REPRESENTATIONS OF
Finitt and Lie Groups
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Charles B Thomas University of Cambridge, UK
Imperial College Press
Published by Imperial College Press 57 Shelton Street Covent Garden London WC2H 9HE Distributed by World Scientific Publishing Co. Re. Ltd.
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REPRESENTATIONS OF FINITE AND LIE GROUPS Copyright 0 2004 by Imperial College Press All rights reserved. This book, or parts thereox may not be reproduced in any form or by any means, electronic or mechanical, includingphotocopying, recording or any information storage and retrieval system now known or to be invented, without written permissionfrom the Publisher.
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ISBN 1-86094-482-5 ISBN 1-86094-484-1 (pbk)
Printed in Singapore by World Scientific Printers (S)Pte Ltd
C.B. Thomas In memory of Ali Frohlich (1916-2001)
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Preface
Good grief, not another book on representation theory! A cursory inspection of the small, if select, library at the Max-Planck-Institut in Bonn yields at least eight good introductory texts. These include the elegant book by J.P. Serre [J-P. Serre], against which all others should be judged. Beyond that the choice is perhaps a matter of taste - what particular slant does the author give to the subject, has she or he any special concerns? The approach chosen here is to present the elementary representation theory of finite groups in characteristic zero in a way which generalises immediately to compact topological groups. The only fresh ingredient needed is an invariant integral, which replaces taking the average by means of the sum over the group elements divided by its order. The parallel development is summarised at the end of Chapter 6 ; with finite groups as a special case of compact groups there is an inner product on the space of class functions under which the irreducible characters form a normal orthogonal set spanning a dense subspace. Two other topics receive special attention, exterior powers and the finite algebraic groups SL2(Pp). I have long believed that the A-structure of the representation ring R ( G ) is a much under-used tool. Some indication of this is given in the exercises devoted to the symmetric groups S,, but the applications are much wider, extending not only to the various families of simple groups of Lie type, but also to the 26 sporadic groups. As a topologist I have long been interested in SLz(lFp),and Chapter 8 is intended to illustrate the general principle that in characteristic zero the representation theory of a finite algebraic group has the flavour of the theory for the corresponding group defined over R or C. In contrast in the natural characteristic p the model is that of a maximal compact subgroup in the complexification.
vii
viii
Preface
The exercises are an important part of the text, and should be attempted, not just for their own sake, but also because in a few cases the results are used in a later chapter. The book concludes with an uneven collection of hints, worked solutions and additional references. The bibliography is short and contains no more than the rival books, which I have consulted, and references to theorems mentioned in the text but not proved. The starred sections (*) may be omitted at a first reading. The book has grown out of various sets of notes for a course of 16 or 24 lectures at the senior year level at Cambridge. My thanks are due to the generations of students who have attended, and interrupted, these lectures and to those who I have individually supervised. Their comments are a reminder of what a privilege it is to work in a great university. Errors inevitably remain, and are solely my responsibility. I wrote the final version during sabbatical leave from Cambridge at the University of California at Santa Cruz, Stanford University and the MaxPlanck-Institut in Bonn. I am grateful to all three institutions for their hospitality and support. I also thank Laurent Chaminade and Gabriella Frescura at Imperial College Press for their help, and most of all Michele Bailey for typing and producing the camera-ready text. Bonn, Michaelmas 2003
Contents
Preface
vii
1. Introduction
1
2. Basic Representation Theory - I
11
3. Basic Representation Theory - I1
25
.... ..... .........
30 31
4. Induced Representations and their Characters
35
5. Multilinear Algebra
47
Representations of GI x GZ * Real Representations . . .
Alternating and Symmetric Products . . . . . . . The Representation Ring R(G) and its A-structure * Representations of SZz(F,) in Characteristic p .
6. Representations of Compact Groups Induced Representations . . . . . Irreducible Representations of SU2
52 55 57 63
.., ..... ...... . .. ..... . .......... .
7. Lie Groups
67 69 75
Representations of the Lie algebra
8. SLz(IW)
.....
,
... ......
84
89 ix
*
.
Contents
Principal Series for SL2(IF,) . . . . . . . . . . . . . . . . . . . . Discrete Series for SL2(P,) . . . . . . . . . . . . . . . . . . . . . The Non-compact Lie Group SL2(R) . . . . . . . . . . . . . . Principal Series . . . . . . . . . . . . . . . . . . . . . . . . . . . Discrete Series . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
91 92 95 97 97
Appendix A Integration over Topological Groups
101
Appendix B Rings with Minimal Condition
107
Appendix C Modular Representations
115
Solutions and Hints for the Exercises
121
Bibliography
143
Index
145
Chapter 1
Introduction
Our topic is the representation theory of finite and, more generally, of compact topological groups . The latter will be defined formally later; for the moment the reader can think of a topological group as a set carrying both a topology and a group structure, which are compatible in the sense that multiplication and inversion are continuous. Examples are SL2 (R)(noncompact) and the special unitary groups Sun (compact), both of which are important in theoretical physics. A representation of G is a homomorphism of G into Autc(V), the group of linear automorphisms of a finite dimensional vector space over the complex number @. By choosing a basis {el . . . en} of V such a representation determines a homomorphism
p :G
-
GLn(@).
If G carries a topology, we give GLn(C)its topology as an open subset of C n 2 ,and require the homomorphism p to be continuous.
Examples. (i) Let CF and let
{ a : a‘ = 1) be a cyclic group of order T generated by a , some primitive r t h root of unity. The homomorphism aq : C, .+ Ul C @* = C - (0) which maps a to CQ is a 1-dimensional representation of the group. Note that aq is injective (we say that aq is ‘faithful’) if and only the greatest common divisor (gcd = ( r ,4 ) ) of r and q equals 1. We will see later that every representation of a finite abelian group A is built up from ‘irreducible’ representations of this kind. (ii) Let G = Qs, the quaternion group with presentation =
C be
{ a , b : a4
=
1, a 2 = b2, b-lab = a-I}. 1
2
Representations of Finite and Lie Groups
Such a 'presentation' can be regarded as a contracted multiplication table in that it tells us that each element can be written as a product aiP, that there are eight such distinct products, and that they can be multiplied using the rule ba-' = ab repeatedly. [Exercise- Write out the multiplication table, and check that it corresponds to that of the basic quaternions {fl, f i , fj,fk} under the rule a H i , b j.] The map p
-
a-
(i"i)
b-
(yi')
extends to a homomorphism of Q8 into SU2. If W denotes the algebra of quaternions, note that W is 2-dimensional over C and that SU2 may be identified with the quaternions of unit length, using the representation just defined. The group Qs also has 1dimensional representations, obtained by composing the projection homomorphism
having Ker(?r) equal to the subgroup generated by a', with any one of the four 1-dimensional representations mapping a, b to f l . We label these as l,a,pand
[email protected] the multiplication table we see that Q8 has five conjugacy classes of elements (1) (a2)(a, a - l ) ( b , b-l)(ab, a-lb).
Taking the trace of the representating matrices, and noting that the trace is constant on conjugacy classes, we obtain the following table
We ask the reader to check three things about this square array. First and most importantly, each row can be associated unambiguously with one of the representations described. Secondly, if we add the first four entries in each column to twice the fifth we obtain 8 (equals the order 1Q81) for column one and zero otherwise. Thirdly, even though the matrices describing p are
Chapter 1: Introduction
3
complex, the entries in the table are real. We shall see later that these are special cases of important general phenomena. Before leaving Q g let us present the generalised quaternion group of order 4t, which will be useful later, Q4t = { a , b : u~~=
l , a t = b2, b-lab = U-'}.
In terms of the quaternion algebra W a can be identified with eKiltand b with j . Given a homorphism of G into Aut@(V)we can think of G as acting on the vector space V via the map g.v = p ( g ) v for all v E V . More directly we can define such an action as a (continuous) map
G x V -V (g,v) - g . v
satisfying g l ( g 2 o ) = ( 9 1 9 2 ) ~and 1 . v = w for all g1,g2 E G and o E V . At least when G is finite, the G-action and C-action (scalar product) on V can be combined as a C[G]-action, where @[GIdenotes the so-called group algebra of the finite group G. Definition. The ring C[G] consists of formal linear combinations
It is a straightforward and tedious exercise to check that, with these definitions of addition and multiplication, @[GI is a ring, which is commutative if and only if G is an abelian group. Although we are primarily interested in the complex group algebra, it is important to note that the same definition holds with A equal to any commutative ring rather than A = C. We then obtain the group ring A[G], which has been much-studied both by topologists and number theorists. Examples. (i) Write out the multiplication table for the group ring
lF2[C2], where IF2 is the finite field with 2 elements, and Cz is a cyclic group of order 2.
Representations of Finite and Lie Groups
4
(ii) If Cg is a cyclic group of order p ( p = prime), and [ is a primitive show that the map root of unity (for example e extends to a homomorphism of rings Z[Cp] -+ Z(<)L) Q(<), and identify its kernel.
<
The discussion above shows that, for finite groups G, complex representation theory is equivalent to the theory of the structure of finitelygenerated C[G]-modules. One of our first results will be to show that C[G] belongs to a very special class of semisimple rings, and, although we shall not adopt this approach, the whole of classical representation theory can be read as a special case of that of semisimple rings. We develop this more abstract algebra in Appendix B. Now for some basic definitions, which can be formulated either in the language of G-spaces or of C[G]-modules. (1) The G-spaces VI and V2 are said to be equivalent if there exists a Clinear isomorphism f : V1 -+ V2, compatible with the G-actions. This means that f(pl(g)v) = p2(g)f(v) for all g E G and w E VI. Note that we must distinguish carefully between Homc(V1, V2) and Homc[~l(Vl, VZ), and that equivalence is expressed in terms of the latter family of homomorphisms. Note also that a group homomorphism from G into Autc(V) extends to an algebra homomorphism @[GI -+ Homc(V, V), and that conversely, given such an algebra homomorphism, we can recover p by restricting to the elements of G. However injectivity of the group homomorhpism does not necessarily imply injectivity of the algebra homomorphism. (2) The ring of linear maps Homc(V, V) is itself a vector space over C and can be given the structure of a G-space using the definition gf(v) = g(f(g-'w)). Clearly the same holds when V is replaced by a pair of G-spaces V1 and V2 , and Homcp](Vl , Vz) coincides with the subset of invariant elements. Notation: For any G-space V, VG = {w E V : gw = w for all g E G}. Thus HomC[G](Vl,b)= HomC(V11 v2)'. If G acts trivially on V, i.e. the image of p in Autc(V) equals I,, the identity map, we write VG = V. (3) As with ordinary vector spaces we can define sub-objects and quotient objects. A G-subspace of V is thus a C[G]-submodule, or a sub-vector space W C V such that gw E W for all g E G and w E W . Given
Chapter 1: Introduction
5
a G-homomorphism f : V1 + V2 we note that the kernel of f is a Gsubspace of V1 and that the quotient vector space V 2 / f ( & )admits a G-action with respect to which the projection map V2 4 V2/f (V1)is a G-homomorphism.
Definition. The G-space V is said to be irreducible if the only Ginvariant subspaces are {0} and V itself. The space V is indecomposable if there is no non-trivial splitting of V as the direct sum of G-subspaces w @ W‘. Proposition 1.1 (Schur’s Lemma) Let V1 and V2 be irreducible Gspaces. A G-linear map f : V1 fi is either trivial or an isomorphism. In particular if v = & = v2, Hom@[~](v, v) is a division ring. --$
Proof.
Consider the kernel of a G-linear map f . Since VI is irreducible this either equals VI, in which case f is trivial, or equals { 0 } , in which case & maps injectively onto its image. This is a submodule of V2, hence either trivial or equal to V2. Hence a non-trivial map f must be bijective. The second claim is an immediate consequence of this. 0
--
(4) If V is a finite-dimensional G-space we can define a C-linear map
T~G :v
V Gby &G gv*
The image is invariant, since pre-multiplication by h E G does no more than permute the elements gv among themselves. As an important special case note that, if f E Horn@(&,h),then T r ~ ( f )E Hom@[G](&> fi). More generally consider the composition
---
v,)
v 1
‘p
f
v 2
dJ
v;,
where cp, f and 7c, are all @-linear and cp, 7c, are G-maps. Then T~G($J . f ’ = $ ’ n G ( f ) * q.
Proposition 1.2 (Maschke’s Theorem) Let W be a G-subspace of the finite-dimensional G-space V over the complex numbers. There exists a complementary G-subspace W’ such that V E W @ W‘.
6
Representations of Finite and Lie Groups
Proof. Decompose V as a direct sum W @ W’ over the complex numbers C, and let T : V W be the C-linear projection map onto W . Thus ~ ( v= ) v for all v E W . We can average T over the elements of the finite group G by setting
obtaining a pair of G-homomorphisms
such that j is the inclusion, and cp . j = Idw. It is now easy to see that there is a G-splitting of V as
V as W @I ker(cp), so that we can take W’ = ker(cp). From the definitions, jWnker(cp) = {0}, and for an arbitrary element v E V,v - cp(v) E ker(cp). 0
Simple though its proof is, Maschke’s Theorem is fundamental to representation theory. We note in passing that it holds for representation spaces over an arbitrary field k provided either that the characteristic of k does not divide \GI, the order of G, or that the characteristic equals 0. We may use it for example to study representations of a finite pgroup of order p t over an extension field of lF,(q # p ) .
Definition. Let R be a ring. The R-module V is said to be semisimple if every R-submodule is a direct summand. The ring R itself is said to be semisimple, if 1 # 0, and R is semisimple as a left module over itself. We state the next group of results in terms of semisimple rings rather than in terms of the special case C[G]. Throughout we assume that all the rings R which we consider contain a multiplicative identity 1 = 1 ~Although . not strictly necessary we will also make this assumption in Appendix B.
Proposition 1.3 T h e following two conditions are equivalent o n a (finitely generated) R-module V : (1) V is a (finite) direct s u m of irreducible R-modules,
(2) Every submodule of V is a direct summand.
Chapter 1: Introduction
Proof.
Let V = @ V,, W iEI
7
C V and J
be a maximal subset of the indexing
+ j @€ J V,
is direct. We claim that this sum
set I such that the sum W
V* equals V , that is contains each summand V,, i E I . The intersection V * n V , = (0) or V , because V , is irreducible. In the former case we can adjoin i to J , and J is not maximal. Hence 5 V * . We next show that if (2) holds, V is a sum of irreducible R-modules, with the sum not necessarily direct. An intermediate result is that every non-zero R-module contains an irreducible R-module. Let v E V,v # 0, and consider the submodule Rv. The kernel of the homomorphism R --t Rv is a left ideal L in R, contained in a maximal ideal M # R. Then MIL is a maximal submodule of R / L , and hence M u is a maximal submodule of Rv, not equal to Rv. We have isomorphisms
Rv M/L 1 - Mu, RIL -
and since (2) holds V = Mu @ M’ for some submodule MI. Furthermore
Rv
= Mu @ (M’ n R v ) ,
+
because every element x E Rv decomposes uniquely as x = au x‘, with a E M , x’ E MI, x’ = x - av E Rv. The maximality of M u in Rv implies that M’ n Rv is irreducible. Let VOC V be the sum of all irreducible submodules of V . If VO# V, V = VO@ W with W # {0}, and there exists an irreducible submodule of W , contradicting the definition of VO. Passage from sum to direct sum is achieved by the same trick as in the proof that (1) + (2), i.e. take the maximal direct sum in V and show that 0 it contains each V,. We leave this as an exercise.
Proposition 1.4 Every submodule and every quotient module of a semisimple module is semisimple; also every ,5nitely generated R-module over a semisimple ring R is semisimple. Let U be a submodule of W a submodule of V . Since V is semisimple V splits as U @ U’. If w E W w decomposes uniquely as w = u + u‘ with u E U , u‘ E U’. But u‘ = w - u also belongs to W , so that W = U @ (U’ n W ) . Hence U is a direct summand of W . For the quotient module V/W we have a splitting V = W @ W’, with W’ semisimple and isomorphic to VIW. Finally every module over a semisimple ring is semisimple, since it may be expressed as the quotient of a free module
Proof.
Representations of Finite and Lie Groups
8
F over R. As a sum of copies of the underlying ring F is semisimple by 0 Proposition 1.3. Our aim is to show that an irreducible module V can give rise to a kind of duality between the original ground ring R and the ring S = EndR(V) =
HOmR(V, V). This is sometimes called the double centraliser condition. Let V be a semisimple R-module, i.e. a direct sum of irreducible (hence "simple") R-modules. As above let S = EndR(V). We can regard v as an S-module with scalar multiplication defined by (cpl
v)
cp(v)
for cp E S and v E V. Each a E R induces an S-homomorphism f a : V V via the rule fa(.) = av. This is an S-map, because of the module condition cp(av)= acp(v>.Hence there is a homomorphism of rings -+
R 0
--
Ends (V)
fa,
and we would like to know the size of the image.
s
Proposition 1.5 Let v be semisimple over R, = EndR(V) and f E Ends(V). For an arbitrary element v E V we can find a E R such that a21 = f(v).
Proof. Using the semi-simplicity of V write V = Rv@W with projection map 7r : V Rv, 7r E HomR(V, V) = S. Since f is given to be an S-map, f (v) = f ( x u ) = 7r f (v),which must belong to the submodule Rv. 0 -+
Using a diagonal trick one can generalise 1.5 from one to finitely many elements of V. Proposition 1.6 Let V be irreducible over R, S = EndR(V), f E Ends(V). Let v 1 . . . v, be elements of V. Then we can find an element a E R such that avi = f (vi)for i = 1 , 2 , . . . ,n.
Proof.
Consider the product map f'"' : V" --+ V" (Vl. * . V") H
(f(v1). . . f (Vn)),
and write S' = EndR(V"). As in elementary linear algebra, the ring S' can be identified with the ring of n x n matrices with coefficients in S. The
Chapter 1 : Introduction
9
map f commutes with elements of S, hence f(") E Endst(V"). So by the previous Proposition 1.5 we can find Q E R with ( a w l , . . . 1 awn) = ( f ( v 1 ) .
. . f(vn)).
0
Remark: This proposition actually holds for V = V1@. . $V, with each V , irreducible, and the proof above can be adapted to handle the more general case (double indices, block matrices...). As we will only make use of the simple case, we leave the details to the determined reader.
Corollary 1.7 (Burnside) Let V be a finite-dimensional space over a n algebraically closed field K , and let R be a subalgebra of EndK(V). If V is a n irreducible R-module, t h e n R = EndK (V). Note that this corollary applies when R is the subalgebra of Endc(V) obtained by extending a representation p : G --f AutcV to one of C[G].
Proof. We claim that EndR(V) = K . It is certainly the case that EndR(V) = K' is a division ring (Schur's Lemma) containing K , and that the elements of K commute with the elements of K". Let a E K'. Then K ( Q )is a field. The extension K* embeds in EndK(V) as a K-subspace, so that dimK(K*) < m, implying the finiteness of the extension K ( a ) . Since K is algebraically closed, it follows that K ( a ) = K for all a , and EndR(V) = K . Let {vl,.. . , v,} be a K-basis for V and let A E EndK(V). By Proposition 1.6 we can find a E R such that avi
= Avi
for i = 1,.. . ,n.
Since A is determined by its action on basis elements, R = EndK(V).
0
Dropping the assumption that K is algebraically closed the argument shows
Corollary 1.8 (Wedderburn ). L e t R be a ring and V a n irreducible R-module. L e t D denote the division ring EndR(V). If the m a p R 4 EndD(V) is injective and V is finite-dimensional over D , t h e n R = EndD (V). Proof. Let ( ~ 1 ...v,} be a basis for V with respect to the division ring D. Given A E EndK(V), Proposition 1.6 again implies that we can find Q E R with avi = Avi for i = 1,. . . ,n. Hence R maps onto EndK (V). We 0 assume that R + EndD(V) is injective, so R must equal EndDV.
Representations of Finite and Lie Groups
10
These corollaries are a first step in the classification of semisimple rings, the details of which are given in Appendix B. We first express R as a direct sum of irreducible left ideals, and then group together those left ideals which are themselves isomorphic as a ring Ri. As a ring in its own right Ri is simple, that is the only two-sided ideals are (0) and Ri itself. Rephrased Corollary 1.8 then shows that if Ri = @jnY1Rijand D = EndR(&j) then R, 2 M,(D), the ring of n x n matrices over the division ring D. In the next chapter we shall carry out this process for C[G]. For the group C, we note that C[G]is isomorphic to T copies of the complex numbers C.For Q8, if K is a subfield of R,then K[&] E 4K @ D where D is the quaternion algebra over K .
Exercises. 1. Let p be a representation of the group G. Show that det(p) is a 1dimensional representation of G. 2. Let 8 : G 4 C* be a l-dimensional representation of G and p : G 4 Aut(V) another representation. Show that 8p : z H 8(z)p(z)is also a representation, which is irreducible if and only if p is irreducible. 3. The symmetric group S3 acts on R2 by permuting the vertices of an equilateral triangle centred at the origin. Choose a basis for R2 and for each g E 5’3 write down the matrix of g with respect to this basis. 4. Construct the 5 x 5 ‘character table’ for the dihedral group 0 8 = { a , b : a4 = b2 = 1, b-lab = a - l } by listing the representations of dimensions one and two over C. Note that it coincides with the table for Q8. 5 . Use the representations from Exercise 4 to describe the group algebra @.[&I as a sum of matrix rings. *6. The following exercise is included for those who already know some group theory. Doing it, although not essential, will be a considerable help in understanding later chapters. The table below gives the numbers of finite groups of small order up to isomorphism. Describe the groups in each case.
n Number ofgroups
1 2 3 4 1 1 1 2
5 6 1 2
7 8 1 5
9 2
10 2
11 12 1 5
Chapter 2
Basic Representation Theory - I
From Chapter 1 we collect together the results already proved:
- C[G] is a semisimple ring, which splits as a finite direct sum of irreducible C[G]-modules. Note that the finiteness of the sum is guaranteed by the fact that dim@C[G] = number of elements in G. -
If V is a C[G]-module of finite dimension over C, the space of linear maps into a second such space W (Hom@(V,W ) )can be given the structure of a @[GI-modulevia gcp(v) = g(cp(g-'v)). Furthermore Hom@[G](V, W )= Homc(V, W)G.
The first of these results extends to any finitely-generated C[G]-module; the argument used also implies that any representation of G in GL,(@) is equivalent to a representation in the compact subgroup U,. To see this we first average the inner product over G, that is define a new inner product by (2,y)' = CSEG(gx,gy). We can then choose a normal orthogonal (NO) basis for C n with respect to ( , )'. The construction of the group action on Homc(V, W )is only an example of a whole family of representations associated with an initial homomorAut(V). Other examples are the conjugate representation phism p : G V and the dual representation V* = Homc(V, C ) , in which we suppose that G acts trivially on C. Note that in the presence of an invariant, positive definite Hermitian form on V, we may identify the representation modules This corresponds to the canonical isomorphism over C between V* and 7. V and V * induced by an inner product on V . We reformulate Proposition 1.3 as
&
--f
Proposition 2.1 Every representation space for the finite group G (every finitely generated C[G]-module)is a direct s u m of irreducible subspaces. 11
Representations of Finite and L i e G T O U ~ S
12
Alternative proof: Argue by induction on dimCV. Assuming that dim@V 3 1, either V is irreducible already or V = V1 @ V2 by Maschke’s 0 Theorem with dim@V, < dim@V ( j = 1 , 2 ) . Tensor products (Utility Version) Let V1 and V2 be vector spaces. A space W equipped with a map (vi,212) H ul €3 212 of Vl x V2 into W is called the tensor product of V1 and V2 if the following conditions are satisfied: (i) v1 €3 v2 is linear in each of the variables v1 and 712, (ii) If {ei : 1 6 i 6 m} and {fj : 1 6 j 6 n } are bases of V1 and V2 respectively, then {ei €3 fj : ( 1 , l ) 6 i , j ) 6 (m,n)}is a basis of W . (Order these basis elements according to some fixed convention, say lexicographically (1,I), (1,2), . . . , ( l , n ) , (2, I ) , . . . , (m,n).)It is not hard to see that such a space exists, and is unique up to C-linear isomorphism. We label it V1 €3 V2 and observe that its dimension is m . n.
Definition. The tensor product of two representations p l and Aut(V1) and Aut(V2) respectively is given by
p2
of G in
P ( g ) ( v l €3 v2) = (P1 €3 P2)(g)(vl€3 v2) = P l ( 9 ) ( V l )(8P2(9)(v2) In term of matrices we proceed as follows: Let aih(g) represent p l ( g ) with respect to the basis { e i } , and let b j k ( g ) represent p 2 ( 9 ) with respect to the basis {fj}. The formulae g ( e h ) =
xi
uih(g)ei and g ( f k ) =
g(eh €3 fk)
=
C
Cjb j k ( g ) f j imply that
aih(g)bjk(g)ei€3 e j ,
ij
once more with the elements of A @B ordered compatibly with the ordering of the basis elements. The following example shows that, even if V1 and VZ are irreducible, the same may not be true for their product V1 @ V2. Take V = Vi = V2 with basis { e i } and switching map
8 : V €3 V ei
€3 e j
--f
H
V €3 V given by ej €3 ei.
The map 8 is independent of the choice of basis, and 82 = 1. Furthermore V @ V Sym2( V )@ Alt’ ( V ) ,where {ei @ ej e j @ ei}i<j and
=
+
Chapter 2: Basic Representation Theory
-I
13
{ ei €3 ej - e j €3 ei}i<j form bases of the symmetric and alternating subspaces respectively. Both subspaces are invariant under G and have dimensions
T. Note that the sum of the two dimensions
equal to and equals m2,as expected.
Definition. Let p : G --+ Aut(V) be a representation of the finite group G. For each g E G put
the character of p at the group element g . Note that if we work in terms of matrices, and define x P ( g ) as the sum of the elements on the principal diagonal of p ( g ) , this definition is independent of bases. Since g has finite order m, so has p ( g ) . The representing matrix is periodic, and its eigenvalues are all mth roots of unity. The minimum polynomial divides Am - 1, and so p ( g ) is diagonalisable.
Proposition 2.2
(i) x(1)= 72 = A m @ ( P ) t (zi) x ( 9 - l ) = x ( g ) , where 7 denotes complex conjugation, (iii) X ( h g h - ' ) = x ( g ) , for all 9 , h E G, (iv) X P l + P Z ( 9 ) = X P l ( 9 ) + X P Z ( g ) , (v) X P l @ P Z ( 9 ) = X P l (g)XPz ( 9 ) . Proof. (i) is obvious. (ii) follows from the remark that the representing matrix is periodic. Therefore
Xb)=
wm
=
E X= p i ' i
= 'Pr
(P(d-')
=
n ( P ( g - ' ) ) = x(9-l).
i
(iii) follows from the properties of the trace function. [A function f : G -+ C which is constant on conjugacy classes is called a class function.]
(iv) is obvious, and for (v) we have the character of the product representation equal to
14
Representations of Finite and Lie Groups
Note from a utilitarian point of view that it is much easier to work with the character of the tensor product of two representations, rather than with the representations themselves. Similar arguments apply to the alternating and symmetric squares . We shorten the notation from Sym2 and Alt2 to S2 and A'. The use of the latter will become clear in Chapter 5 .
Proposition 2.3 (2)
xszp =
i ( X p ( d 2 +xp(g2))
(Zi)
XA2p =
$(xp(g)2- xp(g2)).
Proof. Note that for a fixed element G the automorphism p ( g ) can be represented by a unitary matrix, and hence by a diagonal matrix with eigenvalues down the principal diagonal. If x ( g ) = then x ( g 2 ) =
ci
xi&,
On the other hand
+
(pi @ p2)(ei 8 ej ej @ ei) = XiXj(ei @ ej (pi @ pz)(ei @ ej - ej @ ei) = XiXj(ei 8 ej
+ ej 8 ei)and -
e3. 8 e i ) .
Therefore xsz(g) = &g) =
X i X j = $(Xi %<3 X i X j = + ( X I
c. ,
+ . .. + X,)2 + ;(A:
+ . . + X,)2
- ;(A::
*
+ . . . + X i ) , and + +Xi). *
'
Once more note that Proposition 2.2 checks with Proposition 2.1 (v), namely x' = xsz xA2. Note also that X,,Z is the second elementary symmetric function in the eigenvalues Xi.
+
Proposition 2.4
A s G-spaces we have an isomorphism Homc(V, W ) S V* 8 W,
c
where V* = Homc(V,@) is given the module structure ( g f ) ( v ) = f ( g - l v ) .
Proof.
Map the right-hand side to the left by means of
v* 8 w
H 'p
with 'p(v) = v*(v)w.
We note that u*(v)is a scalar and that the map is bijective on bases. Note that if e f , ei and fj are basis elements for V*, V and W respectively, then ef @ fj --+ pij with Vij(ei)fj = fj. The dimension of both sides equals dim(V) dim(W), and the map has been chosen to be compatible with the G-action. 0
Chapter 2: Basic Representation Theory
-I
15
Proposition 2.5 If p : G Aut(V) is a representation, then I = I G l - l C p ( g ) E Homc(V, V) is an idempotent, that is I 2 = I . The image of 9
I equals the invariant subspace V G . Proof.
If I(v) =
cg.v , then h I ( v ) 9
=
z h g ( v ) = I(v), since summing 9
over g or over hg gives the same answer. Therefore Im(I) G VG. Clearly IIVG = J G Jx Identity, which explains why we modify the definition of I by dividing out by ]GI. Therefore (with this change - which will drop out in the compact case), Im(I) =
vG.
We have the following complement to Schur’s Lemma.
Lemma 2.6 Let f : V 4 V be an element in the division ring HOmc[~](v, V), where V is irreducible. Then f (v) = Xw for some constant X E C. In particular, zf
v 2 W, HOmc[q(v,W )= 0, V
and if 1.
,
E W , dim@HO%[G~(V, W ) =
Proof. The map f : V is C-linear, hence has an eigenvalue A. The map f’ = f - X . Id is @[GI-linear and has non-trivial kernel. This can only be 0 all of V. One consequence of Lemma 2.6 is that an irreducible representation of an abelian group A over the complex numbers must be 1-dimensional. This follows since the C-linear map p(a) : V -+ V is compatible with the group action, provided that ag = ga. Hence a 1-dimensional subspace is A-invariant and exhausts V. Note that we have the following relation on characters
This relation follows from the definition of the idempotent I above, since the left-hand side equals the trace of I .
Theorem 2.7
(Orthogonality relations between characters)
(i) Let V,W be finite dimensional C[G]-modulesfor the finite group G. Then &CK(g)x,( 9 ) = dim@Homc[q (V, W ) = d (say). 9
Representations of Finite and Lie Groups
16
(ii) If V and W are both irreducible, then V 9 W d=l. Proof.
Write
+ d = 0 , and V E W +
H = Hom@(V,W ) with its given G-action. We have d i m q ~Horn( ] V, W ) = dim@HG = &iCx,(s) 9
This proves (i) and part (ii) follows from Schur's Lemma.
Definition. The inner product of the two characters
0
xv xw is given by I
Theorem 2.8 Let V be a representation module with character cp and suppose that V decomposes into a direct sum of irreducible representations
If xw denotes the character of W , the number of summands Wi in the decomposition above isomorphic to W equals (cp, x,).
+ + + +
Proof. Decompose cp as cp = x1 . . . x k , and use the linearity of the inner product (9,x) = ( X I , x ) . . ( x k ,x). Each term on the right hand 0 side equals 0 or 1 by Theorem 2.7 and the result follows. Corollary A: The number of summands isomorphic to W in a decomposition of V into irreducible submodules is independent of the decomposition. Corollary B: If V and W are two representation modules for G, then if and only if V E W .
xv = xw
These corollaries show that the study of representation modules for G amounts to the study of their characters. If x1 . . . x k are the distinct irreducible characters of G (we will show in a moment that k is finite) and
Chapter 2: Basic Representation Theory
-I
17
if w 1 . . . w k represent the corresponding equivalence classes of representations, each representation module V is isomorphic to a direct sum
v
with mi 2 0. The character cp of = mlXl + . . . + m k x k , where mi = (cp, x i ) . AS a special case the character x i x j of Wi @ Wj decomposes as xixj
=Crnl',Xk k
with m:j 2 0. An example of this will be given in Chapter 6 for the group SU, (Clebsch-Gordan formula). The orthogonality relations imply that k
i=l
Corollary C: For any representation module V with character cp, the inner product (cpl cp) is a positive integer, and (cp, cp) = 1 if and only if cp is irreducible.
Proof. The sum of integral squares above can equal 1 if and only if all but one m: vanish, and this last one equals 1. From now on, we standardise the notation: the degrees of the irreducible characters X I . . . X I : equal 7 2 1 . . . n k , that is ni = X i ( l ) l and we let Preg be the representation carried by the group algebra @[GI itself. We call Preg the regular representation. With respect to the obvious basis {el = i d , e 2 , . . . e p l } o f @[GIwe have preg(g)eh= egh, i.e. the elements of G permute the basis elements. If g # 1 , g h # h for all h, so that trace (preg(g))= 0. On the other hand trace ( p r e g ( l ) ) = dimc@[G]= /GI. Th'is proves Proposition 2.9
The character of the regular representation is given by
X P & ) (1) =
IGll
X P r e g ( 9 ) = 0 othe7-wise.
Corollary A: The regular representation contains ni = dimc Wi copies of the irreducible representation Wi.
Representations of Finite and Lie Groups
18
Proof.
Compute
k
Corollary B: (a) E n :
=
IGI
i=l
k
(ii) If G 3 g # 1, then Cnixi(g) = 0. i=l
k
Proof. g
We know that
# 1 to obtain (ii).
xPreg= CniXi.
Evaluate at 1 to obtain ( i )and
i=l
0
The fact that each Wi occurs in the finite dimensional representation module C[G] shows that the number k of irreducible isomorphism classes is finite. Its value will be determined in the corollary to Theorem 2.11 below. Each degree ni actually divides the order of G; this useful result follows from an even stronger one to be proved later.
Proposition 2.10 Let f E C&(G) p : G + Aut(V) be a representation. fk7)Pb). Define Pf :=
=
CI(G) be a class function, and
c
SEG
If the representation module V is irreducible of degree n and the linear map pf equals X . 1 with
x = XV,
then
Proof. Direct calculation shows that p ( h ) - l p f p ( h ) = p f , i.e. that pf commutes with the action of G on V , and is thus a C[G]-map. Since V is irreducible, Schur's Lemma implies that pf = 1 . X for some scalar A. Taking traces gives trace ( A . 1) = n . X = C f ( g ) x ( g ) = IGI(7, x). 9
0
Theorem 2.11 The irreducible characters XI . . . Xk of the finite group G form a normal orthogonal (NO) basis for the complex vector space of class functions CI(G).
Chapter 2: Basic Representation Theory - I
19
Proof. We have already shown that { X I . . . xk} is an NO-system, and hence in particular is linearly independent. Let the class function f be orthogonal to each conjugate character xi,Proposition 2.10 shows that p f = 0 so long as p is irreducible. The same holds for an arbitrary representation p by taking direct sums and using linearity. Take p = preg, and compute the image of el (the basis element corresponding to the identity element) under p f . Then Pfel=
C f(g)Preg(g)(el)= C f(g)eg. gEG
gEG
The elements { e g }are linearly independent, so f ( g ) = 0 , i.e. f the orthogonal complement to (XI,.. . , Xk) is trivial.
5
0, and
0
Corollary Up to C[G]-equivalencethe number of irreducible representations of G equals the number of conjugacy classes of elements in G.
Proof. This follows from 2.11 and the fact that dima: CC(G)clearly equals the number of conjugacy classes. 0 We next sum over characters
xi rather than over elements g of G.
Proposition 2.12 Let c(g) = number of elements in the conjugacy class represented by the element g .
k
(ii) If the element h E G is not conjugate to g , then C z ( g ) x i ( h )= 0. k l
Proof. Let f g be the class function taking the value 1 on the class of g and the value 0 on all other conjugacy classes. Then, since the xi form a spanning set for Cl(G),
k
Therefore f g ( h )=
C #E(g)Xi(h)
i=l
=
m c ( gx ) - i ( g ) by the definition off,. 1 if h E class of g 0 otherwise.
0
In Chapter 1 we introduced informally the character table for the quaternion group Q 8 by mapping irreducible characters against their values on conjugacy classes. Using the numerical relations arising from orthogonality
20
Representations of Finite and Lie Groups
and the decomposition of the character of the regular representation we can now consider groups of larger order in a similar way. Examples. (1) Let T 2 A4 be the tetrahedral group (E alternating group on 4 symbols) of order 12. Then T contains a normal subgroup of order 4, consisting of the identity and the three pairs of transpositions (call two of these p , q ) . There is a complementary subgroup (x) of order 3 , isomorphic to the quotient group A4/(p, q ) , and one easily sees that the degrees of the irreducible representations are 1 (3 representations factoring through the quotient) and 3. The regular representation thus splits as xo X I x 2 3p (Proposition 2.9, Corollary A), and our earlier work shows that the character table must be
:1.::
+ + +
1
x2
1
w2 w
w2 w
0
0
3 - 1
p
(2) The octahedral group 0 2 S4 can be handled similarly. The number of conjugacy classes equals the number of ‘disjoint cycle types’ (this holds for any symmetric group Sn),which is 5. There are three obvious irreducible representations xo (trivial) E (equals +1 on even cycles, -1 on odd) and the permutation representation 7r (minus a copy of xo) of degree 3. This can be multiplied by E without loosing irreducibility. Since 24 - (1 1) - (9 9) = 22, there is one other irreducible representation of degree 2, whose character can be calculated from the characters already constructed and that of the regular representation. It appears as 9 in the table below
+
+
1
(12)
(12)(34)
(123)
(1234)
7r
1 1 2 3
&T
3
1 -1 0 1 -1
1 1 2 -1 -1
1 1 -1 0 0
1 -1 0 -1 1
xo E
e
Chapter 8: Basic Representation Theory - I
21
The irreducibility of A follows by counting the number of elements in each conjugacy class l(1) 6(12) 3(12)(34) 8(123) and 6(1234). We then have ( A , A ) = & (9 6 3 0 6) = 1. This argument generalises from 4 to any value of n. Let us look at S4 in a slightly different way, which is again illuminating for higher order symmetric groups. An application of Proposition 2.2 shows that the characters of the symmetric and alternating squares of T are
+ + + +
A2r 3-1-1
0 1
P&-mT* Hence A27r = ~ 7 r(irreducible) and the ‘virtual character’ (negative terms allowed) S 2 -~T - xo = 8. Anticipating a later result that the determinant E can also be regarded as an alternating power, we see that the entire character table can be built up from T through use of the alternating and symmetric power constructions. For the group S5 we have characters T (of dimension 4) and A 2 (of ~ dimension 6), and both are irreducible. As before we may add E and ET to these, and note that 120 - (2.1 2 2.16 36) = 50 = 2.25. There are two remaining irreducible characters 8 and ~8 of dimension 5. The values taken by their sum 8 + a3 can be worked out using the regular character, and then split using the values taken by E . (3) In the same spirit the reader may like to tackle the icosahedral group 1 E As (order = 60). Note that in this case the group is simple, and the only 1-dimensional representation is trivial.
+ +
+
The character table turns out to be:
1
(12)(34)
(123)
(12345)
1 0
1 1
1
4 5
1
-1 0
xo A ‘
p
$1 :
-1 -1
0
(13452) Q
-1 0
-1 0
Q
P
P
6)
= $(1+
p = i(1-6)
a
Hint (non-compulsory): the rotations of a regular icosahedron represent A5 in SO3. The permutation representation of S5 restricts to As
Representations of Finite and Lie Groups
22
giving an irreducible 4-dimensional representation. There are actually two 3-dimensional representations differing only on their values on 5cycles, which result from the splitting of a conjugacy class in the full symmetric group. This leaves (p, for which a numerical check similar to that for S5 gives the dimension 5 and the values on the non-identity conjugacy classes. To see that this route is not the only one possible, turn to the character table for SLZ(P5) in Chapter 8. The relation between the characters of 5's and A5 illustrates a general pattern. A conjugacy class in S, either survives in A, or splits into two distinct classes. Two distinct characters on S, may be equal on A,(xo and E for example), or may survive as distinct irreducible characters. The third possibility, corresponding to the splitting of conjugacy classes, is that a character becomes reducible on An; thus when n = 5, A'T splits as II, 3.
+
Exercises.
1. Let G be a group of odd order. By considering G - (1) as the union of ([GI - 1)/2 pairs {xi,xL1} show that no conjugacy class other than 1 is real, i.e. equal to its inverse. Deduce that if x is an irreducible character distinct from the trivial character X O , then x # X. 2. Let G be a non-abelian group of order p q where p and q are distinct prime numbers with p > q and q dividing p - 1. Show that the commutator subgroup [G,G] generated by all elements x-ly-lzy has order p , and that the number of conjugacy classes equals q ( If p = 7 and q = 3 the group has a presentation'
+
y).
{ a , b : a7 = b3 = 1, b-lab = a'},
Describe the 5 conjugacy classes of elements. Show that the group has 3 irreducible characters of degree 1 and 2 of degree 3. In Chapter 4 we generalise this result to other pairs ( p , 4). 3. Let G be a subgroup of order 18 in the symmetric group given by
s(3
G = ((123), (456), (23)(56)). Show that G has a normal subgroup of order 9 and four normal subgroups of order 3. By considering quotients show that G has two repre-
Chapter 3: Basic Representation Theory - I1
23
sentations of degree one and four inequivalent irreducible representations of degree 2, none of which is injective. 4. A group of order 720 has 11 conjugacy classes. Two representations of this group are known and have characters a and p. The table below gives the sizes of the conjugacy classes in the group and the values taken by a and p on them: 1 5 40 90 45 120 144 120 90 15 40 a 6 2 0 0 2 2 1 1 0 - 2 3 p211-3-1 1 1 1 0-1-3 0
Prove that the group has an irreducible representation of degree 16 and write down the values taken by the corresponding character on the conjugacy classes. 5. A group of order 168 has 6 conjugacy classes. Three representations of this group are known and have corresponding characters a ,,B and y. The table below gives the sizes of the conjugacy classes and the values taken by a ,p and y on them 1 21 42 56 24 24 0 1 4 2 0-1 0 0 p 15 -1 -1 0 1 1 y16 0 0 4 2 2
Construct the character table of the group. 6. How can you use the character table of the finite group G to detect the existence of a normal subgroup in G? *7. (a) The group studied in Question 5 is the projective linear group PSLz(lF7) of non-singular 2 x 2 matrices over the field of 7 elements, modulo the scalar matrices. The group acts as a permutation group of degree 8 on the set of l-dimensional subspaces of the vector space IF:. Obtain the character of this action, and decompose it into irreducible characters. (b) Show that PSLz(lF7) is generated by an element of order 2, and an element of order 3, whose product has order 7.
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Chapter 3
Basic Representation Theory - I1
This chapter is a simple continuation of its predecessor, and is devoted to expanding on and illustrating the results obtained using the orthogonality relations between the irreducible characters. The topics covered include the uniqueness of the decomposition of a representation module into irreducible summands, the relation of the values xi(1) = ni with the order of G, and the characters of the direct product G I x Gz. We present this last topic in what may seem an over-complicated way, but one which applies word for word to compact groups in Chapter 6. Given a decomposition V = vl @ . . . @ v k , with each V , E @Wji) and j
each isomorphic to some representative irreducible @[GI-moduleW ( i ) , to what extent do we have uniqueness? Theorem 3.1
(i) T h e decomposition V = V1@.. .@ v k does n o t depend o n the particular decomposition of V into irreducible summands. (ii) The projection m a p pi : V + V , associated with this decomposition is given by pi = xi(g)p(g), where p is the homomorphism describing the G-action on V .
5 c,
Proof. It suffices to show that pi has the properties of a projection map, i.e. that pi restricted to the summand Vj is either zero or the identity, depending on whether i # j or i = j . For this it is enough to restrict pi to the irreducible summand W j . Apply Proposition 2.10 with p x = CgEC Fi(g)p(g). The corresponding value of X equals
0
as required. 25
26
Representations of Finite and Lie Groups
Warning: the decomposition of V , into a sum of copies of W(2)is no more unique than is the decomposition of a vector space into a direct sum of 1-dimensional subspaces. The uniquely determined summands V,, which are not in general irreducible, are called isotypic . The next four propositions are concerned with the role of abelian subgroups and quotients in representation theory. Proposition 3.2 The finite group G is abelian if and only i f each irreducible representation of G has dimension 1 over C. Proof. We use the numerical relation \GI = nf+. . .+n& where ni equals the degree of an irreducible representation. If each ni = 1, I; = \GI and each conjugacy class consists of a single element. Conversely,-if this holds, each ni must equal 1. See also the remark following Lemma 2.6. 0
Corollary Let A be an abelian subgroup of G. Each irreducible representation of G has degree no greater than the index [G : A] = lGl/lA\. Proof. Let p : G 4 Aut(V) be irreducible; consider its restriction A 4 Aut(V). Let W be an irreducible C[A]-summand of V; by 3.2 it has degree 1. Let V’ be the subspace of V generated by all translates of W under elements of G. Since V’ is obviously a representation space for G, V’ = V by irreducibility. But for g E G and a E A we have gaW = gW, hence the number of distinct copies of W in V’ is bounded above by lGl/lAl.
PA :
0 It should be obvious by now that the determination of the 1-dimensional representations is an important step in the construction of the character table of G. As previously, let [G,GI denote the normal subgroup of G generated by elements x-ly-’xy. Proposition 3.3 Composition with the projection map T : G -+ A = G/[G, G] defines a (1- 1) correspondence between the 1-dimensional representations of G and the irreducible representations of A. Proof. The quotient A is maximal in the sense that if 7r’ : G 4 B is a homomorphism into any abelian group, then [G,G] LKer(n’). Hence if 0 : G 4 @* is a 1-dimensional representation 0 factors through A. Conversely if r is an irreducible representation of A , necessarily 1-dimensional by 3.2, r ~ is’ a 1-dimensional representation of G. o
Chapter 3: Basic Representation Theory
- II
27
Remark. Part of the problem in dealing with simple groups, of which As is the smallest non-abelian example, is that there are no non-trivial quotients to provide part of the character table. In particular the only 1-dimensional representation is the trivial one. Definition. The centre Z(G) of the group G equals { z gz for all g E G}.
:
zg
=
From its definition we see that Z(G) is an abelian normal subgroup of G, and so the corollary to Theorem 3.2 implies that the degree of an irreducible representation is bounded above by the index [G : Z(G)]. Our next aim is to show that the degree actually divides this index, proving also that it divides [G : {l}]= IGI. First a result on the centre of the ring @[GI,again equal to those elements which commute multiplicatively with all z E @[GI.
Proposition 3.4 Let Ci equal the sum of elements in each conjugacy class (xi) of elements in G, i = 1,.. . ,k. Then the Ci are k @-linearly independent elements which span the centre Z(@[G]). Proof. This is an easy exercise. Note that it provides an alternative proof that k equals the number of irreducible representations of G. This follows from the description of the centre @[GIas a sum of matrix rings for each of which the centre consists of scalar matrices (see Appendix B). 0 We recall that if the complex number Q belongs to a finite extension of the rational numbers Q, a is said to be an algebraic integer if it is a zero of a monic polynomial with integral coefficients. We need one easily proved property of algebraic integers.
Lemma 3.5. If a
EQ
is an algebraic integer, a
E
Z.
Proof. Let Q = q = in lowest terms. The rational number q is a zero of the polynomial z - q and is assumed to satisfy the monic polynomial h ( z ) E Z[z]. Therefore h ( z ) = (z - q)g(z)with g ( z ) E Q[z]. Clearing fractions write z - q = (l/b)(bz- a ) and g ( z ) = (c/d)G(z)with both (bz - u ) and G(z) primitive, i.e. having no prime p 2 2 dividing the set of coefficients of each. Reduction modulo p shows that (bz - a ) G ( z ) is still primitive. We have bdh(z) = c(bz - u)G(z), and by primitivity, 0 h ( s ) = (bz - a)G(z). Since h ( z ) is monic, b = 1 and q E Z.
Representations of Finite and Lie Groups
28
Lemma 3.6. Let ( 2 1 ) .. . (Zk) be the conjugacy classes of G , ci = I ( z i ) l , x1 . . . x k be the irreducible characters, and n i = degree xi. T h e n the numbers
are algebraic integers. Proof.
Let Ci be the sum of the elements in the class
(xi)
as in 3.4.
k
Multiplying out we must have
CiCj
=
C dijmCm. m=l
Let X e be the character of the representation cpe : G extend cpe to all of @[GIby @-linearity. Then
--t
Aut
(h)and
* Since cpe(Ci) is central, cpe(Ci)xw = xcpe(Ci)w, for all w E by Schur's Lemma, since & is irreducible, p e ( C i ) = wie
h.Therefore
(id)
for some scalar wit E @. Taking traces we obtain c i x e ( z i ) = wiene, explaining the choice of in the statement of the lemma. Equation (*) becomes wiewje =
C
dijmwme,
wit
or
m
F
** m=l
Taking i = 1,so that (XI)= 1 and c1 = 1,we have wle = = 1. Fix e, and write (**) as 0 = C m ( d i j m - 6jmwie)ym. Since wig = 1 # 0, the system has a non-trivial solution, and hence
i and
0 = det(dij, - c5jmwie).
It follows that
wit is an eigenvalue of the integral matrix (dij,) (remember that only j and m vary). The characteristic polynomial is monic and belongs to Z [ y ] ,so wit is algebraic. 0
Chapter 3: Basic Representation Theory
Proposition 3.5 divides IGI.
The degree nt
=
- II
29
xt(1) of the irreducible character Xt
Proof. As before write ci = I(xi)l. Since the sum and product of two algebraic integers are again algebraic integers (Exercise), and Xt (xi') is algebraic, so is
Therefore IGl/nt E (algebraic integers) n Q = Z.
0
A more refined argument leads to a sharper version of 3.5:
*Theorem 3.8 If x is the character of the irreducible representation cp G + Aut (V), then n = x(1) divides the index [G : Z(G)].
:
Proof. (Glauberman) Write 2 = Z(G), and use the same notation for conjugacy classes as before. If z is central (zj) ++ ( z z j ) = (xi) defines an equivalence relation, and if (xj) (xi)the orders c j and ci are equal. Write {(xi) . . . (xlzl)}.. . { ( x ( t - l ) l Z l , .. . , (xtlzl)}for all the equivalence classes containing 121distinct members. If j > tlZl the class containing (xj)is smaller, and for some central element z ( z j ) = (zxj). Let cp : G -+ Aut(V) have irreducible character x, and assume first of all that cp is faithful, i.e. that cp is injective. The image cp(z) = azln, so ~ ( x j = ) ~ ( z z j= ) trace(cp(z)cp(xj))= trace(a,cp(zj)) = a,x(xj). The character x is faithful, so az # 1 and x ( x j ) must equal zero. Now let x E G and z E 2 be arbitrary. Repeat the previous argument: ( x ( z z )=~ la,x(z)( = lcy,JJx(~)I = Jx(z)J so Ix(z)Jis constant not just on a conjugacy class, but on an equivalence class of conjugacy classes. And N
Representations of Finite and Lie Groups
30
neglecting the value zero, each such class contains 121members. Now count:
i=l
xEG
i=l
t i=l t i=l
But by Lemma 3.6 cilz~x(zilzl = x ( l ) w i , where wiis an algebraic integer. t
Therefore from the last equation [G : Z ] / x ( l )= C w i X ( z i l z l ) is a sum i=l
of products of algebraic integers, hence also an algebraic integer. Since [G : Z ] / x ( l )is rational, it must belong to Z, and x ( 1 ) divides [G : 21. If cp is not faithful and has kernel K , then x ( 1 ) divides [ G / K : Z ( G / K ) ] .But Z ( K ) / K 2 Z ( G / K ) ,so x(1) divides [ G / K , Z ( K ) / K = ] [G : Z ( K ) ]which divides [G : 21. 0
Theorem 3.8 can be further sharpened and the centre Z ( G ) replaced by any (sub)normal abelian subgroup S . (The subgroup S is subnormal if there exists a chain S = S, a . . . a SO= G.) See [L. Dornhoff, $221.
Representations of GI x
G2
The irreducible representations of the product G I x G2 of two groups are determined by those of G1 and G2. In terms of characters we let x(') correspond to a representation p(') of G,(r = 1 , 2 ) , and define a character of G1 x G2 by ~ ( ' ) ( g l ) x ( ~ ) ( g Then 2). Proposition 3.7
(i) If x(") i s a n irreducible character of G,(r = 1 , 2 ) t h e n x = x ( l ). x ( ~ i s an irreducible character of G1 x G2. (ii) All t h e irreducible characters of G I x G2 are of t h e t y p e x(l) . x ( ~ ) . Proof.
Part ( i ) follows from the equation 2 C I~(~)(g~)x(~) IGlIlG2191,Q2
= 1 Part (ii) follows by orthogonality. Consider the class function
f on
Chapter 3: Basic Representation Theory - II
GI x
G2
orthogonal to characters of the type
c
31
x(l). x ( ~ )We . have
~~
f(g1’S2)X(1)(g1)x ( 2 ) ( 9 2 )= 0.
91 292
Put f’(g1) = c f ( g 1 , g 2 ) Z ( ~ ) ( g 2 and ) , note that C f ‘ ( g l ) x ( l ) ( g l ) = 0 for 92
91
all characters x(’) of GI. The character spanning property implies that f’(g1) = 0. Interchanging the roles of x(l) and x ( ~it) follows that 0 = C f ( g i , g 2 ) X ( 2 ) ( g 2 showing ) that f ( g 1 , g a ) = 0 for each pair (g1,92).
,,
92
For a finite group G the proof above can be considerably shortened using the results of Chapter 2. We have given it in a form which translates without change to the compact groups considered in Chapter 6.
* Real Representations It is sometimes of interest to know if a representation p : G 4 GL,(C) is equivalent to a representation po : G 4 GLn(R). In general, as the examples of Q 8 and Ds show, it is sometimes but not always sufficient that the character of p take real values. Before stating a general result let us note two general constructions: (1) A complex representation always determines a real representation by composing p with the natural map GL,(C) 4 GLz,(R). Denote the image real representation space by VIWG. (2) A real representation can always be ‘complexified’ by extending the scalars from R to @. This corresponds to composing po with the natural inclusion map GL,(R) 4 GL,(C). Denote this by h H C @3 VO.
It will not be hard to see that
C
(VIWG) E V
@
7and
w
that (C @
V0)IiP.G = 2h. We will say that the C[G]-module V has an R-form if V some R[G]-module h. The basic result is the following:
@ @3 VOfor
w
Theorem 3.8 (Frobenius-Schur) Let V be a finitely generated @[GImodule with character x. (a)
x
takes values in R i f and only if V admits a non-degenerate Ginvariant bilinear form,
Representations of Finite and Lie Groups
32
(ii) V has an W-form if and only if V admits a non-degenerate symmetric G-invariant bilinear form (over C).
Proof. For (i) x is real-valued @ V E V* = Homc(V, C ) . By elementary linear algebra such an isomorphism exists if and only if a bilinear form exists as in ( 2 ) . For (ii) assume that V has an R-form VO.By the usual averaging argument (Exercise 5 below) VOadmits the required non-degenerate symmetric G-invariant bilinear form, which survives extension of scalars. Conversely w] be the usual Hermitian assume that B(u,v) is such a form and let [u, form on V . Another exercise in linear algebra shows that there exists an additive homomorphism cp : V --f V such that B ( u , v ) = [cp(u),v]. The map cp is semi-linear in the sense that cp(zu) = .Cp(u).Therefore b 2 U , 7JI
= B(cp(u),v) = B(v,c p ( U ) ) = [cp(v),cp(U)l
for all u,vE V . But [u,v] = [v,,SO
1% cp2(v)l = [cp2(v),4= [cp(U),cp(41 = [cp(v),cp(u)l = [ c p 2 w J 1 . Therefore cp2 is self-adjoint. It is positive definite because [cp"(.>,
4 = [cp(.), cp(4l
> 0 for U # 0.
Again appealing to elementary linear algebra we can extract a positive definite self-adjoint square root u of cp2 and write u = 9 u - l . The square root u can be expressed as a polynomial in cp2, so that u and cp commute. Therefore u2 = (cpu-')(cpu-') = cp2u-2 = 1. It follows that the C[G]module V splits as VO@ Vi,where VO= {v : uv = v} and V1 = {v : uv = -v}. The semi-linearity of u implies that VOand V1 are W-subgroups of V such that Vi = i h ( i = Since cp2 and u are both G-maps, so is u 0 and Vo is the required real form of V .
a).
With
x regarded as a complex valued function the character of V ~ WisG
x+x. Theorem 3.8 shows that there are three possibilities for an irreducible, finitely generated, @[GI-moduleV . (i) The character x is not real-valued in which case V ( Ris ~irreducible over RG, has twice the dimension of V , and character x y. (ii) The character x is real-valued, and an R-form Vo exists. In this case the complexified module is Vo @ &.
+
Chapter 3: Basic Representation Theory - II
33
(iii) The character x is real-valued, but there is no R-form VO.In this case V ~ WisGagain irreducible of twice the dimension of V ,and the character is 2x. For more on this subject, see Exercise 6 below.
Exercises. 1. Let p : G -+ GL,(C) be a representation with character x. Show that Ker('p) = {x E G : x(x) = n}. Show further that for any x E G Ix(x)l n, and equality holds if and only if p(z) = XI, for some
x E c.
<
2. Let x be a character of G and z E G. If x has order 2 show that x(x) = X(l)(mod 2). If G # C2 is simple (that is the only normal subgroups of G are itself and {l}),show further that x(x) E X(l)(mod 4). 3. Show that the 1-dimensional complex characters of G form a group, denoted by E. Show that, if x E G , the map x H x(g) from 6 to C is a
-
h
character of
6,and hence an element of2. Describe the kernel of G 2.
e e
4
h
If G is abelian show that the natural map G 4 is an isomorphism, and deduce that for a finite abelian group G and are also isomorphic. Let 'p : GI 4Ga be a homomorphism of finite abelian groups. Define a
2,
4.
5.
6.
* 7.
dual map cp* : 6 2 -+ and show that 'p* is surjective if and only if cp is injective. (This is one reason why it is not a good idea to identify G and 6.) Show that the cyclic group C3 has only one non-trivial irreducible representation over the real numbers R. If the representation space is V show that dimw Homc, (V,V ) = 2. Why does this not contradict Schur's Lemma? If p : G -+ GL,(R) is a real representation show that there is a nonsingular matrix P such that Pp(x)P-' is an orthogonal matrix for each x E G. Determine all finite groups which have a faithful representation of dimension 2 over R. Use Proposition 3.9 to describe the characters of the direct product C3 x S3 of order 18. Is this group isomorphic to the group considered in Chapter 2, Exercise 3? If ( p , V ) is an irreducible complex representation for G with character x, recall the characters of V 8 V , S2V and A2V from Chapter 2. Use
34
Representations of Finite and Lie Groups
these to show that
c IGI 1
-
xEG
x ( x 2 )=
x is not real-valued, f l , if x is real-valued.
0, if
Show further that, if the sum equals +l, ( p , V ) has a real form (PO, G). (The sign is called the Frobenius-Schur indicator.)
Chapter 4
Induced Representations and their Characters We start with a representation p : G + Aut(V), and write i * p : H 4 Aut(V) for its restriction to some subgroup H 5 G. When it is necessary to specify the subgroup concerned we will replace i* by i& or ResH. Let W be a subspace of V stable under the action of H, and so corresponding to a subrepresentation 8 : H 4 Aut(W) . For each left coset gH, represented by g, it makes sense to define W, 5 V to be p ( g ) W , where this subspace depends on the coset and not on the representative chosen. Action by the elements of G permutes the subspaces W, among themselves, and their sum C W, is a G-subrepresentation of V. gH
Definition The representation p of G is induced from the representation 8 of H (written p = i,8 or p = ig8) if V equals the direct sum of the subspaces W, as g runs through a family of representatives for the left cosets gH. Note that dim@V = [G : HI dim@W. Furthermore it is clear that i,(& 8,) = i,81 i,&, and that if W1 is H-stable in W , then V1 = C gW1, is G-stable in V. We shall see later
+
+
9H
that i , is not a ring homomorphism, but a module homomorphism, when an H-action is regarded as a G-action via the restriction map i*. Examples.
(i) If 1~ is the trivial representation of H, i * 1 is~the permutation representation (xH ++ gxH) of G on the set of cosets G I H . (ii) The induced representation of the regular representation of H is the regular representation of G. To see this let {e, : g E G} be the natural basis of @[GI and define W to be the subspace generated by 35
36
Representations of Finite and Lie Groups {eh : h E
H} (equal to C[H]). The definition of W, above shows that
z*Preg,H = Preg,G.
To avoid overloading the notation in the next proposition we write for a coset gH in GIH.
(T
Proposition 4.1 For each representation 8 : H -+ Aut(W) there exists a representation i,8 = p : G -+ AutV, which is unique up to equivalence.
Proof. Let g,, E CT be a family of coset representatives (transversal) in G, and suppose that 1 represents the coset H . If v is an element of V, v = c g D w , . Assuming that p = i,8 exists we first show that it must be ”
unique. If x is an arbitrary element of G, xg, must then have
= grh
for some h E H . We
from which uniqueness follows. This step gives the clue for existence. Write V = $W, where W, = W,,, and define a bijection between W = W1 and L7
W, by 1 ’ w + g , * w. We note that u E V has a unique decomposition as C g a w a and define g(g,w,) = g,(hw,). We check that this is a G-map: U
let g’g7 = g,h’, then g’(g(g,w,)) = (g’g)(g,,w,), since by associativity of group multiplication (g’g)g, = g’(gTh) = g,(h’h). 0 The two immediate consequences of the proposition are (i) The induction map is compatible with direct sums, i.e. i,(& i,O1 i,&, and (ii) Induction is transitive, i.e. if K C H C G
+
iZ(W) = iz :i
+ 02) =
(W).
Another property is most easily understood in terms of characters: (iii) If U and W are G- and H-representations respectively, then U@i,W = Z,(i*U@W).
Chapter
4:
Induced Representations and their Characters
A variant of ( i i i ) goes as follows: if
U is an H-map, then by means of the diagram below we can 'transfer' cp to a map W , 4 U :
w-u
cp :
W
37
4
cp
+,
The definition of is independent of the coset representative chosen because of H-linearity, and g;l is the inverse to the identification of W with W, introduced in the proposition. Summing over u we obtain the extension p : V + U which is a G-map. This construction sets up a (1- 1) correspondence
HOmH(W,i*U) Homc(i,W, u). A variant of the construction which we have given for p = i,8 reduces it to the example of the regular representation. Compatibility with direct sums follows from the requirements we impose in i,, so it is enough to assume that W is irreducible. As such it is a summand of p r e g , ~and V = i,W is the image of W in p r e g , ~Uniqueness . follows by the method used to extend cp to p, now providing a G-bijection between two induced representation spaces V and V', restricting to the identity on W . This is the method used in [J-P. Serre]. Now for characters: Proposition 4.2 The character of the induced representation V = i, W can be computed from
(The divisor IHI occurs in the second sum to compensate for the fact that we s u m over all group elements rather than over coset representatives).
Proof. We examine carefully the action of a group element x on the direct sum V of the translates gW. Here in the first instance g runs through a family of coset representatives.
38
Representations of Finite and Lie Groups
The formula which we have to prove expresses xP(x)as a linear combination of values of xw = X,g on Hn(e1ements conjugate to 2 in G}. Changing the notation slightly and writing V = @, g,W, where g1 . . .g[G:H]are coset representatives we see that p(x) permutes the g,W among themselves. If xg, = g7h, then x sends g,W to g,W with u H T the permutation of the coset representatives associated with the element x. In order to determine the trace we take a basis of V consisting of a union of bases of the subspaces g,W. If u # T , the restriction of p to g,W makes zero contribution to the terms on the principal diagonal. The others give the trace of x on g,W; note that 0 = T implies that g;lxg, = h E H , and this h is the element which defines the action of G. To be quite precise we use g i l to map g,W back on to W, and use the action of h as an element of Aut(W). This gives the formula XJX) =
c
xe(g;lxgu),
U
gz1xgC EH
and the second formula follows by noting that all elements k of G in the left coset g H satisfy ~,g(k-'zk) = X,g(g-lxg) (invariance of 0 under Hconjugacy). 0 Direct calculation allows us to prove: Proposition 4.3 (Frobenius reciprocity) Let U and W be G-and Hrepresentations respectively. Then (Xi,W,XU)G = (XWIXi*U)H.
Proof. Assume that W and U are irreducible. The left-hand side counts the number of times U appears in i,W, which equals dim(HomG(i,W, V ) ) . The right-hand side counts the number of times W appears in i*U, which equals dim(HomH(W, z*U)). The two spaces of homomorphisms have been identified using the map 'p t+ (p. 0 In the exercises at the end of this chapter we give a step-by-step approach to G. Mackey's criterion for an induced representation to be irreducible. Given the importance of induction in constructing the character table of many finite groups the reader is strongly recommended to look at this, see also the book by J.P. Serre, $7.3-7.4. At this point it is also useful to give the matrix version of induction. For the reader who knows about tensor products over non-commutative
Chapter 4: Induced Representations and their Characters
39
rings, this is encapsulated in the formula V = @[GIgCHW , where @[GI is regarded as being a left module over itself and as a right module over @[HI. More explicitly let W have a basis w 1 . . . w,, so that in terms of this basis O(h) = (Oij(h))with hwj = Oijwi. If V is given the @-basis
Ci
{
9
~
'
8
~
~
~
~
~
~
~
~
~
m
~
~
~
~
~
~andggi=geh, ~ ~ ~ ~
then m
g ( g i @ w j )= C O k j ( ( S e ' g g i ) g l '8 wk. k=l
e
Define on G by e ( h ) = ( O i j ) if h E H and e ( g ) = 0, (the zero m x m matrix) if g $ H . Then the block matrix representing p ( g ) = i*O(g) is
Note that in this block array only one block in each row and each column is non-zero. Evaluation of the trace checks the character formula in 4.2.
Examples. For an arbitrary prime p 2 3 consider the dihedral group
DzP = { a , b : ap = b2 = 1,b-lab = a-'}. The element a generates a normal subgroup of index 2; let a H [ define a 1-dimensional representation, with C = e 2 ? r i / P . Inducing up to the group DZP itself gives a 2-dimensional irreducible representation described by the matrices
The matrix representing b carries the information that multiplying by b switches the cosets < a > and b < a >. The entry in the (2,2) position in a reflects the relation F l a b = a-1 in the given presentation. Note that the normality of < a > implies that the matrix representing a is diagonal. This is a useful general property, which will occur again for more complicated groups.
[-'
40
Representations of Finite and Lie Groups
Replacing D2p by the quaternion group Qdp introduces the relations 1,aP = b2 in place of the first two above. The matrices are almost 0 -1 the same; a is mapped to q!l) with q = eTi/P and b to . The
a2P =
(:
( )
-1 in the (1,2) position reflects the fact that the extension of < a > by < b > is no longer split, i.e. that b4 = 1 rather than b2. We next prove a very useful result (Blichfeldt's Theorem) on the representations of groups of order p t ( p = prime).
Lemma 4.4 Let G be a non-abelian group of orderpt. Then there exists a normal abelian subgroup A, which contains the centre Z(G) properly.
Proof. First note that the centre Z(G) of a p-group is always larger than the identity. This follows by counting the conjugacy classes, the number of elements in each of equals the index of a centralising subgroup, and hence equals 1 or a power of p . Hence the identity cannot be the only element to belong to a singleton conjugacy class. The quotient group G/Z(G)still has order equal to a power of p , so that its centre Z ( G / Z ( G ) )is non-trivial. Pull back a cyclic subgroup of order p in this centre to G, obtaining an element a which commutes: with Z ( G ) ,and take A = (a,Z(G)). Theorem 4.5 Let V be an irreducible @[GI-module.If dim@V > 1 there exists a proper subgroup H of G and an irreducible @[HI-moduleW such that V = i,W.
Proof. First suppose that V is faithful, i.e. p : G -+ Aut(V) is injective. Under this assumption we claim that there exists HI C G such that HI contains the subgroup A from Lemma 4.4,and an irreducible @[H1]-space W with V = i,W. Consider V as an A-space. As such it is a sum of irreducible 1dimensional A-spaces. Let v generate one of these (with character $). If w generates another 1-dimensional A-space and has the same character $J,then with Q E A , a(X1v X2w) = $(Q)(X~ZI X2w). If W$ C V consists of all 1-dimensional A-spaces with character $J, then
+
+
Let W = W$ for some fixed character $. Assume that W exhausts V, i.e. that we need only one character. For v # 0 E V and g E G we would have that g-'v describes a 1-dimensionalA-space, which would again have
Chapter 4: Induced Representations and their Characters
41
to have character $. But for a E A a(g-'v)
= $(a)g-'v,
and
(gag-')v
= g$(a)g-'v
= $(a)v (because
$ ( a ) is a scalar).
This shows that a and gag-' have the same effect on v for all g E G. But A is strictly larger than the centre Z ( G ) of G, and for some pair of elements g , a gag-' # a. This would contradict V being faithful, and we conclude that more than one character $ is needed to describe the A-restriction of V. Indeed G permutes the W, transitively. We argue as follows: Let v E W,,a E A and g E G. Then 4 w ) = g(g-lag)v = g$(g-lag)v = $g(a)gv, where & ( a ) = $(g-lw). Thus g : W , H W,, and 9-l : W,, --t W,. The action is transitive, since we can sum over all g translates, obtaining V' = CW+,( for some &,) C V. Irreducibility shows that V' = V. For 9
the final step we fix W = W$,, (for some $1) and let H1 stabilise W , i.e. hW = W . be the subgroup containing A with the property that h E HI Note that H1 is properly contained in G since V # W+l. As in the original discussion of induced representations (compare the construction of V1 above), V is certainly obtained from W by translating this subspace by means of a family of H1-coset representatives. This sum is direct by choice of H1 as the isotopy subgroup of the action of G on the {W+}.The subspace W used in the induction must be irreducible, since if it were not, V = i,W would be reducible also. It remains to remove the restriction that V is faithful. Let GO= Ker(p : G -+ AutV) and = GIGO. The space V is certainly faithful for the quotient group GIGO. If V is not 1-dimensional, G is not abelian, and there exists H c G with V = i, W for some fi-space W . Let H be the inverse image of fi in G. As a subgroup H contains Go and W is an irreducible H-space. The definition of fi as an isotropy group implies that H has the same property, and copying the argument for the quotient group we see that V = a, W as a representation induced up from W using the cosets G I H . Recall that 0 elementary group theory shows that GIH 2 G/Go/H/Go.
*
h
h
h
Note as a complement to the argument that by repeating the construction we eventually arrive at a 1-dimensional representation of some proper subgroup of the p-group G, from which to start the induction process.
42
Representations of Finite and Lie Groups
Non-abelian groups of order p 3 provide beautiful examples of both Blichfeldt’s Theorem and of the explicit construction of induced representation spaces. We recall that, up to isomorphism, there are two non-abelian groups of order p 3 , which for convenience we refer to as ‘metacyclic’and ‘elementary’. We again assume that p is odd; the case p = 2 (when G 2 0 8 or Qs) has been considered in Chapter 1. Denote the two groups concerned by P+ and P- with presentations
P- = { a ,b : up2= bp = 1,b-lub = u1+P}, and P+ = { a , b , c : ap = bp = cP = [c,u]= [c,b]= 1,[a,b]= c}. In both cases the centre equals the commutator subgroup of order p , giving an abelianised group isomorphic to the direct product of two cyclic groups of order p . Hence there are p 2 1-dimensional representations; the remaining irreducible representations are obtained by induction up from a normal subgroup of index p . As a check we note that P+ contains p 2 ( p - 1) conjugacy classes of elements. By direct calculation we see that in terms of matrices model p-dimensional representations are given by
+
and
Here C = e2.rri/Pand 7 = e2.rri/P2 are primitive roots, in the case of Pwe induce up from the subgroup generated by a, and in the case of P+ from (b,c). The remaining representations are obtained by taking powers of 17 and C respectively. The p-dimensional representations for P* are examples of so-called monomial representations. A group which satisfies the conclusion of Theo-
Chapter 4: Induced Representations and their Characters
43
rem 4.5 is called an M-group, a class which is certainly larger than groups of prime power order. Recall that a group is solvable if its derived series terminates:
where Gi = [Gi-l,Gi-l]. It is known that an M-group is solvable, see [L. Dornhoff, $151 but not every solvable group satisfies the M-condition. The best that can be done for an arbitrary group G, using the regular representation, is to show that each irreducible representation of G is contained in an induced representation for some subgroup H G. Our warning example is the so-called binary tetrahedral group T* , with presentation
The group T* maps onto the tetrahedral group T , with characters as described in a previous chapter, with a kernel which is central and generated by p 2 . The order is 24 and each irreducible representation of T also occurs for T*. Simple arithmetic shows that the remaining representations all have degree equal to 2; they are obtained as follows. Identify the special unitary group SU2 with quaternions z of unit length, and R3 as the subspace of EX4 consisting of purely imaginary quaternions y. The map y zyz-' for fixed z is linear and distance preserving, that is defines an element T ( Z ) E so3. As a homomorphism of groups T is two-to-one, with kernel consisting of {flz}. This subgroup of order two is central, and we have a commutative diagram.
T*
T
P
SU2
SO3
P
in which the lower horizontal arrow p denotes the irreducible; 3-dimensional representation of the tetrahedral group. The lifted homomorphism is seen to be irreducible, its restriction to the subgroup Q 8 is the standard representation first considered in the introduction, and restriction to (x) gives a representation with character w + w - l ( w = e2ai/3). (The two other irreducible representations have restricted characters equal to 1 w and 1 w - l respectively.) Since T* has no subgroup of index 2 (why?), the representation cannot be obtained by induction.
+
+
44
Representations of Finite and Lie Groups
For more on the corresponding map n : SU2 compact groups.
t
SO3 see Chapter 6 on
Exercises.
1. Determine the irreducible representations of the dihedral group
and of the quaternion group
2. Let p be an odd prime and P& the two groups of order p3 , whose irreducible representations have been described in the main text. Prove the claim made that P+ and P- both contain p 2 ( p - 1) conjugacy classes, and write down the character table for each of the two groups. What do you notice? 3. Prove that if the order of the group G is a power pt of a prime p , where t 2 2, then the abelianisation G/[G,G] has order equal to at least p 2 . Prove further that every irreducible representation of a group of order p4 bas degree equal to 1 or p . 4. Let G = H x K be the direct product of two subgroups, and let p be a representation of G induced up from a representation 8 of H . Show that p is equivalent to 8 @ p r e g ,where ~ p r e g ,denotes ~ the regular representation of K . 5. Find all the characters of S5 induced from the irreducible characters of S 4 . Hence recover the character table of S5. Repeat, replacing S 4 by the subgroup ((12345),(2354)) of order 20 in S5. *6. (Maclcey 's criterion f o r the irreducibility of a n induced representation). Let W be the representation space for 8 : H + Aut(W) and V the induced representation space for p = i,0. If H, = H"2Hx-l we can distinguish between the representations 8" (conjugate by z) and ResH, (8) of the subgroup H,.. Let 8" : H , -+ Aut(W,). Prove that the space V is irreducible if and only if both the following conditions are satisfied:
+
(a) W is irreducible, and (b) for each x E G - H the representations 0" and ResH, (0)
are disjoint, i.e. have no summands in common.
Chapter
4 : Induced Representations and their Characters
45
[Hint: V is irreducible if and only if the inner product ( X V ,X V ) = 1. By F'robenius reciprocity this equals ( X W , i*i,xw), with the second inner product being taken over H. Now decomposes i*i, W as a direct sum of representation spaces Indg3:W, for a suitable family of elements 5. (This is an application of the so-called double coset formula.) The reader may find it easier to consider first the special case when H is normal in G.)] *7. Let G be the symmetric group S, and let X = { 1 , 2 , . . . ,n}. Write X , for the set of all r-element subsets of X , and let 7rr be the permutation character of the action of G on X , . If T s n / 2 show that G has r+ 1 orbits in its action on X , x X s , and deduce that ( 7 r r , n,) = r+l. It follows that the generalised character 7rr - 7 r r - 1 is irreducible for 1 T n / 2 . *8. Let G be the semidirect product of A and H with A normal in G and the subgroup H 2 G/A. This means that each element g E G can be written uniquely as a product g = ah, and that multiplication is twisted by the action of H on A. The dihedral groups D2, or D2t are examples which we have previously considered. We restrict attention to the special case when A is abelian. The group H acts by conjugation on the group Hom(A, C") of characters of A. If xi represents an orbit of this action, then we can form the product Gi = A . Hi where Hi stabilises xi. As a complex valued class function xi extends to Gi, and by composing with the projection map Gi .+ Hi for any irreducible representation 0 we obtain a representation 0' of Gi. Let pi,@= Ind& 8 0'). Prove
< <
< <
(i) The representations pi,e are irreducible, and equivalent only if the suffixes i are equal and the representations 0 equivalent. (ii) Every irreducible representation of G is equivalent to some pi,e. [Hint: For Part (i) apply Mackey's criterion from Exercise 5, and then restrict to the subgroups A and H. For Part (ii) again restrict the given irreducible representation of G to A, and use the induced H-action to identify a subgroup Hi which fixes one of the summands of the canonical decomposition of the CA-module. Working backwards gives xi 8 0, and by irreducibility the induced representation space for G exhausts the one with which we started.] 9. Apply the method of Exercise 7 to the metacyclic group of order p q with presentation
D,, = { a , b : ap = b* = 1, b-lab = a r ,
T*
= l(mod p ) , qlp - 1).
Here p and q are distinct prime numbers. First recall from Chapter 2 ,
46
Representations of Finite and Lie Groups
+
v.
Exercise 2, that the number of conjugacy classes equals q In the same exercise you have already considered the non-abelian group of order 21.
Chapter 5
Multilinear Algebra
This section is devoted to a more elegant version of the tensor product construction] and leads into a discussion of symmetric and alternating products] generalised from 2 to n. Alternating or exterior powers are important in the general theory of representations; symmetric powers are important for certain particular groups. Let R be a commutative ring - the faint-hearted may assume that R is a field, even restricting themselves to the real or complex numbers without losing much in the following discussion. El . . . ,En, F are R-modules and Ln(E1,.. . ,En;F ) the module (or vector space) of n-multilinear maps f : E l x * . . x En 4 F .
Our aim is to construct a tensor product El 8 EZ 8 . .. 8 En, which will have the following universal property: given a multilinear map f : El x ' . ' x En 4 f , there is a natural map cp : El x . . . x En ---f tensor prodwith mapping (R-linearly of course) uct; and a factorisation of f as the tensor product El 8 ' . ' @ E n into F .
T,cp,
7
Existence Let M be the free R-module generated by the set of all n.tuples (21,.. . ,2,) with xi E Ei, and let N be the submodule generated by elements of the following types:
. . zi + xLl.. . , x n ) - ( ~ 1 , ... ,xi,. . . 2 , ) - ( X I ] . . . xi,.. . , x n ) and ( ( X I , . . . , m i l .. . , x n ) - r ( x l 1 . . 2,) for all xi,xi E Ei and a E R.
(51,.
I
Obviously El x . . . x En maps injectively into M , and we can compose this map with the quotient map onto M I N obtaining 47
Representations of Finite and Lie Groups
48
: El x
. . . x En 4 M / N .
We claim that this multilinear map has the required universal property. Let f : El x . . . x En -+ F be multilinear. Since M is free, we can use the value of f on the n-tuple ( X I , . . . , 2), to define the image of ( 2 1 , .. . x n ) as free generator of M in F . This gives a commutative triangle
and multilinearity implies that the induced map M -+ F takes the value 0 on elements of N . Hence the factorisation : M / N + F exists, and makes the triangle with M / N replacing M commutative in its turn. Since the image of ‘p generates the quotient module, f determines uniquely. As indicated already we write El 8 . . . 8 En for M I N .
7
7
a
Uniqueness The tensor product is unique up to isomorphism, since if were another, the universal property applied to M / N and in turn would produce maps between the two of them. M / N is the quotient of a free module and those maps are inverse to each other. We write ‘p(x1,. . . 2), = 21 8 . . . x,, and if it is necessary to emphasise the coefficient ring R, we write El 8 E2 as El 8 E2. Note that the effect
a
R
of dividing out by N in M is to give the relations ~
1 (x2 8 x1
+ xi) = x1 8 x2 + x1 8 xi and 8 rx2 = r(x1 @ x 2 ) , in
El
8 E2,
with obvious generalisations from 2 to n. Every element of El 8 E2 can be written as a sum of terms x 8 y with x E El and y E E2, because such terms generate El 8 E2 over R and T ( Z 8 y) = rx 8 y for r E R.
Warning example: As a tensor product over Z, Z / m 8 Z/n = 0 if m and n are coprime. More generally Z / m 8 Z/n Z / d with d = g c d ( m , n).
Chapter 5: Multilinear Algebra
49
The first assertion follows from the relations n(x CB y) = x 8 n y = 0 and m ( x €3 y) = mx 8 y = 0. The second is left as an exercise. The next proposition is typical of many.
Proposition 5.1
The tensor product is associative. Thus the map
(x c3 y) €3 z
Hx
8 (y 8 z )
defines an isomorphism between (El8 E2)8 E3 and El 8 (E28 E3). Proof. Uniqueness is clear. We must check that the definition of the map actually makes sense. Let 2 E El. The map
E2 x E3 (Y,
2)
H
(El8 E2)8 E3 .( 8 Y) 8 2
for fixed x is bilinear and factors through E2 €3 E3. Allowing x to vary gives a bilinear map El x (E28E3)into (E18E3)8E3, which factors through El €3 (E2€3 E3),and justifies the notation we have used. Similarly, we have
Proposition 5.2 The tensor product is commutative. The map x 8 y y 8 x defines a unique isomorphism between El 8 E2 and E2 8 El.
H
Proof. As in 5.1 start with the bilinear map (x,y) H y 8 x, factoring as x 8 y H y 8 x,and observe that this map squares to the identity, and thus must be an isomorphism. The construction of the tensor product shows that a family of R-linear maps fi : Ei -+ Ei combine to induce a map
T(f1,. . * fn) = fi 8 f2 8 . . . 8 fn : El 8 ... @ E n 4 Ei 8 . . . €3 EA, whose effect on XI 8 . .. €3 x, is to map it to fi(x1) 8 fi(x2) 8 . . . 8 f n ( x n ) . There are occasions when this notation can be ambiguous - thus f l @ . . . fn can also stand for an element of the module L(E1,Ei)8 - .. 8 L(En, EL).
Proposition 5.3
If E , F and G are R-modules
L(E,L ( F ,G))z L2(E, F ;G)2 L(E 8 F,G). Proof. The first isomorphism is an exercise in linear (or at most quadratic) algebra. The isomorphism L2(E, F ;G) L(E €3 F,G)is given by the map f c f used in the definition of the tensor product. Injectivity comes from
Representations of Finite and Lie Groups
50
uniqueness of the factorisation, surjectivity by composition of g : E@F + G with 'p: E x F + E @ F .
Proposition 5.4 n
Then F @ E
Z
Let E = El
@
' . . @ En be a direct s u m of R-modules.
$ ( F @ Ei). i=l
Proof. The direct sum decomposition of E is equivalent to the existence of projection maps 7ri : E -+
Ei such that n
.
7ri 7ri = 7ri, 7 r i . 7rj = 0
(i # j ) and
c 7 r i
= Id.
i=l
Inspection shows that the maps Id 8 xi : F 8 E + F @ Ei satisfy the same relations, and hence are associated with the required direct sum decomposition of F @ E. We now have a very important special case. Let R be a field and E a 1-dimensional vector space over R with basis {v}. Consider F @ E , in which every element can be written as a sum of terms y 8 v with y E F and r E R. But we know that y 8 ru = r y 8 u,and using linearity on the m
left we see that our typical element of F @ E has the form
( Cy i ) 8 v. i=l
The linear map ( y ,r v ) I+ r y induces a linear map F @ E + F , and (y) H y 8 v induces a linear map F -+F @ E . These maps are inverse to each other, so that each element in F 8 E can actually be written uniquely in the form y 8 v , for y E F. Combined with 5.4 this shows that
Proposition 5.5 Let R be afield and E have afinite R-basis (v1 . . . vn}, then every element in F @ E has a unique expression in the f o r m
c n
yi 8 ui for elements yi E F.
i=l
Corollary Let E and F have R-bases (v1 . . . un} and {wl. . . w,} respectively. Then the vector space E @ F has basis {vi 8 wj} and dimR(E @ f) = (dimR E)(dimR F ) . This corollary brings the definition of a tensor product of vector spaces in this chapter round to the more utilitarian definition in Chapter 2.
Chapter 5: Multilinear Algebra
51
Extension of the coefficients R: Let R -+ R' be a homomorphism of commutative rings (for example the natural inclusion map R 4C ) . Use this map to define an R-scalar product on R', and consider the trilinear map
R' x R' x E 4 R ' B E (a,b, 2 , ) H ab 8 5, inducing an R-linear map R' 8 (R' 8 E ) ++ R' 8 E . We also have an R-bilinear map R' x (R' 8 E ) -+ R' @ E making R' @ E R
into an R'-module. We refer to this construction as extension of coefficients from R to R'. Coefficient extension is illustrated by the complexification of a real vector space, and reduction modulo p (prime) - induced by the ring homomorphism Z --f Z / p . We have met the first in the last section of Chapter 3 (Theorem 3.10). For the second see the discussion of SLz(P,) below and Appendix C.
Basic properties (i) Let E be a free module over R with basis (211 . . ,vn}. Let v,!= 1@vi, then R'@E is a free module over R' with basis {vi, . . . , vk}. 1 .
R
(ii) Extension of coefficients is transitive. Thus given R R' is an isomorphism of R"-modules R" @ E E R" 8 (R' @ E ) . --f
R'
R
-+
R", there
R
The tensor algebra of the R-module E : We start with the general 00 notion of a graded ring R = @ R,, with the grading defined by the nonn=O
negative integers {0,1,2,. . . }. Here each R, is an abelian group and multiplication is such that, if x E R, and y E R,, then xy E R,+,. Note that & is a subring. n If E is an R-module and T 2 0, write T n ( E )= @ E and by convention i=l
define T o ( E )= R. We have already shown how to define T ' ( f ) for any linear map f : E F . Since the tensor product construction is associative (5.1) we have a bilinear map T " ( E ) x T m ( E )-+ Tn+,(E), which is again associative and which can be used to define the graded ring --f
T ( E )= For an R-linear map f : E
T(
4
@.
n=O
Tn(E).
F it is natural to write
. @ zn) = f (31) @ . .
@ f (zn).
52
Representations of Finite and Lie Groups
Now let E be free and finite-dimensional over R.
Definition. The R-algebra P is a non-commutative polynomial algebra over R if there exist elements t l , . . . ,t , E P such the monomials p i ( t ) = ti, . . . ti, (1 6 z j 6 n ) form a basis for P over R. Again by convention, when m = 0, the corresponding monomial is 1 E P. The elements tl . . . t, generate P as an algebra over R, and P = @Pm m
with Pm consisting of R-linear combinations of monomials ti, ti, . . . ti,. We describe t l , . . . , t, as independent non-commutative variables over R.
Proposition 5.6 If E is free and dimR(E) = n, then the tensor algebra T(E) is isomorphic to the algebra P described above. If (211,. . . , v,} is an R-basis for E, then the elements pi(v) = vi, @ . . . @ vim (1 6 ij 6 n) form a basis for T"(E) and every element of T(E) can be uniquely expressed as an R-linear combination of finitely many monomials p i ( v ) . Proof. Apply Proposition 5.5. At least when E is free and finite-dimensional over R, there is an isomorphism of R-algebras between T(EndR(E)), the tensor algebra of the 00
ring of homomorphisms of E into itself, and @ EndR(Tn(E)). n=O
For a pair of linear maps f , g we have f @ g (f, g ) , where T (f , g) has been defined above (between 5.2 and 5.3). One checks that isomorphism holds in each dimension m, and that products are preserved. The details which are mainly a question of keeping the notation straight are left to the reader.
-
Alternating and Symmetric Products The m-multilinear map f : E x . . . x E
4
F is said to be alternating if
m
f ( z i ,. . . , 2,) = 0 whenever xi = xj for some i # j . Let Urnbe the submodule of Tm(E)generated by all elements of type 2 1 @ . . . @ 5, with xi = xj for some i = j. Define A m ( E ) = T"(E)/U,. We have a linear map E x . . . x E -+ Am(E) given by 'p composed m
with the quotient homomorphism, which is alternating, and universal for alternating multilinear maps into some arbitrary module F .
Chapter 5: Multilinear Algebra
53
- . - x E + F factors through
This means that any such map f : E x
m
A"(E). The proof of this is obvious, since one has only to take the factorisation 7 through T m ( E )and note that 7 vanishes on Urn. We denote the image of an m-tuple ($1 , . . . , zm)by z1 A 5 2 A . . . A zm. As with the tensor algebra we define the alternating (or exterior) algebra of E by A(E)=
m=O
Am(E).
A few words about the definition of products: write 2l = and check that 2l is an ideal in the algebra T ( E ) . We can define an R-algebra structure on the graded quotient T ( E ) / U by means of ( ( X 1 A * * . A Z m ) ,( y l A . . . A y n ) ) H z ~ A x ~ A . . . A ~ ~ A E * .Tm+n(E)/Um+n. .A~~
For the 'wedge' or alternating product we note that xAy= -y,-,~, since
(z
+ y ) ~ ( a+: y) = 0. Here dim z = dim y = 1.
I f f : E + F is R-linear, then we can define A(f) : A ( E )-+ A ( F ) in the = f(q)Af(x2) A same way as T(f) and note that A( f ) ( z l A z z A . . . A )z, . . . A f(zm). Proposition 5.7 Let E be a free R-module with dimR E = n. If m > n A ~ =E0 . Let ( ~ 1 , .. . , v,} be an R-basis of E . If 1 m n, then AmE is R-free with basis {ui,A . . A vim; 1 il < i 2 < * . . < im < n}.
<
< <
Proof. Consider first the case when m = n, and for safety's sake let R be a field. (In representation theoretic applications this is the only case which interests us.) Then the subspace lu, admits a 1-dimensional complementary subspace with basis u1 8 . . . 8 u,, mapping to the required basis U l A u 2 / \ . . . ,-,v, of A"E. Now let 1 < m < n and suppose that we have a relation (9
between the generating elements of AmE. Fix some m-tuple ( j ) = {jl . . .jm} and let { j m + l , . . . ,jn} be the complementary ( n - m)-tuple. Take the wedge-product of the sum above with uj,+, A . . . . . . ujn. Each summand except that corresponding to ( j ) will have two copies of some ui in it, and hence will vanish. Up to order, a ( j ) is now the coefficient of v 1 A ... A u,, and by the basis property of this product it follows that a ( j ) = 0. When r = 0, 1 is a basis for R = A0E, and trivially AmE = 0 for m > n. For the dimension count the family of subsets of { 1 , 2 . . .n).
54
Representations of Finite and Lie Groups
By checking first on basis elements we see that A ( E ) is anticommutative in the sense that xy = ( - l ) d i m z ' d i m y Y X . We now turn to the symmetric analogue S ( E ) of A ( E ) . An rnmultilinear map f : E x . . - x E + F is said to be symmetric if
m
f ( ~ 1 , ... ,x,) = f ( x o ( l )., . . , x,,(,)) for all permutations (T E S,. In T m ( E ) let ,6 be the submodule generated by all elements of the form
for all xi E E and all CC
-
(T
E
S,.
Define S"(E) = T m ( E ) / b mand S ( E ) =
@ S m ( E ) ,the symmetric algebra of E. As in the case of alternating
m = n.._
maps the composition E x
.
xE
--t
T"(E)
--t
S m ( E ) is universal for
m
rn-multilinear symmetric maps. For a reason which is about to become obvious we write ~ 1 x 2 .. x., for the image of ( X I , . . . ,x,). Proposition 5.8 Let E be R-free with dimRE = n, and basis ( ~ 1 , ... , vn}. As elements of S 1 ( E )the vi are R-linearly independent, and S ( E ) is isomorphic to the polynomial algebra R [ v l , .. . , v,]. Proof. This is intuitively obvious since modulo b = @b,
m
we can iden-
tify elements which differ only in the order of the components xi, i.e. we can make the variables in Proposition 5.6 commute. More formally let tl . . . tn be independent variables over R and form the polynomial algebra R [ t l , .. . t n ] .Let P, be the submodule of homogeneous polynomials of degree m, and define a map E x . . . x E + P, as follows:
This map is multilinear and symmetric, hence factors through S"(E). The element w 1 . . . w, maps to tl . . . t , and similarly for each wi, . . . wi,. Linear independence of the monomials p ( ( ) ( t implies ) that this map is an isomorphism (obvious for rn = 1). By inspection the map is compatible with multiplication and grading.
Chapter 5: Multilinear Algebra
55
The Representation Ring R ( G ) and its A-structure The set of equivalence classes of @[GI-modules(@-vector spaces with Gaction) admits an addition (direct sum @) and a multiplication (tensor product 8 over C). Because of the results in Chapter 2, in particular Theorem 2.8 and its consequences, we can turn our structured set into a commutative ring R(G), called the (complex) representation ring of G. If it is necessary to emphasise the field of definition we write R(G) = RU(G). The elements of R(G) are formal differences [V]- [W] of equivalence classes of k
miV, with
modules; alternatively we can consider integral combinations i=l
mi E Z,rather than mi E Nu{O}. The real representation ring RO(G) is defined in the same way. These rings have a large amount of additional structure. If V is a @[GI-moduleand g E GI then since p ( g ) E Aute(V) we have an induced automorphism T"(g) : T m ( V )-+ T"(V) for each m. This makes both T"(V) and the direct sum T ( V )= @T"(V) into C[G]m
modules, with similar remarks applying to the symmetric and alternating powers. It is traditional to write Am(V) for the image of the alternating power A"(V), or rather of its equivalence class, in R(G). oc)
C
Notation: & ( V )=
Am(V)tm.
m=O
Strictly speaking one first defines At on the multiplicative monoid of equivalence classes of C[G]-modules, i.e. considers positive representations only, and then extends A t to all of R(G). In the end one obtains an exponential map At : R(G)
+
1
+ tR(G)[[t]]
of the abelian group R(G) into the multiplicative group of formal power series with constant term equal to 1. The fact that At is a homomorphism follows from
Lemma 5.9
@
( E ) @ r \ j ( F )E A ~ ( E @ F ) .
i+j=m
Proof.
0
Compare bases on the two sides. 00
Similarly we can define s t ( V )
=
C sm(V)tm1using
the symmetric
m=O
powers. For any C[G]-module V , we then have the non-trivial relation s , ( V ) L t ( V )= 1.
Representations of Finite and Lie Groups
56
If dimCV = 1, &(V) = 1 relation follows since
+
(1 s l t
+ (Slt)2 +
+ s'(V)t,
* *
since Az(V)
=
0 for i > 1. The
.)(1- s l t ) = St(V)X-,(V)
=
1.
The general case may be reduced to a sum of 1-dimensional representations by means of the so-called 'splitting principle' for A-rings (see [M.F. Atiyah, D.O. Tall] or [D. Knutson]). The author believes that the full power of representation theory only becomes apparent when the alternating structure of R(G) is taken into account. A clue to this is provided by the following result for the symmetric group S, .
The exterior powers A k V of the standard (n - 1)dimensional representation V of S, are all irreducible, 0 6 k 6 n - 1.
Proposition 5.10
Proof. If x denotes the character of the permutation representation (add a trivial summand to V) then V will be irreducible provided that the inner product of x with itself equals 2. The same holds for AkV, since
we again write x for X A k @ n and will prove that (x,x) = 2. Let A equal the set {1,2,. . . ,n}. For a subset B of A with k elements and g E S, let 0 if g B
# B,
1 if g B = B and glB is even, and -1 if g B = B and glB is odd.
Then by looking a t a basis for the kth exterior power we see that x ( g ) =
C{g}B. B Therefore
Chapter 5: Multilinear Algebra
57
Here the sums are taken over subsets B and C of A containing k elements, except that in the last sum we neglect zero terms, and sum over those g with gB = B and gC = C. Such a permutation g splits into permutations of the four subsets B n C , B \ ( B n C ) ,C \ ( B n C) and A \ ( B U C). If C equals the numbers of elements in B n C rewrite the last sum above as
-cc c c c c 1
n!
B
c
aESe b E S k - e
CESk-e
(sgna)2(sgnb)(sgnc)(sgnd)2
dES,-nk+e
( s g n b) B
C
bESk-e
(sgnc)
CESk-e
The last two sums contribute zero unless k - C = 0 or 1,since otherwise the individual summands will cancel in pairs. If k = C, B and C coincide, and the expression reduces to $ C k ! ( n - k ) ! , B
5
which equals ( ; ) k ! ( n- k ) ! = 1. The terms with k - e = 1 similarly add up to 1, and required.
(x,x)= 2, as 0
If we refer back to the examples discussed at the end of Chapter 2 we see that S5 has seven irreducible representations, five of which are given by 1,V, A2V,E and E 18V, where E denotes the 1-dimensional determinant module. We know that V @V splits as A2V@S2V;the latter is a reducible module of dimension 10. This splits as U @ ( E @ U ) say. Assuming the relation st(V)X-t(V) = 1 we see that the character table can be built up from a knowledge of the character of V. In Exercise 3 below we invite the reader to consider SS in a similar way.
* Representations of SL2(P,)
in Characteristic p
As a lead-in to the representations of the compact group SU2 in the next chapter, consider the following application of symmetric powers to the IFpnrepresentation theory of the finite group SL2(IFP).Note that we must expect this to be totally different to anything so far considered, since we can no longer divide an expression by the prime p , and Maschke’s Theorem must be expected to fail. Let K be an algebraic closure of the finite field IF, and G = SLz(IF,), the group of 2 x 2 invertible matrices of determinant 1. We will construct p irreducible K[G]-modules Vm(O< m < p - 1) with dimK V, = m + 1. Let
Representations of Finite and Lie Groups
58
Vm be the K-vector space of homogeneous polynomials of degree m in the independent variables x and y. If A = (aij) E G we put
This gives us a (right) representation module of the correct dimension, which we will show to be irreducible. Let (0) # W C V, and suppose that n
0 # f =zajxjy"-j
E Wl with a,
# 0 and n < m.
j=O
For t E IF, write
S ( t )=
(i I)
n
Then f S ( t ) =
C
uj(z
and T ( t )=
+ ty)jym-j
(: :).
n
=
C
f j ( x l y ) t j E W. Hence by
j=O
j=O
rearrangement we have a family of polynomials such that fo = f and f n = anym. W is closed under scalar multiplication by elements of IF,, so P- 1
C t - l ( fS(t)) E t=l In IF,
Ctt = t=l
I)-
1
t=l
w.
-1 if p - 1 divides i
0 otherwise] so that P-1
n
n
P-1
t=l
j=o
j=o
t=l
=i
-f i for 1
p-llj-2
< i
-f o - f p - l for
i = 0, p - 1 = n = m.
Since fo = f E W it follows that fi E W for all i = 1,.. . n. In particular] since f n = anym,ym E W .
Chapter 5: Multilinear Algebra
59
Similarly we show that W also contains the element
= ‘Ct-qta:
.&j(y”T(t))
+ y)”
t=l
t=l
( 7 ) x j f - j
ZP-’+
yp-l
for 0 < j < m < p - 1, and 0 for m = p - 1 , j = 0.
<j
- 1 = m.
Since yp-’ already belongs to W , it follows that xjym-j E W for all j such that 0 6 j 6 m. Thus W = V, and V, is irreducible for 0 m 6 p - 1. That the family of modules which we have constructed is exhaustive requires a mod p version of the counting argument in Chapter 2. We attempt to make this plausible in Appendix C. In characteristic 0 the number of irreducible representations in AutK ( V ), with K algebraically closed, equals the number of conjugacy classes of elements in G. In characteristic p ( p dividing /GI) we need to count the number of conjugacy classes of elements of order prime to p . Call these p’-elements. In the case of G = SL2(IFp)of order p ( p 2 - 1) we will have shown that the family {V, : 0 m p - 1) is exhaustive if we can show that G contains p p’-classes. This reduces to the
<
<
<
Claim: Trace distinguishes p’-elements, and each element of IF, occurs as a trace. For the second part f 2 are realised by f l 2 . If s
# f 2 write A ( s ) =
( t)
with trace equal to s. We need to show
that the order of A ( s ) is not divisible by p . Any p-component A ( s ) , must be conjugate to B =
(i ‘1) ,
which generates a Sylow p-subgroup.
, so, by an easy calculation The element A ( s ) itself centralises A ( s ) ~and must be conjugate to an element in the centraliser of B. However (see Chapter 8 below) this subgroup consists of matrices of the form f
(; ;>,
all of which have trace equal to f 2 (excluded). Hence A ( s ) is a p’-element. It remains to show that if t r ( A ) = s and A is a p’-element, then A is conjugate to A ( s ) . Write q ~ ( t = ) t2 - st 1 for the characteristic polynomial. If s = f2,p ~ ( t= ) (t f l)’, A has the repeated eigenvalues f l , and A is conjugate in GL2(IFp)to a matrix of the form fB above. But A2P = 1 2 , and since (IAI,p)= 1, A must equal f l 2 .
+
Representations of Finite and Lie Groups
60
If s # f 2 then A is not a scalar multiple of the identity and there exist a pair of vectors { 211,212 = 211 A } which are linearly independent. By the Cayley-Hamilton Theorem 0 = c p ~ ( A= ) A2 - sA 1 2 , and so
212A =
= -211
+
+
S W ~ .
This shows that A
N
A(s) in the general lineargroup GL2(FP),i.e. X-lAX = A ( s )
for some X E GLz(IF,). In order to sharpen this to conjugacy in the special linear group we need Y belonging to the centraliser of A(s) in GL2(FP) with det(XY) = 1. We then have ( X Y ) - l A ( X Y )= Y-lA(s)Y = A(s). For this is enough to show that the determinants of elements in the centraliser take all values in the multiplicative group IF: = IF, - (0). By direct calculation centralisers of A(s) have the form
{
C Yx
:,>
: 2,y E IF,,
+
z2
+ szy + y2 # 0
+
1
Does the quadratic form x2 szy y2 represent all non-zero values in IF,? If p = 2 this is obvious; for p = odd we have x2
1 + sxy + y2 = (z + -sy)2 + (1 - -s41 2)y2 . 2
If 1- as2 is not a square, we are done. Otherwise 1 - as2 is a non-zero square (s # f 2 ) , the theory of binary quadratic forms shows that x 2 szy y2 is equivalent to q2 and represents all values. Our treatment of this example is taken from [N. Blackburn, B. Huppert].
+
+
c2 +
Exercises.
1. Let (p, V) be a representation of G of dimension d. i) Compute the dimension of S"V and A"V for all n. ii) Let g E G and let XI,, . . , Ad be the eigenvalues of g on V . What are the eigenvalues of g on S"V and AnV? iii) Let f(x) = det(g - 51) be the characteristic polynomial of g on V . Describe how to obtain x A n V ( g ) from the coefficients of f ( x ) . iv) * Find a relation between X S n v ( g ) and the polynomial f(x). 2. Calculate xAzp and xszp where p is the irreducible representation of dimension 2 for Dg, and repeat for Q 8 . Are R(Q8)and R(D8) distinct as rings with A-structure?
Chapter 6: Representations of Compact Groups
61
3. Using any of the methods suggested in this and earlier chapters construct the character table for the symmetric group Ss. Which irreducible representations can be constructed using exterior or symmetric powers of the 5-dimensional irreducible summand of the permutation representation?
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Chapter 6
Representations of Compact Groups
A topological group G is a group provided with a topology such that the structural maps
-
p:GxG-G
( 9 ,h )
and gh
-
L:G-G 9
9-I
are continuous. Such a group is said to be compact if it is compact as a topological space, that is every cover by open subsets admits a finite subcover. A Lie group is a topological group which satisfies an additional smoothness condition. Discussion of this is postponed until the next chapter. For the moment we note that the matrix groups we will consider all satisfy the Lie condition. Examples of such compact groups are the matrix groups U,, SU,, 0, and SO,, the first two being defined over the complex, the second two over the real numbers. Thus
U, = { P E GL,(C) : PTP = PpT = I,} SU, c U, and consists of all matrices with determinant 1. On = {Q E GLn(R): QTQ = QQT = In} SO, c 0, and again consists of those Q with det(Q) = 1. On each group the topology is inherited from the topology induced on GL,(R) or GL,(@) by their inclusions as open subsets of Rn2 or Cn2. In both these cases openness follows from their definition as the complements of the hypersurface det = 0. Continuity of multiplication follows from the rule for multiplying matrices, continuity of inversion from the definition of A-' as (det A)-'adj(A). Since the topology is Euclidean a subset is compact if and only if it is closed and bounded. This follows from the definitions for each of the four families of subgroups above. Familiar low 63
Representations of Finite and Lie Groups
64
dimensional examples are UI E SO2 E { z E C : tz = 1) and SU2 = { (z1,z2) E C2 : ~ 1 % ~ 2 %= 1) . The second group can also be identified with the quaternions of norm equal to 1in W.In a later exercise the reader will have a chance to prove that there is a 2-to-1 continuous homomorphism from SU2 onto SO3. In studying the linear representations of a finite group G of order (GI we have frequently achieved invariance by averaging over the elements g E G, i.e. replacing a map f to C or some vector space by C f(g). An
+
6
gEG
analogous operation exists for compact groups, if we replace the sum by an integral JG f(g)dg with respect to some measure dg. More precisely one proves the existence and uniqueness of a measure dg carried by G ,such that f(g)dg = JG f(h-lg)dg = JG f(gh)dg (left and right invariance), and (ii) JG dg = 1 (normalising the integral so that G has volume equal to 1).
(i)
JG
For a general compact, or indeed locally compact, topological group the proof that such a measure exists is relegated to Appendix A. But for the matrix groups which are our concern, and in particular for SU2 , we can proceed as follows. Start with the simplest example IR with addition (+) as group operation, and k(z)dz as line element. For left invariance under g E IR we need
This is achieved with k(z) = z-', so integration is with respect to z-ldz. Generalising this to m x m matrices we use the matrix A-ldA of differential forms. (If A = ( a i j ) ,dA = (daij).) This argument can be applied to
SU2, when A =
(: ib)
with laI2
+ lbI2 = 1, but there is a more geomet-
ric approach, which the reader may find more appealing. Topologically the group SU2 is homeomorphic to the 3-sphere, to which we may assign Cartesian coordinates (z,y, w, z ) such that x 2 y2 w2 z2 = l or hyperspherical coordinates 8,cp,$'. These are related by w = cos8, z = sinecos$, z = sin 8 sin $ cos 4, y = sin 8 sin $ sin cp. Taking 3-dimensional sections through R4 we obtain the familiar area elements on S 2 in terms of spherical polar coordinates as illustrated below.
+ + +
Chapter 6: Representations of Compact Groups
w=o,
65
,g=z2
z = cos$ x = sin $ cos cp y
= sin $ sin 'p
W
z=O,$=K
2
w = case x = sin 8 cos cp y = sin8sin'p X
W
x=o, cp="
2
w = case y = sin 8 sin $ t = sin@cos$
Y
The three area elements sin $d$dcp, sin 8d8dcp and sin 8d8d$ fit together as 27r2dg = sin2 8 sin $dOd$d'p. The use of hyperspherical coordinates implies invariance under rotation, i.e. left and right invariance under the
66
Representations of Finite and Lie Groups
action of the group SU2. The factor 27r2 normalises the volume of S 3 to one. If G is compact but no longer finite a representation of G is a continuous homomorphism p : G t Aut(V). Here V may be an infinite-dimensional complex Hilbert space, but one proves that such a representation is isomorphic to a (Hilbert) direct sum of finite-dimensional representations, each of which may be assumed to be unitary. It is clear how to proceed - go through the arguments in Chapter 2 for finite G replacing bxf(g)by f(g)dg
sG
9
everywhere. For example the inner product of two class functions is defined bY (cp,$,) =
/
G
F(g)$(g)dg*
Schur’s lemma and Maschke’s Theorem continue to hold, so that any finite dimensional representation space is decomposable as a direct sum of irreducible representations V,. The number of summands mi isomorphic to a fixed irreducible representation is again independent of the decomposition, and the character map defines an injection of the representation ring R(G) into the class functions CC(G). This is another way of saying that irreducible representations are distinguished by their characters. Orthogonality holds between irreducible characters, and the irreducible representations of a compact abelian group are all 1-dimensional. For this last result we cannot use the second counting argument but may still use the corollary to Schur’s lemma. We repeat this: because G is abelian, ‘scalar multiplication’ by an element g E G is actually a C[G]-map. But if V is irreducible the elements of H o m q ~ ] ( VV) , are necessarily scalars. Hence dim@V = 1. The arguments in Chapter 2 were framed, as far as possible, to apply to compact topological groups with a finite group as a discrete group of volume one. The exceptions are the arguments which depend on being able to count conjugacy classes, the number of elements in each class, etc. For example, it is necessary to define the regular representation Vreg as the Hilbert space of square integrable functions on G with group action given by
This representation is infinite-dimensional, and it is no longer possible to speak of its character, so propositions using this no longer have content. It can be shown that each irreducible representation of G is still contained in
Chapter 6: Representations of Compact Groups
67
Keg with multiplicity equal to its degree (Corollary 2.9 A). The projection maps used in 3.1 still make sense, i.e. p i = ni JG x i ( g ) p ( g ) , so we can still speak of isotypic summands . Proposition 3.9 on products G1 x G2 is still true. The invariant measure for the product is dglAdg2. So long as H is a closed subgroup of finite index in G, the notion of a representation of G induced up from one of H holds without further change. If the index of H in G is infinite, it is necessary to use a construction which leads to an infinite-dimensional representation space. This deserves a separate subsection. Induced Representations
Let H be a closed subgroup of the compact topological group G, and let W be a representation space for H. We define the underlying topological vector space over C for the induced representation i$W = i,W to consist of all continuous functions f : G -+ W which satisfy f ( g h ) = h - ' f ( g ) for all h E H , g E G. Topological digression: we give the function space i,W c W G the compact-open (CO) topology, which has a subbasis of open sets given by W ( K ,U ) = {f : f ( K ) c U , K compact in G and U open in W } . If the vector space W has a norm, the CO-topology on i,W coincides with the sup-norm topology (G is compact). Both G and W are sufficiently restricted for the expected useful properties to hold. Thus if we define a G-action on i,W by ( g f ) ( z ) = f ( g - ' z ) this action will be continuous. For more on the topology of function spaces see [J.L. Kelley, Chapter 71. Although this definition is easy to use it may not be clear how it is related to the definition given in Chapter 4 for the special case when H has finite index in G. As an alternative we can model the construction of i,W on that of @[GI @ W in the finite case. CH
Let G x W be the quotient space of G x W under the equivalence relation H
( g h , h - l w ) for h E H . This corresponds to the right action of H on C[G]and its left action on W . The projection ( 9 ,w)
N
r:GxW+G/H H
(9,w)
gH
is a continuous G-map, and for each f E i , W we have a section s f : G / H G x W ( r s f = identity) given by s f ( g H ) = ( g , f ( g ) ) . H
-+
Representations of Finite and Lie G T O U ~ S
68
The map f H sf is an isomorphism between i,W and the space of continuous sections of T.
Proposition 6.1
Proof. The naturality of the diagram below allows us to use arbitary sections to construct a section t : G -+ G x W which satisfies Pt = sp. More precisely G x W can be identified with the subset of G x (G x W) H
consisting of pairs ( 9 , { (g', w ) } ) such that p g = n{ ( g I l w ) } (pull-back), and t ( g ) = ( g , s ( g H ) ) . Composing t with the projection of G x W onto W we obtain a map f : G -+ W which lies in i,W. The map s H f s is inverse to f H sf. The group G acts on the space of sections via ( g s ) ( z )= gs(g-'z).
G-
P
GIH
gH O
As a corollary we obtain F'robenius reciprocity. Proposition 6.2
For the H-space W and G-space V we have a natural
isomorphism Homc(V,i,W) E HomH(i*V,W).
Proof.
This is modeled on the isomorphism for finite groups @[GI8 (W 8 VIH) 2 (@[GI8 W ) 8 V. H
H
The notation HomG refers to continuous maps compatible with the Gactions on the spaces concerned. Given F : V 4 i,W let its image in HomH be F composed with evaluation a t the identity. Given f : VIH -+ W define F : V -+ WG by F ( v ) ( g )= f ( g - l v ) . Since
F ( v ) ( g h )= f (h-lg-lv) = h-lf (g-lv) = h-lF(v)g, if f is an H-map F is a G-map taking values in the subspace i,W. We check that if f is continuous so is (v, g ) H f (g-lv), adjointness implies the continuity of F (CO-topology), and our maps are inverse to each other. 0
Chapter 6: Representations of Compact Groups
69
We have devoted some space to this construction, because a variant of it will be used in Chapter 8 to construct irreducible unitary representations of locally but not globally compact topological group SL2(R). Induction will then be up from a semi-direct product R M R* corresponding to Fp >a "1: in the finite group SL2(Fp). Irreducible Representations of SV,
We illustrate these results by the important example of SU2. Recall the notation of the example at the end of Chapter 5 with the complex numbers @ replacing the algebraic closure Fp of IFp. As before VO is the trivial representation on C and Vl the standard representation on C2 (the operation being given by matrix multiplication). For m 3 2 let V, be the space of homogeneous polynomials of degree m in the variables x and y, so that dim@V, = m 1. Viewing polynomials as functions on C2 we obtain an SU2-action via (gP)a:= P(xy), where
+
Since g acts as a homogeneous linear transformation, the subspaces Vm C @[a:, y] are indeed SU2-invariant. As before the polynomials 6 Ic 6 m) form a basis for the vector space V., STEP 1. The representation space V, is irreducible. The argument used to prove Schur's lemma shows that it suffices to show that each linear SU2-equivariant map from V, to itself is a multiple of the identity. Let a be such a map and for each A E U1 set
and gxCYPk = Ck!gxPk = CYA2k-mPk = A2"-"CYPk. Then gxpk = X2",Pk Choose X so that all the powers Ask-,, 0 < Ic m, are distinct. For this is generated by P k . Since X P k belongs the ;\2k-m-eigenspace of gx in to this eigenspace CUPk = C k P k for some Ck E C.
v,
<
Representations of Finite and Lie Groups
70
For t E R let Rt be the rotation matrix
cos t - sin t sin t cos t
By direct computation a:RtPm = a(a:cost+ysint)m
On the other hand Rta:Pm
=
c(y)cod tsinm-k t .
CmPk,
since aP,
=
k
cm Pm . Comparing coefficients shows that ck = c, and a: = C,X
(Identity).
STEP 2. The Vm exhaust the irreducible representations of SU2. We first reduce this step to a 1-dimensional problem. Let e ( t ) =
;(
e"it)
'
Since any element of SU2 is conjugate to a diagonal matrix, any element is conjugate to some e(t). The elements e(s) and e ( t ) are themselves conjugate if and only if s = ft(mod 27r). Hence if f : SU2 + @ is a class function f e : R --+ @(tH f ( e ( t ) ) )is an even 27r-periodic function. Therefore C!(SUz), the space of continuous class functions, may be identified with the space of even 27r-periodic continuous functions R + @. The character xm of Vm takes the value
m
C ei(m-2k)ton e ( t ) . If t is not
k=O an integral multiple of 7r, this sum equals sin(m+l)t . Denote this function by sin K m ( t ) .The addition function for sin, taken from elementary trigonometry, implies that
K m ( t )= cosmt
+ Km-l(t)cost,
from which it follows that the span of the functions {Ko(t).. . . , K m ( t ) } equals that of the functions (1, cost, . . . , cos mt}. Elementary Fourier analysis (see any book on advanced calculus)) shows that the space generated by 1 and cos mt(m E W) is uniformly dense in the space of even 27r-periodic continuous functions f : R -+ @, hence the characters xm are uniformly dense in C!(SU2).
Lemma 6.3
Iff
E C!(SU2),
ssv,f(u)du :s f e ( t )sin2t d t . =
Chapter 6: Representations of Compact Groups
s
71
s
Proof. The V, are irreducible, so that xo = 1 and for m > 0 X , = 0. Since x,(e(t)) sin2 t = sin(m 1)tsint, integration of the function on t.he right hand side gives the same result. Because we have just shown that the irreducible characters X , are dense in CC(G),we must have equality between left hand side and right hand side for a general class function f by continuity. Here integration over SU2 is with respect to the measure introduced earlier in the chapter.
+
Lemma 6.4 Every irreducible unitary representation of SU2 is isomorphic to one of the V,.
Proof. Suppose that there were an irreducible W # V,. By orthogonality (x, , x,) = 0 and (x, , x,) = 1. This is impossible, since the x, generate a dense subspace. The product structure of the representation ring R(SU2) is described by the so-called Clebsch-Gordan Formula.
Proposition 6.5
4
v k @
Ve = ,@ Vk+e-zj with q J=O
= min{lc,C}.
Proof. It is enough to look at characters evaluated on the matrices e ( t ) . Therefore we have only to check the combinatorial identity
(kX&2q (kXG2”) 2(
kyxk+e-2j-2u
=
fi=O
V=O
j=o
u=o
),
and then replace x by eat. Assume, as we may, that C 6 k. Arrange the pairs of indices ( k - 2p,C - 2v) in a rectangular scheme; to each pair there corresponds a summand xk-2~xe-2uin the left hand brackets. We obtain the right-hand sum by first summing over pairs of indices on the individual lines j = constant, and then over j. The process is illustrated below.
Representations of Finite and Lie Groups
72
Remark The representations V k are sometimes enumerated by halfintegers in the literature, say vk = V ( k / 2 ) . The Clebsch-Gordan Formula then reads
=
+
V ( u )c3 V(b) V(lu- bl) a3 V(lu- bl) + 1)CB . . . a3 V ( u b). The argument above has used very little general theory. We do need Schur’s Lemma, that is: if G is compact and V is a C-vector space admitting a continuous G-action with no invariant subspaces other than ( 0 ) and V , then Hom@(V,V ) G2 C. That the left-hand side is a division ring follows from the definitions; that each G-map is multiplication by a complex scalar follows from the algebraic closure of C. The proof that the representation spaces V, are irreducible is done using bare hands. The proof that there are no other irreducibles (equivalently that the irreducible characters are dense in CL(G)) is an easy consequence of Fourier Theory. There is an alternative proof that the family {V, : m 2 0) is exhaustive using Lie algebras rather than analysis. This will be given in the next chapter. We conclude that this chapter with a summary of what is common between representations of finite and compact groups, which should justify our claim that the former are to be regarded as a special case of the latter. 1. W V invariant G-spaces + V W @ W I . There exists dg on G such that (i) f(g)dg = dg = 1. (ii)
sG
sG
sGf(h-lg)dg and
Chapter 6: Representations of Compact Groups
73
2. Irreducible the only G-invariant subspaces are (0) and V. V E W1 @ . . . @ W k , each Wi irreducible. As a special case, if G is abelian, the irreducible representations (over C ) are all 1-dimensional. 3. Schur's Lemma 4. $ m i W i r $ n i W i * m i = n i v i . 2
i
To see this consider HomqG](Wj, @miWi)etc. in the finite case. In the a
compact case the argument is more honestly expressed in terms of what is called the canonical decomposition. For a G-space U define
-
du : HOmG(U, 'p @3
u
v)€4 u + v by
4u).
On the left-hand side the G action is given by
d'p@3
= 'p @3 9u.
Summing over irreducibles gives
which is easily seen to be an isomorphism. [Hint : reduce to the case when V is itself irreducible, when the claim is trivial.] 5. Definition of characters (i) x(e) = n = dim@(V), (ii) x(g-l) = x(s)(existence of an equivalent unitary representation), (iii) x is a class function, (iv) x is compatible with @ and 8 ~ .
6. Inner product, (xi,xj) = 6ij for irreducible characters. Uses: (xv,xv) is a non-negative integer, which equals 1 if and only if V is irreducible. The same inner product counts the multiplicity ni in (4) above. 7. Role of irreducible characters in CI(G). The natural map R(G) 4 C!(G) ( [ V ]H X V ) is injective. If G is compact the subspace spanned by the irreducible characters in the suitably topologised space of class functions is dense. This is one way of stating the so-called Peter-Weyl Theorem. A consequence is the existence of a faithful representation of an arbitrary compact group G in U, for a suitably large value of n. If G is finite, this is easy to prove. First embed G in a symmetric group S, and then S, in U, by means of permutation
Representations of Finite and Lie Groups
74
matrices. For a proof of the Peter-Weyl theorem, see [Th. Brocker, T. tomDieck, 111.31. 8. R(G1 x G2) = R(G1)€3 R(G2)as a tensor product of abelian groups. 9. i, W = CG €3 W ,where CG has a left G and a right H-structure. As it CH
stands this formula is only valid when [G: HI
< 00.
An additional reference for Chapters 6 and 7 is [E.B. Vinberg]. Exercises. 1. Let G = SU2, let V, be the vector space of complex homogeneous polynomials of degree n in the variables x and y. The space V, has the structure of an irreducible representation module for SU2. Using the properties of exterior and symmetric powers, together with the Clebsch-Gordan formula, decompose the following spaces into irreducible G-spaces
(i) K B ~ 3 vP2, , A ~ v S2v3, ~ , (ii) (iii) S2V,, A ~ V ,( n 2 I), S3h.
v?",
2. Let G = SU2 act on the space M 3 ( @ ) of 3 x 3 complex matrices by
A :X where A1 equals
($-) .
H
A1XAT1,
Show that this gives a representation of G and
decompose it into irreducible summands. 3. Either of the following ways can be used to identify SO3 with real projective 3-space RP3, and hence show that S U 2 / { f l 2 } 2 SOB. (i) First project S2 onto its equatorial plane by (z, y, 2) H C = rfiy. 1-Z Show that a rotation of S2corresponds to a transformation of the form H aC+b. Note that with aZi bb = 1 we obtain an element of SUz -bC+ii and that (a,b) and a', b' determine the same transformation if and only if (a', b') = (-a, -b). Now replace SU2 S S3 by the quotient space RP3. (ii) Let Wo = {ai+bj+clc : a, b, c, c R} be the 3-dimensionalspace of pure quaternions, and let the quaternions of unit length Q = {q :I[ q I[= 1) act on Wo by h H qhq-l. Show that this defines a rotation of S2 C_ Ha, so that SU2/{fl2} = Q/{f12} F% SO3. 4. Deduce the structure of the representation rings R(S03) and R(U2) from that of R(SU2). For the former use Exercise 5, for the latter the determinant map to split UZ.
+
Chapter 7
Lie Groups
This chapter provides an introduction to a very important subject, but one which goes beyond algebra, and the methods which we have so far been using. The reader with some knowledge of differential geometry will probably find it easier to follow the arguments; others may be best advised to concentrate on the material towards the end - Lie algebras - and to treat their representations as an independent topic. In the previous chapter we introduced a Lie group as a topological group with differentiable multiplication and inversion. We must now make clear what we mean by 'differentiable', since usually this concept is taken to apply to functions between open subsets of Euclidean space. Now whereas the space underlying a group like Ul S1 or SU2 g S3 is such that each point has a neighbourhood identifiable with the open disc in R",this is obviously not true globally. We therefore define a C" or differentiable manifold to be a Hausdorff, paracompact topological space covered by smoothly overlapping charts homeomorphic to open neighbourhoods of 0 E R". The 'smooth' condition means that if we choose coordinates xi and y j for the charts U and V, then the functions y j = f j ( x 1 . . .x,) describing the coordinate change are infinitely differentiable (C") and smoothly invertible. The family { fj} describes a local diffeomorphism. The subset N" of M" is called a submanifold if locally the pair (M", N") is diffeomorphic to (R",R"),with Rn mapped into R" by the (z, 0). map E We shall also need the union of tangent hyperplanes to such a manifold M m , written T M and called the tangent bundle . At a single point p(x1. . .x,) E M" a tangent vector is a linear combination of basic first order partial derivatives .. . so that as a vector space T,M g Rm. For a chart U M we can form the product U x R" with coordinates ++
& &,
75
Representations of Finite and Lie Groups
76
(XI,.
a , a ) . . , x,, K .. ., K and T M
= UT,M P
is topologised using open sub-
R". Coordinate change in the overlap U n V is extended to UnV x R" by means of the Jacobian matrix operating on a column of
sets in U x
partial derivatives. In a Lie group this identification is easy to understand because globally we still have a product structure in the tangent bundle, T G E G x R". This is a consequence of the fact that we can use the elements of G to translate local coordinates near the identity e E G to an arbitrary point, see Corollary 7.2 below. Global product structure or not there is a smooth projection map r : T M 4 M , and we define a (C") vector field to be a Coo-section s : M 4 T M ( r s = identity). Locally over
U s(x1 . . .)2,
m
=
C ai (g)&
with infinitely differentiable coefficient func-
i=l
tions ai. A CM-map 'p : M union of tangent spaces.
+
N induces T p = 'p.
:
TM
+
T N of the
Definition. The topological group G is a Lie group if its underlying space has the structure of a C"-manifold, and the group operations (x,y) H xy and x H x-l are Cw-maps. A homomorphism 'p : G1 + Gz of Lie groups is a C"-homomorphism. It is a deep result that every connected, locally Euclidean topological group has a compatible differentiable structure. This question was posed by D. Hilbert (1900) and solved by Gleason, Montgomery and Zippin (1952). Elementary examples of Lie groups are provided by R" (under addition), GL,(R), GL,(C) and T" = S1x x S1 Furthermore any countable J .
discrete group can be regarded as a 0-dimensional Lie group. Let x be an arbitrary element of the Lie group G. The symbol L , denotes left translation in G by x,i.e. L,(y) = xy for all y E G. The defining properties of G imply that L , is a diffeomorphism (with inverse L x - l ) , and the derivative T L , = (L,)* : T G -+ T G must also be a diffeomorphism. For each tangent space TyG, T L , : TyG + TxYG is linear isomorphism. The vector field X on G is said to be lefi-invariant if and only if
(-L)*X(Y)= X ( Z Y ) , 01 (L,)* X = X L,, for all points x,y E G. Such a field X is completely determined by its value at the identity X ( e ) . Conversely for every vector V E T,G we can define a (smooth) left-invariant vector field X on G by X ( y ) = L y + ( V ) This . proves
Chapter 7: Lie Groups
77
Lemma 7.1 The map X H X ( e ) is a linear isomorphism between the set of leff-invariant vector fields on G and TeG the tangent space at the identity. Corollary 7.2
If G is a Lie group, then the tangent bundle T G is trivial.
Proof. Let (211,. . . , vn} be a basis for the vector space TeG and write X i ( y ) = Ly*(vi)for all y E G, i = 1,.. . ,n. Then { X I , . . . , X n } is a basis of sections for T G , and provides an isomorphism of T G with the product G x R". 0 Let r ( T M ) denote the family of smooth sections of T M (vector fields). The local definition shows that if f E Co3( M ,R) and X is a vector field then we can regard X f ( x ) as the directional derivative of the function f at the point x . If X and Y are two vector fields, then we define their Lie bracket [ X , Y ]by the formula
1x7 Y l f = X ( Y f )- Y ( X f ) m
for all f E C" ( M ,R). If in terms' of local coordinates X = j
C aj & and =1
m
Y
=
C bi &,then the smoothness of all the functions involved implies that i=l
[ X ,Y ]is again a Coo-vector field. Thus restricting to single components of the sums for X and Y we have:
-a-
db __ df axj dxa
da __ df axi axj
- b-
The bilinearity of [X,Y ]is obvious from its definition. We summarise the simple properties of [ , ] in
Proposition 7.3
If X , Y ,Z E r ( T M ) and f,g
E C m ( M ,R), then
(2) [ X ,YI = -[Y,XI (ii) [ X l + X 2 , Y ] = [ X l , Y ] [X2,YI (222) [fx, sY] = fdx,Y1+ f ( X g ) Y - d y f > X (iv) [ [ XY, ] Z ]+ [[Y, Z ] X ]+ [ [ Z ,X ] Y ]= 0 (Jacobi identity).
+
78
Proof.
Representations of Finite and Lie Groups
Apply the definitions.
0
These properties make I'(TM) into a real Lie algebra . Some other examples are (a) The space R3 with the 'cross product' as operation, (b) The algebra of real n x n matrices with [A,B]= AB - BA, and (c) Any real vector space with all brackets set equal to zero. A Lie algebra of type (c) is called commutative (or abelian). Lie subalgebras and Lie homomorphisms have the obvious definitions.
Definition. Let. cp : M" --t N" be smooth. The vector fields X on M and Y on N are said to be cp-related, if p* . X = Y ' cp. Lemma 7.4 If the pairs of vector fields {Xi, yi : i then [XI, XZ]and [YI , Yz] are also cp-related.
Proof.
=
1,2} are p-related,
For each f E Cm(W,R) we need to show that
[Xl,XZIz(f .cp)
=
[Yl,YZIrp(x)(f)r
which is done by unravelling the definitions.
0
Corollary 7.5 If X and Y are left-invariant vector fields for the Lie group G , then [X,Y] is also left-invariant.
Proof.
Apply 7.4 with cp = Lx.
0
This corollary shows that we may define the Lie algebra of G to be the subalgebra of I ' ( T M ) formed by all left-invariant vector fields. If we denote it by g we have already shown that g is isomorphic as a real vector space to TeG, the tangent space to the group a t the identity. In this way TeG acquires a Lie algebra structure, and clearly dim g = dim TeG = dim G. In showing that [X, Y]is indeed a vector field above we obtained an expression for the bracket operation on vector fields in terms of local coordinates, viz
Examples. 1. Let G = (Rn, +); then g is the abelian Lie algebra Rn with all brackets equal to zero (Exercise).
79
Chapter 7: Lie Groups
2. Let G = GL,(R); then TeG = M,(R) RnZ,which describes g as a vector space. To each X E g associate the (n x n ) matrix A = ( a z j ) of components of X ( e ) , so that X ( e ) =
Cat3(-&Ie),
and write A =
23
p ( X ) . Then by an explicit inspection of components one can show that p [ X ,Y ] = p ( X ) p ( Y )- p ( Y ) p ( X ) , giving the Lie algebra structure on B = eL(R). Here is an alternative definition of the Lie bracket, which can also be used to show that the vector field and matrix constructions agree on TeG. Let c(g) denote conjugation by g, x H gxg-l, and observe that the differential g* induces a linear isomorphism of TeG onto itself. This representation, written Ad : G + Aut (T,G) is called the adjoint representation, and in turn induces a homomorphism of Lie algebras
ad : TeG 4 Tl(Aut (TeG))= End (TeG). Given the use of conjugation in its construction it is not hard to see that the vector X is sent to the linear map Y H [ X ,Y ] ,or that
ad(X)Y = [X,Y]. (The truth of this formula can also be checked locally.) Let 'p : G + H be a homomorphism of Lie groups and denote their Lie algebras by g and b respectively. Define 'p* : g + b as follows: for each X E g, X ( e ) E T,G and the derivative assigns ' p * ( X ( e ) )E TeH to the vector X ( e ) . Here we may use either the definition of g in terms of tangent vectors or left-invariant vector fields.
Proposition 7.6 (i) If p : G -+ H is a homomorphism of Lie groups, then 'p. : g + l~ is a homomorphism of Lie algebras. (ii) the correspondence G H g, cp H cp* is a covariant functor from the category of Lie groups to the category of Lie algebras. Proof.
Granted (7.4) above this is a manipulative exercise.
0
Definition. The subset H of the Lie group G is a Lie Subgroup if (i) H is a subgroup and (ii) H is an immersed submanifold (with respect to some manifold structure).
80
Representations of Finite and Lie Groups
Warning example: Define cp : R + T 2by cp(t)= (e2nit,e2niat), where cr is irrational. Then cp is an injective group homomorphism and an immersion. However cp(R) is a dense subset of T 2and is not embedded. If we require that H is embedded as a submanifold of G, then we have Proposition 7.7 Let H be a subgroup of G as a group, and a submanifold of G as a manifold. Then H is closed in G and is itself a Lie group with respect to the substructures.
Proof. The smoothness of the operations restricted to H is easy and is left as an exercise. In order to show that H is a closed subset of G it suffices to work near the identity e E G, and then use translation. But near e the pair (G, H ) is diffeomorphic to (Rm,Rkx 0), and any sequence of points {h, E H } may be considered as lying in Rk x 0, and hence converging to a limit z in this subspace. The point 5 thus belongs to H , which is closed. There is a harder converse to this proposition which says that if H is a closed subgroup of G, then H also has the structure of an embedded submanifold, and hence must be a Lie subgroup. As a special case the reader may like to try and prove Proposition 7.8 Let cp : G1 + G2 be a homomorphism of Lie groups, then cp has constant rank on GI. It follows that H = K e r cp is a closed, embedded submanifold of GI, and is a Lie subgroup of dimension equal to dim G - rank (cp*).
Hint: The constancy of the rank follows from a left translation argument, and in order to obtain a nice atlas for Ker (cp) we can appeal to a corollary of the inverse function theorem in order to find good local coordinates. Here are some examples of subgroups of GLn(R) and GLn(C), which are also submanifolds and hence closed in the general linear group concerned. The technique is to define the subgroup H in G by means of coordinate functions F i ( z l , . . . ,z,)
=0
(1 6 i
such that the rank of the Jacobian matrix
< m < dimG)
(z)
takes the maximum value
m at each point h E H . If this condition is satisfied, a variant of the inverse
function theorem allows us to choose new local coordinates for G with the vanishing of m of them defining H . The co-dimension of H in G equals m.
Chapter 7: Lie Groups
81
1. The groups SLn(R) and SL,(C) are defined as hypersurfaces (det X = 1) in open subsets of Rn2 and Cn2 respectively. 2. The orthogonal group 0, c GL,(R) is specified by the equations
which, allowing for symmetry, gives $n(n derivatives
+ 1)equations. Taking partial
+
Hence the minor (subdeterminant) of order $n(n 1) corresponding to the variables xst with s t does not vanish at 1,. The two component orthogonal group is a submanifold, and its dimension is - 1). The component which contains 1, is also a submanifold, SO,. A similar argument shows that U , is a submanifold containing SU, as a hypersurface.
<
in(n
The next step is to relate a Lie group with the Lie algebra determined by the tangent space at the identity e E G. We have already established a (1 - 1) correspondence between this and the Lie algebra of left-invariant vector fields. These in turn are in (1- 1)-correspondence with 1-parameter subgroups , defined to be homomorphisms a from the additive group of real numbers R into G. In one direction this map takes the homomorphism a to the tangent direction 0,(0) of the image a t e. In the other direction let the vector X define a left invariant vector field, and integrate this to give a map a : R -+ G with a ( 0 ) = e. The local uniqueness of the solution forces a(s t ) = a ( s ) a ( t ) , i.e. shows that a is a homomorphism.
+
Important special case: Let V be a finite-dimensional vector space with TlAut(V) = End (V). Then the 1-parameter subgroup corresponding to A E End(V) is
a* : R
-+ Aut
(V)
t H exp(tA) =
031 - (tA)' r! r=O
a2
Representations of Finite and Lie Groups
Note that a A has the same initial condition as the exponential, which, together with the homomorphism property, serves to identify it. More generally we define the exponential map for an arbitrary Lie group by exp : T,G = g -, G ( X H a X ( l ) ) . Note that exp, : g --$gl, is the identify. The exponential map has the following properties: (i) Naturality, f o (exp) = expo(f*). (ii) Exp is a local diffeomorphism near 0 E 01,. (iii) If f : G I -+ G2 is a continuous homomorphism between connected Lie groups, then f is determined by f*. This follows since the image of g1 under exp is large enough to generate all of G I . We stay close to the identity, so as to use (ii). Even though exp fails in general to be a homomorphism, the discussion above at least suggests that
G1 2 Gz (locally) eTeG1 = g1 2 g2 = TeG2 (as Lie algebras). The correspondence between Lie group and its tangent Lie algebra extends to representations. A real or complex representation of the Lie algebra g is defined to be a linear map g -+ gl, which preserves Lie brackets. If G is a matrix group a Lie group homomorphism ‘p : G -+GL, induces a map f* : g ---f gl, which can be interpreted as a Lie algebra homomorphism by expressing C ( A , B ) = log(exp(A)exp(B)) in terms of A and B. Here as usual log(1, - A ) = -C$. Expanding C(A,B) as a Taylor series near r
( A ,B ) = (0,O) we have 1 2
C ( A ,B ) = A + B + -b(A, B ) + higher terms. The map is bilinear and skew-symmetric since
C(A, 0) = A, C(0,B) = B and C ( - B , - A ) = -C(A, B ) . Indeed a further calculation shows that b(A,B ) = [A, B] = A B - BA, indicating that f* is bracket preserving. The argument used to show that H closed in G implies that H is a submanifold of G involves the use of ‘log’ to construct a corresponding subalgebra in g. This will also show that i f f : G -+ GL, is irreducible so is f*. Conversely apply the exponential map to a Lie algebra splitting of f* to obtain a Lie group splitting of f. Hence, provided that G is connected,
Chapter 7: Lie Groups
83
there is a (1- 1)correspondence between the irreducible representations of g and G. Remain with the special case of matrix groups, for which we have a quite explicit description of the exponential map exp : gI,(C)
-+
GLn(C).
Let 1, be the identity matrix in the general linear group GL,C, and let A E gI,(@), which we have already identified with the algebra of n x n complex matrices, with bracket operation [A,B] = AB - BA. . , . ; the right hand side converges in the usual Write eA = 1, +A+ topology in Euclidean space, and we have:
+
1.
B ~ A B - I = eBAB-'
2. det(eA) = etraceA.To see this reduce A to a diagonal (or at worst a triangular matrix) and use (1).
It follows from (2) that eA is always non-singular, i.e. the exponential map takes values in GL,(C). 3. If A , B are such that [A,B] = A B - B A = 0 , then eA+B= eAeB.Hence, although in general exp is not a homomorphism from (8,+) to ( G Ix), it does have this property on abelian subalgegras. 4. We have the following correspondences between subgroups of G and subalgebras of g:
--
SL,C V,
skew-Hermitian matrices
0:
skew-symmetric matrices
SU,
matrices of trace 0.
skew-Hermitian matrices of trace 0.
t--)
-
Restricting to the real numbers, SL,(R) real n x n matrices of trace 0 and SO, skew-symmetric matrices of trace 0. The reader should check that the sets of matrices on the right hand side are all closed with respect to [ , 1. The examples illustrate the general principle that, given a subgroup H of GL,, translate the defining equations back to gI, to describe b. As above the translation is done via the inverse to the exponential map (or logarithm).
Representations of Finite rand Lie Groups
84
Representations of the Lie algebra BIZ The results of this section provide an alternative way to prove that the representat,ionspac.esVm(m >, 0) for the group SVz construct,edin the previous chapter a.re irreducible and exhaustive. We do t.his by first. complexifying the algebra SUz. Let a be a. real Lie algebra with basis { e l , . . . en}, and Lie brackets [ei:ej] = C e i j k e k ( e i j k E R). k
As a vector space we hare a@ = C
@, a R
(compare Chapter 5). Extend
the bracket,s a.bove to ac by int.erpret.ingthe coefficients e i j k as complex numbers. The complex Lie algebra ac is called t.he complexification of a. Note that this definition is independent of the actual basis used. A similar argument shows t.hat we can complexify a representation space V for a (defined over t.he complex numbers!) to a representation space for ac. The original and ext,ended represent.ations share the same invariant subspaces. Each element of ac may be written iq(f,, v E a), and the extension of a representat,ion r from a t.o ay follows the rule
<+
TI<
+ 277) =
+iT(V).
7(<)
This makes sense since T takes values in EndcV. The complexification of su2 is the complex Lie algebra s[2 (= complex 2 x 2 matrices wit,h zero trace). The mat.rices
(a
)z:
1
(;
;l)
!
(y ;)
form a basis of 5u2 over W and of s[2 over C. However the latter admits a basis with more convenient commutation relations:
H
=
(I
0 -1
) . E+ (: i) =
and E- =
(y :)
.
We have [H,E+]= 2E+,[ H , E - ] = -2E-, [E+,E-]= H .
Theorem 7.9 Let T he an, rrreduczble representation of 5I2. T h e n there exasts a basw for the representation space V{vl, 2'2.. . . .rn} such that T(N)VZ.'k =
(n - 2 k ) l ; k ;
.r(E-)v,, = V k + l .rjE+)u,,= k(n - k
+ 1)Vk-1.
Chapter 7: Lie Groups
Here
w-1
85
= v,+1 = 0.
Proof. Let v be a eigen vector of the linear map T ( H ) , T ( H ) w= cw (for some number c to be determined later). If 7(E+)w# 0, then T(E+)w is again a eigen vector of T ( H ) with characteristic value c 2. Thus
+
+
r(H)T(E+)v= .([K E+l)w 7(E+)7(H)v = 27(E+)v C T ( E + ) W = (C 2)T(E+)v.
+
+
A similar calculation shows that, provided T ( E - )#~ 0, 7(E-)vis a eigen vector corresponding to the value c - 2. This explains why E+ and E- are sometimes called raising and lowering operators. The linear map T ( H ) has only finitely many distinct cigen values, so after finitely many steps we find a vector wo # 0 with 7(H)110= Co'IJo,
T ( E + ) V o = 0.
Using T ( E _ )and V O , we find a vector which is killcd by T ( E - ) . Write v k = r(E-)'vo(k = 0 , 1 , 2 , . . .), and let n be such that vo,v1, . . . , w, # 0, but ~ , + l= 0. The eigen vector v k corresponds to the value co - 2 k , that is r ( H ) v k = (CO - 2k)vk. We have chosen w k + l so that ~ ( E - ) w k = v k + l and by induction ~ ( E + ) w k= ukvk+l for some ak E @. We write v V 1
=0
and note that T(E+)OO = 0 as required. For k
)
7 (E+ VI
=
1
= 7 (E+)T(E- ) ~ o = .([E+, E q . 0
+ T(E-)T(&)vo
= .r(H)ao = cove.
Assume that we have proved that T ( E + ) ~ , z=- ak-lvk-2, ~ then as in the previous step
+ Uk-1T(E-)Vk-2 = (CO - 2k + 2 + a k - 1 ) ' ~ k - l .
T(E+)Wk = T(H)Wk-l
Herice
0.k
is defined and satisfies the recursion formula ak = U k - 1
4-co - 2 ( k - 1).
The calculation above shows that the subspace spanned by the vectors {wk} is invariant under T . Since V is irreduciblc V = (vo,. . . ,wk). It remains
Representations of Finite and Lie Groups
86
to show that co = n, since the recursion formula for the coefficients ak gives
ak=k~o-2(1+2+.*.+(k-1))=k(~o-k+1). But with k = n
+ 1, a,+l
=0 = (n
+ l)(co
- n). Done.
Corollary 7.10 Two irreducible representations of dimension are isomorphic.
$12
0 having the same
Passing from Lie algebra to Lie group we can recover the result proved in the previous chapter for SU2. First the irreducible representation V, is specified uniquely by its dimension n 1; any other irreducible representation must have a finite dimension, and hence is isomorphic to one of the ones we have constructed. As a final remark note that in the correspondence between Lie algebra and group representations, if S n f i denotes the nth symmetric power of the natural representation of SU2 on C2, the corresponding representation of 512 is given by the action of H , E+ and E- in the statement of Theorem 7.9.
+
Exercises. 1. Compute the Lie algebra of the group of 2 x 2 complex diagonal matrices of determinant 1. 2. If g is a Lie algebra, consider the adjoint representation ad : g + Endg given by a d z (y) = [z,y]. (a) Verify that this is a Lie algebra homomorphism, that is, [adz, ady] = ad[z,y]. (b) If g = 5 [ 2 , show that the image of g under ad is precisely the Lie algebra 503. Part (b) should come as no surprise since the Lie algebra representation comes associated with the adjoint representation of the compact group
so3. 3. Let ( p , V ) and (p’, W ) be two representations of the Lie algebra g. Define a map y : g + End(V @ w) by y(z) = p ( z ) @ I + I @ p’(z). Show that this is a Lie algebra homomorphism, that is, we have defined an action of g on V @ W . Check that the map taking z to p ( z ) @ p’(z) is not a Lie algebra homomorphism. 4. Repeat Chapter 6, Exercise 1 (ii) by looking at the corresponding action of Lie algebras given by X H X @ I @ .. . @ I + . . . I @ I @ . @ X.
+
Chapter 8: SL2(W)
5. Let E+,E-, H be the basis of
d 2
87
given in the text.
(a) Check the following commutation relations
+ n - 1 ) = n ( H - n + l)ET-', [ E n ,E+]= - n E Y 1 ( H - n + 1 ) = -n(H + n - l ) E T 1 ,
[E:, E-] = nE;-'(H [ H ,E:]
= 2nE;4, and [H,E"]= -2nE14.
(b) Define the Casimir operator
C = H 2 + 2(E+E-
+ E-E+) = H2 + 2H + 4E-E+.
Use the commutation relations to show that C commutes with d 2 . (c) Show that C acts as multiplication by m2+m on the representation V, generated by a characteristic vector w of H(Hw = mu). Hence C serves to distinguish the irreducible representations from each other. and W c V a sub-module. (d) (harder) Let V be a representation Use the Casimir operator to show that there exists W' c V complementary to W . 6. The group SL,(C) acts on the vector space & ( C ) of n x n complex matrices of trace 0 by p ( X ) Y = X Y X - l . Show that this is an irreducible n2 - 1 dimensional representation.
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Chapter 8
SL2(r)
At the end of Chapter 5 we described the irreducible representations of the finite group SLz(IF,) in characteristic p . These are symmetric powers in the natural representation of the group as automorphisms of a 2-dimensional vector space, and play the same role in representing the compact group SU2 in characteristic zero. Since SUZ is a maximal compact subgroup in the complexification of SL2(R) it is natural to ask whether there are other results of this kind. The aim of this chapter is to show that, whereas in characteristic p SLz(IF,) behaves like the compact group SU2, in characteristic zero it behaves like the non-compact group SLZ(R). In both cases the ‘regular representation’ contains two families of induced representations, called the principal and discrete series. The former is obtained in a natural way by induction up from the semi-direct product IF, M IF6 or R M R*. For the latter we must slightly vary the construction, obtaining a family conveniently labelled by restriction to a subgroup of order p + 1 in the finite case and to a maximal compact subgroup isomorphic to S1 in the other. For the finite group the construction of the character table is a good application of most of the techniques introduced in this book. We can only sketch the construction for SLz(IR), since the methods go way beyond those of an elementary text. Our hope is to have whetted the reader’s appetite. Let G = SL2(IFP) denote the group of invertible 2 x 2 matrices with determinant one and with entries taken from the prime field IF, ( p = odd). With almost no change the argument given below extends to IF, ( q = p ’ ) . The order of G equals ( p - l)p@ l ) , and we distinguish the following elements:
+
12=
(;;)li=-12= 89
(-; -;)
Representations of Finite and Lie Groups
90
Note that z is central. Let ,O generate IF, as an additive group and (Y generate the multiplicative group of units IF;.
(it)
Writeu= ( oaa -0l ) , c = The subgroup B =
{ (:
")
andd= :
(ky).
a E IF;,,O
E
IFp} has order p ( p - 1)
and can be thought of as a metacyclic extension of IF, by .:FI Let b denote an element of order p 1. To see that such b exist, let T generate the cyclic group of order p 2 - 1 and let b be the linear map z H Tp-ls of a 2-dimensional vector space over IF,. The group G can be thought of as being built up from the metacyclic group B and the cyclic group < b >. For all s E G let (z) denote the conjugacy class containing 5. Numerical notation: E = ( - l ) q , p = a primitive ( p - 1)st root of unity, (T =a primitive ( p 1)st root and 7 = a primitive pth root.
+
IFi2
+
+
Lemma 8.1. G has p 4 conjugacy classes represented by the elements p--3 ~ ~ , z , c , ~ , z c,..., , z ~u, u2 , b ,..., b q . The number of elements in each conjugacy class is given by 12
z
1
1
d
zc
zd ue(l < l <
z ( P - 1)
q) b m ( l < m < q)
P(P + 1)
P(P - 1).
Proof. Let CG and NG denote the centraliser and normaliser of an element or subgroup in G, so that the order of the conjugacy class (x)equals IGl/lC~(z)l.An easy calculation with matrices shows that CG(U)= ( a ) and that
a quaternion group of order 2 ( p - 1). Hence (u) contains &(p 1) elements. A calculation similar in spirit but harder in detail shows that (b) contains $ p ( p - 1) elements. For the remaining classes I (12) I = I (z) I = 1, the conjugacy classes of c, d, zc and zd -1 P are distinct, and all four elements are centralised by and 0 -1).
+
(Z) (
Hence each conjugacy class contains
elements.
Chapter 8: S L n ( 1 )
91
we have accounted for all the elements of G.
0
Since
+
This lemma shows that we are looking for p 4 irreducible representations of which only the trivial one is 1-dimensional.
Exercise. If p 2 5 the group G is perfect, i.e. coincides with its derived subgroup [G, GI. The non-trivial representations are constructed by induction, and fall into two series, which with an eye on SL2(R), we shall refer to as the principal and discrete series. The first is the easier to describe.
Principal Series for SL2 (IFp) Let 8i be the 1-dimensional representation of B which maps the matrix
(; Ll)
to pi and 8f the induced representation of G of degree p
< < 9,
+ 1.
< <
Here 0 i and if 1 i we write xi for the character of Gf. Numerical considerations show that (xi,xi) = 1, so that xi is irreducible. Its values on the conjugacy classes are given by
x
Xi(.)
12 Z c zc p + 1 (-l)i(p+ 1) 1 (-l)i
d 1
ae
zd (-l)i
pie+p-ie
b" 0
These plausible looking entries can be checked using the formula for induced characters. Neither of the extreme characters is irreducible; for the inner product with itself gives 2, and with the trivial representation gives 1. For the second assertion use F'robenius reciprocity. Thus the character of 8g equals 1 $J,and for $J we have
+
x
12
$(x) p
z p
c 0
zc
d
0
0
zd ae b" 0 1 -1.
The remaining presentation GG also splits as a sum of two irreducible 5representations, whose characters we write as & , < 2 , and leave on one side for the moment.
Representations of Finite and Lie Groups
92
Discrete Series for SLz(IF,) For this second series of representations we start with the l-dimensional representation of ( b ) which maps b to oj (1 6 j 6 induct up to G, obtaining a character @ taking the values p ( p - 1) on 1 2 , ( - l ) j p ( p - 1) on z , d m (7-j" on b", and 0 on the remaining conjugacy classes. In order to reduce the degree to p - 1 we form the virtual character
&
q),
+
h
0,
=
(7) - 1)GY - pj".
Note from the previous subsection that GG& still makes sense, and that 2 the degree of Oj equals (p - l)(p 1) - p ( p - 1)= p - 1. For Oj we obtain the following values:
+
x
z
1
Oj(X)
zc
C
p - 1 (-l)j(p- 1) -1
(-1)j+1
d zd ae b" -1 (-l)j+l 0 -(oj"+g-j"
)
< < 9,
If 1 j (Oj,Oj) = 1 and Oj(12) > 0, indicating that 0, is an irreducible character. As with the xi the topmost Opfl splits as ql 772, the sum of two irreducible characters of lower degrei. The hardest step is to understand the splitting of the exceptional representations & & and q q 2 . The value 2 for the squared norm of each implies that we also need to consider the possibilities - & and 71 - 112. But these can be eliminated in parallel with the argument below; the degrees ( p f 1) being forced on us by passing to the projective special linear group PSLz(F,). So far we have the table of values
+
+
X
51 + & 771
1
+
c
Z
p t 1 E(pt1)
+ q 2 p - 1 --E(p-
1 1) -1
zc E E
d zd ae 1 E 2(-l)t -1 E 0
b" 0 2(-l)"+l
Since (& + & ) ( z ) = + & ) 1 2 , & ( z ) = ~ & ( l z ) ( j= 1 , 2 ) , and similarly q j ( z ) = - ~ q j ( 1 2 ) ( j= 1 , 2 ) , sothat thefourcharactersCj,qj arealldistinct. Assume that E = +1, then the irreducible characters of representations which factor through PSLz(F,) = SL2(IFP)/(z)of order $p(p2 - 1) are 1,7 ) , Xeven, Oeven, &, &. Counting squares of dimensions we have Jl(l2)' c ~ ( 1 2 )= ~ !j(p l)', and since < 1 ( 1 2 ) + & ( 1 2 ) = p 1, we must have that each J j ( l 2 ) = i ( p 1). The sums of the squares of the dimensions of the remaining representations must also add up to i p ( p z - l), and so
+
+
+
+
Chapter 8: S&(R)
93
It follows that qj(12) = $ ( p - l ) ( j = 1 , 2 ) . A similar argument holds for E = -1, the roles of <j and qj being interchanged. The splitting for the conjugacy classes represented by ae and b" is clear. For c and d we need a little more work, after which we can use the relation ~ 1 ( 1 2 ) ~ + 1 7 2 ( 1 2= ) ~ $(p-1)'.
with y = c or d. We have (& 52)(c) = 1+ (sum of pth roots of l),and ( ~ 1 q2)(c) = (sum of pth roots of 1 other than 1). Hence the problem is to split the sum of the roots 7, r 2 , . * , 7P-l into two subsums of elements, so as to satisfy the numerical constraints imposed by the group structure. Note that if E = +lo) = 1 mod4) each irreducible complex character of G has to be real. This follows from the fact that the representatives for the conjugacy classes (x) are all conjugate to their inverses (compare Chapter 2, Exercise 1). For E = -l(p = 3mod4) similar considerations imply that and q1 are conjugate to & and q 2 respectively. The two subsums for which we are looking are given by the roots of the quadratic equation x 2 -x+ -&p) = 0. Those knowing a little about the arithmetic of number fields will recognise the similarity of this problem with that of finding a quadratic field inside the cyclotomic field Q(r). The complete character table for G is given in the table below.
+
+ (9)
i(1
Remark. Let p = 5, so that E = +1, and all the characters of G are real. Inspection shows that the character tables of the groups PSL2(lF5) and A5 coincide. Indeed these two groups are isomorphic. The representations with characters ql and q 2 do not factor through the projective quotient; restricting to a 2-Sylow subgroup Q8 we obtain an irreducible 2-dimensional representation (c
H
(k !i)).
The element b2 generates a 3-Sylow sub-
(: ,oa> with 1 + w + w 2 0), and the element a is represented ). The reader knowing some topology may either by (i :!) or by ( o r3
group (b2 H
=
r2
recognise these representations, through which the group acts on the 3sphere, as defining Poincare's homology %sphere. The two orbit spaces are diffeomorphic. Here is an alternative construction of the representations 8j in the discrete series, which may appear to be more natural. As in our discussion of
94
E
10
c3
9)
+E
Y
0
I
4
E
4
+ -
4
v
I
c
4
Representations of Finite and Lie G T O U ~ S
0
+
-
0
*
*
i
Chapter 8: SLz(E%)
95
the representations of SL2(Pp)over the algebraic closure Fp in an earlier chapter, consider the ‘character’ of the natural representation V1 in the automorphisms of a 2-dimensional vector space. On the p’-elements (elements of order not divisible by p) we obtain the Fp-values
x
ae
z
I 1 2
bm
where y denotes an eigenvalue in Fp for the matrix b. Remember that b has been chosen to diagonalise over F P 2 but not over Fp. We now lift this Brauer character to a class function in characteristic zero by mapping Q and y to (p - 1) st and (p 1) st roots of unity respectively, i.e. to powers of p and 0. Note that choice is involved here, contributing to the range of values of i and j in the final character table. We must also make a consistent lift at the prime 2 in view of the relation a 9 = b q . We complete the definition of our class function by mapping the four p-classes in G in the same way as 12. At least when i and j are both odd we now have
+
X
12
z
c
d
xi -,gj
2
-2
2
2
bm
ae pie + p - i e
,im
+,-im
showing that the character 9, can be obtained by subtracting our lifted class function from xi. It remains to show that the class function is a virtual character. This uses a deep theorem of R. Brauer, which asserts that this is so, provided that we have a character on restricting the class function to ‘elementary’ subgroups. For an arbitrary group G these are direct products of cyclic subgroups with subgroups of coprime, prime power order. With G = SL2(Pp) such subgroups are built up from cyclic or quaternionic subgroups, and the answer is immediate. We now proceed to split Bp+l as ~1 r/2 in the same way as before. 2
+
The Non-compact Lie Group SLz(IR) In this section we aim to give some indication of how the representations of SL2(Pp),constructed using finite group theoretical methods, give a guide to the representations of SL2(R). So far as possible we will use the notation
Representations of Finite and Lie Groups
96
already introduced.
D = diagonal subgroup
=
{(;a!!l)
:aER-{O)
U = unipotent subgroup = B = semidirect product U via the rule
>Q
D with D acting on the normal subgroup U
If = { Z E C : Im(z) > 0) denotes the upper half of the complex plane, there is an action of SL2(IR) on b given by
+
az b AZ = - w i t h A = cz d '
+
The matrices
f l 2
(::)
act trivially.
Let K be the isotropy subgroup of i, that is the subgroup of matrices = i. Then ai b = -c d i implies that we can take such that cos 9 - sin 9 a = d = cos9, b = -c = -sine, i.e. K =
+
+
The action of SL2(IR) is thus transitive, and there is a (2 - 1) map
We order the matrices in the product so that a H ca as y H 00 in g. We also allow K to act on the right of the space of cosets B\G, and give the space of complex valued functions L 2 ( K )the usual norm llfllk = J, If(lc)12dlc. Topologically we have shown that G decomposes as U x D x K , corresponding to the set decomposition of SL2(F,) into subsets of order p - 1,p and p 1. G also inherits a product measure from the hyperbolic measure d x d y / y 2 on B and dlc on K Z S1. As a topological space G is isomorphic to
+
0
a product of the open disc D2 and S' . This becomes even more transparent on replacing SL2(R) by the group SU1,l- see the Exercise 2 below.
Chapter 8: SLz(1W)
97
Principal Series We start with the representation of B given by G^,(ud)= a s , s E C. Let H ( s ) be the space of complex valued functions f on G, such that (i) f ( u d l c ) = a s f ' f ( k )and (ii) the restriction o f f to K belongs to L 2 ( K ) . The representation P s = G: is obtained by allowing g to act on H ( s ) on the right
P S ( g ) f ( s= ) f(29). This definition is a variant of that of an induced representation for a finite or compact group, but in this case H ( s ) is an infinite dimensional Hilbert space, and care is needed to show that P s is both bounded and unitary. This is so for s = iv(v # 0), a pure imaginary value, and H ( i v ) contains two irreducible unitary representation modules P+iiv and P-)i". Note the in the finite case. parallel with the splitting 51 + J 2 = GGw 2
For s = 0 there is a 3-fold splitting, only one component P+>Oappears in the principal series. Neglecting P->O leaves a third component, associated with what is called the mock discrete series . Neither this nor the complim e n t a r y series { s E (-1, I), s # 0, positive parity only} contributes to the 'regular representation' L 2 ( G ), with G acting by translation. In addition to Pfgiv we need a series corresponding to the 8-series in the finite case.
Discrete Series On we already have the volume element d p = d x d y / y 2 . Replace this by dpm = ym-2dxdy and let H ( m ) = L&,,(b,pm), m 2 2, be the space of holomorphic functions f : b 4 C which are square integrable with respect to the measure p m . If H ( m ) is given the usual Hermitian inner product via integration, this can be shown to be complete. If A-l
SLz(W) the representation D+i"
: G + Aut
D + ? " ( A ) f ( z )= f
(:I:)
=
(z i)
E
G=
H ( m ) is defined by
- (cz
+ d)-".
This can be shown to be unitary, and D-1" is defined in a similar way to act on the space of complex conjugates. At this point recall that if
I
Representations of Finite and Lie Groups
98
p = 3(mod4) the character O(+) splits as 771 +qz with 'discrete series' is explained by the following result:
The representation D+>rnis irreducible.
Theorem 8.1
(z) n
772 = 5jl.
+
The name
If gn(z)
=
(Z i)-n, gn E H . If Hm+an is the subspace spanned by $ J ~ , Hrn+an is an eigenspace f o r the compact subgroup K , and
It remains to give some explanation of why only the principal and discrete series appear in the regular representation of SL2(W). We do this by giving a measure j on the 'space' SLz(IW)" of irreducible representations. For the positive and negative parts of the discrete series dj = (this point weighting corresponds to a distributional character), and for the principal series d j ( v + ) = t a n h ( y ) d v , dj(v-) = c o t h ( 7 ) d v . The remaining irreducible representations belong to set of measure 0. We illustrate this by the diagram below, and adjoin a corresponding diagram for SLz(F,). In this quick trip through the representation theory of SLz(IW) we have relied mainly on S. Lang's book [S. Lang (1993)l.
&;
&$
measure zero
SL2(W Fig. 2
Exercises 1. Determine the character table of SLZ(F)where F is a finite field containing q = 2t elements. (First show that S L 2 ( F ) contains q 1 conjugacy
+
Appendix A : Integration ower Topological Groups
99
characters of type classes, and determine their orders. Then find 0,. Again xo splits as 1~ $.) characters of type xi and
2. If S U l J = {
(: :)
+
:
a,b E C,laI2 - JbI2 = l}, show that SL2(R)
and SU1,l are isomorphic subgroups of GL2(C) by conjugating with the matrix A =
z
-i).
-,k+%)that SU1,l (2-i)
H
(t
0
Show further, using the Mobius transformation is the group of isometries of the Poincar6 model 0
D2 of the hyperbolic plane. Draw a picture of SL2(R) D2 x S1,and distinguish between elliptic (Itrace1 < 2), parabolic (Itrace1 = 2) and hyperbolic (Itrace1 > 2) subsets. 3. (For those with some knowledge of differential geometry) If exp is the exponential map introduced in the previous chapter, show that exp is surjective for GL2(C) but not for SLz(R). 4. Show that SL2(R) has no non-trivial finite-dimensional unitary representation.
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Appendix A
Integration over Topological Groups
In discussing representations of compact groups we introduced the notion of the Haar integral, proving its existence for our motivating example of SU2. Here we give one of many proofs of its existence in general, using an argument which may remind the reader of the definition of the Riemann integral as the limit of a sequence of approximating finite sums. Another argument, which is often presented in books on Lie groups (see for example [Th. Brocker, T. tomDieck]) starts by integrating a function with support inside some small open subset modelled on R". The change of variable formula from advanced calculus ensures that this is well-defined. A globally supported function can be split by means of a 'partition of unity' into a sum of functions fi, each of which has support inside some Euclidean neighbourhood, and the integral of f is defined as the sum of the integrals of the fi. Group invariance follows if we first define the basic volume element in a neighbourhood of the identity e and then translate this to a neighbourhood of an arbitrary element g. We prefer not to do this, partly on aesthetic grounds, but mainly because generalising the basic results for representations of finite to compact groups does not require the latter to be locally Euclidean. Indeed the most general result known to the author only requires the group G to be locally compact and Hausdorff. Given a compact topological group G we shall show that there exists a unique rule for integrating a continuous function f. The integral should be a linear functional on the space of functions V , and should be non-negative if f is non-negative. In addition it should be left invariant, i.e.
f(h-lg)dg =
/
f(g)dg for all h E G,
G 101
(A.1)
Representations of Finite and Lie Groups
102
and be normalised by
s,
(A.2)
1 . d g = 1.
If in addition it satisfies
we shall call our integral right invariant. Our first proposition is a version of the mean ergodic theorem in functional analysis, see [S. Sternberg].
Let T be a linear transformation of a nonned linear
Proposition A . l space V such that
(i) llTnull < cJJwIIfor some constant c, and (ii) there exists w E V for which the sequence Snw=
1
-
n-l
)
~ T ' W
n (+O
.
possesses a convergent subsequence with limit i5. Then T Z the sequence Snw converges t o a.
Proof.
The subspace ( I d - T ) V = { z E V : Snz
+
=
a and
0). Firstly, if
z = (Id - T)w,then
which tends to zero with increasing n. But ( i ) implies that the subset { z 6 V : Snz 4 0) is closed, so that every element z of the closure of (Id - T ) V satisfies S,z -+ 0. Therefore (Id - T)V { z : Snz 0). Conversely let Snz + 0, so that given E > 0 we can find no such that, for n 2 no ---f
11.2 -
( z - Snz)I\ < E .
But
1
+ (Id - T 2 ) z+ . . + (Id - T"-')z) - T ) { z+ (Id + T ) + . . . + (Id + T + . . . T " - 2 ) ~ ) ,
z - S,Z = -{(Id n =
1
-(Id n
-T)z
which is an element belonging to (Id - T ) V .
*
Appendix A: Integration over Topological Groups
If Sn, w
-+
for some subsequence {n1,722, . . . } of 1
TS,,w-Sn,w=(--)(Tn3w-w)
103
N,then
-+O, s o t h a t m = w .
Now
T"w = T"0 + T"(w - 0 )= Ur + T"(w s,w = w Sn(W - w).
+
By assumption (ii) w
-
S,, w
+w -0
- 0 ) ,so that
and we have proved that
w - Sn, w E ( I d - T ) K Therefore w - w E ( I d - T ) V , so that Sn(w under S, , the proposition is proved.
-
W ) 4 0. Since ;iiT is fixed
We wish to apply this to the space V of continuous complex valued functions on the compact group G, carrying the sup norm, i.e. l l f l l = sup If(g)1. As a compact group G is also totally bounded, i.e. there exists a basic neighbourhood system a t el {Ua : i E I } , and for each i a finite number of elements 91,.. . ,g3(a)such that G C U skuz.Take all the gk'S which arise as i varies and arrange them as some sequence {gk} which will be dense in G. Define an operator T on V by
) If. ( f I
-
f(y)l
< E for all z - l y
E U , where
U is some neighbourhood of el
then
<
l l f l l for all n and f . This is condition Hence T maps V to itself and IIT"fll (i) in the proposition. By direct substitution S,f(g)= ajf(h,g) for some sequence of elements hj in G and real numbers aj with Cjaj = l, aj > 0. Arguing in a similar way as before, for an arbitrary value of E > 0 there exists a neighbourhood U of e such that
Cj
ISnf(z)- Snf(y)l < E for all n if z-'y E U. By the density of the original family {gk} in G it follows that we can choose a convergent subsequence of the values S,f. Hence condition (ii) of the proposition is satisfied for all functions f . Therefore S, f -+ and T7 =
7
7.
104
Representations of Finite and Lie Groups
sG
In order to write 7 = f ( g ) d g we need to show that 7 is a constant. Without loss of generality assume that f is real valued, with M = sup 7. If f is not constant there is some h E G with f(h) < M - 2~ for some E > 0. Since 7 is uniformly continuous (G is compact!) f ( g ) < M - E for all g such that gh-' belongs to some suitably small U . By total boundedness of G, G C UigiU for some finite family (91,. . . ,ge}. If g is arbitrary gh-' E giU for some i smaller than some finite value N . (The gi are taken from the sequence { g k } .) Therefore
gi'gh-'
E U , so that f ( g i l g )
<M
- E.
But then
This contradicts the choice of M as least upper bound. Having shown that f is a constant, linearity, non-negativity for nonnegative f and normalisation (A.2) follow immediately. For any h E G f ( g h ) = f h ( g ) implies that ( T f ) h = T ( f h ) ,from which right invariance follows. The resemblance to the Riemann integral is shown by the formula (valid for any h k )
An analogous formula holds with right replacing left in the argument just given, h i ' g being replaced by gh;, and by a left invariant integral. But with the sum in (*) involving terms of the form h k ' g h ; it should be clear that in the limit we obtain either the left or the right invariant integral. Hence these must be equal. A similar use of approximating sums shows uniqueness. Let us see what the existence of the Haar integral implies for the integration of a class function over the unitary group U,. Thus we assume that f ( g - ' h g ) = f(h) for all h , g E U,, from which it follows that f is a symmetric function of the eigenvalues ( A 1 . . . A), of h.
sG
Appendix A: Integration over Topological Groups
Proposition A.2
I f f : U,
-+
105
C is a class function then the Haar integral
with XI, = e i e k . The argument below depends on the following properties of Lie groups, represented as a class by U,. We may identify the tangent space T,G with the Lie algebra g, on which G acts via the adjoint representation described in the chapter on Lie groups in the main text. We also need to factorise the integration of f : G -+ C into the integration over a maximal torus T and the quotient space G / T of left cosets. The group acts on G/T by left translation (hT H ghT) and the quotient space inherits an invariant measure or volume element from G. There is an open neighbourhood W of the coset eT in G / T over which a local cross-section s : W -+ G is defined. Near T this allows us to describe G topologically as W x T . The volume element dh splits as dtd(hT). We have quite deliberately written this in terms of G and T , even though the proposition is stated in terms of G = U, and T = the subgroup of diagonal matrices D,. Proof of the proposition: for the pair (G = U,,T = D,) consider the map
-
T x (G/T) G ( t ,ST) gtg-l. Let the Jacobian of this map with respect to the invariant volume elements on T, and G/T and G be J ( t ) . Note that this does not depend on the cosets gT. Then for any function f on G we have
The factor lln! is needed to allow for the fact that each conjugacy class is counted n! times. In the language of advanced calculus we have replaced a volume by a repeated integral, and if 7 is constant on conjugacy classes,
Here K
=
volume ( G / T ) / n ! .
106
Representations of Finite and Lie Groups
To calculate J ( t ) use the Lie algebras t and g, noting that T(G/T),T the orthogonal complement of t in g. Perturb t to t(1 5) and g to g ( 1 + 7 ) with 5 E t and 77 E t' respectively, then the change in gtg-l is
+
t',
P Therefore t - l p
(1+ V W + = tE V t - tq =
5)(1 - V ) - t
+ + higher order terms. = 5 + (t-lvt - V ) E t @ 't = g.
This is the tangential change of coordinate formula for our original map, and if Adj(t-') denotes the action o f t on t',
J ( t ) = det(Adj(t-')
-
1).
The complexification of t' has basis { e j k : j # I c } , where e j k is the matrix with 1 in the (j,Ic)-position and 0 elsewhere. These basis elements are characteristic vectors for Adj(t-l) corresponding to the value A k x ; ' , with t = diag(A1,. . . , A n ) . Hence
j#k
j
It remains to determine the volume of G / T , which, since the volume of G is normalised to equal 1, amounts to determining the volume of T = S1 x . . . x S1. Therefore K = l / n ! ( 2 ~ )and ~ , since dt = dB1. . .do,, the product of the line elements for each of the factors S1, the proposition follows.
Appendix B
Rings with Minimal Condition
In this second appendix we have two aims, first to prove Wedderburn's Theorem identifying simple rings with matrix rings over division rings (such as the quaternions W). The second aim is to give a second, less abstract definition of a semi-simple ring. Except for the concrete example of the group algebra C[G],to the study of which much of this book has been'devoted, it may be hard to come to grips with the concept of semi-simplicity. Theorem B.12 below justifies an alternative definition in terms of descending chains of left ideals and the absence of nilpotent elements. The reader should be familiar with the basic definitions and results from non-commutative ring theory, such as left, right and two-sided ideals, the ways in which these can be combined to give other ideals ( An B , A B , AB, etc.), direct sums and Cartesian products, and a t least the definition of a quotient ring. The presentation which we give is based in part of a course of lectures given by Philip Hall a t Cambridge in the 1960s. The author has never seen it bettered. Let R be a not necessarily commutative ring. We will assume that R contains the multiplicative identity 1.
+
Definition. M is a minimal left ideal of R if and only if M only left ideals of R contained in M are 0 and M .
# 0 and the
For example if R is such that R2 # 0 and R is minimal in itself then R is a division ring.
Lemma B.0 If M is a minimal left ideal of R and M 2 # 0, then there exists e E M such that M = Re and e2 = e, e # 0. Such an e is an idempotent.
Proof.
Since M 2
#
0 there exists a E M such that Ma 107
# 0.
But
108
Representations of Finite and Lie Groups
RMa M a C M , so M a is a left ideal contained in M . Since M is minimal M a = M . Therefore there exists e E M such that ea = a, from which it follows that e2 - e E M n S , where S is the left annihilator of a in R. Now S is a left ideal, as is M I = M n S , and since ea = a # 0 , e 6 M I . Again by minimality M I = 0 and e2 = e. Finally M e is a left ideal contained in M , which is non-zero (e2 = e ) and so M e = M . 0 If X is an arbitrary subset of R, M a left ideal and e an indempotent, define N M ( X )= { y : y E M and y z = 0 for all z E X } .
Lemma B . l
(a) N M ( X )is a left ideal in R, (ii) N R ( M ) is an ideal (a) in R, and (ziz) NR(e) = {z - z e : z E R } and R = Re @ NR(e). Proof. (i) N M ( X )= nzEXNM(z), and each N M ( ~=) M n n T ~ ( z )where , NR(z) is easily seen to be a left ideal. (ii) Write N = N R ( M ) . Then N M = 0, and ( N R ) M = N ( R M ) N M = 0, so that N R C N . (iii) Since (z - ze)e = 0, z - ze E NR(e). If y E NR(e),ye = 0 and y = y - ye. Clearly for any z E R, z = ze (z - z e ) and so R = Re NR(e). To see that this sum is direct let y E RenNR(e). Then y = ye, because y E Re, 0 and 0 = ye (because y E NR(e)). Hence y = 0.
+
Definition R is a simple ring if and only if R in R are 0 and R.
+
# 0 and the only ideals
If R is any ring with multiplicative identity 1 let E = E n ( R ) be the ring of n x n square matrices over R.
Lemma B.2
(i) Every ideal of E has the form F = E,(S) where S is an ideal in R. Hence i f R is simple so is E . (ii) If R is a division ring then E is a direct sum of minimal left ideals. Proof. This is an exercise with matrices. For the first part, if F a E let S equal all coefficients of all matrices in F . For the second part define MT = { U = ZE, Xireir, X i , E R}, where ei, denotes the matrix with 1 in
Appendix B: Rings with Minimal Condition
109
the ir-position and zero elsewhere. Then M , is a (minimal) left ideal and E = Mi @ . . . @ M,. 0 We say that the ring R is nilpotent if R" = 0 for some positive integer n. In this case each product ~ 1 x .2. . x, = 0 and, in particular, each x E R is nilpotent.
Definition The radical of a ring R is the sum of all nilpotent ideals of R. The radical is an ideal, but need not itself be nilpotent. Lemma B.3
+ M R is also nilpotent and contained in Rad(R). (ii) A finite sum of nilpotent ideals is also nilpotent. (iii) If R is commutative then the radical equals all nilpotent elements of R. (i) If M is any nilpotent left ideal of R, then M
Proof. (i) This follows from the definitions. (ii) If A and B are left ideals such that A" = B" = 0, in ( A B)"+" every term either has m factors A or n factors B. Using the left ideal property we can group elements from either A or B , from which it follows that A+B is nilpotent. The general result follows by induction. Note that Rad(R) is a possibly infinite sum of such ideals. (iii) The element z belong to Rad(R) if only and if z belongs to some nilpotent ideal of R. Assuming that R (containing 1)is commutative it is enough to show that xn = 0. Shuffling elements inside the nilpotent ideal containing x shows this.
+
0
We say that R satisfies the condition min-l if every non-empty subset of left ideals contains a minimal element. Equivalently any descending sequence of left ideals terminates after finitely many steps. Similarly we can define mas-!. In the next few lemmata we assume that R satisfies min-l and that Rad(R) = 0. Our aim is to show that this property is equivalent to semisimplicity.
Lemma B.4 potent.
Every non-zero left ideal M of R contains a non-zero idem-
Proof. By min-! M contains a minimal left ideal M i of R. M ; since Rad(R) = 0, and by Lemma B.l M , contains an indempotent el
#0 # 0. 0
Representations of Finite and Lie Groups
110
Lemma B.5 Every left ideal M of R is generated b y a n idempotent, M = Re, e2 = e E M .
Proof. Let ?ZJ = all NM(e) with e2 = e E M . The set ?ZJ is non-empty since NM(O)= M E 9. Hence we can find e with N M ( ~minimal ) in ?ZJ. By Lemma B.l (iii)M = Re @ N . Suppose that N is non-zero, by Lemma B.4 there exists el E N , with e; = el # 0. Take e2 = e el - eel, ele = 0 . Therefore e; = ( e el - eel)' = e2 ele - eleel - eele+ eeleel = e2, and e2 is an idempotent in M . Now let N Z = NM(e2). Since e2e = e, if x E N2, x , = xeze = 0 , so that x E N . Hence N2 C N . But el E N , e1e2 = el (from the definition of ez), so e1e2 # 0 and el @ N2. It follows that N2 is not contained in N , a 0 contradiction. Therefore N = 0 and M = Re.
+
+
+
Lemma B.6 Let e2 = e E R and M be a left ideal such that M = Re. Then A4 a R if and only i f e commutes with evey element x E R.
Proof. Sufficiency is immediate. For necessity let S be the right annihilator of M in R. Then S a R, (SnM)' C_ M S = 0 and SnM a R. Since Rad(R) = 0, SnM = 0. By assumption M = Re, SnM is the right annihilator of e in M , and M x = 0 if and only if ex = 0. Using the argument of Lemma B.l (iii) again, M = e R @ (right annihilator) = eR. Hence e is a unit element for M , and is unique, since te = t = et for all t E M . If x is any element of R 0 ex = exe and xe = exe = ex. Lemma B.7 left ideals.
If R # 0 , R is the direct sum of a finite number of minimal
Proof. We can find el such that Re1 is a minimal left ideal. Write R = Re1 @ R1, R1 being NR(e). If R1 # 0, we can find e2 such that Re2 is a minimal left ideal of R and e2 E R1. Thus R1 = Re2 €9 R2, RZ being N ~ ~ ( e 2If) .R2 # 0, we continue the construction, eventually reaching R, = 0. Therefore R = Re1 @ Re2 @ . . . @ Re,. 0 Lemma B.8 R contains only a finite number of minimal (%sided) ideals R1,. . . , R, and R is their direct sum.
Proof. If Ri and Rj are any two distinct minimal ideals of R, RiRj RinRj = (0). Let a = ai E Ri. If a = 0, then, since by Lemma B.5 Ri has a unit
xi ci
Appendix B: Rings with Minimal Condition
111
element ei, 0 = eiO = Cj eiaj = eiai = ai, and this a E R has a unique expansion in terms of elements in the Ri. Hence the sum S is direct; to show that it exhausts R , write R = S @ T , with T = N R ( S )a R. If T # 0 it contains at least one minimal ideal, which could be added to S . Unless T = 0 we have a contradiction. The number of minimal ideals is finite, since otherwise we could construct an infinite descending chain. Note also = @ i E l & , the identity 1 belongs that, if R, the centraliser of a in RZT(a) to some finite sub-sum and R . l = R. 0 Lemma B.9
Using the previous lemma write R = R1 @ . . . @ R,.
(i) Every left ideal M of R has the form M = M I @ ... @ M,, where Mi = Ri n M is a left ideal in Ri. (ii) Conversely any direct sum with Mi a left ideal in Ri is a left ideal of R. (iii) The only ideals in R are the 2" sums of the Ri 's. (iv) Each Ri is a simple ring and inherits the properties man-l and Rad(R) = 0 from R.
Proof.
+ +
E M , x = x1 . . . x,, xi E Ri. Let ei be the multiplicative identity of Ri. We have eix = eixi = xi, because RiRj = 0 for j # i. It follows that M is the direct sum of the intersections M n Ri, each of which is easily seen to be a left ideal in its Ri. Each left ideal Mi in Ri is a left ideal in R, RMi = Cj RjMi =
(i) Let x
(ii)
RiMi
CM.
CEl Mi where Mi a Ri by (i) and (i)' for right ideals. This is a corollary of (ii). Every ideal in Ri is an ideal in R, Ri is minimal as an ideal, and so Ri is simple as a ring.
(iii) Every ideal of R has the form (iv)
0
If a E R the centraliser of a in R, &.(a) = { x : x E Definition R , Z R ( A ) = { x : x E R and xu = R and xu = ax}. For a subset A ax for all a E A } = naEAZR(a).The centre of R equals ZR(R).
Lemma B.10
(i) The centre of R = 2 1 @ . . . @ Z,, where Zi equals the centre of Ri. (ii) If R is simple then its centre is a field. In particular each Zi in ( i ) is a field.
Representations of Finite and Lie Groups
112
Proof.
+ + +
(i) Let z E 2,z = z1 zz . . . z,, and let x E R, x = x1 x2 . . + x,, where the elements xi and zi belong to &. Then if z x = x z , we have C i z i x i = C i x i z i . By minimality, and because the sum is direct, corresponding terms are equal. Therefore z E 2 if and only if zi E Zi for each i. (ii) Let 2 denote the centre of the simple ring R. As always we assume that R contains 1. If 0 # z E 2,Z R= R z a R. The element z E z R # 0 , so the simplicity of R implies that z R = R, i.e. there exists z’ with zz’ = z’z = 1. Therefore z-l exists in R and centralises x if z does. It follows that 2 is a field. 0
+ +
We are now ready to characterise semisimple rings . Theorem B . l l ring R.
The following conditions are equivalent for an arbitrary
(a) R satisfies R min-L and Rad(R) = 0 , (ii) R contains an identity 1 and is a direct sum of minimal left ideals, (iii) R contains an identity 1, every left ideal S is a direct summand in R, . R = S’ @ S , and S’ is also a left ideal. (iv) Every left ideal in R has the form R e with e2 = e .
Proof. (i) + (ii) follows from Lemma B.7. (ii) + (iii). Write R = M1 @ . . . @ M , with each Mi a minimal left ideal in R. If S is a left ideal, put S = So and define inductively a sequence of left ideals {S,} as follows. Given S,-l R let i, be the first s u f h for which Mi, $ ST-l, and define S, = Sr-l + Mi,. F’rom the definition il < i2 < . + . ,so that for some finite value n, R = S,. Since Mi, is minimal S,-l n Mi, = 0 and S, = S,-l @ Mi,. Therefore R = S’ @ S with S’ = Mi, @ . . . @Mi,. (iii) =+- (iw). Let S be a given left ideal and split R as R = S @ S’. Write 1 = e e’, e E S , e’ E S’. If x E S , x = xl = z e xe’, where xe and xe’ belong to S and S’ respectively. By the usual argument x = xe and e2 = e, so that S = S e R e G S. (iw) (2). Trivially Rad(R) = 0. Suppose that M I M2 2 . . . is a descending sequence of left ideals in R. By assumption Mi = Rei with e: = ei. Define Ni = N ~ , ( e i + l )so , that Mi = Ni @ Mi+l. The left ideal Ni # 0, because Mi+l is contained in Mi properly. Consider the infinite direct sum S = C,”=, Ni of left ideals in R. By (iw) S is generated by an
c
+
+
c
Appendix B: Rings with Minimal Condition
113
indempotent which belongs to some finite sub-sum of S. This leads to a contradict ion. 0 It remains to characterise the simple rings from which R is built up. We use the language of left R-modules, noting that if R is regarded as a left module over itself, then the submodules are the left ideals M . Furthermore, the quotient abelian group R I M inherits a left R-module structure. In the main text we have already noted how scalar multiplication can be used to define a ring homomorphism from R into the endomorphisms of the R-module M . The module axiom 1z = z for all z E M ensures that 1 E R is taken to the identity map of M . Notation: ER = EndR(M) = set of all R-linear maps from M to itself. Suppose M = M I @ . . . @ M,, each Mi being an R-module.
Lemma B.12 The ring E R ( M ) is isomorphic to the ring H of n x n square matrices ( ~ i j where ) qij E HomR(Mi, M j ) . Proof. One first checks that addition and multiplication are well-defined in H and gives it the structure of a ring. Note in particular that the composition v i k c k j E Hij = HomR(Mj, M i ) . Each element z in M decomposes uniquely as a sum z1 . . . z, with xi E Mi. Define a map ~i : M + Mi by E ~ = Z z i for all 17: E M . This is R-linear and the identity map 1 of M equals ~1 . . . E ~ . For any 77 E E R ( M ) , 77 = 1171 = C233 . . E ~ v E ~ .The maps ~i play the role j 0 if i # of idempotents, i.e. ~i acts as the identity on Mi and ~ i M = j. The composition ~ i r ] maps ~ j Mj into Mi and all Mk to O(k # j ) . Hence it determines a unique 'matrix element' r]ij E Hij = HomR(Mj, Mi). Conversely, by splitting z into its components, we can use the matrix ( ~ i j ) t o build up an endomorphism 77 of M which will satisfy E ~ Q E=~ qij. The inverse maps 77 H ( v i j ) are both compatible with the ring structure; this is linear algebra. 0
+ +
+ +
Corollary B.13 If the modules Mi are all isomorphic, then E R ( M ) E,(S) where S = E R ( M ~ ) . Proof. Let O i l be any module isomorphism of M1 onto Mi, and suppose that ell is the identity. Map Mi to M j by O j i = ejleZ;', so that B i j = 0;' and t?ki = e k j e j i . Use these maps to define ring isomorphisms between Hij and H11 = S , namely
114
Representations of Finite and Lie Groups
Using Schur’s Lemma we see that if M is a minimal left ideal of R and a E R, then either M a = 0 or M a is a minimal left ideal isomorphic to M .
Theorem B.14 (Wedderburn) Let R be a simple ring which satisfies min-!. Then R Z E,(D) for some division ring D . Indeed D is isomorphic with E R ( M ) ,M being any minimal left ideal of R, and n equals the number of minimal left ideals of which R is the direct sum. Proof. The radical of R equals 0, and from Lemma B.5 every left ideal of R has the form Re with e2 = e. Hence R also satisfies max-!, for if there exists an ascending sequence of left ideals M1 & M2 & . . . , with M = UzMi,M = Re and e belongs to some M z . The sequence terminates at i. It follows from this that R cannot contain any infinite direct sum of minimal left ideals. If M I is any minimal left ideal of R, M1 M I R a R, and by simplicity M1R = R. Therefore R = CaER M1a and, by Schur’s Lemma (see above) all the relevant M l a are minimal left ideals isomorphic to M I . If M I is properly contained in R, choose M2 = M1a2 $ M I . We must have M I n M2 = (0) and can write S2 = M1 @ M2. If S2 is properly contained in R, choose M3 = M1a3 $ S2 with M3 n 5’2 = (0) and S3 = M I @ M2 @ M3. Continue in this way until we reach R = M I CBM2 @ ’ . ’ M , with each M z isomorphic to M I . By Corollary B.13 E R ( M ) Z E,(D), where D = E R ( M ~ )By . Schur’s Lemma once again, since M1 is irreducible as an R-module, D is a division ring. We have already proved in Lemma B.l that the ring of n x n matrices over a division ring is simple, so that Wedderburn’s Theorem shows that this class in exhaustive. 0
Appendix C
Modular Representations
The example of SL2(P,) in Chapters 5 and 8 shows the value of considering representations not just over C or R,but also over a finite field F,(q = p t ) . To this we now give the briefest of introductions. Suppose that the representation p of the finite group G is equivalent to one by integral matrices; an example is provided by the permutation representation of the symmetric group S,. If we reduce the entries of the matrix p(g) modulo p we obtain a representation p : G + GL,(F,), for which we can use the trace to define a character and ask about the irreducibility, or indecomposability of the IF,[G] module corresponding to ;ii. It is important to note at the outset that, if p is irreducible in characteristic zero it does not follow that is irreducible over extensions F, of IF,. Furthermore Maschke's Theorem fails if p divides ]GI, since we cannot average over the group elements, and hence reducible no longer implies decomposable for representation modules. Not every representation is integral in the elementary sense above, and it is in general necessary to adjoin roots of unity to the rational numbers in order to obtain values for the character table. The non-abelian groups of order p 3 , for which we use induction to construct the irreducible representations in Chapter 4,provide explicit examples of this. However, if we work with the analogue of the rational integers Z in some finite extension field of the rational numbers it is still possible to reduce module a prime, obtaining a representation over some extension field P,. An explicit example is provided by the group S3 = { a , b : u3 = b2 = l,b-'ub = u - ' } , for which over the complex number C there are two 1-dimensional, and one 2-dimensional, irreducible representations. Label these 1, ,B and p, where in matrix form
a,'
115
Representations of Finite and Lie Groups
116
Reducing modulo 3 gives the matrices Now conjugate using the matrix P =
P-lij(b)P =
(i i)
and P-'ij(u)P
=
(:;)
(i 1>
(i ;) ,
and
a);;(
in GLz(P3). We obtain
with character X(b)
=
0,
X(u) = 2. Bearing in mind that irreducible # indecomposable, so that we cannot remove the 1 in the lower left-hand corner of each matrix, we have shown that (modulo 3) x p is equivalent to the sum of the two 1-dimensional characters, x p = x1 + x p . Note also that in cutting down the number of irreducible representations we can also cut down the number of conjugacy classes to the identity and ((12)). This motivates the next definition. Definition. Let G be a finite group of order n = p a h , where p is prime and p h. An element of G is said to be p-regular (or to be a p'-element) if its order rn is coprime to p.
+
Lemma C . l If the order of g equals pbk(p 1 k), g = xy = yx, where x is p-regular, y has order equal to a power of p, and both x and y are unique. The element x is called the p-regular factor of 9 .
+
Proof. If rpb s k = 1, write x = g T p b and y = gsk. . If xy = xlyl are two decompositions of 9 , xr'x = yly-', and both sides have order one since the greatest common divisor of pb and k equals 1. Note that x commutes with x1 since both commute with g and similarly for y and y1. 0 The following result gives the clue to much of Brauer character theory for the prime p.
Lemma C.2 If x is the p-regular factor of g = xy and p is a representation of G in characteristic p, then the matrices p(g) and p(x) have the . same eigenvalues.
Proof. From linear algebra we know that p ( g ) is conjugate via P to a triangular matrix with the eigenvalues A1 . . . Ad down the principal diagonal. Let 9' = x,gt =s y and the order of y be p s . Then matrix multiplication shows that (Xi). = 1. But in a field of characteristic p , 0 = (A:)."
- 1 = (A: - l).', so
that At
=
1.
Appendix C: Modular Representations
117
The matrix p(y) = p ( g t ) conjugates to a matrix having entries X i = 1 down the principal diagonal. Hence P - l p ( g ) P = P-lp(zy)P = P-lp(z)PP-lp(y)P
Therefore A! = Xi, and since the Xy are eigenvalues for gT = z, the lemma follows. 0 We now lift back from characteristic p to characteristic 0. As above let IGI = p a h , and let p be a matrix representation of G in an algebraic closure F, of IF,. The eigenvalues of the matrix p(g) equal those of p ( z ) , and since z h = 1 these eigenvalues are hth roots of unity. Let 77 be such a primitive root in Fpl so that all other hth roots are powers of r ] , and let C be a primitive root in C.
Definition The Brauer character $ corresponding to p is defined (on p-regular elements only) by $ ( g ) = ECk1where the sum is taken over the exponents k of every eigenvalue vk of p ( g ) . This definition is justified by
Lemma C.3 Two representations of G in P p have the same Brauer character if and only if they have the same irreducible constituents.
Proof. One way round the lemma is immediate. If the Brauer characters of p1 and p 2 are the same we have an equation of the form cia1
+ ., .+ Ciad
=~
i b+ i
. .. + ~ i b c
derived from the eigenvalues of p l ( g i ) and p2(gi). Consider the representations in characteristic zero for the cyclic group generated by g: p;(gi) = diag(cial,. . . c a d ) ] pi(gi) = diag(
118
Representations of Finite and Lie Groups
Proposition C.4 (i) If p1 and p 2 are representations defined over an algebraically closed field of characteristic p and the irreducible representation u occurs with multiplicities a and b in p1 and p 2 respectively, then a = b (modulo p ) if and only if x ( p 1 ) = x ( p 2 ) . (ii) The representations p1 and p2 have the same irreducible constituents i j and only if p I ( g ) and p 2 ( g ) have the same eigenvalues for each element g in G.
Proof. The point behind the proof of (i) is that taking sums in F can introduce multiples of p , which will contribute zero to the characters. One implication in (ii)is obvious, for the other assuming that the eigenvalues are equal, we argue by induction on the common dimension of p1 and p 2 . Since the characters are equal we can use part (i) of the proposition. If for some irreducible representation u we have a # 0, b # 0, we can remove one copy from each of the direct sums corresponding to p 1 and p 2 , and appeal to the inductive assumption to show that a - 1 = b - 1. Note that replacing p by a direct sum does not change the value of a or b. Otherwise, for each irreducible representation ui,pal = ai and pb: = bi. The representations Caiai and Cb6.i have smaller common dimension, i
a
and as before the multiplicities a: and b: must be equal.
0
It is clear that .a Brauer character extends to a class function on G. As we have illustrated in Chapter 8 this class function turns out to be a virtual character, i.e. the formal difference of two characters of honest representation spaces over @. In general this follows because the restriction to a preferred class of subgroups gives a character, as is clear for cyclic subgroups by the previous lemma. The relation between irreducible representations in characteristic p and characteristic zero is summarised in the so-called decomposition matrix D. If D = ( d i j ) the decomposition number dij is the multiplicity of the irreducible representation uj in the mod p reduction of the irreducible @ representation p i . We suppose here that pi may be taken in integral form. The number of rows of D equals the number of conjugacy classes in G; the number of columns is determined by the result to which we appealed in our discussion of SL2(lFP)in Chapter 5. Proposition (2.5 The number of irreducible representations of the finite group G in IF, equals the number of p-regular (p’) conjugacy classes in G.
Proof. There are several ways of proving this. Here is a sketch of one of them, using the methods and terminology of Appendix B. Let K be the
Appendix C: Modular Representations
119
finite field IF, for a sufficiently large power q of p. Define the Jacobson radical of an arbitrary ring R to be the intersection of all maximal left ideals in R. This coincides with the intersection of all maximal right ideals, and is an ideal, so that we can define the quotient ring K[G]/J(K[G]). For the quotient there'is an analogue of Wedderburn's theorem, i.e. a decomposition into a sum of matrix rings over division rings. Each division ring is isomorphic to HomKrcl,J(V,,V,) where V , is irreducible. As for the complex numbers we can now count dimensions and find that the total number of irreducible modules V, equals the number of p'-conjugacy classes. Dividing out by J corresponds to cutting down from all conjugacy classes to this smaller subfamily. Note the Jacobson radical is at least as big as the radical defined in Appendix B. At least when R is commutative the latter (sometimes called the nilradical) equals the intersection of all prime 0 as opposed to maximal ideals in R (Exercise). This summary is taken from
[N.Blackburn, B. Huppert].
We conclude with two easy examples to emphasise the difference between characteristic p and characteristic 0. 1. Let x generate a cyclic group of order p , where p is prime.
x
H
(k t)
Then
defines a 2-dimensional representation of C; over P, which
is irreducible. It cannot be decomposed as a direct sum of 1-dimensional subspaces, because there is only one eigenvalue (equal to 1) corresponding to a less than 2-dimensional eigenspace. (This show that p { [GI is a necessary assumption in Maschke's Theorem.) 2. Let G be a group and F a field of characteristic p. Suppose that p I (GI. Show that the nilradical of &"GI is non-trivial, and hence that F[G] is not semisimple. Hint: consider powers ( C g ) k of the sum of the group 9
elements.
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Solutions and Hints for the Exercises
Chapter 1 1. The determinant takes values in C* 2 GLl(C), and det(p) defines a homomorphism, since det(p(gl)p(gz)) = det p(g1) det p(g2). 2. Op(zy) = Op(z)Op(y), because scalar multiplication commutes with the operation of matrices. For the same reason a subspace W C V is p invariant if and only if it is Op invariant. 3. The group S3 is generated by a 3-cycle a = (123) and a 2-cycle b = (12).
If a is mapped to the rotation matrix reflection matrix
(; t) , the relation
($;: )::--
a-l
and b to the
= bab is satisfied. All six
elements of S3 can be described in terms of these together with
(i
12
=
;)a
4. The 1-dimensional representations agree with those for Qs. A distinct 2-dimensional representation is given by a
H
(i")
andb-
(!;).
Note that b2 rather than b4 equals 1. It is now easy to see that the characters for Dg and Qg coincide. 5 . One clean result is that, if K is a field of characteristic not equal to 2, then the group ring K[&] = 4K @ Mz(K). The first four summands correspond to the 1-dimensional representations in Exercise 4, the last to the 2-dimensional representation.
121
Solutions and Hints f o r the Exercises
122
6. We have the following table: 1
5
6
7
3
4
c 3
c 4
c 5 c 6 c 7
C2 X C2
S3
2
(1) c 2
8
9
CErC4 X c 2 C2 X c2 X c 2 Ds. QR
c 9 C3 X
10
ClO C3 Dio
11
12
c 1 1
C121c2 X c 6 C2 X D6(2 D12) &12,
A4
The entries can be justified using the Sylow theorems. For example a group of order 2p (p = odd prime) is generated by elements a and b of orders p and 2 respectively. The subgroup generated by a is normal and the only possible rules for conjugation by b are b-lab = a or b-lab = a-l. For non-abelian groups of order 12 a 3-Sylow subgroup is isomorphic to C3 and a 2-Sylow subgroup to C4 or C2 x C2. Because of the smallness of the order one of these subgroups must be normal. Writing the normal subgroup first the possibilities are (C3,C4) giving Q12, (C3,C2 x C2) giving 0 1 2 and (C2 x C2,C3) giving Aq. The pair (C4, C3) cannot correspond to a non-abelian group, since the only possible conjugation rule is b-lab = a. Put more pedanticly C3 only maps trivially into the automorphism group of C4.
Chapter 2 1. Given the division of the non-identity elements of G into pairs, if x = X C , x(z) = x(1) 2 a , where a is a sum of eigenvalues. The inner product (x,xo) = x(z), so that (x,xo) = i$x(l) 20).
+
&xEG c
+
Unless x = xo the inner product equals 0, since x is irreducible. This implies that a = a non-integral rational number which contradicts the definition of a above. The claim also follows from the more general result that the number of irreducible real representations equals the number of conjugacy classes which equal their own inverses. See [G. James, M. Liebeck, Theorem 23.11. This depends on being able to invert the character table X of G as a matrix, a result which depends on the orthogonality relations (sum over g) and Proposition 2.12 (sum over x). With the obvious notation we have P X = and X Q = for promutation matrices P and Q , from which it follows that Q = X - l P X , so that trace ( P ) = trace(Q). But in each case the trace is the number of points fixed by the permutation, and trace ( P ) = number of irreducible real characters, trace ( Q ) = number of self-inverse conjugacy classes. We actually appeal to this equality in the discussion of the characters of
-q,
x
x
Solutions and Hints f o r the Exercises
123
SL2(Fp)in Chapter 8. 2. The Sylow theorems imply that G has a presentation { a , b : u p = bq = 1, b-l ab = a', rq f l(modp}. Here the initial and most important step is to note that the subgroup generated by a of order p must be normal, since the total number of its conjugates both divides p q and is congruent to 1 modulo p . The condition qlp 7 1 is needed in order to obtain a non-trivial solution to the congruence rq = l(modp). It is now clear that [G,G] = ( a ) . The conjugacy classes are represented by powers of a (giving ( distinct classes), powers of b (giving ( q - 1) classes with p elements in each) and the class (1) of the identity. Check: 1 ( q - 1)p (y)q = pq. When p = 7 and q = 3, we can take T = 2, and list the conjugacy classes as (l),( a ,a', a4), (a3,a5,a 6 ) , (b,ba, ba2,. . .), (b', b2a,. . .). The group G has 3 one-dimensional representations, and two more irreducible representations. These both have dimension equal to 3 by Corollary B to Proposition 2.9. In the light of Exercise 1 above we may denote their characters x and y. If C denotes a primitive 7th root of unity we have x ( a ) = C Cz and x ( a ) z C6 C3. If w = e2.lriI3, then ~ ( b =) z ( b ) = 1 w w2 = 0. To see this use the l-dimensional characters plus xregto determine x+X on the elements a and b, and then split in the obvious way. For a more elegant approach we will use induced representations in Chapter 4. This will also apply to primes q > 3. For the moment note that our calculations suggest that there should be q l-dimensional representations and a family of q-dimensional irreducible +.. .), ~ ( b=) representations, with typical character x ( a ) = (C+<'+C'' 0, where C is now a primitive pth root. 3. The subgroup generated by the two 3-cycles (123) and (456) is isomorphic to C3 x C3 and must be normal, since it has index 2. Composing permutations shows that if a = (123) and b = (456) then the subgroups of C3 x C3 generated by l),(1,b), ( a ,b) and ( a ,b-l) are also normal. For example ((23)(56))(123)((23)(56))= (312). Dividing out by a normal subgroup of order 3 gives a quotient group isomorphic to S,. Each of the four possible quotients provides a 2-dimensional irreducible representation for G, and this list exhausts the possibilities, since 2 4.2' = 18. Each irreducible representation has a non-trivial kernel. 4. The obvious candidate for the character of a 16-dimensional representation is provided by x = /3 - (Y 1. This takes values
y)
+
+
+ c5 +
+ + c4 + +
(a,
+
+
124
Solutions and Hints for the Exercises
P-a+l
1 15 40 16 0 -2
90 0
45 0
120 0
144 120 90 1 0 0
15 40 0 -2.
+
Taking the inner product of x with itself we obtain 256 + 160 + 144 160 = 720. The group in question is s 6 with its 11 conjugacy classes determined by cycle type. Granted the existence of the 16-dimensional irreducible representation above it is possible to construct the character table in the same way as for S5 . Six irreducible representations have characters XO, E , T ,E T , A 2 r and E A2 T (same notation as in the chapter). The symmetric square S 2 r reduces to xo T 8 with O(1) = 9, and we can add the pair (8,a3)to our list. Counting squares of dimensions gives
+ +
2
+ 2.25 + 2.100 + 2.81 + 256 = 670.
Hence there are two more irreducible characters cp, ~ c p ,both corresponding to representations of dimension 5, which can be determined using the numerical information above together with the character of the regular representation. (This discussion provides a hint for Exercise 3 in Chapter 5). 5. Taking inner products we find that ( a , a )= (P,P) = 2 and (y,y) = 8. It follows that a and P are the sums of two distinct irreducible characters, and since y/2 makes sense, it seems reasonable to try y/2 = 1 x7 also. Since x(1) divides 168 for each irreducible character, by looking at the small factors 2, 3, 4,6, 7, 8, . . . it seems reasonable to look for a solution of the form
+
a
= x8
+ x6
P = x7 + X8
712 = xo + x7,
where xo is the trivial character and otherwise the suffix corresponds to ~ ( 1 )Note . that 168 = 1 2.9 36 49 64. The equations above give all the characters except those of degree 3:
+
1 x o 1 X6 6 x7 7 ~8 8
21 1 2 -1 0
+ + + 42 1 0
-1 0
56 1 0 1 -1
24 1 -1 0 1
24 1 -1 0 1.
Solutions and Hints for the Exercises
125
Check that each inner product (x,x) = 1 and the the four constructed are orthogonal. Call the remaining pair x3 and xg. This agrees with the discussion of Exercise 1, since the first four conjugacy classes are ‘real’ and the last two ‘conjugate’ to each other. The usual comparison with xregshows that x3+5 takes the values (6 - 2 2 0 - 1 - l),and the remaining problem is to split -1 as E $, with each summand equal to the sum of three primitive 7th roots of unity. If C denotes one such root, =C 2 = C’ C5 C3 (compare Exercise 2). Note that [ then equals (-1 i f i ) / 2 . The final two entries in the character table are
+
+ <’ + c4, +
+ +
x 3 3 - 1 1 0 5 ; j xg 3 -1 1 0 5.
r
The reader may like to revise this exercise before reading Chapter 8. 6. If x is an irreducible character of G then Ker(X) = { x E G : x ( x ) = ~ ( 1 will be a normal subgroup. More generally we have Proposition A If N is normal in G ( N a G ) then there exist irreducible
characters X I , . . . , xr such that N =
,fi Ker(Xi).
2=1
Proof. First observe that the intersection of all the kernels of the irreducible characters equals { 1). This is a consequence of the result that the irreducible characters span the space of class functions. If N a G the characters x1 . , . xr of G describe the characters of the quotient group GIN. By the previous observation h equals {identity coset in G I N ) , i=l from which the Proposition follows. We also have Proposition B The group G is not simple if and only if x(g) = ~ ( 1 for some non-trivial irreducible character x and some element g(# 1) of G.
Proof. If the condition is satisfied, the irreducible and non-trivial properties of x combined with the assumption that g # 1 show that (1) Kerix) 4 G. Conversely, if G is not simple, some irreducible character of GIN lifts back to G and we can find the required pair (x, 9 ) . Note that the Proposition B plus the calculation in Exercise 5 show that 0 there is a simple group of order 168.
4
Solutions and Hints for the Exercises
126
7. Assuming that the group G may be identified with PSLz(IF7) we use the character table above without further comment. Identify the set of 1-dimensional subspaces of F$ with the projective line F7P1, containing eight points with non-homogeneous coordinates 0, 1, . . . , 6, 00. The group acts via Mobius transformations of the type z H The matrices A =
(i :>
and B =
(i :)
g.
have orders 7 and 3 respectively.
A fixes co and acts as a 7-cycle on the remaining points, while B fixes 0, co and acts as a product of 3-cycles (124)(365) on the remaining six points. Hence the trace of the A-action equals 1, and that of the Baction equals 2. Dimension 8 implies that the possible decompositions of the permutation character into irreducible characters are xo x 7 , 2x0 xs, 2x0 x 3 x3. In order to distinguish between them, we
+
+
+ +
look at the action of
(! ;I),
which squares to
-12,
and decomposes as
the product of four transpositions (0~)(16)(23)(45). This gives a trace equal to 0, corresponding to xo x7. For the last part use the formula that the number of pairs of elements conjugate to z and y respectively (zy being conjugate to t ) equals
+
Here the sum is taken over all irreducible characters of G, and = IG12(ICG(z)1
IcG(Y)I
ICG(t)l)-l.
Chapter 3 1. Let XI,. . . ,Ad be the eigenvalues of the linear transformation g , which we may assume to be represented by a unitary matrix. Then 1x(g)1 = 1x1 . . . Xdl 6 1x1I . . . = d. There is equality if and only if the X i are all equal, i.e. p(g) = X l d . Removing the absolute value sign: g E Kerp H { x ( g ) = d } . See also [G. James, M. Liebeck, Proposition 13.4 (2)]. 2. If g has order 2 the eigenvalues of a representing unitary matrix equal f l , from which it follows that x(g) = X(l)(mod2). For the second part consider those elements g such that x(g) = l(mod4), and check that they determine a normal subgroup. Hence, if G is simple (and # C Z ) , there can be no elements of order 2 with x(g) = 4k 3.
+ +
+ +
+
Solutions and Hints for the Exercises
127
A similar argument shows that if g has order 3 and is conjugate to 9-l (so that x(g) involves both e2?ri/3and e-2?ri/3),then x(g) = l(mod3). See also [G. James, M. Liebeck, Corollary 22.251. 3. Write Hom(G, C * ) for the 1-dimensional representations/characters of G. The group operation is given by tensor product x1 8 x2. Both G and = Hom(G, C*) are abelian, so that the 'class function' condition is satisfied trivially, and compatibility with addition is checked in the
e h
same way as for dual vector spaces. We obtain a map G -+ via g (x ++ x(g)). An element g belongs to the kernel of $ H x(g) = ++
II,
1 for all x, i.e. g belongs to the intersection of the kernels of all 1dimensional representations H g E [G,G]. If G is finite abelian then by the classification theorem G 2 Hom(G, C " ) , since this holds for the
e. h
The definition of cp* is again based cyclic group C,. Hence G 2 2 on that of a dual linear map cp"xZ(g1) = ~2(cpg1).The proof that cp* is surjective H cp is injective is purely formal. 4. Denote the irreducible complex representations of C3 by l , w , Z . Then the irreducible real representations are 1 and w +W, the latter of which is 2-dimensional. No contradiction of Schur's lemma arises, since the real numbers W are not algebraically closed. 5. The first part is proved exactly as for complex representations by using the averaging process to construct a real G-invariant inner product. The matrix P describes the basis change from the initially given one to an orthonormal one. The second part reduces to describing all finite subgroups of 0 2 , which are seen to be cyclic or dihedral. Note that 0 2 is an extension of the circle group SO2 by a cyclic group of order 2. An extension of this exercise is to classify all finite subgroups of SO3; these include the three polyhedral groups isomorphic to A4, 5'4 and As. 6. Applying the theorem on representations of direct products to Cs x S3 we see that there are 6 irreducible representations of degree 1 and 3 of degree 2. (Check: 6 + 4.3 = 18.) In contrast the group ((123)' (456)' (23)(56)) C s6 considered in Chapter 2 has two representations of degree 1 and 4 of degree 2. Hence the two groups are not isomorphic. 7. From the definition of the inner product
Therefore (x2,1c) = 0
(x not
real) or = f l
(x real).
But
x2 is the
sohtions and Hints for the Exercises
128
character of the tensor product V @ V , so that x2 = xS2+ x A 2 . Hence if (x2,1 ~=)1 precisely one of xs2 and x A 2 contains the trivial character xo as a constituent. Now define the indicator function ix by
ix=
{
0 1 -1
XO
@ xs2
01 X A 2
xo E xsz 2 0 E XAZ.
Thus ix # 0 H the character x is real. In term of the indicator function we prove (i) V admits a non-degenerate G-invariant bilinear form B H i x # 0, (ii) which is symmetric H i x = 1, and (iii) skew-symmetric H ix = -1. For (i) if the trivial character xo E x , we can project V @ V onto a summand C, obtaining the required bilinear form p. If ,L3 exists we can incorporate a summand C into V @ V . For part (ii) and (iii) either project S2V or A2V onto C, obtaining a symmetric or skew-symmetric form. Combining this discussion with the characterisation of real representations in the main text we see that 0, 1, -
1,
x is not real x can be realised over R x is real but x cannot be realised over R.
Representations of the third type are said to be symplectic. For a more extended discussion see [G. James, M. Liebeck, pages 272-2741. Chapter 4 1. In both cases there are four 1-dimensional representations, the abelianised group being isomorphic to C2 x C2. The remaining irreducible representations are 2-dimensional and induced up from a cyclic subgroup of index 2. Let be a primitive 2t-1th root of unity. For Dzt we have
<
a
I+
('0 ) , (1 6 i < 2t-1, 2t-2 c-i
- 1 distinct
representations),
Solutions and Hints for the Exercises
For Q2t we have a H
(z(yi)
, b ++
0 (-1)i
(1
129
) , with the matrix for b
explained by the relation = b2. Similar consideration apply to Dzm and Q4m, although care must be taken to distinguish between odd and even values of m. 2. In order to count the conjugacy classes in P- use the typical relations b-'ab = a'+, and aha-' = bal+pa-' = baP. The first type gives p 2 conjugacy classes, the second ( p - 1). Similarly for P+ use [a,b1b-l = cb-' and b-lab = ac. As claimed in the text the irreducible representations of P* are either one- or pdimensional. In both cases the 1-dimensional representations are those of C, x C,. Taking the traces of the given matrices shows that (as for Q s and 0 8 ) the character tables coincide. Note that the centre of the metacyclic group P- is generated by up(++ diag (C . . . C)), and that we have the isomorphisms P - / ( a p ) CF x CE, P+/(c) CF x C:. (The 'bar' refers to the coset containing the element concerned.) 3. If t = 2 G is abelian, if t = 3 G is either abelian or isomorphic to PA. In both cases the abelianised group is isomorhic to C, x C,. For higher powers the argument is similar. Let H of order p s be the centre of G of order p t . Let K be the subgroup of G mapping to the centre of G I H ; G I K again has a non-trivial centre. Proceeding in this way we eventually reach a quotient group which is its own centre, i.e. abelian. Therefore G contains a series of normal subgroups Ho = G , H I ,H2,. . . ,Hn+l = {l},such that for each i Hi-lIHi is abelian and the centre of GIHi. Furthermore GIH1 cannot be cyclic. If this were the case, then for some value a , we could arrange the elements of G / H z as { H l / H z ,z H l I H 2 , . . . zPa-'H1/H2}. But the elements of H1/H2 are central in,G/H2, so that GIH2 would be the abelian group generated by x and H1IH2, contradicting our assumption. It follows that IG/HlI 2 p2, and hence that the same is true for G / [ G ,GI. For the second part of the question we must show that any non-abelian group of order p4 contains a normal abelian subgroup of order p 3 , and then appeal to the results of Chapter 3 (3.7 plus corollary to 3.2). Either this is immediate or G contains a normal abelian subgroup of order p 2 with quotient isomorphic to C, x C,. Arguing as in the first part of the question we can find the required abelian subgroup of index p . This is a special case of a more general result: if J G J= p t G contains a normal abelian subgroup of order p a with a(. 1) 2 2 t .
+
130
Solutions and Hints for the Exercises
4. Given the description of the representations of H x K in terms of the representations of the two factors (Proposition 3.9) the question reduces to transfering the trivial representation up to K . We have already seen . the induced representation space in terms that this gives p r e g , ~Defining of tensor products, i,W = C[G] €4 W , makes this almost obvious. @[HI
5. In what follows denote a representation of Sn(n = 4 or 5) by its dimension, multiplied if necessary by E , to indicate twisting by the odd/even 1-dimensional representation. Thus for n = 4 we have { 1,3,~ 3E ,,2) and for n = 5{1,4,6, ~ 4E ,,5, €5) with the characters given in the tables at the end of Chapter 2. Inducting from 5’4 up to 5’5 gives two representations of dimension 5, two of dimension 15 and one of dimension 10. With the obvious notation we have the scheme ind(1)
4, ind ( ~ 1 3 ) €4
}’
ind(s3) ind(3) ind(2) = 10 = 5
+~ 5 ,
from which the character table for 5’5 can be recovered. Less conclusive information is acquired by induction up from the second subgroup of order 20. With a = (12345), b = (2354), we obtain the presentation { a , b : a5 = b4 = 1, b-’ ab = a3}, which, because the action of b on the subgroup generated by a is faithful, is distinct from Q 2 0 . There are four 1-dimensional representations of ( a ,b) and one 4-dimensional irreducible representation obtained by induction up from ( a ) . Inducing one step further to S5 gives representations of dimensions 6 and 24 = 6 2(4+5). To distinguish between the four 6-dimensional representations restrict the action to the subgroups generated by a and b. The first argument illustrates a general method for constructing representations of S,. For n = 4,5 we associate the irreducible representations with the partitions of (1234) and (12345) illustrated by the (Young) diagrams below.
+
Each partition can also be matched with a subgroup of 5’5, and by inducting up from the trivial representation we obtain a representation of S5 containing the one labelled by the same partition. Thus the trivthe ial representation of S5 corresponds to 5’5 itself
(m
permutation representation (4) to
S4 x
(1)
(Fyq
(5) to
Solutions and Hints for the Exercises
n=4
131
n=5
EF3E F E€P5 H2 4
E l
5'3
x 5'2
(CFF'),
etc. For very readable introductions to this
elegant classical theory see [W. Fulton, J. Harris] and [S. Sternberg]. 6 . For a general version of the double coset formula let cg, denote conjugation by the coset representative g T , Kgr = gTKgF1and L, = K n K'gr.
Solutions and Hints for the Exercises
132
Make restriction and induction precise by specifying the subgroups concerned: for example i&+Kdenotes restriction from G to the subgroup K . Then
with K = K’ condition (b) is precisely what is needed to ensure that ( X W ,~;+&H+GXW) is no larger than 1 . 7. This is a special case of a more general result: if an arbitrary finite group G acts on the finite sets X and Y with permutation characters 7r and T respectively, then the inner product ( T , T ) equals the number of orbits of the action of G on the product set X x Y . This follows by first noting that ( K , T ) expands as a sum of terms (a,a)= 1 with a a common irreducible component of 7r and T , while an orbit of the G-action on X x Y also corresponds to an irreducible component. For the action of the symmetric group S, on { r - subsets} x {s - subsets} we must show that there are r 1 orbits. Note that S, acts transitively on the sets of r- and s-subsets independently, and that we can count the orbits by matching the r-subset A against the s-subset B in ( A ,B ) . For example, if A and B are singletons there are two orbits, one represented by ( A ,A ) and the other by (A,B ) with A # B. For the last part
+
This exercise can be repeated with G = GL,(P,) acting on the set of r-dimensional subspaces of an n-dimensional vector space over Q. Compare Chapter 3, Exercise 7. 8. For an expansion of the hint see [S. Sternberg, 53.81, or [J-P.Serre, $71. The construction is called the method of “little groups” of Wigner and Mackey . 9. As with the other metacyclic groups we have considered the conjugacy classes are obtained using the relation b-’ab = ar. There are q 1dimensional representations and q-dimensional representations obtained by induction up from the normal cyclic subgroup generated by a. (Check: q q2(y) = p q . ) If C is a primitive p t h root of unity an
(y)
+
Solutions and Hints for the Exercises
133
example of the latter type is given by
Irreducibility can either be checked directly or deduced from Mackey’s criterion above.
Chapter 5 1. The clue to the characters of AnV and S n V is given by the calculation for n = 2 given in Chapter 2 and based on the splitting V @ V = A2V@S2V. If the eigenvalues for p(g) are A 1 . . . Xd then the eigenvalues for Anp(g) are the products X i , Xi, . . . Xi, (1 6 i l < 22 < .. . < in 6 d). Note that sum-
ming over all such gives the n t h elementary symmetric function in the variables XI.. . Ad. For Snp(g) we obtain products of the same form, but with repetitions allowed (see Proposition 5.8). In each case the dimension follows by counting the possible monomials, for example dim(AnV) equals the binomial coefficient For part (iii) g(x) = det(g - XI), which, up to sign, factorises as (x - Xl)(x - X2) . . . (x - Ad). Hence xAnv(g) = f(coefficient of xdPn in f(~)).The corresponding formula for XSnv(g) is more complicated; we use the relation st(V)X-t(V) = 1 together with L t ( V ) = 1- A l t A2t2 - .. . (-l)dAd td = The simple relation between A2V, S 2 V and V @ V does not extend to higher values of n. For n = 3 we may first define V(2y1)to be the quotient space of A2V@ V by W ,where W is generated by all vectors of the form ( u A w ) @ 20 - ( w A w ) @ u - ( U A W ) @ W. Then
(t). +
A2V @ v
Ez!
A3V @I
&tdf(i
+
v(2i1)
S 2 V @ V Ez! S3V @ V(231),and Vg3
E A3V @
2v(2*1) @ s3v.
+
From this the dimension of S 3 V can be read off as d2 (!). 2. The representation rings of &8 and D8 may be distinguished by the second exterior power (or determinant) of the 2-dimensional irreducible representation. For Q8 we have ( a H 1, b H 1) and for D8 ( a H 1, b H -1).
Solutions and Hints for the Exercises
134
3. The character table of Ss with exterior powers marked is as shown in the figure opposite. The permutation representation (-1) is denoted by 7r.
The 16-dimensional representation was pulled out of a hat in an earlier exercise. It arises by induction from the sugbroup S3 x S2 x (1) associated
p mm
with the (symmetric) partition 3
1
in the same way as we have
indicated for Ss (Chapter 4,Exercise 5). In general the dimension of such a representation is given by the ‘hook length formula’, here 16 = The characters p and EP can either be calculated using the regular repreand 3
sentation or from the pair of partitions
&.
2 . In the
first case we obtain the values of p f ~ pas (10,0, -2,2,0,0,0,0, -2,4,0). The character shows the significance of the exterior powers of 7r for the representation ring R(S6). For the family of all symmetric groups S, we have the following ‘stable’ result. Let R ( S )= C,R(S,) be given the structure of a (graded) ring by means of the natural product R(S,) x R(S,) -+ R(S,+,), and let A be the free A-ring on a single generator. As a ring A = Z[Q, 22,.. .], polynomials in infinitely many variables, with exterior powers given by A n ( q ) = 2,. Then it is possible to show that A Z R ( S ) ,and so to grade A that A, E R(S,). For the bare bones of the argument see [M.F. Atiyah]; for a more leisurely discussion see [D. Knutson].
Chapter 6 Using the method suggested, or doing a direct calculation on characters, we obtain 1. (i)
(ii) Vpn is isomorphic to the sum over k of
I
(i)=
k-1
(:)I
vn-2k, where e=o for n = even we sum from k = 0 to n/2, and for n = odd from k = 0 to ( n - 1)/2. See also Chapter 7, Exercise 4.
136
Solutions and Hints for the Exercises
(iii) If n is even s2Vn E
chk
n/z
=
VO @ V, CB V, @ . . . , and if n is odd,
k=O (n--1)/2
S2Vn
c
h k + 2 = V2 @ & @
k=O
n
n
Vn C3 Vn 2 j 2 0 & n - 2 j
= @ &n-Zj j=O
. . . . Clebsch-Gordan implies that n
= @
k=O
Kk. Combine this with the
previous calculation to obtain A2Vnfor both n = even and n = odd. For S3& recall from Chapter 5, Exercise 1 that, if g E G and A1 . . . Ad are the eigenvalues of g acting on the representation space V , then the eigenvalues of g acting on SnV are {A:. . .A% : i l + . . . + i , = a}.Granted this let the eigenvalues of g E SU2 acting on 6 be z 2 ,1,Z - ~ ( I Z I = 1). Then the eigenvalues for S3& are z*~,z*~,z*' (twice) and 1 (twice), giving a decomposition
S3& 2 & @ &. 2.
Therefore A E Aut(M3(@)).
A B ( X ) = (AB1)X(AB)L1 A o B ( X )= AIBIXBrlAT1 and (AB)I = A1B1. Therefore A B ( X )= ( Ao B ) ( X ) . Now let
then
+ + +
+ + +
From this the character of the A-action equals z 2 22 3 22-1 z - ~ , and M3((c) E V2 @ 2V1@2Vo. As a dimensional check we have 3 4 2 = 9. 3. Additional reference: [Th. Brocker, T. tomDieck, Chapter I, Section 61. 4. The irreducible representations of SO3 are in (1- 1)correspondence with the irreducible representations V, of SU2 such that - 1 2 acts as the identity, i.e. m = even. An analogue of the Clebsch-Gordan formula holds. For U2 there is a surjective map U1 x SU2 -+ U2 with kernel { ( 1 , 1 2 ) ,
Solutions and Hints for the Exercises
137
(-1, -12)) and for UI x SU2 we can use the analogue of Proposition 3.9 for compact groups.
Chapter 7 1. The Lie algebra consists of 2 x 2 complex diagonal matrices of trace equal to 0. [adz,z d y ] ( z ) = (adzady - adyadz)(z) 2. (i) = adz[y, z] - ady[z,z ]
41 “z,41,
41 I. 4 + “z,Z1,YI
- [Y,[z, = Yl - “Ylzl, = “z,zl, Yl f “z, Y1, = “X>Yl, = [z, [Y,
4
= a d k , Yl(Z).
[adz,ady] = ad[z,y].
Therefore (ii) Let e =
(: i) ,f (: :) ,h (i =
=
1 :)
. Then iil2
= ( e ,f , h, ) and
[ h , e ]= 2e, [ h , f ]= -2f and [ e , f ] = h. So
ade(e) = 0 , a d f ( e ) = -h,
a d e ( f ) = h, adf(f) = 0, a d h ( f )= - 2 f ,
adh(e) = 2e,
ade(h) = -2e, adf(h) = 2 f , adh(h) = 0.
Hence the matrices representing ade,adf and adh with respect to the basis { e , f , h} are respectively
0 0 -2
0 0-0
2 0 0
Call these matrices E , F and H . Now 503 = ( A ,B , C , ) ,where
0 -1 0
0 0 -1
00 0
and [A,B] = C , [ A ,C] = -B and [B,C] = A. It is sufficient to find a matrix P E GL3(@)with the mapping (pp : { E ,F, H } 4 A&(@), X P - l X P having { A ,B , C} as its image. 3. Let {vi} and {wj}be bases for V and W respectively. ---f
Solutions and Hints for the Exercises
138
+
+
(a) We show that y(s y) = y(z) y(y) by applying the left-hand side to the generic basis element vi 8 wj. A similar argument applies to
Y(XX). (b) [y(z),y(y)l = (ZY - YZ) 8 I 8 ( 2 ’ ~ ’- ~ ’ 2 ’= ) y([a,y]) by formal manipulation. (c) Write cp : g + End(V 8 W) for the map z H p(z) 8 p ’ ( y ) , and take p = p1 and p’ = p 2 to be the usual representations of degrees 2,3 of 5I2. Then with the same notation as in Exercise 2,
By calculating the effect of p(H), p(E) o p(F) and p(F) 0 cp(E) on the basis element
(i)
8
(!)
we see that
Hence cp is not a homomorphism of Lie algebras.
4. Given the solution of Chapter 6, Exercise 2, and the pairing of representations of SU2 and 5I2 in Corollary 7.10, we have only to note that the algebra and group representations are compatible with the exponential map. Thus
n times
5. Parts (i) and (ii) are routine calculation. For part (iii) we note that, by Schur’s lemma, the Casimir operator is constant on V,, and this constant may be computed by applying C to v. Here v is a ‘highest weight vector’, that is an eigenvector for H with maximal eigenvaluelweight. * (iv) We use the Casimir operator C to show that V is a direct sum of irreducibles. Assume that C has only one eigenvalue on V .
Solutions and Hints for the Exercises
139
Let V * be a subspace of V that is a direct sum of irreducibles of maximal possible dimension. If V * # V find an irreducible subspace of V/V'. Because of the eigenvalue assumption for C, and the irreducible summands of V * are all isomorphic to some Vd. Here we use the same notation as in Chapters 6 and 7, together with elementary facts about semisimple modules from Chapter 1. Suppose that = U/V*. Then E$+'V = 0, so that E$+'U c V * . On the other hand E$+'V* = 0, hence E+ is nilpotent on U , say E?+' = 0, E$ # 0 on U . Thus k 2 d. Since [E;", E-] = (Ic -t1)(H- Ic)E$,Ic must be an eigenvalue of H , so that k < d. Therefore k = d. Now choose u E U such that ;ii is a highest weight vector in Let w = Edu and u' = E f w . Then 21' is a non-zero multiple of ii. On the other hand E+u' = 0, and we have just seen that u' = E f w is an eigenvector of H . It follows that u' generates a copy of V d not contained in V " ,a contradiction. See also [W. Fulton, J. Harris, page 4811. 6. p : SL,(@) --+ Aut(sI,(@)). (i) For all X E SL,(C), Y E sI,(C), truce ( X Y X - l ) = trace(Y). Moreover p ( X ) is linear, so p ( X ) E Aut(sI,(@)). Let X,2 E SL,(@), then under p ( X 2 ) Y is mapped to X Z Y Z - l X - l , so that p ( X 2 ) = p ( X ) o p ( Z ) ,i.e. p is a group homomorphism. (ii) As a vector space sK,(C) has dimension n2 - 1, since the condition This equals the dimension of the trace = 0 gives a hypersurface in PZ. domain group SL,(@), and if the representation were to split we would have a contradiction. (Any such splitting would give a representation of lower dimension than p , hence one with non-trivial kernel.) For more on the adjoint representation see [Th. Brocker, T. tomDieck] or [W. F'ulton, J. Harris].
v
v
u.
Chapter 8
1. With the same notation as for odd primes p write
where CY is a generator of the cyclic group F - (0). As before by passing to a field with q2 elements we can find a group element b of order q f 1.
Solutions and Hints for the Exercises
140
The conjugacy classes and their orders are given by
Check: and o again denote primitive (q )– 1)st (q+1) + 1)st roots of unity the character table may be summarised as
I lcl $
12 1
xa
Q q+ 1
9,
q-1
I
I
ae 1 1 1 0 1 1 pie +p-ie -1 0
b" 1
C
-1 0 +j"
+ (T-j" )
As the entries betray, the construction proceeds as for p = odd. In splits as 1 + and for the family particular the induced character we use the induced charater and the relation
2. With
by direct calcula-
tion. Properties of Mobius transformations show that the spaces map upper half-planc goes to the boundary point as claimed; note that For the picture we have
SL2(R)
o2x S1, with copies of D2 normal to the page.
Solutions and Hints for the Exercises
141
3. From the diagram above one has that the elliptic elements equal the union of all subgroups of SLz(R) which are isomorphic to the circle group S1. Those elements at the ‘top’ of the diagram (trace 2 2) form the union of all subgroups isomorphic to R. The remaining elements (trace 6 -2) do not belong to any 1-parameter subgroup, with the exception of -12. To see that the exponential map is surjective for GLz(cC) (indeed for GL,(C)) use the Jordan normal form. The diagonalisability of a unitary matrix can be used to show that ‘exp’ is surjective for U,. Like other properties this generalises to other compact Lie groups; any geodesic passing through the identity is infinitely extendable, and any two points can be joined by a geodesic. 4. The claim follows from the following three steps:
(i) With m any natural number and A ( t ) =
:(
A(t)
(%’
0)
= A(m2t)= A@)”’
(ii) Let cp be a finite-dimensional unitary representation, cp : SL2(R) -+ U,. By step (i) the eigenvalues of cpA(t)are a promutation of their m2-powers for any value of m and hence are roots of unity. They must therefore all equal 1. (iii) The representation cp is trivial on the normal subgroup generated by the matrices A@),which coincides with SLz(R) itself.
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Bibliography
M.F. Atiyah: Power operations in K-theory, Quarterly J. Math. 17 (1966) pp. 165-193. M.F. Atiyah, D.O. Tall: Group representations,X-rings and the J-homomorphism, Topology 8 (1969), pp. 253-297. N. Blackburn, B. Huppert: Finite groups I1 (Grundlehren, no. 242) SpringerVerlag (1982). Th. Brocker, T. tomDieck: Representations of compact Lie groups, SpringerVerlag (1985) M. Burrow: Representation theory of finite groups’, Academic Press-NY (1965). R. Carter, G.B. Segal, I. Macdonald: Lectures on Lie groups and Lie algebras (LMS Student Text, no. 32), Cambridge University Press (1995). L. Dornhoff Group representation theory, Part A, Marcel Dekker-NY (1971). W. Fulton, J. Harris: Representation theory, a first course, Springer-Verlag (1991). G. James, M. Liebeck: Representations and characters of groups, Cambridge University Press (1993). J.L. Kelley: General topology, Springer-Verlag (Reprint of 1955 edition). D. Knutson: A-Rings and the representation theory of the symmetric group, Springer LN no. 308 (1973). S. Lang: Algebra, Springer-Verlag (3rd edition). S. Lang: SLz(W), Springer-Verlag (1993). J-P. Serre: Linear representations of finite groups, Springer-Verlag (1977). S. Sternberg: Group theory and physics, Cambridge University Press (1994). E.B. Vinberg: Linear representations of groups, Birkhauser-Base1 (1989).
143
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Index
group ring, 3 Groups C,., cyclic, 1 Lhrn,dihedral, 10, 39, 44 I , icosahedral ( A 5 ) ,21 0, octahedral ( S 4 ) , 20 PSLz(lF7),23, 126 P*, order equals p3, 42, 44 Q 4 t , quaternion, 44 SLz(Iw), special linear, 1, 91 SLz(F,), characteristic 0 representations, 89, 92, 95, 123 SLz(F,), characteristic p representations, 57 SUz, special unitary, 2, 43, 64, 84, 86 5’5, symmetric, 44, 124 SS,symmetric, 127 T , tetrahedral (A4), 20 T ’ , binary tetrahedral (SL2(lF3)), 43 Un,unitary, 63, 81, 104, 105
pregular (p’-element), 116 A-structure, 55 1-parameter subgroups, 81 abelian group, 15, 26 adjoint representation, 79, 86 algebraic integer, 27 alternating product, 52 alternating square, 14 Brauer character, 116 theorem, 95 Burnside’s theorem, 9 Casimir operator, 87 character, 13 table, 19-21, 23, 93, 124 class function, 13, 18 Clebsch-Gordan formula, 71 complementary series, 97 complexification, 84 differential manifold, 75 discrete series, 92, 97 double centraliser condition, 8 double coset formula, 45, 131
Haar integral et seq., 101 indecomposable, 5 induced representation SLZ(W),97 finite, 35 compact, 67 irreducible, 1
exponential map, 82 F’robenius reciprocity, 38 145
146
isotypic, 26, 67 Jacobson radical, 119 Lie algebra, 78 5 [ 2 , (representations), 84 Lie group, 75 Mackey’s criterion, 38 Maschke’s theorem, 5 metacyclic group, 42 Method of little groups (Mackey-Wigner), 132 minimal condition, 107, 109 mock discrete series, 97 modular representation, 115 orthogonality relations, 15 principal series, 91, 97 product representation (GI x G z ) , 30 radical, 109 radical (nilradical), 119 real representation, 31, 33 F’robenius-Schur Theorem, 34
Index
regular representation, 17 representation, 1 representation ring R(G),55 representations of SU2, 57 Schur’s lemma, 5 semisimple module/ring, 4, 112 simple ring, 6, 108, 114 symmetric group S, exterior powers, 56 permutation representation, 20 symmetric product, 52 symmetric square, 14 tangent bundle/vector, 75 tensor product (advanced), 47 (utility), 12 topological group, 1, 63 Wedderburn’s theorem, 9 Young diagram, 130
