 2 1C this must be an uninterrupted string.) Indeed if we choose p and m ~ 0 so that (14.10) is the string corresponding to nonzero summands in (14.9), then the string of integers

fJ(H,.), ({J

+ «)(H,.) = {J(H.) + 2, ... , (p + moc)(H.) = {J(H.. ) + 2m

must be symmetric about zero, so {J(H.) = - m. In particular,

W.(P + koc) = P + (- {J(H.) - k)oc = P+ (m - k)oc. Note also that by the same analysis the multiplicities of the weights are invariant under the Weyl group. We should mention one other fact about the Weyl group, whose proof we :also postpone: Note that by the nondegeneracy assertion (ii) or Fact 14.6, the line C·« does not lie in the , hyperplane n•. Recall that ( , ) is the pairing between~ and 1) 0 , so (H., {I) = {1(11.).

-' 1

202

14. The General Set-up: Analyzing the Structure

Fact 14.11. Every element of the Weyl group is induced by an auto111orplrism of the Ue algebra g carrying I) to itself. We can even say what automorphism of odoes the trick: to get the involution ~. take the adjoint action of the exponential exp(ntU.) E G, where G is any group with Lie algebra g and is a suitable element of the direct sum of the root spaces g. and 9-o· To prove that Ad(exp(niU.)) actually does this requires more knowledge of g than we currently possess; but it would be an excellent exercise to verify this assertion directly in each of the cases studied below. (For the general case see (23.20) and (26.15).) Step 6. Draw the picttlre (optional). While there is no logical need to do so at this point, it will be much easier to think about what is going on in ~* if we introduce the appropriate inner product, called the Killing form, on g (hence by restriction on ~. and hence on 1)*). Since the introduction of the Killing form is, logically, a digression, we will defer until later in this lecture a discussion of its various definitions and properties. It will suffice for now to mention the characteristic property of the induced inner product on~*: up to scalars it is the unique inner product on~· preserved by the Weyl group, i.e., in terms of which the Weyl group acts as a group of orthogonal transformations. Equivalently, it is the unique inner product (up to scalars) such that the line spanned by each root a e I)* is actually perpendicular to the plane n. (so that the involution ~ is just a refiection in that hyperplane). Indeed, in practice this is most often how we will compute it. In terms of the Killing form, then, we can say that tlte Weyl group is just the group generated by the reflections itl the hyperplanes perpendicular to the roots of the Lie algebra. Step 7. Choose a direction in lJ*. By this we mean a real linear functional/ on the lattice A11. irrational with respect to this lattice. This gives us a decomposition of the set

u.

(14.12) where n+ = {a: /(a)> 0} (the IX E R+ are called the positive roots, those in Rnegatlve); this decomposition is called an orderl11g of tlte roots. For most purposes, the only aspect of I that matters is the associated ordering of the roots. The point of choosing a direction-and thereby an ordering of the roots R = R + v R- -is, of course, to mimic the noiion of highest weight vector that was so crucial in the cases of si 2 1C and sl 3 1C. Specifically, we make the Definition. Let V be any representation of g. A nonzero vector v E V that is both an eigenvector for the action of~ and in the kernel of g. for alia e R+ is called a highest weight vector of V. Just as in the previous cases, we then have Proposition 14.13. For any semisirnple complex Lie algebra g, (i) every finite-climensional repre.~entatim• V (~{ !l possesses a higlrest weiglrt

vector;

§14.1. Analyzing Simple Lie Algebras in General

203

(ii) the subspace W of V generated by the images of a highest weight vector v under successive app/icatimrs of root spaces !lp for PeR- is an irreducible subrepresentation; (iii) an irreducible representation possesses a unique higltest weight vector up to scalars. PROOF. Part (i) is immediate: we just take o: to be the weight appearing in V for which the value l(o:) is maximal and choose v any nonzero vector in the weight space v;,. Since V.+ll = (0) for all fJ e R+,such a vector v will necessarily be in the kernel of all root spaces g, corresponding to positive roots p. Part (ii) may be proved by the same argument as in the two cases we have already discussed: we let W,. be the subspace spanned by allw. · v where w. is a word of length at most n in elements of g11 for negative (J. We then claim that for any X in any positive root space, X· W,. c:: w•. To see this, write a generator of W, in the form Y · w, we W,_ 1 , and use the commutation relation X· Y · w = Y·X·w +[X, Y]·w; the claim follows by induction, since [X, Y] is always in 1). The subspace W c:: V which is a union of all the W,'s is thus a subrepresentation; to see that it is irreducible; note that if we write W = W' $ W", then either W' or W' will have to contain the one-dimensional weight space W,., and so will have to equal W. The uniqueness of the highest weight vector of an irreducible representation follows immediately: if v e V. and we V11 were two such, not scalar multiples of each other, we would have l(o:) > l(P) and vice versa. 0 Exercise 14.14. Show that in (ii) one need only apply those g11 for which g11 • v +- 0. (Note: with W,. defined using only these g11 , and X in any root space, the same inductive argument shows that X· JV., c:: W..+t· On the other hand, if one uses all g11 with fJ negative and primitive, as in Observation 14.16, then X· W, c:: W.-t· One cannot combine these, however: V may not be generated by successively applying those g11 with Pnegative, primitive, and n11 ·v +- 0, e.g., the standard representation of si 3 C.) The weight o: of the highest weight vector of an irreducible representation will be called, not unreasonably, the ltigilest weight of that representation; the term dominant weight is also common. We can refine part (ii) of this proposition slightly in another direction; this is not crucial but will be useful later on in estimating multiplicities of various representations. This refinement is based on Exercise 14.15*. (a) Let o: 1 ,

••• ,

o:~

be roots of a semisimple Lie algebra g and

o., c:: g the corresponding root spaces. Show that the subalgebra of g generated by the Cartan subalgebra ~ together with the o«, is exactly the direct sum I)$ (ffig,.), where the direct sum is over the intersection of the set R of roots of g with the semigroup N {o: 1 , ••• , o:k} c:: I) generated by the o:1• · ; (b) Similarly, let o: 1 , ••• , o:k be negative roots of a semi simple Lie algebm !1 and g,., c:: othe corresponding root spaces. Show that the subalgebra of g gene-

14. The General Set-up: Analyzing the Structure

204

E9

fia• where the direct sum is over the rated by then~, is exactly the direct sum intersection of the set R of roots of g with the semi group N {oc 1 , ... , «t} c I) generated by the oc1• (Note that by the description of the adjoint action of a Lie algebra on itself we have an obvious inclusion; the problem here is to show-given the facts above-that if oc + p E R, then [g~, g11 ] .f. 0.)

From this exercise, it is clear that generating a subrepresentation W of a given representation V by successive applications of root spaces g11 for PeRto a highest weight vector v is inefficient; we need only apply the root spaces g11 corresponding to a set of roots p generating R- as a semigroup. We accordingly introduce another piece of terminology: we say that a positive (resp., negative) root oc e R is primitive or simple if it cannot be epxressed as a sum of two positive (resp. negative) roots. (Note that, since there are only finitely many roots, every positive root can be written as a sum of primitive positive roots.) We then have Observation 14.16. A11y Irreducible representation Vis generated by the images of its highest weight vector v under successive applications of root spaces g11 where (I ranges over the primitive negative roots. We have already seen one example of this in the case of si 3 1C, where we observed (in the proof of Claim 12.10 and in the analysis of Sym 2 V® V* in Lecture 13) that any irreducible representation was generated by applying the two elements E 2, 1 e 9L,-t., and E 3 , 2 e 9L,-L, to a highest weight vector. To return to our description of the weights of an irreducible representation V, we observe next that in fact every vertex of tlte cmtvex hull of the weights of V must be cmtjugate to oc under the Weyl group. To see this, note that by the above the set of weights is contained in the cone oc + c;, where c; is the positive real cone spanned by the roots (I e R- such that g11 (v) #< 0-that is, such that oc(/111 ) #- 0. Conversely, the weights of V will contain the string of weights IX,

0(

+ p, + zp, ... ' + (-oc(H,))P 0(

0(

(14.17)

for any PeR-. Thus, any vertex of the convex hull of the set of weights of V adjacent to oc must be of the form

a - a(Hp)p

= »j(oc)

for some (I; applying the same analysis to each successive vertex gives the statement. From the above, we deduce that the set of weights of V will lie in the convex hull of the images of a under the Weyl group. Since, moreover, we know that the intersection of this set with any set of weights of the form {P + ny} will be a connected string, it follows that the set of weights of V will be exactly the weights that are congruent to oc modulo the root lattice i\R and tlrat lie i11 the con11ex hull of the images of a Ull(/er the Weyl grmiJJ.

§14.1. Analyzing Simple Lie Algebras in General

205

One more bit of terminology, and then we are done. By what we have seen (cf. (14.17)), the highest weight of any representation of V will be a weight a satisfying a(H7 ) ~ 0 for every y e R +. The locus if", in the real span of the roots, of points satisfying these inequalities-in terms of the Killing form, making an acute or right angle with each of the positive roots-is called the (closed) Weyl chamber associated to the ordering of the roots. A Weyl chamber could also be described as the closure of a connected component of the complement of the union of the hyperplanes n•. The Weyl group acts simply transitively on the set of Weyl chambers and likewise on the set of orderings of the roots. As usual, these statements will be easy to see in the cases we study, while the abstract proofs are postponed (to Appendix D). Step 8. Classify the irreducible, finite-dimensional representations of g. Where all the above is leading should be pretty clear; it is expressed in the fundamental existence and uniqueness theorem:

Theorem 14.18. For any a in the Intersection of tl1e Weyl cltamber· if" associated to the ordering ~f tlte roots with tire weight lattice 1\w, there exists a rmique irreducible, finite-dimensional represelltatioll r~ of 9 with highest weight a; tit is gives a bijection between ir· n A 11• and the set of irredllcible representations of g. The weights of r. will consist of those elements of the weight lattice congruent to !X modulo the root lattice 1\ R and lying in the convex lrull of the set of points in I)* conjugate to a under the Weyl group. HALF-PROOF. We will give here just the proof of uniqueness, which is easy. The existence part we will demonstrate explicitly in each example in turn; and later ott we will sketch some of the constructions that can be made in general. The uniqueness part is exactly the same as for s1 3 C. If V and W are two irreducible, finite-dimensional representations of g with highest weight vectors v and w, respectively, both having weight a, then the vector (v, w) e V ~ W will again be a highest weight vector of weight a in that representation. Let U c V~ W be the subrepresentation generated by (v, w); since U will again be irreducible the projection maps n 1 : U-> V and n 2 : U -> W, being nonzero, will have to be isomorphisms. 0

Another fact which we will see as we go along-and eventually prove in general-is that there are always fundamental weights oJ 1 , ••• , w. with the property that any dominant weight can be expressed uniquely as a nonnegative integral linear combination of them. They can be characterized geometrically as the first weights met along the edges of the Weyl chamber, or algebraically as those element5 w 1 in I)* such that w,(H.) = J1, 1, where a1, ••. , a. are the simple roots (in some order). When we have found them, we often write ra,, .... a. for the irreducible representation with highest weight a1w 1 + · ·· + a.w.; i.e., As with most of the material in this section, general proofs will be found in Lecture 21 and /\ppendix D.

206

14. The General Set-up: Analyzing the Structure

One basic point we want to repeat here (and that we hope to demonstrate in succeeding lectures) is this: that actually carrying out this process in practice is completely elementary and straightforward. Any mathematician, stranded on a desert island with only these ideas and the definition of a particular Lie algebra g such as sl.C, so.c, or sp 2.C, would in short order have a complete description of all the objects defined above in the case of g. We should say as well, however, that at the conclusion of this procedure we are left without one vital piece of information about the representations of g, without which we will be unable to analyze completely, for example, tensor products of known representations; this is, of course, a description of the multiplicities of the basic representations r... As we said, we will, in fact, describe and prove such a formula (the Weyl character formula); but it is of a much less straightforward character (our hypothetical shipwrecked mathematician would have to have what could only be described as a pretty good day to come up with the idea) and will be left until later. For now, we will conclude this lecture with the promised introduction to the Killing form.

§14.2. About the Killing Form As we said, the Killing form is an inner product (symmetric bilinear form) on the Lie algebra g; abusing our notation, we will denote by B both the Killing form and the induced inner products on l) and ~·· B can be defined in several ways; the most common is by associating to a pair of elements X, Y E g the trace of the composition of their adjoint actions on g, i.e., B(X, Y) = Tr(ad(X) o ad(Y): 9-+ g).

(14.19)

As we will see, the Killing form may be computed in practice either from this definition, or (up to scalars) by using its invariance under the group of automorphisms of g. We remark that this definition is not as opaque as it may seem at first. For one thing, the description of the adjoint action of the root space g.. as a "translation" of the root diagram-that is, carrying each root space g11 into 9.. +11 -tells us immediately that !k is perpendicular to g11 for all Pother than -ex; in other words, the decomposition

EB

>)

g = '' E11 ( «aR""
(14.20)

is orthogonal. As for the restriction of B to~. this is more subtle, but it is not hard to write down: if X, Yare in~. and z.. generates g., then ad(X) o ad( Y)(Z.) = cx(X)cx(Y)Z., so B(X, Y) = I;cx(X)cx(Y), the sum over the roots; viewing Bl~ as an element of the symmetric square Sym 2 (1J*), we have

(14.21)

§14.2. About the Killing Form

~·

207

A key fact following from this-one that, if nothing else, makes picturing with the inner product B involve less eyestrain-is

(14.22) B is positive definite on the real s11bspace {11 a E R}.

of~

spanned by tile vectors

11 :

Indeed, all roots take on real values on this space (since all a(H11 ) E 7l c R), so for H in this real subspace of~. B(H, II) is non-negative, and is zero only when all a(H) = 0, which implies H = 0, since the roots span 1)•. To see that the Killing form is nondegenerate on all of g, we need the useful identity: (14.23) B([X, Y], Z) = B(X, [Y, Z]) for all X, Y, Z in g. This follows from the identity Trace((XY- YX)Z) = Trace(X(YZ - ZY)) for any endomorphisms X, Y, Z of a vector space. And this, in turn, follows from Trace(YXZ- XZY) =Trace([ f. XZJ) = o. An immediate consequence of (14.23) is that if a is any ideal in a Lie algebra g, then its orthogonal complement a.i with respect to B is also an ideal. In particular, if g is simple, the kernel of B is zero (note that the kernel cannot be g since it does not contain l)). Since the Killing form of a direct sum is the sum of the Killing forms of the factors, it follows that the Killing form is nondegenerate on a semisimple Lie algebra g. One ofthe reasons the Killing form helps to picture~· is the fact mentioned above:

Proposition 14.24. With respect to B, the line spanned by each root a is perpendicular to the hyperplane n•. As we observed, this is equivalent to saying that the involutions W. above are simply reflections in hyperplanes, and in turn to saying that the whole Weyl group is orthogonal. Note also that Proposition 14.24 thereby follows immediately from the Fact 14.11: from the definition of B above, it is clearly invariant under any automorphism of g. Nevertheless, we would prefer not to rely on this fact; and anyway giving· a direct proof of the proposition is not hard, in terms of the picture we have of the adjoint action of g on itself. To prove the assertion a .1. n.., it suffices to prove the dual assertion that H .1. H. for all H in the annihilator of a. But now by construction 11. is the commutator [X., Y,] of an element x .. e g. and an element Y. e 9-.· Using (14.23) we have for any II in ~. B(H.. , H) = B([X., Y,], H) = B(X., [Y,, H]) = B(X., a(H) Y.) = a(H)B(X., f.),

which vanishes since a(I/) = 0.

(14.25)

14. The General Set-up: Analyzing the Structure

208

Note that as a consequence of this, we can characterize the Weyl chamber associated to an ordering of the roots as exactly those vectors in the real span ofthe roots forming an acute angle with all the positive roots (or, equivalently, with all the primitive ones); the Weyl chamber is thus the cone whose faces lie in the hyperplanes perpendicular to the primitive positive roots. Equation (14.25) leads to a formula for the isomorphism of I) with I)* determined by the Killing form. First note that for H = H~~. it gives B(I-111 , H~~.) = 2B(X11 , Y.) #- 0,

for if B(X~~., Y.) were zero we would have B(H11 , H) = 0 for all H, contradicting the nondegeneracy of Bon~· The element 7;. ofT) which corresponds to ex e I)* by the Killing form is by definition the element of lJ that satisfies the condition B(1;., H) = ex(/i)

for all H

E

(14.26)

IJ.

Looking at (14.25), we see that 7;. = H~~./B(X11 , Y.) = 2H~~./B(H11 , H11 ). This proves Corollary 14.27. The isomorphism of IJ• and IJ determined by rite Kllli11g form B carries ex to 7;. = (2/B(H11 , H~~.))· J-1 11 • The Killing form on

r,• is defined by B(ex, P) = B(T., 7p).

Exercise 14.28. Show that the inverse isomorphism from I) to (2/B(ex, ex))· ex.

I,• takes I-1~~. to

The orthogonality of W.. can be expressed by the formula W.(f1) = p _ 2B(p, ex) ex. B(ex, IX)

Comparing with (14.8) this says: Corollary 14.29. If IX and Pare roots, tlren 2B(P, 1X)/B(1X, IX) = P(H~~.)

is an integer.

By the above identification of I) with I)*, (14.22) translates to Corollary 14.30. Tile Kilfing form B is positive definite on the real vector space spanned by the root lattice AR. Note that it follows immediately from (14.22) that the Weyl group liD is finite, being simultaneously discrete (!ill preserves the set R of roots of g and hence the lattice AR; it follows that !ill can be realized as a subgroup of GL.Z)

§14.2. About the Killing Form

209

and compact (lffi preserves the Killing form, and hence is a subgroup of the orthogonal group O.R.) Alternatively, lffi is a subgroup of the permutation group of the set of roots. As we observed, the Killing form on l)* is preserved by the Weyl group. In fact, in case g is simple, the Killing form is, up to scalars, the unique inner product preserved by the Weyl group. This will follow from Proposition 14.31. The space 1)* is an irreducible represelltation of tire Weyl group m. PROOF. Suppose that 3 c ~· were preserved by the action of !lB. This means that every root ex e ()*of g will either lie in the subspace 3 or be perpendicular to it, i.e., for every ex e 3 and f1 ~ 3 we will have fJ(H«) = 0. We claim then that the subspace g' of g spanned by the subalgebl'as {!;«}«~~will be a11 ideal in g. Clearly it will be a subalgebra; the space spanned by the distinguished subalgebras !;«corresponding to the set of roots lying in any subspace of~· will be. To see that it is in fact an ideal, let Y e g11 be an element of a root space. Then for any ex e 3, we have

[Y,Z]eg«+/1=0 since ex

+ f1 is neither in 3 nor perpendicular to it, and so cannot be a root; and [Y, I-1«] = -[H«, Y] = fJ(H«)· Y = 0.

Thus, ad(Y) kills g'; since, of course, all of H itself will preserve g', it follows that g' is an ideal. Thus, either all the roots lie in 3 and so 3 =~·.or all roots 0 are perpendicular to 3 and correspondingly 3 = (0). Note that given Fact 14.11, we can also express the last statement by saying that (in case g is simple) the Killing form on I) is the unique form preserved by every automorphism of the Lie algebra g carrying l) to itself. As we will see, in practice this is most often how we will first describe the Killing form. Exercise 14.32. Find the Killing form on the Lie algebras si 2 C and sl 3 C by explicit computation, and verify the statements made above in these cases. Exercise 14.33*. If a semisimple Lie algebra is a direct sum of simple sub· algebras, then its Kiliing form is the orthogonal sum of the Killing forms of the factors. Show that, conversely, if the roots of a semisimple Lie algebra lie in a collection of mutually perpendicular subspaces, then the Lie algebra decomposes accordingly. Exercise 14.34"'. Suppose g is a Lie algebra that has an abelian subalgebra 1) such that g has a decomposition (14.1), satisfying the conditions of Facts 14.2 and 14.6. Show that g is semisimple, and l) is a Cartan subalgebra.

210

14. The General Set-up: Analyzing the Structure

The preceding exercise can be used instead of Weyl's unitary trick or any abstract theory to verify that the algebras we meet in the next few lectures are all semisimple. It is tempting to call such a Lie algebra "visibly sernisimple." The discussion of the geometry of the roots of a semisimple Lie algebra will be continued in Lecture 21 and completed in Appendix D. The Killing form becomes particularly useful in the general theory; for example, solvability and semisimplicity can both be characterized by properties of the Killing form (see Appendix C).

EB.>o

Exercise 14.Js•.Show that b = IJ Ee g. is a maximal solvablcsubalgcbra of g; b is called a Borel subalgebra. Show thai E£).> 0 g. is a maximal nilpotent subalgebra of g. These will be discussed in Lecture 25. Exercise 14.36•. Show that the Killing form on the Lie algebra gl., is given by the formula B(X, f) = 2m Tr(X

o

f) - 2 Tr(X) Tr( f).

Find similar formulas for sl.,, so.,, and sp.,, showing in each case that B(X, f) is a constant multiple of Tr(X o f). · Exercise 14.37. If G is a real Lie group, the Killing form on its Lie algebra 9 = T.G may.not be positive definite. When it is, it determines, by lefi translation, a Riemannian metric on G. Show that the Killing form is positive definite for G = SO.IR, but not for SL.IR.

LECTURE 15

sl 4 C and sl" C

In this lecture, we will illustrate the general paradigm of the previous lecture by applying it to the Lie algebras ti.C; this is typical of the analyses of specific Lie algebras carried out in this Part. We start in §15.1 by describing the Cartan subalgebra, roots, root spaces, etc., for si.C in general. We then give in §15.2 a detailed account of the representations of sl 4 C, which generalizes direclly to sl.C; in particular, we deduce the existence part of Theorem 14.18 for si;,C. In §15.3 we give an explicit construction of the irreducible representations of si.C using the Weyl construction introduced in Lecture 6; analogous constructions of the irreducible representations of the remaining classical Lie algebras will be given in §17.3 and §19.5. This section presupposes familiarity with Lecture 6 and Appendix A, but can be skipped by those willing to forego §17.3 and 19.5 as well. Section 15.4 requires essentially the same degree of knowledge of classical algebraic geometry as§§ 11.3 and 13.4 (it does not presuppose §15.3~ but can also be skipped. Finally, §15.5 describes representations ofGL.C; this appears to involve the Weyl construction but in fact the main statement, Proposition 15.47 (and even its proof) can be understood without the preceding two sections. §15.1: §15.2: §15.3: §15.4: §15.5:

Analyzing si.C Representations of si 4 C and si.C Weyl's construction and tensor products Some more geometry Representations ofGL.C

§15.1. Analyzing sine To begin with, we have to locate a Cartan subalgebra, and this is not hard; as in the case of 1l 2 C and sl 3 C the subalgebra of diagonal matrices will work line. Writing H 1 for the diagonal matrix E 1• 1 that takes e 1 to itself and kills e1

212

for j #: i, we have I)= {a 1 H 1

+ a 2 H 2 + ... + a.H.: a 1 + a 2 + ... +a.= 0};

note that H 1 is not in I). We can correspondingly write I)• = C{L, L 2 ,

... ,

+ L 2 + ... + L. =

L.}/(L 1

0),

where L 1(H1 ) = /51•1. We often write L 1 for the image of L, in I)•. We have already seen how the diagonal matrices act on the space of all traceless matrices: if E 1,1 is the endomorphism of C" carrying e1 to e1 and killing e~ for all k #: j, then we have

ad(a 1 H 1

+ a 2 /1 2 + .. · + a.II.)(E1, 1) =

(a, - a1) • E1•1;

(15.1)

or, in other words, E1,1 is an eigenvector for the action of ~ with eigenval11e L 1 - L1; in particular, the roots of si.C are just the fJairwise differences oft he L 1• Before we try to visualize anything taking place in 1) or I)•, let us take a moment out and describe the Killing form. To this end, note that the automorphism (p ofC" sending e1 to e1, e1 to -e1and fixing ek for all k t: I, j induces an automorphism Ad(q7) of the Lie algebra si.C (or even gl.(C)) that carries I) to itself, exchanges H, and H1, and fixes all the other H~. Since the Killing form on I) must be invariant under all these automorphisms, it must satisfy B(L, L 1) = B(L1, L1) for all i and j and B(L,. L~) = B(L1, Lk) for all i, j and k #: i, j; it follows that on I) it must be a linear combination of the forms

B'(L a H;, L h H 1

=

1)

1

La b

1 1

and

B"(L a,H, L h H 1

{L

L

I:b

'l.J,.

1)

=

l: 1,. 1 a,h1.

On the space a1H1: a 1 = 0}, however, we have 0 =
=

b2

= ···

=b. implies that it must be a

(15.2)

=

I I: a,h, - If-I: aA I

1,)

We may, of course, also calculate the Killing form directly from the definition. By (14.21), since the roots ofsi.C are {L1 - L1 },n, we have B(L a,H,

L bJI

1)

=

I:

1,. 1 (a 1 -

a1)(b1 -

= Lt l: 1,. 1 (a,b,

b1)

+ a1b1 -

a 1b1 - a1b1).

Noting that l:1 ,., a1 = - a1 and, similarly, L: 1.. , h1 = - h1, this simplifies to

§15.1. Analyzing si.C

213

B(L a;Hit L b1H1) = 2n L aA

(15.3)

It follows with a little calculation that the dual form on !)• is

B(L a1 L 1, L b1L 1) = (l/2n)(I 1 a,b1 - (1/n) L 1•1 a1b1 ).

(15.4)

It is probably simpler just to think of this as the form, unique up to scalars, invariant under the symmetric group 6. of permutations of {I, 2, ... , u}. The L, therefore, all have the same length, and the angles between all pairs are the same. To picture the roots in ()•, then, we should think of the points L 1 as situated at the vertices of a regular (11 - I )-simplex !1, with the origin located at the barycenter of that simplex. This picture is easiest to visualize in the special case 11 = 4, where the L 1 will be located at every other vertex of a unit cube centered at the origin:

(15.5)

Now, as we said, the roots of si.C are now just the pairwise differences of the L 1• The root lattice 1\R they generate can thus be described as 1\R

=

{L a1L 1: a1 E Z, La, = 0}/(L L 1 = 0).

Both the roots and the root lattice can be drawn in the case of s1 4 C: if we think of the vectors L 1 E !)• as four of the vertices of a cube centered at the origin, the roots will comprise all the mid points of the edges of a second cube whose linear dimensions are twice the dimensions of the first:

_..-----··------,., LI

-

L2

214

15. sl 4 C and si.C

The next step, finding the distinguished subalgebras s., is also very easy. The root space 91.,-L, corresponding to the root L 1 - L1 is generated by E1,1, so the subalgebra sL,-L, is generated by E1, 1,

E1, 1,

and

[E 1,1, E1,,] = H1 - H1 •

The eigenvalue of H1 - H1 acting on E1, 1is (L1 - L1)(111 - H1) = 2, so that the corresponding distinguished element H 1.,-L, in I) must be just H 1 - H1. The annihilator, of course, is the hyperplane 01.,-1., = f2: a 1L1: a1 = a1 }; note that this is indeed perpendicular to the root L 1 - L 1 with respect to the Killing form Bas described above. Knowing the H. we know the weight lattice: in order for a linear functional 1L 1 E l)* to have integral values on all the distinguished elements, it is clearly necessary and sufficient that all the a1 be congruent to one another modulo 7L. Since 2:L1 = 0 in l)*, this means that the weight lattice is given as

L:a

A,.= 7L{L., ... , L"}/(L L, = 0). In sum, then, the weight lattice of si.C may be realized as tire lattice generated by the vertices of a regular (n - I)-simplex I! centered at the origin; and the roots as the pairwise dijfere11ces of these vertices. While we are at it, having determined ARandA,. we might as well compute the quotient A,./AR. This is pretty easy: since the lattice A,. can be generated by AR together with any of the vertices L 1 of our simplex, the quotient AwlAR will be cyclic, generated by any L 1; since, modulo AR, 0=

L

(L1 - l,1)

= nL1 - L

L1 = nL 1.

we see that L 1 has order dividing ninA,. fAR. Exercise i5.7. Show that L, has order exactly ninA,. fAR, so that A,./AR ~ 7Lfn7L. From the above we can also say what the Weyl group is: the reflection in the hyperplane perpendicular to the root L 1 - L 1 will exchange L 1 and L 1 E ~· and leave the other Lt alone, so that the Weyl group llD is just the group 6., acting as tire symmetric group on the ge11erators L 1 of~·· Note that we have already verilied that these automorphisms off)* do <;orne from automorphisms of the whole Lie algebra sl.C preserving 1). To continue, let us choose a direction, and describe the corresponding Weyl chamber. We can write our linear functional/ as

I(L a L

1 1)

=

Lca

1 1

with I;c1 = 0; let us suppose that c 1 > c 2 >···>c•. The corresponding ordering of the roots will then be

R+ and

=

{L,- L1: i < j}

§15.1. Analyzing9I.C

215

The primitive negative roots for this ordering are simply the roots L1+1 - L;. (Note that the ordering of the roots depends only on the relative sizes of the c, so that the Weyl group acts simply transitively on the set of orderings.) The (closed) Weyl chamber associated to this ordering will then be the set if/=

{L a L

1 1:

a 1 ~ a 2 ~···~a.}.

One way to describe this geometrically is to say that if we take the barycentric subdivision of the faces of the simplex !1, the Weyl chamber will be the cone over one (n - 2)-simplex of the barycentric subdivision: e.g., in the case 11 = 4

L

1 +L 2 ~-L 3 ·L•

It may be easier to visualize the case 11 = 4 if we introduce the associated cubes: in terms of the cube with vertices at the points ± L 1, we can draw the Weyl chamber as

(15.8)

.1------

/

/

/

216

Alternatively, in terms of the slightly larger cube with vertices at the points ±2L~o we can draw if' as

(15.9)

I

' I

I I

From the first of these pictures we see that the edges of the Weyl chamber are the rays generated by the vectors L 1 , L 1 + L 2 , and L 1 + L 2 + L 3 ; and that the faces of the Weyl chamber are the planes orthogonal to the primitive negative roots L 2 - L,, i 3 - L 2 , and L 4 - L 3 . The picture in general is analogous: for sr.c, the Weyl chamber will be the cone over an (11 - 2)simplex, with edges generated by the vectors L 1,

L 1 + L 2,

L 1 + L2 + L 3,

••• ,

L 1 + ···

+ L.- 1 =

-L•.

The faces of if/" will thus be the hyperplanes fl 1.,-L.. ,

= fL a1L1: a1 = al+ 1 }

perpendicular to the primitive negative roots LrH - L1• Note the important phenomenon: the intersection of the closed Weyl chamber with the lattice A". will be a free semigroup N"- 1 generated by the fundamental weightsw1 = L, + ·· · + L1 occurringalong theedgesofthe Weyl chamber. One aspect of its significance that is immediate is that it allows us to index the irreducible representations sl.C nicely: for an arbitrary (n - I)tuple of natural numbers (a 1 , ••• , a._ 1 ) e N"- 1 we will denote by ra, ..... a._, the irreducible representation of si.C with highest weight a 1 L 1 + a 2 (L 1 + L 2 ) + · ·· + a._,(L 1 + ·· · + L.-d = (a 1 + · ·· + a.-dL 1 + (a 2 + · ·· + a._ 1 )L 2 + ... +an-I L.-1:

This also has the nice consequence that once we have located the irreducible representations V(ll with highest weight L 1 + ··· + L 1, the general irreducible

§15.2. Representations of s1 4 C and sr.c

representation ra,, ... ,•• _, with highest weight Lai(L, inside the tensor product of symmetric powers

217

+ ... + Ll)

will occur

Syma• V( 11 ® Syma, vm ® .. · ® Sym•·-• v(•-o of these representations. Thus, the existence part of the basic Theorem 14.18 is reduced to finding the basic representations V(l); we will do this in due course, though at this point it is probably not too hard an exercise to guess what they are.

§15.2. Representations of sl4 C and sl,C We begin as usual with the standard representation of sl 4 C on V = C 4 . The standard basis vectors e1 of C 4 are eigenvectors for the action of ~. with eigenvalues L, so that the weight diagram looks like

or, with the reference cube drawn as well,

218

The dual representation V* of course has weights - L, corresponding to the vectors of the dual basis for V*, so that the weight diagram, with its reference cube, looks like

er

Note that the highest weight for this representation is - L 4 , which lies along the bottom edge of the Weyl chamber, as depicted in Diagram (15.8). Note also that the weights of the representation /\ 3 V-the triple sums L 1 + L 2 + L 3 , L 1 + L 2 + L 4 , L 1 + L 3 + L 4 , and L 2 + L 3 + l- 4 of distinct weights of V-are the same as those of V*, reflecting the isomorphism of these two representations. This suggests that we look next at the second exterior power NV. This is a six-dimensional representation, with weights L 1 + L1 the pairwise sums of distinct weights of V; its weight diagram, in its reference cube, looks like

The diagram shows clearly that NV is irreducible since it is not the nontrivial union of two configurations invariant under the Weyl group 6 4 (and all weights occur with multiplicity 1). Note also that the weights are symmetric about the origin, reflecting the isomorphism of NV with (NV)*= N(V*). Note that the highest weight L 1 + L 2 of the representation NV is the primitive vector along the front edge of the Weyl chamber if!" as pictured in Diagram (15.8). Now, we have already seen that the intersection of the closed

§ 15.2. Representations of 91 4 C and 91.C

219

Weyl chamber with the weight lattice is a free semigroup generated by the primitive vectors along the three edges of if'-that is, every vector in if/ n Aw is a non-negative integral linear combination of the three vectors L 1 , L 1 + L 2 , and L 1 + L 2 + L 3 . As we remarked at the end of the first section of this lecture, it follows that we have proved the existence half of tire general existence and uniqueness theorem (14.18) in the case of the Lie algebra sl4 C. Explicitly, since V, NV, and NV= V* have highest weight vectors with weights L 1 , L 1 + L 2 , and L 1 + L 2 + L 3 , respectively, it follows that the representation Sym"V ® Sym 6 (N V) ® Sym'(N V) contains a highest weight vector with weight aL 1 + b(L 1 + L 2 ) + c(L 1 + L 2 + L3 ), and hence a copy of the irreducible representation r•. &,c with this highest weight. Let us continue our examination of representations of sl 4 C with a pair of tensor products of the three basic representations: V ®NV and V ®NV. As for the first of these, its weights are easy to lind: they consist of the sums 2L1 + L1 (which occur once, as the sum of L1 and L1 + L1) and L1 + L1 + L• (which occur three times). The diagram of these weights looks like

(We have drawn only the vertices of the convex hull of this diagram, thus omitting the weights L 1 + L1 + L.; they are located at the centers of the hexagonal faces of this polyhedron.) Now, the representation V ®NV cannot be irreducible, for at least a couple of reasons. First off, just by looking at weights, we see that the irreducible representation W = r 1,1.o with highest weight 2L 1 + L 2 can have multiplicity at most 2 on the weight L 1 + L 2 + L 3 : by Observation 14.16, the weight space WL,+L,+L, is generated by the images oft he highest weight vector v e W2 L,+L, by successive applications of the primitive negative root spaces gL,-L,• gL,-L,• and gL.-L,· But L 1 + L 2 + L 3 is uniquely expressible as a sum of 2L 1 + L 2 and the primitive negative roots: L 1 + L 2 + L 3 = 2L 1 + L 2 + (L 2

-

L.) + (L 3

-

L 2 );

220

so that V,., +L,+ 1., is generated by the subspaces flt.,-r.,(!'lr.,-r.,(v)) and flr.,-r.,(Or.,-r.,(v)). We can in fact check that the representation rr.r.o takes on the weight L 1 + L 2 + L 3 with multiplicity 2 by writing out these generators explicitly and checking that they are independent: for example, we have Qr.,- 1., (9r.,- 1.,(v)) = C · E2 , 1 (E 3 , 2 (e 1 ® (e 1

"

e 2 )))

=C·E 2 , 1 (e 1 ®(e 1 "e 3 )) = C · (e 2

® (e 1

"

e3)

+ e 1 ® (e 2

"

e 3 )).

This is in fact what is called for in Exercise 15.10. Alternatively, forgetting weights entirely, we can see from standard multilinear algebra that the representation V ® tVV cannot be irreducible: we have a natural map of representations cp: V®NV-+NV

which is obviously surjective. The kernel of this map is a representation with the same set of weights as V ®NV (but taking on the weights L1 + L1 + L• with multiplicity 2 rather than 3), and so must contain the irreducible representation r~,,, 0 with highest weight 2L 1 + L 2 . Exercise 15.10. Prove that the kernel of cp is indeed the irreducible representation r,,r,o· Finally, consider the tensor product V ®NV. This has weights 2L1 + L• + L 1 = L 1 - L1, each occurring once, and 0, occurring four times. Its weight diagrams thus look like

This we may recognize as simply a direct sum of the adjoint representation with a copy of the trivial; this corresponds to the kernel and image of the obvious contraction (or trace) map

221

§I 5.2. Representations or sl 4 C and gf.C

V ®NV = V ®

v• = Hom(V, V)--+ C.

(Note that the adjoint representation is the irreducible representation with highest weight 2L 1 + L 2 + L 3 , or in other words the representation r 1 • 0 , 1 .) Exercise 15.11. Describe the weights of the representations Sym"V, and deduce that they are all irreducible. Exercise 15.12. Describe the weights of the representations Sym"(N V), and deduce that they are not irreducible. Describe maps qJ.:

Sym"(NV)---> Sym"- 2 (NV)

and show that the kernel of qJ. is the irreducible representation with highest weight n(L 1 + L 2 ). Exercise 15.13. The irreducible representation r~, 1, 1 with highest weight 3L 1 + 2L 2 + L 3 occurs as a subrepresentation of the tensor product V ®NV® NV lying in the kernel of each of the three maps

V®NV®NV->NV®NV V®NV®NV--+NV®NV~

NV

V®NV®NV~ V®NV*® V*--+ V®NV* ~ V® V

obtained by wedging two of the three factors. Is it equal to the intersection of these kernels? To test your graphic abilities, draw a diagram of the weights (ignoring multiplicities) of this representation.

Representations of sine Once the case ofsl 4 Cis digested, the case oft he special linear group in general offers no surprises; the main difference in the general case is just the absence of pictures. Of course, the standard representation V of !!I.e has highest weight L 1 , and similarly the exterior power NV is irreducible with highest weight L 1 + ·· · + Ll. It follows that the irreducible representation r., ..... •.-• with highest weight (a 1 + .. · + a._.)L 1 + ·.. + a._ 1 L0 _ 1 will appear inside the tensor product Sym•• V ® Sym">(N V) ® · · · ® Sym•·-•(t\"- 1 V), demonstrating the existence theorem (14.18) for representations of si.C. Exercise J 5.14. Verify that the exterior powers ofthe standard representations of sl.C are indeed irreducible (though this is not necessary for the truth of the last sentence).

222

§15.3. Weyl's Construction and Tensor Products At the end of the preceding section, we saw that the irreducible representation r., ..... •.-• ofsi.C with highest weight(a 1 + ... + a.-dL 1 + ... + a.- 1 L.- 1 will appear as a subspace of the tensor product Sym•• V ® Sym"•(N V) ® · · · ® Sym•·-•(t\"- 1 V), or equivalently as a subspace of the dth tensor power V® 4 of the standard representation V. The natural question is, how can we describe this subspace? We have seen the answer in one case already (two cases, if you count the trivial answer r. = Sym"V in the case n = 2): the representation r.,b of si 3C can be realized as the kernel of the contraction map Sym"V ® Symb(NV)-+ Sym•-t V ® Symb- 1 (N V).

This raises the question of whether the representation r. can in general be described as a subspace of the tensor power (g)(Sym"•(A'V)) by intersecting kernels of such contraction/wedge product maps. Specifically, fori and j with i + j ~ n we can define maps Sym•• V ® Sym"•(N V) ® · · · ® Sym•·-•(t\"- 1 V) -+NV® Nv ® Sym•• V ® · · · ® Sym••- 1 (A'V) ® · · ·

® Sym•r 1 (NV)®···® Sym•·-• (t\"- 1 V) and we have similar maps fori < j with i + j ~ nand i even with 2i ~ n; there are likewise analogously defined maps in which we split off three or more factors. The representation r., ..... •.-• is in the kernel of all such maps; and we may ask whether the intersection of all such kernels is equal to r •. The answer, it turns out, is no. (It is a worthwhile exercise to lind an example of a representation r. that cannot be realized in this way.) There is, however, another way of describing r. as a subspace of V® 4 : in fact, we have already met these representations in Lecture 6, under the guise of Sclrur fimclors or We}•l modules. In fact, at the end of this lecture we will see how to describe them explicitly as subspaces of the above spaces (g)(Sym"•(/\'V)). Recall that for V = c• an n-dimensional vector space, and any partition A: A1

~

A1

~ .. · ~

A•

~

0,

we can apply the Schur functor §A to V to obtain a representation §AV = §A(C") ofGL(V) = GL.(C).Ifd = I,A, this was realized as §A V = y®• · cA

= V®d ®cs. VA,

where cA is the Young symmetrizer corresponding to A, and VA is the irreducible representation of 6 4 corresponding to A. We saw in Lecture 6 that §AVis an irreducible representation of G L.C. It follows immediately that §A Vremains irreducible as a representation ofSL.C,

223

§15.3. Weyl's Construction and Tensor Products

since any element of GL.C is a scalar multiple of an element of SL.C. In particular, it determines an irreducible representation of the Lie algebra st. C. Proposition 15.15. Tire representation §A(C") is the irreducible representation of sl.C with higl1est weigl1t .t 1 L 1 + .t1 L 2 + · · · + .t.L•. Jn particular,§ A(C") and § 11 (C") are isomorphic representations of sl.C if and only if A., - 111 is constant, independent of i. To relate this to our earlier notation, we may say that the irreducible representation r., ..... •.-• of sl.C with highestweighta 1 L 1 + a 1 (L 1 + L 2 ) + ··· + a.-dL 1 + ··· + L._.)isobtained by applying the Schur functor §A to the standard representation V, where .t = (a 1 + ··· + a._ 1 , a 1 + · .. + a._ 1 ,

••• ,

a._ 1 , 0).

(If we want a unique Schur functor for each representation, we can restrict to those ,t with .t. = 0.) In terms of the Young diagram for .t, the coefficients a1 = .t, - ,tl+ 1 are the differences of lengths of rows. For example, if = 6,

n

I I I I I I 1'--- a_...! I

'-a_; 2

-

~a _...I 4

is the Young diagram corresponding to f 3 , 2 , o. 3 , 1 • PROOF OF THE PROPOSITION. In Theorem 6.3 we calculated that the trace of a diagonal matrix with entries x., ... , x. on § A(C") is the Schur polynomial SA(x 1 , .•. , x.). By Equation (A.19), when the Schur polynomial is written out it takes the form

SA(x 1 , ... ,x.)=MA+

L

KA 1,Mjt,

(15.16)

tt< A

where MJt is the sum of the monomial X"= x~·x~•· ... · x:• and all distinct monomials obtained from it by permuting the variables, and the KAJt are certain non-negative integers called Kostka numbers. When §A(C") is diagonalized wit!t respect to the group of diagonal matrices in GL.(C), it is also diagonalized with respect to I) c sl.(C). There is one monomial in the displayed equation for each one-dimensional eigenspace. The weights of §A(C") as a representation of sl.(C) therefore consist of all p. 1 L 1 + p. 2 L 2 + · .. + Jt.L.,

each occurring as often as it does in the monomial X'' in the polynomial

224

••• , x.). Since the sum is over those partitions p for which the first nonzero A1 - p 1 is positive, the highest weight that appears is A1 L 1 + A2 L 2 + ··· + ,l..L., which concludes the proof. [In fact one can describe an explicit basis of eigenvectors for §l(C") which correspond to the monomials that appear in (15.16), cf. Problem 6.15 or Proposition 15.55.] 0

SA(x 1 ,

In particular, we have (by Theorem 6.3) formulas for the dimension of the representation with given highest weight. Explicitly, one formula says that dim(f

)=

cri, ... ,CI"-1

1J

(al

+ ... + ai-d + j

1 ~ <}S.n

- i

j- i

(15.17)

As we saw in the proof, this proposition also gives the multiplicities of all weight spaces as the integers K A,. that appear in (15.16), which have a simple combinatorial description (p. 456): the dirne1rsion of the weight space with weight p in the representation §l(C") is the number of ways one can fill the Young diagram of Awit II I' 1 l 's, p 2 2's, ... , 11. 11's, in sue/• a way tlrat the enJries in each row are nondecreasing a11d those in each column are strictly increasing. Exercise 15.18. Use the formula in case n = 4 to calculate the dimensions of the irreducible representations rl.l.o and f 1• 1, 1 ofsl 4 C. In the former case, use this to redo Exercise 15.1 0; in the latter case, to do Exercise 15.13. Exercise 15.19*. Use this formula to show that the dimension orthe irreducible b of sl 3 with highest weight aL 1 + b(L 1 + L 2 ) is representation (a+ b + l)(a + l)(b + l)/2. This is the same as the dimension of the kernel of the contraction map

r..

'•.b: Sym"V ® Symbv•-+ Sym•-l V ® Symb-l v•. Use this to give another proof of the assertion made in Claim 13.4 that r. b is this kernel. · Exercise 15.20*. As an application of the above formula, show that if Vis the standard representation of si.C, then the kernel of the wedge product map

·v®Nv ..... N+ 1 v is the irreducible representation fi,O, .... O,I,O,... With highest weight 2L 1 + L 2 + ··· + Ll; and that the irreducible representation rl-1.1.o .... with highest weight k · L 1 + L 2 is the kernel of the product map

V ® Syml V ..... SymH 1 V. Exercise 15.21*. Show that the only nontrivial irreducible representations of sl. C of dimension less than or equal to n are V and v•. One important consequence of the fact that the irreducible representations of si.C are obtained by applying Schur functors to the standard representation

225

§15.3. Weyl's Construction and Tensor Products

is that identities among the Schur- Weyl functors give rise to identities among representations of GL. (and hence SL. and sl.), as we saw in Lecture 6. For example, the representation (15.22) is a direct sum of representations §A(V) E9 (f),. K,.A§,.(V), where K,.A is the coefficient described above. The particular application of this principle that we will use most frequently in the sequel, however, is the consequence that one knows tile decomposition of a tensor product of any two irreducible representations of !!l.C: specifically, the tensor power §A(V) ® §,.(V) decomposes into a direct sum of irreducible representations §A(V) ® §,.(V) =

EB N.l,..§.(V), •

(15.23)

where the coefficients NA,.. are given by the Littlewood- Richardson rule, which is a formula in terms of the number of ways to fill the Young diagram between .t and v with J.!J 1's, Jlz 2's, ... , 11. n's, satisfying a certain combinatorial condition described in (A.8). Exercise 15.24. Use the Littlewood-Richardson rule to show that the representation r., +b, ... , •• _, H._, occurs exactly once in the tensor product

r., .....•.-• ® rb,,. ..,b._,. A special case of this is the analogue of Pieri's formula, which allows us to decompose the tensor product of an arbitrary irreducible representation with either Sym~JI = r~.o or the fundamental representation NV= ro .... , 1.o ..... o• (where the 1 occurs in the kth place):

..... o

Proposition 15.25. (i) The tensor prodr1ci of decomposes illlo a direct sum:

r., .....•.-•

with Sym~V =

r •. o..... o

r., ..... •.-· ® r...... o = Ef>rb,,. ... b._,, the sum over all (b 1 , •.• , b._ 1 ) for which there are non-negative integers c 1 , ..• , c. whose sum is k, with cl+ 1 ~ a 1 for I ~ i ~ 11- 1, and with b1 = a1 + c;- cl+r for 1 ~ i ~ n - I. (ii) The tensor product of with Nv = O,I,O, .•.• o decomposes into a direct sum:

r., ... ,••_,

ro.....

the sum over all (b 1 , ••• , b._ 1) for which there is a subset S of { l, ... , 11} of cardinality k, suclt that if i ¢Sandi+ 1 e S, then a1 > 0, with b, =

a,- 1 {a + l 1

a1

if i ¢Sandi+ 1 e S if i e S and i + 1 ¢ S otherwise.

226

PROOF. This is simply a matter of translating the prescriptions of (6.8) and (6.9), which describe the decompositions in terms of adding boxes to the Young diagrams. In (i), the c1 are the number of boxes added to the ith row, and in (ii), S is the set of rows to which a box is added. 0 Exercise 15.26. Verify the descriptions in Section 2 of this lecture of V ®NV and V ®NV. where Vis the standard representation of sl 4 C. Exercise 15.27. Use Pieri's formula (with n = 4) twice to find the decomposition into irreducibles or v ® N v ® N v, where vis the standard representation of s1 4 C. Use this to redo Exercise 15.13. Exercise 15.28. Use Pieri's formula to prove ( 13.5). You may also want to look around in Lecture 13 to see which other of the decompositions found there by hand may be deduced from these formulas. Exercise 15.29. Verify that the statement of Exercise 15.20 follows directly from Pieri's formula. In the following exercises, V = c• is the standard representation of st. c. Exercise 15.30. Consider now tensor products of the form NV® 1\'V, with, say, k ~ I. Show that there is a natural map NV®A'V _.N+ 1 V®N- 1 V given by contraction with the element "trace" (or "identity") in End(V). Explicitly, this map may be given by (v 1 A .. • A vk)

® (w1

A .. • A

w1)

I

,. . . I

(-I)'(v 1

A ... A

vk

v ® v• =

A

w1)®{w 1

A ... A

'W; A

... A

w1).

1=1

What is the image of this map? Show that the kernel is the irreducible representation ro ..... o, 1 ,o ..... o, 1 ,o, ... with highest weight 2L 1 + ... + 2L1 +

L,H + ... +Lt. Exercise 15.31"'. Carry out an analysis similar to that of the preceding exercise for the maps SymkV ® Sym'V __. Symu 1 V ® Sym 1- 1 V defined analogously. Exercise 15.32"'. As a special case of Pieri's formula, we see that if V is the standard representation of st.c, the tensor product

22

§15.4. Some More Geometry

Nv ® Nv =

ffi § 12, ... ,2,J, ...••• o....,(V)

-- ffi w

r.o..... o. 1,0, ... ,0,1.0, ....

where in the ith factor the l's occur in the (k- i)th and (k the same time, of course, we know that

+ i)th

places. A•

NV® NV= Sym 2 (NV) $ N(NV). If we denote the ith term on the right-hand side of the first displayed equation for NV® Nv bye,, show that

Sym 2 (NV)

= ffi 8

21

and

N(NV) = ffi 82i+l·

Exercise 15.33•. As another special case of Pieri's formula, we see that the tensor product

ffi § 1u,,l_,(V) = ffi rll,l-1,0 ... 0·

Symlv ® SymlV =

At the same time, of course, we know that SymlV ® SymlV = Sym 2 (SymlV) $ N(SymlV). Which of the factors appearing in the first decomposition lie in Sym 2 (SymlV), and which in N(Sym1 V)? It follows from the Littlewood-Richardson rule that if l, 11. and l' all have at most two rows, then the coefficient NA~· is zero or one (and it is easy to say which occurs). In particular, for the Lie algebras sl 2 C and si 3 C, the decomposition of the tensor product of two irreducible representations is always multiplicity free. Groups whose representations have this property, such as SU(2), SU(3), and S0(3) which are so important in physics, are called "simply reducible," cf. [Mack].

§15.4. Some More Geometry Let V be ann-dimensional vector space, and G(k, n) = G(k, V) = Grassl V the Grassmannian of k-planes in V. Grassl Vis embedded as a subvariety of the projective space P(NV) by the Plucker embedding: p: Grassl V

c.. P(NV)

sending the plane W spanned by vectors v1 , ••• , vl to the alternating tensor v1 A··· A vl. Equivalently, noting that if W c Vis a k-dimensional subspace, then NW is a line in NV, we may write this simply as p: W1-+NW

15. si 4 C and si.C

228

This embedding is compatible with the action of the general linear group: PSL.C = Aut(P(V)) = {a e Aut(P(NV)): u(G(k, V)) = G(k,

VW.

This follows from a fact in algebraic geometry ([Ha]): all automorphisms of the Grassmannian are induced by automorphisms of V, unless n = 2k, in which case we can choose an arbitrary isomorphism of V with V* and compose these with the automorphism that takes W to (cn/W)*. Here the superscript • denotes the connected component of the identity. As in previous lectures, if we want symmetric powers to correspond to homogeneous polynomials on projective space, we should consider the dual situation: G = Grass•v is the Grassmannian of k-dimensional quotient spaces of V, and the Plucker embedding embeds Gin the projective space P(NV*) of one-dimensional quotients of NV. The space of all homogeneous polynomials of degree m on P(NV*) is naturally the symmetric power Sym"'(NV). Let I(G).., denote the subspace of those polynomials of degree m on IP'(NV*) that vanish on G. Each /(G).., is a representation of slnC: 0-+ /(G)..,-+ Sym'"(NV)-+ W, -+0, where W,, denotes the restrictions to G of the polynomials of degree m on the ambient space IP(NV*). We shall see later that W., is the irreducible representation ro ..... o,m,o, ... with highest weight m(L 1 + · ·· + L.) (the case m = 2 will be dealt with below). In the following discussion, we consider the problem of describing the quadratic part I(Gh of the ideal as a representation ofsi.C. Exercise 15.34. Consider the first case of a Grassmannian that is not a projective space, that is, k = 2. The ideal of the Grassmannian G(2, V) of 2-planes in a vector space is easy to describe: a tensor qJ EN Vis decomposable if and only if q1 1\ rp = 0 (equivalently, if we think of qJ as given by a skewsymmetric n x 11 matrix, if and only if the Pfaffians of symmetric 4 x 4 minors all vanish); and indeed the quadratic relations we get in this way generate the ideal of the Grassmann ian. We, thus, have an isomorphism J(Gh ~NV and correspondingly a decomposition into irreducibles Sym 2 (NV) ~NV E9 ro, 2 ,o..... o• where ro.z.o ..... o is, as above, the irreducible representation with highest weight2(L 1 + L 2 ), cf. Exercise 15.32. Exercise 15.35. When k = 2 and 11 = 4, G is a quadric hypersurface in IP'\ so polynomials vanishing on G are simply those divisible the quadratic polynomial that defines G. Deduce an isomorphism. /(G)..,= Sym'"- 2 (NV).

§15.4. Some More Geometry

229

The first case of a Grassmanniau that is not a projective space or of the form G(2, V) is, of course, G(3, 6), and this yields an interesting example. Exercise 15.36. Let V be six dimensional. By examining weights, show that the space /(G)z of quadratic polynomials vanishing on the Grassmannian G(3, V) c P(NV) is isomorphic to the adjoint representation of si 6 C, i.e., that we have a map rp: Sym 2 (N V) _. V ® V*

with image the space of traceless matrices. Exercise 15.37. Find explicitly the map rp or the preceding exercise. Exercise 15.38. Again, let V be six dimensional. Show that the representation Sym 4 (NV) has a trivial direct summand, corresponding to the hypersurface in P(NV*) dual to the Grassmannian G = G(3, V) c P(N V). In general, the ideal /(G) = EJj /(G)m is generated by the famous PlUcker equations. These are homogeneous polynomials or degree two, and may be written down explicitly, cf. (15.53), [H-PJ, or [Ha]. In the following exercises, we will give a more intrinsic description of these relations, which will allow us to identify the space /(Gh they span as a representation on si.C (and to

see the general pattern of which the above are special cases). Exercise 15.39. For a given tensor I\ EN V, we introduce two associated subspaces: W = {v E V: v

1\

I\

=

0} c V

and W* = {v*

E

V"': v*

1\

I\* =0} c V*,

where, abusing notation slightly, I\"' is the tensor I\ viewed as an element of NV= 1\"-kv•. Show that the dimensions of Wand W* are at most k and 11 - k, respectively, and that I\ is decomposable if and only if W has dimension k or W* has dimension n - k; and deduce that I\ is decomposable if and only if the annihilator W' of W* is equal to W. Exercise 15.40. Now let 8

E

N+l v•

111 :

=

1\"-k-l v.

Wedge product gives a map

Nv--. /\"- v = v•. 1

Using the preceding exercise, show that I\ is decomposable if and only if 1!!(/\) 1\

for all:S E N+ 1 v•.

I\

= 0 E N-l v

230

Exercise 15.41. Observe that in the preceding exercise we construct a map

/\u 1 V* ® Sym 2 (NV)--+ N- 1 V, or, by duality, a map N+~ V*

® N- 1 V*--+ Sym 2 (NV*)

{15.42)

whose image is a vector space of quadrics on IP(NV) whose common zeros are exactly the locus of decomposable vectors, that is, the Grassmannian G(k, V). Show that this image is exactly the span of the Pliicker relations above. Exercise 15.43. Show that the map ( 15.42) of the preceding exercise is just the dual of the map constructed in Exercise t 5.30, with k = I and restricted to the symmetric product. Combining this with the result of Exercise 15.32 (and assuming the statement that the Pllicker relations do indeed span /(Gh), deduce that in terms of the description Sym 2 (NV) =

E9 9 21

of the symmetric square of NV, we have

w2 =

E>o

= ro, ... ,0,2,0, ...

(the irreducible representation with highest weight 2(£ 1 /(G)2

+ .. · + Lt)), and

= Ef> E>u. I:. I

Hard Exercise 15.44. Show that in the last equation the sub-direct sum

/(/) = ffiE>u 1;,1 is just the quadratic part of the ideal of the restricted chordal variety of the Grassmannian: that is, the union of the chords LM joining pairs of points in G corresponding to pairs of planes L and M meeting in a subspace of dimension at least k - 2/ + l. (Q1,1estion: What is the actual zero locus of these quadrics?) Exercise 15.45. Carry out an analysis similar to the above to relate the ideal of a Veronese variety PV* c IP(Symt V*) to the decomposition given in Exercise 15.33 of Sym 2 (SymtV). For which k do the quadratic polynomials vanishing the Veronese give an irreducible representation? Exercise 15.46. (For algebraic geometers and/or commutative algebraists.) Just as the group PGL.C acts on the ringS of polynomials on projective space P"', preserving the ideal of the Veronese variety, so it acts on that space of relations on the ideal (that is, inasmuch as the ideal is generated by quadrics, the kernel of the multiplication map Ix(2) ® S--+ S), and likewise on the entire minimal resolution of the ideal of X. Show that this resolution has the form

§15.5. Representations or GL.C

· · · -+

231

R 1 ® S-+ R 1 ® S-+ Ix(2) ® S,

where all the R, are finite-dimensional representations of PGL.C, and identify the representations R1 in the specific cases of (i) the rational normal curve in P 3 , (ii) the rational normal curve in P 4 , and (iii) the Veronese surface in P~.

§15.5. Representations of GLnC We have said that there is little difference between representations of GL.c and those of the subgroup SL.c of matrices of determinant I. Our object here is to record the difference, which, naturally enough, comes from the determinant: if V = C" is the standard representation, 1\"V is trivial for SL.C but not for GL.C. Similarly, V and /\"- 1 V* are isomorphic for SL.C but not for GL.C. To relate representations ofSL.c and GL.C, we first need to define some representations of GL.c. To begin with, let Dt denote the one-dimensional representation of GL.C given by the kth power of the determinant. When k is non-negative, Dt = (1\"V)®t; D_t is the dual (Dt)* of Dt. Next, note that the irreducible representations of SL.C may be lifted to representations ofGL.C in two ways. First, for any index a = (a 1, ... , a.) of length n we may take !Jl. to be the subrepresentation of the tensor product Sym"• V ® · · · ® Sym•·-•(1\"- 1 V) ® Sym"•(/\"V) spanned by the highest weight vector with weight a 1 L 1 .. · + a.-dL 1 + · .. + L._r)-that is, the vector

v = (e 1 t• ·(e 1

1\

e1 )"' • ... · (e 1

1\ .. · 1\

+ a 2{L 1 + L 2 ) +

e.)"·.

This restricts to SL.C to give the representation r•. , where a' = (a 1 , ... , a._r); taking different values of a. amounts to tensoring the representation with different factors Sym"•(/\"V) = (1\"V)®•. =D••. In particular, we have

lll., ..... •.+t = !~>.,. ....•• ® Dt, which allows us to extend the definition of Ill. to indices a with a. < 0: we simply set for large k. Alternatively, we may consider the Schur functor§~ applied to the standard representation V ofGL.C, where

l

=

(a 1

+ ·· · + a., a1 + ··· + a., ... , a._, + a., a.).

We will denote this representation

§A V of GL.C

by '1\; note that

232

which likewise allows us to define '1'1 for any index 1 with 1 1 ~ 1 2 ~ even if some of the 11 are negative: we simply take '1'1, ..... 1.

••• ~

1.,

= '1'1,H .... ,1.H®D-~

for any sufficiently large k. As is not hard to see, tire two representati011s II>. and '1'1 are isomorphic as representations ofGL.C: by §15.3 their restrictions to SL,.C agree, so it suffices to check their restrictions to the center C* c GL.. C, where each acts by multiplication by zLA, = zL'a•). It is even clearer that there are no coincidences among the q,• (i.e., q,• will be isomorphic to II> •. if and only if a =a'): if II>.~ II> •. , we must have a1 =a; fori= l, ... , n- I, so the statement follows from the nontriviality of D. fork -F 0. Thus, to complete our description of the irreducible finite-dimensional representations of GL.. c, we just have to check that we have found them all. We may then express the completed result as

Proposition 15.47. Every irreducible complex representation of GL.C is isomol'pllic to '1'1 for a unique index 1 = 11 , ••. , l,. witlt 1 1 ~ 1 2 ~ • • • ;;::: 1. (equivalently, to II>. for a unique index a = a 1 , ••• , a. wit I! a 1, ••• , a._, ~ 0). PROOF. We start by going back to the corresponding Lie algebras. The scalar matrices form a one-dimensional ideal C in gi,.C, and in fact gl,.C is a product of Lie algebras:

(15.48) In particular, Cis the radical of gi.C, and si.C is the semisimple part. It follows from Proposition 9.17 that every irreducible representation of gi,.C is a tensor product of an irreducible representation of si.C and a one-dimensional representation. More precisely, let W1 = § 1 {C") be the representation of si.C determined by the partition 1 (extended to si,.C x C by making the second factor act trivially). For wE C, let L(w) be the one-dimensional representation of sl.C x C which is zero on the first factor and multiplication by w on the second; the proof of Proposition 9.17 shows that any irreducible representation of si.C x C is isomorphic to a tensor product W1 ® L{w). The same is therefore true for the simply connected 1 group SL,.C x C with this Lie algebra. · We write GL,.C as a quotient modulo a discrete subgroup of the center of SL,.C XC: I --> Ker(p)--> SL,.C x C

!.. GL.. C--> I,

(15.49)

where p(g x z) = e• · g, so the kernel of p is generated by e• ·I x (- s), where s = 2ni/n. Our task is simply to see which of the representations W;. ® L(w) of SL.. C x Care trivial on the kernel of p. Now e'·I acts on § 1 C" by multi1

For a proof that SL.C is simply connected, see §23.1.

§15.5. Representations of GL.C

233

plication by e'4, whered = I 11; indeed, this is true on the entire representation (C")® 4 which contains §AC". And -s acts on L(w) by multiplication by e-•w, so e• · I x (- s) acts on the tensor product by multiplication by e•d-sw. The tensor product is, therdore, trivial on the kernel of p precisely when sd - SW E 2nil., i.e., when for some integer k. We claim finally that any representation ~ ® L(w) satisfying this condition is the pullback via p of a representation 'I' on GL.C. In fact, it is not hard to see that it is the pullback of the representation 'I' 4 , H, ... ,A.H: the two clearly restrict to the same representation on SL.C, and their restrictions to C are just multiplication by ew• = e
More Remarks on Weyl's Construction We close out this lecture by looking once more at the Weyl construction of these representations of GL( V). This will include a realization "by generators and relations," as well as giving a natural basis for each representation. First, it may be illuminating-and it will be useful later-to look more closely at how § 4 V sits in V® 4• We want to realize § 4 V as a subspace of the subspace Sym"•(NV) ® Sym"•-•(N- 1 V) ® · · · ® Sym"•(V) c y® 4, where a1 is the number of columns of the Young diagram of l of length i (and k is the number of rows). This space is embedded in y®d in the natural way: from left to right, a factor Sym"(NV) is embedded in the corresponding y®•" · by mapping a symmetric product of exterior products (vt,t

1\ Vz,t 1\ ...

"v,,tl·(vt,l"

"2.2 1\ ...

"v,,z)· ...

• (VI. a 1\ Vz.a II " · II vb,a)

to

I

sgn(q)(v,, 111 ,p{tl ® · · · ® v,, 1, 1,p{ 11 ) ® · · · ® (v,.111 .PI•l ® · · · ® v,.lhi.PI•l),

the sum over p E 6. and q = (q 1, ... , q.) E 6, x · · · x 6,. In other words. one

234

first symmetrizes by permuting columns ofthe same length, and then performs an alternating symmetrizer on each column. Letting a = (a 1 , ••• , a ... ), Jet A"(V) denote this tensor product of symmetric powers of exterior powers, i.e., set

A'V = Syma•(NV)® Syma•-•(N- 1 V) ® ··· ® Syma•(V). We want to realize § 1 Vas a subspace of A"V. To do this we use the construction of § 1 Vas v®•. c1 , where c1 is a Young symmetrizer; to get compatibility with the embedding of A"V we have just made, we use the tableau which numbers the columns from top to bottom, then left to right.

I

4

6

sJ

2

5

7

a,

3

a2 = 2

:~.,

=I

=4

:1. 2 = 3

'--

Wetake.u = ).' = (fl 1 ~ .. • ~ l'r > 0) to betheconjugateof..t Thesymmetrizer c1 is a product a1 • b1 , where a1 = l:e,., the sum over all p in the subgroup P = 6 1 , x · · · x 6 1• of e 4 preserving the rows, b1 = l:sgn(q)q, the sum over the subgroup Q = e,, x · · · x e,, preserving the columns, as described in Lecture 4. The symmetrizing by rows can be done in two steps as follows. There is a subgroup

R=

e •• x

.. · x

e.,

of P, which consists of permutations that move all entries of each column to the same position in some column of the same length; in other words, permutations in Rare determined by permuting columns which have the same length. (In the illustration, R = { 1, (46)(57) }.) Set

L e,

a~ =

in Ced·

reR

Now if we define a~ to be L eP, where the sum is over any set of representatives in P Cor the left cosets P/R, then the row symmetrizer a1 is the product of ai and a~. So §

1

(V)

= (v®• ·a:\)· aA ·b)..

The point is that, by what we have just seen, V®d

·aA ·b1 = A"V.

§15.5. Represenlalions of GL.C

235

Since V®" ·a~ is a subspace of V®", its image§ 1 (V) by a~· b1 is a subspace of A "(V), as we claimed. There is a simple way to construct all the representations §AV ofGL(V) at once. In fact, the direct sum of all the representations § 1 V, over all (nonnegative) partitions l, can be made into a commutative, graded ring, which we denote by§" or §"(V), with simple generators and relations. This is similar to the fact that the symmetric algebra Sym"V = ffiSymtv and the exterior algebra (\"V = ffiNVare easier to describe than the individual graded pieces, and it has some of the similar advantages for studying all the representations at once. This algebra has appeared and reappeared frequently, cf. [H-P]; the construction we give is essentially that of Towber [Tow 1]. To construct §"(V), start with the symmetric algebra on the sum of all the positive exterior products of V: set A"(V) = Sym"(VED NVED NVED ... E9 NV) =

ffi

Sym"•(NV) ® · · · ® Sym">(N V) ® Sym••(V),

the sum over all n-tuples a1 , ... , a" of non-negative integers. So A"( V) is the direct sum of the A"(V)just considered. The ring§"= §"(V) is defined to be the quotient of this ring A"(V) modulo the graded, two-sided ideal f generated by all elements ("Plucker relations") of the form (v 1

11 .. • A

vp) · (w1

- f (v

11 .. • 11

1 11 .. • A

v1_ 1

w9) 11

w1

A

v1+1

11 .. • 11

vp) · (v1 A w2

11 .. · 11

w9)

1~1

(15.53)

for all p ~ q ~ 1 and all v" ... , vP, W~o ... , w9 E V. (If p = q, this is an element ofSym 2 (NV); if p > q, it is in NV®NV = Sym 1 (NV)®Sym 1 (NV). Note that the multiplication in§"( V) comes entirely from its being a symmetric algebra and does not involve the wedge products in A"V.) Exercise 15.54*. Show that I" contains all elements of the form (v 1

11 ... 11

vp)·(w1

-"(v L., I

11 ... 11

w9 )

1\'''1\WI 1\'"'1\W• 1\"''1\V) p

for all p ~ q ~ r ~ I and all v1 , ••• , vP, w1 , ••• , w9 E V, where the sum is over all1 ~ i 1 < i 2 < · · · < i,:;;; p, and the elements w1 , ••• , w, are inserted at the corresponding places in v1 11 • · • 11 vP.

Remark. You can avoid this exercise by simply taking the elements in the exercise as defining generators for the ideal J". When p = q = r, the calcula-

236

lion of Exercise 15.54 shows that the relation (v 1 1\ • • • 1\ t'p) · (w 1 1\ • · • 1\ wp) ={w 1 1\ ••• 1\ wp)·(v 1 1\ ••• 1\ vp)followsfromthegeneratingequationsforf'. In particular, this commutativity shows that one could define §'(V) to be the full tensor algebra on V $NV$··· $1\"V modulo the ideal generated by the same generators. The algebra §'(V) is the direct sum of the images §"(V) of the summands A"(V). Let e~o ... , e. be a basis for V. We will construct a basis for §"(V), with

a basis element er for every semistandard tableau Ton the partition A which corresponds to a. Recall that a semistandard tableau is a numbering of the boxes of the Young diagram with the integers l, ... , 11, in such a way that the entries in each row are nondecreasing, and the entries in each column are strictly increasing. Let T(i, j) be the entry of T in the ith row and the jth column. Define er to be the image in §"(V) of the element I

J]

eT(I,Jl 1\

er12 •1,

1\ .. • 1\

er11,1 , 1, E Sym"•(I\"V) ® · · · ® Sym"•(V),

i.e., wedge together the basis elements corresponding to the entries in the columns, and multiply the results in §'(V). Proposition 15.55. (l) The projectioll from A"(V) to §"(V) maps tire subspace § 4 {V) isornorphically onto §"(V).

(2) The er for T a semistandard tableau on A form a basis for §"( V). PROOF. We show first that the elements er span §"(V). It is clear that the er span if we allow all tableaux T that number the boxes of ..1. with integers between l and n with strictly increasing columns, for such elements span before dividing by the ideal f. We order such tableaux by listing their entries column by column, from left to right and top to bottom, and using the reverse lexicographic order: T' > T ifthe last entry where they differ has a larger entry forT' than for T.lf Tis not semistandard, there will be two successive columns ofT, say thejth and (j + l)st, in which we have T(r,j) > T(r,j + I) for some r. It suffices to show how to use relations in I' to write er as a linear combination of elements er· with T' > T. Foe this we use the relation in Exercise 15.54, with v1 = en1•1, for I :::;; is p = 1-LJ• and w1 = eru,J+Il for l s i :::;; q = J.lJ+r, to interchange the first r of the {w1} with subsets of r of the {v1}. The terms on the right-hand side of the relation will all correspond to tableaux T' in which the r first entries in the (j + t)st column of Tare replaced by r of the en ties in the jth column, and are not otherwise changed beyond the jth column. All ofthese are larger than Tin the ordering, which proves the assertion. It is possible to give a direct proof that the er corresponding to semistandard tableaux Tare linearly independent (see [Tow!]), but we can get by with less. Among the semistandard tableaux on A there is a smallest one To whose ith row is filled with the integer i. We need to know that ero is not zero

§15.5. Representations of GL.C

237

in §". This is easy to see directly. In fact, the relations among the er in f n A"(V) are spanned by those obtained by substituting r elements from some column of some T to an earlier column, as in the preceding paragraph. Such will never involve the generator er. unless the T that is used is T0 , and in this case, the resulting element off is zero. Since ero occurs in no nontrivial relation, its image in §"cannot vanish. Since ero comes from §A(V), it follows that the projection from §A(V) to §"(V) is not zero. Since this projection is a mapping of representations of SL(V), it follows that §"(V) must contain a copy of the irreducible representation § A(V). We know from Theorem 6.3 and Exercise A.31 that the dimension of §A(V) is the number of semistandard tableaux on ..l. Since we have proved that the dimension of§•(V) is at most this number, the projection from §A(V) to§"( V) must be surjective, and since §A(V) is irreducible, it must be injective as well, and the er for T a semistandard tableau on l must form a basis, as asserted. 0 Note that this proposition gives another description of the representations § 1 (V), as the quotient of the space A"{V) by the subspace generated by the "Pliicker" relations (15.53). Exercise 15.56. Show that, if the factor NV is omitted from the construction, the resulting algebra is the direct sum of all irreducible representations of SL(V) = SL.C. It is remarkable that all the representations §A(C") ofGL.C were written down by Deruyts (following Clebsch) a century ago, before representation theory was born, as in the following exercise.

Exercise 15.57•. Let X = (x 1, 1) be an n x n matrix of indeterminants. The group G = GL.C acts on the polynomial ring C[x1,1] by g · x 1,1 = 1 at,;Xt,J forg = (a 1, 1) e GL.C. For any tableau Ton the Young diagram of ..lconsisting of the integers from 1 to 11, strictly increasing in the columns, let er be the product of minors constructed from X, one for each column, as follows: if the column ofT has length Jt1, form the minor using the first Jli columns, and use the rows that are numbered by the entries of the column of T. Let DA be the subspace of C[x 1, 1] spanned by these er, where dis the number partitioned by ..l. Show that: (i) D1 is preserved by GL.C; (ii) the er, where Tis semistandard, form a basis for D1 ; (iii) DA is isomorphic to § A(C").

Il=

LECTURE 16

Symplectic Lie Algebras

In this lecture we do for the symplectic Lie algebras exactly what we did for the special linear ones in §15.1 and most or §15.2: we will first describe in general the structure or a symplectic Lie algebra (that Is; give a Cartan subalgebra, find the roots, describe the Killing form, and so on). We will then work out in some detail the representations or the specific algebra 11p4 C. As in the case or the corresponding analysis or the special linear Lie algebras, this is completely elementary. §16.1: The structure or Sp2 .C and sp 2.C §16.2 Representations or 11p4 C

§16.1. The Structure of Sp 2 "C and sp 2nC Let V be a 2n-dimensional complex vector space, and . Q:

v x v -c.

a nondegenerate, skew-symmetric bilinear form on V. The symplectic Lie group Sp 2 .C is then defined to be the group of automorphisms A of V preserving Q-that is, such that Q(Av, Aw) = Q(v, w) for all v, we V-and the symplectic Lie algebra sp 2 .C correspondingly consists of endomorphisms A: V _. V satisfying Q(Av, w)

+ Q(v, Aw) = 0

for all v and we V. Clearly, the isomorphism classes of the abstract group and Lie algebra do not depend on the particular choice of Q; but in order to be able to write down elements of both explicitly we will, for the remainder of our discussion, take Q to be the bilinear form given, in terms of a basis e 1 , • •• ,

239

§16.1. The Structure of Sp 2 .C and tp 2.C e 2 • for Jl, by

Q(e1, ei+.)

= I,

Q(ei+n• e 1) = -1,

and

Q(e 1, e1) = 0

if j .f i ± n.

The bilinear form Q may be expressed as

Q(x,y)='x·M·y, where M is the 2n x 2n matrix given in block form as

M=(~I. ~} the group Sp 1.C is thus the group of 2n x 2n matrices A satisfying·

M='A·M·A and the Lie algebra sp 1 .C correspondingly the space of matrices X satisfying the relation 'X·M

+ M·X == 0.

(16.1)

Writing a 2n x 2n matrix X in block form as

X=(~ ~) we have

'X·M==(-'C

'A)

-'D 'B

and

M·X=(:A

~B)

so that this relation is equivalent to saying that tire off-diagonal blocks Band C of X are symmetric, and tire diagonal blocks A and D of X are negative transposes of each other. With this said, there is certainly an obvious candidate for Cartan subalgebra () in sp 2 .C, namely the subalgebra of matrices diagonal in this representation; in fact, this works, as we shall see shortly. The subalgebra ~ is thus spanned by then 2n x 2n matrices H1 = E1, 1 - E. +I.• +I whose action on Vis to fix e1, send en+l to its negative, and kill all the remaining basis vectors; we will correspondingly take as basis for the dual vector space ~· the dual basis L1, where (L1, H1) = h1,r We have already seen how the diagonal matrices act on the algebra of all matrices, so that it is easy to describe the action of~ on g. For example, for

16. Symplectic Lie Algebras

240

1 ~ i, j ~ n the matrix E1,1 E gi 2.C is carried into itself under the adjoint action of H1, into minus itself by the action of H1, and to 0 by all the other HA; and the same is true of the matrix E•+J,nH· The element

x,.} =

E,,J - En+},n+l E sp2.c

is thus an eigenvector for the action of~. with eigenvalue L 1 - L1. Similarly, fori,;. j we see that the matrices E,,.+J and E1••+1 are carried into themselves by H 1 and H1 and killed by all the other Hk; and likewise E.H.J and E•+J,I are each carried into their negatives by 111 and H1 and killed by the others. Thus, the elements

¥,, 1 = E,,.+J + EJ,•+I and Z 1,1 = E.H.J + E•+J,,

are eigenvectors for the action of 1), with eigenvalues L 1 + L 1 and - L, - L1, respectively. Finally, when f = j the same calculation shows that E1,n+1 is doubled by H, and killed by all other H1; and likewise E. H. I is sent to minus twice itself by H 1 and to 0 by the others. Thus, the elements U, = El,n+l

and

V,

= E.H,I

are eigenvectors with eigenvalues 2L1 and - 2L1, respectively. In sum, then, the roots of the Lie algebra !ip 2.C are the vectors ± L 1 ± L1 E ~·· In the first case n = l, of course we just get the root diagram of s[ 2 C, which is the same algebra as sp 2C. In case n = 2, we have the diagram

(16.2)

As in the case of the special linear Lie algebras, probably the easiest way to determine the Killing form on !IP 2 .C (at least up to scalars) is to use its

§16.1. The Structure of Sp 2.C and

11p 2.C

24

invariance under the automorphisms of sp 2.C preserving (J. For example, w have the automorphisms of sp 2.C induced by permutations of the basi vectors e, of V: for any permutation q of { l, 2, ... , n} we can define a1 automorphism of V preserving Q by sending e 1 to e~ 111 and e.H to e,;+~ 111 , an• this induces an automorphism of sp 2.C preserving IJ and carrying 111 to /1"111 Also, for any i we can define an involution of V-and thereby of sp 2 .C-b: sending e1 to e•H• e.+l to -e1, and all the other basis vectors to themselves this will have the effect of sending H, to - H, and preserving all the other Hi Now, the Killing form on I) must be invariant under these automorphisms from the first batch it follows that for some pair of constants ex and fJ we mus have

and B(H1,

H.J) = fJ

for i "' j;

from the second batch it follows that, in fact, P= 0. Thus, B is just a multiplf of the standard quadratic form B(H1, H1) = h1, 1, and the dual form correspond· ingly a multiple of B(L, L1) = b1,1; so that the angles in the diagram above arf correct. Also as in the case of sl.C, one can also compute the Killing form direct!) from the definition: B(l/, H') = I cx(H)cx(H'), the sum over all roots ex. F01 H = L,a,H1 and H' = L,b1H, this gives B(ll, H') as a sum

I

(a,

+ a1)(b1 + b1) + 2 I

(2a,)(2bi)

I

It-]

+ L (a, - a1)(b1 - b1) l,;,j

which simplifies to B(H, H') = (4n

+ 4HI a,b,).

(16.3)

Our next job is to locate the distinguished copies s. of si 2 C, and the corresponding elements 11. E IJ. This is completely straightforward. We start with the eigenvalues L 1 - L1 and L1 - L 1 corresponding to the elements X,.i and X1,,; we have [X1, 1, X1,,] = [E,, 1 - E•+J,•+" EJ.i- E•+i,n+J]

+ [E•+J,•+" E.H.•+J] E,_,- EJ.J + E•+J.•+J- E.+l,n+l

= [E,,,, E,,,J =

= H,-H1.

Thus, the distinguished element HL,-l.1 is a multiple of H,- 111. To see what multiple, recall that HL,-L1 should act on X 1•1 by multiplication by 2 and on X1, 1 by multiplication by - 2; since we have ad(/11 - H1)(X1)

=

((L1 - L1)(H1 - Hi))· X 1,1

=

2x,_ 1,

16. Symplectic Lie Algebras

242

we conclude that HL,-LJ

= H,- H).

Next consider the pair of opposite eigenvalues L, + L1 and - L, - L1, corresponding to the eigenvectors lf.J and Z 1, 1. We have

[l'f, 1, Z 1, 1] = [E1,n+J + EJ.n+l• En+I,J

+ En+J,;]

= tEI,n+J• En+},;]+ [Ej,n+l• En+l,j] =

E,,,- E•+J,n+J + EJ.J-

= I-1,

E.H.nH

+ Hr

We calculate then ad(H,

+ 1/1)(1';) = ((L, + L1)(H, + 111))·1';, 1 =

2·lf.J·

so we have and similarly

H_ 1_,-t1 =

-

H1 - H1.

Finally, we look at the pair of eigenvalues ±2L 1 coming from the eigen· vectors u, and "'·To complete the span of U1 and "'to a copy of s1 2 C we add [U,, V,] = [E 1.•+" E.+l,,]

=H,. Since ad(H1)(U1) = (2L1(H1))· U, = 2·U,

we conclude that the distinguished element HlL, is H1, and likewise H_ 2 t, = -111• Thus, the distinguished elements {H~} c [) are { ±H 1 ± H1; ± Ht}; in particular, the weight lattice Aw of linear forms on r, integral on all the H4 is exactly the lattice of integral linear combinations of the L,. In Diagram (16.21 for example, this is just the lattice of intersections ofthe horizontal and vertical lines drawn; observe that for all n the index [Aw: t\R] of the root lattice in the weight lattice is just 2. Next we consider the group of symmetries of the weights of an arbitrary representation of sp 2 .C. For each root a we let W,. be the involution in I)'" fixing the hyperplane n~ given by (H4 , L) = 0 and acting as -1 on the line spanned by a; we observe in this case that, as we claimed will be true in general, the line generated by a is perpendicular to the hyperplane n•. so that the involution is just a reflection in this plane. In the case n = 2, for example,

§16.1. The Structure ofSp 2.C and tpz.C

243

we get the dihedral group generated by reflections around the four lines drawn through the origin:

so that the weight diagram of a representation of sp 4 C will look like an octagon in general, or (in some cases) a square. In general, reflection in the plane llu., given by (H1, L) = 0 will simply reverse the sign of L1 while leaving the other L1 fixed; reflection in the plane (If,- H1, L) = 0 will exchange L 1 and L1 and leave the remaining Lt alone. The Weyl group liD acts as the full automorphism group of the lines spanned by the L 1 and !its into a sequence

l

-+ (Z/2Z)" -+

liD __. 6.

-+

l.

Note that the sequence splits: liD is a semidirect product of s. and (Z/2Z)". (This is a special case of a wreath product.) In particular the order of liD is 2"n!. We can choose a positive direction as before:

l([a 1L1) = c 1 a 1

+ ... + c.a.,

C1

>

C2

> ''' >

Cn

> 0.

The positive roots are then

R+

= {L 1 + L1},,. 1 v {L1 - L1}1<J•

(16.4)

with primitive positive roots {L, - Ll+ 1 } l=t ...... -• and 2L•. The corresponding (closed) Weyl chamber is 1r

= {a 1 L 1 + a 2 L2 + ... + a.L.: a 1 ~ a2 ~ ... ~a.~ 0};

(16.5)

note that the walls of this chamber-the cones

{.L, a1L 1: a 1 > · · · > a1 =

a 1+1 > · · · > a. > 0}

and

{I, a 1L1: a 1 > a2 > · · · > a. = 0} ·Jie in the hyperplanes llL,-L,., and ll2L. perpendicular to the primitive positive or negative roots, as expected.

244

16. Symplectic Lie Algebras

§16.2. Representations of sp 4 C Let us consider now the representations ofthe algebra sp 4 C specifically. Recall that, with the choice of Weyl chamber as above, there is a unique irreducible representation r~ of sp 4 C with highest weight IX for any IX in the intersection of the closed Weyl chamber "'r with the weight lattice: that is, for each lattice vector in the shaded region in the diagram

Any such highest weight vector can be written as a non-negative integral linear combination of L 1 and L 1 + L 2 ; for simplicity we will just write r•. b for the irreducible representation r.L, +b(L, +L,) with highest Weight aL 1 + b(L 1 + L 2 ) =(a+ b)L 1 + bL 2 • To begin with, we have the standard representation as the algebra of endomorphisms of the four-dimensional vector space V; the four standard basis vectors e., e2 , e 3 , and e4 are eigenvectors with eigenvalues L 1 , L 2 , - L., and - L 2 , respectively, so that the weight diagram of Vis

§16.2. Representations of sp 4 C

245

Vis just the representation r •. o in the notation above. Note that the dual of this representation is isomorphic to it, which we can see either from the symmetry of the weight diagram, or directly from the fact that the corresponding group representation preserves a bilinear form V x V-+ C giving an identification of v with v•. The next representation to consider is the exterior square NV. The weights of NV, the pairwise sums of distinct weights of V, are just the linear forms ±L1 ± L1 {each appearing once) and 0 {appearing twice, as L 1 - L 1 and L 2 - L 2 ), so that its weight diagram looks like

Clearly this representation is not irreducible. We can see this from the weight diagram, using Observation 14.16: there is only one way of getting to the weight space 0 from the highest weight L 1 + L 2 by successive applications of the primitive negative root spaces 9-L,+L, (spanned by xl,l = E2,1 - E3,4) and 9-lL, {spanned by V2 = E4 , 2 )-that is, by applying first V2 , which takes you to the weight space of L 1 - L 2 , and then X 2 , 1 -and so the dimension of the zero weight space in the irreducible representation r 0 • 1 with highest weight L 1 + L 2 must be one. Of course, we know in any event that /\ 2 V cannot be irreducible: the corresponding group action of Sp 4 C on V by definition preserves the skew form Q e NV• ~NV. Either way, we conclude that we have a direct sum decomposition

NV=

w~c.

where W is the irreducible, five-dimensional representation of sp 4 C with highest weight L 1 + L 2-in our notation, r 0 , 1 -and weight diagram

16. Symplectic Lie Algebras

246

Let us consider next some degree 2 tensors in V and W To begin with, we can write down the weight diagram for the representation Sym 2 V; the weights being just the pairwise sums of the weights of V, the diagram is

This looks like the weight diagram of the adjoint representation, and indeed that is what it is: in terms of the identification of V and V* given by the skew form Q, the relation (16.1) defining 'he symplectic Lie algebra says that the subspace

sp 4 C c Hom(V, V) = V®

v• =

V® V

is just the subspace Sym 2 V c V® V. In particular, Sym 2 V is the irreducible representation r 2 , 0 with highest weight 2L 1 • Next, consider the symmetric square Sym 2 W, which has weight diagram

247

§16.2. Representations of !1p 4 C

To see if this is irreducible we first look at the weight diagram: this time there are three ways of getting from the weight space with highest weight2L 1 + 2L 2 to the space of weight 0 by successively applying X2.1 = E 2 , 1 - E 3 , 4 and V2 = E4 , 2 , so if we want to proceed by this method we are forced to do a little calculation, which we leave as Exercise 16.7. Alternatively, we can see directly that Sym 2 W decomposes: the natural map given by wedge product NV(i9NV -+NV=

c

is symmetric, and so factors to give a map Sym 2 (NV))-+ C. Moreover, since this map is well defined up to scalars-in particular, it does not depend on the choice of skew form Q-it cannot contain the subspace Sym 2 W c Sym 2 (/VV)) in its kernel, so that it restricts to give a surjection qJ:

Sym 2 W ...... c.

This approach would appear to leave two possibilities open: either the kernel of this map is irreducible, or it is the direct sum of an irreducible representation and a further trivial summand. In fact, however, from the principle that an irreducible representation cannot have two independent invariant bilinear forms, we see that Sym 2 W can contain at most one trivial summand, and so the former alternative must hold, i.e., we have (16.6)

Exercise 16.7•. Prove (16.6) directly, by showing that if v is a highest weight vector, then the three vectors X 2 • 1 V2 X 1 , 1 V1 v, X 2 , 1 X 1 , 1 V2 V2 t', and J.iX2 , 1 X 2 , 1 V1 v span a two-dimensional subspace ofthe kernel of lf'· Exercise 16.8. Verify that !VW ~ Sym 2 V. The significance of this isomorphism will be developed further in Lecture 18.

248

16. Symplectic Lie Algebras

Lastly, consider the tensor product V ® W. First, its weight diagram:

This obviously must contain the irreducible representation r,, 1 with highest weight 2L 1 + L 2 ; but it eannot be irreducible, for either of two reasons. First, looking at the weight diagram, we see that r 1 , 1can take on the eigenvalues ± L 1 with multiplicity at most 2, so that V ® W must contain at least one copy of the representation V. Alternatively, we have a natural map given by wedge product

": V®NV-+NV= V* = V; and since this map does not depend on the choice of skew form Q, it must restrict to give a nonzero (and hence surjective) map rp: V® W-+ V.

Exercise 16.9. Show that the kernel of this map is irreducible, and hence that we have

v ® w = r1.1 $ v. What about more general tensors? To begin with, note that we have established the existence half of the standard existence and uniqueness theorem (14.18) in the case of sp 4 C: the irreducible representation r•. b may be found somewhere in the tensor product Sym•v ® SymbW. The question that remains is, where? In other words, we would like to be able to say how these tensor products decompose. This will be, as it was in the case of sl 3 C, nearly tantamount (modulo the combinatorics needed to count the multiplicity with which the tensor product Sym"V ® SymbW assumes each of its eigenvalues) to specifying the multiplicities of the irreducible representations r •. b· Let us start with the simplest case, namely, the representations Sym"V. These have weight diagram a sequence of nested diamonds D; with vertices at al. 1 , (a - 2)L 1 , etc.:

§16.2. Representations ofsp 4 C

249

Moreover, it is not hard to calculate the multiplicities of Sym"V: the multiplicity on the outer diamond D1 is one, of course; and then the ·multiplicities will increase by one on successive rings, so that the multiplicity along the diamond D, will be i. Exercise 16.10. Using the techniques of Lecture 13, show that the representations Sym"V are irreducible. The next simplest representations, naturally enough, are the symmetric powers SymbW of W These have eigenvalue diagrams in the shape of a sequence of squares S1 with vertices at b(L 1 + L 2 ), (b- l)(L 1 + L 2 ), and so on:

---·---·---.

I_·I

..___.____ _

. ·---

Here, however, the multiplicities increase in a rather strange way: they grow quadratically, but only on every other ring. Explicitly, the multiplicity will be one on the outer two rings, then 3 on the next two rings, 6 on the next two; in general, it will be i(i + l)/2 on the (2i- I)st and (2i)th squares S21 _ 1 and

s2,·

250

16. Symplectic Lie Algebras

Exercise 16.ll. Show that contraction with the skew form rp E Sym 2 W* introduced in the discussion of Sym 2 W above determines a surjection from SymbW onto Symb-lW, and that the kernel of this map is the irreducible representation ro.b with highest weight b(L 1 + L 2 ). Show that the multiplicities of ro,b are ion the squares s21-1 and sll described above. We will finish by analyzing, naively and in detail, one example of a representation r •. b with a and b both nonzero, namely, r 2 , 1; one thing we may observe on the basis of this example is that there is not a similarly simple pattern to the multiplicities ofthe representations r.,b with general a and IJ. To carry out our analysis, we start of course with the product Sym 2 V ® W. We can readily draw the weight diagram for this representation; drawing only one-eighth of the plane and indicating multiplicities by numbers, it is

We know that the representation Sym 2 V ® W contains a copy of the irreducible representation rl.l with highest weight 2L 1 + (L 1 + L 2 ); and we can see immediately from the diagram that it cannot equal this: for example, r 2 , 1 can take the weight 2L 1 with multiplicity at most 2 (if v e r 2 , 1 is its highest weight vector, the corresponding weight space (r2 • 1hL, c r 2 , 1 will be spanned by the two vectors X 2 , 1{V2 (v)) and V2 (X 2 ,.(v))); since it cannot contain a copy of the representation r 0 , 2 (the multiplicity of the weight 2(L 1 + L 2 ) being just one) it follows that Sym 2 V ® W must contain a copy of the representation rl,O = SymlV. We can, in this way, narrow down the list of possibilities a good deal. For example, r 2 , 1 cannot have multiplicity just one at each of the weights 2L 1 and L 1 + L 2 : if it did, Sym 2 V ® W would have to contain two copies of Sym 2 V and a further two copies of W to make up the multiplicity at L 1 + L 2 ; but since 0 must appear as a weight of r 2 , 1, this would give a total multiplicity of at least 7 for the weight 0 in Sym 2 V ® W. Similarly, r 2 , 1 cannot have multiplicity I at 2L 1 and 2 at L 1 + L 2 : we would then have two copies of Sym 2 V and one of W in Sym 2 V ® W; and since the multiplicity of 0 in r 2 • 1 will in this case be at least 2 (being greater than or equal to the multiplicity of L 1 + L 2 ), this would again imply a multiplicity of at least 7 for the weight 0

251

§16.2. Representations of sp4 C

in Sym 2 V ® WIt follows that Sym 2 V ® W must contain exactly one copy of Sym 2 V; and since the multiplicity of L 1 + L 2 in r 2 , 1 is at most 3, it follows that Sym 2 V ® W will contain at least one copy of ro, 1 = Was well.

Exercise 16.12. Prove, independently of the above analysis, that Sym 2 V ® W must contain a copy of Sym 2 V and a copy of W by looking at the map cp: Sym 2 V ® W-+ V ® V

obtained by sending

u·v®(w A z)Hu®Q(v A w A z)

+ v®Q(u A w A z),

where we are identifying NV with the dual space v• and denoting by V the isomorphism induced by the skew form Q on V. Specifically, show that the image of this map is complementary to the line spanned by the element Q e Nv• =NV c V® V.

Q: V*-+

The above leaves us with exactly two possibilities for the weights of r 2 , 1 : we know that the multiplicity of 2L 1 in rl.l is exactly 2; so either the multiplicities of L 1 + L 2 and 0 in r 2 , 1 are both 3 and we have Sym 2 V® W = or the multiplicities of L 1 Sym 2

r2 , 1 $

Sym 2 V$ W;

+ L 2 and 0 in r 2 , 1 are both 2 and we have

V® W =

r 2 , 1 $Sym 2 V$

W«l2 •

Exercise 16.13. Show that the former of these two possibilities actually occurs, by (a) Showing that if v is the highest weight vector in r 2 , 1 c Sym 2 V ® W, then the images (X2, d2V2(v), X 2, 1 V2X 2, 1(v), and V2 (X 2 , .) 2v are independent; and (redundantly) (b) Showing that the representation Sym 2 V ® W contains only one highest weight vector of weight L 1 + L 2 . The weight diagram of r2 , 1 is therefore

252

16. Symplectic Lie Algebras

We see from all this that, in particular, the weights of the irreducible representations of sp 4 Care not constant on the rings of their weight diagrams. Exercise 16.14. Analyze the representation V ® Sym 2 W of sp 4 C. Find in particular the multiplicities of the representation r l , l ' Exercise 16.15. Analyze the representation Sym 2 V®Sym 2 W ofsp 4 C. Find in particular the multiplicities of the representation 1 2 • 2 •

LECTURE 17

sp 6 C and sp 211 C

In the first two sections of this lecture we complete our classification of the representations of the symplectic Lie algebras: we describe in detail the example of sp 6 C, then sketch the representation theory of symplectic Lie algebras in general, in particular proving the existence part ofTheorem 14.18 for sp 2.C.In the final section we describe an analog for the symplectic algebras of the construction given in§ 15.3 of the irreducible representations of the special linear algebras via Weyl's construction, though we postpone giving analogous formulas for the decomposition of tensor products of irreducible representations. Sections 17.1 and 17.2 are completely elementary, given the by now standard multilinear algebra of Appendix B. Section 17.3, like §15.3, requires familiarity with the contents of Lecture 6 and Appendix A; but, like that section, it can be skipped without affecting most of the rest of the book.

§17.1: Representations of 1ip6 C §17.2: Representations of the symplectic Lie algebras in general §17.3: Weyl's construction for symplectic groups

§17.1. Representations ofsp 6 C As we have seen, the Carlan algebra I) of sp 6 C is three-dimensional, with the linear functionals L 1 , L 2 , and L 3 forming an orthonormal basis in terms of the Killing form; and the roots of sp 6 C are then the 18 vectors ± L 1 ± L1• We can draw this in terms of a "reference cube" in I}* with faces centered at the points ± L 1; the vectors ± L 1 ± L1 with i 'f j are then the midpoints of edges of this reference cube and the vectors ± 2L 1 the midpoints of the faces of a cube twice as large. Alternatively, we can draw a reference octahedron with vertices at the vectors ± 2L1; the roots ± L 1 ± L 1 with i # j will then be the

254

midpoints of the edges of this octahedron:

n,

or, if we include the reference cube as well, as

(17.1)

This last diagram, however ineptly drawn, suggests a comparison with the root diagram of sl 4 C; in fact the 12 roots of sp 6 C of the form ± L 1 ± L1 for i ¥- j are congruent to the l2 roots of sl4 C. In particular, the Weyl group of !!p6 C will be generated by the Weyl group ofsi 4 C, plus any of the additional three reflections in the planes perpendicular to the L 1 (i.e., the planes parallel to the faces of the reference cube in the root diagram of either Lie algebra). We can indicate the planes perpendicular to the roots of sp 6 C by drawing where they cross the visible part of the reference cube:

§17.1. Representations or sp 6 C

255

We see from tl~is that the effect or the additional reflections in the Weyl group of sp 6 C on the Weyl chamber of sl 4 Cis simply to cut it in half; whereas the Weyl chamber ofsi 4 C looked like

the Weyl chamber of sp 6 C will look like just the upper half of this region:

In terms of the reference octahedron, this is the cone over one part of the barycentric subdivision or a face:

256

or, if we rotate 90° around the vertical axis in an attempt to make the picture clearer,

(17.2)

We should remark before proceeding that the comparison between the root systems of the special linear algebra sl4 C and the symplectic algebra sp 6 C is peculiar to this case; in general, the root systems of sl•+t C and sp 2.C will bear no such similarity. As we saw in the preceding lecture, the weight lattice of sp 6 C consists simply of the integral linear combinations of the weights L 1• In particular, the intersection of the weight lattice with the closed Weyl chamber chosen above will consist exactly of integral linear combinations a 1 L 1 + a 2 L 2 + a 3 L 3 with a1 ;;;::.: a2 ;;;::.: a3 ;;;::.: 0. By our general existence and uniqueness theorem, then, for every triple (a, b, c) of non-negative integers there will exist a unique irreducible representation of sp 6 C with highest weight aL 1 + b(L 1 + L 2 ) +

§17.1. Representations ofsp 6 C

2

c(L 1 + L 2 + L 3 ) =(a + b + c)L 1 + (b + c)L 2 + cL 3 ; we will denote this re resentation by r •. b,r and will demonstrate its existence in the following. We start by considering the standard representation of !5p 6 C on V = { The eigenvectors of the action off} on V are just the standard basis vecto e1, and these have eigenvalues ± L 1, so that the weight diagram of V looks Iii the midpoints of the faces of the reference cube (or the vertices of ; octahedron one-half the size of the reference octahedron):

-LI

In particular, Vis the representation f 1 ,o,o· Since we are going to want to find a representation with highest weigl L 1 + L 2 , the natural thing to look at next is the second exterior power N of the standard representation. This will have weights the pairwise sum 1 distinct weights of V, or in other words the 12 weights ± L 1 ± L1 with i .f. and the weight 0 taken three times. This is not irreducible: by definition th action of sp 6 C on the standard representation preserves a skew form, so th! the representation on NV will have a trivial summand. On the other han< the skew form on V preserved by sp 6 C, and hence that trivial summand ( NV, is unique; and since all the nonzero weights of NV occur with mull plicity l and are conjugate under the Weyl group, it follows that the compl< ment W of the trivial representation inN Vis irreducible. So W = ro.t.o· As in previous examples, we can also see that NV is not irreducible b using the fact (Observation 14.16) that the irreducible representation f 0 , 1• with highest weight L 1 + L 2 will be generated by applying to a single highe~ weight vector v the root spaces St,-I.,• 9I.,-L,• and 9-lL, corresponding t• primitive negative roots. We can then verify that in the irreducible represent a tion W with highest weight L 1 + L 2 , there are only three ways of going fron the highest weight space to the zero weight space by successive application o these roots spaces: we can go L1

+ L 2 -+ L 1 + L 3 -+ L 1 - L 3 -+ L 1 - L 2 ~-~ ~ L 2 + L 3 -+ L 2 - L 3 -+ 0

258

Exercise 17.3. Verify this, and also verify that the lower two routes to the zero-weight space in NV yield the same nonzero vector, and that the upper route yields an independent element of NV, so that 0 does indeed occur with multiplicity 2 as a weight of ro. 1 , 0 • To continue, we look next at the third exterior power NV of the standard representation; we know that this will contain a copy of the irreducible representation ro.o,t with highest weight L 1 + L 2 + L 3 . The weights of NV are of two kinds: we have the eight sums ±L 1 ± L 2 ± L 3 , corresponding to the vertices of the reference cube and each occurring once; and we have the weights ± L 1 each occurring twice (as ± L 1 + L1 - Li and ± L 1 + L~ - L~). The weight diagram thus looks like the vertices of the reference cube together with the midpoints of its faces:

Now, the weights ± L 1 must occur in the representation r 0 , 0 , 1 with highest weight L 1 + L 2 + L 3 , since they are congruent to L 1 + L 2 + L 3 modulo the root lattice and lie in the convex hull of the translates of L 1 + L 2 + L 3 under the Weyl group (that is, they lie in the closed reference cube). But they cannot occur with multiplicity greater than 1: for example, the only way to get from the point L 1 + L 2 + L 3 to the point L 1 by translations by the basic vectors L 2 - L 1 , L 3 - L. 2 , and -2L 3 pictured in Diagram (17.1) above (while staying inside the reference cube) is by translation by - 2L 3 1irst, and then by L 3 - L 2 • it follows that the multiplicities of the weights ± L 1 in ro.o.t are l. On the other hand, we have a natura! map

NV-+ V obtained by contracting with the element of NV* preserved by the action of sp 6 C, and the kernel of this map, which must contain the representation ro.o. 1 , will have exactly these weights. The kernel of ({J is thus the irreducible representation with highest weight L 1 + L2 + L 3 ; we will call this representation U for now. At this point, we have established the existence theorem for representations of sp 6 C: the irreducible representation r. h c with highest weight (a+ b + c)L 1 +(a + b)L 2 + cL 3 will occur inside th~ ~epresentation

Sym"V ® SymbW ® Sym< U.

§ 17.2. Representations of sp 2 • C in General

259

For example, suppose we want to find the irreducible representation r 1 , 1 • 0 with highest weight 2L 1 + L 2 • The weights of this representation will be the 24 weights ± 2L1 ± L1, each taken with multiplicity l; the 8 weights ± L 1 ± L 2 ± L 3 , taken with a multiplicity we do not a priori know (but that the reader can verify must be either I or 2), and the weights ± L 1 taken with some other multiplicity. At the same time, the representation V ® W, which contains rr.t.o• will take on these weights, with multiplicities I, 3, and 6, respectively. In particular, it follows that V ® W will contain a copy of the irreducible representation U with highest weight L 1 + L 2 + L 3 as well; alternatively, we can see this directly by observing that the wedge product map

V®NV-.NV factors to give a map

V®W-+U and that r,,,,o must lie in the kernel of this map. To say more about the location of r 1 , 1 , 0 inside V ® W, and its exact weights, would require either explicit calculation or something like the Weyl character formula. We will see in Lecture 24 how the latter can be used to solve the problem; for the time being we leave this as Exercise 17.4. Verify by direct calculation that the multiplicities of the weights ofr1 , 1 , 0 are I, 2, and 5, and hence that the kernel of the map qJ above is exactly the representation ri,J,O•

§17.2. Representations of sp 2 nC in General The general picture for representations of the symplectic Lie algebras offers no further surprises. As we have seen, the weight lattice consists simply of integral linear combinations of the L 1• And our typical Weyl chamber is a cone over a simplex in n-space, with edges the rays defined by a1

= a2 = · · · = a1 > ai+ 1 = · · · = a. =

0.

The primitive lattice element on the ith ray is the weight w 1 = L 1 + ··· + L 1, and we may observe that, similarly to the case of the special linear Lie algebras, these n fundamental weights generate as a semigroup the intersection of the closed Weyl chamber with the lattice. Thus, our basic existence and uniqueness theorem asserts that for an arbitrary n-tuple of natural numbers (a to ••• , a.) e N" there will be a unique irreducible representation with highest weight

= (a 1 + ··· + a.)L 1 + (a 2 + ··· + a.)L 2 + ··· + a.L•.

260

As before, we denote this by r ......... :

These exhaust all irreducible representations of sp 2,C. We can find the irreducible representation V(k) = ro, .... t, .... o with highest weight L 1 + ··· + Lt easily enough. Clearly, it will be contained in the kth exterior power NV of the standard representation. Moreover, we have a natural contraction map

defined by ll'k(v 1 1\

.. • 1\

v1 ) =

L Q(v,. vJ)( -tt+J- 1 v1 1\ .. • 1\ 01 1\ .. • 1\ OJ 1\ • • • 1\ v1 I<)

(see §8.3 of Appendix B for an intrinsic definition and explanation). Since the representation N- 2 v does not have the weight L 1 + ···+Lt. the irreducible representation with this highest weight will have to be contained in the kernel of this map. We claim now that conversely Theorem 17.5. For I ::;;; k :s;; 11, tfte kernel of the map ll'k i.~ exactly the irreducible representation v(t) = ro, ... ,O,I,O, ... ,o with highest weight Ll + ... +Lt. PROOF. Clearly, it is enough to show that the kernel of ll'k is an irreducible representation of sp 2,C. We wiii do this by restricting to a subalgebra of sp 2.C isomorphic to si,C, and using what we have learned about representations of si,C. To describe this copy of sl.C inside sp 2.C, consider the subgroup G c Sp2.C of transformations of the space v = C 2 " preserving the skew form Q introduced in Lecture 16 and preserving as well the decomposition V = C {e 1 , ••• , e.} (!) C{en+l, ... , e 2 ,}. These can act arbitrarily on the first factor, as long as they do the opposite on the second; in coordinates, they are the matrices

G=

{(~

,;_ 1 ) ,

X e GL,c}.

We have, correspondingly, a subalgebra

s=

{(~ -~A)' Aesl,c} c

sp2,C

isomorphic to sl, C. Now, denote by W the standard representation of si.C. The restriction of the representation V of sp 2,C to the subalgebra s then splits V= W$W* into a direct sum of Wand its dual; and we have, correspondingly,

§17.2. Representations of sp 1.C in General

NV=

EB

261

(1\"W®NW*).

•+b=l

How does the tensor product NW ® Nw• decompose as a representation ofsl.C7 We know the answer to this from the discussion in Lecture 15 (see Exercise 15.30): we have contraction maps

'l'•. b: NW®Nw•--+ N- 1 W®/\H w•; and

the kernel of 'l'.,b is the irreducible representation

w(•.bJ

=

ro... .,O,J,O,. .. ,O,I,O, ... With(if,say,a ~ 11- b) highest Weight2Lr + '" +2L. +

L.+J + · · · + L.-b· The restriction of NV to

EB

NV=

9

is thus given by

w(o,h)

•+bsk

a+bek(l)

and by the same token, Ker(~PA) = .~t

W'•.bJ.

Note that the actual highest weight factor in the summand

Wl•.b)

c Ker(~PA) c

NV is the vector

Exercise 17.6. Show that more generally the highest weight vector in any summand W 1•.b) c NV is the vector W(a,b)

= el

A ..• A

e.

A eln-k+a+l A ... A

= el

A ... A

e.

A

ezn-t+a+J

eln

A Qlt-a-b)/2

A ..• A el• A

(L (e, A e.H))(t-o-b)/2.

By the above, any subspace of Ker(~PA) invariant under sp 2 .C must be a direct sum, over a subset of pairs (a, b) with a + b = k, of subspaces w 1•.bJ. But now (supposing for the moment that k < n) we observe that the element

z•.• -b = Ez.-b,o + E.+ •. n-b E S~'l.c carries the vector

wl•.bJ

into

Y..+t,n-b+l

carries

w 1•,b)

to

wl•-l.b+J)

= Eo+l,2n-b+l + E,.-b+J,n+o+l E 9Vz,.C

wl•+l.b-1).

In case a + b = k

v., = carries the vector

wi•.&J

and, likewise,

into

E•+•.• E

wl•-t,b+l),

= 11, we see similarly that

svz.C

and

u.+l = E.+J,n+a+l E

sv2.c

carries wl•.bl to wl•+t,b-ll. Thus, any representation of sp 2.C contained in Ker(rpt) and containing any one of the factors WI•. b) will contain them all, and we are done. 0

262

Exercise 17.7. Another way to conclude this proof would be to remark that, inasmuch as all the wl•.bJ above are eigenvectors of different weights, any highest weight vector for the action of sp 2 .C on ker(f/\) c NV would have to be (up to scalars) one of the wl•.hl.Jt would thus be sufficient to find, for each (a, b) with a + b = k other than (a, b) = (k, 0), a positive root a such that g.(w 1d,hJ) ~ 0. Do this. Note that, having found the irreducible representations VllJ = f 0 ... ., 1 , ... ,o with highest weight L 1 + ... + L., any other representation of sp 2 .C will occur in a tensor product of these; specifically, the irreducible representation r., ....... with highest weight a 1 L 1 + ... + a.(L 1 + .. · + L.) will occur in the product Sym•• V ® Sym•• Vl2l ® · · · ® Sym"" VI•J. One further remark is that there exist geometric interpretations of the action of si 2.C on the fundamental representations v••J. We have said before that the group PSp 2 .C may be characterized as the subgroup of PGL 2.C carrying isotropic subspaces of V into isotropic subspaces. At the same time, PGL 2.C acts on the projective space P(NV) as the connected component of the identity in the group of motions of this space carrying the Grassmannian G = G(k, V) c P(NV) into itself. Now, the subset GL c G of k-dimensional isotropic subspaces of V is exactly the intersection of the Grassmannian G with the subspace P(V 1.,) associated to the kernel of the map ((J above; so that PSp 2.C will act on P(V't1) carrying GL into itself and indeed when 1 < k ~ n may be characterized as the connected component of the identity in the group of motions of P( V1tl) preserving the variety GL. Exercise 17.8. Show that if k >

11

the contraction f/\ is injective.

§17.3. Weyl's Construction for Symplectic Groups We have just seen how the basic representations for sp 2.C can be obtained by taking certain basic representations of the larger Lie algebra sl 2 .C-in this case, NV fork~ n-and intersecting with the kernel of a contraction constructed from the symplectic form. In fact, all the representations of the symplectic Lie algebras can be given a similar conrete realization, by intersecting certain of the irreducible representations of sl 2 .C with the intersections of the kernels of all such contractions. Recall from Lectures 6 and 15 that the irreducible representations of sl 2.C are given by Schur functors§ .l V, where 1 = (A 1 ;;:: • • • ;;:: ..t 2 • ;;:: 0) is a partition of some integer d = L: Aj, and V = C 2 ". This representation is realized as the image of a corresponding Young symmetrizer cA acting on the d-fold tensor product space V 04 . For each pair I = {p < q} of integers between l and d, the symplectic form Q determines a contraction ~,:

V 0 "-+ V 0 '"_ 2 ,,

(17.9)

§ 17.3. Weyl's Construction for Symplectic Groups

263

Let v<•> c V 04 denote the intersection of the kernels of all these contractions. These subspaces is mapped to itself by permutations, so v< 4 >is a subrepresentation of V 04 as a representation of the symmetric group 6 4 . Now let 1 §(A)V= V< 4>n§AV,

(17.10)

This space is a representation of the symplectic group Sp 2.C of Q, since and § 1 (V) are subrepresentations of V 04 over Sp 2.C.

v< 4 >

Theorem 17.11. The ,,pace §< 1 >(V) is nonzero if and only if the Young diagram of). has at most n rows, i.e., ).•+I = 0. In this ca.,e, §< 1>(V) is the irreducible representation of s~l 2 .C witlt highest weight l 1 L 1 + · ·· + l.L•. In other words, for an n-tuple (a 1 ,

•.. ,

a.) of non-negative integers

r., ....... =

s

v,

where). is the partition (a 1 + a2 + ···+a., a2 +···+a., ... , a.). The proof follows the patlern for the general linear group given in §6.2, but we will have to call on a basic result from invariant theory in place of the simple Lemma 6.23. We first show how to lind a complement to v< 4 >in V 04 . For example, if d = 2, then

v® 2 = v< 2 >$C·t/l, where 1/J is the element of V ® V corresponding to the quadratic form Q. In terms of our canonical basis, 1/1 = L(e1 ® e.+l- e.H ® e1). In general, for any I= {p < q} define 'I' I: v@«d-2) -+ v®• by inserting 1/J in the p, q factors. Note that 11>1 o '1'1 is multiplication by 2n =dim Von V014 - 1 1. We claim that (17.12)

To prove this, put the standard Hermitian metric ( , ) on V = C 2 ", using the given e1 as a basis, so that (ae, be1) = ~,1 ab. This extends to give a Hermitian metric on each v0•. We claim that the displayed equation is a perpendicular direct sum. This follows from the following exercise. Exercise 17.13. (i) Verify that for v, we V, (1/J, v ® w) = Q(v, w). (ii) Use (i) to show that Ker(ll> 1) = lm('l'1).l for each /. Now define F: c V 04 to be the intersection of the kernels of all r-fold contractions 11>1 , o · · · o 11> 1,, and set vd~i.

=I '1'1,

0 ... 0

'I'I,(V®(d-lr>).

(17.14)

1 This follows a classical notation of using ( ) for the symplectic group and [ ] for I he orthogonal group (all hough we have omitted the corresponding notation { ) for the general linear group).

264

Lemma 17.15. Tf1e te11Sor power V~"' decomposes into a direct sum

v®" = v<"> ED V/·!i ED

v..n ED··· ED V..
with p = [d/2], and, for all r ~ I,

F: = v<"> ED V..~i $

·· · $ V..~l.-1-2·

Exercise 17.16. (i) Show as in the preceding exercise that there is a perpendicular decomposition

V®"

= F.", WLT\"' !J 'I'/ 1

0 ••• 0

'I'1,. (V@Id-2rl) •

F:+J.

(ii) Verify that 'I',{F:- 2 ) c (iii) Show by induction that V®" is the sum of the spaces V..~f,. (iv) Finish the proof of the lemma, using (i) and (ii) to deduce that both 0 sums are orthogonal splittings. All the subspaces in these splittings are invariant by the action of the symplectic group Sp 2.C, as well as the action of the symmetric group 6,. In particular, we see that

(17.17) Exercise 17.18*. (i) Show that if s > n, then NV® V®!d-•1 is contained in

Lr 'I',(V®!"- 21 ), and deduce that §(V) = 0 if ..t.+J is not 0. (ii) Show that §(V) is not zero if ..t•+I

= 0.

For any pair of integers I from {I, ... , d}, define

a, = '1', o <1> 1 : v@"--+ v@d. From what we have seen, v<"> is the intersection of the kernels of all these endomorphisms. Note that the endomorphism of v®" determined by any symplectic automorphism of V not only commutes with all permutations of the factors 6 4 but also commutes with the operators 81 • We need a fact which is proved in Appendix F.2: Invariant Theory Fact 17.19. Any er!domorpl1ism of V®" that commutes with all permutations in 6 4 and all the operators 8 1 is a finite C-lir1ear com!Jination of operators of tl1e form A ® · · · ® A, for A E Sp 2.C. Now let B be the algebra of all endomorphisms of the space v<"> that areC-Iinearcombinationsofoperators of the form A®···® A, for A e Sp 2.C. Proposition 17.20. Tire algebra B is precisely the algebra of all endomorpl!isms of v<"> commuting with all permutations in 6 4 • PRooF. IfF is an endomorphism of v<"> commuting with all permutations of •· -·- -- •'-~- •ho on and

§17.3. Weyl's Construction for Symplectic Groups

265

Lt

zero on the complementary summand '1'1 (V@ 14 - 21 ) is an endomorphism that commutes with all permutations and all operators .91 . The fact that f. is a linear combination of operators from the symplectic group (which we know from Fact 17.19) implies the same for F. 0 Corollary 17.21. Tlae representations §(V) are irreducible representation.~ of Spz.C. PROOF. Since B is the commutator algebra to A = C[6,] acting on the space v< 4 >, Lemma 6.22 implies that (v<•>). CA is an irreducible B-module. But we have seen that (V<">) · cA = § 0 >V, and the proposition shows that being irreducible over B is the same as being irreducible over Sp 2.C. 0 Exercise 17.22*. Show that the multiplicity with which §(V) occurs in v< 4 > is the dimension rnA of the corresponding representation VA of 6 4 • As was the case for the Weyl construction over GL.C, there are general formulas for decomposing tensor products of these representations, as well as restrictions to subgroups Sp 2 ._ 2 C, and for their dimensions and multiplicities of weight spaces. We postpone these questions to Lecture 25, when we will have the Weyl character formula at our disposal. As we saw in Lecture 15 for GL.C, it is possibie to make a commutative algebra which we denote by s<·> = s<·>(v) out of the sum of all the irreducible representations ofSp 2.C, where V = C 2 " is the standard representation. Probably the simplest way to do this, given what we have proved so far, is to start with the ring A'(V, n) = Sym'(V E9 NV$ NV$···$ 1\"V) =

EB Sym"•(/\"V) ® · · · ® Sym"'(N V) ® Sym"•(V), tr., ...• ci,.

the sum over all 11-tuples a = (a 1 , ••• , a.) of non-negative integers. Define a ring §'(V, n) to be the quotient of A'(V, 11) by the ideal generated by the same relations as in (J 5.53). By the argument in §15.5, the ring §'(V, 11) Is the direct sum of all the representations §A(V) of GL(V), as). varies over all partitions with at most n parts. The decomposition V@ 4 = v< 4 >$ w< 4 > of (17.12) determines a decomposition V@d' CA = V(of) 'Cl $ W(d)' CA, Which is a decomposition §A(V) = §(V) $ .T (V) of representations of Sp 2 .C. We claim that the sum J<·> = EJJ1 J (V) is an ideal in §'(V, n) = EBA §A(V). This is easy to see using weights, since J (V) is the sum of all the representations in §A(V) whose highest weight is strictly smaller than).. This implies that the image of J (V) ® §p(V) in §A+p(V) is a sum of representations whose highest weights are less than ). + Jl, so they must be in J(V). The quotient ring is, therefore, the ring §<·>(V) we were looking for:

266

§<·> = §'(V, n)p<·> =

E9 §(V). A

In fact, the ideal;<·> is generated by elements oftheform x 1\ rp, wherex eN V, rp is the element in Nv corresponding to the skew form Q. An outline of the proof is sketched at the end of Lecture 25. The calculations, as well as other constructions of the ring, can be found in [L-T], where one can also find a discussion of functorial properties of the construction. For bases, see (DC-P], (L-M-S], and (M-S].

i ~ n- 2, and

LECTURE 18

Orthogonal Lie Algebras

In this and the following two lectures we carry out for the orthogonal Lie algebras what we have already done in the special linear and symplectic cases. A..q in those cases, we start by working out in general the structure of the orthogonal Lie algebras, describing the roots, root spaces, Weyl group, etc., and then go to work on lowdimensional examples. There is one new phenomenon here: as it turns out, all three of the Lie algebras we deal with in §18.2 are isomorphic to symplectic or special linear Lie algebras we have already analyzed (this will be true oflo 6 C as well, but of no other orthogonal Lie algebra). As in the previous cases, the analysis or the Lie algebras and their representation theory will be completely elementary. Algebraic geometry does intrude into the discussion, however: we have described the isomorphisms between the orthogonal Lie algebras discussed and special linear and symplectic ones in terms of projective geometry, since that is what seems to us most natural. This should not be a problem; there are many other ways or describing these isomorphisms, and readers who disagree with our choice can substitute their own. §18.1: SO,.C and so,.C §18.2: Representations of so 3 C, so 4 C, and so 5 C

We will take up now the analysis of the Lie algebras of orthogonal groups. Here there is, as we will see very shortly, a very big difference in behavior between the so-called "even" orthogonal Lie algebras so 2 .C and the "odd" orthogonal Lie algebras so 2.+ 1 C. Interestingly enough, the latter seem at first glance to be more complicated, especially in terms of notation; but when we analyze their representations we see that in fact they behave more regularly .than the even ones. In any event, we will try to carry out the analysis in parallel

§<·> = §"(V, n)j.J<·> =

EB §< 1 >(V). l

In fact, the ideau<·> is generated by elements of the form x " 1/J, where x E NV, 11- 2, and 1/1 is the element in NV corresponding to the skew form Q. An outline of the proof is sketched at the end of Lecture 25. The calculations, as welL.as other constructions of the ring, can be found in [L·T], where one can also find a discussion of functorial properties of the construction. For bases, see (DC-P], [L-M-S], and [M-S]. i:.:::;

LECTURE 18

Orthogonal Lie Algebras

In this and the following tw.1 lectures we carry out for the orthogonal Lie algebras what we have already don" it· the special linear and symplectic cases. As in those cases, we start by working out iu general the structure of the orthogonal Lie algebras, describing the roots, root spaces, Weyl group, etc., and then go to work on lowdimensional examples. There is one new phenomenon here: as it turns out, all three of the Lie algebras we deal with in §18.2 are isomorphic to symplectic or special linear Lie algebras we have already analyzed (this will be true of so 6 C as well, but of no other orthogonal Lie algebra). As in the previous cases, the analysis of the Lie algebras and their representatio11 theory will be completely elementary. Algebraic geometry does intrude into the dis.:ussion, however: we have described the isomorphisms between the orthogonal Lie algebras discussed and special linear and symplect;c ones in terms of projective geometry, since that is what seems to us most natural. This should not be a problem; there are many other ways of describing these isomorphisms, and readers who disagree with our choice can substitute their own. §18.1: SO,.C and 3n.,C §18.2: Representations of so 3 C, so 4 C, and so 5 C

§18.l. SO,.C anrl so,.C We will take up now the analysis of the Lie algebras of orthogonal groups. Here there is. as we will see very shortly, a very big difference in behavior between the so-called "even" orthogonal Li~; alt.;c.Lws so 2 .1C and the "odd" orthogonal Lie algebras !:.0 2 ,+ 1 1C. Interestingly enough, the latter seem at first glance to be more cm!lpitcated, especially in terms of notation; but when we analyze their rcpresemations we sec · !1at in fact tht y beh:ue more regularly than the even ones. [n any event, we w'!l try to carry uut the analysis in parallel

18. Orthogonal Lie Algebras

268

fashion for as long as is feasible; when it becomes necessary to split up into cases, we will usually look at the even orthogonal Lie algebras first and then consider the odd. Let V be a m-dimensional complex vector space, and Q: V XV-+

e

a nondegenerate, symmetric bilinear form on V. The orthogonal group SOme is then defined to be the group of automorphisms A of V of determinant 1 preserving Q-that is, such that Q(Av, Aw) = Q(v, w) for all v, we V-and the orthogonal Lie algebra some correspondingly consists of endomorphisms A: V-+ V satisfying Q(Av, w)

+ Q(o, Aw) = 0

(18.1)

for all v and wE V. As in the case of the symplectic Lie algebras, to carry out our analysis we want to write Q explicitly in terms of a basis for V, and here is where the cases of even and bdd 111 first separate. In case m = 211 is even, we will choose a basis for V in terms of which the quadratic form Q is given by Q(eh el+.) = Q(e;+n• e;)

=I

and

Q(e;, e1)

=0

if j ::1 i

± n.

The bilinear form Q may be expressed as Q(x, y) == 'x · M · y, where M is the 2n x 2n matrix given in block form as

M

=(~ ~}

the group sol.e is thus the group of 2n

X

2n matrices A satisfying

M ='A·M·A and the Lie algebra so 2.e correspondingly the space of matrices X satisfying the relation

'X·M

+ M·X = 0.

Writing a 2n x 2n matrix X in block form as

X=(~ ~) we have

'A)

'X·M= ( 'C 'D 'B and

269

§18.1. SO,.C and so,.C

M·X=(~ ~) so that this relation is equivalent to saying that tlte off-diagonal blocks Band C of X are skew-symmetric, and the diagonal blocks A and D of X are negative transposes of each other. Exercise 18.2. Show that with this choice of basis,

and so 2 C =C. The situation in case the dimension m of Vis odd is similar, if a little messier. To begin with, we will take Q to be expressible, in terms of a basis e 1 , ••• , e 2 ~+I for V, by Q(e1, e 1+,.)

= Q(e;+,, e 1) =

1 for 1 ~ i ~ n;

Q(e 2 ~+1• e 2 ,.+I)

= 1;

and Q(e 1, e)

=0

for all other pairs i, j.

The bilinear form Q may be expressed as Q(x, y) where M is the (2n

+

1) x (2n

+

M=

= 'x · M · y,

1) matrix

-**

( ~)

(the diagonal blocks here having widths 11, n, and 1). The Lie algebra so 2 ~+• C is correspondingly the space of matrices X satisfying the relation 'X· M + M ·X = 0; if we write X in block form as

then this is equivalent to saying that, as in the previous case, B and C are skew-symmetric and A and D negati11e transposes of each other; and in addition

E= -'H,F= -'G,andJ=O. With these choices, we may take as Cartan subalg<.obra-in both the even and odd cases-the subalgebra of matrices diagonal in this representation. 1 1 Note that if we had taken the simpler choice of {l, with M the identity matrix, the Lie algebra would have consisted of skew-symmetric matrices, and there would have been no nonzero diai!Onal matrices in the Lie ah1ebra.

18. Orthogonal Lie Algebras

270

The subalgebra I) is thus generated by the n 2n x 2n matrices H1 = E1, 1 E.+i,n+t whose action on Vis to fix e1, send en+i to its negative, and kill all the remaining basis vectors; note that this is the same whether m = 2n or 2n + 1. We will correspondingly take as basis for the dual vector space I)* the dual basis L1, where (L1, H1 ) = ~l.i' Given that the Cartan subalgebra of so 2 .C coincides, as a subspace of si 2 .C, with the Cartan subalgebra of sp 2 .C, we can use much of the description of the roots of sp 2.C to help locate the roots and root spaces of so 2 .C. For example, we saw in Lecture 16 that the endomorphism X 1,j

= E 1, j - En+j,n+i E SPzniC

is an eigenvector for the action of I) with eigenvalue L 1 - Li. Since X 1, 1 is also an element of so 2.C, we see that L1 - L 1 is likewise a root of so 2 .C, with root space generated by X 1, 1. Less directly but using the same analysis, we find that the endomorphisms

Y,,j =

Ei,n+j- Ej,n+i

and

are eigenvectors for the action of 1), with eigenvalues L 1 + L1 and - L 1 - L1, respectively (note that Y,, 1 and Z 1•1 do not coincide with their definitions in Lecture 16). In sum, then, the roots of tire Lie algebra so 2 .C are the vectors

{±L1 ± L1};., 1 c 1)*. The case of the algebra so 2.+ 1 Cis similar; indeed, ;,il the eigenvectors for the action of I) found above in so 2 .1C, viewed as endomorph isms of IC 2 "+ 1 , are likewise eigenvectors for the action of I) on so2n+riC. In addition, we have the endomorphisms

U;

= Ei,2n+l

Jli

= En+I,Zn+l- £2n+l,i

-

£2n+l,n+i

and which are eigenvectors with eigenvalues + L 1 and - L;. respectively. The roots of so 2 •+l C are thus the roots ± L 1 ± L1 of so 2.1C, together with additional roots ±L1• We note that we could have arrived at these statements without decomposing the Lie algebras some: the description (18.1) of the orthogonal Lie algebra may be interpreted as saying that, in terms of the identification of V with V* given by the form Q, somiC i:. just the Lie algebra of skew-symmetric enrlvmorphisms of V (an endom(•rphism being skew ..symmetric if it is equal to minus its transpose). That is, the adjoint representation uf som IC is isomorphic to the wedge product NV. In the even case m = 211, since the weights of V are ± L1 (inasmuch as the subalgebras I) c End( V) coincide, the weights of V must likewise be the same for so 2 .1C as for sp 2 .1C), it follows that the roots of so 2 .1C

§18.1. SO,.C and so,.C

271

are just the pairwise distinct sums ± L; ± Li. In the odd case m = 2n + 1, we see that e 2 ,~+ 1 E V is an eigenvector for the action of f) with eigenvalue 0, so that the weights of the standard representation V are { ± L;} u {0} and the weights of the adjoint representation correspondingly { ± L; ± L 1} u { ± LJ}. Exercise 18.3. Use a similar analysis to find the roots of sp 2 ,.C without explicit calculation. To make a comp:tdson with the Lie algebra sp 2 ,1C, we can say that the root diagram of so 2.C looks like that of sp 2.C with the roots ±2L; removed, whereas the root diagram of so 2 ,.+ 1 1C looks like that of sp 2 ,.C with the roots ±2L; replaced t>y ±L;. Note that this immediately tells us what the Weyl groups are: first, in the case of so 2 ,.+ 1 1C, the Weyl group is the same as that of SP2niC: . 1 -+ (7!./2)" -+

w•••.• ,c -+ 6,. -+ 1.

In the case of so 2 ,.C, the Weyl group is the subgroup of the Weyl group of sp 1 ,.C generated by reflection in the hyperplanes perpendicular to the roots ±L1 ± L1, without the additional g• nerator given by reflection in the roots ± L 1• This subgroup still acts as the full symmetric group on the set of coordinate axes in f)*; but the kernel of this action, instead of acting as ±I on each of the coordinate axes independently, will consist of transformations of determinant I; i.e., will act as - 1 on an even number of axes. (That every such transformation is indeed in the Weyl group is easy to see: for example, reflection in the plane perpendicular to L; + Li followed by reflection in the plane perpendicular to L; - Li will send L 1 to - L;, Lito - L1, and Lk to Lk for k # i, j.) Another way to say this is that the Weyl group is the subgroup of the Weyl group of sp 2 ,.1C consisting of transformations who~e determinant agrees with the sign of the induced permutation of the coordinate axes; so that while the Weyl grot:p of sp 1 ,.1C fits into the exact sequence l -+ (7!./2)"-+

w.p •• c-+ 6,.-+ l,

the Weyl group of so 2 ,.C has instead the sequence } -+ (7l./2)"~J -+ lill 502 .c-+ 6,.-+ 1.

We can likewise describe the Weyl chambers of so 2 .1C and so 2,.+ 1 C by direct comparison with sp 2 ,.1C. To start, to choose an ordering of the roots we take as linear functional on I>* a form 1= c 1 H 1 + · .. + c,.H,., where c 1 > c2 > · · · > c,. > 0. The positive roots in the case of so 2 ,.+ 1 1C are then

R+

=

{L 1 + L1L< 1 u {L;- L1};< 1 u {L;};,

whereas in the case of so 2 ,.1C we have

R+

=

{L;

+ Li};
The primitive positive roots are

18. Orthogonal Lie Algebras

272

for L1

-

L 2, L 2

L 3,

-

L.- 1

••• ,

-

L., L.- 1

+ Ln

SOzn+l

C;

for so 2 .C.

In the first case, the Weyl chamber is exactly the same as for sp 2 .C, namely, form= 2n + 1, "'f" =

{L a L

1 1:

a 1 ~ a2 ~

•• •

~ a.~

0}

since the roots are the same except for the factor of 2 on some. In the case of so 2 .C, since there is no root along the line spanned by the L 1, the equality a.= 0 does not describe a face of the Weyl chamber; however, since L._ 1 + L. is still a root (and a positive one) we still have the inequality a._ 1 +a. ~ 0 in iY, so that we can write, form = 2n, "'f"

=

{L a L

1 1:

a 1 ~ a2 ~

• ·'

~ a.- 1 ~ Ia. I}.

(Note that in the case of so 2.C we could have chosen our linear functional I= c 1 H 1 + .. · + c.H. with c 1 > c 2 > ... > -c.> 0; the ordering of the roots, and consequently the Weyl chamber, would still be the same.) As for the Killing form, the same considerations as for the symplectic case show that it must be, up to scalars, the standard quadratic form: B(H1, Hi)= b1.i. (This was implicit in the above description of the Weyl group.) The explicit calculation is no more difficult, and we leave it as an exercise:

~

~b

L

) _ {(4n - 2) a1b1 if m = 2n + I (4n - 4) L a1b1 if m = 2n.

B(L.. a,H,, L.. ,H, -

Next, to describe the representations of the orthogonal Lie algebras we have to determine the weight lattice in~*; and to do this we must, as before, locate the copies s~ of sl 2 C corresponding to the root pairs ±o:, and the corresponding distinguished elements Ha of(). This is so similar to the case of sp 2.C that we will leave the actual calculations as an exercise; we will simply state here the results that in so.,C for any m, (i) the distinguished copy sL,-L; of sl 2C associated to the root L 1 - L1 is the span of the root spaces !1L,-L1 = IC · X 1.i, !1-I,,-I.; = IC ·Xi, I and their commutator [Xi.i• Xi.;]= E 1,1 - E1.i + E•+i.•+i- En+i.n+i• with distinguished element HL,-L1 = H 1 - Hi (this is exactly as in the case of sp 2 .C); (ii) the distinguished copy sL,+L1 of sl 2 C associated to the root L, + L1 is the span of the root spaces !1L,H1 = C · ¥;, 1, !1-L,-I.1 = IC · Z 1, 1 and their commutator [Y;.i, Z 1.i] = -E1, 1 + Ei.i- E•+i.•+i + En+i.n+i = -H1 - H1, with distinguished element HL,+L1 = H 1 + H1 (so that we have also H-L,-L1 = - H 1 - Hi); and in the case of so 2 n+J C, (iii) the distinguished copy sL, of .:;l 2C associated to the root L 1 is the span of the root spaces !lL, = C · U1, 9-L, = C · V; and their commutator [U" Vj] = [E;,zn+t- Ezn+l.n+i• £.+ 1• 2 .+ 1 - E 2 .+ 1 ,;] = - H,, with distinguished element H 1,, = 2H1 (so that H-L, = -2H1 as well). Exercise 18.4. Verify the cornputations made here.

§18.2. Representations ofso 3 C, so 4 C, and so 5 C

273

Again, the configuration of distinguished elements rrsembles that of sp 2"C closely; that of so 2 n+I C differs from it by the substitul:on of ± 2H; for ± H;, whereas that of so 2 "C differs by the removal of the ±H;. The effect on the weight lattice is the same in either case: for both even and odd orthogonal Lie algebras, the weight lattice Aw is the lattice generated by the L; together with the element (L 1 + .. · + Ln)/2. Exercise 18.5. Show that

71./2 AwfAR

if m if m if m

= { 71./4 71./2 Ef) 71./2

=

2n

+1

= 2n and n is odd

= 2n and n is even.

§18.2. Representations of so 3 C, so 4 C, aud so 5 C To give some examples, start with the case n = 1. Of course, so 2 C ~Cis not semisimple. The root system of so 3 C, on the other hand, looks like that of sl 2 C:

•

0

•

This is because, in fact, the two Lie algebras are isomorphic. Indeed, like the symplectic group, the quotient PSOmC of the orthogonal group by its center can be realized as the motions of the projective space P V preserving isotropic subspaces for the quadratic form Q; in particular, this means we can realize PSO,.C as the group of motions of PV = pm-I carrying the quadric hypersurface

Q=

{[v]: Q(v, v)

= 0}

into itself. In the first case of this, we see that the group PS0 3 C is the group of motions of the projective plane P 2 carrying a conic curve C c P 2 into itself. But we have seen before that this group is also PGL 2 C (the conic curve is itself isomorphic toP 1, and the group acts as its full group of automorphisms), giving us the isomorphism so 3 C ~ sl 2 C. One thing to note here is that the "standard" representation of so 3 C is not the standard representation of s1 2 C, but rather its symmetric square. In fact, the irreducible representation with highest weight t L 1 is not contained in tensor powers of the standard representation of so 3 C. This will turn out to be significant: the standard representation of sl 2 C, viewed as a representation of so 3 C, is the first example of a spin representation of an orthogonal Lie algebra. The next examples involve two-dimensional Cartan algebras. First we have so 4 C, whose root diagram looks like

274

18. Orthogonal Lie Algebras

Note one thing about this diagram: the roots are located on the union of two complementary lines. This says, by Exercise 14.33, that the Lie algebra 50 4 Cis decomposable, and in fact should be the sum of two algebras each of whose root diagrams looks like that of 5l 2 C; f'xplicitly, 5o 4 C is the direct sum of the two algebras 5«, for IX = L 1 + L 2 and rx = L 1 - L 2 • In fact, we can see this isomorphism

(18.6) as in the previous example, geometrically. Precisely, we may realize the group PS04 C = S04 IC/{ ± l} as the connected component of the identity in the group of motions of projective three-space 1? 3 carrying a quadric hypersurface Q into itself. But a quadric hypersurface in IP' 3 has two rulings by lines, and these two rulings give an isomorphism of Qwith a product I? 1 x IP' 1

PS04 1C thus acts on the product I? 1 x IP' 1 ; and since the connected component of the identity in the automorphism group of this variety is just the product PGL 2 1C x PGL 2 C, we get an inclusion PS04 1C--+ PGL 2 1C x PGL 2 1C. Another way of saying this is to remark that PS04 1C acts on the variety of isotropic 2-planes for the quadratic form Q on V; and this variety is just the disjoint union of two copies of IP' 1 . To see in this case that the map is an

§18.2. Representations of so 3 C, so 4 C, and

so~ C

275

isomorphism, consider the tensor product V = U ® W of the pullbacks to si 2 C x si 2 C of the standard representations of the two factors. Clearly the action on IP'(U ® W) will preserve the points corresponding to decomposable tensors (that is, points of the form [u ® w]); but the locus of such points is just a quadric hypersurface, giving us the inverse inclusion of PGL 2 C x PGL 2 C in PS04 C. In fact, all of this will fall out of the analysis of the representations of so 4 C, if we just pursue it as usual. To begin with, the Weyl chamber we have selected looks like

Now, the standard representation has, as noted above, weight diagram

with highest weight L 1 (note that the highest weight of the standard representation lies in this case in the interior of the Weyl chamber, something of an anomaly). Its second exterior power will have weights ± L 1 ± [, 2 and 0 (occurring with multiplicity 2), i.e., diagram

276

18. Orthogonal Lie Algebras

We see one thing about this representation right away, namely, that it cannot be irreducible. Indeed, the images of the highest weight L 1 + L 2 under the Weyl group consist just of ±(L 1 + L 2 ), so that the diagram of the irreducible representation with this highest weight is

We see from this that the second exterior power ;VV of the standard representation of so 4 C must be the direct sum of the irreducible representations WI = rL,+Lz and w2 = rL,-L, with highest weights Ll + L2 and L 1 - L 2 . Since NV is at the same time the adjoint representation, this says that so 4 C itself must be a product of Lie algebras with adjoint representations rL,+L, and ~~.,-L,·. One way to derive the picture of the ruling of the quadric in IP' 3 from this decomposition is to view so 4 C as a subalgebra of sl 4 C, and the action of PS04 C on IP'(N V) as a subgroup of the group of motions of IP' 2 (N V) = IP' 5 preserving the Grassmannian G = G(2, V) of lines in IP' 3 • In fact, we see from the above that the action of PS04 on IP' 5 will preserve a pair of complementary 2-planes IP'W1 and IP'W2 ; it follows that this action must carry into themselves

§18.2. Representations of so 3 C, so 4 C, and su 5 C

277

the intersections of these 2-planes with the Grassmannian. These intersections are conic curves, corresponding to one-parameter families of lines sweeping out a quadric surface (necessarily the same quadric, since the action of so4 c on V preserves a unique quadratic form); thus, the two rulings of the quadric.

Note one more aspect of this example: as in the case of so 3 C ~ sl 2 C, the weights of the standard representation of so 4 C do not generate the weight lattice, but rather a sublattice 1'{L 1 , L 2 } of index 2 in Aw. Thus, there is no way of constructing all the representations of so 4 C by applying linear- or multilinear-algebraic constructions to the standard representation; it is only after we are aware of the isomorphism so 4 C ~ si 2 C x si 2 C that we can construct, for example, the representation r(L, +L 1>12 with highest weight (L 1 + L 2 )/2 (of course, this is just the pullback from the first factor of sl 2 C x sl 2 C of the standard representatk"l of si 2 C). We come now to the case of so 5 C, which is more interesting. The root diagram in this case looks like

278

18. Orthogonal Lie Algebras

(as in the preceding example, the weight lattice is the lattice of intersections of all the lines drawn). The first thing we should notice about this diagram is that it is isomorphic to the weight diagram of the Lie algebra sp 4 C; the diagram just appears here rotated through an angle of n/4. Indeed, this is not accidental; the two Lie algebras sp 4 C and so 5 C are isomorphic, and it is not hard to construct this isomorphism explicitly. To see the isomorphism geometrically, we simply have to recall the identification, made in Lecture 14, of the group PSp4 C with a group of motions of IP 4 . There, we saw that the larger group PGL 4 C could be identified with the automorphisms of the projective space P(NV) = 1? 5 preserving the Grassmannian G = G(2, 4) c IP'(NV). The subgroup PSp 4 C c PGL 4 C thus·preserves both the Grassmannian G, which is a quadric hypersurface in 1? 5 , and the decomposition of NV into the span IC · Q of the skew form Q E NV*~ NV and its complement W, and so acts on IP'W carrying the intersection GL = G n IP'W into itself. We thus saw that PSp 4 C was a subgroup of the group of motions of projective space IP' 4 preserving a quadric hypersurface, and asserted that in fact it was the whole group. (To see the reverse inclusion directly, we can invoke a little algebraic geometry, which tells us that the locus of isotropic lines for a quadric in IP' 4 is isomorphic to IP' 3 , so that PS0 5 C acts on IP 3 • Moreover, this action preserves the subset of pairs of points in IP 3 whose corresponding lines in IP' 4 intersect, which, for a suitably defined skew-symmetric bilinear form Q, is exactly the set of pairs ([v], [w]) such that Q(JJ, w) = 0, so that we have an inclusion of PS0 5 C in PSp 4 C.) Let us proceed to analyze the representations of so 5 1C as we would ordinarily, bearing in mind the is<,morphism with sp 4 1C. To begin with, we draw the Weyl chamber picked out above in 1)*:

As for the representations of so 5 C, we have to begin with the standard, which has weight diagram

§18.2. Representations ofso 3 C, so.C, and so 5 C

279

This we see corresponds to the representation W = N V/C · Q of sp 4 C. Next, the second exterior power of the standard representation of so 5 C has weights

This is of course the adjoint representation of so 5 C; it is the irreducible representation with highest weight L 1 + L 2 • Note that it corresponds to the symmetric square Sym 2 V of the standard representation of sp 4 C (see Exercise 16.8). Exercise 18.7. Show that contraction with the quadratic form Q E Sym 2 V* preserved by the action of so 5 C induces maps qJ:

Sym"V-+ Sym"- 2 V.

Show that the kernel of this coutraction is exactly the irreducible representation with highest weight a· L 1 . Compare this with the analysis in Exercise 16.11.

18. Orthogonal Lie Algebras

280

Exercise 18.8. Examine the symmetric power Syma(N V) of the representation NV. This will contain a copy of the irreducible representation facL, +L,>; what else will it contain? Interpret these other factors in light of the isomorphism so 5 C ~ sp 4 C.

Exercise 18.9. For an example of a "mixed" tensor, consider the irreducible representation f 2 L, +L,· Show that this is contained in the kernels of the wedge product map
V®NV -+NV

and the composition

ql: V®NV-+ V*®NV-+ V, where the first map is induced by the isomorphism Q: V---+ V* and the second is the contraction V* ®NV---+ V. Is it equal to the intersection of these kernels? Show that the weight diagram of this representation is

After you arc done with this analysis, compare with the analysis given of the corresponding representation in Lecture 16. Note that, as in the case of the other orthogonal Lie algeLras studied so far (and as is the case for all somiC), the wei{Ats of the standard representation do not generme the weight lattice, but only tire suhlattice of i11dex two generated by the L1. Thus, the tensor algebra of the standard representation will contain only one-half of all the irreducible representations of so 5 C. Now, we do know that there are others, and even something about them-for example, we see in the following exercise that the irreducible representation of so 5 C with highest weight (L 1 + L 2 )/2 is a sort of"symmetric square root" of the adjoint representation: Exercise 18.10. Show, using only root and weight diagrams for so 5 C, that the exterior square NV of the standard representation of so 5 C is actually the syt.lmetric square of an irreducible representation.

§18.2. Representations of so 3 C, so 4 C, and so 5 C

281

We can also describe this irreducible representation via the isomorphism of SOsC with sp4 C: it is just the standard representation of sp4 con C 4. We do not at this point have, however, a way of constructing this representation without invoking the isomorphism. This representation, the representation of so 3 C with highest weight Ltf2, and the representation of so 4 C with highest weight (L 1 + L 2 )/2 discussed above are called spin representations of the corresponding Lie algebras and will be the subject matter of Lecture 20.

LECTURE 19

so 6 C, so 7 C,

and some

This lecture is analogous in content (and prerequisites) to Lecture 17: we do some more low-dimensional examples and then describe the general picture of the representations of the orthogonal Lie algebras. One difference is that only half the irreducible representations of so.,C lie in the tensor alg.;bra of the standard; to complete the picture of the representation theory we have to construct the spin representations, which is the subject matter of the following lecture. The first four sections are completely elementary (except possibly for the discussion of the isomorphism so 6 C ~ si 4 C in §19.1); the last section assumes a knowledge of Lecture 6 and §15.3, but can be skipped by those who did not read those sections.

§19.1: Representations of so 6 C §19.2: Representations of the even orthogonal algebras §19.3: Representations of so 1 C §19.4: Representations of the odd orthogonal algebras §19.5: Weyl's construction for orthogonal groups

§19.1. Representations of so 6 C We continue our discussion of orthogonal Lie algebras with the example of so 6 C. First, its root diagram:

§19.1. Representations of so 6 C

283

Once more (and for the last time), we notice a coincidence between this and the root diagram of a Lie algebra already studied, namely, si4 C. In fact, the two Lie algebras are isomorphic. The isomorphism is one we have already observed, in a sense: in the preceding lecture we noted that if V is a fourdimensional vector space, then the group PG L4 C may be realized as the connected component of the identity in the group of motions of P(NV) = P 5 carrying the Grassmannian G = G(2, 4) c IP'(NV) into itself, and PSp4 C c PGL 4 C the subgroup fixing a hyperplane IP'W = IP' 4 c P 5 . We used this to identify the subgroup PSp 4 C with the orthogonal group PS05 C; at the same time it gives an identification of the larger group PGL 4 C with the orthogonal group PS06 C. Even though so 6 Cis isomorphic to a Lie algebra we have already examined, it is worth going through the analysis of its representations for what amounts to a second time, partly so as to understand the isomorphism better, but mainly because we will see clearly in the case of so 6 C a number of phenomena that will hold true of the even orthogonal groups In general. To start, we draw the Weyl chamber in()*:

As usual, we begin with the standard representation, which has weights

± L;, corresponding to the centers of the faces of the cube:

19. so 6 C, so 7 C, and so.. C

284

4' ,' '," .. , , I ' --- e .

........ - ' -·, ·.·.L ,•,I' .. r•.-- ..... : , r. , , , ........ ~ ---;'-..-: ---

I

...

...

_.,./·

Note that the highest weight L 1 once more lies on an edge of the Weyl chamber (the front edge, in the diagram on the preceding page). Observe that the standard representation of so 6 C corresponds, as we have already pointed out, to the exterior square of the standard representation of sl 4 C. Next, we look at the exterior square NV of the standard representation of so 6 C. This will have weights ±L 1 ± Li (of course, it is the adjoint representation) and so will have weight diagram

Note that the highest weight vector L 1 + ], 2 of this representation does not lie on an edge of the Weyl chamber, but rather in the interior of a face (the back face, in the diagram above). In order to generate all the representations, we still need to find the irreducible representations with highest weight along the remaining two edges of the Weyl chamber. We look next at the exterior cuhe /\ 3 V of the standard representation. The weights here are the eight weights ±L 1 ± L 2 ± L 3 , each taken wi1 h multiplicity one, and the six weights ± L 1, each taken with multiplicity 2, as in the diagram

285

§19.1. Representations of so 6 C

Now, we notice something very interesting: this cannot be an irreducible representation. We can see this in a number of ways: the images of the weight L 1 + L 2 + L 3 under the Weyl group, for example, consist of every other vertex of the reference cube; in particular, their convex hull does not· contain the remaining four vertices including L 1 + L 2 - L 3 • Equivalently, there is no way to go from L 1 + L 2 + L 3 to L 1 + L 2 - L 3 by translation by negative root vectors. The representation /\ 3 V will thus contain copies of the irreducible representations rL,+L,+r.J and rL,+L,-L, with highest weightsLl + Lz + L3 and L 1 + L 2 - L 3 , with weight diagrams

·• I ·.I .

...

;~·.-__ / ' /

·

and

Since the weight diagram of each of these is a tetrahedron containing the weights ± Li, we have accounted for all the weights of /\ 3 V and so must have a direct sum decomposition

f\ 3 V

= rL,+L

2

+L1

$ rL,+L,-L,·

We can relate this direct sum decomposition to a geometric feature of a quadric hypersurface in IPS, analogous to the presence of two rulings on a quadric in IP 3 • We saw before that the locus of lines lying on a quadric surface in IP 3 turns out to be disconnected, consisting of two components

286

19. so 6 C, so 7 C, and so,.C

each isomorphic to IP' 1 (and embedded, via the Plucker embedding of the Grassmannian G = G(2, 4) of lines in IP' 3 in IP'(NIC) = IP' 5, as a pair of conic curves lying in complementary 2-planes in IP' 5 ).1n a similar fashion, the variety of2-planes lying on a quadric hypersurface in IP' 5 turns out to be disconnected, consisting of two components that, under the PlUcker embedding of G(3, 6) in IP'(NC 6 ) = IP' 19, span two complementary 9-planes IP'W1 and IP'W2 ; these two planes give the direct sum decomposition of NV as an so 6 C-module. In fact, if we think of a quadric hypersurface in IP' 5 as the Grassmann ian G = G(2, 4) of lines in IP' 3 , we can see explicitly what these two families of 2-planes are: for every point p E IP' 3 the locus oflines passing through p forms a 2-plane on G, and for every plane~ H c IP' 3 the locus of lines lying in H is a 2-plane in G. These are the two families; indeed, in this case we can go two steps further. First, we see from this that each of these families is parametrized by IP'l, so that the connected component PS06 C of the identity in the group of motions of IP' 5 preserving the Grassmannian acts on IP' 3 , giving us the inverse inclusion PS06 C c PG L 4 C. Second, under the PlUcker embedding each of these families is carried into a copy of the quadratic Veronese embedding of IP' 3 into IP' 9 , giving us the identification of the direct sum factors of the third exterior power of the standard representation of so 6 C with the symmetric square of the standard representation of sl 4 C.

Exercise 19.1. Verify, without using the isomorphism with so 6 C and the analysis above, that the standard representation V of sl 4 C satisfies N(NV) ~ Sym 2 V EB Sym 2 V*. Note that we have now identified, in terms of tensor powers of the standard one, irreducible representations of so 6 C with highest weight vectors L 1 , L 1 + L 2 + L 3 and L 1 + L 2 - L 3 lying along the edge of the Weyl chamber, as well as one with highest weight L 1 + L 2 lying in a face. We can thus find irreducible representations with highest weighty, if not for every yin Awn "'Y, at least for every weight y in the intersection of #~ with a sublattice of index 2in Aw.

§19.2. Representations of the Even Orthogonal Algebras We will not examine any further representations of so 6 C per se, leaving it as an exercise to do so (and to compare the .csults to the corresponding analysis for s1 4 if:). Instead, we can now describe the general pattern for representations of the even orthogonal Lie algebras so 2 .C. The complete story will have to wait until the following lecture, since at present we cannot construct all the representations of so 20 C {as we have pointed out, we have been able to do so in the cases n = 2 and 3 studied so far only by virtue of isomorphisms with

287

§19.2. Representations of the Even Orthogonal Algebras

other Lie algebras; and there are no more such isomorphisms from this point on). We will nonetheless give as much of the picture as we can. To begin with, recall that the weight lattice. of so 2 .C is generated by L 1 , ... , L. together with the further vector (L 1 + ·· · + L.)/2. The Weyl chamber, on the other hand, is the cone . "'I'=

{L a;L;: a 1 ;::: a 2

;::: • · ·;:::

±a.}.

Note that the Weyl chamber is a simplicial cone, with faces corresponding to the n planes a 1 = a2 , •.• , a,_ 1 = a. and a._ 1 = -a.; the edges of the Weyl chamber are thus the rays generated by the vectors Lt> L 1 + L 2 , ••• , L 1 + · ·· + L.- 2 , L 1 + ··· + L. and L 1 + ·· · + L._ 1 - L. (note that L 1 + ··· + L._ 1 is not on an edge of the Weyl chamber). We see from this that, as in every previous case, the intersection of the weight lattice with the closed Weyl cone is a free semigroup generated by fundamental weights, in this case the vectors L 1 , L 1 + L 2 , ••• , L 1 + · ·· + L._ 2 and the vectors 1

As before, the obvious place to start to look for irreducible representations is among the exterior powers of the standard representation. This almost works: we have

Theorem 19.2. (i) The exterior powers NV of the standard representation V of so 2 ,C are irreducible fork = I, 2, ... , n- 1; and (ii) The exterior power 1\"V has exactly two irreducible factors. PROOF. The proof will follow the same lines as that of the analogous theorem for the symplectic Lie algebras in Lecture 17; in particular, we will start by considering the restriction to the same subalgebra as in the case of sp 2.C. Recall that the group Sp 2 .C c SL 2 .C of automorphisms preserving the skew form Q introduced in Lecture 16 contains the subgroup G of attfomorphisms of the space v = IC 2 " preserving the decomposition V = C{e 1 , ••• ,e.}$ C{en+t• ... , e 2 .}, acting as an arbitrary automorphism on the first factor and as the inverse transpose of that automorphism on the second factor; in matrices

In fact, the subgroup S0 2,C c SL 2 .C also contains the same subgroup; we have, correspondingly a subalgebra To conform to standard conventions, with simple roots a, = L,- Lr+ 1 for I :;:; l:;:; n- I, and + L., to have w,(H,,) = {)1•1, the fundamental weights w, should be put in the order: w, = L 1 + ··· + L, for I :::; i :;:; 11- 2, :uJd 1

a.= L._ 1

Wn-1

=

P=(I~ I + "· + L._ 1 -

L.)/2,

w.

=

a = (L 1

+ .. · + L.)/2.

288

19. so 6 C, so 7 C, and so,.C

isomorphic to si.C. Denote by W the standard representation of si.C. As in the previous case, the restriction of the standard representation V of so 2 .C to the subalgebra s then splits

V =WEB W* into a direct sum of Wand its dual; and we have, correspondingly,

N

v = EB

a+b:k

(1\"W ® NW*).

We also can say how each factor on the right-hand side of this expression decomposes as a representation of sl.C: we have contraction maps 'l'... b: 1\"W ® NW*

---+

1\"- 1 W ® N- 1 W*;

and the kernel of 'l'a,b is the irreducible representation w
NV=

EB

a+b:s;k

w
aH=kt2)

where the actual highest weight factor in the summand w<•.bl c NV is the vector wl•.bl

= e1

" ... "

e.. " e2n-b+t " ... " e2, " Q
= e1

" ... "

e.. " ezn-b+l " ... " ez, "

(L (e; "

e.+;))tk-a-b)/2.

Now, all the vectors w1"·b1 have distinct weights; and it follows, as in Exercise 17.7, that any highest weight vector for the action of so 2 .C 011 NV wiil 11e a scalar multiple of one of the w<•.bl. It will thus suffice, in order to show :hat NV is irreducible as representation of so 2.C fork < 11, to exhibit for each (a, b) with a + b :S k < n other than (k, 0) a positive root a such that the image f1 2 (w<•.bl) 'i= 0. This is simplest in the case a + b = k < n (so there is no factor of Q in wt•.b 1): just as in the case of sp 2.C we have . ¥.+ 1,n-b+ 1 (w(a,b)) = (E.+I,2n-b+1 - En-b+l,,~·a+l)(el

" ... " e." e1n-b+l " ... "

= w<•+l,b-11 and ¥;, 1 is the generator of the positive root space !lL,+L1 . In case a + b < k < 11, we observe first that for any i and j

e2H;

§19.2. Representations of the Even Orthogonal Algebras

289

Y;,J(Q) = (E;,n+ i - Ej,nHHL (ep A en+p))

= 2·e1 A so that whenever a < i, j

:$;

e1

n - b,

Y;,j(w(a,bl)

= Y;,j{e 1 A = el

··· A

A .. · A ea A

= (k- a-· b). (el

e.

A

e 2n-b+t

A •· • A

e2n-b+J A .. · A ·. ·

A e.

A

e2n

A e} A el

e 2• A

A

Q(k-a-bl/2)

Y,,J{(L (epA

A ezn-1>+1

en+p))lk-a-blfl)

A ··· A

e2n A Q(k-a-b-ll/2)

,.1:0.

It is always possible to find a pair (i, j) satisfying the conditions a < i, j :$; n - b since we are assuming a + b < k < n; this concludes the proof of part (i). The proof of part (ii) requires only one further step: we have to check the vectors w(a,bj with a + b = k = n to see if any of them might be highest weight vectors for so 2 .C. In fact (as the statement of the theorem implies), two of them are: It is not hard to check that, in fact, wCn.ol and wCn-t.l) are killed by every positive root space 9L,+LJ· To see that no other vector wca.n-al is, look at the action of Ya+t,a+z E 9L.+t+L.+,: we have Y,+l,a+ z(W(a,n-al)

- e1

A · ·. A e. A ea+2 A en+a+2 A ..• 1\ e2n

,.1:0.

0

Remarks. (i) This theorem will be a consequence of the Weyl character formula, which will tell us a priori that the dimension of the irreducible representation of so 2.C with highest weight L 1

+ ··· + Lk has dimension(~)

if k < n, and half that if k = n. (ii) Note also that by the above, NV is the direct sum of the two irreducible representations f 2 ,. and f 211 with highest weights 2a = L 1 + ··· + L. and 2P = L 1 + ··· + L._ 1 - L •. Indeed, the inclusion f 2" 61 f 211 c NV can be seen just from the weight diagram:;\" V possesses a highest weight vector with highest weight L 1 + · ·· + L., and so contains a copy of f 2 "; but this representation does not possess the weight 2/3, and so NV must contain f 211 as well. (Alternatively, we observed in the preceding lecture that in choosing an ordering of the roots we could have chosen our linear functional I = c 1 H 1 + .. · + c.H. with c 1 > c 2 > .. · > -c. > 0 without altering the positive

19. so 6 C, so 7 C, and so,.C

290

roots or the Weyl chamber; in this case the weight A. of NV with /(A.) maximal would be 2p, showing that r 211 c NY.) (iii) If we want to avoid weight diagrams altogether, we can still see that NV must be reducible, because the action of so 2 .C preserves two bilinear forms: first, we have the bilinear form induced on NV by the form Q on V; and second we have the wedge product q>:

NV

X

NV--+ t'?"V = c,

the last map taking e 1 " · · · " e 2 • to I. It follows that /\"Vis reducible; indeed, if we want to see the direc1 sum decomposition asserted in the statement of the theorem we can look at the composition r: NV--+ NV*--+ NV,

where the first map is the isomorphism given by Q and the second is the isomorphism given by q>. The square of this map is the identity, and decomposing NV into + I and - I eigenspaces for this map gives two subrepresentations. Exercise 19.3*. Part (i) of Theorem 19.2 can also be proved by showing that for any nonzero vector wE NV, the linear span of the vectors X(w), for X E some, is all of N v. For these purposes take, instead of the basis we have been using, an orthonormal basis v,' ... ' vm for v = em, 111 = 2n, so Q(v;, vj) = Ji,J· The vectors v1 = V;, " · • • " v;., I= {i 1 < · · · < ik}, form a basis for NV, and some has a basis consisting of endomorphisms vp,q• p < q, which takes v4 to vP, vP to - v4 , and takes the other v; to zero. Compute the images Vp. 4 (v1), and prove the claim, first, when w = v1 for some I, and then by induction on the number of nonzero coefficients in the expression w = I,a 1 v1 • For (ii) a similar argument shows that NV is an irreducible representation of the group OnC, and the ideas of§S.I (cf. §19.5) can be used to see how it decomposes over the subgroup SOniC of index two. We return now to our analysis of the representations of so 2,C. By the theorem, the exterior powers V, NV, ... , N- 2 V provide us with the irreducible representations with highest weight the fundamental weight along the first n- 2 edges of the Weyl chamber (of course, the exterior power N- 1 V is irreducible as well, but as we have observed, L 1 + ··· + L 11 _ 1 is not on an edge of the Weyl chamber, and ,;o N- 1 Vis not as useful for our purposes). For the remaining two edges, we have found irreducible representations with highest weights located there, numely the two direct sum factors of NV; but the highest weights of these two representations are not primitive ones; they are divisible by 2. Thus, given the theorem above, we see that we have constructed exactly one-half the irreducible representations of so 211 C, n.uuely, those whose highest weight lies in the sublattice 1' {I.;, ... , Ln} c ,\ '"' Explicitly, any weight}' in lhe closed Weyl chamber can bee ·pressed (uniquely) in the form

§19.2. Representations of the Even Orthogonal Algebras

291

+ ··· + a._ 2 (L 1 + ··· + L._ 2 ) + a.- 1(L 1 + ··· + Ln-l- 1-.)/2 + a.(J- 1 + · ·· + L.)/2 with a1 E 1'\1. If a._ 1 + a. is even, with a._ 1 ~ a. we see that the representation ')I=

a1L 1

Sym•• V ® · · · ® Sym•·-z(N- 2 V) ® Sym""(N- 1 V) ® Sym 1•·-·-··''2 (r211 ) will contain an irreducible representation fy with highest weighty; whereas if a. ~ an-I' we will find ry inside Sym•• V ® · · · ® Sym•·-z(N- 2 V) ® Sym•· •(/\"- 1 V) ® Symln.-•.-dl2(r2 .). There remains the problem of constructing irreducible representations rY whose highest weight 'l' involves an odd number of a's and {3's. To do this, we clearly have to exhibit irreducible representations r. and r, with highest weights a and {3. These exist, and are called the spin representations of so 2 .C; we will study them in detail in the following lecture. We see from the above that once we exhibit the two representations r. and r 11 , we will have·constructed all the representations of so 2 .C. The representation fy with highest weight ')I written above will be found in the tensor product Sym•• V ® · · · ® Sym••- 2 (/\"- 2 V) ® Symn·-•(f11 ) ® Sym•·(r.). For the time being, we will assume the existence ofthe spin representations of so 2 .C; there is a good deal we can say about these representations just on the basis of their weight diagrams. Exercise 19.4*. Find the weights (with multiplicities) of the representations NV, and also off2., f 211 , r., and r 11 • Exercise 19.5. Using the above, show that r. and r11 are dual to one another when n is odd, and that they are self-dual when 11 is even. Exercise 19.6. Give the complete decomposition into irreducible representa-' tions of Sym 2 r. and /\ 2 r•. Show that

r. ® r. =

f 2 • ® 1\"-zv ElJ N- 4 v E9 N- 6 v ® · · ·.

Exercise 19.7. Show that

r,. ® r11 = N- 1 v E91\"- 3 v E91\"- 5 V E9 .... Exercise 19.8. Verify directly the above statements in the case of so 6 C, using the isomorphism with s! 4 C. Exercise 19.9. Show that the automorphism of C 2 " that interchanges e. and e2 , leaving the other e1 fixed, determines an automorphism of so 2 .C that

preserves the n- 2 roots L 1 - L 2 , ••• , L._ 2 - L._ 1 and interchanges L. and L._ 1 + L •. This automorphism takes the representation V to itself, but interchanges r. and r/1.

L._ 1

292

§19.3. Representations of so 7 C While we might reasonably be apprehensive about the prospect of a family of Lie algebras even more strangely behaved than the even orthogonal algebras, there is some good news: even though the roots systems of the odd Lie algebras appear more complicated than those of the even, the·representation theory of the odd algebras is somewhat tamer. We will describe these representations, starting with the example of so 7 C; we begin, as always, with a picture of the root diagram:

As we said, this looks like the root diagram for sp 6 C, except that the roots ± 2Li have been shortened to ± Li. Unlike the case of so 5 C, however, where the long and short roots could be confused and the root diagram was correspondingly congruent to that of sp 4 C, in the present circumstance the root diagram is not similar to any other; the Lie algebra so 7 C, in fact, is not isomorphic to any of the others we have studied. Next, the Weyl chamber:

c.,

Again, the Weyl chamber itself looks just like that of sp 6 C; the difference in this picture is in the weight lattice, which contains the additional vector (L1 + L 2 + L 3 )/2.

§19.3. Representations of so 7 C

293

As usual, we start our study of the representations of so 7 C with the standard representation, whose weights are ± L; and 0:

Note that the highest weight L 1 of this representation lies along the front edge of the Weyl chamber. Next, the weights of the exterior square I\ 2 V are ± L; ± Li, ± L;, and 0 (taken three times); this, of course, is just the adjoint representation. Note that the highest weight L 1 + L 2 of this representation is the same as that of the exterior square of the standard representation for so 6 C, but because of the smaller Weyl chamber this weight does indeed lie on an edge of the chamber. Next, consider the third exterior power NV of the standard. This has weights ±L 1 ± L 2 ± L 3 , ±L; ± Li, ±L; (with multiplicity 2) and 0 (with multiplicity 3), i.e., at the midpoints of all the vertices, edges, and faces of the cube:

I·· .

.

.··• ; .. -it·.. .:·.-·1··:····; ··:~_eL ··.· .. I :

·-.r~-

1· ·.

/

.

/

/

-· - .. -:-.-. . •· .

LI+L2 I

.

~~·~;~0----...

--·

It is not obvious, from the weight diagram alone, that this is an irreducible representation; it could be that NV contains a copy of the standard representation Vand that the irreducible representation rL, +Lz+LJ thlis has multiplicity I on the weights ± L; and multiplicity 2 (or I) at 0. We can rule out this possibility by direct calculation: for example, if this were the case, then NV would contain a highest weight vector with weight L 1 • The weight space with

eigenvalue L 1 in NV is spanned by the tensors e 1 " e 2 " e5 and e 1 " e 3 " e6 , however, and if we apply to these the generators X 1 , 2 = E 1 , 2 - E 5 , 4 , X 2 , 3 = E 2 , 3 - E 6 , 5 , and U3 = £ 3 , 7 - £ 7 • 6 of the root spaces corresponding to the positive roots L 1 - L 2 , L 2 - L 3 , and L 3 , we see that X 2 , 3 (e 1 "e 3

"

e6 ) = e 1 "e 2 "ei;,

U3 (e 1

"

e3

"

e6 )

X2 , 3 (e 1

"

e2

"

e5 )

U3 (e t

"

e2

"

=

e1

= e1 e 5 ) = 0.

"

e3

"

e7

"

e2

"

e6 ,

:f.

0;

There is thus no linear combination of e 1 " e2 " e5 and e 1 " e 3 " e6 killed by both U3 and X 2 , 3 , showing that NV has no highest weight vector of weight Lt. Exercise 19.10. Verify that NV does not contain the trivial representation. We have thus found irreducible representations of so 7 C with highest weight vectors along the three edges of the Weyl chamber, and as in the case of so 6 C we have thereby established the existence of the irreducible representations of so 7 C with highest weight in the sublattice l{L 1 , L 2 , L 3 }. To complete the description, we need to know that the representation rll with highest weight 11. = (L 1 + L 2 + L 3 )/2 exists, and what it looks like, and this time there is no isomorphism to provide this; we will have to wait until the following lecture. In the meantime, we can still have fun playing around both with the representations we do know exist, and also with those whose existence is simply asserted. Exercise 19.1 t. Find the decomposition into irreducible representations of the tensor product V ®NV; in particular lind the multiplicities of the irreducible representation rn,+L, with highest weight 2L 1 + L 2 . Exercise 19.12. Show that the symmetric square of the representation r~ decomposes into a copy of NV and a trivial one-dimensional representation. Exercise 1913. Find the decomposition into irreducible representations · :f

J\2r•.

§19.4. Representations of the Odd Orthogonal Algebras We will now describe as much as we can of the general pattern for representations of the odd orthogonal Lie algebras so 2 .+ 1 C. As in the case of the even orthogonal Lie algebras, the proof of the existence part of the basic theorem (14.18) (that is, the construction of the irreducible representation with given

295

§19.4. Representations of the Odd Orthogonal Algebras

highest weight) will not be complete until the following lecture, but we can work around this pretty well. To begin with, recall that the weight lattice of so 2.+t Cis, like that of so 2.C, generated by L 1 , ••. , L. together with !Le further vector (L 1 + ·· · + L.)/2. The Weyl chamber, on the other hand, is tlie cone 1

'lr

=

0:: a L 1

1:

a1

~

a2

~

.. • ;;::::

a.

~ 0}.

The Weyl chamber is as we have pointed out the same as for sp 2 ,C, that is, it is a simplicial cone with faces corresponding to the n planes a 1 = a 2 , ... , a._ 1 = a. and a.= 0. The edges of the Weyl chamber are thus the rays generated by the vectors L 1 , L 1 + L 2 , ••• , L 1 + · · · + Ln-t and L 1 + · · · + L. (note that L 1 + · · · + L._ 1 is on an edge of the Weyl chamber). Again, the intersection of the weight lattice with the closed Weyl cone is a free semigroup, in this case generated by the fundamental weights W 1 = L 1 , w 2 = L 1 + L 2 , ... , w.- 1 = L 1 + · · · + L.- 1 and the weight w. = a = (L 1 + · · · + L.)/2. Moreover, as we saw in the cases of so 5 C and so 7 C, the exterior powers of the standard representation do serve to generate all the irreducible representations whose highest weights are in the sublattice 71. {L 1 , ••• , L.}: in general we have the following theorem. Theorem 19.14. For k = I, ... , n, the exterior power NV of the standard represemation V of so 2 .+I Cis the irreducible representation with highest weight Lt + ... + Lk. PROOF. We will leave this as an exercise; the proof is essentially the same as in the case of so 2.C, with enough of .1 difference to make it interesting. 0

We have thus constructed one-half of the irreducible representations of so 2 .+1 C: any weighty in the closed Weyl chamber can be written

y = a1 L 1

+ a 2 (L 1 + L 2 ) + .. · + a._ 1(L 1 + .. · + L._Jl + a.(L 1 + ... + L.)/2

with a1 E N; and if a. is even, the representation Sym"' V ® · · · ® Sym•·-'(/\"- 1 V) ® Sym"·'2 (/\"V) will contain an irreducible representation rY with highest weighty. We are still missing, however, any representation whose weights involve odd multiples of a; to construct these, we clearly ha ·.:e to exhibit an irreducible representation r. with highest weight a. This exi~ts and is called (as in the case of the even orthogonal Lie algebras) the spin representation of so 2.+1 C. We see from the above that once we exhibit the spin representation r., we will have constructed all the representations of so 2n+I C; for any y as above the tensor Sym"'V ® ·· · ® Sym•·-•(/\"- 1 V) ® Sym""(f..) will contain a copy of fY. As in the case of the spin representation r. of the even orthogonal Lie algebras, we can say some things about r. even in advance of its explicit construction; for example, we can do the following exercises.

296 Exercise 19.15. Find the weights (with multiplicities) of the representations

N v, and also of r«. Exercise 19.16. Give the complete decomposition into irreducible representations of Sym 2 fa and J\2 r... Show that

r« ® ra = J\"V EfJ J\"- 1 V EfJ /\"- 2 V EfJ · · • EfJ NV Et> NV. Exercise 19.17. Verify directly the above statements in the case of so 5 e, using the isomorphism with sp 4 C.

§19.5. Weyl's Construction for Orthogonal Groups The same procedure we saw in the symplectic case can be used to construct representations of the orthogonal groups, this time generalizing what we saw directly for NV in §§19.2 and 19.4. For the symmetric form Q on V =em, the same formula ( 17.9) determines contractions from y®J to V®(d- 21 • Denote the intersection of the kernels of all these contractions by yiJI. For any partition A= (A 1 :C: · · • :C: Am :C: O) of d, let

(19.18) As before, this is a representation of the orthogonal group Ome of Q. Theorem 19.19. Tire space § 141 Vis an irreducible representation of Ome; §!AI V nonzero if and only if tlte sum of the lengtlrs of the first two columns of tire Young diagram of A is at most m. The tensor power V®J decomposes exactly as in Lemma 17.15, with everything the same but replacing the symbol (d) by [tl]. In particular, § 141

v=

yldl. c.<=

lm(c.<:

J.lldJ-> yiJI).

Exercise 19.20. Verify that §p 1V is zero when the sum of the lengths of the first two columns is greater than m by showing that NV® NV® y<J-a-b) is contained in I,, '1' 1(V®<J-l)) when a+ b > m. ~;how that §p 1V is not zero when the sum of the lengths of the first two colt•.nns is at most 111. Exercise 19.21*. (i) Show that the kernel of the contraction from SymJV to Symd- 2 Vis the irreducible representation § 1d1V of some \':ith highest weight clL 1 • (ii) Show that Sym4 V = § 1d1V Ef) § 1d_ 21 V Ef) • · · Ef) § 1J- 2 p 1 r', where p is the largest integer

~

d/2.

'"§19.5. Weyl's Construction for Orthogonal Groups

297

The proof of the theorem proceeds exactly as in §17.3. The fundamental fact from invariant theory is the same statement as (17.19), with, of course, the operators 91 = '1'1 o
+ 1, and A.= (A.t ~ · · · ~A.. ~ 0), then § 141 Vis the irreducible representation of some with highest weight A.t Lt + .. · + A..L•. (ii) If m = 2n, and A.= (At ~ · · · ~ An-t ~ 0), then §!AI V is the irreducible representation of some with highest weight A. 1 L 1 + ·· · + A..L •. (iii) If m = 2n, and A. = (At ~ · · · ~ A..-t ~ A.. > 0), then § 141 Vis the sum of two irreducible representations of some with highest weights A.t Lt + ·· · + A..L. and A. 1 Lt + ... + A..-tLn-t- A..L•.

Theorem 19.22. (i) If m = 2n

Exercise 19.23. When m is odd, show that ome =some X {±I}. Show that if A. and 11 are associated, then 11 = A.® e, where e is the sign of the detnminant. We postpone to Lecture 25 all discussion of multiplicities of weigh! 3paces, or decomposing tensor products or restrictions to subgroups. As we saw in Lecture 15 for GL.e and in Lecture 17 for Sp:,.,e, it is possible to make a commutative algebra §1'1 = §l'l(V) out of the sum of ~n the irreducible representations of some, where v =em is the standard representation. First suppose m = 2n + 1 is odd. Define the ring §'(V, n) as in §15.5, which is a sum of all the representations § A(V) of GL(V) where A. runs over all partitions with at most 11 parts. As in the symplt:ctic case, there is a canonical decomposition §A(V)

and the direct sum Jl'l

= §pJ(V) ffi J!AJ(V),

= EJjA Jp,1(V) is an ideal in §'(V, 11). The quotient ring

§f'I(V) = A'(V, n)fJI'I

= EB §IAI(V) 4

is a commutative graded ring which contains each irreducible representation of so2n+te one.~.

298

19. so 6 C, so 7 C, and so,.C

If m = 2n is even, the above quotient will contain each representation twice if A. has n rows. To cut it down so there is only one of each, one can add to J 1"1 relations of the form x - r(x), for x e NV, where r: NV~ NV is the isomorphism described in the remark (iii) after the proof of Theorem 19.2. For a detailed discussion, with explicit generators for the ideas, see [L-T]. § 1 .~ 1 (V)

LECTURE 20

Spin Representations of some

In this lecture we complete the picture of lhe rep;·esentations of the orthogonal Lie algebras hy constructing the spin representations s± of some; this also yields a description of the spin groups Spinme. Since the representation-theoretic analysis of the spaces s± was carried out in the preceding lecture, we are concerned here primarily with the algebra involved in their construction. Thus, §20.1 and §20.2, while elementary, involve some fairly serious algebra. Section 20.3, where we briefly sketch the notion of triality, may seem mysterious to the reader (this is at least in part because it is so to the authors); if so, it may be skipped. Finally, we should say that the subject of the spin representations of some is a very rich one, and one that accommodates many different points of view; the reader who is interested is encouraged to try some of the other approach.~s that may be found in the literature. §20.1: Clifford algebras and spin representations of somC §20.2: The spin groups SpinmC and Spin,.IR §20.3: Spin 8 e and triality

§20.1. Clifford Algebras and Spin Representations of some We begin this section by trying to motivate the definition of Clifford algebras. We may begin by asking, why were we able to find all the representations of SLne or Sp 2 ,e inside tensor powers of the standard representation, but only half the representations of SOme arise this way? One difference that points in this direction lies in the topology of these groups: SL,e and Sp 2 ,C are simply connected, while some has fundamental group 71./2 for m > 2 (for proof·; see §23.1). Therefore SOme has a double covering, the spin group Spinme· (Form~ 6, these covering;' could also be extracted from our identification::;

:lU. Spm Keprescntations of so,.C

jl)\J

of the adjoint group PSOmiC with the adjoint group of other simply connected groups; e.g. the double cover of S03 1C is SL 2 1C.) We will see that th(; missing representations are those representations of SpinmiC that do not come from representations of some. This double covering may be most readily visible,. and probably familiar, for the case of the real subgroup S0 3 1R of rotations: a rotation is specified by an axis to rotate about, given by a unit vector 11, and an angle of rotation about u; the two choices ±u of unit vector give a two-sheeted covering. In other words, if D 3 is the unit ball in IR 3 , there is a double covering

S3 = D3 /iJD 3

--+

S0 3 1R,

which sends a vector v in D 3 to rotation by the angle 2nll,,ll about the unit vector v/llvll (the origin and the unit sphere oD 3 are sent to the identity transformation).

This covering is even easier to see for the entire orthogonal group 0 3 1R, which is generated by reflections Ru in unit vectors v (with ± v determining the same reflection): we can describe the double cover of 0 3 1R as the group generated by unit vectors v, with relations V 1 • ••• • V 0

whenever the compositions of the whenever

= 11' 1 • ••• • Wm

corn~sponding

reflections are equal, i.e.,

and also relations (-v)·(-w)

= v·w

for all pairs of unit vectors v and w. (Note that if we restricted ourselves to products of even numbers of the generators v E iJD 3 we would get back the double cover of the special orthogonal group S0 3 1C.) How should we generalize this? The answer is not obYiou::. For one thing, for various reasons we will not try to construct directly a group that covers the orthogonal group in general. Instead, given a vector space V (real or complex) and a quadratic form Q on V, we will first construct an algebra ClifT( V, Q), called the Clifford algebra. The algebra ClifT( V, Q) will then turn

§20.1.

Clifford Algebras and Spin Representations of so.,C

301

out to contain in its multiplicative group a subgroup which is a double cover of the orthogonal group O(V, Q) of automorphisms of V preserving Q. By analogy with the construction of the double cover of S0 3 1R, the Clifford algebra Cliff(V, Q) associated to the pair (V, Q) is an associative algebra containing and generated by V. (When we want to describe the spin group inside Cliff(V, Q) we will restrict ourselves to products of even numbers of elements of V having a fixed norm Q(v, :•); if odd products are allowed as well, we get a group called "Pin" which is a double covering of the whole orthogonal group.) To motivate the definition, we would likl' Cliff(V, Q) to be the algebra generated by V subject to relations analogous to those above for the double cover of the orthogonal group. In particular, for :my vector v with Q(v, v) = 1, since the reflection Rv in the hyperplane perpendicular to v is an involution, we want

v· v = 1 in CliiT(V, Q). By polarization, this is the same as imposing the relation

+ w·v = 2Q(v, w) for all v and w in V. In particular, w · v = - v · w if v and w are perpendicular. v·w

In fact, the Clifford algebra 1 will be defined below to be the associative algebra generated by V and subject to the equation v · v = Q(v, v). Looking ahead, we will see later in this section that each complex Clifford algebra contains an orthogonal Lie algebra as a subalgebra. The key theorem is then that CliiT(V, Q) is isomorphic either to a matrix algebra or to a sum of two matrix algebras. This in turn determines either one or two representations of the orthogonal Lie algebras, which turn out to be the representations which were needed to complete the story in the last lecture. Just as in the special linear and symplectic cases, the corresponding Lie groups are not really needed to construct the representations; they can be written down directly from the Lie algebra. In this section we do this, using the Clifford algebras to construct these representations of some directly, and verify that they give the missing spin representations. In the second section of this lecture we will show how the spin groups sit as subgroups in their multiplicative groups.

Clifford Algebras Given a symmetric bilinear form Q on a vector space V, the Clifford algebra C = C(Q) = CliiT(V, Q) is an associative algebra with unit 1, which contains and is generated by V, with v · v = Q(v, v) · I for all v E V. Equivalently, we have the equation

v · w + w · v = 2Q(v, w),

(20.1)

1 The mathematical world seems to be about evenly divided about the choice of signs here, and one must translate from Q to -Q to go from one side to the other.

20. Spin

jUL

Rcpres~:ulations

of so,C

for all v and w in V. The Clifford algebra can be delined to be the universal algebra with this property: if E is any associative algebra with unit, and a linear mapping j: V--+ E is given such that j(v) 2 = Q(v, v) · l for all v E V, or equivalently j(v)'j(w)

+ j(w)·j(v) = 2Q(v, w)·l

(20.2)

for all v, wE V, then there should be a unique homomorphism of algebras from C(Q) to E extending j. The Clifford algebra can be constructed ctuickly by taking the tensor algebra T'(V) =

EB y®• = CEt> VEt>(V® V)Et>(V® V® V)Ef>"', n~O

and setting C(Q) = T'(V)/I(Q), where /(Q) is the two-sided ideal generated by all elements of the form v ® v - Q(v, v) · 1. Tt is automatic that this C(Q) satisfies the required universal property. The facts that the dimension of Cis 2m, where m = dim(V), and that the canonical mapping from V to C is an embedding, are part of the following lemma:

Lemma 20.3. ~f e 1 , ... , e, form a basis for V, then tire proclucts e1 = e1, • e1, • ••• • e1•• for I= {i 1 < i2 < · · · < id, ancl with e; = 1, form a hasis for C( Q) = Cliff( v, Q). PRoor. From the equations e1 • e1 + e( e1 = 2Q(e1, e1) it follows immediately that the elements e1 generate C(Q). Their independence is not hard to verify directly; it also follows by seeing that the images in the matrix algebras under the mappings constructed below are independent. For another proof, note that when Q = 0, the Clifford algebra is just the exterior algebra J\'V. In general, the Clifford algebra can be filtered by subspaces Fk, consisting of those elements which can be written as sums of at most k products of elements in V; one checks that the associated graded space Fk/Fk+ 1 is NV. For a third proof, one can verify that the Clifford algebra of the direct sum of two orthogonal spaces is the skew commutative tensor product of the Clifford algebras of the two spaces (cf. Exercise B.9), which reduces one to the trivial 0 case where dim V = 1.

Since the ideal I (Q) c T( V) is generated by elements of even degree, the Clifford algebra inherit~ a 7L/27L grading:

ceven EB codd = c+ EB c-' with c+. c+ c c+, c+. c- c c-, c-. c+ c c-, c- ·c- c c+; c+ is spanned by products of an even number of elements in Vand c- is spanned by products of an odd number. In particular, ceven is a subalgebra of dimension 2m- 1. C=

Since C(Q) is an associative algebra, it determines a Lie algebra, with bracket [a, b] = a· b - b ·a. From now on we assume Q is nondegenerate. The new representations of some will be found in two steps:

§20.1. Clifford Algebras and Spin Rep1 esentations of so,. C

303

(i) embedding the Lie algebra so(Q\ =some inside the Lie algebra of the even part of the Clifford algebra C(Q); (ii) identifying the Clifford algebras with one or two copies of matrix algebras. To carry out the first step we make explicit the isomorphism of I\2 V with so(Q) that we have discussed before. Recall that so(Q)

= {X e End(V): Q(Xv, w) + Q(v, X w) =

0 for all v, .win V}.

The isomorphi~:m is given by NV~ so(Q) c End(V),

a " b f-+ q>.,.,b,

for a and b in V, where q>a"b is defined by

q>.,.,b(v) = 2(Q(b, v)a- Q(a, v)b).

(20.4)

It is a simple verification that q>.,.,b is in so(Q). One sees that the natural bases correspond up to scalars, e.g., e1 " en+J maps to 2(E1, 1 - En+J,n+i), so the map is an isomorphism. (The choice of scalar factor is unimportant here; it was chosen to simplify later formulas.) One calculates what the bracket on NV must be to make this an isomorphism of Lie algebras:

[cp.,.,b, q>c,.,JJ (v)

= q>.,.,b =

0

q>,,.,Av) - q>,,.,J 0 q>.,.,b(v)

2q>."b(Q(d, v)c- Q(c, v)d) - 2q>,,., 4(Q(b, v)a - Q(a, v)b)

= 4Q(d, v)(Q(b, c)a- Q(a, c)b) - 4Q(c, v)(Q(b, d)a - Q(a, d)b) - 4Q(b, v)(Q(d, a)c -- Q(c, a)d)

+ 4Q(a, v)(Q(d, b)c -

= 2Q(b, c)cp.,.,Av) - 2Q(a, d)q>,,.,b(v)

Q(c, b)d)

2Q(b, d)q).,.,,(v)

+ 2Q(a, c)q>d,.,b(v).

This gives an explicit formula for the bracket on NV:

[a " b, c " d] = 2Q(b, c)a " d - 2Q(b, d)a " c - 2Q(a, d)c " b + 2Q(a, c)d

1\

b.

(20.5)

On the other hand, the bracket in the Clifford algebra satisfies

[a·b, c·d] = a·b·c·d- c·d·a·b

= (2Q(b, c)a · d - a· c · b ·d) - (2Q(a, d)c · b - c ·a· d ·b) = 2Q(b, c) a· d - 2Q(a, d)c·b

= 2Q(b, c)a · d-

(2Q(b, d)a · c - a· c · d ·b)

+ (2Q(a, c)·d·b- a·c·d·b) 2Q(b, d)a · c - 2Q(a, d)c · b + 2Q(a, c)· d ·b.

20. Spin Representations of so.,C

304

ifr: NV-+ Cliff(V, Q) defined by 1\ b) = t(a · b - b ·a) = a· b - Q(a, b)

It follows that the map ljl(a

(..:0.6)

is a map2 of Lie algebras, and by looking at basis elemen.ts again one sees that it is an embedding. This proves:

Lemma 20.7. The mapping 1jJ subalgebra of C(Q)•••n.

o

cp- 1 : so(Q)-+ C(Q)•ven embeds so(Q) as a Lie

Exercise 20.8. Show that the image of ljJ is

F2 n C(Q)even n Ker(trace), where F2 is the subspace of C(Q) spanned by products of at most two elements of V, and the trace of an element of C(Q) is the trace of left multiplication by that element on C(Q). We consider first the even case: write V = W E9 W', where Wand W' are n-dimensional isotropic spaces for Q. (Recall that a space is isotropic when Q restricts to the zero form on it.) With our choice of standard Q on V = C 2 ", W can be taken to be the space spanned by the first n basis vectors, W' by the last n.

Lemma 20.9. The decomposition V = WEB W' determines an isomorpllism of algebras C(Q)

~

End(/\'W),

where 1\'W = Nw E9 ·· · E9 1\"W PROOF. Mapping C(Q) to the algebra E = Fnd(/\'W) is the same as defining a linear mapping from V to E, satisfying (20.2). We must construct maps 1: W-+ E and 1': W'-+ E such that

l(w) 2

= 0,

l'(w') 2 = 0,

(20.10)

and /(w)

o

l'(w') + l'(w')

o

/(w) = 2Q(w, w')/

for any wE W, w' E W'. For each wE W, let Lw E E be left multiplication by won the exterior algebra 1\'W:

Lw(e)

= w 1\ e. eE 1\'W

For 9 E W*, let D9 E E be the derivation of 1\'W such that D9 (l) = 0, D9(w) = .9(w)ENW = C for we W = NW, and 1

Note that the bilinear form Y, given by (20.6) is alternating since Y,(a linear map on Nv.

1\

a) = 0, so it defines a

§20.1. Clifford Algebras and Spin Representations of som C

Ds
l(w)

+ ( -l)dea(')C

305

A D11(().

.L< -l)i-1.9(w;)(Wl A ... A wi A ... A w.). Now

= Lw,

l'(w')

= D 9,

(20.11)

where .9 E W* is defined by the identity .9(w) = 2Q(w, w') for all wE W Therequired equations (20.10) are straightforward verifications: one checks directly on elements in W = N W, and then that, if they hold on Cand ~. they hold on C A ~. Finally, one may see that the resulting map is an isomorphism by 0 looking at what happens to a basis. Exercise 20.12. The left C(Q)-module J\'W is isomorphic to a left ideal in C(Q). Show that iff is a generator for NW', then C(Q) · f = J\'W· f, and the map CH C•f gives an isomorphism

l\W -+J\'W·f = C(Q)·f ofleft C(Q)-modules. Now we have a decomposition J\'W = N•••w E9 Nddw into the sum of even and odd exterior powers, and C(W)•••• respects this splitting. We deduce from Lemma 20.9 an isomorphism

(20.13) Combining with Lemma 20.7, we now have an embedding of Lie algebras:

(20.14) and hence we have two representations of so(Q) = so 2 .C, which we denote by

s+

=

J\evenw and s- = J\Oddw

Proposition 20.15. The representations S± are the irreducible representations of so 2.C with highest weights IX = !(L 1 + · · · + L.) and {3 = !(L 1 + ··· + Ln-~ L.). More precisely,

s+ = r« and s- = lp if n is even; s+ = lp and s- = r« if n is odd. PROOF. We show that the natural basis vectors e1 = e1, A • • • A e1k for J\'W are weight vectors. Tracing through the isomorphisms established above, we see that Hi= Ei.i- En+i,n+l in I) c so 2 .C corresponds to !(e1 A e.+ 1) in A2 V, which corresponds to !{ei · e,.+i - l) in C(Q), which maps to

f(Le, o D2 .r- /) A simple calculation shows fhat

= L,, o Der - !J E End(J\'W).

20. Spin Representations of somC

306

ifi e I t I.

ifi

Therefore, e1 spans a weight space with weight 1 L 1 - LJ -~ LJ All such weights with given III mod 2 are congruent by the Weyl group, so each of s+ = N•••w+ and s- = f\Oddw must be an irreducible representation. The highest weights are easy to read off. For example, the highest weight for N••• W is t L L 1 = a if n is even, while if n is odd, its highest weight is p. 0

!<2:,.

These two representations s+ and s- are usually called the half-spin representations of so 2.C, while their sum S = s+ $ s- = 1\"W is called the spin representation. Frequently, especially when we speak of the even and odd cases together, we call them all simply "spin representations." Elements of S are called spinors. For other proofs of the proposition see Exercises 20.34 and

20.35. For the odd case, write V = W E9 W' E9 U, where W and W' are ndimensional isotropic subs paces, and U is a one-dimensional space perpendicular to them. For our standard Q on C 2 "+ 1, these are spanned by the first n, the second n, and the last basis vector. Lemma 20.16. Tlte decomposition V = W $ W' $ U determines an isomorphism of algebras C(Q);:;;; End(I\"W) E9 End(/\"W').

PRooF. Proceeding as in the even ca:;e, to map V toE = End(/\"W), map we W to Lw, w' e W' to D8 , where ,9(w) = 2Q(w, w') as before. Let u0 be the element in U such that Q(u 0 , u0 ) = I, and send u0 to the endomorphism that is the identity on N••nw, and minus the identity on NddW. Since this involution skew commuteS with all Lw and D3 , the resulting map from V = W $ W' E9 U to E determines an algebra homomorphism from C(Q) to E. The map to End(/\"W') is defined similarly, reversing the roles of Wand W'. Again one 0 checks that the map is an isomorphism by looking at bases. Exercise 20.17*. Find a generator for a left ideal of C(Q) that is isomorphic to 1\"W. The subalgebra C(Q)evc" of C(Q) is mapped isomorphically onto either of the factors by the isomorphism of the lemma, so we have ah isomorphism in the odd case: C(Qt••• ~ End(/\"W). (20.18) As before, this gives a representation S = i\"W of Lie algebras: so 2 .+ 1 C

= so(Q) c

C(Q)''"'' ~ gl(/\"W)

= gl(S).

(20.19)

SLU.L. I he :Spin Uroups :-spm,,.IL and :Spin., If.!

Proposition 20.20. The representationS of so 2 n+I C with higllest weight

307

= 1\"W is the irreducible representation

a = !(L 1 + ·· · + L.). Exactly as in the even case, each ei is an eigenvector with with weight LuI Li). This time all such weights are congruent by the Weyl group, so this must be an irreducible representation, and the highest weight is clearly !(L 1 + · · · + L.). . 0 PROOF.

!
As we saw in Lecture 19, the construction of this spin representation S finishes the proof of the existence theorem for representations of some, and hence for all of the classical complex semisimple Lie algebras. Exercise 20.21 *. Use the above iJtntification of the Clifford algebras with matrix algebras (or direct calculation) to compute their centers. In particular, show that the intersection of the cenler of C with the even subalgebra c•••n is always the one-dimensional space<'' scalars. Show similarly that if xis in cood and x · v = - v · x for all v in V, then x = 0. Exercise 20.22*. For X e so(Q) and v e V, we have X· v E V by the standard action of so(Q) on V. On the other hand, w~ have identified so(Q) and Vas subspaces of the Clifford algebra C, so we can compute the commutator [X, v]. Show that these agree: X· v =[X, v] e V c C.

Problem 20.23*. Let C(p, q) be the real Clifford algebra corresponding to the quadratic form with p positive and q negative eigenvalues. Lemmas 20.9 and 20.16 actually construct isomorphisms of C(n, n) with a real matrix algebra, and of C(n + 1, n) with a product of two real matrix algebras. Compute C(p, q) for other p and q. All are products of one or two matrix algebr~s over IR, C, or IHI.

§20.2. The Spin Groups Spin"' C and Spin"' Ol The Clifford algebra C = C(Q) is generated by the subspace V has an anti-involution x 1-+ x*, determined by

(v 1 · ... · v,)*

=em, and

C

= (- 1)'v, · ... · v1

for any v1 , .•• , v, in V. This operation *• sometimes called the conjugation, is the composite of: the main antiautomorphism or reversing map r: C-+ C determined by

t(v 1 · ... · v,)

= v, · ... · v1

(20.24)

the main involution a. which is the identity on ce••n and minus the identity on C'dd, i.e., a.(v 1 • ... • v,)

= (-l)'v 1 · ... · v,.

(20.25)

Note that (x · y)* = y* · x*, which comes from the identities t(x · y) = t(y) · t(x) and a.(x · y) = a.(x) · a.(y). Exercise 20.26. Use the universal property for C to verify that these are well defined: show that a. is a homomorphism from C to C and t is a well-defined homomorphism from C to the opposite algebra of C (the algebra with the same vector space structure, but with reversed multiplication: x 7 y = y · x). Instead of defining the spin group as the set of products of certain elements of V, it will be convenient to start with a more abstract definition. Set Spin(Q) = {x E C(Q)•••n: x · x* = l and x · V · x* c V}.

(20.27)

We see from this definition that Spin(Q) forms a closed subgroup of the group of units in the (even) Clifford algebra. Any x in Spin(Q) determines an endomorphism p(x) of V by

p(x)(v)=x·v·x*, Proposition 20.28. For x

E

veV.

Spin(Q), p(x) is in SO(Q). The mapping

p: Spin(Q)-+ SO(Q) is a homomorphism, making Spin(Q) a connected two-sheeted covering of SO(Q). The kernel of pis {l, -1}. PROOF. We will prove something more. Define a larger subgroup, this time of the multiplicative group of C(Q), by Pin(Q) = {x E C(Q): x · x*

= I and x · V · x* c

V},

(20.29)

and define a homomorphism p: Pin(Q)-+ O(Q),

J'(x)(v) = a(x)·v·x*,

(20.30)

where a: C(Q)-+ C(Q) is the main invrhtion. To see that p(x) preserves the quad, o.1tic form Q, we use the fact that for w in V, Q(w, w) = w· w = - w· w*, and calculate:

Q(p(x)(v), p(x)(v))

= - a(x) · v · x* · (a.(x) · v · x*)* = - a(x) · v · x* · x · v* · a(x)*

= - a(x) · v · v* · a(x*) = Q(v, v)a(x) · a(x*) = Q(v, v)a(x · x*) = Q(v, v).

We claim next that p is surjective. This follows from the standard fact (see Exercise 20.32) that the orthogonal group O(Q) is generated by reflections. Indeed, if Rw is the reflection in the hyperplane perpendicular to a vector w, normalized so that Q(w, w) = -1, it is easy to see that w is in Pin(Q) and p(W) = Rw; in fact, w·w*

= w·(-w) = -Q(w, w) = 1,

and so p(w)(w)

= a(w) · w ·w* = -

w ·1

=-

w;

and if Q(w, v) = 0, p(w)(v)

= a(w)· v· w* =

-w· v· w*

=

v· w· w*

= v.

The next claim is that the kernel of p Cdl the larger group Pin(Q) is ± 1. Suppose X is in the kernel, and write X = Xo + X I with Xo E ceven and X I E codd. Then x 0 · v = v· x 0 for all v e V, so x 0 is in the center of C. And x 1 • v = -v· x 1 for all v e V. By Exercise 20.21, x 0 is inC ·1, and x 1 = 0. Sox= x 0 is inC and x 2 = 1; sox=

± l.

It follows !hat if R e O(Q) is written as a product of reflections Rw, o ... o Rw,• then the two elements in p- 1(R) are ± w1 • ••• • w,. In particular, we get another description of the spin groups: Spin(Q)

= Pin(Q) n

C(Q)even

= {±w1 • ••. • w2k:

= p- 1 (SO(Q))

w1 e V, Q(w1, w1) =

-1}.

(20.31)

Since -I = v · v for any v with Q(v, v) = - 1, we see that the spin group consists of even products of such elements. To complete the proof, we must check that Spin(Q) is connected or, equivalently, that the two elements in the kernel of p can be connected by a path. We leave this now as an exercise, since much more will be seen shortly. 0 Exercise 20.32*. Let Q be a nondegenerate symmetric bilinear form on a real or complex vector space V. (a) Show that if v and w are vectors in V with Q(v, v) = Q(w, w) =I 0, then there is either a reflection or a product of two reflections that takes v into w. (b) Deduce that every element of the orthogonal group of Q can be written as the product of at most 2·dim(V) reflections. Exercise 20.33*. Since Spin(Q) is a subgroup of the multiplicative group of C(Q), its Lie algebra is a subalgebra of C(Q) with its usual bracket. Verify that this subalgebra is the subalgebra so(Q) that was constructed in §20.1.

Exercise 20.34. The fact that t\"W (and t\"W' in the odd case) is an irreducible module over C(Q) is equivalent to the fact that jt is an irreducible representation of the group Pin(Q) since the linear span of Pin(Q) is dense in C(Q). (a) Apply the analysis of §5. 1 to the subgroup Spin(Q) c Pin(Q) of index two. In the odd case, t\"W and 1\'W' are conjugate representations, so their restrictions to Spin(Q) are isomorphic and irreducible: this is the spin representation. In the even case, 1\"W is self-conjugate, and its restriction to Spin(Q) is a sum of two conjugate irreducible representations, which are the two half-spin representations. (b) Of the representations of Spin(Q) (i.e., the representations of so'"C), which induce irreducible representations of Pin(Q) and which are restrictions of irreducible representations of Pin(Q)? Exercise 20.35. Deduce the irreducibility of the spin and half-spin representations from the fact that their restrictions to the 2-groups of Exercise 3.9 are irreducible representations of these finite groups. Exercise 20.36*. Show that the center of Spin'"(C) is p- 1 (I)= {±I} ifm is odd. If m is even show that the center is p- 1 (± 1)

= {± 1, ±w},

where, in terms of our standard basis, w

ie ·en+t- ien+t ·e

ie.·e

.-

ie .·en

1 1 2 = -----2-----·-· ... ·- ---2 -2--·--

Exercise 20.37*. Show that the spin representation Spin(Q)---+ GL(S) maps into the special linear group SL(S). Show that for m = 2n and n even, the half-spin representations also map into the special linear groups SL(S+) and SL(S-). Exercise 20.38*. Construct a nondegenerate bilinear pairing p on the spinor spaceS= 1\'W by choosing an isomorphism of NW with IC and letting il(s, t) be the image of r(s) " t e /\' W by the projection to NW = IC, where r is the main antiautomorphism). (a) When m = 2n, show that fJ can also be defined by the identity fJ(s, t)f = r(s ·f)· t · f for an appropriate generator f of 1\"W'. Deduce that the action of Spin(Q) on S respects the bilinear form p. (b) Show that p is symmetric if n is congruent to 0 or 3 modulo 4, and skew-symmetric otherwise. So the spin representation is a homomorphism

=0, 3 (4),

Spin2n+IIC-+ so2 . 1C

if n

Spin 2n+tl[---+ Sp 2 niC

if n :: I, 2 (4).

§20.2. The Spin Groups SpinmC and

Spinm~

311

(c) If m = 2n, the restrictions of Pto s+ and s- are zero if n is odd. For n even, deduce that the half-spin representations are homomorphisms Spin 2.C ..... S02.-,C

if n

=0 (4),

Spin 2.C ..... Sp2.-• C

if n

= 2 (4).

Note in particular that Spin 8 C has two maps to S0 8 C in addition to the original covering. "Triality," which we discuss in the next section, describes the relation among these three homomorphisms. Exercise 20.39. Show that the spin and half-spin representations giv·e the isomorphisms we have seen before:

= GL 1 C = C*,

Spin 2 C

~

Gl.(s+)

Spin 3 C

~

SL(S)

Spin 4 C

~

SL(S+) x SL(S-) = SL2 C x SL 2 C,

Spin 5 C

~

Sp(S) = Sp 4 C,

Spin 6 C

~

SL(S+)

= SL 2 C,

= SL4 C.

Exercise 20.40. Let Cm denote the Clifford algebra of the vector space em with our standard quadratic form Qm. (a) Th ·embedding ofC 2 " = W $ W' in 1[;2 "+1 = W Ei?J W' E9 U as indicated induces an embedding of c2n in c2n+l' and corresponding embedding of Spin 2,C i!l Spin2.+ 1 C and of S02.C in S02 n+t C. Show that the spin representationS ofSpin 2 .+ 1 C restricts to the spin representations+ EE> s- ofSpin 2 .C. (b) Similarly there is an embedding of Spin 2 .+ 1 C in Spin 2.+ 2C coming from an embedding ofC 2 "+ 1 = W ~ W' ® U in C 2 "+ 2 = W ® W' ® U1 ® U2 ; here U1 EE> U2

= C $C

with the quadratic form

embedded in U 1 Ei?J U2 by sending l to

(~ ~).

and U = C is

(~. ~). Show that each of the

half-spin representations of Spin 2.+ 2 Q' restricts to the spin representation of Spin2n+l C. Very little of the above discussion needs to be changed to construct the real spin groups Spinm(IR), which are double coverings of the real orthogonal groups SOm(IR). One uses the real Clifford algebra Cliff(!Rm, Q) associated to the real quadratic form Q = - Qm, where Qm is the standard positive definite quadratic form on /Rm. If v; are an orthonormal basis, the products in this Clifford algebra are given by and

V( V;

= - t.

20. Spin Representations of somC

312

The same definitions can be given as in the complex case, giving rise to coverings Pinm(IR) of Om(IR) and Spinm(IR) of SOm(IR). Exercise 20.41. Show that SpinmiR is connected by showing that if v and ware any two perpendicular elements in V with Q(v, v) = Q(w, w) = -I, the path H-+(cos(t)v

+ sin(t)w)·(cos(t)v- sin(t)w),

0

~

t

~

n/2

connects - 1 to l. Exercise 20.42. Show that i 1-+ v2 • v3 , j 1-+ v3 · v1 , k 1-+ v 1 • v2 determines an isomorphism of the quaternions DD onto the even part of Cliff(IR 3 , - Q 3 ), such that conjugation - in H corresponds to the conjugation * in the Clifford algebra. Show that this maps Sp(2) = {q E HI qq = l} isomorphically onto Spin 3 1R, and that this isomorphism is compatible with the map to S03 1R defined in Exercise 7.15. More generally, ifQ is a quadratic form on IRm with p positive and q negative eigenvalues, we get a group Spin+(p, q) in the Clifford algebra C(p, q) = Cliff(IRm, Q), with double coverings Spin+(p, q)-+ so+(p, q). Exercise 20.43*. Show that Spin+(p, q} is connected if p and q are positive, except for the case p = q = l, when it has two components. Show that if in the definition of spin groups one relaxes the condition x · x* = 1 to the condition x · x* = ± 1, one gets coverings Spin(p, q) of SO(p, q).

§20.3. Spin 8 C and Triality When m is even, there is always an outer automorphism of Spinm(C) that interchanges the two spin representations s+ and s-, while preserving the basic representation V = em (cf. Exercise 19.9). In case m = 8, all three of these representations V, s+, .u.J s- are eight dimensional. One basic expression of triality is the fact that there are automorphisms of Spin 8 e or so 8 C that permute these three representations arbitrarily. (In fact, the group of outer automorphisms modulo inner automorphisms is the symmetric group on three elements.) We give a brief discussion of this phenomenon in this section, in the form of an extended exercise. To see where these automorphisms might come from, consider the four simple roots: a1

= L1

-

L 2,

a2

=

L2

-

L 3,

a3

=L3 -

L4,

a4 = L 3

+ L4.

Note that a 1 , a 3 , and a 4 are mutually perpendicular, and that each makes an angle of 120° with a 2 :

§20.3. Spin 8 C and Triality

313

Exercise 20.44*. For each of the six permutations of {a 1 , a 3 , a 4 } find the orthogonal automorphism of the root space which fixes a2 and realizes the permutation of a 1 , a 3 ; and a 4 . Each automorphism of this exercise corresponds to an automorphism of the Carlan subalge!Jra ~. In the next lecture we will see that such automorphisms can be extended (nonuniquely) to automorphisms of the Lie algebra so 8 (C). (For explicit formulas sec [Ca2].) There is also a purely geometric notion of triality. Recall that an evendimensional quadric Q can contain linear spaces A of at most half the dimension of Q, and that there are two families of linear spaces of this maximal dimension (cf. [G-H], [Ha]). In case Q is six-dimensional, each of these families can themselves be realized as six-dimensional quadrics, which we may denote by Q+ and Q- (see below). Moreover, there are correspondences that assign to a point of a11y one of these quadrics a 3-plane in each of the others: Point in Q

----+

3-plane in / Q-

""

Point in Q-

~

Point in Q+

J·plane in Q+ (20.45)

/

----+

3-plane in Q

Given P E Q, {A E Q+: A contains P} is a 3-plane in Q+, and {A E Q-: A contains P} :sa 3-plane in Q-. Given A E Q+, A itself is a 3-plane in Q, and {r e Q-: rnA is a 2-plane} is a 3-plane in Q-. Given A E Q-, A itself is a 3-plane in Q, and {r E Q+: rnA is a 2-plane} is a 3-plane in Q+. To relate these two notions of triality, take Q to be our standard quadric in P 7 = P(V), with V = WEB W' with our usual quadratic space, and let s+ = N•enw and s- = NddW be the two spin representations. In Exercise 20.38 we constructed quadratic forms on s+ and s-, by choosing an isomorphism of Nw with C. This gives us two quadrics Q+ and Q- in P(S+) and P(S-). To identify Q+ and Q- with the families of 3-planes in Q, recall the action of V on S = i\'W = s+ EB s- which gave rise to the isomorphism of the Clifford algebra with End(S) (cf. Lemma 20.9). This in fact maps s+ to s-

20. Spin Representations of somC

314

and

s- to s+; so we have bilinear maps V X s+ -'+ s- and

V

X

s- -'+ S+.

(20.46)

Exercise 20.47. Show that for each point in Q+, represented by a vectors E s+, { v E V: v · s = 0} is an isotropic 4-plane in V, and hence determines a projective 3-plane in Q. Similarly, each point in Q- determines a 3-plane in Q. Show that every 3-plane in Q arises uniquely in one of these ways. Let ( , >v denote the symmetric form corresponding to the quadratic form in V, and similarly for s+ and s-. Define a product

s+

x

s--'+ v,

S X

t I-+ S • t,

(20.48)

by requiring that (v, s · t)v = (v · s, t)s- for alit• E V. Exercise 20.49. Use this product, together with those in (20.46), to show that the other four arrows in the hexagon (20.45) for geometric triality can be described as in the preceding exercise. This leads to an algebraic version of triality, which we sketch following [Ch2]. The above products determine a commutative but nonassociative product on the direct sum A = V ffi s+ $ s-.. The operation

(v, s,

t)~-+

(v ·s, t>s-

determines a cubic form on A, which by polarization determines a symmetric trilinear form on A. Exercise 20.50*. One can construct an automorphism J of A of order three that sends v to s+' s+ to s-' and s- to v, preserving their quadratic forms, and compatible with the cubic form. The definition of J depends on the choice of an element v1 E V and s 1 E s+ with (v 1 , v1 )v = (s 1 , s 1 >s• = l; set t 1 = v1 ·s 1 , so that (t 1 , t 1 ) 5 - =las well. The map J is defined to be the composite JL o v of two involutions JL and v, which are determined by the following: (i) Jl interchanges s+ and s-, and mapS V to itself, With Jl(S) Jl(V) = 2(v, v 1 )vv 1 - v for v E V. (ii) v interchanges v and s-' maps s+ to itself, with v(v) v(s) = 2(s, S1 )s+S 1 - S for S E S+.

= Vt · S for S E S+; = v. St

for v E V;

Show that this J satisfies the asserted properties. E)(ercise 20.5l*.ln this algebraic form, triality can be expressed by the assertion that there is an automorphism j of Spin 8 C of order 3 compatible with J, i.e., such that for all x E Spin 8 C, the following diagrams commute:

§20.3. Spin 8 C and Triality

315

v __.!____. s+ ---+ s- ---+ v J

]"''

]··'"'"

V---+ J

J

j•-
]••I

s----+ v s+ ----J J

If j': so 8 C-+ so 8 C is the map induced by j, the fact that j is compatible with the trilinear form
for X

E

so 8 C, Y

=

+
j'(X), Z

j'(Y).

=

Ys, t)

+
PART IV

LIE THEORY

The purpose of this final part of the book is threefold. First of all, we want to complete the program stated in the introduction to Part II. We have completed the first two steps of this program, showing in Part II how the analysis of representations of Lie groups could be reduced to the study of representations of complex Lie algebras, of which the most important are the semisimple; and carrying out in Part III such an analysis for the classical Lie algebras sf.C, sp 2 .C, and some. To finish the story, we want now to translate our answers back into the tenns of the original problem. In particular, we want to deal with representations of Lie groups as well as Lie algebras, and real groups and algebras as well as complex. The passage back to groups is described in Lecture 21, and the analysis of the real case in Lecture 26. Another goal of this Part is to establish a framework for some of the results of the preceding lectures-to describe the general theory of semisimple Lie algebras and Lie groups. The key point here is the introduction of the Dynkin diagram and its use in classifying all semisimple Lie algebras over C. From one point of view, the impact of the classification theorem is not great: it just tells us that we have in fact already analyzed all but five of the simple Lie algebras in existence. Beyond that, however, it provides a picture and a language for the description of the general Lie algebra. This both yields a description of the five remaining simple Lie algebras and allows us to give uniform descriptions of associated objects: for example, the compact homogeneous spaces associated to simple Lie groups, or the characters of their representations. The classification theory of semisimple Lie algebras is given in Lecture 21; the description in these terms of their representations and characters is given in Lecture 23. The five exceptional simple Lie algebras, whose existence is revealed from the Dynkin diagrams, are studied in Lecture

318

PartlY. Lie Theory

22; we give a fairly detailed account of one of them (!h), with only brief descriptions or the others. Third, all this general theory makes it possible to answer the main outstanding problem left over from Part II 1: a description of the multiplicities or the weights in the irreducible representations of the simple Lie algebras. We give in Lectures 24 and 25 a number of fonnulas for these multiplicities. This, it should be said, represents in some ways a shift in style. In the previous lectures we would typically analyze special cases first and deduce general patterns from these cases; here, for example, the Weyl character formula is stated and proved in general, then specialized to the various individual cases (this is the approach more often taken in the literature on the subject). In some ways, this is a fourth goal of Part IV: to provide a bridge between the naive exploration of Lie theory undertaken in Parts II and III, and the more general theory readers will find elsewhere when they pursue the subject further. Finally, we should repeat here the disclaimer made in the Preface. This part of the book, to the extent that it is successful, will introduce the reader to the rich and varied world of Lie theory; but it certainly undertakes no serious exploration of that world. We do not, for example, touch on such basic constructions as the universal enveloping algebra, Verma modules, Tits buildings; and we do not even hint at the fascinating subject of (infinitedimensional) unitary representations. The reader is encouraged to sample these and other topics, as well as those inct~;~ded here, according to background and interest.

LECTURE 21

The Classification of Complex Simple Lie Algebras

In the first section of this lecture we introduce the Dynkin diagram associated to a semisimple Lie algebra g. This is an amazingly efficient way of conveying the structure of g: it is a simple diagram that not only determines g up to isomorphism in theory, but in practice exhibits many of the properties of g. The main use of Dynkin diagrams In this lecture, however, will be to provide a framework for the basic classification theorem, which says that with exactly five exceptions the Lie algebras discussed so far in these lectures are all the simple Lie algebras. To do this, in §21.2 we show how to list all diagrams that arise from semisimple Lie algebras. In §21.3 we show how to recover such a Lie algebra from the data of its diagram, completing the proof of the classification theorem. All three sections are completely elementary, though §21.3 gels a little complicated; It may be useful to read it in conjunction with §22.1, where the process described is carried out in detail for the exceptional algebra g2 • (Note that neither §21.3 or §22.1 is a prerequisite for §22.3, where another description of g2 will be given.) §21.1: Dynkin diagrams associated to semisimple Lie algebras §21.2: Classifying Dynkln diagrams §21.2: Recovering a Lie algebra from its Dynkin diagram

§21.1. Dynkin Diagrams Associated to Semisimple Lie Algebras For the following, we will let 9 be a semisimple Lie algebra; as usual, a Cartan subalgebra I) of 9 will be fixed throughout. As we have seen, the roots R of 9 span a real subspace of I)* on which the Killing form is positive definite. We denote this Euclidean space here by E, and the Killing form on E simply by

21. The Classification of Complex Simple Lie Algebras

320

( , ) instead of B( , ). The geometry of how R sits in IE is very rigid, as indicated by the pictures we have seen for the classical Lie algebras. In this section we will classify the possible configurations, up to rotation and multiplication by a positive scalar in IE. In the next section we will see that this geometry completely determines the Lie algebra. The following four properties of the root system are all that are needed:

(I) R is a finite set spanning IE. (2) a E R =- -a E R, but k ·a is not in R if k is any real number other than (3) For a E R, the reflection W.. in the hyperplane a.L maps R to Itself. (4) For a, pER, the real number

± l.

Is an Integer.

Except perhaps for the second part of (2), these properties have been seen in Lecture 14. For example, (4) is Corollary 14.29. Note that "fl« = {1(11«), and

(21.1) For (2), consider the representation i = ffitS.« of the Lie algebras«~ si 2 C. Note that all the nonzero factors but I)= Ao are one dimensional. We may assume a is the smallest nonzero root that appears in the string. Now, decompose i as an !.-module:

i = s«®i'. By the hypothesis that a is the smallest nonzero root that appears in the string, i' is a representation of s. having no eigenspace with eigenvalue I or 2 for H•. It follows that I' must be trivial, i.e., 9-. = (0) fork ""0 or ± 1. Any set R of elements in a Euclidean space E satisfying conditions (I) to (4) may be called an (abstract) root system. Property (4) puts very strong restrictions on the geometry of the roots. If .9 is the angle between a and {J, we have

n _

IIPII·

~·- 2 cos(.9)w.

(21.2)

In particular, (21.3)

is an integer between 0 and 4. The case when this integer is 4 occurs when cos(.9) = ± 1, i.e. p = ±a. Omitting this trivial case, the only possibilities are therefore those given in the following table. Here we have ordered the two roots so that IIPII ~ llall, or ln~.l ~ ln.11 1.

321

§2l.l. Dynkin Diagrams Associated to Semlsimple Lie Algebras Table 21.4 cos(.9)

.}3/2

.}2/2

.9

n.,n,.

n/6

n/4

3

2

w

.jj

liP II

1/2 n/3 1 1

./2

0 n/2 0 0

-1/2 2n/3 -1 -1

•

In other words, the relation of any two roots a. and

/

LDf6

/ /_, t

~3

-./212

-.jj/2

3n/4 -2 -1

5n/6

./2

.jj

-3 -1

p is one of

J!:_!_6

&4 ill.!~ \.l. ~ ~ rt/2

'll!_/4

'

The dimension n = dimAE = dim cl) is called the rank (of the Lie algebra, or the root system). It is easy to find all those of smallest ranks. As we write them down, we will label them by the labels (A.), (B.), ... that have become standard. Rank I. The only possibility is

.. which is the root system ofsl 2 C. Rank 2. Note first that by Property (3), the angle between two roots must be the same for any pair of adjacent roots in a two-dimensional root system. As we will see, any of the four angles tr/2, tr/3, tr/4, and tr/6 can occur; once this angle is specified the relative lengths of the roots are determined by Property (4), except in the case of right angles. Thus, up to scalars there are exactly four root systems of dimension two. First we have the case I!J = tt/2,

\.,

+

which is the root system of si 2 C x si 2 C ~ so 4 C. (In general, the orthogonal direct sum of two root systems is a root system;

322

21. The Classification or Complex Simple Lie Algebras

a root system that is not such a sum is called ir·reducibfe. Our task will be to classiry all irreducible root systems.) The other root systems or rank 2 are

(A,)

*

the root system ofsl 3 C;

the root system or so 5 C

~

sp 4 C; and

Although we have not yet seen a Lie algebra with this root system, we will see.. that there is one. Exercise 21.5. Show that these are all the root systems or rank 2. Exercise 21.6. Show that a semisimple Lie algebra is simple ;rand only if its. root system is irreducible. Rank 3. Besides the direct sums or(A 1 ) with one or those or rank 2, we have; the irreducible root systems we have seen; we draw only dots at the ends o£ 1 the vectors, the origins being in the centers of the reference cubes: 'J

§2l.t. Dynkin Diagrams Associated to Semisimple Lie Algebras

323

(A,)

which is the root system of si4 C

~ so 6 C;

the root system of so 7 1C;

(

;,the root system of sp 6 C. ~i' .. ·

!Exercise 21.7. Show that there are no other root systems of rank 3.

~'..

~- We can further reduce the data or a root system by introducing a subset of !the roots, called the simple roots. First, choose as in Lecture 14 a direction

324

21. The Classification of Complex Simple Lie Algebras

R = R+ u n- is a disjoint union of positive and negative roots. Call a positive root simple if it is not the sum of two other positive roots. For the classical Lie algebras, keeping the notations and conventions of Lectures 15-20, the simple roots are (A") &ln+t C Lt - L2, Lz - L3, ... , L._, - L", L. - Ln+l,

1: E-+ R, so that

so 2 n+JC

(B")

L1

-

L2, L2

-

L3 ,

••• ,

L._ 1 - L., L.,

(C.)

L, - L2, L2- L3, ... , L._, - L., 2L",

(D.)

L1

-

L2, Lz- L3, ... , Ln-t - L., L._,

+ L•.

Exercise 21.8. Verify this list, and lind two simple roots for (0 2 ). We next deduce a few consequences of properties (1)-(4), which indicate how strong these axioms are. They will be used in the present classification of abstract systems, as well as in the following section.

fJ are roots witlr the form

(5) 'fa,

p-

fJ #- ±a, tlten the a-string tltrouglt p, i.e., the roots of

p - a, {l, fJ + a, fJ + 2a, ... , p + qa has at most four irr a string, i.e. p + q :-;:; 3; i11 additiorr, p - q = ,.,.,. Indeed, since W.,(fJ + qa) = fJ - pa, and W.(fJ + qa) = (fJ - n,.a) - qa, pa,

p - (p -

I )a, ... ,

we must have p = n11., + q, which is the second equality. For the first, we may take p = 0, and then q = -n,.,, which we have seen is an integer no larger than three. As a consequence of(5) we have (6) Suppose a,

fJ are roots with fJ #-

±a. Then

(fJ, a) > 0 =>a - fJ is a root;

(fJ, a) < 0 =>a + fJ is a root. If (fJ, ex) = 0, tlterr a -

fJ artd a + fJ are simultaneously roots or nonroots.

(7) If a and fJ are distinct simple roots, then a -

fJ and fJ -

a are not roots.

This follows from the definition of simple, since from the equation cannot be in R+, and similarly -(a- fJ) = fJ-IJ. cannot be inn+. From (6) and (7) we deduce that (a, fJ) :-;:; 0, i.e.,

a= fJ +(a- fJ), a- fJ

(8) Tile artgle between two distirrct simple roots carmot be acute.

(9) The simple roots are linearly independerlt. This follows from (8) by Exercise 21.9*. If a set of vectors lies on one side of a hyperplane, with aU mutual angles at least 90°, show that they must be linearly independent. '

§21.2. Classifying Dynkin Diagrams

325

(10) There are precisely " simple roots. Eaclr positive root ca/J be written uniquely as a noll-negative integra/linear cornbi11atiorr of simple roots. Since R spans [, the first statement follows from (9), as does the uniqueness of the second statement. The fact that any positive root can be written as a positive sum of simple roots follows readily from the definition, for if a were a positive root with minimal l(a) that could not be so written, then a is not simple, so a = fJ + y, with fJ and y positive roots with 1(/J), /(y) < I( a). Note that as an immediate corollary of(IO) it follows that no root is a li11ear combination of the simple roots a1 witlr coefficierrts of mixed sign. For example, (7) is just a special case of this. The Dynkin diagram of the root system is drawn by drawing one node 0 for each simple root and joining two nodes by a number of lines depending on the angle /) between them: no lines

0

one line

0-----Q

if 1)

= 2ltl3

two lines

a::p

if 1)

= Jn/4

three lines

a$=D

if1)eSltl6.

0

if1l=ltl2

When there is one line, the roots have the same length; if two or three lines, an arrow is drawn pointing from the /orrger to the shorter root. Exercise 21.10. Show that a root system is irreducible if and only ifits Dynkin diagram is connected. We will see later that the Dynkin diagram of a root system is independent of the choice of direction, i.e., of the decomposition of R into R + and R • .

.§21.2. Classifying Dynkin Diagrams The wonderful thing about Dynkin diagrams is that from this very simple . picture one can reconstruct the entire Lie algebra from which it came. We will · see this in the following section; for now, we ask the complementary question :of which diagrams arise from Lie algebras. Our goal is the following classifica:·uon theorem, which is a result in pure Euclidean geometry. (The subscripts ~ on the labels (An), ... are the number or nodes.) \;Theorem 21.11. Tire Dynkirr diagram.~ of irreducible root systems are preci.~eiJ•:

326

21. The Classification of Complex Simple Lie Algebras

CA.)

o----o-- ....

---0---0

(n:! I)

(B.)

o----o-- ....

----o:::7=o

(n:!2)

(C.)

(}--{}- .... --cr:¢:o

(n:!3)

(D.)

o----o-- ....

(n:!4)

(E6)

-<

~

( E1l

~

(F•)

~

The first four are those belonging to the classical series we have been studying: (A.) (B.) (C.) (D.)

sln+t C so 2 .+1 C

sp2.C so 2 .C

The restrictions on n in these series are to a void repeats, as well as degenerate ; cases. Indeed, the diagrams can be used to recall all the coincidences we have seen: When n = I, all four of the diagrams become one node. The case (D.) is degenerate, since so 2 Cis not semisimple, while the coincidences (C 1 ) = (B.) = (A.) correspond to the isomorphisms sp 2 C

~

so 3 C

~

si 1 C

0.

§21.2. Classifying Dynkin Diagrams

327

For n = 2, (0 2 ) =(Ad x (A,) consists of two disjoint nodes, corresponding to the isomorphism so 4 C

~

si 2 C x si 2 C

0

0 .

The coincidence (C 2 ) = (8 2 ) corresponds to the isomorphism sp 4 C ~ so 5 C

cr¢o = o:::?=o\.

For n = 3, the fact that (0 3 ) = (A 3 ) reflects the isomorphism

PROOF oF THE THEOREM. Our desert-island reader would find this a pleasant pastime. For example, if there are two simple roots with angle 5nf6, the plane of these roots must contain the 0 2 configuration of 12 roots. It is not hard to see that one cannot add another root that is not perpendicular to this plane, without some of the 12 angles and lengths being wrong. This shows that (0 2 ) is the only connected diagram containing a triple line. At the risk of spoiling your fun, we give the general proof of a slightly stronger result. In fact, the angles alone determine the possible diagrams. Such diagrams, without the arrows to indicate relative lengths, are often called Coxeter diagrams (or Coxeter graphs). Deline a diagram of 11 nodes, with each pair connected by 0, I, 2, or 31ines, to be admissible if there are 11 independent unit vectors e1 , ••• , e. in a Euclidean space E with the angle between e1 and e1 being n/2, 2n/3, 3n/4, or Sn/6, according as the number of lines between corresponding nodes is 0, l, 2, or 3. The claim is that the diagrams of the above Dynkin diagrams, ignoring the arrows, are the only connected admissible diagrams. Note that (e;, e1) = 0, -1/2, - ~/2, or - ..fi/2,

(21.12)

according as the number of lines between them is 0, 1, 2, or 3; equivalently, 4(e, e1) 2 =number of lines between e1 and e1.

(21.13)

The steps of the proof are as follows:

(i) Any subdiagram of an admissible diagram, obtained by removing some nodes and all lines to them, will also be admissible. (ii) There are at most 11 - l pairs of nodes tlwt are comaected by lines. The diagram lras 110 cycles (loops). Indeed, if e1 and e1 are connected, 2(e1, e1) s; - t, and 0<

(I e1, I

e,) =

11

+2

I

(e, e1),

l<J

which proves the first statement of(ii). The second follows from the first and (i).

21. The Classification of Complex Simple Lie Algebras

328

(iii) No llode has more tlran tltree lines to it.

By (i), we may assume that e 1 is connected to each of the other nodes; by (ii), no other nodes are connected to each other. We must show that Li= 2 4(e 1 , e1)2 < 4. Since e 2 , ••• ,e. are perpendicular unit vectors, and e 1 is not in their span, I = (e 1 , e.J 2 >

•

I

(e., e1)2,

]=2

as required. (iv) Ill all admissible diagram, any string of nodes connected to eacll otiJer by one line, with none but tire ends of tire string connected to any other nodes, can be collapsed to one node, and res11lting diagram remains admissible:

0

0

0

D

a

0

:o---o-- .... --o---cx a

If e 1 , •.• , e, are the unit vectors corresponding to the string of nodes, then e' = e 1 + ·· · + e, is a unit vector, since (e', e') = r + 2((e 1 , e 2 ) = r-

+ (e 2 , e3 ) + ··· + (e,_ 1, e,))

(r- 1).

Moreover, e' satisfies the same conditions with respect to the other vecton since (e', ei) is either (e 1 , e1) or (e, e1). Now we can rule out the other admissible connected diagrams not on our list. First, from (iii) we see that the diagram (0 2 ) has the only triple edge. Nex~ there cannot be two double lines, or we could find a subdiagram of the form:

and then collapse the middle to get< I ! I ! >,contradicting (iii). Similarly there can be at most one triple node, i.e., a node with single lines to three other nodes, by

>-····-<

X

§21.2. Classirying Dynkin Diagrams

329

By the same reasoning, there cannot be a triple node together with a double line:

o==D--···-< To finish the case with double lines, we musl simply verify that

is not admissible. Consider general vectors v = a 1 e 1 + a5 e5 • We have

+ a1 e1 , and w·= a 3 e 3 +

a4 e4

f.J2.

·and (v, w) = -a 2 a3 We want to choose v and w to contradict the Cauchy -Schwarz inequality (v, w) 2 < Uvll 2 llwll 2 • For this we want la 2 1/llvll and la 3 1/llwll to be as large as possible. ;·

Exercise 21.14. Show that these maxima are achieved by taking a 2 = 2a 1 and

· a3 = 3a 5 , a4

= 2a 5 •

In fact, v = e 1

+ 2e 2 , w = 3e3 + 2e4 + e 5 do give the contradictory

(v, w) 2 = 18, llvll 2 = 3, and llwll 2 = 6. i

Finally, we must show that the strings coming out from a triple node cannot ~be longer that those specified in types (D.), (E 6 ), (E 7 ), or(E 8 ). First, we rule out

;:;!j·

:iW•' ;·"

~};'' ~j . . :..

r,i- . : :1.':

~·:.:'

fConsider the three perpendicular unit vectors:

~~:'· .·

u = (2e2

+ e 3 )/,/3,

v = (2e4

+ e 5 )/,/3,

f~,Then as in (iii), since e 1 is not in the span of them,

w = (2e 6

+ e 7 )/,/3.

21. The Classification of Complex Simple Lie Algebras

330

I = lle 1 ll 2 > (e 1 , u) 2

+ (e 1 , v) 2 + (e 1 , w) 2

= 1/3

+ 1/3 + 1/3 =

I,

a contradiction. Exercise 21.15*. Similarly, rule out

and

(The last few arguments can be amalgamated, by showing that if the legs from a triple node have lengths p, q, and r, then 1/p + 1/q + lfr must be greater than 1.) This finishes the proof of the theorem. 0

§21.3. Recovering a Lie Algebra from Its Dynkin Diagram In this section we will complete the classification theorem for simple Lie algebras by showing how one may recover a simple Lie algebra from the data of its Dynkin diagram. This will proceed in two stages: first, we will see how to reconstruct a root system from its Dynkin diagram (which a priori only tells us the configuration of I he simple roots). Secondly, we will show how to describe the entire Lie algebra in terms of its root system. (In the next. lecture we will do all this explicitly, by hand, and independently of the general discussion here, for the simplest exceptional case (G 2 ); as we have noted, the· reader may find it useful to work through §22.1 before or while reading the, general story described here.) :i To begin with, to recover the root system from the Dynkin diagram, let ... , a. be the simple roots corresponding to the nodes of a connected Dynkin; diagram. We must show which non-negative integral linear combinations~ m1a 1 are roots. Call m1 the level of m1a 1• Those of level one are the simple,; roots. For level two, we see from Property (2) that no 2a1 is a root, and by)

a,;;

I

I

I

§21.3. Recovering a Lie Algebra from Its Dynkin Diagram

331

Property (6) that a 1 + a1 is a root precisely when (a, a1) < 0, i.e., when the . corresponding nodes are joined by a line. Suppose we know all positive roots oflevel at most rn, and let P= Im1a 1 · be any positive root of level rn. We next determine for each simple root a = a1, whether p + a is also a root. Look at the a-string through {J:

p-

p«, ... '

p, ... • p + qa.

We know p by induction (no root is a linear combination of the simple roots ::-;; m1 and p - pa is a positive root). By Property (S), q = p - "p... So fJ + a is a root exactly when

a1 with coefficients of mixed sign, so p

p > np ..

" =2({J,a,-aa)() =I rn 1 r~., ... 1=1

In effect, the additional roots we will find in this way are those pbtained by reOecting a known positive root in the hyperplane perpendicular to a simple root a 1 (and filling in the string if necessary). To finish the proof, we must show that we get all the positive roots in this way. This will follow once from the fact that any positive root of level m + I · can be written in at least one way as a ·sum of a positive root of levelm and . a simple root. If y = I r1a 1 has level m + I, from

0 < (y, y) =

I

r1(y, a 1),

' some (y, a 1) must be positive, with r 1 > 0. By property (6), y - a 1 is a root, as required. Byway of example, consider the rank 2 root systems. In the case of sl 3 C, we · start with a pair of simple roots a~o a 2 with n.. , .• , = I, i.e., at an angle of 2n/3; , as always, we know that fJ = a 1 + a 2 is a root as well.

~-···.

~-<

~r-~

:,i,On the other hand, since fJ- 2a 1 = a 2 - a 1 is not a root, p + a 1 cannot be ~either, and likewise fJ + a 2 is not; so we have all the positive roots. ~~ In the case ofsp 4 C, we have two simple roots a 1 and a2 at an angle of 3n/4; In terms of an orthonormal basis L 1 and L 2 these may be taken to be L 1 and L1 - L 1 , respectively.

~ '

~l'

If ~; fj,:-~:si·

~

a 1 +a 2

2a 1 +a 2

21. The Classilicalion of Complex Simple Lie Algebras

332

We then see that in addition top = a 1 + a 2 , the sum II+ a1 = 2a 1 + a 2 is a root-it is just the renection oral in the plane perpendicular to a.-but II+ a 2 = a 1 + 2a 2 and 3a 1 + a 2 are not because a 1 - a 2 and a 2 - a 1 are not respectively (alternatively, we could note that they would form inadmissible angles with a 1 and a2 respectively). Finally, in the case of (0 2 ), we have two simple roots a 1 , a 2 at an angle of 5n/6, which in terms of an orthonormal basis forE may be taken to be L 1 and ( -3L 1 + j3L 2 )/2 respectively.

Reflecting a 2 in the plane perpendicular to IX 1 yields a string of roots a 2

+ a1,

a 2 + 2a 1 and a 2 + 3a2 • Moreover, reflecting the last of these in the plane perpendicular to IX 2 yields one more root, 2a 2 + 3a 3 • Finally, these are all the positive roots, giving us the root system for the diagram (G 2 ). We state here the results or applying this process to the exceptional diagrams (F4 ), (E6 ), (E 7 ), and (E 8 ) (in addition to (G 2 )). In each case, L 1 , ... , L. is an orthogonal basis forE, the simple roots a, can be taken to be as follows, and the corresponding root systems are given:

a 1 = L 1,

(G 2 )

R+

=

{L

1,

a2 = -

j3L 2 , ±L 1

+

J3

3

2L 1 + TL 2 ;

f

L2,

±~L 1 +

f L2}·

(G 2 ) thus has 6 positive roots.

R + = { L, } u {L 1 + L1}1< 1 u {L 1 - L1 }," 1 u

{L

1

±L2 ± 2

L± L 3

4}

·

In particular, (F4 ) has 24 positive roots.

L 1 - L2 at=

-

L3

-

L4 2

-

L5

+

j3L 6 '

·.~

§21.3. Recovering a Lie Algebra from Jts Dynkin Diagram = L3- L2,

a3 = L2- Lt.

IX4

as= L4- L3,

a6 = Ls- L4;

R+ = {Lt

+ L1}t<Jss v

333

{Lt- L1}1
{ ±Lt ± L2 ± L3 ± L4

± Ls + j3L6}

2

U

nun1ber

or mlnutsians even .

(E 6 ) has 36 positive roots. L1 !Xt

L2

-

L6 +

••• -

2

a6 = L 5

R+

-

=

-

L4,

a7 = L 6

j2L 7

-

'

L5;

= {L, + LJ}t<Js6 v {Lt- LJ}J
U

nun,ber of minus !dans odd •

Thus, (E 7 ) has 63 positive roots.

al =

L 1 - L 2 - .. · - L 7 2 .

+ L8

•

a3 = L2- Ll, a6 =

Ls- L4,

+ LJ}I<Jss v {L,- LJ}J
R+ = {L,

U

2

number o(

mlnu~ 5l~n5 even .

'(E1) has 120 positive roots. i:.i· For (G 2 ) and (F4 ) the simple roots are listed in order reading from left to (right in their Dynkin diagrams

~~--

t~\ r~

!!;'\\

~/ ...

r~in the classical series (A.)-(D.). For (E ~p;-_ ~(,

8 ),

the numbering is

1'1:~,.-

~~~!

~kli

~; ~L-.·

1{,

as

334

21. The Classification of Complex Simple Lie Algebras

while those for (E 7 ) and (E 6 ) are obtained by removing the last one or two nodes. Note that, given the root system of (E 8 ), we can find the root system of(E 7 ) or (E6 ) by taking the subspace spanned by the first seven or six simple roots. Exercise 21.16*. (a) Verify the above lists of roots. (b) In each case, calculate the corresponding fundamental weights. Exercise 21.17*. Show that no two of the root systems of (A.)-(E8 ) are isomorphic, and deduce that the Dynkin diagram of a root system is indepen· dent of choice of positive roots. A more satisfying reason for the last fact is the observation that any two choices of positive roots differ by an element of the Weyl group-the group generated by reflections w.. in the simple roots. This can be seen directly for each of the diagrams (A.)-(E 8 ); for a general proof that two choices differ by an element of the Weyl group, see Proposition D.29. We should mention here another way of conveying the data of a Dynkin diagram. This is simply the n x n matrix of integers (n 1, 1 = , .. ,..). where we take n1• 1 = 2; it is called the Cart an matrix of the Dynkin diagram (or of the Lie algebra). Thus, for example, the Car tan matrix of (A.) is 2

-1

0

-I

2 -1

-1 2

0

0 0

0 0

0 0 0

0 -1

-I

2 -I

-I 2

~:~

These matrices pop up remarkably often, in a variety of seemingly unrelated~ areas of mathematics. They will not play a major role in the present text, bu~ the reader has probably encountered them already in one form or anoth, and will probably do so again. -·~ Exercise 21.18*. Compute the Carlan matrix, and its determinant, for Dynkin diagram.

ea~1

::~

The next task is to see how the root system determines the Lie algebra. V(i' concentrate on the uniqueness, since there are other ways to see the existe indeed, for all but the five exceptions we have already seen the Lie algebrai; We will describe several approaches to this problem, starting with a strai~, forward and computational method and finishing with a slick but abs · approach.

21. The Classification of Complex Simple Lie Algebras

336

X 1 = [X1,, [X,,_,, ... , [X,,, X,.J ... ]],

Pl•• [ Yi,_,, ... , PI,, Yi.J ... JJ. Call/ admissible if each partial sum IX1, + · ·· + IX1• is a root, 1 ~ s ~ r; note Y,

=

that I is admissible exactly when X, is not zero. Lemma 21.21. If I and J are two admissible sequences for wiJiclr IX 1 = a1 , then there is a nonzero rational number q determined by I, J, and the Dyrtkin diagram, such that X 1 = q · X 1 . PROOF. Let k = i, be the last entry in I. If j, = k as well, the result follows by induction on r. We reduce the general case to this case, by maneuvering to,. replace j, by k. We have first

with q 1 a nonzero rational number depending only on J, k, and the Dynkin diagram, since a1 - IX• = a1 - a• is a root; the point is that we know how: s.,, ~ sl 2 acts on the a.-string through a1 as soon as we know the length ofth~; string, and this is Dynkin diagram information. Next, let s be the largestj integer such that j. = k. Then J.; >~

whereK = (j 1 ,

••• ,

j,),since [li,, [X 1, Z]] =[X, [Y., Z]] when i o1 k. Finally,~

'

~-~

,t~

[Jk, [Xh XK]] = q2. X",

with q2 a nonzero rational number depending only on K, k, and the Dynkili~ diagram, since aK + a• is a root. Combining these three equations, we get .~ .<,r. xl = q, q2 · [x., [xi,,. ... [xi• +•' xKJ ... JJ, -ta '"f" which suffices since the sequence for the term on the right ends in the integer k as/.

sami~~

d

Proposition 21.22. Tlte bracket of any two basis elements in (21.20) is a ration multiple of another basis element, that multiple determined from the Dynkl diagram. •.;<• PROOF. This is clear for brac~ts of an H 1 with any basis element. Lemma 21.2 · handles brackets of the form [X,, X 1 ], and those involving only Y's' " similar. For brackets[}/, XI], it suffices inductively to compute [Ya,, a rational multiple of some X"' with K shorter than J (or of H• if J has o·~ term); but this was worked out in the proof of the lemma. · ·· ·

x,r

Exercise 21.23*. (i) Show that in (0 2 ) each positive root can be written ino ,, one way as a sum of simple roots, up to the order of the first two roo~

§21.3. Recovering a Lie Algebra rrom Its Dynkin Diagram

(ii) Work out the multiplication table from the Dynkin diagram. (iii) Ver: that the result is indeed a Lie algebra, which is (visibly) simple. This exercise will be worked out in detail to start the next lecture. Of coun there is nothing but lack of time to keep us from verifying that the other fo exceptional Dynkin diagrams do lead, by the same prescription, to honestl algebras, but doing it by hand gets pretty laborious, and we will describe son of the other methods available. The fact that the multiplication table can be defined with rational coefl cients becomes important when one wants to reduce them modulo prin numbers, which we will not discuss here. The fact that they can be taken 1 · be real, on the other hand, will come up later, when we discuss real forms < · complex Lie algebras and groups. There is a more general and elegant way to proceed, given by Serre [Se3 Write n11 in place of n.,.,. Form the free Lie algebra on generators H 1,

;.f'

... ,H.,X1 , ... ,X.,

Y1 ,

... ,

Y,.,

i.e., form the free (tensor) algebra with this basis, and divide modulo by th '.relations [A, B] + [B, A]= 0 and the Jacobi relation. Then take this free Li ·algebra, and divide by the relations ~.

[X~o

[Hit X1]

Y,] = H 1 (all i);

[X1, lj] = 0 (i of. j);

= n11 X1 (all i, j);

R'~nd, for all i of. j, !1_.'._:····.·_.:_:.·.·.·.·····

~

[Xt, [~:~~~~: ~: ~~: ~~.~;] 0 =

[X.,[X1,[X1,X1]]]=0,

~~~o [Xlt [Xit [X~o X1]]]] = 0,

[Y,,[Y,,[Y;,lj]]]=O

:~::::~I

ifrrf)=-2:

[l't, [l't, [Y,, [ }j, lj]]]] = 0 if n11 = -3.

~\>r

J:xercise 21.24. Verify that if one starts with a semisimple Lie algebra with a i!ven Dynkin diagram, the above equations must hold.

"4' •\.:

•', Serre shows ([Se3, Chap. VI App.], cf. [Hut §18]) that the resulting Lie · gebra is a finite-dimensional semisimple Lie algebra, with Carlan subalgebra :Jenera ted by H 1 , ••• , H. and given root system. In particular, this includes a ;pioof of the existence of all the simple Lie algebras. .~Here is a third approach to uniqueness. Suppose g and g', with given Car tan :···oolgebras ~ and lj', and choice of positive roots, have isomorphic root ~terns. There is an isomorphism I)--+ 1)', taking corresponding H1 to HI. 'fhoose arbitrarily nonzero vectors X1 and x; in the root spaces of g and g' ,.COrresponding to the simple roots.

338

21. The Classification of Complex Simple Lie Algebras

Claim 21.25. Tltere is a unique isomorphism from 9 to 9' extertdlrtg the isomorphism of lJ with I]', arrd mappi11g X, to x; for all i. PROOF. The uniqueness of the isomorphism is easy: the resulting map is determined on the Y; by sl 2 considerations, and the H 1, X, and Y, generate g. For the existence of the isomorphism consider the subalgebra 9 of 9 ~ g' generated by ii, = H, ED n;, X,= X, ED x;, and Y, = Y, ED Y,'. It suffices to prove that the two projections from 9to 9 and 9' are isomorphisms. The kernel of the second projection is fED 0, where f is an ideal in g. Since 9 is simple, f is either 0, as required, or f = g. In the latter case, we must have g = 9 ED g'. To see that this is impossible, consider a maximal positive root p, take nonzero vectors x,, X~ in the corresponding root spaces, and set Xp = X11 ED Xj,, a highest weight vect2r in g. Let W be the subspace of ii obtained by successively applying all Y,'s. Then W is a proper subspace of g, since its weight space w, corresponding to {J is one dimensional. By the argument we have seen several times, g preserves W. Now if !l = 9 ED 9', W would be an ideal in 9 ED n', and this would force X 11 ED 0 to belong to W, making W, two dimensional again. 0

To finish this story, we should show that the simple Lie algebras corre~ sponding to two different Dynkin diagrams cannot be isomorphic, i.e., that the two choices made in going from a semisimple Lie algebra to Dynkin. diagram do not change the answer. The general facts are: (I) Any two Cartan subalgebras of a semisimple Lie algebra are conjugate,; i.e., there is an inner automorphism by an element in the corresponding ., adjoint group, which takes one into the other. (2) Any two decompositions of a root system into positive and negative roots~ differ by an element of the Weyl group. .3 J

These are standard facts which are proved in Appendix D. Both statemen~:: are subsumed in the fact that any two Borel subalgebras of a semisimple L~·:: algebra are conjugate, a Borel subalgebra being the subspace spanned by thej Carlan subalgebra and the root spaces 9cr for positive a. For those readed~ who crave logical completeness but do not want to go through so muc~:) general theory, we observe that most possible coincidences can be ruled outJ by such simple considerations as computing dimensions, and others can be~ ruled out by simple ad hoc methods, cf. Exercise 21.17. ~, Finally, we must also prove the "existence theorem": that there is a sim~J Lie algebra for each Dynkin diagram. Serre's theorem quoted above gives i;~ unified proof of existence. But we have seen and studied the Lie algebras f~~ the classical cases (A.)-(D.), and it is more in keeping with the spirit or th, lectures to at least try to s.ee the five exceptions explicitly. This is the subj~ of the next lecture. :-~

LECTURE 22

g2 and Other Exceptional Lie Algebras

This lecture is mainly about 9z, with just enough discussion of the algebraic constructions of the other exceptional Lie algebras to give the reader a sense or their complexity. g1 , being only 14-dimensional, is different: we can reasonably carry out in practice the process described in §21.3 to arrive at an explicit description of the algebra by specifying a basis and all pairwise products; we do this in §22.1 and verify in §22.2thatthe result really is a Lie algebra. In §22.3 we analyze the representations of 9 2 , and arrive in particular at another description of g1 : it is the algebra of endomorphisms of a seven-dimensional vector space preserving a general trilinear form. (Note that §22.3 may be read independently of either §22. I, §21.2, or §21.3.) Finally, in the fourth section we will sketch some of the more abstract (i.e., coordinate free) approaches to the construction or the live exceptional Lie algebras. 'Yhile the first two sections are completely elementary, the constructions given in §22.4 involve some fairly serious algebra. §22.1: Construction of g 1 from its Dynkin diagram §22.2: Verifying that 91 is a Lie algebra ,§22.3: Representation theory of g 1 'f22.4: Algebraic constructions of the exceptional Lie algebras

:§22.1. Construction of g2 from Its Dynkin Diagram ~}:

~In this section we will carry out explicitly the process described in the preceding

;~tion for the Dynkin diagram (G 2 ), constructing in this way a Lie algebra

?g2 with diagram (0 2 ) (and in particular proving its existence). ~,

The first step is to find the root system from the Dynkin diagram. In the I)* associated

rease of g 2 this is immediate; we may draw the root system R c

22. g2 and Other Exceptional Lie Algebras

340 to the diagram G 2 as follows:

a 2 = (-3{2,/3{2)

Pz

Here the positive roots are denoted a 1, with a 1 and a 2 the simple roots. The coordinate system here has no particular significance (in particular, recall that the configuration of roots a, and {11 is determined only up to a real scalar), but is convenient for calculating inner products. Note that the Weyl group is the dihedral group generated by rotation through an angle of n/3 and reflection in the horizontal; the Weyl chamber associated to the choice of ordering of the roots given is the cone between the roots a 6 and a 4 • As indicated in the preceding section, we start by letting X 1 be any eigen· vector for the action of~ with eigenvalue Ot: 1 , and X 2 any eigenvector for the action ofiJ with eigenvalue a 2 • We similarly let Y1 and Y2 be eigenvectors with eigenvalues /1 1 and /12 and set H 1 = [X 1 , Y1 ]

and

H 2 = [X 2 , Y2 ].

We can choose Y1 and Y2 so that the elements H, e l) satisfy a 1 (H 1 ) = a2 (H 2 ) ==' 2, i.e., [H.,X 1 ]=2·X 1 and It follows that

[H2 ,X2 ]=2·X2 •

'? ·;~

i.e., H,X,,and Yjspan asubalgebra s., ~ si 2 C, with H, X,,and Yj a normalized; --~ basis for this copy of sl 2 C. Now, it is clear from the diagram above that there is a unique way ofwritin&i each positive root a, as a sum of simple roots ex,, + ·· · + ex,. so that the partial) sums a,, + ··· + a 1, are roots for each I ::<;;; k (modulo exchanging the first tWcij terms): we go through the root system by the path ;)

22. !h and Other Exceptional Lie Algebras

342

Similarly, we know that [X, lj] == 0 whenever a,+ this tells us as well that [X., Y2]

= [X1, Y6] = [X2, = [X4, Y2] = [X 5 ,

p1 == a,- a1 is not a

root;

Y1] = [X2, Y4] = [X2, Y5 ] = [X3, Y,] Y2] == [X 5 , Y3] = [X6, Y.J = 0.

The multiplication table thus far looks like

r,

x2

y2

x3

yl

H,

0 2X 1 -2Y1

112

•

•

•

• • x.., • 0 •

•

112

x, r, x2

x,

•

2X

H,

2 -2 r2

x3

0

0

y3 1/2

y2 xl yl x..,

x..,

r..,

x,

"'

• •

• • 0 •

•

"'

•

•

Xs

Y..,

•

"'

0 0

Ys 0 0

•

• •

•

0

x6 0

•

•

r..,

Xs Ys

x6

Ys

x6

~

"' •

• •

• •

•

0 0 0

0 0

•

0 0

•

"' 0

0

•

•

0

"'

0

0 0

y6

0 0 0

0 0

•

0

0

•

• •

•

The next thing to do is to describe the action or H 1 and 1/2 on the various vectors X, and Y;. This can be done using the inner product on ~. but it is perhaps simpler to go back to the basic idea of restriction to the subalgebras 5 41 and 5 41 • For example, if we want to determine the action of H 1 on the various X,, consider how the algebra g = l) EfH~. ED g,,) decomposes as a representation of s«,:

-3

-1

I

3

·------~·~------·~----~;· I

'

'

-2

\

,

I

'

I

"'

tv ;

2

... ,,, .... ··~----:~·~-----1· ' ;

-3 ; ;

•

-I'

•

I

I

\

'I

•

§22.1. Construction of g2 from Its Dynkin Diagram

343

We get two trivial representations (the spans of X 6 and Y6 , as already noted); one copy of the adjoint representation Sym 1 V (the subalgebra s., itself) spanned by X 1 , Y1 , and H1 ; and two copies of the irreducible four-dimensional representation Sym 3 Vspanned by X 1 , X 3 , X 4 , and X 5 and Y5 , Y4 , Y3 , and Y2 . In particular, it follows that X 2 , X 3 , X 4 , and X 5 are eigenvectors for the action of H 1 with eigenvalues of -3, -1, l, and 3, respectively; and likewise Y5 , Y4 , Y3 , and Y1 are eigenvectors with eigenvalues -3, -1, 1, and 3. In similar fashion, we consider the decomposition of g under the action of s., = C{H1 , X 1 , Y1 }: diagrammatically, this looks like

~

: ~~

2

~ ~,:

-2

~-1 Here we have two trivial representations, spanned by X 4 and Y4 , one adjoint (s., itself), and four copies of the standard two-dimensional representation V, spanned by X 6 and X 5 , X 3 and X 1 , Y1 and Y3 , and Y5 and Y6 .lt follows that X6 , X 3 , Y1 , and Y5 are eigenvectors for the action of H 2 with eigenvalue l, and likewise X 5 , X 1 , Y3 , and Y6 are eigenvectors with eigenvalue -l. Including this information, we can fill in the top two rows of the multiplication table:

H1 H2

H2

X1

0

2X 1 -X 1

-2f1 -3X 2 Y1 2X 2

r,

x,

3f2 -2Y1

-X3 X,

Y3 -Y,

X4 0

r.

x,

r,

x.

r,

- f4 0

3X, -X,

-3Y, Y,

0 X6

0 -Y6

Decomposing g1 according to the action of s., and l'i., gives us information about the action of X 1 , X 1 , Y1 , and Y1 on the other basis vectors as well. For example, we saw a moment ago that X 5 and X 6 together span a sub-

22. 92 and Other Exceptional Lie Algebras

344

representation of lh under the action of s.,. with ad(X 2 ) carrying X 5 to X6 • It follows from this that ad(f2 ) must carry X 6 back to X 5 : we have ad(Y2 )(X 6 )

=

ad(Y2 ) ad(X 2 )(X 5 )

= ad(X2 ) ad(Y2 )(X5 ) - ad([X 2 , = 0 - ad(H2 )(X5 )

=

f 2 ])(X5 )

X5•

Similarly, since ad(X 2 ) carries X 1 into -X3 , which together with X 1 spans a copy of the standard two-dimensional representation of s., ~ sl 2 C, it follows that ad(Y2 ) will carry -X 3 back to X 1 • Likewise from the fact that ad(Y2) carries Y1 to- Y3 we see that ad(f2 )(f3 ) = - Y~o and since ad(f2 ): Y5 r-+ Y6 , ad(X2 ): Y6 r-+ Y5 • We can in the same way use the action of 5 41 to determine the values of ad(X 1 ) and ad( Y2 ) on various basis vectors, though because the representation of s,., on 9 2 has larger-dimensional components this is slightly more complicated. To begin with, consider the representation or s,., on the subspace spanned by X 2 , X3 , X4 , and X 5 • We know that ad(X 1 )carries X 2 to X3 ,and since X 2 is an eigenvector for the action of the commutator [X 1 , Y1 ] = H1 with eigenvalue -3, it follows that ad(Yd must carry X3 to 3X2 : we have ad(YtJ(X 3 ) = ad(ftl ad(XJl(X 2 )

= ad(X 1 ) ad(YJl(X 2 ) - ad([X 1 , = 0- ad(HtJ(X 2 )

f 1 ])(X 2 )

= 3X 2 •

Using this, we can next determine the action of Y1 on X4 : ad(Y1 )(X4 )

= ad(Y1 ) ad(X 1 )(X3 ) = ad(Xtl ad(Yd(X3 ) - ad(HtJ(X 3 )

= ad(XJl(3X 2 ) + X 3 = 4X3 , and we calculate likewise that ad(Yd(X 5 ) = 3X4 • Analogously, knowing that··

ad(ftl carries Y2 to f 3 to f 4 to f 5 yields the information that ad(Xd must~ carry f 3 , f 4 , and Y5 to 3Y2 , 4Y3 and, 3Y4 , respectively. Including all this: information in the chart, the next four rows of our multiplication table are :;j

. ·i ~

X4

Y4

X5

Y5

x6

0

3Y• 0 0

0 0

r6

r.··1

o·J

0

0 ·~

.,

V\' ~

x5 oi~ ·~

. ;,/~ : __

;

We next have to find the commutators of the basis elements X 1 and lJ flit;~ i, j ~ 3. We cannot do this by looking at the action of the subalgebrai~

346

22. !h and Other Exceptional Lie Algebras

Table 22.1

H, H,

II,

x,

0

2X 1 -X,

x,

Y,

x,

Y,

x,

Y,

-2Y1

-3X,

3Y1 -2Y1

-X,

x,

Y, - Y,

0

2X 4

-3Y1

-Y,

JX,

-2Y4

0

Y,

r,

H,

Y,

2X1

x, 0

x,

H,

-X,

Y,

x,

x,

Y,

x,

3X,

-3Y, Y,

x.

x. x.

r. - r.

0

0 -2Y,

-x,

JY,

-x.

-3X, 2X,

0

0 -X, 0 0 2Y1 0 -2X 1 JY, 0 211 1 t 311 2 0 0

0 0 11 1 + 311 1 -3X,

Y,

x. r. x, r,

x,

r. 0 0 Y, 0 0 -Y, 0

lf,+lf,

x,

r,

0

-r. 0 0

0

0 0 -X,

r, 0 r,

0

-x. 0 x, 0

x,

0

-r, 0

-r, 0.

H1 t11

There was another good reason for these changes: now each of the brackets

[X, Y,] will be the distinguished element of I) corresponding to the root a,. If ·"

-~

we denote this element by H1, then we read off from the table that

,..;

H3

=

H 1 +3Hz,

114

= 2H 1 +

3Hz,

(22.2) --~

and

.J

111 = [X,, Y,],

[If;, X 1] = 2X1,

[II, }'j] = - 2 Y;,

fori= I, 2, 3, 4, 5, 6.

§22.2. Verifying That g 2 Is a Lie Algebra

(22.3)

·~"

·_.,~ .fi

.:~

· -·~

The calculation of the preceding section gives a complete description of what~:~] the Lie algebra gz must look like, but there is still some work to be done: .-~ unless we know that there is a Lie algebra with diagram (Gz), we do not know~ that the above multiplication table defines a Lie algebra, let alone a simple.~ one. In fact, t~e .simpl!city is not mu~h of a pr~blem (cf. Exercise. ~4.34): b~t ~il to know that 1t IS a L1e algebra requtres knowing that the Jacob1 1denttty IS,~ valid. One could simply check this from the table for all ( 134 ) triples of elementsj -•.,.. from the basis, a rather uninviting task. There is another way, which gives more structure to the preceding calcula~. tions, and which will give a clue for possible constructions of other Liealgebras. The root diagram for (Gz) is made up of two hexagons, one withj long arrows, the other with short This suggests that we should find a copy o-~t-~ the corresponding Lie algebra si 3 C inside g 2 • The subspace spanned by,. and the root spaces corresponding to the six longer roots is clearly closed under brackets, so is the obvious candidate. The long roots are o:,, a: 2 , an. o: 6 = o:, + o: 2 , and their inverses. So we define g0 to be the subspace spanned by the corresponding vectors: ~~

§22.2. Verifying That gz is a Lie Algebra

g0 = C{H5 , H2 , X 5 , Y5 , X2 , Y2 , X 6 , Y6 }.

The multiplication table for g0 is read off from Table 22.1: Hz Hs liz

X, Ys Xz y2 x6

0

x,

Y,

X,

Y,

x6

y6

2X5

-2Y5

-x,

Y, Hs

-Xz

Yz

2X 2

-2Y2

x6 0

0 -Y6 Hz

x6 x6 0 Xz 0 -X,

-Y6 -Y6 - Yz 0 Ys 0 lis +liz

This is exactly the multiplication table for same order):

6[ 3 C,

with its standard basis (in

6l3C = C{Et,t- Ez,z, Ez.z- E3,3• Et,z• Ez.t• Ez,J• EJ,z• Et.J• E3,t}

So we have determined an isomorphism g 0 ~ sl 3 C. (Note right away that this verifies the Jacobi identity for triples taken from! The rest of the Lie algebra must be a representation of the subalgel g0 ~ 6[ 3 C, and we know what this must be: the smaller hexagon is the uniot the two triangles which are the weight diagrams for the standard represen • lion of 6[ 3 and its dual, which we denote here by Wand w•; W is the sum . the root spaces for a:4 , P1 , and P3 , while w• is the sum of those for P4 • ·and a: 3 . ·~.

~' ~:: ~'.

l•

1;:. ~

~;'. ""

~,:

w

~l"

If'~···

';;., a look a! tho tablo •ho"' th:t tho «cto" X,, Y,, and Y, fonn a b• li:.IOr W = C 3 that corresponds to the standard basis e 1 , e 2 , and e 3 , at

348

22. !h and Other Exceptional Lie Algebras

similarly Y4 , X 1 , and X3 form a basis for W* the dual basis er, et, and eJ: we have

02 = OoEf>

=

(C 3 )* that corresponds to

W$ W*.

With these isomorphisms, the brackets Qo

X

W-+ W

and

Qo

X

W* -+ W*

correspond to the standard operations of sl3Con C 3 and (C 3)*. Next we look at brackets of elements in W Note that [W, W] is contained in W*, either by weights or by looking at the table. The table is

X4 Y1

-2X3 0

2X 1 -2Y4

or

-2e1

2et

0

-2ef

Identifying w = c•, w• = (C 3 )* as above, we see that the bracket w• becomes the map

w x w-+ w•

=

N w,

V X WI-+

-2 · V

w X w-+

1\ W.

Similarly for W*, we have [W*, W*] c W, and the bracket is identified with the map

Finally we must look at brackets of elements of W with those of W*, which land in g0 • Here the table is

In terms of the standard bases, [e, ej] = 3E1, 1 mapping

[, ]: w X w• -+si3C c

~IJI.

Intrinsically, this

gi(W)

can be described by the formula [v,
for V,

WE

W and

qJ E

W*.

(224~

, :)'

Exercise 22.5*. Show that [v, q1] is the element of sl 3 C characterized by the' formula ').;

§22.2. Verifying That g2 is a Lie Algebra B([v,q>],Z)= 18q>(Z·v)

fora11Zesi 3 C,

where B is the Killing form on g0 = sf 3 C. In other words, if we write v • the element in g 0 = sf 3 satisfying the identity B(v•q>, Z) = q>(Z · v)

for all Z e g 0 = si 3 C,

then the bracket [v, fP] can be written in the form

[ v, (/)] = 18. v • (/). It is now a relatively painless task to verify the Jacobi identity, since, than having to check it for triples from a basis, it suffices to check it on of arbitrary elements of the three spaces g0 , W, and using the above algebra descriptions for the brackets. We will write out this exercise, sin same reasoning will be used later. For example, for three or two element: g0 , this amounts to the fact that g0 = sl 3 C is a Lie algebra and Wand J-1 representations. · For one element Z in g0 , and two elements v and win W, the Jacobi id for these three elements is equivalent to the identity

w•

Z · (v

A

w)

= (Z · v) 1\ w + v A

(Z · w),

which we know for the action of a Lie algebra on an exterior produc· similarly for one element in g0 and two in w•. The Jacobi identity for Z e g0 , v e W, and IP e w• amounts to

[Z, v•q>]

= (Z · v)•q> + v •(Z ·q>).

Applying B(Y, - ) to both sides, and using the identity B(Y, [Z,) B([ Y, Z], X), this becomes q>([Y, Z]·v) = q>(Y·(Z·v))

·Since q>([Y, Z]"v)

+ (.Z·q>)(Y·v).

= q>(Y·(Z·v)) -q>(Z·(Y·v)), this reduces to (Z ·q>)(w) = -q>(Z · w),

for w = Y· v, which comes from the fact that W and

w•

are

~ representations. >y·

For triples u, v, win W, the Jacobi identity is similarly reduced to the idl (u

,.·

1\

v)(Z · w)

+ (v 1\

w)(Z · u)

+ (w A u)(Z · v) = 0

}'for all z E g 0 , which amounts to ~~'

r;'

u " v " (Z. w)

i~-

+u "

=Z·(uAvAw)=O

(~·and similarly for triples from w•. :;._

For v, wE

IV

inNW=C;

w, and (/) E w•. noting that

ll'f·· if[[v, w],q>] = ~:;:

(Z. v) " w + (Z. u) " v "

-2·[v A w, q>] = -4·(v A w) A IP = -4·(q>(v)w

{the Jacobi identity for these elements reads

-4p(11

22. g2 and Other Exceptional Lie Algebras

350

-4·((/)(v)w- (/)(w)v)

= -[w, (i)](v) + [v, (i)](w).

(22.8)

The right-hand side is

-[w, (i)](v)

+ [v, (i)](w) =

-(3(/)(v)w- (/)(w)v)

+ (3(/)(w)v- (/)(v)w),

which proves this case. (This last line was the only place where we needed to use the definition (22.4) in place of the fancier (22.7).) The last case is for one element v in W and two elements (/) and 1/J in w•. This time identity to be proved comes down to

-4 ·(1/J(v)(/)- (i)(v)I/J) = [v, (/)] · 1/1- [v, 1/Jl (/). Applying both sides to an element w in W, this becomes

-4 · (1/J(v)(/)(W) - (/)(v)I/J(w)) = (i)([v, 1/J] · w) - 1/J([v, (/)) · w). If we apply 1/J to the previous case (22.8) we have -4 · ((i)(v)I/J(w)- (/)(W)I/J(v)) = -1/J([w, (/)) · v) + 1/1([1', (/)) • w). And these are the same, using the symmetry of the Killing form:

18 · (i)([v, 1/J ]- w)

= B([v, 1/J], [w, (i)]) = B([w, (i)], [v, 1/J ]) = 181/J([w, (/)l v).

This completes the proof that the algebra with multiplication table (22.1) is a Lie algebra. With the hindsight derived from working all this out, of course, we see that there is a quicker way to construct g2 , without any multiplication table: simply start with sl 3 C E9 W E9 w•, and define products according to the above rules.

§22.3. Representations of g2 We would now like to use the standard procedure, outlined in Lecture 14 (and carried out for the classical Lie algebras in Lectures 15-20) to say something about the representations of g 2 • One nice aspect of this is that, working simply from the root system of g 2 and analyzing its representations, we will arrive at what is perhaps the simplest description of the algebra: we will see that g2 is the algebra of endomorph isms of a seven-dimensional vector space preserving a general trilinear form. The first step is to find the weight lattice for g2 • This is the lattice A, c ~· dual to the lattice c ~generated by the six distinguished elements H,. By (22.2), r w is generated by HI and H 2. Since the values of the eigenvalues Ill and o: 2 on H 1 and H 2 are given by

r,

o: 1 (H 1 ) = 2,

o: 1 (l/ 2 ) = -I,

O:z(Htl = -3,

O:z(Hz)

= 2,

it follows that the weight lattice is generated by the eigenvalues o: 1 and o: 2 (a11d in particular the weight lattice A, is equal to the root lattice AR)· The picture is thus .!

§22.3. Representations of 91

As in the case of the classical Lie algebras, the intersection of the (do Weyl chamber if/' with the weight lattice is a free semigroup on the fundamental weights

w1

= 2o: 1 + o: 2

and

w 2 = 3o: 1 + 2o: 2 •

Any irreducible representation of 9 2 will thus have a highest weight vect• which is a non-negative linear combination of these two. As usual, we "' r•.• for the irreducible representation with highest weight aco, + bw2. Let us consider first the representation r,, 0 with highest weight w 1 • Tn lating w1 around by the action of the Weyl group, we see that the wei ~iagram or rt,o looks like

352

22. !h and Other Exceptional Lie Algebras

Since there is only one way of getting from the weight w 1 to the weight 0 by subtraction of simple positive roots, the multiplicity of the weight 0 in f1.0 must be I. f~, 0 is thus a seven-dimensional representation. It is the smallest of the representations of g 2 , and moreover has the property (as we will verify below) that every irreducible representation of 9 2 appears in its tensor algebra;· we will therefore call it the standard representation of g 2 and denote it V. The next smallest representation of 9 2 is the representation f 0 • 1 with highest weight w 2 ; this is just the adjoint representation, with weight diagram

.,,,,

·~

; ... ~

Note that the multiplicity of 0 as a weight of f 0 • 1 is 2, and the dimension of~ f 0 , 1 is 14. ::~ Consider next the exterior square NV of the standard representation~ V = f~, 0 of 9 2 . Its weight diagram looks like :.:.;

§22.3. Representations of g2

3:

Jrom which we may deduce that NV~ f 0 , 1 E9 V.

:'rn particular, since the adjoint representation f of g is contained in NJ ~:and the irreducible representation r. b with highest weight aw, + bw i 0•1

2

2

~-contained in the tensor product Sym"V.® Symbf0 , 1 , we see that every irreduc :· lble representation of g 2 appears in some tensor power V 0 '" of tlte standar( ·representation, as stated above.

.·

Next, look at the symmetric square Sym 2 Vof the standard representation

Ut has weight diagram

22. g2 and Other Exceptional Lie Algebras

354

Clearly, this contains a copy of the irreducible representation f 2 ,o or g2 with highest weight 2w 1 • Depending on the multiplicities of this representa· tion, it may also contain a copy of V itself, of the trivial representation, or both; or it may be irreducible. To see which is in fact the case, we need to know more about the action of 9 2 on the standard representation V. We will do this in two ways, first by direct calculation, and second using the decomposition of g 2 into sl 3 ffi W (9 W*. Although the second approach is· shorter, the first illustrates how one can calculate for the exceptional Lie algebras very much as we have been doing in the classical cases. To describe V explicitly, start with a highest weight vector for V, i.e., any nonzero element v4 of the eigenspace V4 c V for the action ofiJ with eigenvalue : IX 4 • The image of v4 under the root vector Y1 will then be a nonzero element ; of the eigenspace V3 With eigenvalue IX3 (this folloWS from the fact that the : direct sum V3 (9 V4 , as a representation of the subalgebra s~, c g, is a copy or'; the standard representation ofs., ~ si 2 C). Similarly, the image of v3 under l2 <' is a generator v1 of the eigenspace V1 with eigenvalue IX 1 , the image of v1 under Y1 is a generator of the eigenspace V0 with eigenvalue 0, and so on. We may : thus choose as a basis for V the vectors

v4 , w1

v3 = Y1 (v),

=! Y1 (u),

v 1 = - Y2 (v 3 ),

w3 = Y2 (wt),

and

~ ~j

where v1 (resp. w1) is an eigenvector with eigenvalue IX1 (resp. {11). (The signs and.~ coefficient ! in the definition or w1 are there for reasons of symmetry-see~ Exercise 22.10.) Diagrammatically, the action of 9 2 may be represented by the·~ ~M

; ·-·~

;~

j

4~

l ..•

·:~

:,! ·.~~1

~xercise 22.9. (i)

~~

Verify that the ve~tors .v1, w{l and ~~•. as. defi?ed above, a~~ mdeed generators of the correspondmg etgenspaces. (n) Ftnd, m terms of thiS,:1 basis for V, the images of v4 under the elements Y3 , Y4 , Y5 , and Y6 •

·?i

vecto~t1

Exercise 22.10. Show that the elements X 1 and Y; e 9 2 all carry basis v1 and w1 into other basis vectors, up to sign (or to zero, of course), and carry·:·) u to twice basis vectors, that is, X1u = 2v1 and Y;u = 2w1 fori= I, 3, 4. {~~

§22.3. Representations of g1

35~

Now, the representation Sym 2 V has, as basis, the pairwise products ol the basis vectors for V; and the subrepresentation f 2 , 0 is just the subspace generated by the images of the highest weight vector vl under (repeated applications of) the generators 1'11 Y2 of the negative root spaces of g 2 • Thus, for example, the eigenspace in Sym 2 V with eigenvalue o: 4 is the span of the products U • v4 and V3 'V 1 ; the part of this lying in f 2 , 0 will be the span of the two vectors Y2 Y1 Y1 (v!) and Y1 Y2 Y1 (v!). We calculate: Y2 Y1 Y1 (vi) = Y2 Y1 (2v3·v 4 ) = Y2(2v~) = -4v 1 ·v3

and = -2rJ 1 ·v 3 -2u·v4 • We see, in other words, that f 2 , 0 assumes the weight tX 4 with multiplicity 2, so that in particular Sym 2 V does not contain a copy of V. , Similarly, to see whether or not Sym 2 V contains a copy of the trivial representation, we have to calculate the multiplicity of the weight 0 in f 2 , 0 . Since any path in the weight lattice from the eigenvalue 2o: 4 to 0 obtained by subtracting o: 1 and o: 2 must pass through o: 4 , we can do this by evaluating the products of Y1 and Y2 on the generators v1 • v3 and u · v4 of the eigenspace with .eigenvalue tX 4 : we have

Y1 Y1 Y2 (v 1 v3 )== -Y1 Y1 (vn= -Yt(2u·v 1 ) == -4w 1 ·v 1 - 2u 2 ; Y1 Y1 Y2 (u·v 4 )

= 0;

Y1 Y2 Y1 (v 1 v3) = Y1 Y2 (u·v 3 ) = - Y1 (u·vt)

356

22. !h and Other Exceptional Lie Algebras

In particular, we deduce that the action of 9 2 on the sta11dard representation V = C 7 preserves a quadratic form; and correspondingly that the subalgebra 9 2 c si(V) = s;I 7 C is actually contained in the algebra so 7 C. We will see this again in the following section, where we will give alternative descriptions of the exceptional Lie algebras, and again in §23.3 where we describe compact homogeneous spaces for Lie groups. Exercise 22.1 1. Analyze in general the symmetric powers Symt V of the stan· dard representation V of 9 2 • Finally, consider the exterior cube N Vof the standard representation. The weight diagram is

and after we remove one copy of the representation r 2 , 0 with highest weight\ 2w 1 (this is the sum of the three highest weights tx 4 , tx 3 , and tx 1 of V), we are~ left with ·~

§22.3. Representations of g 2

357

·=·.

Jhis, by what we have seen, can only be the direct sum of the standard representation V with the trivial representation C. In sum, then, we conclude that NV~

r 1, 0 $

V$C.

Note in particular that, as a corollary, tile action ~f g 1 on the standard representation preserves a skew-symmetric trilinear form w on V. It is not hard to write down this form: it is a linear combination of the live vectors w3 " u " v3 , v4 " u 1\ w4 , w1 1\ u " v" v1 1\ v3 1\ w4 , and w1 " w3 ·" v4 ; and the fact that it is preserved by X 1 and X 2 is enough to deterriline the coefficients: . we have I

•p.

+ 2v 1 1\

v3

1\

w4

+ 2w1 "

w3

1\

v4 •

(i,,, The fact that the action of g 1 on

V preserves the skew-symmetric cubic takes on additional significance when we make a naive dimension ttount. The space NV of all such alternating forms has dimension 35, while ~ihe algebra gi(V) of endomorphisms of V has dimension 49; the difference is ~l~xactly the dimension of the algebra g 2 • In fact, we can check directly that the ~Unear map :,=i;. ~-':··· cp: gi(V) _. Nv ~·-.'

1.torm w

~

[sending A e End(V) to A(w) is surjective. We deduce that w is a general cubic 1ilternating form [i.e., an open dense subset of NV corresponds to forms 1equivalent tow under Aut(V)], and hence that

F,.·.

[Proposition 22.12. The algebra g2 is exact/}' tile algebra of endomorphisms of ~ir seven-dimensional vector space V preserving a general skew-symmetric cubic :'.'form w 011 V.

358

22. g 2 and Other Exceptional Lie Algebras

Exercise 22.13•. Verify that the map q1 above is surjective by direct calculation of the action of gi(V) on wE NV. Exercise 22.14. As an alternative to the preceding exercise, analyze skewsymmetric trilinear forms on c• to show that for 11 ~ 7 there are only finitely many such forms, up to the action of GL.C. Verify that the form w above is general in NC 7 • (In fact, there are only finitely many cubic alternating forms on C 8 as well, though this is fairly complicated; for n ~ 9 a simple dimension count shows that there is a continuously varying family of such forms.) Note that the cubic form w preserved by the action of 9 2 gives us explicitly the inclusion

Vc...NV deduced earlier from their weight diagrams: this is just the map v•-> NV given by contraction/wedge product with m, composed with the isomorphism of v with v•. ' Exercise 22.15•. Find the algebra of endomorphisms of a six-dimensional vector space preserving a general skew-symmetric trilinear form. We will see the form w again when we describe g 2 in the following section. These calculations using the table amount to using all the information that can be extracted from the subalgebras s. ;;;r si 2 C of g 2 • Using the copy of si 3 C that we found in the second section can make some of this more. transparent. Make the identification '· .:

92

= 9o E9 W E9 w• =

si 3 C E9 W

E9

~

w•.

As a representation of sl 3 C, the seven-dimensional representation V must be the sum of W, w•, and the trivial representation C. If we make this identification,

v=

WEFJ

w• m c,

it is not hard to work out how the rest of 9 2 acts. This is given in the followini: ~J table:

:i~

w w Qo

w

w•

w•

"'

c z

X

X·w

X·Y,

0

v

-VA W

2z·v

qJ

rp(w)

Y,(v) rpAY,

With this identification, we have u = I in C, and

2z · rp

!':~~ --~~

359

§22.4. Algebraic Constructions of the Exceptional Lie Algebras

w1

=

e2 ,

v1 =

ef,

Conversely, it is not hard to verify that the above table defines a representation of 92• by checking the various cases of the identity 'I]. y = (17" y)IJ·(e · y) fore, '1 in 9 2 andy in V. Note that the cubic form w becomes

[e.

e.

3

w=

L e," u" e: + 2(e, 1=1

" e2 " e3

+ ef " er "

en

This description of V can be used to verify the calculations made earlier, and also to study its symmetric and exterior powers. For example, Sym 2 V decomposes over si 3 C into

w• ® C EB W® w• W* EB C EB WEB w• EB si C EB C.

Sym 2 WEB Sym 2 W• EB Sym 2 C EB W® C EB = Sym 2 WEB Sym 2

3

To get the weights around the outside ring, the irreducible representation rl.O must include Sym 2 W, Sym 2 W*, and si 3 C. Checking that Wc:9 2 maps Sym 2 W• nontrivially to w• shows that it must also include Wand w•. To finish it suffices to compute the part killed by g 2 , which must lie in the sum ·of the two components which are trivial for si 3 C; checking that this is one dimensional, one recovers the decomposition Sym 2v = r2,0 EfJ c. Exercise 22.16. Use this method to decompose

NV and Sym 3 V.

§22.4. Algebraic Constructions of the Exceptional Lie Algebras :In this section we will sketch a few of the abstract approaches to the construe; lion of the five exceptional Lie algebras. The constructions are not as easy as jyou might wish: although the exceptional Lie groups and their Lie algebras r:. have a remarkable way of showing up unexpectedly in many areas of mathe~~atics and physics, they do not have such simple descriptions as the classical ~.series. Indeed, they were not discovered until the classification theorem forced fmathematicians to look for them. i1' To begin with, the method we used to construct 9 2 in the second section ~~f this lecture can be generalized to construct other Lie algebras. This is the ~Construction of Freudenthal, which we do first. It can be used to construct the ~;Lie algebra e8 for the diagram (E 8 ). From e8 it is possible to construct e7 and ~t 6 and f4 • Then we will present (or at least sketch) several other approaches their construction. Since it is a rather technical subject, probably not really

no

22. !h and Other Exceptional Lie Algebras

360

suited for a first course, we will touch on several approaches rather than give a detailed discussion of one. The construction of g 2 as a sum g 0 ffi W $ W* that we found in the second section works more generally, with very little change. Suppose flo is a semi· simple Lie algebra, and W is a representation of g 0 ; let w• be the dual representation, and set fl =ooE9W$ W*.

We also need maps

": I\ 2 W _.

w•

and

":

Nw• _. W

of representations of flo· We assume these are given by trilinear maps T: NW-> C and T': Nw•-> C, which means that (u

1\

v)(w) = J'(u, v, w)

and

,9(qJ

1\

t/1) = T'(qJ, t/J, .9).

We assume these maps are related by the identity:

(v " w)" qJ = qJ(v)w- qJ(w)v

(22.11)

w•.

for all v, we Wand qJ e We can then define a bracket on g by the same· rules as in the second section. To describe it, we let X, Y, Z, ... denote arbitrary, elements of flo• 11, v, w, ... elements of W, and qJ, t/J, .9, ... elements of W*. The bracket in g is determined by setting: ,·_; (i) [X, Y] = [X, Y]

(ii) (iii) (iv) (v) (vi)

[X, v] = X·v [X, (f'] = X·qJ [v, w] = a·(v" w) [qJ,

1/t] = b·(qJ" 1/t)

[v, (f']

= c·(v • qJ)

(the given bracket in g 0 ), (the action of g 0 on W), (the canonical action of g 0 on W*), (for a scalar a to be determined), (for a scalar b to be determined) (for a scalar c to be determined).

As before, v • qJ is the element of g 0 such that B(v • (p, Z) = qJ(Z · v)

for all Z e g0 ,

•;

···.1

'it

cz

;~

tij .·~

··~

where B is the Killing form on g 0 • The rules (i)-(vi) determine a bilinear~ product [ , ] on all of g, and the fact that it is skew follows from the factS· that [X, X]= 0, [v, v] = 0, and [(f', qJ] == 0. "' The argument that we gave showing that g 2 satisfies the Jacobi identi works in this general case without essential change, except for the one case of a triple with two elements in Wand one in where we had to do an expli •''. calculation. The identity we need in the general case, in place of (22.7), is the identity: : ::~

w•,

ab · (qJ(v)w- qJ(w)v) = c · ((v • qJ) · w- (w • qJ) · v),

which is equivalent to the Jacobi identity for v, w, and qJ. Again, the simpliciij, of the resulting Lie algebra is easy to see, provided all the weight spaces one dimensional, using Exercise 14.34, so we have:

ari

§22.4. Algebraic Constructions or the Exceptional Lie Algebras

3

Proposition 22.19 (Freudenthal). Given a represerttation W of a serni~;imr Lie algebra g0 and trilinear forms T and T' inducing maps N W--+ W* 01 N W"' --+ W, sucl! that (22.17) and (22.18) are satisfied, tlse above products makf g = g0

e3 W$ W"'

into a Lie algebra. If the weiglst spaces of W are all one dimensional, and tl weiglsts of W. W"', and tlse roots of g 0 are all distinct, and abc # 0, tis en 9 semisimple, with tire same Carton subalgebra as g0 • Exercise 22.20"'. (a) Show that the trilinear map T determines a map t NW--+ W"' of representations if and only if it satisfies the identity

T(X ·11, v, w)

+ T(u, X· v, w) + T(u, v, X· w) =

0

VXeg 0 ,

and similarly for T'. (b) Show that (22.18) is equivalent to the identity

ab·(v

1\

w)(rp

1\

t/1) = c·(B(w•t/f, v•rp)- B(w•rp, v•t/t)).

The Lie algebra e 8 for (E8 ) can be constructed by this method. This tim

.9o is taken to be the Lie algebra slg C; if v = C 9 is the standard representatiOI ,of sl 9 C, let W =NV, so W"' = NV•; the trilinear map is the usual wedg• ,product '.·J

NV® NV® NV--+ NV=

c.

and similarly for NV•. We leave the verifications to the reader:

·ij_, il'

},

;Exercise 22.21"'. (i) Verify (22. t 7), and check the conditions on the roots of sl~ :,and the weights of N Vand N v•. (ii) Use the fact that B(X, Y) = 18 · Tr(X Y :for s1 9 to show that (22.18) holds precisely if c = 18ab. (iii) Show that the fJ>ynkin diagram of the resulting Lie algebra is (E 8 ).

~j;

~]j.Note tm !j·

that the dimension of sl 9 C is 80, and that of Wand W"' is 84, so the has dimension 248, as predicted by the root system of (E 8 ). Once the Lie algebra e8 is constructed, e 7 and e 6 can be found as sub._ifgebras, as follows. Note that removing one or two nodes from the long arm ~the Dynkin diagram of (E8 ) leads to the Dynkin diagrams (E 7 ) and (E 6 ). ~ hi general, if g is a simple Lie algebra, with Dynkin diagram D, consider a tbdiagram Do of D obtained by removing some subset of nodes, together ~th all the lines meeting these nodes. 1 Then we can construct a semisimple ~balgebra go of g with Do as its Dynkin diagram. In fact, go is the subalgebra ~jmerated by all the root spaces 9±«• where IX is a root in D 0 •

tc.~ere

are doubie or triple lines between two nodes, both nodes should be removed or kept

22. !h and Other Exceptional Lie Algebras

364

Exercise 22.21•. The center of 0 is R · I, which is preserved by G. Let Y be orthogonal space to R · 1 with respect to the quadratic form N. Then G is imbedded in the group SO(Y) of orthogonal transformations of Y. (a) Deline a "cross product" x on Y by the formula v x w = v · w + P(v, w)·l. Show that G can be identified with the group of orthogonal transforma· . tions of Y that preserve the cross product. (b) Show that G = Aut(O) acts transitively on the 6-sphere

S6

={I r,e,: I

rr

= I},

and the subgroup K that fixes i = e1 is mapped onto the 5-sphere in ef by the map g~->g-j. Conclude from this that G is 14-dimensional and simply connected. (c) Show that {De Der(O): D(i) = 0} is isomorphic to su 3 . (d) Verify that the Lie algebra of derivations of the complex octonians is the simple Lie algebra of type (0 2 ). · Exercise 22.28•. The octonions can also be constructed from the Clifford algebra of an eight-dimensional vector space with a nondegenerate quadratic form. With v, s+, and s- as in §20.3, with v, E v, s, E s+, t I = v, . s, e s·. chosen so the values of the quadratic forms are l on each of them as in Exercise 20.50, define a product V x V-+ V, (v, w)~-> v o w by the formula

v. ,;

Note that v· t, E s+' w. s, E s-, so their product (v· r,)· (w ·sd is back in (a) Show that V with this product is isomorphic to the complex octoniaos 0, with unit v 1 , with the map v~-> -p(vd(v) corresponding to conjugationi~ 0. 1 Conversely, starting with the complex octonians 0, one can reconstrud, the algebra of §20.3: define A = 0 ~ 0 $ 0, define an automorphism J Of order 3 of A by J(x, y, z) = (z, x, y), and define a product · from each su~' sion of two factors to the third by the formulas x · y = x o y, y · z =Jill:~

z. tb

)s~~:·

:;~

that A is isomorphic to the algebra described in §20.3. (c) Identifying SOsC with the space of skew linear transformations or).: show that for each A in so 8 C there are unique Band C in so 8 C such that;: A(x o y)

= B(x) o y + x o C(y)

~.·:f

::~

,.,.,.

for all complex octonions x andy. Equivalently, if one defines a trilinear fo~ ( , , ) on the octonions by (x, y, z) = Tr((x o y) o z) = Tr(x o (yo z)), ,J: (Ax, y, z)

+ (x, By, z) + (x, y, Cz) = 0

Jj ._,.

·,i

for all x, y, z. Show that this trilinear form agrees with that defined in Exe~ 20.49, and the mapping A ..... B determines the triality automorphism}' ofto19 of order three described in Exercise 20.51. ··iii

§22.4. Algebraic Constructions of the Exceptional Lie Algebras

365

Exercise 22.29. Define three homomorphisms from the real Clifford algebra C7 "" C(O, 7) to Endn(O) by sending v e R7 = L Re1 to the maps L., R., and .~ T, defined by L.,(x) = v o x, R.,(x) = x o v, and T,(x) = v o (x o v) = (v o x) o v. :. ' (a) Show that these do determine maps of the Clifford algebra, and that the ~ induced maps

'are the two spin representations and the standard representation, respectively . .- (b) Verify that T,(x o y) = L.,(x) · L.(y) for all v, x, y, and use this to verify ::the triality formula in (c) of the preceding exercise. 1'.•

K:

li': - The algebra f4 can be realized as the derivation algebra of the complexificaHion of a 27-dimensional Jordan algebra J. This can be constructed as the set ~;of matrices of the form

(a-

~~..;.

f:i

~j'·;

01:

~~

if

![i

~-~tb a, b, c scalars, and .,.._,.

~: ~Y,bere

01:,

~ 01:

r

a)

I'

r ' c

{J, r in 0. The product o in J is given by x oy

= !(xy + yx),

the products on the right-hand side are defined by usual matrix multifPlication. This algebra is commutative but not associative, and satisfies the ~identity ((x ox) o y) ox = (x ox) o (yo x). In factj (F4 ) is the group of auto';iJOrphisms of this 27-dimensional space that preserve the scalar product ·~ y) = Tr(x o y) and the scalar triple product (x, y, z) = Tr((x o y) o z). The kernel of the trace map is an irreducible 26-dimensional representation of f4 • 'For details see (Ch-S], [To], [Pos]. .;In addition, there is a cubic form "det'' on J such that the linear auto- rphisms of J that preserve this form is a group of type (E6 ). This again f4 as a subalgebra ofe 6 . . \The other exceptional Lie algebras can also be constructed as derivations ,Of appropriate algebras. We refer for this to [Ti2], [Dr], [Fr2], (Jac2], and ihe references found in these sources. Other constructions were given by Witt, ,tf.[Wa]. The simple Lie algebras are also constructed explicitly in [S-K, §l]. $ee also [Ch-S], [Frl], and [Sc]. ;~What little we will have to say about the representations of the four ~Ceptional Lie algebras besides g 2 can wait until we have the Weyl character

ows

LECTURE 23

Complex Lie Groups; Characters

This lecture serves two functions. First and foremost, we make the transition back from Lie algebras to Lie groups: in §23.1 we classify the groups having a given semisimple Lie algebra, and say which representations of the Lie algebra, as described in the preceding lectures, lift to which groups. Secondly, we introduce in §23.2 the notion of c/Jaracter in the context of Lie theory; this gives us another way of describing the representations of the classical groups, and also provides a necessary framework for the results of the following two lectures. Then in §23.3 we sketch the beautiruJ interrelationships among Dynkin diagrams, compact homogeneous spaces and the irreducible representations of a Lie group. The first two sections are elementary modulo a little topology needed to calculate the fundamental groups of the classical groups in §23.1. The third section, by contra~!, may appear impossible: it involves, at various points, projective algebraic geometry, holomorphic line bundles, and th~ir cohomology. In fact, a good deal of §23.3 can be understood without these notions; the reader is encouraged to read as much of the section as seems intelligible. A final section §23.4 gives a very brief introduction to the related Bruha! decomposition, which is included because of its ubiquity in the literature. §23.1: §23.2: §23.3: §23.4:

Representations of complex simple groups Representation rings and characters Homogeneous spaces Bruha! decompositions

§23.1. Representations of Complex Simple Lie Groups. In Lecture 21 we classified all simple Lie algebras over C. This in turn yields a classification of simple complex Lie groups: as we saw in Lecture 7, for any Lie algebra g there is a unique simply connected group G, and all other (connected) complex Lie groups with Lie algebra g are quotients of G by

§23.1. Representations of Complex Simple Gronps

367

discrete subgroups of the center Z(G). In this section, we will first describe the groups associated to the classical Lie algebras, and then proceed to describe which of the representations of the classical algebras we have described in Part III lift to which of the groups. We start with Proposition 23.1. For all" ~ I, the Lie grmtps SL.C and Sp 2 .C are connected and simply connected. For 11 ~ I, SO.C is connected, with n 1 (S0 2 C) = l., and n1(SO.C) = l./2 for 11 ~ 3. PROOF. The main tool needed from topology is the long exact homotopy sequence of a fibration. If the Lie group G acts transitively on a manifold M, and His the isotropy group of a point P0 of M, then G/H = M, and the map G--+ M by g t-> g • P0 is a fibration with fiber H. The resulting long exact sequence is, assuming the spaces are connected,

· · ·--+ n2 (M)-+ n 1 (H)-+ n 1 (G)--+ n 1 (M)-+ {I}.

(23.2)

(The base points, which are omitted in this notation, can be taken to be the identity elements of Hand G, and the point P0 in M.) In practice we will know M and Hare connected, from which it follows that G is also connected. From this exact sequence, if M and H are also simply connected, the same follows for G. To apply the long exact homotopy sequence in our present circumstance we argue by induction, noting first that SL 1 C = S0 1 C = f I}. Now consider the action of G = SL.C on the manifold M = C"\{0}. The subgroup H fixing the vector P0 =(I, 0, ... , 0) consists of matrices whose first column is (I, 0, ... , 0) and whose lower right (n- I) by (n- I) matrix is in SL._ 1 C; it follows that as topological spaces H ~ SL.-Ic X c•-l. Since M is simply , connected for n ~ 2 (having the sphere S2"- 1 as a deformation retract), and H ·• has SL._ 1 C as a deformation retract, the claim for SL.C follows from (23.2) : by induction on n. ': The group S0 2 C is isomorphic to the multiplicative group c•, which has ~the circle as a deformation retract, so n 1 (S0 2 C) = l.. The group G = SO.C .1 acts transitively on M = {v E c•: Q(v, v) = I}. where Q is the symmetric bi~linear form preserved by G. (The transitivity of the action is more or less ;,equivalent to knowing that all nondegenerate symmetric bilinear forms are !·equivalent.) For explicit calculations take the standard Q for which the ; standard basis {e1} of C" is an orthonormal basis. This time the subgroup H , fixing e 1 is S0._ 1 C. From the following exercise, it follows that M has the :. sphere s•-• as a deformation retract. By (23.2) the map '.i·

~ 1tt(SOn-tC)-+7tt(SOnC) ~~ ~is an isomorphism for n ~ 4. So it suffices to look at S0 3 C. This could be

i done by looking at the maps in the sme exact sequence, but we saw in Lecture

~IOthat S0 3 C has a two-sheeted covering by SL 2 C, which is simply connected

t,by the preceding paragraph, so nt(S0 3 C) = l./2, as required.

23. Complex Lie Groups; Characters

368

The group G = Sp 2 .C acts transitively on

M = { (v, w) e C2 " x C2 ": Q(v, w) = I}, where Q is the skew form preserved by G, and the isotropy group is Sp 2._ 2 C. Since Sp 2 C = SL 2 C, the first case is known. By the following exercise, since M is defined in C 4 " by a nondegenerate quadratic form, M has s••-• as a 0 deformation retract, so we conclude again by induction.

L

Exercise 23.3*. Show that {(z 1 , ••• , z.) e C": zf = I} is homeomorphic to the tangent bundle to the (n - I)-sphere, i.e., to

Ts·-· = {(u, v) E s•-l

X

IR": u. v = 0}.

Using the exact sequence { l}-+ SL.C-+ GL.C-+ c•-+ {I} we deduce from the proposition and (23.2) that n 1 (GL.C)

= l.

(23.4)

Exercise 23.5. Show that for all the above groups G, the second homotopy, groups n 2 (G) are trivial. We digress a moment here to mention a famous fact. Each of the above groups G has an associated compact subgroup: SU(11) c SL.C, Sp(n) c Sp2.C," and SO(n) c SO. C. In fact, each of these subgroups is connected, and these inclusions induce isomorphisms of their fundamental groups. :;

Exercise 23.6. Prove these assertions by finding compatible actions of the subgroups on appropriate manifolds. Alternatively, observe that in each case the compact subgroup in question is just the subgroup of G preserving a Hermitian form on C" or C 2", and use Gram-Schmidt to give a retraction of," G onto the subgroup. j.

;~'

Now, by Proposition 23.1 the simply-connected complex Lie groups corrc·i sponding to the Lie algebras g = si.C, sp 2 .C, and so'"C are ,f ·~::

'~

We also know the center Z(G) of each of these groups. From Lecture 7 weJ also know the other connected groups with these Lie algebras: )~ • The complex Lie groups with Lie algebra slnC are SL,.C and quotients Of~ SL.C by subgroups of the form {e2 " 11''" ·/} 1 for m dividing 11 (in particular1t~ if n is prime the only such groups are SL.C and PSL.C). ';~ • The complex Lie groups with Lie algebra sp 2 .C are Sp 2 .C and PSp2.C. ;;~ • The complex Lie groups with Lie algebra so 2 .HC are Spin 2.+tC and,_~ so2n+l c.

.·,_':!"'

and _:;,:.,.'11 • The complex Lie groups with Lie algebra so 2 .C are Spin 2 .C, S0 2 .C a~~ PS02 .C; in addition, ifn is even, there are two other groups covered doubli~ by Spin 2.C and covering doubly PS02 .C [cf. Exercise 20.36]. :~·

§23. t. Representations of Complex Simple Groups

3

These are called the classical groups. In the cases where we have observ• coincidences of Lie algebras, we have the following isomorphisms of grour Spin 3 C

~

SL 2 C and

Spin 4 C

~

SL 2 C x SL 2 C

Spin~C ~

Sp4 C

and

S03 C

~

and

S05 C

PSL 2 C; PS04 C

~

~

PSL 2 C x PSL 2 C;

PSp4 C;

and Spin6 C

~

SL 4 C

and

PS06 C

~

PSL 4 C.

Note that in the first case n = 4 where there is an intermediate subgrou between SL.C and PSL.C, the subgroup in question is interesting: it tun out to be S06 C. In general, however, these intermediate groups seldom aris· Consider now representations of these classical groups. According to th basic result of Lecture 7, representations of a complex Lie ·algebra g wi correspond exactly to representations of the associated simply connected Li group G: specifically, for any representation p: g---+ gi(V)

of g, setting p(exp(X))

= exp(p(X))

determines a well-defined homomorphism

p: G---> GL(V). For any other group with algebra g, given as the quotient GfC ·of G by l subgroup C c Z(G), the representations of G are simply the representation: of G trivial on C. It is therefore enough to see which of the representation: of the classical Lie algebras described in Part III are trivial on which subgroups C c Z(G). , ' This turns out to be very straightforward. To begin with, we observe that the center of each group G with Lie algebra g lies in the image of the chosen ~Carlan subalgebra ~ c g under the exponential map. It will therefore be enough to know when exp(p(X)) = I for X E I); and since the representations p of g are particularly simple on l) this presents no difficulty. ;p What we do have to do first is to describe the restriction of the exponential ~map to~. so that we can say which elements of~ exponentiate to elements ol ~Z(G). For the groups that are given as matrix groups, this will all be perfectly [_obvious, but for the spin groups we will need to do a little calculation. We ~Will also want to describe the Cal'fan subgro11p H of each of the classical groups ~. which is the connected subgroup whose Lie algebra is the Carlan subf~Jgebra ~of g. For G = SL.C, His just the diagonal matrices in G, i.e., h:.'

~;,

H = {diag(z 1 ,

... ,

z.): z 1 • ... · z. = 1}.

;similarly in Sp 2 .C or S0 2.C, If= {diag(z 1 , .•• , z•• z( 1 , .•. , z; 1 )}, whereas in .S02.+ 1 C, I-1 = {diag(z 1 , ... ,z.,z( 1 , ... ,z; 1, 1)}. In each of these cases the

23. Complex Lie Groups; Characters

370

exponential mapping from I) to II is just the usual exponentiation of diagonal matrices. To calculate the exponential mapping for Spin,.C, we need to describe the elements in Spin,.C that lie over the diagonal matrices in SO,.C. This is not a difficult task. Calculating as in §20.2, we find that for any nonzero complex number z and any I :<;;; j :<;;; n, and with m = 2n + I or 111 = 2n, the elements

wj(z) =

~(ze)·e·+J + z- e•+(eJ) = 1

z-l

+

e-2z-~)ej·e·+J

(23.7)

in the Clifford algebra are in fact elements of Spin,.C. Moreover, if p: Spin,.C-+ SO,.C is the covering, the image p(lo\j(z)) is the diagonal matrix whose jth entry is z 2 , (11 + j)th entry is z- 2 , and other diagonal entries are I. These elements w1(z) also commute with each other, so for any nonzero complex numbers z 1 , ••. , z. we can define

(23.8) Then p(w(z 1 , ... , z.)) = diag(z~, ... , z~, zj" 2 , ... , z; 2 ) if m = 211, while if m = 2n + I, we get the same diagonal matrix but with a l at the end. Let /11 = E 1, 1 - E•+l.•+" the usual basis for I) c to,.C. Lemma 23.9. For any complex numbers a 1 ,

exp(a 1 /1 1

+ ··· +a. H.) =

••• ,

a.,

w(e"• 12 ,

••• ,

e••i2)

in Spin,. C.

PRooF. Since the map exp: ij-+ Spin.,C is determined by the facts that it is continuous, it takes 0 to 1, and its composite with p is the exponential for SO,.C, this follows from the preceding formulas. 0; Exercise 23.10*. Show that exp(}::01H1) = I if and only if each a1 is in 2niZ,':: and }.:a1 e 4niZ. ·.~· '~ ~1

We see also that exp(l)) contains the center of Spin,.C. lndeed,'j - l = w(- l, I, ... , l), and if m is even, the other central elements are ±OJ,~ with w = w(i, ... , i), as we calculated in Exercise 20.36. (This, of course, al!O,j co~tains. the fact that there is a path between l and -l, proving again t~tJ Spm,.C 1s connected.) ·~~

exp(l)l~~

Exercise 23.11*. Verify for all the classical groups G that: (i) H = a closed subgroup of G that contains the center or G; (ii) the map of fundao~ mental groups n 1 (H, e)-+ n 1 (G, e) is surjective; (iii) for any connected coverinll n: G' -+ G, n- 1 (H) is connected and is the Carlan subgroup or G'. ·';;},~ ':i;jj "·tij

Now let G = G/C be a semisimple Lie group with Lie algebra g and C~ subalgebra 1). Choose an ordering of the roots, and let rA be the irreduci~j representation of !1 with highest weight .t. The basic fact that we need is r:~

§23.1. Representations of Complex Simple Groups

Lemma 23.12. Tile representation only if

r 1 is a representation of G = G/C if a whenever exp(X) e C.

.l(X) e 2nil.

PROOF. The representation rA is a representation of G when g. v = v for ! g e C, where v is a highest weight vector in rA. Since exp(~) contains C, th says exp(X)· v = v for all X e ~such that exp(X) e C. Now by the naturali oftheexponential map, and since X· v = .l(X)v for X e ~.we have exp(X)· v' eA(XIv. Hence the condition is that eA(XIv = v, or that eA(Xl = I if exp(X) e < which is the displayed criterion. [

Let us work this out explicitly for each of the classical groups. It may hel to introduce a notation for the irreducible representations which, among otht virtues, allows some common terminology in the various cases. Note that fo each of s(•+l• 5p 2., so 2 ., and so 2 .+t the root space~· is spanned by weight we have called L 1 , ••• , L., so a weight can be written uniquely in forr; .l 1 L 1 + ·· · + .t.L•. We may sometimes write .l in place of the weigh .l1L 1+ ··· + .t.L•. In the rest of this lecture at least, we write rA for th· irreducible representation with highest weight .l 1 L1 + ··· + .t.L•. Note tha by our choice of Weyl chambers the highest weights .l = (.l 1 , ••• , .t.) that aris• satisfy .l 1 ~ ~ 2 ~

· ••

~

.t. ~ 0 for sln+t• sp 2., and so 2•+1•

where the .t, are all integers in the first two cases, and for 5o 2 .+t they are eithe1

· all integers or all half-integers; and At ~ .lz ~ · · · ~

.t.-t

~ lA. I ~ 0

for 5o 2.,

{with the .l1 all integers or all half-integers.

-

'c·'·,

' ;,.Proposition 23.13. For each subgi'Oup C ~{ tl1e cellfer of G, tlte representation l~f1 is a represemation of GfC precisely under the following conditions: ~· f,~ (i) <2 = SL.+t C, C ltas order m dividing rt + l: ~1 = 0 mod(m). ~' (ii) G = Sp 2 .C, C = {± 1}: I .t1 is even. ~.(iii) G= Spin 2 .C or Spin 2.+t C, C = {± 1}: all .l1 are integers. ~(lv) ~ = Spin 2 ,. C, C = {± I, ± w}: all .l1 are integers, I .t1 is even. ~1·(v) G = Spin 2 .C, n even, C = { 1, w}: I .t1 is an even integer; and for C = ';i> {l, - w}: I .l1 - n/2 Is an odd integer.

-

L

~i~.

In particular, representations of PSL.+ 1 C are given by partitions .l with ~["'.! = 0 mod(n + l), and those for PSp 2.C have I~1 even. Case (iii) verifies what we saw in Lecture 19 about representations of SOmC. Representations ~C PSOmC correspond to integral partitions .l with .t1 even .

: I

.

I

;!·

PRooF. With the preceding lemma and the explicit description of everything in sight, the calculations are routine. In case (i), for example, a generator for

23. Complex Lie Groups; Characters

372

Cis of the form exp(X), with

X = (2ni/m)(

I

J=l

EJ.J- nEn+l.n+t)•

and so ..t(X) = (2nifm)(L 11) will be a multiple of 2ni exactly when "f.~ is divisible by 111. For Sp 2.C, exp(X) = -I when X = ni(L H1), so ..t(X) = ni ).).1, and (ii) follows. The calculations are similar for Spin'"C, noting that exp(2ni(/:lt) = -I and exp(ni("f. H1)) = w. 0 By way of an example, recall that any irreducible representation of si 2 C is of the form Symtv, where Vis the standard two-dimensional representation. Any such representation, of course, lifts to the group SL 2 C; but it lifts to PSL 2 C ~ S03 C if and only if k is even (in particular, the "standard" representation of S03 C on C 3 is the symmetric square Sym 2 V). For another example, we have seen that any irreducible representation of sp 4 C may be found in a tensor product Symtv ® Sym 1W, where Vis the standard four· dimensional representation of sp 4 C and W c NV the complement of the trivial one-dimensional representation. All such representations lift to Sp4 C, but they lift to PSp 4 C ~ S05 C if and only if k is even-equivalently, if they are contained in a representation of the form Sym 1W ® Symk(N W), where W is the "standard" representation of S05 C. Exercise 23.14. Show that each of these semisimple complex Lie groups G has a finite-dimensional faithful representation.

The result of the proposition can be put in a more formal setting, which brings out a feature that our alert reader has surely noticed: the center of the simply-connected form of g is isomorphic to the quotient group Aw/A11 of the weight lattice modulo the root lattice. We note first that this abelian group Aw/11. 11 is finite. We have seen this for the classical Lie algebras. In general, we have Lemma 23.15. Tlte group Aw/11. 11 is finite, of order eq11al to the determinant of the Carton matrix. . PROOF. The simple roots a form a basis for the root lattice 11. 11 • The correspond~: ing elements H. form a basis for ,.

a lattice in ~; this is proved in Appendix D.4. Since Aw is defined to be tbci; lattice Of elementS Of!)* that take integra) values OD f 11 , the determinant ~;_

is the index [Aw: Anl

§23.1. Representations of Complex Simple Groups

In particular, for the exceptional groups, 1\". f/\R is trivial for (G 2 ), (F4 ), a (E 8 ), and cyclic of order two for (E 7 ) and order three for (E 6 ). In fact, the center of the simply-connected group is naturally isomorpl to the dual of 1\wf/\R. To express this, consider the natural dual of this h group. The lattice defined in the preceding proof is a sublattice of the latti

rR

f 1v ={X e lr 1X(X) e 71. for all IX E R}. Note that Aw was defined to be the lattice of elements of !)• that take integ1 values on rR. It follows formally from the definitions and the fact that Aw/1 is finite that we have a perfect pairing (X, 1X)H1X(X).

The claim is that there is a natural isomorphism from fw/fR to the cent of G, which is given by the exponential. More precisely, let ea: I)-+ H c G I the homomorphism defined by · ea(X) = exp(2niX).

We claim that when G

= Gis the simply-conneeted group, Ker(ea) = rR atl

ea(fw) is the center of G, from which it follows that eii induces an isomorphisn fw/fR More generally, for any G =

9if

Z(G).

GfC, define a lattice f(G) between rR and r"' 1: f(G) = Ker(ea).

Then ea determines an isomorphism

fw/f(G)

.·,

~

Z(G).

; We may thus state our result as ~·

~Theorem

23.16. There is a one-to-one cor·respondence between connected Li

,.; groups G with the Lie algebra g and lattices 1\ c

()• such that

f'The correspondence is givell by associating to a group G the lattice dual to til ) kernel of the exponential map exp: g-+ G; in particular, the largest lattice /\ 1 )..corresponds to tile simply-connected group, the smallest 1\R to the adjoi11t gro111 ~With no center. In terms of this correspondence, tire irreducible representatim ~'·yA of g with llighest weight .t e l)• will l!ft to a representation of the group < ~torresponding to 1\ c I)• if ancl only if). e 1\. ~),

~r Note also that

~-

·

~-With n

H = l>ff(G)

= dime I) copies of c•.

~ c• x ... x c•,

23. Complex Lie Groups; Characters

374

Exercise 23.17*. Show that these claims follow formally from what we have seen: that the image of the exponential map contains the center, and that for any weight IX there is a representation V of g whose weight space v. is not zero. Show also that e6 determines an isomorphism r(G)/fR ~ n 1 (G). In diagram form, Go U Center(G) l G r(G)}

fw} U

l

n 1 (G)

G

r"

Exercise 23.18. Find the kernels of each of the spin and half-spin representations SpinmC _. GL(S) and SpinmC _. GL(S±). Exercise 23.19*. Classify the irreducible representations of the full orthogonal groupOmC. Note that by our analysis of the Lie algebra 9 2 there is a unique group G2 with this Lie algebra, which is simultaneously the simply-connected and adjoint forms; the representations of this group are exactly those of the algebra 9 2 • The same is true for the Lie algebras of type (F4 ) and (E 8 ), while (E7) and (E 6 ) each have two associated groups, an adjoint one with fundamental group Z/2 and Z/3, and a simply-connected form with center Z/2 and Z/3 respectively. It may be worth pointing out that each complex simple Lie group G can be realized as a closed subgroup defined by polynomial equations in some general linear group, i.e., that G is an affine algebraic group. Every irreducible representation G _. GL(V) is also defined by polynomials in appropriate coordinates. This explains why the whole subject can be developed from the point of view of algebraic groups, as in [Borl] and [Hu2]. The Weyl group 9.13, which we defined as a subgroup of Aut(l)*), can be interpreted in terms of any connected Lie group G with Lie algebra g. Let H· be the Carlan subgroup corresponding to ~. and let N(H) be the normalizer:

N(H) = {g e G: gHg- 1 = H}. We have homomorphisms: •::;

N(H) _.Aut( H)_.

Aut(~)_. Aut(~*),

the first defined by conjugation, the second by differentiation at the identity; and the third using the identification of l) and l)* via the Killing form. Fact' 14.11 can be sharpened to the claim that this map determines an isomorphism

N(H)/N ~!!.D.

(23.20)

When G is the adjoint form of the Lie algebra, this isomorphism is proved in Appendix D. The general case follows, using:

376

23. Complex Lie Groups; Characters

spaces obtained by reflecting in walls of the Weyl chambers all have the same dimension. Let w 1 , ••. , w. be a set of fundamental weights; as we have seen, these are the first weights along edges of a Weyl chamber, and they are free generators for the lattice A. Let r,, ... , r. be the classes in R(g) of the irreducible representations with highest weights w 1 , ••• , w •. Theorem 23.24. (a) The representation ring R(g) is a polynomial ring on the variables r.' ... ' r•. (b) The homomorphism R(g)-+ Z[A] m is an isomorpltism. In particular, this says that Z[A]m is a polynomial ring on the variables Char(r1 ), ••. , Char(r.). In fact, the theorem is equivalent to this assertion, since if we take variables U1 , ••• , U. and map the polynomial ring on the U1 to R(g) by sending U1 to r, we have

Z[U1 , ... , U.]--+ R(g)--+Z[Af 0• If the composite is an isomorphism, the second being injective, both must be

isomorphisms, which is what the theorem says. In spite of its fancy appearance, we will see that the theorem follows quite easily from what we know about the action of the Weyl group !ID on the weights. For any P e Z[A] let us say that a is a ltighest weight for P if the coefficient of e(a) in P is nonzero, and, with a chosen ordering of weights as before, a is the largest such weight. We first observe that if P is invariant under !ID, then the highest weight for P is in "/f/" 11 A, where "/f/" is our chosen (closed) Weyl chamber. In general, weights in 71/" 11 A are often referred to as domi11ant weights. Now suppose {PA} is any collection of elements in Z[A]m, one for each dominant weight ..1., such that Pl has highest weight ..1. and the coefficient of e(A.) is I. We claim that the Pl form an additive basis for Z[A]u' over Z. This is easy to see and is the same argument used in the theory of symmetric polynomials in any algebra text: given P with highest weight ..1., if the coefficient of e(A.) ism, then P - mPA is invariant whose highest weight is lower, and one continues inductively until one reaches weight zero, i.e., the constants. Let P1 = Char(r,), which has highest weight w, and suppose the coefficient of e(w 1) is 1. Since any weight ..1. e "'Il' 11 A can be uniquely expressed in the form ..1. = 1w 1, for some non-negative integers m., and the highest weight of (P,)'"' is it follows that the monomials (P,)'"' in PI' ... ' P. form an additive basis for Z[A]m. This says precisely that Z[/'1 , ... , P.] = Z[AJ", and completes the proof. 0

n

Im Im,w,,

n

Let us work this out concretely for each of our cases tl.+l C, sp,.C, so 2 .+1C, and so 2 .C. Each lattice A contains weights we have called L 1 , ••• , L.; in the first case we also have L.+t with L 1 + ··· + L.+t = 0. We set

§23.2. Representation Rings and Characters

x, Note that in case L 1 ,

l[A] = l[x 1 ,

= e(L1), ••• ,

...

377

xj 1 = e(-L1)eZ[A].

(23.25)

L. is a basis for A, then

,x.,x!1 ,

...

,x; 1 ] = l[x 1 ,

...

,x., (x 1 · ... ·x.)- 1 ]

as a subring of the field Q(x 1 , ••• , x.). (A.) For sl•+t C, fundamental weights are L 1,

L1

+ L 2,

L1

+ L 2 + L 3 , .•• , L 1 + ··· + L.,

corresponding to the irreducible representations V, NV, ... , NV, with V = c•+l the standard representation. The character of NV is e(tX), the sum over all tX that are sums of k different L 1 for 1 :s;; i :s;; n + l. So Char(N V) = At. where Ak is the kth elementary symmetric function of x 1 , ... , x.+t· The Weyl group is the symmetric group 6.+1, acting by permutation on the indices, so the theorem in this case says that

L

(23.26) Note that Z[A]=Z[x., ... ,x.,x.H]/(x 1 · ... ·x.+J-1), so .Z[A] has an additive basis consisting of all monomials x•, with tX ann-tuple of non-negative integers, but with not all tX 1 positive. (C.) For sp 2.C, the lattice A and fundamental weights have the same description as in the preceding case. The corresponding irreducible representations are the kernels Vlkl of the contraction maps NV-+ N- 2 v, with now v = C 2" the standard representation, k = 1, ... ' II. The character of N v is e(tX), the sum over all tX that are sums of k different ± L 1 for 1 :s;; i :s;; 11. The character Char(NV) is thus the elementary symmetric polynomial Ck in the variables x 1 , xl\ x 2 , x]\ ... ,x., x,;- 1 • The theorem then says that

l:

R(sp 2.C) = Z[AJfil = Z[C1, C2- t, C3=

Z[C1, C2 ,

c., ... , c.- c.-2J

c3, ... , c.].

(23.27)

(B.) For so 2 .+ 1 C, A is spanned by the L 1 together with !(L 1 + · · · + L.). The fundamental representations are V, NV, ... , N- 1 V, and the spin representation S. The character of NV is the kth elementary symmetric function of the 21l + 1 elements x 1 , x 1- 1, ... , x., x; 1, and 1; denote this by Bk. 112 • ••• • x! 112 , where The character of S, which we denote by B, is the sum

Exr

xi 112 =

e(L1/2),

x~ 1 '2 = e(- L1/2).

SoB is the nth elementary symmetric polynomial in the variables x: 1' 2

(23.28)

+ x;- 1' 2 •

Therefore,

(23.29) , (D.) For so 2.C, A and Z[A] are the same as in the preceding case. The fundamental representations are V, NV, ... , N- 2 V, and the half-spin representations s+ and s-. The character of N v, denoted Dk, is the kth elementary symmetric function of the 2n elements x 1 , r.l 1, •.• , x., x;•. The

LJ.

Lulnplc>. Lll; \.JI UlljJ~, Llllllil~l~IS

character D± of s± is the sum L,xr 1' 2 • ••• • x! 1' 2 , where the number of plus signs is even or odd according to the sign. We have (23.30) Exercise 23.31•. (a) Prove the following relation in R(so 2 .H C): 8 2 = B.

+ ··· + B1 + l,

corresponding to the isomorphism

S ® S ~ NV$···$NV$1\0 V. This describes R(so 2 n+t C) as a quadratic extension of the ring Z[B 1 , ••• , B.]. (b) Let (respectively, D;) be the character of the representation whose· highest weight is twice that of v+ (resp., D-), so that, for example, the sum of the representations and D; is 1\"V. Prove the relations in R(so 2.C):

v:

v:

v+ ·v+ = v: + D.- 2 + D.-4 + · · ·,

v- · v- = D;; + D.- 2 + D.- 4 + ···, D+ ·D- = D._ 1

+ D._ 3 + D.-s + ···.

We can likewise describe the representation ring for g 2 • Here, we may take as generators for the weight lattice the weights L 1 and L 2 as pictured in the diagram

.• ·f

·.:;

and correspondingly write Z[A] as Z[x 1 , x! 1, x 2 , x] 1 ], where x 1 = e(L1).1() will be a little more symmetric to introduce L 3 = - L 1 - L 2 as pictured and:; x 3 = xj" 1 • x} 1 = e(L 3 ), and write '.,;~i .

~,::.::

In these t e n s tlie Weyl group is the group 9U generated by the symmetric group B, permuting the variablesx, and the involution sending each xi to x;' The standard representation has weights L, and 0, and so has character

*

Similarly, the adjoint representation has weights fL,, f(L, - Lj), and (taken twice); its character is

(

B = A(x19x 2 9 ~ 3 )+ A ( ~ 1 1 ~~21x3. 2, ~31x1). The theorem thus implies in this case the equality (23.32

R(gz) = Z[AIm = B[A, B].

Exercise23.33.Verify directly thestatement that any element ofZ[x,, x,, x,], (xlx,x3 - 1) invariant under the group as described is in fact a polynomia in A and B. Similarly we can define the representation ring R(G) of a semisimple grour G. When G is the simply-connected form of its Lie algebra g, R(G) = R(g) SO R(SLnC), R(Sp,,C), R(Spin,,+,C), and R(Spin,,C) are given by (23.26) (23.27), (23.29), and (23.30). In general, R(G) is a subring of R(g); we can reac ofl which subring by looking at Proposition 23.13. We have, in fact,

with D: and D; as in Exercise 23.31. But this time there is one relation:

,

Exercise 23.36*.

(a) Prove (23.34) (b) Show that the relation in (23.35) comes from Exercise 23.31(b). Sltow t h a ~ R(S02,C) is the polynomial ring in the n 1 generators shown, modulc the ideal generated by the one polynomial indicated. , (c) Describe the representation rings for the other groups with these simplc Lie algebras. ' (d) Prove the isomorphism

+

R(GL,C) = Z[E,,

..., En, Eil],

where the E, are the elementary symmetric functiolls o f x , , ..., x,.

: Exercise 23.37*. (a) Show that the image of R(0,C) r b

nomial ring Z[B,, . . ., B,,] if m = 2n

+ 1, and Z[D,,

in R(S0,C) is tlie poly i f r t ~= 211.

.. . ,D,]

23. Complex Lie Groups; Characters

380

(b) Show that R(02•+1 C)= R(S02n+l C)® R(7L/2) = 7Llilt, ... ,B•• B2n+I)/((B2n+d 2 __.:I)

and R(0 2.C) = 7L[D 1 , where I is the ideal generated by (D,.) 2

... ,

D., D2.]/I,

I and D.D 2 •

-

-

D•.

Exercise 23.38*. The mapping that takes a representation V to its dual V* induces an involution of the representation ring: [V]• = [ v•]. The ring 7L[A] has an involution determined by (e(A))* = e(- A). Show that the character homomorphism commutes with these involutions. Show that for sl•+l• (A.)• = A•+l-t; for sol•+l C, and sp,.C, and so 2 .C for 11 even, the involution is the identity; while for solNC with II odd, (D.)* = Dt, (D+)• = D-' (D-)*= v+. Deduce that all representations of all symplectic and orthogonal groups are self-dual. Note that when • is the identity, all representations are self-dual. In the other cases, compute the duals of irreducible representations with given highest weight. The following exercise deals with a special property of the representation rings of semisimple Lie groups and algebras. Exercise 23.39*. The representation rings R = R(g) and R(G) have another important structure: they are A-rings. There are operators

A1: R(G)--+ R(G),

i = 0, I, 2, ... ,

determined by A ([V]) =[NV] for any representation V. (a) Show that this determines well-defined maps, satisfying A0 and 1

A'(x

+ y) =

L

=

I, A1 = ld,

A'(x)·Al(y)

l+}~t

for any x and y in R. In fa:ct, R is what is called a special A-ring: there are formulas for A1(x · y) and A1(Ai(x)), valid as if x andy could be written as sums of one-dimensional representations (see, e.g., [A-T]). (b) Show that A1 extends to 7L[A],and use this to verify that R(G) is a special A-ring. Deline Adams operators If/: R--+ R by !f/(x) = Pt(A 1 x, ... , A"x), where Pt is the expression for the kth power sum (cf. Exercise A.32) in terms of the elementary symmetric functions, 11 ~ k. Equivalently, 1/tt(x)- 1/tH(x)A 1 (x)

+ .. · + ( -l)tHt(x) =

0.

(c) Show that, regarding R as the ring of functions on the group G, (1/,tx)(g) = x(l). Equivalently, 1/tt(e(A)) = e(kA.).

§23.2. Representation Rings and Characters

38'

(d) Show that each t/Jt is a ring homomorphism, and t/Jt (e) Show that for a representation V,

o

t/J 1 = t/JH'.

= t Char(V) 2 + lt/1 2 (Char(V)), Char(NV) =! Char(V) 2 - !t/1 2 (Char(V)).

Char(Sym 2 V)

Show that Char(SymdV) and Char(NV) can be written as polynomials in .pt(Char(V)), l ~ k :s;; d.

Formal Characters and Actual Characters Let G be a Lie group with Lie algebra g. For any representation V of fl, the image of [V] E R(g) in Z[A] is called the formal character of V. As it turns out, this formal character can be identified with the honest character of the corresponding representation of the group G, restricted to the Carlan subgroup H:

L m«e(a) is the formal character, a11d exp(X) is an eleme111 of H, then tlte trace of exp(X) on Vis Em«e«!Xl_

(23.40) JfChar( V) =

This is simply because exp(X) acts on the weight space V,. by multiplication by e~
x:'

Exercise 23.42*. If g 1 and !h are two semisimple Lie algebras, show that R(gt x f!2) = R(!Jt) ® R(g2).

Exercise 23.43*. (a) For the natural inclusion si.C c st. +I C, restriction of representations gives a homomorphism R(£11.+ 1 C)_. R(si.C), which can be

23. Complex Lie Groups; Characters

382

described by saying what happens to the polynomial generators. Since N(C" Et> C)= N(C") $ N- 1(C"), this is At~->

At+ At- 1 •

Give the analogous descriptions for the following inclusions:

si.C c sp 2.C, sp 2.C c si 2.C,

ill.C c so 2.+ 1 C,

si.C c ilo 2 .C;

ilo 2.+ 1 C c sl 2.+t C,

so 2.C c si 2 .C.

(b) The inclusion ill.C x slmC c sl•+mC determines a restriction homomorphism R(sl•+mC) _. R(iii.C x slmC) = R(si.C) ® R(slmC), which takes polynomial generators At to At® l + At_ 1 ®A 1 + · ·· + I ®At. Compute analogously for

Which of these inclusions correspond to removing nodes from the Dynkin diagrams'! Exercise 23.44. Compute the isomorphisms of representation rings corre· sponding to the isomorphisms sl 2 C;;:;;; !\lo 3 C, so 5 C;;;;; !!p 4 C, and !!1 4 C;;:;;; !!o6 C.

§23.3. Homogeneous Spaces In this section we will introduce and describe the compact homogeneous spaces associated to the classical groups. As we will see, these are classified neatly in terms of Dynkin diagrams, and are, in turn, closely related to the representation theory of the groups acting on them. Unfortunately, we are unable to give here more than the barest outline of this beautiful subject; but we will at least try to say what the principal objects are, and what connections among them exist. In particular, we give at the end of the section a diagram (23.58) depicting these objects and correspondences to which the reader can refer while reading this section. We begin by introducing the notion of Borel subalgebras and Borel subgroups. Recall first that a choice of Cart an subalgebra I) in a semisimple Lie algebra g determines, as we have seen, a decomposition g = l) G) ffim., R 94 - To each choice of ordering of the root system R = R + v R-, we can associate a subalgebra b= ~®

EB

)•eR•

g.,

called a Borel subalgebra. Note that b is solvable, since ~b c ffig 1 , ~ 2 b c E99m+-• etc. In fact, b is a maximal solvable subalgebra (Exercise 14.35).

§23.3. Homogeneous Spaces

383

If G is a Lie group with semisimple Lie algebra A. the connected subgroup B of G with Lie algebra b is called a Bol"e( subgroup.

Claim 23.45. B is a closed subgroup of G, and the quotient G/B is compact. PROOF. Consider the adjoint representation of Gong. The action of the Borel subalgebra b obviously preserves the subspace b c g, and, in fact, b is just the inverse image of the subalgebra of gl(g) preserving this subspace: if X = L xm is any element of g with Xm e gm and x. :1- 0 for some IX e R-, we could find an element H of~ c b with ad(X)(H) ¢ b-any H not in the annihilator of IX e I)* would do. B is thus (the connected component of the identity in) the inverse image in G of the subgroup of GL(g) carrying b into itself. It follows that B is closed; and the quotient GfB is contained in a Grassmannian and hence compact. (Alternatively, we could consider the action of G on the projective space P(g), and observe that B is just the connected component <1f the identity in the stabilizer of the point corresponding to the highest weight space g. c g.)

In fact, in the case of the classical groups, it is easy to describe the Borel subgroups and the corresponding quotients. For G = SL..... e, B is the group of all upper-triangular matrices in G, i.e., those automorphisms preserving the standard flag. It follows that G/B is the usual (complete) flag manifold, i.e., the variety of all flags

G/B = {0 c V1

c ·· · c

V.

c

C"+t}

of subspaces with dim( V,.) = r. For G = S02.e or S02.+ 1 e the orthogonal group of automorphisms of em= e 2 " or e 2 •+1 preserving a quadratic form Q, B is the subgroup of automorphisms which preserve a fixed flag V1 c · · · c V. of isotropic subspaces with dim(V,.) = r. All such flags being conjugate, GfB is the variety of all such flags, i.e.,

G/B = {0

c

V1 c .. · c V, c em: Q(V,, V,)

= 0}.

Note that B automatically preserves the flag of orthogonal subspaces, so that we could also characterize G/B as the space of complete flags equal to their orthogonal complements, i.e.,

G/B

= {0 c

V1

c ··· c

Vm-t

c

Vm

= em: Q(Jt;,

Vm_ 1)

= 0}.

Finally, the same holds for Sp 2 .C: the Borel subgroups B c Sp 2 .e are just the subgroups preserving a half-fl11g of isotropic subspaces, or equivalently a full flag of pairwise complementary subspaces; and the quotient G/B is correspondingly the variety of all such flags. Exercise 23.46. With our choice of basis {e 1}, let V,. be the subspace spanned by the first r basic vectors. If B is defined to be the subgroup that preserves V, for l :s; r :s; 11, verify that the Lie algebra of B is spanned by the Carlan subalgebra and the positive root spaces described in Lectures 17 and 19.

23. Complex Lie Groups; Characters

384

We now want to consider more general quotients of a semisimple complex group G. To begin with, we say that a (connected 1 ) subgroup P of G is parabolic if the quotient G/P is compact. Of course, a Borel subgroup B is parabolic, and so is any closed subgroup containing a Borel subgroup. The following claim is a converse to this: it asserts that the Borel subgroups are exactly the minimal parabolic subgroups. Claim 23.47. If B is a Borel subgroup and P a paralwlic subgroup of G, then there is all x E G wilft

We will not prove this here, but will remark that it is closely related to Claim 23.48. If P c G is a parabolic subgroup, tlzen the quotient G/P may be realized a.~ a11 orbit of tile action Q{ G on PV for some representatioll V of G (in particular, G/P is a projective algebraic variety). The first claim follows from the second by a version of Borel's fixed poinl tlteorem: if B is a connected solvable group, V a representation of B ·and X c P V a projective variety carried into itself under the action of B on PV, then B must have a fixed point on X. This is straightforward: we observe (by Lie's theorem (9.11)) that the action of the solvable group B on V must preserve a flag of subspaces OcV1 c···cV.=V

with dim(V,) = i. We can thus find a subspace V. c V fixed by B such that X intersects P V, in a finite collection of points, which must then be fixed points for the action of Bon X. In the other direction, we will soon see directly how G/P is a projective variety whenever Pis a subgroup containing B. We can now completely classify the parabolic subgroups of a simple group, . up to conjugacy. By the above, we may assume that P contains a Borel subgroup B. Correspondingly, its Lie algebra p is a subspace of g containing b and invariant under the action of Bon g; i.e., it is a direct sum

for some subset T of R that contains all positive roots. Now, in order for p to ' be a subalgebra of g, the subset T must be closed under addition (that is, if two roots are in T, then either their sum is in T or is not a root). Since, in addition, T contains all the positive roots, we may observe that if a, p, andy are positive roots with a = P + y, then we must have

-a E T 1

= - Pe T and - y e 7:

It is a general ract that P must be connected ir G/P is compact.

§23.3. Homogeneous Spaces

385

Clearly, any such subset T must be generated by R + together with the negatives of a subset E of the set of simple roots. Thus, if for each subset E of the set of simple roots we let T(E) consist of all roots which can be written as sums of negatives of the roots in E, together with all positive roots, and form the subalgebra (23.49) then p(E) is a parabolic subalgebra, the corresponding Lie group P(E) is a parabolic subgroup containing 8, and we obtain in this way all the parabolic subgroups of G. We can express this as the observation that, up to conjugacy,

parabolic subgroups of the simple group G are in one-to-one correspondence with subsets of the nodes of the Dynkin diagram, i.e., with subsets of the set of simple roots. Examples. In the case of si 3 C, there is a symmetry in the Dynkin diagram, so that there is only one parabolic subgroup other than the Borel, corresponding to the diagram

0

•

This, in turn, gives the subset of the root system

R+

corresponding to the subgroup

and the homogeneous space

G/P = P 2 • In the case of sp 4 C, there are two subdiagrams of the Dynkin diagram:

• <

I >

and

=··

tQ!C::::::!<~~

these correspond to the subsets of the root system

23. Complex Lie Groups; Characters

386

and

(Here we are using a black dot to indicate an omitted simple root, a white dot to indicate an included one.) The corresponding subgroups of Sp 4 C are those preserving the vector e 1 , and preserving the subspace spanned by e1 and e2 , respectively. The quotients G/B are thus the variety of one-dimensional iso· tropic subspaces (i.e., the variety P 3 of all the one-dimensional spaces) and the variety of two-dimensional isotropic subspaces. Exercise 23.50. Interpret the diagrams above as giving rise to parabolic subgroups ofthe group S05 C of automorphisms of C 5 preserving a symmetric bilinear form. Show that the corresponding homogeneous spaces are the variety of isotropic planes and lines in C 5 , respectively. In particular, deduce the classical algebraic geometry facts that: (i) The variety of isotropic 2-planes for a nondegenerate skew-symmetric bilinear form on C4 is isomorphic to a quadric hypersurface in P 4 . (ii) The variety of isotropic 2-planes for a nondegenerate symmetric bilinear form on C 5 (equivalently, lines on a smooth quadric hypersurface in fJ4) is isomorphic to P 3 . In general, it is not hard to see that any parabolic subgroup P in a classical group G may be described as the subgroup that preserves a partial flag in the standard representation. In particular, a maximal parabolic subgroup, corresponding to omitting one node orthe Dynkin diagram, may be described as the subgroup of G preserving a single subspace. Thus, for G = SL.,C, the kth node of the Dynkin diagram

corresponds to the Grassmannian G(k, m) of k-dimensional subs paces of C"'. (Note that the symmetry of the diagram reflects the isomorphism of the Grassmannians G(k, m) and G(m - k, m).) For Sp2.C, the kth node of the Dynkin diagram

§23.3. Homogeneous Spaces

387

corresponds to the Lagrmtgian Grassmannimt of isotropic k-planes, for k = I, 2, ... , tr. Similarly, for G = S02.+ 1 C, the kth node of the Dynkin diagram corresponds to the ortlwgonal Gras.~mannian of isotropic k-planes in C 2•+1. Finally, forS0 2.C,for k = I, 2, ... , 11- 2 the kth node of the Dynkin diagram

yields the orthogonal Grassmannian of isotropic k-planes in C 2 ", but there is one anomaly: either of the last two nodes

gives one of the two connected components of the Grassmann ian of isotropic n-planes. Exercise 23.51•. Compute p(I:) directly for each of the classical groups, and ' verify the above statements. Why is the orthogonal Grassmannian of isotropic (n- I)-planes in C 2 " not included on the list? As we saw already in Exercise 23.50, the low-dimensional coincidences between Dynkin diagrams can be used to recover some facts we have seen • before. For example, the coincidence (0 2 ) =(Ad x (Ad identifies the two family of lines on a quadratic surface in IP 3 with two copies of P 1• The coincidence (A 3 ) = (0 3 )

< <

gives rise to two identifications of marked diagrams: we have

23. Complex Lie Groups; Characters

388

corresponding to the isomorphism between the Grassmann varieties P 3 = G(l, 4), P 3 = G(3, 4) and the two components of the rumily or 2-planes on a quadric hypersurface Q in P 5 ; and

o-e----D

corresponding to the isomorphism of the Grassmannian G(2, 4) with the quadric hypersurface Q itself. Finally, an observation that is not quite so elementary, but which we saw in §20.3: the identification of the diagrams

·-< says that either connected componem of the variety of 3-planes on a smooth

quadric ltypersllrface Q in P 7 is isomorphic to tire quadric Q itself. There is another way to realize the compact homogeneous spaces associated to a simple group G. Let V = rA be an irreducible representation of G with highest weight A., and consider the action of G on the projective space PV. Let p e PV be the point corresponding to the eigenspace with eigenvalue .t We· have then Claim 23.52. The orbit G ·pis the unique closed orbit of the action of G on PV. PROOF. The point pis fixed under the Borel subgroup B, so that the stabilizer· of p is a parabolic subgroup PJ..; the orbit GfPA is thus compact and hence' closed. Conversely, by the Borel fixed point theorem, any closed orbit G· contains a fixed point for the action of B; but pis the unique point in PV fixed byB. O;

or

In fact, it is not hard to say which parabolic subgroup PJ.. is, in terms of the·: classification above: it is tire parabolic subgroup correspmrding to tire subset of' simple roots tlrat are perpendicular to tire weight 1. Now, sets :r. of simple•: roots correspond to faces of the Weyl chamber, namely, the face that is the:·. intersection of all hyperplanes perpendicular to all roots in :r.. ;) We thus have a correspondence between faces of the Weyl chamber and: parabolic subgroups P, such that if V = r 1 is the irreducible representation~' with highest weight 1, then the unique closed orbit of the action of G on PY,"

§23.3. Homogeneous Spaces

3f

is of the form G/P, where Pis the parabolic subgroup corresponding to th open face of 1Y containing A. In particular, weights in the interior of the We~ chamber correspond to P1 = B, and so determine the full Oag manifold G/l whereas weights on the edges give rise to the quotients of G by maxim~ parabolics. Note that we do obtain in this way all compact homogeneou spaces for G. For example, we have the representations of SL 3 C: as we have seen, th representations SymlV and Symlv•, with highest weights on the boundarie of the Weyl chamber, have closed orbits {v•}u~v and W}r~v·· isomorphic I• PV and PV*. By contrast, the adjoint representation-the complement o the trivial representation in Hom(V, V) = V ® V*-has as closed orbit th• variety of traceless rank I homomorphisms, which is isomorphic to the Oa1 manifold via the map sending a homomorphism cp to the pair (lm cp, Ker cp) The picture is

representations Sym'v• have closed orbit F 2 ' -

/

•

/

adjoint representation has dosed orbit the nog rnonifold (=hyperplane seclion of rv x rv• c J>
=

•

- representations - I ....--Synt'V have dosed orbil f 2

, In general, if V is the standard representation of SL.C, in the representa. lions of SL.C of the form W = SymlV we saw that the vectors of the form : {v•} •• v formed a closed orbit in PW, called the Veronese embedding of p•-t. ·.Likewise, in representations of the form W = NV the decomposable vectors {v 1 1\ v1 1\ • • · 1\ vl} formed a closed orbit in PW; this is the Pliicker embedding of the Grassmannian. Similarly, we may identify the closed orbits in representations of Sp4 C. Recall here that the basic representations of Sp4 C are the standard representation V ~ C'' and the complement W of the trivial representation in the exterior square NV; all other representations are contained in a tensor product of symmetric powers of these. Now, Sp 4 C acts transitively on PV; the closed orbit is all of P 3 • In general, in P(Sym• V) the closed orbit is just the set of vectors {v•} •• v ~ P 3 • By contrast, the closed orbit in PW is just the intersection of the hyperplane PW c P(N V) with the locus of decomposable vectors {v 1\ w}.,.w .. v; this is the variety

23. Complex Lie Groups; Characters

390

X= {v

1\

w: Q(v, w) = 0}

of isotropic 2-planes A c V for the skew form Q.

ttprt!!enUIIion W has closed orbil a quadric hypersurrace in FW

• •

•

•

representation~ Sym' V ---I -----

hove closed orbil F'

For the group Spin 2.+ 1 C, the closed orbit of the spin representationS is the orthogonal Grassmannian of n-dimensional isotropic subs paces of C 2" ... The corresponding subvariety

G/P <=+ P(S) is a variety of dimension (11 + I )n/2 in P"', N = 2" - I, called the spinor variety, or the variety of pure spinors. Similarly for Spin 2 .C, the two spin representa· tions s+ and s- give embeddings of the two components of the orthogonal Grassmannian of n-dimensional isotropic subspaces of C 2 ", one in P(S+), one in P(S-). These spinor varieties have dimension n(11 - I )/2 in projective spaCtS of dimension 2"- 1 - 1. Exercise 23.53. Show that the spinor variety for Spin 2 ._ 1 C is isomorphic to each of the spinor varieties for Spin 2 .C. In fact they are projectively equivalent: as subvarieties of projective space P"', N = 2"- 1 - t. It follows that, form~ 8, the spinor varieties for SpinmC are isomorphic to homogeneous spaces we have described by other means. The first new OhC is the 10-dimensional variety in pB, which comes from Spin 9 C or Spin 10 C. It is worth going back to interpret some of the "geometric plethysm" of earlier lectures (e.g., Exercises 11.36 and 13.24) in this light. Finally, we can describe (at least one of) the compact homogeneous spaces for the group G2 in this way. To begin with, G2 has two maximal parabolic subgroups, corresponding to the diagrams ()

<•

and

• <

0

391

§23.3. Homogeneous Spaces

These are the groups whose Lie algebras are the parabolic subalgebras spanned by the Carlan subalgebra l) c 9 together with the root spaces corresponding · to the roots in the diagrams

and

~;:

r

r ~'

r::_..

·}.'{l:":"ij1 ~P~'·

~~-~--

~·;··

~In

particular, each of these parabolic subgroups will have dimension 9, so

t~that both the corresponding homogeneous spaces will be live-dimensional

'Yarieties. We can use this to identify one of these spaces: if Vis the standard ~}:even-dimensional representation of G 2 , the closed orbit in IP V ~ IP 6 will be !·~{a hypersurface, which (since it is homogeneous) can only be a quadric hyper~~Grface. Thus, the homogeneous space for G2 corresponding to the diagram

1;'~:i

• <

(J

t;.i·

W'~isaquadric hypersurface in IP 6 • In particular, we see again that the action of ~ G1 on V preserves a nondegenerate bilinear form, i.e., we have an inclusion

~;;·

Gz <=+ so7c. ~t•The other homogeneous space Y of 92 is less readily described. One way

m:to describe it is to use the fact that the adjoint representation W of 9

2

is

23. Comple11: Lie Groups; Characters

392

contained in the exterior square NV of the standard. Since the Grassmannian G(l, 7) c P(NV) of lines in PV is closed and invariant in IP(NV), it follows that Y is contained in the intersection of G with the subspace IP W c P(/\.2 V~ In other words, in terms of the skew-symmetric trilinear form won V preserved by the action of G2 , we can say that Y is contained in the locus

I:= {A c V: w(t\, t\, ·) = 0} c G(2, V). Problem 13.54. Is Y = I:? Exercise 13.55. Show that the representation of £ 6 whose highest weight is the first fundamental weight w 1 determines a 16-dimensional homogeneous space in P 26 • These homogeneous spaces have an amazing way of showing up as extremal examples of subvarieties of projective spaces, starting with a discovery of Severi that the Veronese surface in IP~ is the only surface in IP~ (nonsingular and not contained in a hyperplane) whose chords do not lillup IP~. For recent work along these lines, see [L-VdV], with its appendix by Zak on interesting projective varieties that arise from representation theory. Although we have described homogeneous spaces only for semisimple Lie· groups, this is no real loss of generality: any irreducible representation V of a Lie group G comes from a representation of its semisimple quotient, up to multiplying by a character (see Proposition 9.17), and this character does not change the orbits in IP(V). It is possible to take this whole correspondence one step further and, use it to give a construction of the irreducible representations of G; this the modern approach to constructing the irreducible representations, due: primarily to Borel, Wei!, Bott, and, in a more general setting, Schmid. We do· not have the means to do this in detail in the present circumstances, but we.' will sketch the construction. ·~ The idea is very straightforward. We have just seen that for every irredu~;l ible representation V of G there is a unique closed orbit X= G/P of the action. : of G on IP V. We obtain in this way from V a projective variety X together witb'~ a line bundle L on X invariant under the action of G (the restriction of tboJ universal bundle from P V). In fact, we may recover V from this data simply~ as the vector space of holomorphic sections of the line bundle L on X. Wha~] ties this all together is the fact that this gives us a one-to-one correspondence~ between irreducible representations of G and ample (positive) line bundles on) compact homogeneous spaces G(P. More generally, using the projection maps;) GfB-+ G/P, we may pull back all these line bundles to line bundles on GJB~~ This then extends to give an isomorphism between the weight lattice of 9 and~ the group of line bundles on G/B, with the wonderful property that ror· dominant weights A, the space of holomorphic sections of the associated bundle LA is the irreducible representation of G with highest weight A. )i:£

is:

lioe

§23.3. Homogeneous Spaces

J9

The point of all this, apart from its intrinsic beauty, is that we can g· backward: starting with just the group G, we can construct the homogeneou space G/B, and then realize all the irreducible representations of G as co homology groups of line bundles on G/B. To carry this out, start with a weigh Ae ~·for g. We have seen that Aexponentiates to a homomorphism H --> c~ i.e., it gives a one-dimensional representation CA of H. We want to induce thi representation from H to G. If H c B c G is a Borel subgroup, the represenla tion extends trivially to B, since B is a semidirect product of H and th• nilpotent subgroup N whose Lie algebra is the direct sum of those g. fo; positive roots a:. Then we can form Ll = G x 8 CA = (G

X

CA)/{(g, V)""' (gx, x- 1 v),

X

E 8},

which, with its natural projection to G/B, is a holomorphic line bundle on th( projective variety G/B. The cohomology groups of such a line bundle are finite dimensional, and since G acts on LA, these cohomology groups are represent a· lions of G. We have Bolt's theorem for the vanishing of the cohomology of this line bundle: Claim 23.56. H1(G/B, LA)= 0 fori :f. i(A), where i(A.) is an integer depending on which Weyl chamber A belongs to. If A. is

a dominant weight (i.e., belongs to the closure of the positive Weyl chamber for the choice of positive roots used in defining B), then i(- A.) = 0. In this case the sections H 0 (G/B, L_A) are a finite-dimensional vector space, on which G acts. Oaim 23.57. For Aa clomilla11t weiglzt, tile space of sectio11s H 0 (G/B, L_l) i.~ tlte Irreducible represe11tatio11 willt IJigltest weight A.. :, In this context the Riemann-Roch theorem can be applied to give a for'mula for the dimension of the irreducible representation. In fact, the dimension :part ofWeyl's character formula can be proved this way. More refined analysis, using the Woods Hole fixed point theorem, can be used to get the full char(acterformula (cf. [A-B]). For a very readable introduction to this, see [Bot]. :!: We conclude this discussion by giving a diagram showing the relationships rlmong the various objects associated to an irreducible representation of a semi'simple Lie algebra g. The objects and maps in diagram (23.58) are explained

:next. '"

~' First of all, as we have indicated, the term "Grassmannians" means the !ordinary Grassmannians in the case of the groups SL.C, and the Lagrangian ':orassmannians and the orthogonal Grassmannians of isotropic subs paces in cases of Sp 2.C and SO.,C, respectively. Likewise, "Oag manifolds" refers iio the spaces parametrizing nested sequences of such subspaces. In the cases

rlhe

interior ofWeyl chamber

~

(')

a

;

::I"

.!'! (j

.,

0c

!i' Cl

r

(i"

~

..

3

0

(j

Diagram 23.58

....

...··

"0

.~........

......... /

F ........ /.F'

...................···

~

{ partial flag } manifolds

{ Irreducible representations } Vofg orO

~

..

,/

// E'

..

''

'

.

{ parabolic } subgroups

. , { Grassmanmans J

full flag maniford

{ maximal parabolic } subgroups -

Borel subgroup

~

{ open faces of } Weyl chamber

~\

--+--

( subsets of nodes of } \ the Dynkin diagram

\

diagram

all of the Dynkin

{ single nodes of the } . _ {edges of Weyl} _ Dynkin diagram chamber

§23.4. Bruhat Decompositions

395

of the exceptional Lie algebras, the term "Grassmannian" should just be ignored; except for the quotient of G2 by one ofits two maximal parabolic subgroups, the homogeneous spaces for the exceptional groups are not varieties with which we are likely to be a priori familiar. With this said, we may describe the maps A, B, etc., as follows:

A, A': the map A associates to a subset of the nodes of the Dynkin diagram (equivalently, a subset S of the set of simple roots) the face of the Weyl chamber described by 71'. 5

= {..tY·• a)> 0, Va E S;}• (..1., a) = 0, Va ¢

S

where ( , ) is the Killing form; the inverse is clear. "'f~ of the Weyl chamber the subalgebra · g5 spanned by the Cartan subalgebra l), the positive root spaces. g., a E R -t-, and the root spaces g_ 11 corresponding to those positive roots a perpendicular to "'ff"s. Equivalently, in terms of the corresponding subset S of the simple roots, g5 will be generated by the Borel subalgebra, together with the root spaces 9-o for a ¢ S. Again, since every parabolic subalgebra is conjugate to one of this form, the inverse map is clear. C, C': The map C simply associates to a parabolic subalgebra v c 9 the -~•.. quotient GjP of G by the corresponding parabolic subgroup P c G. In the ··~ other direction, given the homogeneous space X= G/P, with the action of .. G, the group Pis just the stabilizer of a point in X. Note that the connected :~ component of the identity in the automorphism group of G/P may be ·::: strictly larger: for example, IP 2 "- 1 is a compact homogeneous space for Sp 2.C, and we have seen that a quadric hypersurface in IP 6 is a homoge),, neous space for G2 • :D, D': The map D associates to the irreducible representation V of g with > highest weight ..1. the open face of the Weyl chamber containing ..1.. In the other direction, given an open face "'ffS of "'F, choose a lattice point ~- . 1. E iYs II Aw and take v = rA. ~E: We send the representation V to the subalgebra or subgroup lixing the j:: highest weight vector v e V. {F, F': We associate to the representation V the (unique) closed orbit of the f~%rcorresponding action ofthe group G on the projective space IPV. Going in ;;. the other direction, we have to choose an ample line bundle Lon the space . \ G/P, and then take its vector space of holomorphic sections.

B, B': the map B associates to a face

t'

n

~···

r3.4. Bruha! Decompositions le •i1C·

end this lecture with a brief introduction to the Bruhat decomposition of

[~ scmisimple complex Lie group G, and the related Brultat cells in the flag

f)i_ianifold GfB. These ideas are not used in this course, but they appear so often ~etsewhere that it may be useful to describe them in the language we have

23. Complex Lie Groups; Characters

396

developed in this lecture. We will give the general statements, but verify them only for the classical groups. General proofs can be found in [Bor I] or [Hu2]. As we have seen, a choice of positive roots determines a Borel subgroup B and Cartan subgroup H, with normalizer N(H), so N(H)/H is identified with the Weyl group !lB. For each We !ID fix a representative nw in N(H). The double coset B · nw · B is clearly independent of choice of n11., and will be denoted B · W · B. Theorem 23.59 (Bruhat Decomposition). Tire group G is a disjoi11t IIJ.DI double co.~ets B · W · B, as W varies ot>er tire Weyl group.

W1i011

of the

Let us first see this explicitly for G = SLmC. Here N(H) consists of all · monomial matrices in SLmC, i.e., matrices with exactly one nonzero entry in each row and each column, and !ID = 6.,; a monomial matrix with nonzero •• entry in the a(j)th row ofthejth column maps to the permutation a. To sec' that the double cosets cover G, given g e G, use elementary row operations by : left multiplication by elements in B to get an element b · g- 1, with be B chosen:: so that the total number of zeros appearing at the left in the rows in b·g- 1 is·,; as large as possible. If two rows of b · g- 1 had the same number of zeros at the ; left, one could increase the total by an elementary row operation. Since all the .' rows of b · g- 1 start with different numbers of zeros, this matrix can be i put in upper-triangular form by left multiplication by a monomial matrix;'~ therefore, there is a permutation a so that b' = n6 • b · g- 1 is upper triangular, ti i.e., g = (b'f 1 ·n,. · b is in B ·a· B. To see that the double cosets are disjoin~':l suppose ,,. = ll· n., · b for some b and b' in B. From the equation b :s c~ (n6 ) - 1 • (bT' · 116 • one sees that b must have nonzero entries in each place where~ (n 6 1 '11 6 • does, from which it follows that a' = a. ~i In fact, this can be strengthened as follows. Let U (resp. u-) be the subgroup:~ of G whose Lie algebra is the sum of all root spaces 9~ for all positive (resp;~ negative) roots a. For G = SLmC, U (resp. u-) consists of upper- (resp.lower·l.~ triangular matrices with l's on the diagonal. For Win the Weyl group, defineJ

r

.::1

subgroups U(W)=Unnw·U-·n;;J,

Jl:

U(W)'=Unnw·U·nJ..1

of U, which are again independent of the choice of representative nw for

~3.60.

ele~ent

Corollary Every in B · W · B can be written u · nw · b for elements u 111 U(W) and b 111 B.

w.·}.l •·\\1

u11ique··.~.· ·~

To see the existence of such an expression, note first that the Lie algebra~ .-.·: of U(W) is the sum of all root spaces g. for which a is positive and w- 1 («}~. negative; and the Lie algebra of U(W)' is the sum of all root spaces g. ro··, which a and w- 1(a) are positive. One sees from this that U(W) · U(W)' · H if the entire Borel group B. Since II · nw = nw · H and U ( W)' ·llw = llw • U, a~ . 1/ and U are subgroups of B, ·

,:JI

§23.4. Bruhat Decompositions

397

B·nw·B = U(W)· U(W)'·H·nw·B = U(W)·U(W)'·nw·B =

U(W)· nw ·B.

To see the uniqueness, suppose that llw = u ·nw · b for some 11 in U(W) and b in B. Then 11;;f •u · nw is in u- n B = {I}, sou = I, as required. Note in particular that the dimension of U(W) is the cardinality of R+ n W(R-), where R+ and R- are the positive and negative roots; this is also the minimum number l(W) of reflections in simple roots whose product is W, cr. Exercise DJO. It is a general fact, which we will see for the classical groups, that U(W) is isomorphic to an affine space C 1("'1. It follows from the Bruhat decomposition that GfB is a disjoint union of thecosets Xw = B· nw· B/B,again with Wvaryingover the Weyl group. These Xw are called Bruhat cells. From the corollary we see that Xw is isomorphic to the affine space U(W) ~ C'11~'~. ' For G = SL.,C and 11 in !Ill= 6.,, the group U(a) consists of matrices with l's on the diagonal, and zero entry in the i, j place whenever either i > j or 0'-t (i) < a- 1 (j), which is an affine space of dimension I(a) = # {(i, j): i > j and ·CJ(i) < a(j)}. Exercise 23.61. Identifying SL.,C/B with the space of all flags, show that X,. 'consists of those nags 0 c vl c v2 c ... such that the dimensions of inter'RCtions with the standard nag are governed by 11, in the following sense: for :each I ~ k ~ 111, the set of k numbers d such that V, n C4 - 1 ~ V,. n C4 is Lprecisely the set {a( I), a(2), ... , a(k) }. ~·~.

~5 ' We will verify the Bruhat decomposition for Sp 2 .C by regarding it as a tlubgroup of SL 2.C and using what we have just seen for SL 2.C, following f[Ste2]. Our description of Sp 2.C in Lecture 16 amounts to saying that it l~ the fixed point set of the automorphism ({J of SL2 .C given by ({'(A) =

~~- 1 ·'A- 1

•

M, with M =

(~I. ~).The BorelsubgroupofSp .C will be the 2

~intersection of the Borel subgroup B of SL 2 • C with Sp 2 • C, provided we change

!.the order of the basis of C

2"

to e 1 ,

... ,

e., e2 ., ... , e.+t, so that B consists of

[~trices whose upper left block is upper triangular, whose lower left block is r~ro, and whose lower right block is lower triangular. The automorphism ~aps

({J

this B to itself, and also preserves the diagonal subgroup H and its ''ormalizer N(H), and the groups U and u-. The Weyl group of Sp 2.C can • .· identified with the permutations in 6 2 • such that a(n + i) = a(i) ± n for iii I :s; i :s; 11, and it is exactly for these a for which one can choose a monomial ..'presentative 11 11 in Sp 2.C. Now if g is any element in Sp 2.C, write g = 11'11 h rding to the above corollary. Then 11 '

g = ({'(g) = ({'(11). ((1(11!1). (p(h),

398

23. Complex Lie Groups; Characters

and by uniqueness of the decomposition we must have fl'(ll) = 11, qJ(n,.) = "• · h, II E II, and qJ(b) = ,,-• ·b. It follows that u belongs to the Wcyl group ofSp1.C. This gives the Bruhat decomposition, and, moreover, a unique decomposition of g E Sp 2.C into u · 11., • b, with u in U(u) n Sp 2.C. Since this latter is an affine space, this shows that the corresponding Bruhat cell in the symplectic nag manifold is an affine space. Exactly the same idea works for the orthogonal groups SO.,C, by realizing them as fixed points ofautomorphisms ofSL.,C of the form AI-+ M- 1 ·'A- 1 • M, with M the matrix giving the quadratic form. Note linally that if W' is the element in the Weyl group that takes each root to its negative, then 8 · W' · 8 is a dense open subset of G, a fact which is evident for the classical groups by the above discussion. The corresponding Bruhat cell Xw· is the image of u- in GfB, which is also a dense open set. It follows that a function or section of a line bundle on G/B is determined by its values on u-. For treatises developing representation theory via functions on u-, see [N-S] or [Zel]. The following exercise uses these ideas to sketch a proof of Claim 23.57 that the sections of the bundle L_~ on GfB form the irreducible representation with highest weight A.: Exercise 23.62*. (a) Show that sections s of L_A are all of the form s(gB)"' (g,f(g)), where f is a holomorphic function on G satisfying

f(g · x) = A(x)f(g) for all x

E

B.

(b) Let n' E N(I-1) be a representative of the element W' in the Weyl group which takes each element to its negative. Show that f is determined by its_' value at 11'. (c) Show that any highest weight for f must be A, and conclude that· I-1°(G/B, L __t) is the irreducible representation f'A with highest weight A. ·.,·. The holomorphic functions f of this exercise are functions on the space:· G/U. In other words, all irreducible representations of G can be found in spaceS of functions on G/U. This is one common approach to the study of representa~.· tions, especially by the Soviet school, cf. [N-S], [Zel]. ( Functions on G/U form a commutative ring, which indicates how to make'.: the sum of all the irreducible representations into a commutative ring. In fac~'~ for the classical groups, these rings are the algebras §', §<'>, and §1'1 coO-:j structed in Lectures 15, 17, and 19, cf. [L-T]. They are also coordinate ring$';' for natural em beddings of nag manifolds in products of projective spaces.

?;

LECfURE 24

Weyl Character Formula

This lecture is pretty straightforward: we simply state the Weyl character formula in §24.1, then show how it may be worked out in specific examples in §24.2.1n particular, we derive in the case of the classical algebras formulas for the character of a given irreducible representation as a polynomial in the characters of certain basic ones (either lhe alternating or the symmetric powers of the standard representation for si.C and their analogues for sp 2 .C and so.,C). The proofs of the formula are deferred to the following two lectures. The techniques involved here are elementary, though the determinantal formulas are fairly complex, involving all the algebra of Appendix A. §24.1: The Weyl character formula §24.2: Applications to classical Lie algebras and groups

§24.1. The Weyl Character Formula ,We have already seen the Weyl character formula in the case of si.C, and it :is one reason why we were able to calculate so many more representations in )hat case. We saw in Lectures 6 and 15 that for the repre~entation rA = §A c· '~r SL.C with highest weight A. = l 1L 1, the trace of the action of a diagonal inatrix A e S L. C with entries x 1 •••• , x. is the symmetric function called the :Schur polynomial SA(x 1 , ••• , x.). This included a formula for the multiplicities, :which are the coefficients of the monomials in these variables. In order to extend this formula to the other Lie algebras, let us try to rewrite .this Schur polynomial in a way that may generalize. The Schur polynomial is ~defined to be a quotient of two alternating polynomials:

I

~l :

24. Weyl Character Fonnula

400

These determinants can be expanded as usual as a sum over the symmetric group s., which is the Weyl group !lB. Writing x 1 = e(L1) in Z[A] as in the preceding lecture, and writing ( -l)w for sgn(W) = det(W) for Win the Weyl group, the numerator may be expanded in the form

where we write A. for I:A. 1L 1 and we set p = I:(n - i)L 1• Our formula therefore takes the form Ch

(r) =

l) -l)we(W(A. + p))

A

I:( -l)we(W(p))

ar

The denominator is the discriminant

This can be written in terms of the positive roots L 1 - L1, i

l\(x 1 ,

••• ,

x.)

= 0 (e(i(L1 -

< j, as

L1 ) - e( -!(L,- L1))).

I<}

Note also that p

= I:(n 1

i)L, = L 1

= -2 I<} L (LI -

+ (L 1 + L 2 ) + ··· + (L 1 + · ·· + L._.)

Lj),

which is the sum of tlte fundamental weights, and half the sum of the positivi··

roots. These are the formulas that generalize to the other semisimple Lie algebras:. For any weight Jl, define A1, e Z[A] by ;

A,. =

L

(-I )we(W(Jl)).

(24.1)

We!IJ

Note that A,. is not invariant by the Weyl group, but is alternating: W(A,.) .. ,_ (- l)w A,. for We !lB. The ratio of two alternating polynomials will be._ invariant. Theorem 24.2 (Weyl Character Formula). Let p be ltaif tlte sum of the posili~~t; roots. Then pis a weight, and AP ¥- il. Tlte cltaracter of the irreducible reprt-:

sentation

rA WitiJ /tjg/test Weig/tt Ais

·' (WCFr

§24.1. The Weyl Character Formula

401

The assertions about pare part of the following lemma and exercise, which

will also be useful in the applications: Lemma 24.3. The denominator AP of Weyl's formula is AP

=

n (e(a/2)- e( -a/2))

IIER ...

= e(p)

0

(I - e( -a))

cre.R+

= e(-p)

n (e(a)- 1). liE

.R+

n

_ PROOF. Since e(p) = e([a/2) = e(a/2), the equality of the three displayed expressions is evident; denote these expressions temporarily by A. The key point is to see that A is alternating. For this, it suffices to see that A ·changes sign when a reflection in a hyperplane perpendicular to one of the simple roots •· is applied to it, since these reflections generate the Weyl group. This follows immediately from the first expression for A and (a) in Exercise 24.4 below. Now, by the second displayed expression, the highest weight term that appears in A is e(p), which is the same as that appearing in AP. Calculating 1/A formally as in (24.5) below, we see that APIA is a formal sum Im,.e(IJ) that is invariant by the Weyl group, and, using part (c) of the following exercise, it has weight 0. As in Theorem 23.24 it follows that APIA is constant; and, since A and AP have the same leading term e(p), we must have Av =A. 0 :_ Exercise 24.4•. (a) If W = W.., is the reflection in the hyperplane perpendicular : loa simple root a., show that W(a 1) = -a;, and W permutes the other positive roots. (b) With Was in (a), show that W(p) = p - a1• Deduce that p is the ele· , ment in ~· such that p(H«,) = 2(p, a 1)/(a1, a 1) = I for each simple root a,. ; Equivalently, p is the sum of the fundamental weights. In particular, p is a ; weight. (c) For any W # 1 in the Weyl group, show that p - W(p) is a sum of ~distinct positive roots. Deduce that W(p) is not in the closure of the positive ,,.: Weyl chamber. ( · Proofs of the character formula will be given in §25.2 and again in §26.2. i For now we should at least verify that it is plausible, i.e., that AHvfAv is in ~: Z[A] 9 and that the highest weight that occurs is A.. Note that since the ;c numerator and denominator are alternating, the ratio is invariant. The fact i" that AP is not zero follows from the second expression in the preceding lemma. f'To see that the ratio is actually in Z[A], however, we must verify that it has - only a finite number of nonzero coefficients. Write

24. Weyl Character Formula

402

n (I- e(-«)r

I

- = e(-p) Ap

1

= e(-p)

aeR+

n L"' e(-n«). •

(24.5)

n=O

When this is multiplied by Aap = l:( -l)we(W(..t + p)), we get a formal sum where the highest weight that occurs is the weight ..t. This means in particular that there are only a finite number of nonzero terms corresponding to weights in the fundamental (positive) Weyl chamber "'I'. But since the ratio is invariant by the Weyl group, the same is true for all Weyl chambers, so AA+p/AP is in Z[A]m, and has highest weight..t. II follows in particular that the AA+p/AP, as . ..t varies over "'f/ 11 A, form an additive basis for l[A]m. Before considering the proof or any other special cases, we apply (WCF) to '• give a formula for the dimension of r.~,: Corollary 24.6. The dimension of tlte irreducible representation . d 1m

r -

.~,-

!] «E

wltere («,

p)

= «(Hp)

+

<..t + p, «) -(p, «)

= 2(«, p)/(P, P) and (

!]

(..t +

«E

r.~, is

+ p, «) ,

(p, «)

, ) is the Killing form.

r.~,

-~

PRooF. The dimension of is obtained by adding the coefficients of all e(«) in Char(r.~,), i.e., computing the image ofChar(r.~,) by the homomorphism from .i: l[A] to C which sends each e(«) to I. However, as in the case of the Schur} polynomial, the denominator vanishes if we try to do this directly. To get}j around this, we factor this homomorphism through the ring of power series:·;,; T

l[A]-+ C[[t]]-+ C,

.

:•i

where the second homomorphism sets the variable t equal to zero, i.e., picks~ off the constant term of the power series, and the first homomorphism 'I' takes:~ e(«) to e1P·«1'. More generally, for any weight JL define a homomorphism ',~ I'

'I'P:

Z[A]-+ C[[t]],

e(«) H e'P·«''·

• :~j

We claim that 'l'p(A.~,) = 'I'.~,(A,,) for all..t and tt. This is a simple consequence'~ of the invariance of the metric ( , ) under the Weyl group: ·~-~

L (-I)"'e
'I'P(A.~,) =

1

= 'I'.~,(A,,).

Therefore,

.::~

ii; ....l

,:~

.;,

;·~

·~·~ :':~

§24.2 Applications to Classical Lie Algebras and Groups

=

C!J. ()., oc)}"''R'I +terms of higher degree in t.

Hence, 'I'(A.l.+,/ Ap)

= 'I'(A.l.+p)/'I'(Ap) + p, oc) f .. d . = n(). (p, oc) + terms 0 pOSitive egree In f,

n

which finishes the proof.

0

Exercise 24.7. In the case of sf.C, verify that the above corollary gives the dimension we found in Lecture 6. Exercise 24.8. Verify directly that the right-hand side of the formula for the dimension is positive. Since X.t = AHp/AP is the character of a virtual representation which takes on a positive value at the identity, as in the case of finite groups, to prove that it is the character of an irreducible representation, it suffices to show that Ja X.tXA = I for an appropriate compact group G. This was the original approach of Weyl, which we will describe in the last lecture. Since the highest weight appearing is )., we will know then that this irreducible representation must be r.t. Exercise 24.9. Use Corollary 24.6 to show that if ). is a dominant weight (i.e., in the closure of the positive Weyl chamber), and OJ is a fundamental Weight, then the dimension Of r.l.+to iS greater than the dimension Of rA. Conclude that the nontrivial representations of smallest dimension must be among the 11 representations r., with w a fundamental weight.

§24.2. Applications to Classical Lie Algebras and Groups In the case of the general linear group GL.C, the character• of the representation rA is the Schur polynomial

1 We use the representation of GL,.C instead of its restriction to SL.C, since the Iauer would require the product of the variables x 1 to be I.

24. Weyl Character Formula

404

which has several expressions in terms of simpler symmetric functions. Note that the character of the dth symmetric power of the standard representa· tion is the dth complete symmetric polynomial JJ4 in 11 variables (Appendix

A. I):

The first "Giambelli" or determinantal formula (A.5) of Appendix A gives the character of the representation with highest weight ). = (). 1 ~ • • • ~ )., > 0) as an r x r determinant:

(24.10)

Equivalently, this expresses a general element rAe R(G) of the representation ring as a polynomial in the representations Sym 4 (C"). A second determinantal formula, from (A.6), expresses rA in terms of the basic representations N(C"), whose characters are the elementary symmetric polynomials

E4

= Char(N(C")).

This formula is, with 11 the conjugate partition to ).,

E,.,-1+ 1 •••

E,.,

In this section we work out the character formula for the other classical Lie algebras, including analogues of these determinantal formulas. l11e analogues of the first determinantal formula (24.10) were given by Weyl, but tbc analogues of(24.11) were found only recently ([Ko-Te]). We also pay, at least by way of exercises, the debts to (WCF) that we owe from earlier lectures.

The Symplectic Case The weights for sp 1 .C are integral linear combinations of L 1 , •.• , L •. We often write 11 = (/1 1 , .•• , 11.) for the weight 11 1 L 1 + ··· + 11.L•. The positive roots are {L,- L1 },
=I (n + I -

i)L, = L 1

i.e., p = (n, n- I, ... , 1).

+ (L 1 + L 2 ) + ··· + (L 1 + ··· + L.),

(24.12)

§24.2. Applications to Classical Lie Algebras and Groups

405

As we saw in Lecture 16, an element in the Weyl group can be written uniquely as a product eu, where u is a permutation of {L 1 , ••• , L. }, and e = (e 1 , ••• , r..), with e1 = ± l. Hence AP =

I(-I)"I(-t)•e(t tl

r. 1111L.,111);

(24.13)

i=l

'

here the sign (- I)' is the product of the c1• Now with x, = e(L1), this can be written

or

(24.14) where

la1, 11denotes the determinant of the rt x n matrix (a,). In particular, (24.15) A,.= lxr•+~ - xj'"- 1+1 11.

From (24.14) or Exercise A.52 we have

A,.= ~(x 1

+ X! 1, ... , x. + x; 1)·(x 1 -

x! 1)· ... ·(x.- x;'), (24.16)

where ~is the discriminant. Exercise 24.17. Show that

A,.=

0 (x,- x1)(x,x1 I<J

1) · 0 (xf - 1)/(x 1 • ••• • x.)". I

The character of the irreducible representation r~ with highest weight

A= IA.,L,, A., :2 ·· · :2 A.. :2 0, is therefore:

24. Weyl Character Formula

406

lxl,+n-1+1 - x-ll,+n-1+1)1 J I " i+t J I• i+T!I-

Char(rA) =

xJ

(24.18)

- x1

The dimension of f 1 is easily worked out from Corollary 24.6: dim(fl) = =

where 11 = ).1 + n - i

n(/; - IJ).j) n(2n +(I;2-+ IJ)i- j) n
I<]

(j-

I<J

(mf- m])

(24.19)

1

+ l and m1 = 11 - i + I.

Exercise 24.20. Show that, setting 1; = A.1 + n - i,

n<': - 'iH'; + 'i + n(l; + 2).

dim(f1 ) =

i<J

1)

1

(211 - I)!. (211 - 3)! ..... I!

These formulas give the dimension of the irreducible representation r., ....••• with highest weight a 1 w 1 + ··· + a.w., where the w1 are the fundamental weights, using the relation ). 1 = a 1 + ··· + a•. Exercise 24.21. Use Exercise 24.20 to verify that for ). = L 1

+ ·· · + L1 ,

dimension of rl is 2n if k = I, and

2. Use this to

C:)- (k :· 2)

if k 2

the

give another proof that the kernel of the contraction from NV to N- 2 v is irreducible. The first determinantal formula for the symplectic group goes as follows. Let J 4(x 1, ... , x.) = H4 (x 1 ,

... ,

x., xl 1 ,

... ,

x; 1),

where Jld is the dth complete symmetric polynomial in 2n variables. In other words, J4 is the character of the representation Sym 4 (C 2 ") of sp 2 .C. From Proposition A.50 of Appendix A ·we have Proposition 24.22. If). = (). 1 2 · · · 2 )., > 0), the cltaracter of f 1 is tire determinant of tlte r x r matrix whose ith row is (Jl,-1+1

Jl,-1+2

+ Jl,-1

Jl,-1+3

+ Jl,-i-1

·· •

Jl,-l+r

+ Jl,-l-r+2).

For example, for).= (d), i.e.,).= dL 1 , we have Char(f141 ) = J4 , which is the character of Sym4 (C 2 "). In particular, this verifies that the kth symmetric powers Sym 1 (C 2 ") of tl1e standard representation are all irreducible. (This, of course, is a special case of the general description given in §17.3, since all the contraction maps vanish on the symmetric powers.)

407

§24.2. Applications to Classical Lie Algebras and Groups

Exercise 24.23. (i) Find the character of the representation of sp 4 C with highest weight w 1 + w2 = 2L 1 + L 2 , verifying that the multiplicities are as we found in §16.2. (ii) Find the character of the representation of sp 6 C with highest weight w 1 + w 2 , thus verifying the assertion of Exercise 17.4. The second Giambelli formula in the symplectic case expresses r.l. in terms of the basic representations

r.,. =

Ker(N(C 2")-> N- 2 (C 2 "))

which are the kernels of the contractions. The character of r ... is E0= 1, E'1 = E1 = x1 + ··· + x" + x(' + ·· · + x;', and

fork is

~

E~,

where

2, where Ek is the kth elementary symmetric polynomial. The formula

Corollary 24.24. Let Jl = (Jl 1 , ••• , 111) be tlte corljugate partition to A.. Tlte cltaracter of r.l. is equal to the determinallt of tlte I X I matrix wltose ith row is (E~,-1+-1

E~,-1-1·2

+ E~,-1

E~,-1+3

+ E;,,-1-1

···

E~,-1+1

+ E~,-i-1+2).

PRooF. This follows from the proposition and Proposition A.44, which equates the two determinants before specializing the variables. 0 There is also a simple formula for the character in terms of the characters

E1 of N(C 2 "), which also follows from Proposition A.44: (24.25) Note that EnH = E.-k (corresponding to the isomorphism NHC 2 " ~ A"-kC 2 ") and E~H = - E~-k+l· In particular, Corollary 24.24 expresses Char(f'.l.) as a polynomial in the characters of the basic representations r.,,, ••• , r(l)n"

· The Odd Orthogonal Case For so 2 n+I C the weights are LJl1L, I' = (Jl 1 , •.• , Jl.), with all111 integers or all half"integers. The positive roots are {L1 - L1} 1<J• {L1 + L1 } 1<J• and {L1}, sop is !{L 1 + ··· + L.) less than in the case for sp 2.: p =

L (n + i -

(24.26)

1)L1,

,.or p

= (n- !. n- !, ... , !).

With x(' = e(±L1) and x['fl before [(24.14)] for A,.

= e(±L,/2), we

have the same formula as

24. Weyl Character Formula

408

Exercise 24.27•. Show that A,= lxj-1+1/2 - XJ-In-1+1/2)1 = ~(Xt

). 1

+ xj•, ... , Xn + x;t)·(xl/2- xjti2)· ... ·(x~/2- x;;-•12).

If rA is the irreducible representation with highest weight A. = _LA.,L1, 2 · · · ~ ).. ~ 0, then the character formula can be written lxA,h-1+1/2 _ x-IA,h-i+l(2) 1

Jlx)

Char(rA) =

/+IJ2 -

~ I•

l+t/2JI

(24.28)

Similarly, dim(f'A)

=

[1 (1,-J]. [1-_Q~JL_

I<J(j-i) lsJ(2rt+l-i-j)

- [l (1,2 - If) . [l I,

- ,<1 (m?- mJ) where 11 = ).1 +

11-

i +!,and m1 =

Exercise 24.30. Show that, with

t; =

). 1 +

[l (I;- lj)(l; dim(f'A) =

l<j

i+

11-

(24.29)

, m,'

!.

n - i,

+ lj + I)· [l (2/; + I) I

(2n- l)!·(2n- 3)1· ... ·1!

These formulas give the dimension of the irreducible representation r., ....... with highest weight a 1 w 1 + ··· + a"w., where the w1 are the fundamental weights, using the equations ).1 =

a1 + · ·· + a._ 1 +!a•.

Exercise 24.31. Use the dimension formula to verify that for A.= L 1 + · · · + L,, the dimensi-on of rA is

C":

1). Use this to give another proof that

N vis

irreducible for I s; k s; n. Verify that the dimension of the spin representation is 2", thus reproving that it is irreducible. Exercise 24.32. Use the dimension formula to verify that the kernel of the contraction · 2 2 2 Symd(C "H)-+ Symd- (C "+1) is an irreducible representation with highest weight dL 1 • In case the representation is a representation of S02.+t C, i.e., the J., are all' integral, there is a first determinantaJ formula that expresSeS fA in terms of the kernels of the contractions Ker(Symd(C2n+t)-+ Symd-2(C2n+l )).

§24.2. Applications to Classical Lie Algebras and Groups

409

Let Kd denote the character of this kernel, so K 0 = I, K 1 = x 1 + · ·· + x. + x( 1 + ··· + x;' + l, and K 4 = H4 (x 1 ,

••• ,

x., x( 1,

... ,

x;•, l)- H4 _ 2 (x 1 ,

... ,

x., x(l, ... , x;;•, 1),

where H4 is the dth complete symmetric polynomial. From Proposition A.60 we have Proposition 24.33. If ). = (). 1 ~ • • • ~ )., > 0), witlt tlte ). 1 integral, tlten the character of r, is the determinant of the r x r matrix whose ith row is

In particular, for). = (d), the character is K 4 , which verifies that the kernel ofSym 4 (C 2 "+ 1 ) - . Sym 4 - 2 (C 2 "+t) is irreducible. Exercise 24.34. Use the character formula to verify that the multiplicities of the representation r2L, +Lz of 10o 5 C are as specifted in Exercise 18.9. The second determinantal formula for S0 2.+ 1 C writes r, in terms of the representations N(C 2 "+1 ), whose characters are

E" = E~c(x" ... , x., x( 1,

... ,

x;;•, l ).

Applying Proposition 24.33 with Corollary A.46, we have Corollary 24.35. Let J1 = (J1 1 , ... , p 1) be the conjugate partition to ).. Tlte character of r, is equal to the determinant of the l x l matrix whose ith row is

Since EnH = E.+J-t (corresponding to the isomorphism /\"HC 2 "+ 1 ;;:;; f\"+l-lcC 2"+ 1 ), this expresses Char(r,) as a polynomial in E 1 , ... , E., with E4 = Char(NC 2 "+ 1 ).

The Even Orthogonal Case For so 2 .C the weights are the same as in the preceding case. This time the {L1} are not positive roots, however, so pis !(L 1 + ··· + L.) less than in the case of !Oo 2.+ 1 C, or L 1 + .. · + L.less than in the case of !Op 2 .C: p =

L (n -

i)L 1,

(24.36)

or p = (11 - l, II - 2, ... , 0).

The calculation of A,. is similar, but using only those s of positive sign. This time

24. Weyl Character Formula

410

This leads to

(24.37) Note that the second determinant term vanishes when any 111 is zero. In particular, (24.38) AP = !lx;-' + x1- 1"_ 1,1. From (24.14) or Exercise A.66, AP

= ~(x 1

+ x; 1, ••• , x. + :c;;- 1 ).

(24.39)

This gives, with rA the irreducible representation with highest weight A= L),L,, At ;;:: ... ;;:: lA. I 2 0,

(24.40) where 11 = A1 + n - i. As before, dim(rA)

= n (I, i <J

IJ). (I, + 11 ) (j - i) (2n - i - j)

=nJ.IpL (m mj)' I<J

(24.41)

1 -

where 11 = A1 + 11 - i and m1 = 11 - i. Note that, as expected, the two represen· tat ions with weights (A 1 , ••• , A._ 1 , ±A.) have the same dimensions. Exercise 24.42. Show that

These formulas give the dimension of the irreducible representation r., ....... with highest weight a 1 w 1 + ... + a.w., where the w 1 are the fundamental weights, using the equations

..t, = a1 + ·· · + a._ 2 + !(a._ 1 + a.), I

~i~

n - 2,

Exercise 24.43. Use the dimension formula to verify that for w = L 1 + ·· · + Lk,

k
C:).

so N(C 2") is irreducible.

..t = L I + ... + L.-1 ± L., the dimension is

~

e:)'

Fo~

so /\"(C 2 ") is the sum of

the two corresponding irreducible representations. Verify that the dimension of the two spin representations are 2"- 1, proving irreducibility again.

§24.2. Applications to Classical Lie Algebras and Groups

411

Note that the second term in the numerator in (24.40) changes sign when

.t. is replaced by -.l..; in particular, it vanishes when .l.. = 0. When ).. = 0, the representation lA is a representation of the orthogonal group 0 2 .C. When ).. ,P 0, the direct sum of the two representations with highest weights ± .l..) is an irreducible representation of 0 2• C. (See Exercises 23.19 and 23.37.) Let Ld be the character of Ker(Symd(C 2 ") ~ Symd- 2 (C 2 ")), i.e., Ld = Hd(x 1 , ... , x., xl'. ... , x;")- Hd_ 2 (x 1 , ••• , x., xl 1 , ... , x;"). In either case, Proposition A.64 applies to give the first determinantal formula:

(.t 1 , ••• ,

Proposition 24.44. Given integers ). 1 ~ .. • ~ )., > 0, the clwracter of the irreducible representation of 0 2 .C with highest weight ). = (). 1 , ..• , .l.,) is the determinant of the r x r matrix whose ith row is (LA,-1+1

LA,-1+2

+ LA,-1

···

Ll,-i+r

+ Ll,-1-r+z).

Again, for ). = (d), this verifies that the kernel of the contraction from Symd(C 2 ") to Symd- 2 (C 2 ") is irreducible. The second determinantal formula is the same as in the odd case, but with E.= E.(x 1 , ..• , x., xl 1, ... , x;"): Corollary 24.45. Let J.l = (Itt, ... , p 1) be the conjugate partition to .l.. Tlte character of lA is equal to the determinant of the I x I matrix whose ith row is

(Ep,-l+t

Ep,-1+2

Using the fact that E.H

+ Ep,-1

···

Ep,-1+1

+ E~,-1-1+2).

= E•-•• this expresses Char(IA) as a polynomial in

E" ... , E., with Ed = Char(NC 2 "). Exercise 24.46"'. For each of the orthogonal groups O.,C, show that the character oft he irreducible representation with highest weight). can be written in the form Char(IA) = lhA,-I+J- ltA,-1-JI• where '•• is the character of Sym•(C"'). Another formula for the dimension of

fA is obtained by substituting (:) for

lti

in this determinant.

There are other formulas expressing the characters of general representations in terms of simpler ones. Abramsky, Jahn, and King [A-J-K] give one that can be expressed by the same formula for the general linear, symplectic, and orthogonal groups. The general irreducible representations are given by partitions ). or Young diagrams, and in their formula the simpler representations are those corresponding to hooks. To express it, let (a • b) denote the hook with horizontal leg of length a + l and vertical leg of length b + I, i.e., the partition (a+ 1, I, ... , 1), with b !'s. More generally, given a= (a 1 >···>a,~ O) and b = (b 1 > · · · > b, ~ 0) with a, or f,, nonzero, let (a • b) denote the partition whose Young diagram has legs of these lengths to

24. Weyl Character Formula

412

the right of and below the r diagonal boxes (cf. Frobenius's notation, Exercise 4.17). Let xf•, ~>denote the character of the corresponding irreducible represen· tation. Their formula is (24.47)

Taking the degree of both sides gives new formulas for the dimensions of the irreducible representations. These formulas are particularly useful if the rank r of the partition is small.

Exceptional Cases We will, as a last example, work out the Weyl character formula for the exceptional Lie algebra g 2 , and thereby verify some of the analysis of its representations given in Lecture 22. The remaining four exceptional Lie algebras we will leave as exercises. To begin with, the value of pis easily seen to be 2L 1 + 3L 2 , in terms of the basis L~o L 2 for the weight lattice introduced in Lecture 22.

Now, for any weight AP =

I

..,c e3

Jl

= pL 1 + qL 2 + rL 3 , we have

x:01 • x:(2) · x~m-

I s1 x;f~J · x.:;-1~ 1 • x.:;(31

(lfE

§24.2. Applications to Classical Lie Algebras and Groups

413

where we write x for (x 1 , x 2 , x 3 ) and x- 1 for (xj" 1 , x2 1 , x] 1 ), ll. is the discrimXI = 1 we can also inant, and sp,q,r the Schur function. Using the relation write this as

n

= .t\(x)-(Sp. 4 ,,(x) -

S.,-p,m- 4 ,.,_,(x))

for any 111 ~ max(p, q, r). To make this notation agree with the standard notation for Schur polynomials from Appendix (A.4), note that sp. 4,, is the Schur polynomial S1,,,1 for the partition (s, t), s ;;::-: t, where s is two less than the difference between the largest and smallest of p, q, and r, while t is one less than the difference between the second largest and the smallest; if p, q, and r are not distinct, Sp,q,, = 0. Thus, for example,

AP = .1\(x) · (S0 , 11 (x) - S111 (x)) = ll.(x)·(x 1 x 2

+ x 1x 3 + x2 x 3

-

x1

-

x2

-

x 3 ).

Now, any irreducible representation rA of 92 has highest weight A. = awl + bw 2 , where o> 1 = L 1 + L 2 and w 2 = L 1 + 2L 2 are the two fundamental weights, and a and bare non-negative integers. Then A. + p =(a + b + 2)L 1 + (a+ 2b + 3)L 2 • The Weyl character formula in this case becomes ;froposition 24.48. The character of tlte repre.~entation of !h witlt higltest weigltt aw 1 + bw 2 is

Exercise 24.49. In the case of the standard representation r, 0 , the adjoint representation f 0 , 1 , and the representation f 2 , 0 , use this formuia to verify the multiplicities found in Lecture 22. We can also work out the dimension formula explicitly in this case. The two fundamental weights w 1 and w 2 have inner products and ro 1 and w 2 are among the positive roots of g 2 , and in terms of these the remaining positive roots are 2w 1 - w 2, 3ro 1 - w 2, w 2 - ro 1 , and 2w 2 - 3w 1 . The weight pis the sum of the fundamental weights w 1 and w 2 , so that for an arbitrary weight A.= aw 1 + bw 2 we have the following table ofinner products:

2rol - w2 3rol - OJ2 OJI (1)2 -OJ.+ 0)2 -3w 1 + 2ro 2

(·, p)

(·,A)

(",A+ p)

1/2 3 5/2 9/2 2 3/2

af2 3a/2 + 3b/2 a+ 3b/2 3af2 + 3b a{2 + Jb/2 3b/2

(a+ 1)/2 3(a + b + 2)/2 (2a + 3b + 5)/2 3(a + 2b + 3)/2 (a + 3[, + 4)/2 3(b + 1)/2

24. Weyl Character Formula

414

We conclude that tlte dimension of tlte irreducible representation lliglrest weight).= aw 1 + bw 2 is

d' tm(ra.b) =

ra.h

of g2 with

(a+ l)(a + b + 2)(2a + 3b + 5)(a + 2b + 3)(a + 3b + 4)(b + 1) 120 .

We can check this in the cases a= 1, b = 0 and a= 0, b = 1, getting the dimensions 7 and t 4 of the standard and adjoint representations, respectively. In case a = 2, b = 0 we may verify the result of the explicit calculation in Lecture 22, finding that dim(r2 , 0 ) and, therefore, deducing that NV=

= 27

r 2 .o E9 V E9 C and Sym 2 V = r 2 , 0 E9 C.

Exercise 24.50. Show that Sym~V = Efjl."~~ r•-lk.o· We leave the analogous computations for the remaining four Lie algebras as exercises, using the description of the root systems found in Exercise 21.16. Since we have not said much about the Weyl group in the exceptional cases the formula (WCF) cannot be used directly-not to mention the fact that the orders of these Weyl groups are: 27 · 32 = 1152 for f4 ; 2 7 • 34 • 5 = 51,840 for 1! 6 , 2 10 ·3 4 ·5·7=2,903,040 for e 7 , and 2 14 ·3 5 ·5 2 ·7=696,729,600 for e8 • However, the dimension formula is available. Exercise 24.51"'. For each of the four remaining exceptional Lie algebras, compute p = half the sum of the positive roots. For each of the fundamental weights w, at least for f4 , compute the dimension of the irreducible represen· tation with highest weight w. In particular, lind the nontrivial representation of minimal dimension. Use this to verify that (E 6 ) is not isomorphic to (8 6 ) or (C 6 ), i.e., that e6 is not isomorphic to so 13 C or sp 12 C. Exercise 24.52"'. List all irreducible representations V of simple Lie algebras g such that dim V :-:;; dim g. Note that these include all cases where the corresponding group representation has a Zariski dense orbit, or a finite number of orbits.

LECTURE 25

More Character Formulas

In this lecture we give two more formulas for the multiplicities of an irreducible representation of a semisimple Lie algebra or group. First, Freudenthal's formula (§25.1) gives a straightforward way of calculating the multiplicity of a given weight once we know the multiplicity of all higher ones. This in turn allows us to prove in §25.2 the Weyl character formula, as well as another multiplicity formula due to Kostant. Finally, in §25.3 we give Steinberg's formula for the decomposition of the tensor product oftwo arbitrary irreducible representations of a semisimple Lie algebra, and also give formulas for some pairs I) c g for the decomposition of the restriction to I) of irreducible representations of g. §25.1: Freudenthal's multiplicity formula §25.2: Proof of (WSF); the Kostant multiplicity formula §25.3: Tensor products and restrictions to subgroups

§25.1. Freudenthal's Multiplicity Formula Freudenthal's formula gives a general way of computing the multiplicities of a representation, i.e., the dimensions of its weight spaces, by working down successively from the highest weight. The result is similar to (but more complicated than) what we did for i~l 3 C in Lecture 13, where we found the multiplicities along successive concentric hexagons in the weight diagram. Let rl be the irreducible representation with highest weight A., which will be fixed throughout this discussion. Let tt,. = n,,(rl) be the dimension of the weight space 1 of weight Jl in r 1 , i.e., Char(r1 ) = In,.e(/t). Freudenthal gives a formula for n,. in terms of multiplicities of weights that are higher than Jl. 1

In the literature, these multiplicities n. are often rererred to as "inner multiplicities."

25. More Character Formulas

416

Proposition 25.1 (Freudenthal's Multiplicity Formula). Wltlr tlte abot'e notation,

Here llflll 2 = (fJ, /J), ( , ) is the Killing form, and pis halfthe sum of the positive roots. Exercise 25.2*. Verify that c(/l) is positive if ll -:F .l. and rt~ > 0. The proof of Freudenthal's formula uses a Casimir operator, denoted C. This is an endomorphism of any representation V of the semisimple Lie algebra g, and is constructed as follows. Take any basis U1 , ••. , U, for g, and let Ul, ... , be the dual basis with respect to the Killing form on g. Set

u:

c = u. u; + ··· + u,u:. i.e., for any v e V, C(v) =I U1 ·(U/ · v). Exercise 25.3. Verify that Cis independent of the choice of basis 2 • The key fact is Exercise 25.4*. Show that C commutes with every operation in g, i.e.,

C(X · v) = X· C(v) for all X e g, v e V. The idea is to use a special basis for the construction of C, so that each term U1 Uj will act as multiplication by a constant on any weight space, and this constant can be calculated in terms of multiplicities. Then Schur's lemma can be applied to know that, in case Vis irreducible, C itself is multiplication by a scalar. Taking traces will lead to a relation among multiplicities, and a little algebraic manipulation will give Freudenthal's formula. The basis for g to use is a natural one: Choose the basis H 1 , ••• , H. for the Cartan subalgebra 1), where H1 = H«, corresponds to the simple root Ot;, and let H; be the dual basis for the restriction of the Killing form to 1). For each root Ot, choose a nonzero X« e g«. The dual basis wiii then have X~ in 9-«· In fact, if we let Y. E 9-« be the usual element so that X«, f., and H« = [X«, Y.J are the canonical basis for the subalgebra §« ~ !51 2 C that they span, then X~=

((at, at)/2) Y,..

(25.5)

Exercise 25.6*. Verify (25.5) by showing that (X«, Y«) = 2/(at, at). 2 In rancy language, Cis an element or the universal enveloping algebra or g, but we do not nml this.

417

§25.1. Freudenthal's Multiplicity Formula Now we have the Casimir operator

C= L:H,H;

+I

x~x~.

«ER

and we analyze the action of Con the weight space V,, corresponding to weight

p for any representation V. Let n,. = dim(V,.). First we have

L H,H; acts on V,. by multiplication by (/l, 11) =

ll!lll 2 .

(25.7)

L L:r;w;,

r1w 1, Indeed, H,H; acts by multiplication by !l(H1)/l(Hj). If we write /l = where the 1 are the fundamental weights, then lt(H1) = r1, and if It= with w; the dual basis to w 1, then similarly !l(H;) = r;. Hence L/l(H1)1t(H;) = L,r,r; = (/l, It), as asserted. Now consider the action of x.x; = ((a, a)/2)X« Y.. on Vw Restricting to the VP+I• corresponding to subalgebra s« ~ s1 1 and to the subrepresentation the a-string through 11, we are in a situation which we know very well. Suppose this string is

w

ffi,

v, E9 v,_. EB · •· E9 v,_.,«, so m = fl(H.) [cf. (14.10)], and let k be the integer such that /l = p- ka. We assume for now that k ~ m/2. On the first term v,, X. Y« acts by multiplication by m = fJ(H.) = 2(/J, 'a)/(a, a), so x.x~ acts by multiplication by (/1, a). In general, on the part of V,-l• which is the image of v, by multiplication by (Y.f, we know (cf. (11.5)] that X. Y. acts by multiplication by (k + l)(m -- k). This gives us a subspace of V,. of dimension on which X«X; acts by multiplication by

n,

(k + 1)((/l, a)- k(a, a)/2) = (k + 1)((/l. a)+ k(a, a)/2).

v,_. v,,

Now peel off the subrepresentation (overs.) of V spanned by and apply the same reasoning to what is left. We have a subspace of of dimension to which the same analysis can be made. From this we get a subspace of V,. of dimension "II-• - n, on which X~X~ acts by multiplication by

."--· -n,

(k)((lt, a)+ (k- !)(a, a)/2). Continuing to peel off subrepresentations, the space V,, is decomposed into pieces on which x.x; acts by multiplication by a scalar. The trace of X~X; : on V,. is therefore the sum n,-(k + l)((lt, a)+ k(a, a)/2) + (n,_.- n,)·(k)((ll, a)+ (k- l)(a, a)/2) + · · · + ((n,_l~- n11 _ 1l-l)•) · (1)((/l, a)+ (O)(a, a)/2). ·' Canceling in successive terms, this simplifies to k

~(·:;,

Trace(X.X~Iv) ,.

= L (/l +

ia, a)n,.H•·

(25.8)

1=0

/One pleasant fact about this sum is that it may be extended to all i ;;:.: 0, since 0 fori> k.

t.1n..... =

418

25. More Character Formulas

In case k ~ m/2, the computation is similar, peeling off representations from the other end, starting with Vp-m~· The only difference is that the action of Xs Y, on 1111 _.., is zero. The result is

'rrace(X.X~Iv, ) =

-.

I

(25.9)

(p - iiX, 1X)n,,_,•.

, 1=1

Exercise 25.10. Show that x.x~ =X -.x~. +({IX, 1X)/2)H., and deduce (25.9) directly from (25.8) by replacing IX by -IX. In fact, (25.8) is valid for all Jl and

f

(Jl

l=-rn

IX,

as we see from the identity

+ iiX, IX)Ilp+lo =

(25.11)

0.

Exercise 25.12*. Verify (25.11) by using the symmetry of (he IX-string through p. Now we add the assumption that V is irreducible, so C is multiplication by some scalar c. Taking the trace of Con V,. and adding, we get en,.= (p, p)n,.

+L L

(p

CII!R i~O

+ iiX, 1X)r~,.+l•·

(25.13)

Note that when i = 0 the two terms for IX and -IX cancel each other, so the summation can begin at i = I instead. Rewriting this in terms of the positive weights, and using (25.11) the sums become

I I"' aeR+ i=l «eR+ f:t Summarizing, and observing that (Jt - iiX, IX)n,_,. LuR• (p, IX)= (p, 2p), we have cnP = ((p, JL) + (p, 2p))n,. + 2 L L"' (Jl + iiX, 1X)n ,+1•· 1 a:e R+ 1=1 Note that (JL, JL) + (JL, 2p) =(It + p, Jt + p)- (p, p) = IIJL + Pll 2 - llpll 2 • To evaluate the constant we evaluate on the highest weight space VA, where n1 = I and 11a 1s = 0 fori> 0. Hence, c =(.I., .1.) + (.1., 2p) = 11.1. + Pll 2 - IIPII 2 - (25.14) Combining the preceding two equations yields Freudenthal's formula. 0 Exercise 25.15. Apply Freudenthal's formula to the representations of si 3 C considered in §13.2, verifying again that the multiplicities are as prescribed on. the hexagons and triangles. §25.2. Proof of(WCF); the Kostant Multiplicity Formula 419 Exercise 25.16. Use Freudenthal's formula to calculate multiplicities for the representationS ft,O• fo,lo and f2,0 Of (gz). §25.2. Proof of (WCF); the Kostant Multiplicity Formula It is not unreasonable to anticipate that Weyl's character formula can be deduced from Freudenthal's inductive formula, but some algebraic manipulation is certainly required. Let XA = Char(fA) = L npe(11) be the character of the irreducible representation with highest weight ).. Freudenthal's formula, in form (25.13), reads 3 · c· XA = Lp (Jt, ll)ttpe(J.I) + Lp oeR L 1=0 I (J.I + ia, a)llp+ioe(fl), where c = Q). + pll 2 - IIPII 2 • To get this to look anything like Weyl's formula, we must get rid or the inside sums over i. If a is fixed, they will disappear if \we m\lltiply by e(IX) - l, as successive terms cancel: (e(a) - I)· L p <X> L (II + ia, 1X)IlwH•e(J1) = Lp (fl, IX)Itpe(JI + a). 1=0 Let P = n.cR(e(a)- l) = (e(IX)- l)· P., where P. = n,,.(e(p)- 1). The preceding two formulas give c · P · XA = P · L (Jt, J.l)tt1,e(f1) + L (JI, a)P.nAfl + a). p (25.17) ~.a Note also that P = (-l)'A,.· AP, where r is the number of positive roots, so at least the formula now involves ' the ingredients that go into (WCF). :· We want to prove (WCF): AP · XA = AA+p· We have seen in §24.1 that both i sides of this equation are alternating, and that both have highest weight term ~; e(). + p), with coefficient 1. On the right-hand side the only terms that appear ~- are those of the form ±e(W(). + p)), for W in the Weyl group. To prove (, (WCF), it suffices to prove that the only terms appearing with nonzero ::.:coefficients in AP · XA are these same e(W(A. + p)), for then the alternating ~•. property and the knowledge of the coefficient or e(). + p) determine all the ~.coefficients. This can be expressed as: :, ~-. w ~ 1 1n this section we work in the ring C[A] or finite sums L;m.e(J•) with complex coefficients m•. ~~- .. 420 25. More Character Formulas Claim. Tl1e only terms e(l') occurring In tlwse witl1 II vii = Ill+ Pll. AP • XA witl1 IIOilzei'O coe.fficiellt are To see that this is equivalent, note that by definition of AP and lA• the terms in AP • Y.A are all of the form ± e( v), where v = 11 + W(p), for JL a weight of fA and W in the Weyl group. But if lltt + W(p)ll = Ill + pll, since the metric is invariant by the Weyl group, this gives II w- 1 (Jl) +pi! = Ill + Pll- But we saw in Exercise 25.2 that this cannot happen unless Jl = W(l), as required. We are thus reduced to proving the claim. This suggests looking at the "Laplacian" operator that maps e(J.l) to IIJ.111 2e(Jl), that is, the map 6: C[A]-+ C[A] defined by ML m,,e(J.l)) = L (Jl, Jl)m,.e(Jl). The claim is equivalent to the assertion that F = equation" 6(F) =Ill+ AP • XA satisfies the "differential PII 2 F. From the definition 6(XA) = L(Jl, Jl)n,.e(J.l). And 6(Ap) since II W(tx)ll = lltxll for all We liD, 6(A«) = IIPII 2AP.In general, = L (-l)wll W(tx)ll 2 e(W(tx)) = II1XII 2A«. So we would be in good shape if we had a formula for 6 of a product of two functions. One expects such a formula to take the form 6(/g) = 6(f)g + 2(VJ, Vg) + .f6(g), (25.18) where V is a "gradient," and ( , ) is an "inner product." Taking f = e(p~ g = e(l'), we see that we need to have (Ve(Jt), Ve(v)) = (J.l, v)e(J.l + 1•). There is indeed such a gradient and inner product. Define a homomorphism V: C[A]-+ IJ* ® C[A] = Hom(IJ, C[A]) by the formula V(e(Jl)} = J.l' e(Jl), and define the bilinear form ( , ) on I)'''® C[A] by the formula (txe(Jl), Pe(v)) = (tx, p)e(J.l + v), where (tx, Pl is the Killing form on 1)'1'. Exercise 25.19. With these definitions, verify that (25.18) is satisfied, as well as the Leibnitz rule V(.fg) = V(f)g + .fV(g). For example, V(XA) = L ,1lpJl' e(J.l), and, by the Leibnitz rule, V(P) = L P«tx · e(tx). 1 «ER But now look at formula (25. I 7). This reads §25.2. Proof of (WCF); the Kostant Multiplicity Formula c· PXA = P6(X.<) 421 + (VP, VX.<)· Since, also by the exercise, V(P) = 2( -I)'A,. V(A,.), we may cancel ( -l)'A,. from each term in the equation, getting c · A,.XA = A,.6(XA) + 2(V A,., VX.<)· By the identity (25.18), the right-hand side of this equation is 6(A,.XA)- 6(A,.)XA = 6(A,.XA)Sincec = ll.l. the proof. npii 2A,.XA· + pll 2 - llpll 2, this gives ll.l. + pii 2A,X.< = 6(A,.XA), which finishes 0 We conclude this section with a proof of another general multiplicity formula, discovered by Kostant. It gives an elegant closed formula for the multiplicities, but at the expense of summing over the entire Weyl group (although as we will indicate below, there are many interesting cases where all but a few terms of the sum vanish). It also involves a kind of partition counting function. For each weight J.l, let P(p) be the number of ways to write JL as a sum of positive roots; set P(O) = 1. Equivalently, I n -- 1 ( )= a!A+ ect L~ P(J.l)e(J.l). (25.20) Proposition 25.21. (Kostant's Multiplicity Formula). Tile multiplicity n11 (1A) of rA is given by L (-1)w P(W(). + p)- (J.l + p)), weigllt ll in the Irreducible representation n11 (1A) = W&'IU where p is ltalf tlte surn of tl1e positive roots. PRooF. Write (A,.f 1 = e(- p)/TI (1 - e( -IX))= XA = AH,(A,r' = = I. P(v)e(- v- p). By (WCF), L (-l)we(W(.-1. + p)P(v)e(-v- p) w.• L (-l)w P(v)e(W(). + p)- (v + p)) w.• =f.. (-l)wP(W(). + p)- (II+ p))e(Jt), ·" as seen by writing 11 = W(). + p)- (v + p). 0 In fact, the proof shows that Kostant's formula is equivalent to Weyl's formula, cf. [Cart]. One way to interpret Kostant's formula, at least for weights ll close to the · highest weight). of lA, is as a sort of converse to Proposition 14.13(ii). Recall ' that this says that r,. will be generated by the images of its highest weight vector v under successive applications of the generators of the negative root ! spaces; in practice, we used this fact to bound from above the multiplicities of 25. More Character Formulas 422 various weights Jl close to A by counting the number of ways of getting from Ato Jl by adding negative roots. The problem in making this precise was always that we did not know how many relations there were among these images, if any. Kostant's formula gives an answer: for example, if the difference A - 11 is small relative to A, we see that the only nonzero term in the sum is the principle term, corresponding to W = 1; in this case the answer is that there are no relations other than the trivial ones X(Y(v))- Y(X(v)) =[X, Y](v). Whenp gets somewhat smaller, other terms appear corresponding to single renections Win the walls ofthe Weyl chamber for which W(A + p) is higher than Jl + p; we can think of these terms, which all appear with sign -1, as correction terms indicating the presence of relations. As Jl gets smaller still, of course, more terms appear of both signs, and this viewpoint breaks down. To see how this works in practice, the reader can for example carry out the analysis of the example at the end of §13.1. Exercise 25.11"' (Kostant). Prove the following formula for the function P, which can be used to calculate it inductively: P(O) = 1, and, for Jl # 0, P(p) = - r (-1)~~'P(Jl + W(p)- p). "'"' Exercise 25.23"' (Racah). Deduce from Kostant's formula and the preceding exercise the following inductive formula for the multiplicities n~ of 11 in r 1 : n,. = 1 if 11 =A, and if Jl is any other weight of r 1 , then n,, = r (-t)~~',,,.+p-ll'!p)• - "'"' Show, in fact, that for any weight 11 L (-1)~~'n,.+p-W!pl = L (-1)"'', r wem where the second sum is over those W' E llD such that W'(A + p) = 11 + p. Note that Kostant's formula, more than any of the others, shows us directly the pattern of multiplicities in the irreducible representations of si 3 C. For one thing, it is easy to represent the function P diagrammatically: in the weight lattice of si 3 C, the function P(Jl) will be a constant I on the rays {aL 2 - aL 1 }e~o and {aL 3 - aL 2 }.~ 0 through the origin in the direction of the two simple positive roots L 2 - L 1 and L3 - L 2 • It will have value 2 on, the translates {aL 2 - (a+ 3)Ld.~-l and {aL 3 -(a- 3)L 2}.,. 2 of these two· rays by the third positive root L 3 - L 1 : for example, the first of these can be · written as aL 2 -(a + 3)L 1 = (a + 1) · (L2 - =(a+ 2)·(L 1 - Ld + L3 - L1 Ld + L3- L2; and correspondingly its value will increase by 1 on each successive translate.·. of these rays by L 3 - L 1 • The picture is thus 423 §25.2. Proor or(WCF); the Kostant Multiplicity Formula ... Now, the prescription given in the Kostant formula for the multiplicities is to take six copies of this function nipped about the origin, translated so that the vertex of the outer shell lies at the points w(..1. + 11) - p and take their alternating sum. Superimposing the six pictures we arrive at · which shows us clearly the hexagonal pattern of the multiplicities . . Exercise 25.24*. A nonzero dominant weight . 1. of a simple Lie algebra is called minuscule if ..1.(H«) = 0 or I for each positive root a. (a) Show that if . 1. is minuscule, then every weight space of dimensional. r1 is one 25. More Character Formulas 424 (b) Show that ). is minuscule if and only if all the weights of r~ are conjugate under the Weyl group. (c) Show that a minuscule weight must be one of the fundamental weights. Find the minuscule weights for each simple Lie algebra. §25.3. Tensor Products and Restrictions To Subgroups In the case of the general or special linear groups, we saw general formulas for describing hOW the tensor product fA® fp Of tWO irreducible representa· tions decomposes: r~ ® r,. = EB N~,•• r•. • In these cases the multiplicities N~,•• can be described by a combinatorial formula: the Littlewood- Richardson rule. In general, such a decomposition is equivalent to writing (25.25) in l[A], where XA = Char(f~) denotes the character. 4 By Weyl's character formula, these multiplicities N~,.. are determined by the identity (25.26) This formula gives an effective procedure for calculating the coefficients N~P•' if one that is tedious in practice: we can peel otT highest weights, i.e., successively subtract from AHP · A,.+p multiples of AP · A.+pfor the highest v that appears. There are some explicit formulas for the other classical groups. R. C. King [Ki2] has showed that for both the symplectic or orthogonal groups, the multiplicities N~,.. are given by the formula N~,.. = (,fJ',t L M~~~ · M~,,. · M<~tv• (25.27) where theM's denote the Littlewood-Richardson multiplicities, i.e., the corresponding numbers for the general linear group, a1id the sum is over all partitions (, u, t. For other formulas for the classical groups, see [Murl], [Wei, p. 230]. Exercise 25.28*. For so 4 C, show that all the nonzero multiplicities N~P• are l's, and these occur for v in a rectangle with sides making 45° angles to the axes; Describe this rectangle. • In lhe tileralure lhese mulliplicilies N1•• are oflen called "ouler muhiplicilies," and I he probklll of finding lhem, or decomposing lhe lensor producl, I he "Ciebsch-Gordon" problem. §25.3. Tensor Products and Restrictions to Subgroups 425 Steinberg has also given a general formula for the multiplicities N~~·· Since it involves a double summation over the Weyl group, using it in a concrete situation may be a challenge. Proposition 25.29 (Steinberg's Formula). The multiplicity of N~~· = r. ill r.t ® r,, is L (-l)ww·P(W(A + p) + W'(11 + p)- 11- 2p), W.ll" where tile sum is over pairs W, W' E W, and Pis tl1e counting jimctiou appearing In Kostant's multiplicity formula. Exercise 25.30*. Prove Steinberg's formula by multiplying (25.25) by A,, using (WCF) to get x~A"+' = L N~,,A,+p· Write out both sides, using Kostant's formula for x~. and compute the coefficient of the term e((J + p) on each side, for ar.y (J. This gives L (-l)ww· P(W(A + p) + W'(11 + p)- (J- 2p) = L (-l)w N~.,•. Wfl+p)-p• w ~r Show that for (J = vall the terms on the right are zero but N~,••. Exercise 25.31 (Racah). Use the Steinberg and Kostant formulas to show that I \ The following is the generalization of something we have seen several times: Exercise 25.32. If l and p are dominant weights, and a is a simple root with .l.(H.) and Jr(H.) not zero, show that A + p - 11. is a dominant weight and r~ ® r" contains the irreducible representation rA+~-· with multiplicity one. So r,, = rA+, $ rA+,-o $others. In case p = A, with A( H.) ~ 0, Sym 2 (r.t) contains rA+"' while N(rl) contains r~ ® r1+p-o• . Exercise 25.33. If A + { is a dominant weight for each weight { of r,, show that the irreducible representations appearing in rl ® r,, are exactly the rA+C· . In fact, with no assumptions, every component of rA ® r, always has this form . .One can show that N~,.. is the dimension of {v E (r.t>.-~: H/•+ 1 (v) = 0, l ~ i ~ n, 11 = p(H,)}. •For this, see [Zel, §131]. ! For other general formulas for the multiplicities NA,. see [Kern], [K-N], JLi], and [Kuml], [Kum2]. :' We have also seen a formula for decomposing the representation r~ of 25. More Character Formulas 426 GLme when restricted to the subgroup GLm-l e.In this case the multiplicities of the irreducible components again have a simple combinatorial description. There are similar formulas for other classical groups In the literature, such formulas are often called "branching formulas," or "modification rules." We will just state the analogues of this formula for the symplectic and orthogonal cases: For so 2 .e c so 2.+1 e, and r 1 the irreducible representation of so 2.+ 1C given by A. = (A. 1 ~ • · • ~ A.. ~ 0), the restriction is (25.34) the sum over all X= (I1 , ••• , X,) with A.1 ~ X. ~ A.2 ~ X2 ~ · · · ~ I.-t ~ A.. ~ II. I, with the I 1 and A., simultaneously all integers or all half integers. For so 2._ 1 e c so 2.e, and r 1 the irreducible representation of so 2 .e given by A.= (A. 1 ~ • • • ~ lA.. I), (25.35) Res::::~.drl) = ri. the sum over all X= (II, ... , x.-d with EB .t, ~ xl ~ A.2 ~ x2 ~ · · · ~ x._. ~ lA.. I, with the I 1 and A.1 simultaneously all integers or all half integers. For sp 2. - 2 e c sp 2 .e, and r 1 the irreducible representation of sp 2 .e given by A.= (A. 1 ~···~A..~ 0), the restriction is (25.36) x._.) x.-1 the sum over all X= (XI' ... ' with II ~ ... ~ ~ 0, and the multiplicity N1 I is the number of sequences p 1 , ••• , p. of integers satisfying At ~ Pt ~ A2 ~ P2 ~···~A..~ P. ~ 0 and Pt ~ X1 ~ P2 · · · ~ Pn-1 ~I.-I ~ fJ •• As in the case of GL.C, these formulas are equivalent to identities among symmetric polynomials. The reader may enjoy trying to work them out from this point of view, cr. Exercise 23.43 and [Doe]. A less computational approach is given in [Zel]. · As we saw in the case of the general linear group, these branching rules can be used inductively to compute the dimensions of the weight spaces. For example, for some consider the chain some=> som_ 1 e => som_ 2 e => ... => so 3 e. Decomposing a representation successively from one layer to the next wiD finally write it as a sum of one-dimensional weight spaces, and the dimension can be read off from the number of "partitions" in chains that start with the given A.. The representations can be constructed from these chains, as described by Gelfand and Zetlin, cf. [Zel, §I 0]. §25.3. Tensor Products and Restrictions to Subgroups 427 Similarly, one can ask for formulas for decomposing restrictions for other inclusions, such as the natural embeddings: Sp 2"C c SLhC, SO.,C c SL.,C, GL.,C x GL"C c GLm+nC, GL.,C x GL"C c GL.,"C, SL.C c Sp 2 .C, SL"C c S02 "+1 C, SL"C c S02"C, to mention just a few. Such formulas are determined in principle by computing what happens to generators of the representation rings, which is not hard: one need only decompose exterior or symmetric products of standard representations, cf. Exercise 23.31. A few closed formulas for decomposing more general representations can also be found in the literature. We state what happens when the irreducible representations ofGL.,C are restricted to the orthogonal or symplectic subgroups, referring to [Lit3] for the proofs: For O.,C c GL.,C, with m = 2n or 2n + 1, given A.= (A. 1 ~ • • • ~ A..,~ 0), Resg~cc(fA) = the sum over all ffi NAirl. (25.37) X= (I1 ~ • · • ~ X, ~ 0), where NA1 = I:N~H· ~ with NuA the Littlewood-Richardson coefficient, and the sum over all 6 = (d 1 ~ d2 ~···)with all d1 even. Exercise 23.38. Show that the representation ~ 2 • 21 of GL.,C restricts to the direct sum r(2,2J $ I(2, $ I(o, over O.,C. (This decomposition is important in differential geometry: the Riemann-Christoffel tensor has type (2, 2), and the above three components of its decomposition are the conformal curvature tensor, the Ricci tensor, and the scalar curvature, respectively.) Similarly for Sp 2 .C c GL 2.C, Res~P~:cc(f).) = the sum over all ffiNAirJ, (25.39) I= (I1 ~···~I"~ 0), where NA1 = L N~lA· ~ . N,n is the Littlewood- Richardson coefficient, and the sum is over all 11 = ('1 1 = '1 2 ~ '1 3 = '14 ~ • • ·) with each part occurring an even number of times. ' It is perhaps worth pointing out the the decomposition of tensor products . is a special case of the decomposition of restrictions: the exterior tensor product rA lEI rl' of two irreducible representations of G is an irreducible . representation of G x G, and the restriction of this to the diagonal embedding 'of Gin G X G is the usual tensor product r). ® r,.. There are also some general formulas, valid whenever jJ is a semisimple Lie 428 25. More Character Formulas subalgebra of a semisimple Lie algebra g. Assume that the Cart an subalgebra ij is a subalgebra of 1), so we have a restriction from I)* to ij•, and we assume the half-spaces determining positive roots are compatible. We write ji for weights of g, and we write 11! ji to mean that a weight J1 of g restricts to ji. Similarly write Wfor a typical element of the Weyl group of g, and ji for half the sum of its positive weights. If A (resp. 1) is a dominant weight for g (resp. g), let N11 denote the multiplicity with which r 1 appears in the restriction Of rA to ij, i.e., Exercise 25.40*. Show that, for any dominant weight A of g and any weightji ofg, Exercise 25.41* (Kiimyk). Show that NAI = 2: (-1)w W 2: n,.(rA)· ,.U+;;-Wr;;l Exercise 25.42. Show that if the formula of the preceding exercise is applied to the diagonal embedding of g in g x g, then the Racah formula of Exercise 25.31 results. For additional formulas of a similar vein, as well as discussions of how they can be implemented on a computer, there are several articles inS/ AM J. Appl. Math. 25, 1973. Finally, we note that it is possible, for any semisimple Lie algebra g, to make the direct sum of all its irreducible representations into a commutative algebra, generalizing constructions we saw in Lectures 15, §17, and §19. Let r., be the irreducible representations corresponding to the fundam~ntal weights w 1 , ••• , w •. Let r:., , ... , A' = Sym'(r.. , EB · · · EB rruJ This is a commutative graded algebra, the direct sum of pieces A" = EB Sym•• (r.. ,) ® · · · ® Sym"•(rru.), 111 1, ... ,a" where a= (a" ... , a.) is an n-tuple of non-negative integers. Then A' is the direct sum of the irreducible representation r 1 whose highest weight is A= 1w 1, and a sum J" of representations whose highest weight is strictly J" is an ideal in smaller. As before, weight considerations show that J" = A', so the quotient · Ia EB. A'/J' = §25.3. Tensor Products and Restrictions to Subgroups 429 is the direct sum of all the irreducible representations. The product rl ® r;. __. rH,. in this ring is often called Carton multiplication; note that the fact that rHp occurs once in the tensor product determines such a projection, but only up to multiplication by a scalar. Using ideas of §25.1, it is possible to give generators for the ideal J'. If Cis the Casimir operator, we know that C acts on all representations and is multiplication by the constant c~ = ()., ..1.) + (2..1., p)onthe irreducible representation with highest weight A.. Therefore, if A. = L a1w 1, the endomorphism C- ell of A" vanishes on the factor rl, and on each of the representations rp of lower weight /-1 it is multiplication by cp- cl ¥- 0 (cf. (25.2)]. It follows that J" = lmage(C- ell: A"--> A"). Exercise 25.43*. Write C = L U1 u; as in §25. I. Show that for v 1 , ••• , vm vectors in the fundamental weight spaces, with vj E and = a,w, the element (C- c~I)(v 1 • v2 • ••• • vm) is the sum over all pairs j, k, with l ~ j < k $ m, of the terms r.J La.} L ( L (U,(vl). UI(v,.) + u:(vj). U,(v~))- 2(a.J, a.,)vj. v,). nJ.' v,. I .. I From this exercise follows a theorem of Kostant: J' is generated by the elements L (V,(v)· u;(w) + U;(v)· u,(w))- 2(a., (J)v· w I a. for v Er., IllEr,, with and (J fundamental roots. For the classical Lie algebras, this formula can be used to lind concrete realizations of the ring. If one wants a similar ring for a semisimple Lie group, one has the same ring, of course, when the group is simply connected; this leads to the ring described in Lectures 15 and 17 for SL"C and Sp 2.C. For SOmC, little change is needed when m is odd, but there is more work for m even. Details can be found in [L-1]. LECTURE 26 Real Lie Algebras and Lie Groups In this lecture we indicate how to complete the last step in the process outlined at the beginning of Part II: to take our knowledge of the classification and representation theory of complex algebras and groups and deduce the corresponding statements in the real case. We do this in the first section, giving a Jist of the simple classical real Lie algebras and saying a few words about the corresponding groups and their (complex) representations. The existence or a compact group whose Lie algebra has as complexi· fication a given semisimple complex Lie algebra makes it possible to give another (indeed, the original) way to prove the Weyl character formula; we sketch this In §26.2. Finally, we can ask in regard to real Lie groups G a question analogous to one asked for the representations or finite groups in §3.5: which or the complex representations V of G actually come from real ones. We answer this in the most commonly en· countered cases in §26.3. In this final lecture, proofs, when we attempt them, are generally only sketched and may require more than the usual fortitude from the reader. §26.1: Classification of real simple Lie algebras and groups §26.2: Second proof of Weyl's character formula §26.3: Real, complex, and quaternionlc representations §26.1. Classification of Real Simple Lie Algebras and Groups Having described the semisimple complex Lie algebras, we now address the analogous problem for real Lie algebras. Since the complexification !Jo ®1 C. of a semisimple real Lie algebra g 0 is a semisimple complex Lie algebra and. we have classified those, we are reduced to the problem of describing the rtGl fo,.ms of the complex semisimple Lie algebras: that is, for a given complex Lie; algebra g, finding all real Lie algebras g 0 with >~ ) 431 §26.1. Classification or Real Simple Lie Algebras and Groups 9o®RC F !1· We saw many of the real forms of the classical complex Lie groups and algebras back in Lectures 7 and 8. In this section we will indicate one way to approach the question systematically, but we will only include sketches of proofs. To get the idea of what to expect, let us work out real forms of si 2 C in detail. To do this, suppose g0 is any real Lie subalgebra of si 2 C, with g0 ®R C = sl 2 C. The natural thing to do is to try to carry out our analysis of semisimple Lie algebras for the real Lie algebra g0 : that is, find an element HE g0 such that ad(H) acts semis imply on g 0 , decompose !lo into eigenspaces, and so on. The first part of this presents no problem: since the subset of si 2 C ofnon-semisimple matrices is a proper algebraic subvariety, it cannot contain the real subspace g 0 c si 2 C, so that we can certainly find a semisimple H E g0 • The next thing is to consider the eigenspaces of ad(H) acting on g. Of course, ad(H) has one eigenvalue 0, corresponding to the eigenspace 1) 0 = R · H spanned by H. The remaining two eigenvalues must then sum to zero, which leaves just two possibilities: (i) ad(H) has eigenvalues A and -A, for A a nonzero real number; multiplying H by a real scalar, we can take A = 2. In this case we obtain a dec~mposition of the vector space g0 into one-dimensional eigenspaces !lo = l)o $ !12 $ !l-2· We can then choose X E g 2 and Y E g_ 2 ; the standard argument then shows that the bracket [X, Y] is a nonzero multiple of H, which we may take to be ; I by rechoosing X andY. We thus have the real form si 2 R, with the basis y = (~ ~). (ii) ad(H) has eigenvalues iA and - iA for Asome nonzero real number; again, ; adjusting H by a real scalar we may take A = 1. In this case, of course, there /are no real eigenvectors for the·action of ad( H) on g0 ; but we can decompose l9o into the direct sum of 1) 0 and the two-dimensional subspace !l(l, -i} corre!:sponding to the pair of eigenvalues i and - i. We may then choose a basis B ~.and C for 9{I,-IJ with ~F [H, B] =C and [H, C] =-B. ~:The commutator [B, C] will then be a nonzero multiple of H, which we may ttake to be either H or -H (we can multiply B and C simultaneously by a ~icalar p, which multiplies the commutator [B, C] by 11 2 ). In the latter case, we tee that g0 is isomorphic to si 2 R again: these are the relations we get if we lake as basis for sl 2 C the three vectors l . H =( -~ ~). B = G~). and C = G_~). 26. Real Lie Algebras and Lie Groups 432 Finally, if the commutator [B, C] =H. we do get a new example: g0 is in this case isomorphic to the algebra su 2 ={A: 'A= -A and trace(A) = 0} c si 2 C, which has as basis H= c~2 -~/2)' B= (- ~/2 1/2) , 0 and C = ( i/02 i/2) 0 . Exercise 26.1. Carry out this analysis for the real Lie algebras so 3 R and so 2, 1 R. In particular, give an isomorphism of each with either si 2 R or su 2 • This completes our analysis of the real forms of sl 2 C. In the general case, we can try to apply a similar analysis, and indeed at least one aspect generalizes: given a real form g0 c g of the complex semisimple Lie algebra g, we can find a real subalgebra 1) 0 c g0 such that ~ 0 ® C is a Carlan sub· algebra of g = g0 ® C; this is called a Carta/! subalgebra of g0 • There is a further complication in the case of Lie algebras of rank 2 or more: the values on ~ 0 of a root IX E R of g need not be either all real or all purely imaginary. We, thus, need to consider the root spaces g.., !lii• !l-s• and g_1 , and the subalgebra they generate, at the same time. Moreover, as we saw in the above example, whether the values of the roots IX e R of g on the real subspace 1Jo are real, purely imaginary, or neither will in general depend on the choice of ~o· Exercise 26.2*. In the case of g0 = si 3 R c !1 = si 3 C, suppose we choose as Carlan subalgebra 1) 0 the space spanned over R by the elements 2 0 0) H 1 = ( 0 -I 0 0 0 -1 and H2 = (0 0 0 0 0 -1 Show that this is indeed a Cartan subalgebra, and find the decomposition of, g into eigenspaces for the action of~ = 1) 0 ®C. In particular, find the roots. of gas linear functions on IJ, and describe the corresponding decomposition of !lo· <~ .:*, Judging from these examples, it is probably prudent to resist the temptatio~. to try to carry out an analysis of real semi simple Lie algebras via an analogui,: of the decomposition g = ~ $ (EfJg.) in this case. Rather, in the present boot: we will do two things. First, we will give the statement of the classifica~ theorem for the real forms of the classical algebras-that is, we will list ..~ the simple real Lie algebras whose complexifications are classical algebi., ... · Second, we will focus on two distinguished real forms possessed by any . · semisimple Lie algebra, the split form and the compact form. These are ,, two forms that you see most often; and the existence of the latter in parti~ will be essential in the following section. · !1 §26.1. Classilication of Real Simple Lie Algebras and Groups 433 For the first, it turns out to be enough to work out the complexifications g0 ®A C = g0 $ i · g0 of the real Lie algebras g0 we know. The list is: Real Lie algebra Complexification si.R si.C si.H = 91.3"~/R so,. 9 R so.C sp 2 .R sp 2 .c su,,, up, 9 H u:H si.C si.C x st.c sll.c so,+,C so.C X so.C sp2.c SP2nC X Sp2.C slp+qc sp 21p+,,c so 2.C The last two in the left-hand column are the Lie algebras of the groups Up.lll and V.*H of automorphisms of a quaternionic vector space preserving a Hermitian form with signature (p, q), and a skew-symmetric Hermitian form, respectively. · We should first verify that the algebras on the right are indeed the complexili~tions of those on the left. Some are obvious, such as the complexification (st.R)c = sl.R $ i· si.R = si.C. The same goes for sop,,R and sp 2.R. Next, consider the complexification of su. ={A e si.C: 'A= -A} . .To see that si.C = su. $ i · su., let M e st.C, and write M = !(M - 'M) + !(M + 'M) = !A + !B; then A e su., iB e su., and M =!A- i(i/2)B. The general case of sttp,q c slp+qC is similar: if the form is given by (x, y) = ~fQy, then sup,q ={A: 'AQ = -QA}. Writing Me si,+,C in the form ~-~ = !(M- Q·'M·Q)- i·(!(iM + iQ·'M·Q)) :•nd using Q = 'Q = Q- 1 = Q, one sees that Me su,,, $ i · su,,,. :,_;:· M ~'··· For the complexification of sl,C, embed slmC in slmC x slmC by A 1-+ (A, A). i,Oiven any pair (B, C), write &.,.; tJ ~:-t.~. (B, C) if(•; = ! ~·For the quaternionic Lie algebra, from the description ofGL.IH we saw in ~lure 7, we have ~~;· 9I.IH ={A e gl 2.c: AJ = JA}, 26. Real Lie Algebras and Lie Groups 434 with J = ( ~ 1 ~). As before, forM e gi 2.C, we can write M = !(M- J·M·J)- i·(!(iM to see that gl.H ®R + iJ ·iM·J)) C = gl 2.C. Exercise 26.3. Verify the rest of the list. The theorem, which also goes back to Carlan, is that this includes tht complete list of sl1irp/e real Lie algebras associated to the classical complex typts (A.)-(D.). In fact, there are an additional 17 simple real Lie algebras asso· ciated with the five exceptional Lie algebras. The proof of this theorem is rather long, and we refer to the literature (cf. [H-S], [Hel], [Ar]) for it. Split Forms and Compact Forms Rather than try to classify in gencrul the real forms g0 of a semisimple Lie algebra g, we would like to focus here on two particular forms that are possessed by every semisimple Lie algebra and that are by far the most commonly dealt with in practice: the split form and the compact form. :;· These represent the two extremes of behavior of the decomposition 9 ~; l) $ (E9g.) with respect to the real subalgebra g0 c g. To begin with, the spill form of o is a form g0 such that there exists a Carlan subalgebra [) 0 c 9o (thai is, a subalgebra whose complexification ~ = 1) 0 ® C c g0 ® C = g is a Cartal subalgebra of g) whose action on g0 has all real eigenvalues-i.e., such that all the roots ~ e R c I)* of g (with respect to the Carlan subalgebra ~ !<, 1) 0 ® C c g) assume all real values on the subspace 1) 0 • In this case we have I direct sum decomposition 'I ·A _.:~; of g 0 into 1) 0 and one-dimensional eigenspaces j« for the action of 1)0 (each~ will just be the intersection of the root space g« c g with g0 ); each pair 1. ~ i-« will generate a subalgebra isomorphic to sl2 R As we will see momentarilfJ :01 this uniquely characterizes the real form g0 of g. By contrast, in the compact form all the roots IX e R c 1)* of g (with respect; to the Cartan subalgebra 1) = 1)0 ® C c g) assume all purely imaginary vatuci: on the subspace [) 0 • We accordingly have a direct sum decomposition .,~j ·;\1 '~ .{~ (eadtl d1 of g0 into 1)0 and two-ditnensional spaces on which 1) 0 acts by rotation 1. will just be the intersection of the root space g« $ 9-.. with g0 ); each 11 generate a subalgebra isomorphic to su 2 • ,_:;~ The existence of the split form of a semisimple complex Lie algebm vial already established in Lecture 21: one way to construct a real-even rationa'l.: §26.1. Classification of Real Simple Lie Algebras and Groups 435 -form g0 of a semisimple Lie algebra g is by starting with any generator X«, for the root space for each positive simple root cx 1, completing it to standard basis X«,, f.,, and H1 = [X«,• Y.,] for the corresponding s«, = si 2 C, and taking g0 to be the real subalgebra generated by these elements. Choosing a way to write each positive root as a sum of simple roots even determined a basis {H1 e 1), X« E 9«, Y. E 9-«} for 90 , as in (21.20). The Carlan subalgebta 1) 0 of 9o is the real span of these H1• Note that once I) is fixed for g, the real subalgebra ~ 0 is uniquely determined as the span of the H« for all roots ex. The algebra g 0 is determined up to isomorphism; it is sometimes called the natural real form of g. Note that this also demonstrates the uniqueness of the split form: it is -_. the only real form g0 of g that has a Carlan subalgebra 1}0 acting on g 0 with -- all real eigenvalues. As for the compact form of a semisimple Lie algebra, it owes much of its : significance (as well as its name) to the last condition in · Proposition 26.4. Suppo.~e 11 is any complex sellrisimple Lie algebra a11tl Oo c 11 a real form of !1· Let 1) 0 be a Carton suhalgebra of !lo. I)= 1)0 ® C the corre: sponding Carton .~rrbalgel11·a 1?{ !1· 1'11e following are equivalent: ;,:~ (i) Each root ex E R c 1)'1' of g assumes pr4rely imaginary values on 1]0 , and ~ti' f.or eac/1 root IX the subalgebra of !lo generated by the intersection 10 of ~)"' (g.$ g_«) with g0 is isomorphic to su 2 ; ~:~~) The restl'iction to g0 of the Killing form of. g is ltegative d~finite; ~·(Ill) fo,. ~;;'' The real Lie group G0 with Lie algebra g0 r.~ compact. In (iii), G0 can be taken to be the adjoint form of g 0 • However, a theorem ~\of Weyl ensures that the fundamental group of any such G0 is finite, so the ~Condition is independent of the choice of G0 • Note also that, by the equivalence ~with (ii) and (iii), the condition (i) must be independent of the choice of Carlan V:lubalgebra 1)0 • This is in contrast with the split case, where we require only ~''that there exist a Car tan subalgebra whose action on 9 has all real eigenvalues; fiswesaw in the case of si 2 1R, in the split case a different 1)0 may have imaginary ,eigenvalues. 1., ~hooF. We start by showing that the first condition implies the second; this 'a·WI_·n follow from direct observation. To begin with, the value of the Killing Jorm on H e 1)0 is visibly ~;' >~f; . t:. .,t B(H, H)= L (cx(H)) 2 < 0. '.;,Next, the subspaces I« are orthogonal to one another with respect to B, so it Jremains only to verify B(Z, Z) < 0 for a general member Z e I., To do this, \~l X and Y be generators of g.. and 9-« c g respectively, chosen so as to form, :(ogether with their commutator H = [X, Y] a standard basis for si 2 C. By the ,inalysis of real forms of si 2 C above, we may take as generators of the algebra · icnerated by 1. the elements iH, U = X - Y and V = IX + i Y. If we set ·p Z =aU+ bV =(a+ ih)· X+ (-a+ ib)· l', 436 26. Real Lie Algebras and Lie Groups then we have ad(Z) o ad(Z) = (a - + iW ad( X) o ad(X) (a 2 + b 2 )(ad(X) o ad(Y) + ad(Y) o ad(X)) -1- (a- Now, ad(X) iW ad(Y) o ad(Y). ad(X) and ad(Y) o ad(Y) have no trace, so we can write trace(ad(Z) o ad(Z)) = -2 · (a 2 + b 2 ) • trace(ad(X) o ad(Y)). (26.5) o By direct examination, in the representation Sym"V of si 2 C, ad(X) o ad( f) acts by multiplication by (n- A}(ll +A- 2)/4 ~ 0 on the A-eigenspace for H, from which we deduce that the right-hand side of (26.5) is negative. Next, we show that the second condition implies the third. This is immediate: the adjoint form G0 is the connected component of the identity of the group Aut(n 0 ). In particular, it is a closed subgroup of the adjoint group of g, and it acts faithfully on the real vector space !ln• preserving the bilinear form B. lf B is negative definite it follows that G0 is a closed subgroup of the orthogonal group SOmR, which is compact. Finally, if we know that G0 is compact, by averaging we can construct a positive definite inner product on g 0 invariant under the action of G0 • For any X in Slo• ad(X) is represented by a skew-symmetric matrix A = (a 1,1) with respect to an orthonormal basis of g0 (cf. (14.23)), so B(X, X)= Tr(A o A)= 'f.r. 1 aua1, 1 = al, 1 ~ 0. Itt particular, the eigenvalues of ad(X) must be purely imaginary. Therefore a(l) 0 ) c iR and a= -a for any root a, from which (i) follows. 0 -I We now claim that every semislmple complex Lie algebra has a w1iqut compact form. To see this we need an algebraic notion which is, in fact, crucial to the classification theorem mentioned above: that of conjugate linear involu· tiou. If g = g0 ®R C is the complexification of a real Lie algebra g0 , there is 1. map 11: g -> n which takes x ® z to x ® z for x E g0 and z E C; it is conjugate : linear, preserves Lie brackets, and 11 2 is the identity. The real algebra g0 is the fixed subalgebra of 11, and conversely, given such a conjugate linear involution·· u of a complex Lie algebra g, its fixed algebra g" is a real form of g. To prove the·: claim, we start with the split, or natural form, as constructed in Lecture 21 : and referred to above. With a basis for g chosen as in this construction, it is not hard to show that there is a unique Lie algebra automorphism q1 of g that takes each element of~ to its negative and takes each X~ to Y,. (this follows· from Claim 21.25). This automorphism tp is a complex linear involution which·. preserves the real subalgebra g 0 . This automorphism commutes with the· associated conjugate linear 11. The composite l1((1 = q111 is a conjugate linear: involution, from which it follows that its fixed part !Jc = gn"' is another real form of g. This has Carlan subalgebra ~< = It• = i · ~ 0 . We have seen that the, restriction of the Killing form to 1) 0 is positive definite. It follows that its' restriction to IJc is negative definite, and hence that g., is a compact form of g.. Finally, this construction of !lc from g 0 is reversible, and from this one can.· deduce the uniqueness of the compact form. §26.1. Classification of Real Simple Lie Algebras and Groups 437 We may see directly from this construction that Exercise 26.6. Verify that {A1 = i · H1: 1 ~ j ~ n} is a basis for ~<> { B« = X«i ·(X. + Y,.)} is a basis for r., and the action is given by c. = [A 1,B.]=p·C. and Y,., [A1,C.]= -p·B., where pis the integer !1.(H1). In particular, 1), acts by rotations on the planes 1•. Our classical Lie algebras 9 all came equipped with a natural real form !lo• and with a basis of the above type. These split forms are: Complex simple Lie algebra Split form si.+IC soln+IC spl.c 1:1o2.c sl.+t R SOn+ I.• sp 2.R so•.• Exercise 26.7. For each of these split forms, find the corresponding compact . form g<. Exercise 26.8. Let 9o be a real semisimple Lie algebra. Show that a subalgebra ~ 0 of g0 is a Cartan subalgebra if and only ifit is a maximal abelian subalgebra and the adjoint action on 9o is semisimple. ·; Exercise 26.9*. Starting with a real form 9o of 9 with associated conjugation (a, show that one can always find a compact form 9r of 9 such that u(g,) = g<, ~ and such that go= t EEl p, i;. where t = 1)0 = 9o n g., and p = g0 n (i ·g.). Such a decomposition is called a VCartan decomposition of g0 • It is unique up to inner automorphism. I;:- ~- Exercise 26.10*. For any real form g 0 of g, given by a conjugation u, show t\ that there is a Cartan ~ubalgebra I) of g that is preserved by u, so g 0 n I) is }i a Cartan subalgebra of g0 • ·.·... ~ ~-; . { • Naturally, the various special isomorphisms between complex Lie algebras :r (sl 2C ;;;;; so 3C ;;;;; sp 2 C, etc.) give rise to special isomorphisms among their real ~;~ forms. For example, we have already seen that /;:i .-. ~:·while n:._•. 438 26. Real Lie Algebras and Lie Groups (cf. Exercise 26.1 ). Similarly, each of the remaining three special isomorphisms of complex semisimple Lie algebras gives rise to isomorphisms between their real forms, as follows: so 4 C ~ si 2 C x si 2 C compact forms: so 4 R ;;::f su 2 x su 2 split forms: so 2 , 2 ~ si 2 R x si 2 1R others: so 3 , 1 ~ si 2 C, u~H ;;::f su 2 x si 2 R. (ii) sp 4 C ~ so 5 C compact forms: u 2 D-I ~ so 5 R split forms: sp 4 R ~ so 3 • 2 other: u 1 , 1 D-1 ;;::f so 4 , 1 • (iii) s1 4 C ~ so 6 C compact forms: su 4 ~ so 6 R split forms: sl 4 R ~ so 3 , 3 others: su 2 , 2 ~ so 4 • 2 ; su 3 , 1 ;;::f uJ H; sl 2 H (i) ~ so 5 , 1 . In addition, the extra automorphism of so 8 C coming from triality gives rise to an isomorphism u:1-1 ~ so 6 , 2 • Exercise 26.11. Verify some of the isomorphisms above. (Of course, in the case of compact and split forms, these are implied by the corresponding iso· morphisms of complex Lie algebras, but it is worthwhile to see them directly in any case.) Real Groups We turn now to problem of describing the real Lie groups with these Lie algebras. Let G be the adjoint form of the semisimple complex Lie algebra g. If g0 is a real form of g, the associated conjugate linear involution u of g that fixes g 0 lifts to an involution ii of G. (This follows from the functorial nature of the adjoint form, noting that G is regarded now as a real Lie group.) The fixed points G" of this involution then form a closed subgroup of G; its connected component of the identity G0 is a real Lie group whose Lie algebra is g0 • G is calied the complexification of G0 • We have seen in §23.1 that if r = r w is the lattice of those elements in I) on which all roots take integral values, then 2nir is the kernel of the exponential mapping exp: I)~ G to the adjoint form. If 1)0 is a Carlan subalgebra of g0 ; T = exp(l) 0 ) will be compact precisely when the intersection of 1)0 with the kernel2nir is a lattice of maximal rank. In this case, Twill be a product of n copies of the circle S 1 , 11 = dim(!)), and, since the Killing form on 1)0 is negative. definite, the corresponding real group G0 will also be compact. Such a G0 wiU.: be a maximal compact subgroup of G. When G0 c G is a maximal compact subgroup, they have the same irreduc· ible complex representations. Indeed, for any complex group G', each comple~ §26.1. Classification or Real Simple Lie Algebras and Groups 439 homomorphism from G toG' is the extension of a unique real homomorphism from G0 to G'. This follows from the corresponding fact for Lie algebras and the fact that G0 and G have the same fundamental group. This is another general fact, which implies the finiteness of the fundamental group of G0 ; we omit the proof, noting only that it can be seen directly in the classical cases: Exercise 26.12*. Prove that 1t 1 (G0 )-+ 1t 1 (G) is an isomorphism for each of the classical adjoint groups. Exercise 26.13*. The special isomorphisms of real Lie algebras listed above give rise to special isomorphisms of real Lie groups. Can you find these? It is another general fact that any compact (connected) Lie group is a quotient (G, X Gl X ••. X G, X T)/Z, where the G; are simple compact Lie groups, T;:; (S 1 )tis a torus, and Z is a discrete subgroup of the center. In particular, its Lie algebra is the direct sum of a semisimple compact Lie algebra and an abelian Lie algebra. This provides another reason why the classification of irreducible representations in the real mmpact case and the semisimple complex case are essentially the same. Representations of Real Lie Algebras Finally, we should say a word here about the irreducible representations (always here in complex vector spaces!) of simple real Lie algebras. In some cases these are easily described in terms of the complex case: for example, the irreducible representations of sum or slmR are the same as those for si"'C, i.e., they are the restrictions of the irreducible representations r l = § l C"' corresponding to partitions or Young diagrams ..l.. This is the situation in general whenever the complexification g = g 0 ® C of the real Lie algebra g0 is still simple: the representations of g0 on complex vector spaces are exactly the representations of g. The situation is slightly different when we have a simple real Lie algebra whose complexilication is not simple: for example, the irreduc. ible representations of si,.C, regarded as a real Lie algebra, are of the form fl ® fP, where f" is the conjugate representation of f,,. The situation in general is expressed in the following Exercise 26.14. Show that if g0 is a simple real Lie algebra whose complexificalion g is simple, its irreducible representations are the restrictions of (uniquely determined) irreducible representations of g. If !lois the underlying real algebra of a simple complex Lie algebra, show that the irreducible representations of · 9o are of the form V ® W, where V and Ware (uniquely determined) irreduc• ible representations of the complex Lie algebra. 440 26. Real Lie Algebras and Lie Groups §26.2. Second Proof of Weyl's Character Formula The title of this section is perhaps inaccurate: what we will give here is actually a sketch of the first proof of the Weyl character formula. Weyl, in his original proof, used what he called the "unitarian trick," which is to say he introduces the compact form of a given semisimple Lie algebra and uses integration on the corresponding compact group G. (This trick was already described in §9.3, in the context of proving complete reducibility of representations of a semi· simple algebra.) Indeed, the main reason for including this section (which is, after all, logically unnecessary) is to acquaint the reader with the "classical" treatment of Lie groups via their compact forms. This treatment follows very much the same lines as the representation theory of finite groups. To begin with, we replace the average (1/IGI) }:;,., 6 /(g) by the integral 6 f(g) dp., the volume element dp chosen to be translation invariant and such that 6 dp = I. If p: G-> Aut(V) is a finite-dimensional representation, with character J J Xv(B) = Trace(p(g)), J then 6 p(g) dJI E Hom(V, V) is idempotent, and it is the projection onto the invariant subspace V 6 . So 6 Xv(B) dp = dim(V 6 ). Applied to Hom(V, W) as before, since Xuom(Y, w1 = Xv Xw. it follows t_hat J L f XvXw dp = dim(Hom 6 (J.-: W)). So if V and W are irreducible, 6 XvXw dlt = { l 0 ifV~ W otherwise. Up to now, everything is completely analogous to the case of finite groups, and is proved in exactly the same way. The last general fact, analogous to the basic Proposition 2.30, is harder in the compact case: Peter-Weyl Theorem. The characters of irreducible representations span a dense subspace of the space of continuous class functions. It is, moreover, the case that the coordinate functions of the irreducible matrix representations span a dense subspace of all continuous (or U) functions on G. For the proof of these statements we refer to [Ad] or [8-tD]. Given the fundamental role that (2.30) played in the analysis of representations of finite groups, it is not surprising that the Peter- Weyl theorem is the cornerstone most treatments of compact groups, even though it has played no role so far in this book. We now proceed to indicate how the original proof of the Weyl character ot §26.2. Second Proof of Weyl's Character Formula 441 formula went in this setting. In this section, G will denote a fixed compact group, whose Lie algebra n is a real form of the semisimple complex Lie algebra 9c = g ®R C. We have seen that g = IJ $ E9 1., tt•R+ compatible with the usual decomposition !lc = ~c $ E9(g. $g .• ) when complexilied. The real Carlan algebra ~ acts by rotations on the planes 1•. Now let T = exp(~) c G. As before we have chosen ~ so that it contains the lattice 2n:if which is the kernel of the exponential map from l)c to the simply-connected form of nc. so T ~ (S 1 )" is a compact torus. In this compact case we can realize the Weyl group on the group level again: Claim 26.15. N(T)/T ~ m. PROOF. For each pair of roots IX, -IX, we have a subalgebra s. ~ si 2 C c fie, with a corresponding su 2 c g. Exponentiating gives a subgroup SU(2) c G. The element ( _ ~ ~)acts by Ad, taking H to -H, X to Y, and Y to X. It i,s,in N(T), and, with Bas in the preceding section, ( _ Then exp G ~ ~) = expGn:iB). n:iB) e g acts by renection in the hyperplane IX.l c 1). Note that lffi acting on IJ takes the lattice 2n:if to itself, so T = ~/2n:if by conjugation. 0 1m acts on Theorem 26.16. Every element of G is conjugate to a11 element ofT. A ge11eral element is conjugate to IWI sucll elements of T. Sketch of a proof Note that G acts by left multiplication on the left coset space -+ X which takes yT to zyT. The claim is that f. must have a fixed point, i.e., there is a y such that y· 1zy e T. Since all f. are homotopic, and X is compact, the Lefschetz number of f. is the topological Euler characteristic of X. The first statement follows from the claim that this Euler characteristic is not zero. This is a good exercise for the classical groups; see [Bor2] for a general proof. For another proof see Remark 26.20 below. For the second assertion, check first that any element that commutes with every element of T is in T. Take an "irrational" element x in T so that its multiples are dense in T. Then for any y e G, yxy·• e T-= yTy- 1 = T, and ·yxy- 1 = x-= y e T. This gives precisely IWI conjugates of x that are in T. X= G/T. For any z e G, consider the map f.: X Corollary 26.17. The class functions on G are tile liD-invariant fullctions 011 T. 26. Real Lie Algebras and Lie Groups 442 Suppose G is a real form of the complex semisimple group Gc. i.e., G is a real analytic closed subgroup of Gc, and the Lie algebra of Gc is 9c· The characters on Gc can be written n,.e 2 "'P, the sum over ll in the weight lattice A; they are invariant under the Weyl group. From what we have seen, they can be identified with !lll-invariant functions on the torus T. Let us work this out for the classical groups: L Case (Aw): G = SU(n matrices, + t). The Lie algebra su•+t consists of skew-Hermitian IJ = su.+J n sl.+t n= {imaginary diagonal matrices of trace 0}, L and T = { diag(e 2 d 3 •, .•. , e 2 "' 9··• ): ,91 = 0}. In this case, the Weyl group 211 is the symmetric group 6•+t, represented by permutation matrices (with one entry ±I on each row and column, other entries O) modulo T. Let z1: T .... S1 correspond to the ith diagonal entry e 2 "w•. So characters on Tare symmetric polynomials in z 1 , ••. , z.+r modulo the relation z 1 • •.• ·z.+ 1 =I. Therefore, characters on SU(n + l) are symmetric polynomials in z 1, ••• , z.+t· Case (B.): G = S0(2n form + 1). I) consists of matrices with ( cos(2n.9 1) sin(2n,9 1) n 2 x 2 blocks of the - sin(2n.9 1) ) cos(2n.9 1) along the diagonal, and one l in the lower right corner. Again we see that T = (S' )". This time N(T) will have block permutations to interchange the blocks, and also matrices with some blocks(~ ~)in the squares along the diagonal, with the other blocks 2 x 2 identity matrices, with a ±I in the corner to make the determinant positive; these take .9 1 to -.9 1 for each i where : a block is G~).This again realizes the Weyl group as a semidirect product •: of 6. and (l./2)". With z, identified with e2 "' 9• again, we see that the character$ are the symmetric polynomials in the variables z1 + z~ 1 , i.e., in cos(2n.9 1), ... , cos(2n.9.). ·1 . ··;~. ·_\_ Case (D.): G = S0(2n).l) is as in the preceding case, but with no lower comet; i Since we have no corner to put a -1 in, there can be only an even number or.,~ blocks of the form (~ ~), reflecting the fact that !lll is a semi direct produii of(l./2)"- and 6 •. This time the invariants are symmetric polynomials in tbci\ z1 + z;- 1, and one additional TI 1(z1 - z;- 1 ). ··!;?~ 1 cons~' Case (C.): G = Sp(2n). IJ consists of imaginary diagonal matrices, T of diagonal matrices with entries e 2 "w'. The Weyl group in generated bj] §26.2. Second Proof of Weyl's Character Formula 443 permutation matrices and diagonal matrices with entries which are l's and quaternionic j's: ~D is a semidirect product of (lL/2)" and 6 •. The invariants are symmetric polynomials in the z1 + z;- 1 • The key to Weyl's analysis is to calculate the integral of a class function f on Gas a suitable integral over the torus T. For this, consider the map n: G/T x T--. G, By what we said earlier, sheets. It follows that f G is a generically finite-sheeted covering, with 7t f n(x1: y) = xyx- 1• dp = l"~l :u> J G/T> llll\1 n*(f)n*dp. Now n*(f)(xT, y) = f(y) since f is a class function. To calculate n*dp, consider the induced map on tangent spaces n. = dn: g/[) x I)--. g. At the point (x 0 T, y 0 ) E G/T x T, We want to calculate . which is xo(XYo + YoY- Yox)xo 1(XoYo 1Xo 1) = . Now y 0 yy01 = y since Yo E xo(x + YoYYo 1 - YoXYo 1 )Xo 1• Tandy E 1). To calculate the determinant of n. '~we can ignore the volume-preserving transformation x 0 ( )x 01 • If we identify e- f g with g/() x 1), the matrix becomes ~d(yo) ~). ~So the determinant of n• is det(J ~acts as t.• ::.~<': det(n.) = ~J.~ ru e\ (l- elft'•), (26.18) a function on T alone, independent of the factor G/T. This gives Weyl's ~ifnlegration ~Vi n 11... ;;::,., ~.+':'. Ad(y0 )). Now (g/IJ)c = Efjg., and Ad(y 0 ) e 2 d•(Yol on g•. Hence formula: ff G dpG = - 1- I!IDI f T f(y) n (l :.lt eZdo(y)) dJlr. (26.19) (j~mark 26.20. The same argument gives another proof of the theorem that G !iis covered by conjugates of T. This amounts to the a~scrtion I hat the map 26. Real Lie Algebras and Lie Grou(IS 444 n: G/T x T-+ G of compact manifolds is surjective. By what we saw above, for a generic point y 0 e T there are exactly lliDI points in n- 1(y 0 ), and at each of these the Jacobian determinant is the same (nonzero) number. It follows that the topological degree of the map nisI liD I, so the map must be surjective. Now (l - e 2.,")(1 - e-hl•) = (e"'" - e-ft 1")(e" 1"- e Ki•), so if we set A= n (e" 1" - e-"1"), «Eft+ then det(n.) =At\. As we saw in Lemma 24.3, A= of the positive roots and, for any weight Jl, AP = A,, where pis half the sum L (-l)wel.tW(PI. Wo!D Now we can complete the second proof of Weyl's character formula: the character of the representation with highest weight A is A J.+p/ Ar Since we saw in §24.1 that AJ.+,IA, has highest weight A and (see Corollary 24.6) its value at the identity is positive, it suffices to show that the integral of Ja XX= I, where x = AJ.+,!A,. By Weyl's integration formula, f xx If- l f a = !WI r xxAA = lliDI r AH,AA+, =-'-f L (-l)we2"1W(H,,, L (-J)""e-2•iW(J.+,,= l, !liD I r w~ m w~ m which concludes the proof. §26.3. Real, Complex, and Quaternionic Representations The final topic we want to take up is the classification of irreducible complu representations ofsemisimple Lie groups or algebras into those of real, quate~ ionic, or complex type. To define our terms, given a real semi simple Lie group G0 or its Lie algebra g0 and a representation of G0 or g0 on a complex vector space V we say that the representation V is real, or of real type, if it comeS from a representation of G0 or g0 on a real vector space V0 by extension ci( scalars ( V = V0 ®R C); this is equivalent to saying that it has a conjugate endomorphism whose square is the identity. It is quaternionic if it comes rroni a quaternionic representation by restriction of scalars, or equivalently ifil hQ a conjugate linear endomorphism whose square is minus the identity. FinaUJ: we say that the representation is complex if it is neither of these. (CompaN. with Theorem 3.37 for finite groups.) :t-M Having completely classified the irreducible representations of the class~ complex Lie algebras, and having described all the real forms of these J.JI lineai §26.3. Real, Complex, and Quaternionic Representations 445 algebras, we have a clear-cut problem: to detemine the type of the restriction of each representation to each real form. Rather than try to answer this in every case, however, we will instead mention some of the ideas that allow us to answer this question, and then focus on the cases of the split forms (where the answer is easy) and the compact forms (where the answer is more interesting, and where we have more tools to play with). We assume the complexification g of g0 is simple, so irreducible representations of g0 are restrictions of unique irreducible representations of g (cf. (26.14)); in particular, we have the classification of irreducible representations by dominant weights. To begin with, the tensor products of two real, or two quaternionic, or of a pair of complex conjugate representations is always real; and exterior powers of real and quaternionic representations are equally easy to analyze, as for finite groups (see Exercise 3.43). Such tensor and exterior powers may not be irreducible, but the following criterion can often be used to describe an irreducible component of highest weight that occurs inside them: Exercise 26.21*. Suppose W is a representation of a semisimple group G that is real or quaternionic, and suppose W has a highest weight A that occurs with multiplicity l. Show that the irreducible representations r 1 with highest weight Ahas the same type as W. We may apply this in particular to the tensor product r1 ® r,. of the irreducible representations of g with highest weights Aand /.l; since the irreducible representation rA+,, with highest weight A + ll appears once in this tensor product, we deduce , Exercise 26.22*. (i) If r1and r, are both real or both quaternionic, then rH,. is real. (ii) If rl is real and r,. is qua tern ionic, then rl+p iS quaternionic. (iii) If 'r1 and r, are COmplex and conjugate, then rl+p is real. The last two exercises almost completely answer the question of the repre_,:sentations of the split forms of the classical groups: we have s··;··. ~Proposition 26.23. Every irredJ4cible representation of tl1e split forms sl.+ 1 R, ~~o+t,.R, sv 2.R, and so•.• R of tile classical Lie algebras is real. ~~:: ·. J.PitooF.In each ofthese cases, the standard representation Vis real, from which !itrollows that the exterior powers NV are real, from which it follows that the rh ~:symmetric powers Sym«•(NV) are real. Now, in the cases of sl.+t nand sp 2 .n, ~.we have seen that the highest weights wl of the representations NV fork= ~1, ... , n form a set offundamental weights: that is, every irreducible representa~~on r has highest weight ak • Wt for some non-negative integers a 1 , ••• , a •. ht follows that r appears once in the tensor product L ~;' Sym"•V® Sym"•(NV) ® ··· ® Sym•·(I\"V) 446 26. Real Lie Algebras and Lie Groups and so is real. (Alternatively, Weyl's construction produces real representations when applied to real vector spaces.) The only difference in the orthogonal case is that some of the exterior powers NV of the standard representation must be replaced in this description by the spin representation(s). That the spin representations are real follows from the construction in Lecture 20, cf. Exercise 20.23; the result in this case then follows as before. 0 The Compact Case We turn now to the compact forms of the classical Lie algebras. In this case, the theory behaves very much like that of finite groups, discussed in Lecture 5. Specifically, any action of a compact group G0 on a complex vector space V preserves a non degenerate Hermitian inner product (obtained, for example, by choosing one arbitrarily and averaging its translates under the action ofG0 ). It follows that the dual of V is isomorphic to its conjugate, so that V will be either real or quaternionic exactly when it is isomorphic to its dual V*. (In terms of characters, this says that the character Char(V) is invariant under the automorphism of Z[A] which takes e(J•) to e( -11); for groups, this says the character is real.) More precisely, an irreducible representation of a compact group/Lie algebra will be real (resp. quaternionic) if and only if it has an invariant nondegenerate symmetric (resp. skew-symmetric) bilinear form. In other words, the classification of an irreducible Vis determined by whether contains the trivial representation, and, if so, in which factor. So determining which type a representation belongs to is a very special case of the general plethysm problem of decomposing such representations. With this said, we consider in turn the algebras su", u.H, and so'"R. Let rA be the irreducible representation of si.C with highest weight A• a.- w 1, where ro 1 = L 1 + · · · + L 1, i = I, ... , n- I are the fundamental weights of si.C. The dual of r will have highest weight w,, so that r will be real or quaternionic if and only if a 1 = a._ 1 for all i. We now distinguish three '~i cases: (i) If n is odd, then the sublattice of weights A. = a 1 • w 1 with a1 = a._, for all i is freely generated by the sums ro 1 + fori = l, ... , (n - l)/2. Now, is the highest weight of the exterior power NV, so that the irreducible rep..; will appea,r once in the tensor product' sentation with highest weight ro 1 + I I w._, a._,· I w._, NV®I\"-'V=(NV)®(NV)*, oii ,J j~ .~-~ which by Exercise 26.21 above is real. It follows that for any weight: A.= a1 • ro 1 with a 1 =a._, for all i, the irreducible representation rA is reaL'~ (iia) If n = 2k is even, then the sublattice of weights A.= :La1·w1 willa: L ,,:_: §26.3. Real, Complex, and Quaternionic Representations 447 a1 = a._ 1 for all i is freely generated by the sums ro 1 + w.-r fori = I, ... , k - 1, together with the weight ro1.- As before, the irreducible representations with highest weight ro 1 + ro.-r are all real. Moreover, in case 11 is divisible by 4 the representation NV is real as well, since NV admits a symmetric bilinear form NV®NV-4(\ltV= C given by wedge product. It follows then as before that for any weight ,l. = ~::0 1 ·ro1 with a 1 = a.-r for all i, the irreducible representation rl is real. (iib) In case 11 is congruent to 2 mod 4, the analysis is similar to the last case except that wedge product gives a skew-symmetric bilinear pairing on NV. The representation N Vis thus quaternionic, and it follows that for any weight A.= 2:a1 · ro 1 with a 1 = a._ 1 for all i, the irreducible representation rl is real if aA is even, quaternionic if aA is odd. In sum, then, we have Proposition 26.24. For any weigltt ,l. = La.·w. of su., the irreducible representation rl witlt llighest weight A. is: complex if a1 '# a.-r for any i; real if a1 = a._, for all i a11d 11 is odd, or 11 = 4k, or 11 = 4k + 2 and a 2l+l is even; and quaternio11ic if a1 = a._, for all i a11d n = 4k + 2 a11d a 2H 1 is odd. Next, we consider the case of the compact form u.HI of sp 2.C. To begin with, we note that since the restriction to u.IHI of the standard representation of sp 2.C on V;;:;; C 2 " is quaternionic, the exterior power NV is real fork even and quaternionic fork odd. Since the highest weights rot of NV fork = 1, ... , 11 form a set of fundamental weights, this completely determines the type of the irreducible representations of u. HI: we have Proposition 26.25. For any weight A. = L a 1 • ro 1 of u. IHI, the Irreducible repreSelltation rl with lligl1est weight A. is real if a 1 is even for all odd i, and quaternionic if a 1 is odd for any odd i. Next, we consider the odd orthogonal algebras. Part of this is easy: since 'the restriction to so 2.+ 1 R of the standard representation V of so 2 .+ 1 Cis real, so are all its exterior powers; and it follows that any representation of so 2 .+ 1 n ,whose highest weight lies in the sublattice of index two generated by the highest weights of these exterior powers is real. It remains, then, to describe ,·the type of the spin representation; the answer, whose verification we leave as :Ellercise 26.28 below, is that the spin representation r~ of !.'llz•+~ c (that 'is, the irreducible representation whose highest weight is one-half the highest weight of 1\"V) is real when n = 0 or 3 mod 4, and quaternionic if 11 = l or ;;2mod 4. This yields tProposition 26.26. Let w1 be the highest weight of tlte representatloll NV of ~lo 2 .+ 1 C. For any weight A. = a 1 ro 1 + · ·· + a._ 1 ro._ 1 + a.ro./2 of so 2 .+1 R, the firreducible representation rl with highest weight A. Is real if a. is even, or if 11 is 448 26. Real Lie Algebras and Lie Groups congr.~ent to 0 or 3 mod 4; if a. is odd and n = l or 2 mod 4, then quaternionic. rA Is (Note that, in each of the last two cases, the fact that every representation is either real or quaternionic follows from the observation that the Weyl group action on the Cartan subalgebra ~ c g includes multiplication by -1.) Finally, we have the even orthogonal Lie algebras. As before, the exterior powers of the standard representation V are all real, but we now have two spin representations to deal with, with highest vectors (in the notation of Lecture 19) IX= (L 1 + .. · + L.)/2 and {J = (L 1 + ... + L._ 1 - Ln)/2. The first question is whether these two are self-conjugate or conjugate to each other. In case 11 is even, as in the case of the symplectic and odd orthogonal algebras, the Weyl group action on the Carlan subalgebra contains multiplication by - I (the Weyl group contains the automorphism of ~· reversing the sign of any even number of the basis elements L 1), so that and 11 will be isomorphic to their duals; if 11 is odd, on the other hand, we see that r. will have - {J as a weight, so that r. and r, will be complex representations dual to each other. We consider these cases in turn. · .,. '~l." (i) Suppose first that n is odd, and say A. is any weight, written as r. r .·, A.= a 1 w 1 + .. · + a._ 2 w._ 2 + a.-d1 + a.a. rA a., the representation with highest weight A. will not be isomorphic to its dual, and so will be complex. On the other hand, r.+ll appears once ill r. ® r, = End(f.), and so is real; thus, if an-I = a., the representation rl w~. be real. ':.; (ii) If, by contrast, n is even then all representations of so 2 nll will be either real or quaternionic. The half-spin representations r. and r, are real if II!! o. (mod 4), quaternionic if 11 = 2 (mod 4), a fact that we leave as Exercise 26.28.' It follows that, with A. as above, rl will be real if either n is divisible by 4, if a._ 1 + a. is even; if 11 = 2 mod 4 and a._ 1 + a. is odd, fA will be quaternionK::.: In sum, then, we have :;: If an-t -:/= or: ' '\l rA of so 2 .n wit It higl1est weight .t Jj a 1w 1 + ... + a._ 2 w._ 2 + a._ 1 {J + a.a will be complex if 11 is odd and a.-i ~·. a.; it will be quaternionic if n = 2 mod 4 and a0 _ 1 + a. is odd; and it will bi real otl1erwise. . .4; Proposition 26.27. The representation ;~.}~ Exercise 26.28*. Verify the statements made above about the types of I~ spin representation r. of the orthogonal Lie algebras, i.e., that the spi~; representation r. of so 2.+ 1 R is real when n = 0 or 3 (mod 4); and quater~ nionic if n = 1 or 2 (mod 4), and that the half-spin representations of to 2~R: are real if 11 ~ 0 (mod 4) and quaternionic if n = 2 (mod 4). Show, in fact, thaf the even Clifford algebras c~··• c C., = C(O, m) are products of one or I~ copies of matrix algebras over R, C, or H, with R occurring for m = 0 o~ ± l mod 8, C occurring for 111 ± 2 mod 8, and 11-1 for m ± 3 or 4 mod L\~ = = §26.3. Real, Complex, and Qualernionic Representations 449 Exercise 26.29. Show that for a representation V of a compact group G, if Vis complex ifVisreal. if Vis quaternionic. Exercise 26.30*. Show that for a representation V of a compact group, the number of irreducible real components it contains, minus the number of quaternionic representations, is the number of times the trivial representation occurs in lji 2 V in the representation ring, where 1/1 2 is the Adams operation (cf. Exercise 23.39). APPENDICES :.nese appendices contain proofs of some of the general Lie algebra facts that Ytere postponed during the course, as well as some results from algebra and Invariant theory which were used particularly in the "Weyl constructionSchur functor" descriptions of representations. ;i The first appendix is a fairly serious excursion in polynomial algebra. It proves some basic facts about symmetric functions, especially the Schur &;olynomials, which occur as characters of representations of GL. or SL., and pves determinantal formulas for them in terms of other basic symmetric polynomials. The last section of Appendix A includes some new identities lmong symmetric polynomials, which, when the variables are specialized, ~press characters of representations of Sp 2• and SO., as determinants in the ~racters of basic representations. ~; Appendix 8 gives a short summary of some basic multilinear facts about ~lerior and symmetric powers. The first two sections can be used as a 'leference for the conventions and notations we have followed; the third 't:Ontains a general discussion of constructions such as contractions, many :i~Jecial cases of which were discussed in the main text. ir The next three appendices conclude our discussion of the theory of Lie ·Nebras, which began in Lectures 9, 14, and 21. Proofs are given, by standard ;,liethods, of the promised general results on semisimplicity, the theorem on 'Conjugacy of Carlan subalgebras, facts about the Weyl group, Ado's theorem ~that every Lie algebra has a faithful representation, and Levi's theorem that :"'its the map from a Lie algebra to its semisimple quotient. ~\-The last appendix develops just enough classical invariant theory to lind :I!Jc polynomial invariants for SL.C, Sp 2.C, and so. c. This was the key to ~r proof that Weyl's construction gives the irreducible representations of the :iJmplectic and orthogonal groups. APPENDIX A On Symmetric Functions ; fA. I: Basic symmetric polynomials and relations among them §A.2: Proofs or the determinantal identities 'fA.J: Other determinantal identities •§A.l. Basic Symmetric Polynomials and Relations among Them c ; The vector space of homogeneous symmetric polynomials of degree d in k ·variables x 1 , ••. , x~ has several important bases, usually indexed by the .:partitions A.= (A. 1 ~ A. 2 ~···~A.~~ 0) of d into at most k parts, or by Young '·diagrams with at most k rows (see §4.1). We list four of these bases, which ·.are all valid for polynomials with integer coefficients, or coefficients in any 1 commutative ring. First we have the monomials in the complete symmetric polynomials: HA = HA,. HA, .... ·IIA.. (A. I) where H1 is the jth complete symmetric polynomial, i.e., the sum of all distinct i monomials of degree j; equivalently, ~'For example, with three variables, ::~ . ll0 ,t 1 = (xt H12 •01 = x~ + X2 + x3)2, +xi+ x; + x 1 x 2 + x 1 x 3 + x 2 x 3 • 454 A. On Symmetric Functions Next are the monomial symmetric polynomials: MA =I x·. (A1) the sum over all distinct permutations a = (a 1, ..• , aA) of (A 1, ... , A,); here x• = xj' · ... · x:•. For example, M 0 , 1> = x 1 x 2 + x 1x 3 + x 2 x 3, =X~+ X~+ X~. M(l,OJ The third are the monomials in the elementary symmetric functions. Unlike~ the first two, these are parametrized by partitions /.l of din integers no larger: than k, i.e., k ~ p 1 ~ • • • ~ ft1 ~ 0. These are exactly the partitions that are: conjugate to a partition of d into at most k parts. (The conjugate to a partitioa: A is the partition whose Young diagram is obtained from that of A by inter~, changing rows and columns. We denote the conjugate of A by).', although the' notation 1 is also common.) For such p set · E~ = E~, · Ep, · ... · Ep,• (A.l). where E1 is the jth elementary symmetric polynomial, i.e., For example, + x 2 + x 3) 2 , E(l,or = x 1 x 2 + x 1 x 3 + x 2 x3 . £ 0 , 11 = (x 1 The fourth are the Schur polynomials, which may be the most important, although they are less often met in modern algebra courses: lxf•H-11 8). = lx/'H-11 lxj 'I =-A-, (A.4) n,< where A= 1(x,- x1) is the discriminant, and la1, 1i denotes the deter~ minant of a k x k matrix. For example, + XrXJ + XzXJ, S(I.IJ = X1X2 S(l,OJ =X~ +X~+ X~+ XrXz + XtXJ + X2X3. The first task of this appendix is to describe some relations among these symmetric polynomials. For example, one sees quickly that Su,,, = E(l.o, = Hf - H2. S(l,or = H,z,o> = E~ - Ez, su.o 1· S1,,o1 = s 0 • 0 + S12.o1. These are special cases of three important formulas involving Schur polynomials, which we state next. The lirst two are known as determiltantal §A. I. Basic Symmetric Polynomials and Relations among Them 455 fonnulas. The first is also known as the Jacobi- Tmdy identity. From geometry, the first two are sometimes called Giambelli's formulas, and the third is Pieri's formula. The proofs will be given in the next section. (A.5) , Note that if A.p+t = ·· · = A.k = 0, the determinant on the right is the same as the determinant of the upper left p x p corner. The second is E,,, .·., SA =-IEp,+j-1 I = E,,,- 1 E,,,+J ... E,.,+l-t E,., ... E,.,-l+t . . . (A.6) E,., where Jt = (p 1, •.. , Jt 1) is the conjugate partition to A.. The third "Pieri" formula tells how to multiply a Schur polynomial SA by :a basic Schur polynomial s(lll) = Ilm I: (A.7) the sum over all v whose Young diagram can be obtained from that of A. by adding a total of m boxes to the rows, but with no two boxes in the same :column, i.e., those v = (t•1 , ... , vk) with v, ~ A., ~ v2 ~ A.2 ~ ... ~ ,., ~ A.k ~ 0, . and L v1 = I A.1 + m = d + m. For exanlple, the identity s<2.o·S<2, = s<4·'' + s<3.2) + s(3,1,1) + s(1,2.o .: can be seen from the pictures .:· .. ' · · One can use the Pieri and determinantal formulas to multiply any two ;:Schur polynomials, but there is a more direct formula, which generalizes i Pieri's formula. This Littlewood-Ricltardsmt rule gives a combinatorial formula ·,!_or the coefficients.NA,rv in the expansion of a product as a linear combination :'of Schur polynomtals: \." :j ':?When k is fixed, we orten omit zeros at the end of partitions, so (rrr) denotes the partition il-a.O, ... , 0). A. On Symmetric Functions 456 Here). is a partition of d, p a partition of m, and the sum is over all partitions v of d + 111 (each with at most k par.ts). The Littlewood-Richardson rule says that N11,. is r/re number of ways tire Yormg diagram for ). can be expandttl to tire Young diagram for I' by a strict p-expansion. If ll = (!lr, ... , /'~~ a p-expansion of a Young diagram is obtained by first adding p, 1 boxes, accord· . ing to the above description in Pieri's formula, and putting the integer I in·. each of these 11 1 boxes; then adding similarly !l 2 boxes with a 2, continuini. until finally p,~ boxes are added with the integer k. The expansion is called.' strict if, when the integers in the boxes are listed from right to left, startinj · with the top row and working down, and one looks at the first t entries in this . list (for any t between I and ''• + ·· · + p~), each integer p between I and k- I: occurs at least as many times as the next integer p + I. For example, the equation + s,J,r,r.o + s,2,2,2, + s,2.2,1.u can be seen by listing the strict (2, I)-expansions of the Young diagram ljJ:~ EEEWj ·.-~1 -~ .. [Mac, §1.9]; f~i~1 A proof of the Littlewood-Richardson rule can be found in the other results of this appendix we can get by without using it. Formula (A.7), applied inductively, yields HA = s,A, · s,l,, ·... · s.A., = k KpAs,, ~] 'c' .-.~E (A.9);\ 11 where K,. 1 is tire number of ways one can fill tire boxes of tire Young diagram-~ of I' willr ). 1 t's, A. 2 2's, up to ).1 k's, in such a way that the entries in each ro~~ are norrdecreasing, and those in each column are strictly increasing. Such a};; tableau is called a semistandard tableau Oil f.l of type)., These integers Kpl arej all non-negative, with ·1£] '-~-~~ Ku = 1 and K,. 1 = 0 if). > p, (A.-Iq)~ ·Ar~ i.e., if the first nonvanishing A.,-''' is positive; in addition, KpA = 0 if). hai·~ more nonzero terms than p,. For example, if k = 3, (K,. 1 ) is given by the matril~ §A. I. Basic Symmetric Polynomials and Relations among Them ITIJ I I I 0 I 2 0 0 I 457 -- EP § The integers KP~ are called Kostka numbers. , Exercise A. It. Show that ·'f t~r all i ~ KpA A1 + A2 is nonzero if and only if + .. · +A,~ I'• + 112 + .. · + ll1 l. j:. ' When A= (l, I, ... , 1), Kp(l ..... u is tlte number of standard tableaux on tlte ;illagram qf 11, where a standard tableau is a numbering of the d boxes of a i:)'oung diagram by the integers I through d, increasing in both rows and :Columns. :f · We need one more formula involving Schur polynomials, which comes ! from an identity of Cauchy. Let y 1 , ••• , YA be another set of indeterminates, write P(x) and P(y) for the same polynomial P expressed in terms of ;;variables x 1 , ... , xA and y 1 , ... , YA• respectively. The formula we need is ;lnd :~r~ detl-'--1 = I - x 1yi n~(x)~(y)x (t - __ (A.l2) y1)' 1 1,} ;;The proof is by induction on k. To compute the determinant, first subtract ;;,the first row from each of the other rows, noting that .. , t - - - - - - - - - ~~ - -"'t . __!'j__ l - x 1y1 l - x 1 y1 l - x 1 y1 I - x 1y1 ~.__ i:and factor out common factors. Then subtract the first column from each of rlhe other columns, this time using the equation rr.• ~:: ~; YJ l=-x y1 1 Yt [- x 1y 1 _ Yi- Yt . - ., - x;y; j - X;J'j [to factor out common factors. One is left with a matrix whose first row is 1;(1 0 ... 0), and whose lower right square has the original entries. The formula !Jollows by induction (cf. [Wei, p. 202]). tlY< ('' ?~:. Another form of Cauchy's identity is 0 458 A. On Symmetric Functions (A.I3) the sum over all partitions A. with at most k terms. To prove this, expand the determinant whose i,jentry is (l - x 1y1 = I + x 1y1 + xfyJ +···.One sees that for any I 1 > · · · > /A the coefficient of y\• y~' · ... · J•!• is the determinant lx]'l. By symmetry of the x andy variables we have r' -1 detl-1 1- XIYJ = > ix]•i·iy]'i. 'I' Combining (A.l2) with (A.4) gives (A.l3). (A.I4) 0 Expansion of the left-hand side of(A.l3) gives n (I ~ y) X1 1 = n( I 1 m=O Hm(x)yt) = L H,.(x)M,.(y). .< (A. IS) 1.) Since the polynomials II,_ as well as the MP form a basis for the symmetric polynomials, one can define a bilinear form ( , ) on the space of homo· geneous symmetric polynomials of degree d in k variables, by requiring that (A.l6) where <5,_.,, is I if A. = Jt and 0 otherwise. The basic fact here is that the Schur. polynomials form an orthonormal basis for this pairing: · (A.I7). In particular, this implies that the pairing ( , ) is symmetric. Equation; (A.l7) is easily deduced from the preceding equations, as follows. Write · S1 = I,a,.,II, = I,b,,.M 1 , for some integer matrices a,_, and by 1 • Then (S,_, SP) =I, a,_ 1 bw (A.l8) 1 In order that be equal to I,,H7 (x)M7 (y), which it must by (A.l3) and (A.15), we must have::: This is equivalent to the equation I.,a,_ b 1 1,. =;;,..,,.which by (A.18) implies·:: : -~ (A.J7). Because of this duality, formula (A.9) is equivalent to the equation sp =I Kp,.MA. l -·.--:: (A.I9);; §A.l. Basic Symmetric Polynomials and Relations among Them 459 This gives another formula for these Kostka numbers: KtJ 1 is tl1e coefficient of XA in 8/A, where X 1 = The identities (A.9) and (A.l9) for the basic symmetric polynomials allow us to relate the coefficients of X 1 in any symmetric polynomial P with the coefficients expanding Pas a linear combination of the Schur polynomials. If Pis any homogeneous symmetric polynomial of degree d in k variables, and Ais any partition of d into at most k parts, define numbers t/1 1(P) and w1(P) by x:• ·, .. ·x:k. (A.20) tP1(P) = [P]1. where [P]A denotes the coefficient of X 1 = xt• ·... ·x:k in P, and wA(P) = [A· P] 1, here A= n, <1 I= (J. 1 + k- 1, .l. 2 + k - 2, ... , J.l); (A.21) (x 1 - xl We want to compare these two collections of numbers, as l varies over the partitions. The first numbers t/J1(P) are the coefficients in the expression (A.22) for Pas a linear combination of the monomial symmetric polynomials M 1. The integers w1(P) have a similar interpretation in terms of Schur polynomials: (A.23) . Note from the definition that the coefficient of X1 in A· S1 is I, and that no '"other monomial with strictly decreasing exponents appears in A· S1; from this, , formula (A.23) is evident. In this terminology we may rewrite (A.I9) and ;(A.9) as (A.24) :, and (A.25) ' ~ ~. ; Lemma A.26. For any symmetric polynomial P of degree dink variables, t/11 (P) =I/A KP 1 ·wp(P). !' PRooF. We have } t/11(P)M1 't = P =I wtJ(P)SP = I l,fl p = ~ ( ~ KP w ,(P)) M 1 1 w~(P)KP 1 M1 1, i. and the result follows, since the M1 are independent. D ~'l.':. We want to apply the preceding discussion when the polynomial P is a Jproduct of sums of powers of the variables. Let IJ = x{ + ··· + xl, and for A. On Symmetric Functions 460 i = (i 1 , ... , i.,), ad-tuple of non-negative integers with I,ai. = d, set P(l1 = P/• · P1>· ... · PJ•. These Newto11 or power sum polynomials form a basis for the symmetric functions with rational coefficients, but not with integer coefficients. Let w,(i) = w,(P<''). Equivalently, (A.27) For the proof of Frobenius's formula in Lecture 4 we need a formal lemma about these coefficients w,(l): PROOF. We will use Cauchy's formula (A.IJ). Note that log(n (1- x,y1 )-') = 1,] f !~(x)P1 (y), }=I } so = ~ 1 J'•i,! · ... ·d'•i1 l ~ w,(l)S,(x) ~ w,.(i)S,.(y). Comparing with (A.I3), the conclusion follows. 0 Exercise A.29*. Using the pairing ( , ) of (A.l6), the coefficients wA(i) • w,(P< 11 ) can be written w,(l) = (S,, p<11). (a) Show that the Newton polynomials are orthogonal for this pairing, and (PUI, p<'1) = l'•i 1!211 i2!· ... ·d'•id!. Equivalently, S1 = L z:i) w 1 (i)P''', where the sum is over all partitions i=(i 1 , z(i) = 11 ! 11• · i 2 !2'•· ... · i4 !Ji•. (b) Show that w,(i) ... =I,. (S,, M.) ·(H., pm). ,i1 ) with I,ai.""d, and ·• §A. I. Ba.,ic Symmetric Polynomials and Relations among Them 461 We should remark that we have chosen to write our formulas for a fixed number k of variables, since that often simplifies computations when k is small. It is more usual to require the number of variables to be large, at least as large as the numbers being partitioned-or in the limiting ring with an infinite number of variables, cr. Exercise A.32; the formulas for smaller k are then recovered by setting the variables x 1 = 0 fori> k. For example, if k ~ 2 we haves,~,= s(2) + s(l,l)• which reduces to s,~, = s(2) when k = l. The next two exercises give formulas for the value of the Schur polynomials when the variables x 1 are all set equal to 1; these numbers are the dimensions ofthe corresponding representations. For a formula for s~(l, ... , I) involving hook lengths of the Young diagram of l, see Exercise 6.4. Exercise A.30*. When x 1 = x 1determinants, leading to 1, the numerators in (A.4) are van der Monde (i) Taking the limit as x--> I, one finds (ii) sA(l, ... ,l>=n l i<J 1- l 1 + j- i ... } -I By (A.S) and (A.6) we have also the following two formulas: (iii) (iv) s~c• ..... t) Sl(l, ... ,1) = l''l,+j-11. =I( k. Jl,+J-1 I where I. 111t 1 = 1- - l . - t) <- )I· where(JI 1 , ... ,Jl,) = l'. Exercise A.31*. (a) Show that s,. =I, K,.x·. , the sum over all monomials x• = x'j• · ... · x;•, where, for ally k-tuple a of :non-negative integers, K11 • is the number of ways to number the boxes of the iYoung diagram of Jl with a 1 I 's, a 2 2's, ... , ak k's, with nondecreasing rows ,:ind strictly increasing columns. In particular, the right-hand side is a symmetric polynomial, a fact which is not obvious from the definition. ; ·. (b) Deduce that S,.(l, ... , I) is the number of ways to number the boxes of ''lhe Young diagram of Jl with integers from I to k, with nondecreasing rows )nd strictly increasing columns (i.e., the number of semistandard tableaux). :i \Exercise A.32*. The idea of considering symmetric polynomials in an arbiflrarily large number of variables can be formalized by working in the ring 1.1\,. I~ A(k), where A(k) denotes the ring of symmetric polynomials in k (variables. Then A. On Symmetric Functions 462 A= l[H 1 , 111 , •.. •.• ] = l[E 1 , ..• E1 , ••• ] is a graded polynomial ring, with 111 and E1 of degree i. A ring homomorphism .9: A -+ A can be defined by requiring •9(E1) = 1-11 for all i . (i) Show that ,9 is an involution: 92 = .9. Equivalently, 9(111) = E1• (ii) If A: is the conjugate partition to A., show that 9(Sl) (iii) If~ = =sA·· x{ + ·· · + x~ is the jth power sum, show that 9(~) = ( -1)}-1 ~- (iv) Deduce the formula (v) Deduce a dual form of(A.7): S4 ·S0 •...• 11 = S4 ·E,. =Is.. the sum over all partitions 1t whose Young diagram can be obtained from that of A. by adding 111 boxes, with no two in any row. (vi) Show that where the sums are over all i = (1 1 , z(i) = 11 ! 11' • i 2 !21' • ••• • Vd 1•. Note that •.. , id) with LIXi. = d, and I (-t)'I-I1t 1=(I E~~~r•. §A.2. Proofs of the Determinantal Identities To prove the Jacobi-Trudi identity (A.S), note the identities xf- E 1 xr• + E2 xr 1 - ••• + (-l)1 E1 xr 1 = 0, (AJJ)_ for any l ~ j ~ k, p ~ k. And for any 0 :-;:;; m < k and p ~ k, H,_,.- E1 1-1,_,_ 1 + E2H,_,_l + ··· + (-l)1 E1 H,_,-1 = 0. (AJ4) Both of these follow immediately from the defining power series for the E1 and. 111. Since these two recursion relations are the same, there are universal polynomials A(p, q) in the variables E 1 , ... , E1 such that §A.2. Proofs of the Determinant a! Identities xf 1/11 _, 463 = A(p, l)xt 1 + A(p, 2)x;- 2 + ·· · + A(p, k), = A(p, l)Ht_,_ 1 + A(p, 2)1/t-... - 2 + · · · + A(p, k)IL,.. For any integers ..l 1 , ••• , (A.35) ..lt this leads to matrix identities (xf•+k-1) 11 = (A(..l1 + k - i, r)) 1, • (xt'),1, (HA,+J-i)IJ = (A(..l 1 + k- i, r)),,.(H1_,),1, (A.36) . where ( )pq denotes the k x·k matri.x whose p, q entry is specified between the parentheses. The relations (A.34) ; Jso imply: Lemma (A.37). Tile matrices (Hq-p) and (( -l)q-"Eq-,) are lower-triang11lar matrices with l's along the diagonal, and are Inverses of each otlter. · The identities (A.36) therefore combine to give (X}A,+t-1)I }_- (HA1+p-l )lp . (( _ l)q- "Eq-p )pq . (Xjt-q)qj (A.38) Taking determinants gives (A.5), since the determinant of the matrix in the middle is I. Exercise AJ ~·. Prove the identity n k lx]'l· (I - xjrl J=J =I lxt'l. the sum over all k-tuples (m 1 , ••• , mt) of non-negative integers with m 1 :2': 11 > m2 ~ • • • > mt :2': 1., and deduce Pieri's formula (A.7). To complete the proofs of the assertions in §A.I, we show that the two . determinants appearing in the Giambelli formulas (A.5) and (A.6) are equal, ·. i.e., if ..l = (..l 1 , ••. , ..lt) and p = (p 1 , ••• , p 1) are conjugate partitions, then (A.40) :: Here the H1 and E1 can be any elements (in a commutative ring) satisfying the : identity H1t 1)· -1)1E1t 1) = I, with H0 = E 0 = I and H, = E, = o for :~. f < 0. To prove it, we need a combinatorial characterization of the conjugacy of partitions: x.:· ~-Exercise A.4t •. For ..l = (..l 1 , ;'. show that the sets {..l1 + n + I - i: I ••• , ~ ..lt) and p = (J! 1 , i ~ k} and :) form a disjoint union of the set {1, ... , k t:. {n ... , 111) conjugate partitions, + j - 11/ I ~j ~I} + 1}. ~l· We also need a basic matrix identity which relates minors of a matrix to = (ali) is ar~ r. x r. matrix, ~;and S = (s 1 , ... , St)and T = (t 1 , ••. , tl)are two sequences of k dtstmct mtegers t~; minors of its inverse (or matrix of cofactors). If A 464 A. On Symmetric Functions from {I, ... , r}, let As. r denote the corresponding minor: As. r is the deter· minant of the k k matrix whose i,j entry is x a.,,,. Lemma A.42. Let A and B be r x r matrices whose product is a scalar matrix c · 1,. Let (S, S') arad (T, T') be pernwtatio11s of the seq11ence (1, ... , r), where S and T consists of k integers, S' and T' of r - k. Then c•-• ·As. r = & • det(A) · Br·.s·· where ll is the product of the signs of tlae two permutations. PROOF. By permuting the rows and columns of A, multiplying on the left and right by permutation matrices P and Q corresponding to the two permutations of (I, ... , r), we may take the (S, T) minor to the upper left corner: 1 PAQ = (A A A2) A", 3 As.r=detA 1• Then Q-t Br• = (B•B 882). 3 Br·,s· = det 8 4 • 4 Now taking determinants in the identity = gives the equation det(PAQ)· det(B4 ) det(A 1 )· c•-•. Since ll is the product of the determinants of P and Q, the lemma follows. 0 PROOF OF(A.40). Apply the lemma to A = (Hq_,) and B = (( -1r"Eq_,), with r = k + 1, and + k, l 2 + k- I, ... , ..l, + 1), (k + 1 - p 1 , k + 2 - p 2 , ... , k + I- 11 1), S = (..l 1 S' = T= (k, k- 1, ... , 1), T' = (k + 1, k + 2, ... , k +I). Then Similarly, Br·.s· =I( -1)"'+1-JE,.,+I-Ji = (-1JL'"ril( -J)Lii£,.,+1-JI = ( -l) 4 1E,.,+t-JI• with d = LIIJ = L,..t,. Since e = ( -1)4 , (A.40) follows. 0 · §AJ. Other Determinantal ldentilies 465 : §A.3. Other Determinantal Identities '. · ·. ; In this final section we prove some variations of these formulas which are useful for calculating characters of symplectic and orthogonal groups. We want to compare minors, not of H = (1/1_ 1) and E = (( -l)i-iE1_ 1), but of matrices H+ and E- constructed from them by the following procedures: For an r x r matrix II = (H1•1), and a fixed integer k between I and r, H+ ' denotes the r x r matrix obtained from H by folding H along the kth column, ·'and adding each column to the right of the kth column to the column the ·_-'same distance to the left. That is, H+ _ {H1,1 + H1,2k-J I,J - H,,J ifj < k ifj ;;;:: k · (with the convention that H,,, = 0 if p or q is not between I and r). The matrix E- is obtained by folding E along its kth row, and subtracting rows ·above this row from those below: E- - {EI,J- Elk-1.} I,J- EI,J ifi > k ifi ~ k. Lemma A.43. ~f H and E are lower-triangular matl·ices with I's along tile diagonal, tlrat ar·e inverse to each other, then the same is true for u+ and E-. PRooF. This is a straightforward calculation: the i, j entry of the matrix H+ · E- is l-1 L (H1,, + H ,., 1• 2 ,. = r t_,)E,,J + H1,tEt,J + p•l+l L k-l H1,,.(E,.1 - En-,., 1) ,. I HI,,.E,,J + ,.. I H,,2k_,E,,J- q•l+l I Hl.qEll-q,j• p•l 'The first sum is ~1 • 1 , and the others cancel term by term. Proposition A.44. Let .l. = (..1. 1 , ... , At) arrd Jl = (Jl 1, ... , J1 1) be 0 conjugate parti- tions. Set and E; = E1 fori S. I, Theil tile determinant of the k (HA,-1+1 HA,-1+2 + HA,-1 >< k matrix wltose ith row is HA,-1+3 + HA,-1-1 . .. HA,-I+t + HA,-1-H2) Is eqrlal to tile deterntillarlt of the I x I matrix wltose ith row is (E'p,-1+1 E~,-1+2 + E~,-~ E~,-i+3 + E~,-1-1 ... · Each of these determinants is equal to the determinant IE,.,-i+j- E,.,-1-JI E;,,-1+, + E~,-1-1+2). A. On Symmetric Functions ;~ 466 :;~ :~ ,, altcl to tile determinallt where If;' = H1 fori --~ ~ l, and fori~ 2 Ill' = H, + 11,_2 +. H,_4 + ... + {HI I if i is etJen. if i is odd .:~ .;)I Witl1 H = (H1_ 1)and E = (( -l)q-"Eq_,) we can apply the basiclemmaJ (A.42) to the new matrices A = n+ and B = E-, and the same permutations~~ (S, S') and (T, T') used in the proof of (A.40). This time ;~ PROOF. ,.:'!j: -~ ~i and if j = 2, ...• k if j = I. Similarly, with As before, Lemma A.42 implies that the determinant of the first displayed ·:· matrix of the proposition is equal to that of the third. Noting that l' .., .!-; one can do elementary column operations on the third matrix, subtracting:;' the first column from the third, then the second by the fourth, etc., to see that -, the second and third determinants are equal. Since l-11 = II;' - H;'.. 2 , the same .· argument shows the equality of the first and fourth determinants. 0 Note that in these four formulas, as in the determinantal formulas for Schur . polynomials, if a partition has p nonzero terms, only the upper left r x p ·• subdeterminant needs to be calculated. We denote by Sthe determinant of the proposition: s = IIIA,-1+, HJ.,-1+2 + HA,-• .. . HA,-IH Dually, setH;= If,- lf1_ 2 and E~ = E1 + E1_ 2 + Hl,-f-l+21· +E 1_ 4 + ···. Corollary A.46. Tlte following determinants are equal: (i) IHJ.,-1+1 (ii) IE,,,-f+t HJ.,-1+2 + HJ.,-i Ep,-i+2 + Ep,-1 HJ.,-iH + HJ.,-I-k+2'· E,,,-1+ 1 + E,,,-•-1+21, (A.4S) > 467 §AJ. Other Delerminantal Identities (iii) (iv) Define ~ 41 to be the determinant of this corollary: HJ.,- 1+ 2 + HJ.,- 1 ~AI = IHJ.,-f+l HJ.,-rH + H1,-Hd· (A.47) •. • Exercise A.48•. Let A be the ring of symmetric polynomials, 9: A-+ A the involution of Exercise A.32. Show that 9(S(l)) = ~PI when land J1 are conjugate partitions. For applications to symplectic and orthogonal characters we need to specialize the variables x 1, ••. , Xt· First (for the symplectic group Sp 2 .} take k = 2n, let z 1 , •.• , z. be independent variables, and specialize Set -'J = H1(z 1 , ••• , z., z(I, ... , z;;- 1 ) in the field Q(z 1 , .•. , (A.49) z.) of rational functions. Proposition A.SO. Given illtegers l 1 ;;:.; • • • ;;:>; lzJr+n-1+1 _ 1 n-1+1 Zj where J4 denotes then x 11 - l. ;;:>; 0, we have ZJ-(A 1+n-l+l)l -(n Zj 1+1)1 = IJll• matrix whose ith row is From Proposition A.44 we obtain three other formulas for the right-hand side, e.g., (A.51) where e1 = E1(z 1 , ••• , z., z( 1 , ••• , z;;- 1 ), and Jl is the conjugate partition to l Exercise A.52. Calculate the denominator of the left-hand side: lz;-l+t- ZJ-(.-1+1)1 = 6(e., ... , e.)·(.· ... ·(., Where ej = Z1 + Zl-l and (j PROOF OF PROPOSITION = Zj - z1-l. A.50. Set (A.53) 468 A. On Symmetric Funclions By the same argument that proved the Jacobi-Trudy formula (A.5) via (A.38~ the proposition follows from the following lemma: Lemma A.54. For 1 ~ j ~ 11 and arry Integer I~ 0, {1(/) is tlte prod11ct of tht : I x n, n x n, and 11 x 1 matrices zJ and zT'· and subtracting gives PRooF. From (A.37) we can calculate 2n CJ(/) = L (A.SS) ·~ Jl-2n+,s,, pz( where s, =I~~,(- l)q-"eq_,(j(2rl - q). Multiplying (A.33) by z1-, and sub-. tracting we find CiP)- e• C1(p- I)+···+ ( -1)"-•e,C1(1) = (-1)"+ 1e,+I(j(l) + (- t)"+ 2e,+ 2CP) + ... + e 2.(1(2n- p). (A.56) :; ''J Note also that ( -1)"e, = ( -1) 2 "-"e 2.-,, L( since -l)"e,t" = and (A.57) follows n(1 - z,t)(1 - zj 1 1) = (A.S7) . n(l - e,t + Szn-, = s, = '•-p+l, =I:=,( where r, -lr"eq_,(1(n eludes the proof. +I- t 2 ). From (A.S6) (A.S8) ' q). Combining (A.55) and (A.58) con• 0 Next (for the odd orthogonal groups 0 2•+l) let k = 2n + l, and specialize" the variables x 1 , ••• , x 2• as above, and x 2 -+ 1 HI. We introduce variables z}12 12 , square roots of the variables just considered, and we work in the : and field Q(zl 12 , ... , z!12 ). Set zt K1 = //j(z 1, ... , z., z! 1, ... , z; 1, I) (A:59) where 1/1 is the jth complete symmetric polynomial in 2n Proposition A.60. Given integers ..t, ~ 1 · · · ~ ..t. ~ 0, we have z-0 1+n-i+lf2lj lz~•+•-1+1/2 - 1 lz} 1+1/2 _ zJ + I variables. (n 1+1/llj -IK I A• §A.J. Other Determinantalldentilies where K A is tile n x 11 matrix whose ith row is KA,-1+2 (KA,-1+1 469 + KA,-1 ··· KA,-i+n + KA,-I-n+2)• Corollary A.46 gives three alternative expressions for this determinant, e.g., IKAI wherelr1 = H1(z 1 , •.. , = lltA,-1+] -lrA,-1-j,, (A.61) z.,zl 1, ••• , z;;- 1, 1). Exercise A.62. Calculate the denominator of the left-hand side: lz;-l+l/2 - ZJ-(n-l+l/2)1 =~(e PROOF OF PROPOSITION •.... , e.>·c.w· ... ·C.(~). zJ - z1- 1 and .;1(1) = zj + zT' in A.60. We have ' 1(1) = Q(z: 12 , ... , z~ 12 ) for l an integer or a half integer. First note that ej·,il) = 'P + ~> + 'J. Multiplying the numerator and denominator of the left-hand side of the statement of the proposition by I the numerator becomes "j(..t1 + 11 - i + I) + 'j(..t1 + 11 - i)j, and the denominator becomes ICj(n- i + l) + ' 1(11- i)l = ie1(n- i + l)j. We can, therefore, apply Lemma A.54 to calculate the ratio, getting the determinant of a matrix whose entries are sums of certain .T/s. Note that by direct calculation K1 = J1 + ~-I> so the tenns can be combined, and the ratio is the determinant of the displayed matrix KA. 0 e m..... <. Finally (for the even orthogonal groups 0 2 .), let k = 2n, and specialize the variables x 1 , .•• , x 2 • as above. Set L1 = HJ(z 1 , ••. , z., z1\ ... , z;;- 1) = H1(z 1 , ••• , z., zjl, ... , z;;- 1 ) -/11_2(z 1, ... , z., z! 1, ... , z;;- 1), (A.63) wilh H1 the complete symmetric polynomial in 211 variables. Proposition A.64. Given integers ..t 1 ;:::.: lzf•+•-~ + zj 1A,+n-lll lzj-' + zl (n-Il I where LA is the 11 x 11 (LA,-1+1 ~ 0, we lwve = {tiLAI if ..t. > 0 if ..t. = 0, • • ·;:::.: A.. ILAI matrix wllose ith row is LA,-1+2 + LJ.,-i •· • LJ.,-I+n + LA,-1-n+2). As before, there are other expressions for these determinants, e.g., ILAI where lr1 = H1(z 1 , . .• , = '''A,-I+J- ''A,-1-}'· z., zjl, ... , z;;- 1 ). (A.65) 470 A. On Symmetric Functions Exercise A.66. Calculate the denominator of the left-hand side: lzr• + z;•·-··, = 2. A(~ I' ... '~.). PROOF OF PROPOSITION A.64. Note that C 1 · ~1 (1) = ~j(l + l)- ~1 (1- 1). Multiplying the numerator and denominator by C1 • ••• ·C., the numerator becomes ICP-• + n- i + 1)- C1(A. 1 + n- i - l)! and the denominator becomes 1{1(n- i + l)- C1(n- i - 1)1 = 21C1(n - i + 1)1; this is seen by noting that the bottom row of the matrix on the left is (C1(1)- Cj( -I))= (2C1(1)), and performing row reductions starting from the bottom row. The rest of the proof is the same as in the preceding proposition. The only change is when -t. = 0, in which case the bottom row in the numerator 0 matrix is the same as that in the denominator. Exercise A.67*. Find a similar formula for lzf'+n-l - zJ-tA,h-l)l lz) • + zl t• 'I APPENDIX B On Multilinear Algebra In this appendix we state the basic facts about tensor products and exterior and symmetric powers that are used in the text. It is hoped that a reader with some linear algebra background can fill in details or the proofs. §8.1: Tensor product §8.2: Exterior and symmetric powers §8.3: Duals and contractions §B.l. Tensor Products The tensor product of two vector spaces V and W over a field is a vector space V ® W equipped with a bilinear map VxW-+V®W, V X WI-+V@ W, .which is universal: for any bilinear map fJ: V x W--+ U to a vector space U, there is a unique linear map from V ® W to U that takes v ® w to fl(v, w). This universal property determines the tensor product up to canonical isomorphism. If the ground field K needs to be mentioned, the tensor product is denoted V ®x W. , If {e1 } and {./j} are bases for V and W, the elements {e1 ® .~} form a basis for V ® W. This can be used to construct V ® MZ The construction is functorial: linear maps V--+ V' and W--+ W' determine a linear map from V® Wto V' ® W'. Similarly one has the tensor product V1 ® · · · ® V, of n vector spaces, with its universal multilinear map V1 x · · · x v, __. V1 ® · · · ® v•. B. On Multilinear Algebra 472 taking v1 x · · · x P. to v1 ® · · · ® v•. (Recall that a map from the Cartesian product to a vector space U is multilinear if, when all but one of the factors V, are fixed, the resulting map from V, to U is linear.) The construction of tensor products is commutative: V®W~W®V, v® w~--+w® v; distributive: and associative: ( U ® V) ® W ~ U ® (V ® W) ~ U ® V ® W, by (u® v)® w~--+tl®(v® w)1-+11® v® w. In particular, there are tett.~or powers ve• = V ® · · · ® V of a fixed space V. By convention, yeo is the ground field. If A is an algebra over the ground field, and Vis a right A-module, and W a left A-module, there is a tensor product denoted V ®,.. W, which can be constructed as the quotient of V ® W by the subspace generated by all (v ·a)® w - v ®(a· w) for all v e V, we W, and a e A. The resulting map from V x W to V ®,.. W is universal for bilinear maps fJ from V x W to vector spaces U that satisfy the property that {1(11 ·a, w) = {J(v, a· w). This tensor product is also distributive. §B.2. Exterior and Symmetric Powers The exterior powers NV of a vector space V, sometimes denoted Alt"V, come equipped with an alternating multilinear map vX ••• X v-+ NV, that is universal: for {J: V x · · · x V-+ U an alternating multilinear map, there is a uniquelinearmapfrom/\"Vto U which takesv 1 A··· A v.to{J(v 1, ... ,v.~ Recall that a multilinear map fJ is altemating if fJ(v 1 , ... , v.) = 0 whenever two of the vectors v, are equal. This implies that fJ(v 1 , ••• , v.) changes sign when two of the vectors are interchanged.' It follows that /J(v,(ll• ... , Va(n)) = sgn(u)/J(vt, ... , V 0) for ali U E6 0• The exterior power can be constructed as the quotient space of yen by the subspace generated by all v 1 ® · · · ® v. with two of the vectors equal. We let n: ve"-+1\"V, 1 This follows from the standard polarization.: for two factors. Plw, w) = p(v, w) + p(w, v). P<•• + w, v + w)- P{t•, v)- §8.2 Exterior and Symmetric Powers 473 denote the projection. If {e1} is a basis for V, then {e 1, A e1, e1.: i 1 < i 2 < · · · < t.} A··· A is a basis for 1\"V. Define I\ to be the ground field. If V and W are vector spaces, there is a canonical linear map from NV® ~W to NH(V (9 W), which takes (v 1 A • · • A v.) ® (w 1 A · · · A wb) to v1 A • • · A v. A w1 A • · · A wb. This determines an isomorphism 0V EB" NV® 1\"-·w 1\"(V EB W);:;:;; (B. I) ... o (From this isomorphism the assertion about bases of 1\"V follows by induction on the dimension.) The symmetric powers Sym"V, sometimes denoted S"V, comes with a universal symmetric multilinear map V x ··· x V-+Sym"V, Recall that a multilinear map {J: V x · · · x V-+ U is symmetric if it is unchanged when any two factors are interchanged, or {J(v,.111 , ••• , v.,1• 1) = {J(v 1, ••• , v.) for all q e s .. The symmetric power can be constructed as the quotient space of V 8 " by the subspace generated by all v1 ® · · · ® v.- v.,111 ® · · · ® v. 1• 1, or by those in which q permutes two successive factors. Again we let denote the projection. If {e.} is a basis for V, then {e1, ·e1, • ... ·e1.: i 1 ~ i 2 ~ ... :-;;;; i.} is a basis for Sym"V. So Sym"V can be regarded as the space of homogeneous polynomials of degree n in the variables e1• Define Sym 0 V to be the ground freld. As before, there are canonical isomorphisms n EB Sym•v ® Sym•-•w Sym"(V (9 W) ~ (8.2) .~o The exterior powers 1\"Vand symmetric powers Sym"V can also be realized as subspaces of V 8 ", assuming, as we have throughout, that the ground field has characteristic 0. We will denote the inclusions by 1, so we have The imbedding 1: 1\"V-+ V 8 " is defined by 1(v 1 A··· A v.) = L ffe sgn(q)v.,111 ® .. · ® v., 1• 1• (B.3) e" (This is well defined since the right-hand side is alternating.) The image of r is the space of anti-invariants of the right action of s. on 1' 8 ": B. On Multilinear Algebra 474 (l't ® · · · ® v.)' (1 = Va(ll ® · ·' ® Va(n)• Vr E V, (1 E 6.. (8.4) (The anti-invariants are the vectors z e v®• such that z · q = sgn(q)z for all (1 E s .. ) Moreover, if A = I 0 n, then (1/n!)A is the projection onto this anli· invariant subspace. 2 (Often the coefficient Jfn! is put in front of the formula for r; this makes no essential difference, but leads to awkward formulas for contractions.) Similarly we have 1: Sym"V-+ v®• by r(v, ..... v.) = L ae ''a( I)® .•. ® Va(n)' (8.5) e" The image of 1 is the space of invariants of the right action of 6. on v®•. If A= 1 on, then (1/rr!)A is the projection onto this invariant subspace. The wedge product A determines a product (8.6) (v 1 " .. • " v,.) ® (v,.+t A .. • " v.,+.)H v1 " .. • " v,. " v,.+t " · .. "v,.+•• which is associative and skew-commutative. This product is compatible wilh the projection from the tensor powers onto the exterior powers, but care must be taken for the inclusion of exterior in tensor powers, since for example v " w is sent to v ® w - w ® v [not to !(v ® w- w ® v)] by r. In general, the diagram 1\'"V ® 1\"V " --+ (8.7) V®'" ® v®• --+ V®lm+n> commutes when the bottom horizontal map is defined by the formula (v 1 ® · · · ® v,.) ® (v,.+t ® · · · ® v,.+.) H L sgn(q)va(l) ® ... ® Va(m) ® v.,(m+l) ® ... ® Va(m+n)• (8.8) the sum over all "shuffies," i.e., permutations q of {1, ... , m + 11} that preserve the order of the subsets {1, ... , m} and {m + 1, ... , rn + 11}. Similarly the symmetric powers have a commutative product (v 1 · .•• • v.,) ® (v.,+t · ... · v,.+.) H v1 · ... · v,. · v,.+t · ... · v,.+•• with a similar compatibility. Note that v2 e Sym 2 Vis sent to 2v ® v in V ® V, v" e Sym"V to n!(v ® · · · ® v) in V®•, and generally one has the analogue of (B. 7), changing each "sgn(q)" to "1" in formula (B.8). All these mappings are compatible with linear maps of vector spaces V-+ W, and in particular commute with the left actions of the general linear group GL(V) = Aut(V) of automorphisms, or the algebra End(V)-= Hom(V, V) of endomorphisms, on V®•, 1\"V, and Sym"V. 2 It i~ this factor which limits our present discussion to vector spaces over fields of characterislitO. 475 §BJ. Duals and Contractions It is sometimes convenient to make algebras out of the direct sum of all of the tensor, exterior, or symmetric powers. The tensor algebra T"V is the sum EBn:1:o V 0 ", with product determined by the canonical isomorphism V0 " ® V 0 m-+ V0 1•+m>. The exterior algebra 1\"V is the sum Efj.;, 0 NV, which is the quotient of by the two-sided ideal generated by all v® v in V02 . The symmetric algebra Sym"V is the sum EB.:!: 0 Sym"V, which is the quotient of TV by the two-sided ideal generated by all v ® w - w ® v in V02 . rv Exercise B.9. The algebra Sym·v is a commutative, graded algebra, which satisfies the universal property that any linear map from V to the first graded piece C 1 of a commutative graded algebra C determines a homomorphism Sym"V-+ C of graded algebras. Use this to show that Sym"(V E9 W) ~ Sym·v ® Sym·w, and deduce the isomorphism (8.2). Prove the analogous assertions for 1\"V, in the category of skew-commutative graded algebras. In particular, construct an isomorphism 1\"(V E9 W) ~ 1\"V ® 1\"W, where® denotes the skew-commutative tensor product: it is the usual tensor product additively, but the product has (a® b)· (c ®d)= (- t)d•l(bld••l §B.3. Duals and Contractions Although only a few simple contractions are used in the lectures, and most of these are written out by hand where needed, it may be useful to see the general picture. If v• denotes the dual space to V, there are contraction maps cJ: V®P ® (V*)@q-+ v@(p-1) ® (V*)®(q-1), for any 1 :s;; i ~ p and I :s;; j s q, determined by evaluating the jth coordinate of(V*)®q on the ith coordinate of V 0 ": cj(v 1 ® · · · ® v, ® q1 1 ® · · · ® ({'q) (8.10) = ({'1(v 1)v 1 ® · · · ® (J• ® · · · ® v,. ® ({' 1 ® · · ·® r/J1 ® · · · ® (/Jq· More generally if I= (i., ... , i.) and J = (j 1 , ... , j.) are two sequences of n distinct indices from {1, ... , p} and {1, ... , q}, respectively, there is a contraction d: ve, ® (V*)®q-+ V®lp-•1 ® (V*)®(q-•1 (B.Il) which takes • :.,n v 1 A .. • A v,. ® q1 1 ® (/Jj.(v,.)tJ1 .. · ® (/Jq to ® ... ® 6,, ® ... ® 6,, ® ... ® v, ® (/11 ® ... ® rPJ, ® ... ® (/Jq· For example, if p = q = n and I = J =(I, ... , n), this contraction Jl®• ® (V*) 0 "-+ C identifies ( V*)0 " with the dual space of V 0 ". B. On Multilinear Algebra 476 Now (V 0 ")* consists or n-multilinear rorms on V, and (1\"V)* consists or alternating II multilinear rorms on V; in particular, (1\"V)* is a subspace or (V 0 ")*; this is the inclusion via n*. The composite 1\"(V*)-+ (V*)0 "-+ (V 0 ")*, where the first map is the inclusion 1 and the second is the isomorphism or the preceding paragraph, maps 1\"(V*) isomorphically onto the subspace (1\"V)*. Explicitly, /\"( V*) .! (/\" V)*,

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