Lectures on Quantum Groups (Graduate Studies in Mathematics, V. 6)

Other Titles in This Series Jens Carsten Jantzen, Lectures on quantum groups, 1996 Rick Miranda, Algebraic curves and Riemann surfaces, 1995 4 Russell A. Gordon. The integrals of Lebesgue, Denjoy, Perron, and Henstock, 1994 3 William W. Adams and Philippe Loustaunau, An introduction to GrObner bases. 1994 2 Jack Graver, Brigitte Servatius, and Herman Servatios, Combinatorial rigidity, 1993 I Ethan Akin. The general topology of dynamical systems, 1993 6 5

Lectures on

Quantum Groups

Graduate Studies in Mathematics Volume 6

Lectures on Quantum Groups Jens Carsten Jantzen

American Mathematical Society

Editorial Board James E. Humphreys Lance W. Small 1991 Mathematics Subject Classification. Primary 17B37. ABSTRACT. This book is an introduction to the theory of quantum groups. Its main objects are the quantized enveloping algebras introduced independently by Drinfeld and Jimbo. We study their finite dimensional representations, their centers, and their bases. In particular, we look at the crystal (or canonical) bases discovered independently by Lusztig and Kashiwara. We first look at the quantum analogue of the Lie algebra 512, and then at the quantum analogue of arbitrary finite dimensional complex Lie algebras. The book is directed to anyone who wants to learn the subject and has been introduced to the theory of finite dimensional complex Lie algebras.

Library of Congress Cataloging-in-Publication Data Jantzen, Jens Carsten. Lectures on quantum groups / Jens Carsten Jantzen. p. cm. -(Graduate studies in mathematics, ISSN 1065-7339; v. 6) Includes bibliographical references and index. ISBN 0-8218-0478-2 (alk. paper) 1. Quantum groups. 2. Mathematical physics. QC20.7.G70J36 1995

I. Title. II. Series.

512'.55-dc20

95-25393

CIP

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The American Mathematical Society retains all rights except those granted to the United States Government. The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability.

10987654321

010099989796

Contents Introduction Chapter 0.

Gaussian Binomial Coefficients

Chapter 1.

The Quantized Enveloping Algebra Uq(s[2)

Chapter 2.

Representations of Uq(S[2)

Chapter 3.

Tensor Products or: Uq(s[2) as a Hopf Algebra

Chapter 4.

The Quantized Enveloping Algebra Uq(g)

Chapter 5.

Representations of Uq(g)

Chapter 5A. Examples of Representations Chapter 6.

The Center and Bilinear Forms

Chapter 7.

R-matrices and kq[G]

Chapter 8.

Braid Group Actions and PBW Type Basis

Chapter 8A. Proof of Proposition 8.28 Chapter 9.

Crystal Bases I

Chapter 10.

Crystal Bases II

Chapter 11.

Crystal Bases III

References

Notations Index

Vii

Introduction There is no rigorous, universally accepted definition of the term quantum group.

However, it is generally agreed that this term includes certain deformations of classical objects associated to algebraic groups. Before I go on, let me briefly illustrate by an example what the word deformation means in this context. Take the polynomial algebra k[X, Y] in two variables X and Y over a field k. We can think of k[X, Y] as the (associative) k-algebra with generators X and Y and relation YX = XY. Consider now for each q E k the (associative) k-algebra kq [X, Y] with generators X and Y and relation YX = qXY. One checks easily that all monomials X'Yn are a basis of kq [X, Y] over k (for any q) and that the multiplication of two basis elements is given by

X myn Xrrs = qnr Xm+ryn+s So the kq [X, Y] are a family of algebras (parametrized by q E k) where the multiplication depends in a "nice" way on the parameter q and where we get for q = 1 our old algebra k[X, Y]. This family and its members are then called deformations of k [X, Y].

This example should give you an intuitive feeling of what we mean by a deformation of a given algebra A: It should be a family of algebras depending "nicely" on a parameter q such that we get back A for some special value of q. (Of course, this is not a rigorous definition of a deformation.) In the theory of quantum groups one deals with deformations Uq(g) of the enveloping algebra U(g) of a semisimple Lie algebra g and with deformations kq[G] of the ring k[G] of regular functions on a semisimple algebraic group G. In both cases the original algebras have an additional structure (a comultiplication), and it is essential that the deformations are compatible with this additional structure. The algebras Uq(g) are often called quantized enveloping algebras. They are - strictly speaking - deformations not of U(g) but of a certain covering of U(g). These algebras (or minor modifications thereof) were introduced independently by Drinfel'd and Jimbo around 1985. They were first used to construct solutions to the quantum Yang-Baxter equations, cf. 3.17, 7.6. Since then they have found numerous applications in areas ranging from theoretical physics via symplectic geometry and knot theory to modular representations of reductive algebraic groups. By now, this subject has become so large that any textbook will have to restrict itself to parts of the theory. The choices made here can be explained in part by my personal interests, in part by the circumstances of its genesis. This book has its origin in notes that I wrote for a course on quantum groups in the spring quarter of 1994 at the University of Oregon. (Since then the notes have been revised and the Chapters 5A and 9-11 have been added.) i

2

INTRODUCTION

That course was directed to students who had just been introduced to complex semisimple Lie algebras along the lines of the textbook by Humphreys [H]. The goal of the course (and of the notes) was to make them acquainted with basic parts of the

theory of quantum groups. Later I added to the notes one important application (the canonical or crystal bases) for which there had not been enough time in the actual course. At the time of that course there was only one book on the subject available to us, the one by Lusztig. I felt that my students were not adequately prepared for

his text, and I wrote my notes with the hope to provide this preparation. In the meantime a few other books on the subject have appeared (by Chari & Pressley, by Joseph, and by Kassel). Except for the one by Kassel these books appear to be directed mainly to the research mathematician, leading to the current frontier of the research in these areas. In contrast, this book is meant for the student learning the subject for the first time. So I do not work in the greatest generality possible. I assume that the reader is familiar with the main facts about semisimple Lie algebras (as in [H]), and I try to emphasize similarities and differences with that classical theory. The book by Kassel is also directed to non-experts or not-yet-experts. However, it goes into a direction quite different from the one here. It discusses many beautiful

applications of quantum groups, in particular to knot theory. At the same time it restricts itself mainly to the case of g = sI2 (with a few hints made to the general case).

All these other books can provide useful supplemental or continuing reading. So will the survey article by De Concini and Procesi in Springer Lecture Notes 1565.

Let me now briefly describe how this book is organized. It begins with a short preliminary Chapter (0) collecting basic properties of Gaussian binomial coefficients. Then there are three Chapters (1-3) dealing with U = Uq(g) in the special case g = 512. We see here in an example many of the features to be discussed later on in general (bases, simple modules, Hopf algebra structure). Besides, the results in this special case will be applied in the treatment of the general case. When we classify simple Uq(5I2)-modules, we observe a clear division into two cases: If q is not a root of unity, then the representation theory of Uq(512) looks like that of the Lie algebra sI2 in characteristic 0. If q is a root of unity, then it looks like the representation theory of sI2 in prime characteristic. Both facts generalize

to arbitrary g. While we shall deal with the non-root-of-unity case in general in Chapter 5, we do not treat the second case in this book (for arbitrary g). One can find an introduction to that topic in [De Concini & Procesi). In Chapter 4 we state the definition of Uq(g) for general g and show that there is a "triangular decomposition" of Uq(g). We deal here (and throughout the book) only with finite dimensional g and do not discuss possible generalisations to KacMoody algebras.

We turn then to the representation theory of Uq(g). From now on we assume that q is not a root of unity. We classify (in Chapter 5) the simple finite dimensional Uq(g)-modules and show that all finite dimensional Uq(g)-modules are semisimple (unless we are in characteristic 2). In Chapter 5A we describe several examples of simple modules explicitly. For some additional results in Chapter 5 we assume that k has characteristic 0

and that q is transcendental over Q. This restriction turns out to be unnecessary

INTRODUCTION

3

as we shall see in Chapter 8. But it requires hard work to remove the restriction, and so I give the easy proof in the special case early on. In Chapter 6 we determine the center of Uq (g) and get an analogue of the classical Harish-Chandra theorem on the center of U(g). The proof involves a special feature of Uq(g): a non-degenerate bilinear form on Uq(g) that (in some sense) plays the role of the Killing form on g. In Chapter 8 we construct a basis of Uq(g) similar to the PBW basis of U(g). This basis behaves nicely with respect to the bilinear form just mentioned; that is proved in Chapter 8A. The earlier Chapter 7 discusses the connection between Uq (g) - the deformation

of U(g) - and kq[G], the deformation of the algebra of regular functions on G. Furthermore we look at R-matrices. These are (in our context) isomorphisms M ® M' M' ® M for certain finite dimensional Uq(g)-modules M and M'. They lead to solutions of the so-called quantum Yang-Baxter equations and were the first reason for introducing quantum groups. The last three Chapters (9-11) deal with the crystal or canonical bases. Here we have an example where the investigation of quantum groups has led to new results on the original Lie algebra g: Take a triangular decomposition g = n- ®h ®n+. Then there is a basis B for the enveloping algebra U(n) of n- such that for each simple finite dimensional g-module V with highest weight vector v the uv with

u E B and uv 0 0 are a basis of V. This result is proved by finding first such a basis for the subalgebra U- of Uq(g) that is analogous to U(n-). The theory of quantum groups involves occasionally long computations. I have moved several of them (as well as a few more straightforward calculations) to the end of the corresponding Chapter. The reader may want to do some of them as exercises without looking first at these appendices. I have not tried to attribute credit for all the results in this book. The list of references has been restricted to the sources that I have used, to more advanced books in the area (where one can find a more extensive biliography), and to a few books that are quoted in the text. I want to thank everyone who has pointed out misprints and inaccuracies in the first version of my notes, in particular Patrick Brewer, George McNinch, Jens Gunner Jensen and Jim Humphreys.

CHAPTER 0

Gaussian Binomial Coefficients Many formulas in the theory of quantum groups (more precisely: of quantized enveloping algebras) are analogues of formulas in the theory of enveloping algebras. However, in making this transition one quite often has to replace the familiar binomial coefficients by certain q-analogues that are called Gaussian binomial coefficients. The purpose of this short chapter is to state their definition and to collect a few of their properties.

0.1. Let v be an indeterminate over Q. We are going to work in the fraction field Q(v) of the polynomial ring Q[v]. However, it turns out that all elements considered are actually contained in the subring Z [v, v i] of Q (v). Set for all a E Z [a] - va - V-a (1)

v-v-

0 for a ; 0. Furthermore, [0] = 0 and

We have obviously [0] = 0 and [a] [-a] = -[a] for all a. If a > 0, then va-3 +

[a] = va-t +

... + v-a+3 + v-a+1

This shows for all a (using [-a] _ -[a]) that [a] E Z[v, v

1].

Define the Gaussian binomial coefficients by a

_

n

[1] [2] ... [n]

(2)

for all a, n E Z with n > 0, and by (3)

[a] = [a] and n I = 1, and I a = 0 if 0 < a < n. Using n J [-a] = -[a] one checks easily that L We have obviously

I

I

[n] = (-I)'

I-a+nn- 1

(4)

for all a and n. This shows in particular that

[_1] = (-1)a n

5

(5)

0.

6

GAUSSIAN BINOMIAL COEFFICIENTS

for all n. One defines In]' for integers n > 0 by !

[0]

One has then

=1

[a] _ Ln

and

[n] = [1] [2]

[a]

.

. [n]

for n > 0.

for alla>n>0.

[n]' [a - n]

(6)

(7)

0.2. An easy calculation shows for all a and n with n > 0

La+11=v-n [a]+va-n+lrna 1].

(1)

This implies (by induction on n) that all Gaussian binomial coefficients are contained in Z[v, v J. (For fixed n use induction on a for all a > n; use 0.1(3) to deal with a < 0.) The ring Z[v,v-1] has a unique involutory automorphism that maps v to v-1 and v[a] - 1 to v. Denote it by x --+ T. We have clearly [a] _ [a] for all a E Z, hence also

n

n>0

=

[a] for all a and n. If we apply it to (1), then we get for all a and n [a+11 =vn

+v-a+n-l

[a]

[na 1] .

(2)

An elementary calculation shows for all r > 1 that r

E(-1)1 i V

{]

i=O

0.

(3)

If we apply the automorphism above, then we get also r

r E(-1)iV-i(r-1)

I

1

i I = 0.

(4)

J

i=0

0.3. If k is a ring (with identity) and q E k a unit in k, then there is a unique ring homomorphism Z[v,v-1] -> k with v --+ q (and 1 --+ 1). All [a], [n]', and

[a]

etc.; by abuse of are contained in Z[v, v-1]. We denote their images by n notation we shall often write just [a] = [a]v=q etc., if it is clear which q we consider.

Note that [a]v=i = a, [n];,=1 = n!, and

[a] v=i =

(:)

Warning: The definition of the Gaussian binomial coefficients in 0.1 is not

the traditional one that you can find (e.g.) in [Macdonald], I, §2, ex. 3. You get that old version, if you change the definition of [a] in 0.1(1) to (va -1)/(v - 1). In general, those "classical" Gaussian binomial coefficients (also called q-binomial coefficients) differ from those here by a q power (positive or negative).

APPENDIX

7

Appendix 0.2(1):

landA=lforn=1. We have

+va-_+1 [a])A(_fl[+l]+a_1[])

[n]!(v-' [at]

To get (1) we have to show that the term in the last parentheses is equal to [a + 1]. Well, we have more generally va-b - v-a-b - va+b + va-b v -b [a] + v a [b] =

= [a+ b]

v-V-1

for all integers a and b. 0.2(3):

For r = 1 the left hand side is equal to

v01']-v°[i]=1-1=0. 0 We get for r > 1 using 0.2(l)

E(-l)iVi(r-1)

LT]

i=0

_ E(-1)ava(r-1)(v_ i I r - 1 ] + a

Ylr-i+l

i=0

rr

a-1

r-1

= (-1

i((r-1)-1)

i=O r-1

- E(-1)

(j+l)(r-l)vr-j

1r - 1 ] 3

j=o

The first sum is equal to 0 by induction, the second one is equal to

v2r-I

j r-1

j=o

hence also 0 by induction.

`(_1)jv3((r-1)-1) IT - 1 L

7

] '

CHAPTER 1

The Quantized Enveloping Algebra Uq(5L2) 1.1. Fix a ground field k and an element q E k with q 0 0 and q2 0 1. Then Uq(5(2) is defined as the (associative) algebra (with 1 and over k) with generators E, F, K, and K-1 and relations

(Rl)

KK-1=1=K-1K,

(R2)

KEK-1 = q2E, KFK-1 = q-2F,

(R3)

(R4)

t

- K-1 EF-FE= Kq-q-1

Warning: In some earlier texts (R4) is replaced by

(R4')

- K-2 EF - FE _ K2 q2 - q-2

One has then to require that q4 0 1. This version leads to a very similar, but occasionally more complicated theory.

The algebra Uq(5(2) just defined is supposed to be a quantum analogue of the enveloping algebra U(s(2). The main goal in this chapter is to show that the new algebra shares two main properties with the old one: It has a PB W type basis (1.5) and it has no zero divisors (1.8). In order to prove these results we have to derive several commutator formulas.

1.2. Let us abbreviate U = Uq(5(2). LEMMA. a) There is a unique automorphism w of U with w(E) = F, w(F) = E and w(K) = K-1. It satisfies w2 = 1. b) There is a unique antiautomorphism T of U with r(E) = E, T(F) = F and T(K) = K-1. It satisfies r2 = 1. PROOF. a) We have to check that

(w(E),w(F),w(K) w(K-1)) _ (F E,K-1,K) satisfy the relations (Rl)-(R4). For example, to see that the images under w satisfy (R2), we have to know that K-'FK = q2F; however, that follows from (R3) and (Rl). The other relations are checked in a similar way. The uniqueness is obvious, 9

1. THE QUANTIZED ENVELOPING ALGEBRA Uq(SI2)

10

since E, F, K, K-1 generate U. All these generators are fixed under w2, so clearly

w2=1. b) In this case we have to check that (-r (E), -r (F), T(K), T(K-1)) _ (E, F, K-1, K) satisfy the relations (Rl)-(R4) in the algebra U°PP. Here U°PP is the algebra opposed to U: it is equal to U as a vector space, but the product a b in U°PP is equal to the product ba in U. So to check that the images under T satisfy (R4), we note

E = FE-EF = (K-1 -K)/(q-q-1). The arguments for the other

that

relations are similar. The uniqueness is again obvious; so is the equation T2 = 1. 1.3. Set

Kga - K-iq-a [K; a]

for allaEZ

.

(1)

q - q

Note that the relation (R4) can be written as EF - FE = [K; 01. The [K; a] will occur when we compute more complicated commutators, cf. (5), (6) below. The following formula will be useful in these calculations: One has for all a, b, c E Z [b + c] [K; a] _ [b] [K; a + c] + [c] [K; a - b].

(2)

(We use here [a] = cf. 0.1.) The relations (R2) and (R3) imply easily for all a [K; a]E = E[K; a + 2] and [K; a]F = F[K; a - 2]. (3) The automorphism w from 1.2 satisfies (4) w([K; a]) = -[K; -a] for all a E Z. One checks for all integers s > 0 that

EFs = FSE+ [s]Fs-1[K; 1 - s], and for all integers r > 0 that [r]Er-1 [K; FEr = ErF r - 1].

(5)

(6)

Actually, it is enough to prove (5); we can then apply w to deduce (6). 1.4. LEMMA. The algebra U is spanned as a vector space over k by all monomials FSKnEr with r, s, n E Z, r, s > 0. PROOF. We claim first that the span of these vectors is stable under multipli-

cation by all generators (E, F, K, and K-1) of U. This is trivial for F (where FFSKnEr = Fs+1 KnEr) and an easy consequence of (R3) for K and K-1:

KFSK"Er = q-2sFsKn+'Er and

K -1FsK"Er = g2sFsKn-1Er

Finally, 1.3(5) and (R2) yield

EF5KnEr = FSEKnEr + [s]Fs-1[K; 1 - s]KnEr = q-2nFsKnEr+1 + [s]Fs-1 [K; 1 - s]KnEr, where we drop the last term for s = 0. Since we can write [K; 1 - s]Kn as a polynomial in K and K-1, we get the claim also in this case. Now the claim implies that the span of our monomials is stable under multiplication with any element in U, hence contains U = U 1.

1. THE QUANTIZED ENVELOPING ALGEBRA Uq(512)

11

1.5. THEOREM. The monomials FSK"El with r, s, n E Z, r, s > 0 are a basis of U.

PROOF. We have already shown that these monomials span U; so we have to prove only their linear independence. Consider a polynomial ring k[X, Y, Z] in three indeterminates X, Y, Z and its localization A = k[X,Y, Z, Z-1]. Then

all monomials YsZnXr with r, s, n E Z, r, s > 0 are a basis of A. We define endomorphisms e, f , and h of A (as a vector space over k) by (for all r, s, n)

Ys+l znxr, f (YsZfX r) = e(YsZnXr) = q-2nYsZnXr+1 + [s]Ys-1 Zq1-s - Z I q-q-l

S-1

ZnXr'

h(YsZnXr) = q-2sYsZn+1Xr. Obviously h is bijective with

h-1(YsZnXr) = g2sYsZn-1Xr.

We can now check that (e, f, h, h-1) satisfy the relations (R1)-(R4). Therefore there is a homomorphism U -+ Endk(A) that takes E to e, F to f, and Kt to h'. So it takes a monomial FSKnEr to the monomial f shner. We have f shner(1) = YsZ"X r for all r, s, n. This shows that the f shner are linearly independent, hence that the FSKnET are linearly independent. 1.6. Set U+ resp. U- equal to the subalgebra of U generated by E resp. by F. The Er with r E Z, r > 0 (resp. the FS with s E Z, s > 0) are a basis of U+ (resp. of U-), since these elements are linearly independent by Theorem 1.5. So both U+ and U- are isomorphic to a polynomial algebra over k in one indeterminate. Set U° equal to the subalgebra of U generated by K and K-1. The Kn with n E Z are a basis of U°, and U° is isomorphic to the localization k[Z, Z-1] of the polynomial algebra over k in one indeterminate Z. For all i E Z let -yj : U° -+ U° be the automorphism of the k-algebra U° with 'yi(K) = q'K. We have -yo = 1 (the identity map) and -yj o'yj = ryi+.7 for all i, j E Z. The definition 1.3(1) yields immediately

for all i, a E Z.

-yj ([K; a]) = [K; a + i]

(1)

The relations (R2) and (R3) imply for all h E U°

hE = E'y2(h)

hF = Fry-2(h).

and

1.7. LEMMA. We have for all r, s E Z, r, s > 0

t

min(r,s)

EFS

i

[r] i

IiSi[i]!Fs-Z(fl[K;i-(r+s)+j])E r

i

j=1

(2)

1. THE QUANTIZED ENVELOPING ALGEBRA Uq(512)

12

PROOF. We use induction on r. For r = 0 (or s = 0) the claim is trivial, for r = 1 and s > 0 it is equivalent to 1.3(5). Suppose now that we have for some r > 1 and s > 0 an equation

ErFs =

min(r,s)

Fs-2h,,Er

(1)

i=0

with all hi E U°. We get then [using 1.3(5) and 1.6(2)] min(r,s)

Er+1Fs

= 1` EFs-ihiEr-i i=0 rnin(r,s)

Fs-'EhiEr-i i=0

min(r,s-1) i]Fs-i-1 [K;

[s -

+

1 - s + i]hiEr-

i=0

min(r,s) Y-2(hi)Er-i+1

FS-i i=0

min(r+1,s)

+

[s-i+1]Fs-z[K;i-s]hi-1E,r+1-i

i=1

So we get

Er+1Fs =

min(r+1,s)

Fs-ihiEr+l-i

(2)

i=° with h' =Y-2(ho) and

hi = 'Y-2(hi) + [s - i + 1][K; i - s]hi_1

for i > 0.

(3)

(We set hr+1 =0 in case r + 1 < s.) The argument so far shows that there is always a formula as in (1). It is left to show: If the hi are given by the formula in the lemma, then the hi are given by the analogous formula (with r replaced by r + 1). This is trivial for i = 0 where h° = 1 implies h'0 = 1. For i > 0 the calculations get more complicated and are left to the appendix.

1.8. PROPOSITION. The algebra U has no zero divisors.

PROOF. Any u E U, it # 0 can be written as a sum of a term FshEr (with h E U°, h # 0 and r, s E Z, r, s > 0) and of terms Fs'h'Er' (with h' E U° and r', s' E Z, r', s' > 0) where either s' < s or s' = s and r' < r. In this case we call FshEr the leading term of it. We have by Lemma 1.7 for all h, h' E U° and all integers r, s, p, m > 0 min(r.p)

FshFP-ihiEr-'h'E-

(FshEr)(Fph'E-) = i=o

APPENDIX

13

with suitable hi E U° and h0 = 1. We get from 1.6(2) inin(r.p)

FS+p-'Y2(i-p)(h)hi'Y2(i-r)(h')Er-:+m

(FShEr)(Fph'Em)

(1)

i=O

So, if h # 0 and h' # 0, then the leading term of (FShEr)(Fph'Er) is equal to Fs+pry_2p(h) ry_2r(h')Er+m. (Note that 'Y-2p(h)ry_2r(h') # 0 since the ryi are automorphisms and since U° is an integral domain.) The formula (1) implies also: If u, v E U, u, v # 0 have leading terms FShEr resp. Fph'E', then uv has leading term FS+pry-2p(h)-y-2r(h')Er+m, in particular UV #0.

1.9. The product of two monomials FSKfEr and FPKIEm is by 1.8(1) a linear

combination of monomials FiKhE3 with j - i = (r - s) + (m - p). This shows that we can regard U as a graded algebra where each FSK"Er is homogeneous of degree r - s. However, the existence of this grading is clear from our construction. First of all, we can grade a free algebra by assigning arbitrary degrees to the generators. In our case, we set deg(E) = 1, deg(F) = -1, and deg(K) = 0 = deg(K-1). Then the relations (R1)-(R4) are homogeneous (of degree 0, 1, -1, 0). So they generate a graded ideal in the free algebra, and the factor algebra U inherits a grading. Clearly each FSKfET is homogeneous of degree r - s for this grading. If u E U is homogeneous of degree i, then (Rl)-(R3) imply easily KuK-1 = g2xu.

(1)

If q is not a root of unity, then qi with i E Z are distinct. In that case the graded pieces of U are exactly the eigenspaces of the map u --> KuK-1. If q is a root of unity, then the grading is finer than the eigenspace decomposition.

Appendix 1.3(2):

Multiply both sides with (q - q-1)2. We get on the right hand side (qb -

q-b) (Kqa+c - K-lq-a-c)

+ (qc -

q-c)(Kga-b - K-1q-a+b)

= K(qa+b+c - qa-b+c + qa-b+c - qa-b-c)

+ K-1(-q-a+b-c + q-a-b-c - q-a+b+c + q-a+b-c)

_ (qb+c - q-b-c)(Kga - K-1q-a), i.e., the left hand side. 1.3(3): a

(q - q-1)[K;a]E = (Kq - K-1q-a)E = E(Kq2qa - K-Iq-2q-a) _ (q - q-1)E[K; a + 2]. (Use (R2) for the second equality; otherwise it is just the definition.)

1. THE QUANTIZED ENVELOPING ALGEBRA Uq(S12)

14

1.3(5):

We use induction on s; for s = 1 this is just the relation (R4). Now suppose that s > 1; then

EFS = EFs-1F = Fs-1 EF + Fs-2 [s - 1] [K; 2 - s]F = FS-1FE + Fs-1 [K; 0] + Fs-2F[s - 1] [K; -s] = FSE + Fs-1([K; 0] + [s - 1] [K; -s]).

Now apply 1.3(2) witha=1-s,b=1,andc=s-1. 1.5:

We have to check that (e, f, h, h-') satisfy the relations (Rl)-(R4). Now (Rl) is satisfied by construction and (R3) is more or less obvious. For (R2) note that eh-I(YSZnXr) = g2se(Y$Zn-lXr) = q2s-2n+2YsZn-l xr+l

+

g2s[S]YS-1

Zq'-s - Z-1qs-1 Zn+1Xr,

q-q-1

hence

heh-1(YsZnXr) = q-2n+2YsZnXr+l + g2[S]Ys-l

Zql-s - Z-lqs-l ZnXr q - q-l

= g2e(YsZnXr) Now for (R4): We have ef

(YsZnXr) = q-2nI/s+l ZnXr+l + [s

Zq-S

-

Z-1gsZnXr,

q-q-l

+ 1]YS

and

f e(YSZnX r) = q-2nys+lznxr+l + [S]Ys

Zql

s _ Z-qs l ZnXr, 1

q1

q -

hence

(ef - fe)(YSZnXr) = YsTZnXr, where

T = [s + 1]

Zq-S - Z-1qs

q-q-1

Zq1-S -

Z-lqS-l

q-q-1

In order to get (R4) we have to show that T is equal to

(q-2sZ-g2sZ-1)/(q-q-1)

This is easy to check. (We can also apply 1.3(2) with a = -s, b = s, c = 1, and with K replaced by Z.)

APPENDIX

15

1.7:

Set for all integers r, s, i > 0

hi (r,s)=

r 1 [Z ]

r Is]

[i]!II[K;i-(r+s)+j].

(1)

j=1

Note that hi (r, s) = 0 for i > min(r, s). In order to finish the proof of Lemma 1.7 we have to show that (for all i > 0) hi(r + 1, s) = ry_2(hi(r, s)) + Is - i + 1] [K; i - s]hi_1(r, s).

(2)

Well, we have r

1

'Y- 2 (hi (r, s)) = [r]

r

[i] [i]!fl[K;i-(r+s)+j-2] J

i=1

i- 1 _ [r] [:] [i]!fl[K;i-(r+1+s)+j]. i

i

Using

[:1 [z]e=[s][s-1]...[s-i+2][s-i+1]= [i s1] [i - 1]![s - i+ 1], we can rewrite the last term in (2): Is - i + 1] [K; i - s]hi_1(r, s) ir ] = 1i -1 i,s] [i]![K;i-s]fl[K;i-(r+1+s)+j].

j=1

If [i]! = 0, then both sides in (2) are equal to 0. So we may assume that [i]! # 0. A comparison of the formulas above shows that we have to prove

[][K;i_(r+s+1)1+[.r1][K;i_s] = [r±1][K2. (+ +1)] .

(3)

We know by 1.3(2) [applied to a = 2i - (r + s + 1), b = i, and c = r + 1 - i] that [r + 1][K; 2i - (r + s + 1)] = [i][K; i - s] + [r + 1 - i][K; i - (r + s + 1)]. Multiply (4) by [r] [r - 1]

(4)

[r - i + 2]/[i]! and we get (3).

REMARK. Note that (3) is a special case of

[b+c_1]

[K;a - b] +

[b+c_1] [K;a+c]

[b+c]

[K;a]

(5)

for all integers a, b, c with c > 0 and b > 0. It follows from 1.3(2) by multiplying . with [b + c - 1] [b + c - 2] . . . [c + 1] / [b] (If [b]' = 0 one should first prove (5) as an identity in Z[v, v-1][K, K-1] with v as in 0.1. Then specialize v to q.)

CHAPTER 2

Representations of Uq(s(2) We continue to fix a ground field k and an element q E k with q 0 0 and q2 # 1. We set U = Uq(5C2) and use the notations from Chapter 1.

In this chapter we first look at finite dimensional representations of U and then determine the center of U. In both cases the result depends very much on whether q is a root of unity or not. The general pattern to evolve is this: If q is not a root of unity, then U behaves like the enveloping algebra of 5(2 over a field of characteristic 0; if q is a root of unity, then U behaves like the enveloping algebra of 512 over a field of prime characteristic. And this statement holds independently

of the characteristic - with a small exception if k has characteristic 2. But the really exciting feature of this situation is this: If we take k = C and q equal to a primitive p-th root of unity with p a prime, p > 3, then we get a representation theory over C that looks like the representation theory of 5(2 over an algebraically closed field of characteristic p.

2.1. PROPOSITION. Suppose that q is not a root of unity. Let M be a finite dimensional U-module. There are integers r, s > 0 with ETM = 0 and F5M = 0. PROOF. Denote by k[X] the polynomial ring over k in one indeterminate X. For each irreducible polynomial f E k[X] set

Mifi={mEMI f(K)'m=0 foralln>>0}.

(1)

Then M is the direct sum of the distinct M(f). If f and g are two polynomials with M(f) # 0 and M(g) # 0, then M(f) = M(g) holds if and only if f and g differ by a nonzero factor. We have M(x) = 0, since the action of K on M is invertible.

Let f E k[X] be irreducible with M(f) # 0. For each i E Z set fi equal to the polynomial fi (X) = f (q1X ). This polynomial is again irreducible, since it is the image of f under the unique automorphism of k[X] with X --> q'X. The

formula 1.6(2) shows that f (K)E = E f (q2K) and thus inductively f (K)Er = Er f(g2rK) = ET f2r(K). This implies that ETM(f) C M( f,,) for all r > 0. We want to show that M(hr) = 0 for some r > 0. (This will then imply ErM(f) = 0; since f was arbitrary, the claim follows.) Well, suppose that M( f,,) # 0 for all r > 0. Since the sum of the distinct M(9) is direct and since M is finite dimensional, there have to be integers s > r with M(f2r) = M(f23). Then f2r and f2s have to be proportional. Since they have the same constant term (nonzero, since f # X), they have to be equal. However, if f has degree n, then the leading 17

2. REPRESENTATIONS OF Uq(512)

18

q2(s-r)n coefficients differ by the factor which is not equal to 1, since q is not a root of unity. So we have reached a contradiction. The proof for F is analogous.

REMARK. This proposition as well as 2.3 do not generalize to the case where q is a root of unity, cf. 2.11 below.

2.2. If M is a U-module, then set for all A E k, A

0

MA={mEMI Km=Am},

(1)

i.e., MA is the eigenspace of K acting on M for the eigenvalue A. (It is enough to consider A 0, since K has an inverse in U.) We call MA a weight space of M; the A with MA 0 are called the weights of M. The sum of the Ma is direct. The

relations (R2) and (R3) imply for all A EMACMg2A

FMACMq-2A.

and

(2)

This shows that the sum of the MA is a submodule of M. More precisely, for any A the sum of the Mg2na with n E Z is a submodule. If M is simple and if MA 0,

then M = ®n Mg2na. (If q is not a root of unity, then n runs over all integers, otherwise over a suitable finite subset.) In general, there need not be any nonzero MA. But if k is algebraically closed and if M is finite dimensional, then K has a nonzero eigenspace on M, so there is a A with MA 0. 2.3. PROPOSITION. Suppose that q is not a root of unity and that char(k) 2. Let M be a finite dimensional U-module. Then M is the direct sum of its weight spaces. All weights of M have the form ±qa with a E Z.

PROOF. An endomorphism of a finite dimensional vector space is diagonalizable if and only if its minimal polynomial splits into linear factors, each occurring with multiplicity 1. The eigenvalues are then the roots of the minimal polynomial. So we have to show that the minimal polynomial (in the indeterminate X) of K

acting on M has the form ji(X - A) where the Ai are distinct elements of the form ±qa with a E Z. There is by 2.1 an integer s > 0 such that FSM = 0. Set r-1 hr = [K; r - s + j] for all integers r > 0. j=-(r- 1)

fi

(1)

In particular, ho is the empty product equal to 1. One checks now by induction on

r for 0 < r < s that Fs-rhrM = 0. For r = 0 this holds by our choice of s. The induction step is a longish calculation banished to the appendix. We get for r = s O = hsM = (

s-1

fi

(q - q-11-1gjK-1(K2 _ q-23))M. /J

/

We can drop the nonzero constant factors and apply a suitable power of K; this yields s-1

0

s-1

fi (K2-q-2j))11v7= (j=-(s-1) fi j=-(s-1)

(

(K-q-j)(K+q-r))M.

2. REPRESENTATIONS OF Uq(5(2)

19

So the minimal polynomial of K acting on Al divides fl j=-(s-') (X -q-j)(X+q-j), hence has the desired form. (Note that -qa # +qb for all a, b E Z: Otherwise qb-a = -1 and q2(b-a) = I contradicting the assumption that q is not a root of unity.)

REMARK. In case char(k) = 2 the proof shows that the minimal polynomial of k splits into linear factors and that all weights of M have the form qa with a E Z. If M is simple, then this implies that M is the direct sum of its weight spaces. On the other hand, it is easy to construct examples where M is not simple: Take (1 11

M=k 2 where E and F act as 0 and where K and K-1 both act as

0

1

2.4. For each A E k, A # 0 there is an (infinite dimensional) U-module M(A) with basis mo, m1i m2, ... such that for all i Kmi = Aq-22m" Fmi = mi+1,

Emi =

10, Aqt_

[i]

-

if i = 0, A-1qz-1

q-q

(1)

m,.-,, otherwise.

In order to construct this module one can check that the endomorphisms of M(A) defined by (1) satisfy the relations (R1)-(R4), cf. the proof of 1.5. However, it is easier to observe that we can take

M(A) = U/(UE + U(K - A))

(2)

with mi equal to the coset of P. Then the formulas in (1) follow easily from (R3) and 1.3(5). We get the linear independence of the mi as a corollary to Theorem 1.5.

The description in (2) makes it clear that M(A) has the following universal property: If M is a U-module and m E M a vector with Em = 0 and Km = Am, then there is a unique homomorphism of U-modules o: M(A) -> M with co(ma) = M.

If A is equal to ±qa for some a E Z, then we can simplify the last equation in (1) as follows:

Emi = ±[i] [a + 1 - i]mi_ 1

in case A = ±qa.

(3)

By (1) each mi is contained in M(A)q-2,,\. If q is not a root of unity, then the q-2iA are distinct and we get M(A)q-2z,\ = kmi

for all i > 0.

(4)

If q is a primitive 1-th root of unity, then q- 2'A = q-2jA if and only if l divides 2(j - i). So, if 1 is odd, then the weight spaces in M(A) are the

M(A)q-2,,\ = ®kmi+nt

with 0 < i < 1.

(5)

with 0 < i < 1'.

(6)

71>o

If 1 = 21' is even, then the weight spaces are the M(A)q-2,a =

®kmi+,,,I'

n>o

2. REPRESENTATIONS OF Uq(512)

20

2.5. PROPOSITION. Suppose that q is not a root of unity. Let A E k, A 0 0. If A 0 fqn for all integers n > 0, then the U-module M(A) is simple. If A = fqn for some integer n > 0, then the mi with i > n + 1 span a submodule of M(A) isomorphic to M(q-2(n+I)A); this is the only submodule of M(A) different from 0 and M(A).

PROOF. Let M' be any nonzero submodule of M(A). Since M' is K-stable, it is the direct sum of its weight spaces, i.e., of all M' fl M(A)µ. Now 2.4(4) implies

that M' is spanned by the mi contained in M'. Since we assume M' 0 0, there exists an i with mi E M'. Choose j > 0 minimal with mi E M'. We have then

mi, = Fi-imi E M' for all i > j, so M' is the span of all mi with i > j. If j = 0, then M' = M(A). So let us assume that j > 0. Since Emi E M' is a multiple of m,_1 M' we have Emi = 0, hence Aq1 _i - A - I q-7-1 = 0. This implies A2 = q2(i-1) i.e., A = fq(i-1).

The argument. so far shows that M(A) is simple for A 0 fqn (all n E Z, > 0), whereas in case A = fqn there is at. most one submodule of M(A) different. from 0 and M(A). However, if A = fqn, then 2.4(3) implies Emn+1 = 0. So there is (by M(±q-n-2) -, M(A) the universal property) a homomorphism M(q-2(n+1)A) =

that takes the mo in M(±q-n-2) to the mn+1 in M(A). Since M(±q-n-2) is simple (by what we have already proved), this is an isomorphism onto its image. This implies the claim in the proposition.

2.6. THEOREM. Suppose that Q is not a root of unity. There are for each integer N > 0 a simple U-module L(N, +) with basis mo, m1, ... , mn and a simple U-module L(n, -) with basis mo, mi, ... , m' such that for all i (0 < i < n)

Km = -n-2im q it

n-2imx> Km x = q'-

Fm

mi+I,

if i < n,

0,

ifi=n,

Fm'

=

mi

if i < n,

0,

ifi=n,

[i] [n + 1 - i]mi- 1,

Emi= 1 0,

ifi>0, ifi=0,

[i] [n + 1 - i]m',- i ,

ifi>0,

Emi= 1

-

0,

if i=0,

Each simple U-module of dimension n + I is isomorphic to L(n, +) or to L(n, -). PROOF. We get the existence from Proposition 2.5: Set L(n, ±) = M(fgn)/M' where M' is the submodule spanned by the mi with i > n. Take for the mi resp. mi the images of the mi E M(±qn). The formulas in 2.4 yield the formulas above. The simplicity of L(n, ±) follows from Proposition 2.5. Let M be any finite dimensional simple U-module. We have to show that M

is isomorphic to some L(n, f). We have M = ®a Ma by Proposition 2.3. (Use the remark in 2.3 if char(k) = 2.) Since dim M < oo, the set of A with Ma # 0 is finite. So we can find A with MA 0 0 and M92a = 0. Pick m E MA, m 0 0. We have Em E Mq2A by 2.2(2), hence Em = 0. The universal property implies that there is a nonzero homomorphism p : M(A) -> M. Since M is simple, p is surjective. Since M is finite dimensional, Proposition 2.5 implies that A = fqn for some n E Z, n > 0. Then p induces an isomorphism between L(n, ±) and M.

2. REPRESENTATIONS OF Uq(.512)

21

REMARK. If k has characteristic 2, then clearly L(n, +) and L(n, -) are isomorphic (for each n). If the characteristic of k is not equal to 2, then these modules are not isomorphic: The subspace kmo resp. km'0 is determined as the set of all m with Em = 0, and ±qn is determined as the eigenvalue of K on this subspace. 2.7. Set

C=FE+ Kq +

K-'q-1

(1)

(q - q-1)z

Using (R4) we can rewrite

+Kq+K-'q-1 C=EF- K-K-1 q-q-1 (q-q-1)2 - EF + Kq + K-1 q-1 - (q - q-') (K - K-') (q - q-1)2 hence

C=EF+Kq 1 +K-'q

(2)

w(C) = C = T(C).

(3)

(q - q-1)2 We can express (2) also using the maps w and r from 1.2 as

LEMMA. a) The element C is central in U. b) C acts on each M(A) as scalar multiplication by

(Aq+A-1q-')/(q-q-1)z

c) C acts on M(A) and M(µ) by the same scalar if and only if A = µ or

A = µ-1q-2

PROOF. a) Obviously C is homogeneous of degree 0 for the grading from 1.9.

So 1.9(1) implies that C commutes with K and K-'. An elementary calculation shows that EC = CE. If we apply w to this equation, we get FC = CF. So C commutes with all generators of U, hence it is central.

b) It is clear by 2.4(1) that Cmo = µmo where µ is the scalar in our claim. Since C commutes with F and mi = Fimo for all i, we get also Cmi = µmi for all i, hence the claim. c) We get by b) the same scalar if and only if Aq + A-1q-1 = µq + This is equivalent to µ-'q-

(A - li)q = (p-1 - A-1)q-1 = A-1µ-1G i.e., to A - µ = 0 or q =

li)q-1'

The claim follows.

2.8. Of course C operates also on each homomorphic image of M(A) as scalar multiplication (by the same scalar as on M(A)). This applies in particular to the simple modules introduced in Theorem 2.6 (in case q not a root of unity).

LEMMA. Suppose that q is not a root of unity. Let L and L' be finite dimensional simple U-modules. If C acts on L by the same scalar as on L', then L is isomorphic to L'.

2. REPRESENTATIONS OF Uq(512)

22

PROOF. There are integers n, m > 0 and signs E, E' such that L is a factor module of M(Egn) and L' one of M(E'gm). If C acts by the same scalar on L and L', then also on M(Egn) and M(E'gm). So Lemma 2.7.c implies that Eqn = E'qm (and thus L L' as desired) or that Eqn = E'q-m-2. The second case leads to qn+m+2 = EE' = ±1 which is impossible, since q is not a root of unity. 2.9. THEOREM. Suppose that q is not a root of unity. Let M be a finite dimensional U-module that is the direct sum of its weight spaces. Then M is a semisimple U-module.

PROOF. Let M be a finite dimensional U-module. Pick a composition series 0 =.A/lo C M1 C M2 C ... C M,. = M of M. Each Mi /Mi _ 1 is isomorphic to one of the modules described in 2.6, so C acts by a scalar, say µi, on Mi /Mi- 1. Then fl'= I (C - µi) annihilates M. So the minimal polynomial of C acting on M splits into linear factors, and M is the direct sum of the generalized eigenspaces for C:

M = ED M(,)

where

M(µ) = { m E M I (C - µ)5m = 0 for s >> 0}.

lL

Since C is central in U, each M(N,) is a submodule of M. So it is enough to prove that each M(,) is semisimple.

Let us assume that M = M(N,) for some µ. Then C - µ acts nilpotently on M, hence on each Mi/Mi_1. On the other hand C acts as multiplication by µi on Mi/Mi-1; this implies that µi = µ for all i. Now Lemma 2.8 implies that there is an integer n > 0 and a sign E such that each Mi/Mi_1 is isomorphic to L(n, E). We know by Proposition 2.4 that M is the direct sum of its weight spaces,

M = ® M,,. If N is a submodule of M, then N = ® N and N = N n Al, for all v. This implies that dim M = dim N + dim(M/N),,. If we apply this repeatedly to the composition series, we get

dimM _

dim(MM/Mi-1) =rdimL(n,E),,. i=]

In particular, we get now from 2.6 that dim MA = r where A = Eqn and that Mg2A = 0. For any v E MA we have therefore Ev = 0, so the submodule Uv is a homomorphic image of M(A); since it is finite dimensional, it is isomorphic to L(n, E) (if v 0). Choose a basis v1, v2, ... , v,. of MA. Then M = Ei=1 Uvi, since (M/ Ei_I Uvi)A = 0 and since each composition factor L of M/ E I Uvi is isomorphic to L(n, E), hence satisfies LA 0. By comparing dimensions (dim M = r dim L(n, E) = Ei=1 dim Uvi) we see that M is in fact the direct sum of the Uvi, hence semisimple.

REMARK. In case char(k) 2 the assumption that M is the direct sum of its weight spaces is automatically satisfied by 2.3.

2.10. Let us now turn to the case where q is a root of unity. If q1 = 1 then [1] = 0, hence [i] ! = 0 whenever i > l > 2.

PROPOSITION. If q is a primitive l-th root of unity (l E Z, 1 > 3), then El, FI, K', and K-1 are contained in the center of U.

2. REPRESENTATIONS OF Uq(5(2)

23

PROOF. We get from (R2) and (R3) K1EK-I

= q 21E = E

KIFK-I = q-2IF

and

= F,

so K1 and K-1 are central. We get also

KE1K-1 = g21E1 = E1

KF1K-I = q-21F1 = F1.

and

The formulas 1.3(5),(6) imply that EF1 = F1E and FE1 = E'F since [1] = 0. So also E1 and F1 are central. REMARK. If 1 is even, 1 = 21', then already [1'] = 0 and the argument above

shows that already E", F", K", and K-1' are central in U. In the following subsections I shall restrict myself to the case where q is a primitive l-th root of unity with 1 odd. The even case is similar; usually one just has to replace 1 by 1' = 1/2. Details are left ... 2.11. Suppose that q is a primitive l-th root of unity with 1 odd, 1 > 3. For any b, AEkwith A54 Oset Zb(A) = M(A)/U(m, - bma)

(1)

using the notation from 2.4. We have Em1 = 0 since [1] = 0 and Km1 = Aq-21m, = Amt, hence E(m1 - bmo) = 0 and K(m1 - bmo) = A(m1 - bmo). Therefore U(m1 bmo) is spanned by all Fi(m1 - bmo) = mi+1 - bmi with i > 0. So the images in

Zb(A) of the mj with j < l are a basis of Zb(.\). By abuse of notation we denote the image of mj again by mj. So Zb(A) has basis mo, MI,... , m1-1 such that the action of U is given by Kmi = q-2zAmi,

Frn = r mi+I, bm0,

(2) 0,

Emi = [i]

g

AI-i _ -t i-I g_q_1q

mi-1,

if i > 0.

The q-2iA with 0 < i < l are distinct since q is a primitive l-th root of unity with 1 odd. This implies Zb(A)q-2.a = kmi

for 0 < i < 1.

(3)

We have F1m0 = bma, in fact F1 acts as multiplication by b on Zb(A). This shows that Proposition 2.1 does not extend to the present case: We can choose b 54 0 and then F does not act nilpotently on the finite dimensional U-module Zb(A). We see also that K can have eigenvalues other than ±qa (with a E Z) since we can take a A distinct from these (unless k is finite and -q generates the multiplicative group of k). So at least the last part of Proposition 2.3 does not generalize.

2. REPRESENTATIONS OF Uq(5(2)

24

2.12. PROPOSITION. Suppose that q is a primitive l-th root of unity with 1 odd, 1 > 3. If b 0 or if A21 1, then Zb(A) is a simple U-module. If b = 0 and A = fq'

with 0 < n < 1, then Zb(A) is simple if and only if n = 1 - 1; for n < l - 1 the vj with j > n span a submodule of Zb (A), and this is the only submodule different from 0 and Zb(A). PROOF. We can more or less copy the proof of Proposition 2.5. Let M be any nonzero submodule of Zb(A). Since M is K-stable, it is the direct sum of its weight spaces, i.e., of all M fl Zb(A)µ. Now 2.11(3) implies that M is spanned by the mi contained in M. Since we assume M 0, there exists an i with mi E M. Choose

j > 0 minimal with mj E M. We have then mi = Fi-jmj E M for all i with j < i < 1. If j = 0, then M = Zb(A). So let us assume that j > 0. If b 0 then mo = b-1 Fm1_1 E M, hence j = 0. So we have to have b = 0. Clearly M is the span of all mi with i > j. Since Emu E M' is a multiple of mj-1 M' we have [j](Agl-j - A-1qj-1) = 0. Since 0 < j < 1 our assumption on q Emu = 0, hence implies [j]

0. We get therefore A2 = q2(j-i), hence A21 = 1 and A = fq(j-1).

It is left to show that for b = 0 and A = ±q' with 0 < n < 1 - 1 the span of the mi with i > n is indeed a submodule. That, however, follows easily from Em,+1 = 0. REMARK. The proposition implies that we get for 0 < n < 1 simple U-modules L(n,+) and L(n, -) of dimension n+1 where the formulas in Theorem 2.6 describe the action of U on a suitable basis: Just take the unique simple quotient of Z0 (±q'').

The proposition implies also that the Zo(±q') with 0 < n < 1 - 1 are not semisimple: The submodule spanned by the mi with i > n has no complement. So Theorem 2.9 does not extend to our present situation.

2.13. Suppose that q is a primitive l-th root of unity with 1 odd, 1 > 3. We want to describe all finite dimensional simple U-modules M under the additional assumption that k is algebraically closed. This makes sure that M is the direct sum of its weight spaces. Furthermore, by Schur's lemma the central elements E1, F1, K1, and C have to act as scalars on M. CASE 1. E1 ACTS AS 0 ON M. Then the subspace { m E M I Em = 0} is nonzero. By (R2) it is K-stable, so K has an eigenvector in it. This means that we can find m E M, m #0 and A E k, A OwithEm=0andKm=Am. By

the universal property there is a homomorphism cp : M(A) -> M with cp(mo) = m.

Since M is simple, cp is surjective. There is a scalar b E k such that F1 acts as multiplication by b on M. Then

cp(m1-bmo)=cp(F1mo)-bm=Flm-bm=0. So U(m1 - bmo) is contained in the kernel of co, and p factors through Zb(A). So Proposition 2,12 implies that M is either isomorphic to Zb(A) or to a L(n, ±). CASE 2. F' ACTS AS 0 ON M AND E1 DOES NOT. We use the automorphism w from 1.2 to "twist" U-modules: For any U-module N set wN equal to the U-module that is equal to N as a vector space and where each u E U acts on "N as w(u) acts

on N. It is clear that '-(N) ^- N for all N and that WN is simple if and only if N is simple.

This implies (since w(E1) = F1 and w(Fl) = E1) that WM is a simple module as considered in Case 1 with b 0. So M is isomorphic to some WZb(A). It has

2. REPRESENTATIONS OF U9(5(2)

25

dimension land there is a basis mo, ml , ... , m1_ I such that the action of U is given by

Kmi = g22)t-Imi,

[i]

Emi =

-

I q 1-i

Fmi =

1

_ l q i-]

q - q

mi-1,

(1)

5 mi+ 1, bmo,

CASE 3. BOTH F1 AND El DO NOT ACT AS 0 ON M. Let b E k, b 0 be the scalar through which Fl acts on M. We can find (k is algebraically closed) an eigenvector mo 0 for K in Al. Denote the corresponding eigenvalue by A. Set mi = Fimo for 0 < i < 1. We have F'-imi = F1mo = bmo 0, hence mi 0. The action of K and F on the mi is given by

Kmi = q-2iAmi,

Fmi

mi+l, ifi < l - 1,

J

Sl

(2)

ifi=l-1.

bmo,

The q-2iA with 0 < i < 1 are distinct, so the m2 are lineraly independent as eigenvectors corresponding to distinct eigenvalues. The central element C from 2.7(1) acts on M through a scalar. Since

FEmo = Cmo -

Kq+K- Iq-I (q-q_1)2 m0,

there is an a' E k with FE mo = a'mo. We get then

bEmo = F'Emo = Fl-Ia'mo = a'm1_1i hence Emo = am1_1 with a = a'/b. Now 1.3(5) yields Emi = EFimo = FtEmo + [i]F2-1 [K; 1 - i] mo for all i > 0, hence

ifi = 0,

amt-1,

Emi =

(ab+ (q2 -

- A-1qi-I) ifi > 0. _)mi_1,

q-2)(Ag1-'

(q - q

(3)

This shows in particular that the span of the mi is stable under all generators of U, hence equal to (the simple module) M. So the mi are a basis of M, and (2), (3) describe the module completely.

On the other hand, given a, b, and A we can use (2) and (3) to define a Umodule. (The main thing to be checked is that the relation (R4) is preserved.) If b 0, then the same argument as in the proof of 2.11 shows that the module is q-i)(Agl-i - A-Iqi-1)(q - q- 1)-2 are not equal to 0, simple. If a and all ab+ (qi then El does not act as 0, so we have a module of our present type. It should be noted that the module M uniquely determines b, but does not determine a and A, since we could choose instead of mo any other mi. So we could replace A by q-2i A and a by a + (q2 - q-')(Aql-i - A-1q2-1)(q - q- 1)-2 b-1, and still get an isomorphic module.

2. REPRESENTATIONS OF Uq(5(2)

26

REMARK. For any n with 0 < n < p - 1 the module Z°(q") is a nonsplit extension of L(n, +) and L(p - n - 2, +). This implies that the center of U acts by the same character on these two simple modules. Similarly, it acts by the same

character on L(n, -) and L(p - n - 2, -). Using the classification above one can check that these are the only cases of two simple modules that are not isomorphic, but where the center of U acts via the same character on both. 2.14. We have used in the last subsections that certain elements are central in U. We now want to determine the whole center of U. Recall the grading on U from 1.9. Denote the graded pieces by U,,, with m E Z.

By construction U,,,, is spanned by all F'K'Er with m = r - s. LEMMA. a) If q is not a root of unity, then the center of U is contained in U0. b) If q is a primitive l-th root of unity with l odd, 1 > 3, then the center of U is generated by El, F', and its intersection with U0. PROOF. The center of a graded algebra is graded. So we have to find out for

each m which elements in U,,, are central in U. If u E U,,, with u ; 0 is central, then 1.9(1) implies q2m. = 1. If q is not a root of unity, then this yields m = 0; so a) follows. Suppose now that q is a primitive l-th root of unity with 1 odd, l > 3. Then m has to be a multiple of 1. If m = al with a > 0, then U,,, is spannned by So any u E U,,, can be decomposed u = u'Eal with u' E U0. Then all u is central if and only if u' is central, since E' is central and since U has no zero divisors (1.8). Similarly, if m = -al with a > 0, then the central elements in U" are exactly the products of Fal with central elements in U0. This proves b). FSKnE'+at.

2.15. Any it E U0 can be written uniquely u = Er>0 FrhrEr with all hr E U°, almost all equal to 0. LEMMA. Let u = Er>° FrhrEr E U0 as above. Then it is central in U if and only if for all r > 0. (1) hr - -y-2 (hr) _ [r + 1][K; -r]hr+I PROOF. We have [using 1.3(5) and 1.6(2)]

Eu = E EFrhrEr = > F''EhrEr + r>O

_ r>0 E

r>0

E[r]Fr-'[K-.. 1

- r]hrEr

r>0

F'r'Y-2(h,.)Er+I + E[r + 1]Fr[K; -r]hr+iEr+i r>0

On the other hand uE = Er>0 Frh7.E''+I, so Eu = uE if and only if (1) holds. A similar calculation shows that also Fu = uF is equivalent to (1). Finally, uK = Ku is automatically satisfied by any u E U°.

2.16. Consider the map 7r UI - U° that takes any u = Er>0 FrhrE" as in 2.15 to h0. So 7r is the linear projection from U0 onto the subspace U° with kernel FUOE. We claim that 7r is even an algebra homomorphism. It is :

enough to show that the kernel is a two-sided ideal. So take a typical basis element

FrKfEr (with r, n E Z, r > 0) of the kernel and multiply it with an arbitrary basis element FSKmE' (with s, m E Z, s > 0) of U°. If s > 0, then clearly both (F'KrES)(FrK"Er) and (F"KnEr)(F'K"''E') are in FU0E. Ifs = 0, then q-r?nFrKn+"''Er = (F1 KnE'')Km, so again both products are in FUE.

2. REPRESENTATIONS OF Uq(5I2)

27

LEMMA. If q is not a root of unity, then 7r induces an injective homomorphism from the center of U to U°.

PROOF. We have to prove the inject.ivity. Now Lemma 2.15 says (in its notations) that each hr+1 is determined by hr, since [r + 1] [K; -r] 0 0 (this uses that q is not a root of unity) and since U° is an integral domain. So all hr are inductively determined by ho = 7r(u).

2.17. LEMMA. Suppose that q is not a root of unity. Let u E U be central and write -y_ 1 o 7r(u) _ >jEZ aiK2 with ai E k, almost all zero. Then ai = a_i for all i E Z. PROOF. We have 7r(u) = >iEZ aig1K1, hence

u=

FrhrEr

aig2Ki + r>°

iEZ

with suitable hr E U°. Then u acts on any M(A) as multiplication by >i a,g2Ai since it acts like that on the generator m°. In particular, u acts as multiplication by >i aiq(n+I)i on M(qn) for any n E Z. By Proposition 2.5 there is for n > 0 a M(q-n-2) in M(qn). So u has to act by the same scalar submodule isomorphic to on both modules:

aiq(n+1)i

aiq (n+1)i

=

i

This means aigRi =

aiq-ni

a-igni

(1)

for all n E Z. (For n > 0 this is just the preceding equality, for n < 0 it holds by symmetry, for n = 0 it is trivial.) Denote by '/ii : Z k" the group homomorphism from Z to the multiplicative group k" of our field with y),(n) = qni. The '/ii with i E Z are distinct, since i,L (1) = qi and since q is not a root of unity. So the '0i are linearly independent over k by Artin's theorem on the linear independence of characters. Now (1) can be read as E, (ai - a_i)i,L = 0. So the linear independence implies ai - a_i = 0 for all i, q.e.d.

2.18. PROPOSITION. Suppose that q is not a root of unity. Then the center of U is generated by C as a k-algebra. It is isomorphic to a polynomial ring over k in one indeterminate.

PROOF. Let s : U° U° be the automorphism of U° with s(K) = K-1, i.e., with s(K2) = K for all i E Z. Lemma 2.18 says that ry-1 o 7r maps the center of U to the subalgebra (Uo)s = { h E U° s(h) = h } of fixed points of s in U°. Clearly the Kn + K-n with n > 0 together with I are a basis of (Uo)s, hence so are the (K + K-1)n with n > 0. We have 7_ o 7r(1) = 1 and I

-y-1 o 7r(C) = (q - q-1)-2 (K + K-1). This shows that -y-1 o 7r maps the subalgebra k[C] of the center onto (U°) s. Since 7-1 o 7r is injective on the center, this implies

that 7_I o 7r induces an isomorphism between the center and (U°)' and that the center is equal to k[C]. It is isomorphic to a polynomial ring since the y_1 o7r(Cn) (up to a nonzero scalar factor equal to (K + K- 1)') are linearly independent.

2. REPRESENTATIONS OF U1(5(2)

28

2.19. Suppose that q is a primitive l-th root of unity with 1 odd, 1 > 3. Denote the center of U by Z. Set r 1-1

C U°.

Uo = S E FrhrE' l ho, hl, ... , hl-1 E U°

(1)

r=o

LEMMA. a) The k-algebra Z is generated by E', F', and Z n U. b) The restriction of 7f to Z n UU is injective.

PROOF. a) By Lemma 2.14.b it is enough to look at Z n U0. Any u E Uo can be written u = >r>0 FrhrEr with all hr E U°, almost all zero. We have then I-1

u=EFj'ujEj'

where

uj = E F''hjl+rEr E U.

(2)

r=o

j>0

Since [jl] = 0 for all j, Lemma 2.15 shows that u is central if and only if all uj are central. This implies the claim. b) This follows from 2.15 since [r] 0 for 0 < r < 1. 2.20. PROPOSITION. Suppose that q is a primitive 1-th root of unity with 1

odd, l > 3. Then the center of U is generated by E', F', K', K-', and C. PROOF. By Lemma 2.19.a it is enough to show that Z fl Uo C k[K', K-', C]. All Kj1Cr with j, r E Z and 0 < r < l are contained in Z n U. We have 1-1

1-1

-Y_1 0 7T(E E kK3'Cr) _ E E kKj'(K + K-I)r C _1 0

7r(Z n Uo).

(1)

jEZ r=°

3EZ r=°

By Lemma 2.19.b it is enough to show that we have equality in (1). An elementary

argument shows that all Kj'(K+ K-I)r (with j E Z and 0 < r < l) together with the Ks (with 0 < s < l) are a basis of U°. If we have a strict inclusion in (1), then y-1 0 -(Z n UU) has to intersect 5_I1 kKs nontrivially. So let us take u E Z n U) with -Y-1 o ir(u) E Es-t kKs. If we can show that necessarily u = 0, then we get equality in (1) and the proposition follows. Well, write 1-1

= E FrhrEr U

where

arsKs

hr = sEZ

r=o

with all ars E k. The assumption ry-1 o i(u) E E_1 kKs says that a0s = 0 for all s < 0 and all s > 1. We want to check first by induction on r that this vanishing holds also for all ars: We have on one side

hr - ry-2(hr) _ E ars(1 - q-2s)Ks

(2)

sEZ

on the other side

(Kq-r -

K-lgr)hr+1

= E(ar+1..s-lq-r - ar+I..s+1gr)KS. sEZ

(3)

APPENDIX

29

a,.+1.,,q-rKn+l If n is the largest (resp. m the smallest) s with ar+l.s 0 0, then is the top term on the right hand side in (3), and ar+1.,,,grK'-1 is the bottom term. By 2.15 the sum in (2) is (for r + I < 1) a nonzero multiple of the sum in (3). So necessarily ar.n+I 0 and ar,,,t_ i 0 0. By induction we get n + I < 1 and

m - 1 > 0, hence also n < 1 and m > 0. (Actually, we get a stronger result that we disregard.) We now want to use induction on r from above to show that hr = 0 for all r. (Then u = 0 and the proposition follows.) For r = 1 - 1 (resp. for r < 1 - 1) we get hr - y_2(hr) = 0 from 2.15(1) since [r+ 1] = [1] = 0 (resp. by induction). Then (2) implies for all s that ars = 0 or I = q-2s. Since q is a primitive l-th root of unity, 1 odd, we have q-2s 0 1 for 0 < s < 1, hence ars = 0 for these s. All other ars were already equal to 0, so we get hr = 0. REMARK. It is clear by 1.5 that E1, F1, and K1 are algebraically independent over k. One can show that C is integral over the subalgebra generated by El, F', K' and K-1. One should compare this theorem as well as the classification of the simple modules in 2.13 with the analogous results for 512 in prime characteristic, cf. [Rudakov & Shafarevich].

Appendix 2.3:

We have to prove the induction step: Let r be an integer with 0 < r < s. If Fs-ihiM = 0 for all i with 0 < i < r, then also Fs-rhrM = 0. Well, set r-1

A=ErFSII[K;r-s+j].

(1)

,7=1

We have by Lemma 1.7 r-1

A = > a,Fs- H[K;i - (r+ s) +j] ) ET'-'(rl [K;r - s+ j]) j=1

7=1

i_O

where

ai = [:]

[:] Iii!Using E[K; a] = [K; a - 2]E, cf. 1.6(1), (2), we rewrite this sum r

r-1

i

A=> i=O r

=> i=0 r

aiFs-i(H[K;i-

(r+s)+j])([J[K;-r-s+j+2i])Er-

j=1

j=1

i+r-1

a,Fs-i ( rl j=1

[K; i - (r + s) + j] )

r-i-1

_ > aiFs-ihi ( 11 [K; -s - j] ) i=O

j=o

Er-i

Er-i

2. REPRESENTATIONS OF Uq(5(2)

30

We have F8M = 0, hence AM = 0. The induction yields Fs-ihiM = 0 for all i < r, so the equation above shows that arFs-rhrM = 0. The scalar ar = [s] [s - 1] . . . [s - r + 1] is nonzero, since q is not a root of unity. So Fs-rhrM = 0

as desired. 2.7:

Let us check that CE - EC = 0. We have

(FE)E - E(FE) = -(EF - FE)E _ (q -

q-))-1(K

-

K-1)E.

On the other hand [using 1.6(2)]

Kq +

K-1q-1

(q-q-1)2

+ K-1q-1 E-E Kq(q-q_1)2 1qE Kq(q 1+K = Kq+K-1q-1E(q - q-1)2 - q-1)2

- (K - K-1)(q - q-1)qE. (q - q-1)2

So the claim follows. 2.15:

We have to show that Fu = uF is equivalent to 2.15(1). Well, we have on one side Fu = Er>() Fr+lhrEr, on the other side [using 1.3(6) and 1.6(1),(2)]

FrhrErF =

uF = r>O

Frh,.FEr + j:[r]Frh,.Er-1 [K; r - 1] r>O

r>()

_ E Fr+1,y_2(hr)E'r + E[r]Frhr[K. r - 1 - 2(r r>O

r>O

_ E Fr+1'y-2(hr)Er + E[r + 1]Fr+1hr+1[K; -r]Er r>O

Now compare both sides.

r>O

1)]Er-1

CHAPTER 3

Tensor Products or: Uq(S C2) as a Hopf Algebra We continue to make the same assumptions and use the same notations as in the last two chapters. In particular, we fix a ground field k and an element q E k with q 0 and q2 1. We set U = Uq(sL2). I am going to write ® instead of ®k. We make an additional assumption preceding 3.11 that will be in force throughout 3.11-19.

If G is a group, then representations of G over k are the same thing as modules over the group algebra kG. If g is a Lie algebra over k, then representations of g over k are the same thing as modules over the enveloping algebra U(p). However, in both cases the representation theory has aspects not present for an arbitrary kalgebra A. We can take tensor products of representations of G resp. g, we have a distinguished one dimensional representation (the "trivial" representation) that acts as identity when taking tensor products, and we have natural representations on dual spaces. We want to have the same features for our "quantum group" U. For arbitrary A the tensor product M ® N of two A -modules M and N is just

an (A ® A)-module. In the case A = kG it is made into a kG-module via the homomorphism kG -+ kG ® kG with g " g ® g for all g E G; for A = U(p) one uses U(p) -> U(p) (9 U(9) with X- X ®1 + 1®X for all X E g. For an arbitrary A the dual space M* of a (left) A-module is just a right A-module. In the case A = kG it is made into a left kG-module via the antiautomorphism kG -> kG with

g ti g-' for all g E G; for A = U(p) one uses U(p) --* U(p) with X H -X for all X E p. In this chapter we want to define a homomorphism U -> U 0 U and an antiautomorphism U -> U that will do for U what the maps above do for kG and U(p). We also want to define a homomorphism U -> k that yields the analogue of the trivial representation. 3.1. Recall the grading on U from 1.9. It induces a grading on U (9 U such

that

(U®U)n= ® U®Uj.

(1)

Z+3=n

We get from 1.9(l)

(K(9 K)u(K-'®K-')=q2"u

for alluE(U®U),,.

(2)

LEMMA. There is a unique homomorphism of k-algebras A : U --> U ® U with

0(E)=E®1+K®E, 0(F)=F®K-'+1®F, 0(K)=K®K. 3]

32

3. TENSOR PRODUCTS OR: Uq(5I2) AS A HOPF ALGEBRA

PROOF. We have to show that (E®K-' +1®F, F®K-1 +1®F, K®K, K-' (g K-') satisfy the relations (Rl)-(R4). This is trivial for (RI); for (R2) and (R3) it follows from (2), since A(E) has degree 1 and A(F) has degree -1. Look at (R4):

A(E)A(F) - A(F)A(E) = EF ® K-' + E ® F + KF 0 EK-' + K ® EF

- FE®K-'- FK®K-'E-E®F-K®FE. The two terms E®F cancel, so do KF®EK-' and FK®K-'E, since FK = g2KF by (R3) and EK-' = q2K-'E by (R2). So we are left with (EF - FE) ®K-' + K ® (EF - FE), which is equal to

(K-K-')®K-'+K®(K-K-') _ K®K-K-1 ®K-' q-q-' q-q-1

So (R4) is preserved. REMARK. It is obvious that A(KTh) = Kn ® K"` for all n E Z. Using induction

one checks for all r E Z, r > 0 that

A(FT) = E q'(r-i) i Irl =o

ET-2K2 ®E2,

(3)

r

A(FT) = Eq''(T-i) [ r ] Ft ®FT-i"K-'".

(4)

i=o

3.2. The map A from 3.1 is called the comultiplication on U. If A is a k-algebra and A : A --p A ® A is a homomorphism of k-algebras, then we say that A is coassociative, if the following diagram commutes:

A®A

1®A

A

(1)

A®A Here can is the canonical isomorphism of vector spaces (a ® b) ® c ' --+ a ® (b ® c) for all a, b, c E A.

LEMMA. The comultiplication on U is coassociative.

PROOF. We simply have to check that all the generators of U are mapped both ways to the same image. That is an easy calculation.

3.3. If M and N are U-modules, then M ® N is a (U ® U)-module such that

(u®u')(m®n)=um ®u'n

for all u,u'E U and m E M, n E N.

(1)

Using A we make now M ON into a U-module: We let u E U act as A(u) E U ®U. More explicitly:

If A(u) _

ui 0 ui, then u(m ® n) _

uim 0 u'in.

(2)

3. TENSOR PRODUCTS OR: Uq(5(2) AS A HOPF ALGEBRA

33

The coassociativity of 0 implies for all U-modules MI, M2i and MM733 that the canonical isomorphism of vector spaces

(MI®M2)©M3=>M1®(11MI2(9 M3)

(3)

with

(mI(9

m3)

is an isomorphism of U-modules. Indeed, any u E U acts on the left hand side as (0®1)op(u) E (U®U)®U and on the right hand side as (1®0)o0(u) E U®(U®U). The map in (3) is clearly compatible with the actions of (U®U)®U and U®(U®U) modulo their canonical isomorphism. Consider a tensor product M = MI 0 M® ® . ® Mr of a finite number of U-modules. We can insert parentheses so that M is an iterated tensor product with only two factors at a time. In this way we make M into a U-module. Now (3) implies that the structure as a U-module is independent (modulo canonical identifications) of the way how we introduce the parentheses. Therefore we usually drop the parentheses. Note that we get Lemma 3.2 as a special case of (3) if we take all Mi equal to U, regarded as a U-module under left multiplication. 3.4. LEMMA. There is a unique homomorphism of k-algebras e : U - k with

e(E) = e(F) = 0

e(K) = 1.

and

The following diagrams are commutative

U -A+ U®U

U -A+ U®U tLgE

id l

U

can +

t-&I

idJ

U®k

U

call,

k®U

where can denotes the isomorphism u '--> u ® 1 resp. u --> 1 ® u.

PROOF. It is trivial to see that (0, 0, 1, 1) satisfy (R1)-(R4). So we get e; the commutativity of the diagrams follows easily. 3.5. The homomorphism e from 3.4 is called the counit (or augmentation) of U. It has occurred before (though hidden): The module L(0, +) from 2.6 (resp. from 2.12) has dimension 1 and any u E U acts as multiplication by e(u) on this module. We call this module the trivial one dimensional module and denote it. (usually) by k. The commutativity of the diagrams in 3.4 means that for each U-module M the canonical isomorphisms of vector spaces

M -- M®k

and

M -- k®M

(1)

with m F-> m ® 1 resp. m F+ 10 m are isomorphisms of U-modules. One sets for an arbitrary U-module M

MU={mEAl Ium=e(u)mfor all uEU}.

(2)

This subspace is the sum of all trivial one dimensional submodules of M; it is called the set of fixed points of U in M.

3. TENSOR PRODUCTS OR: Uq(5(2) AS A HOPF ALGEBRA

34

3.6. LEMMA. There is a unique antiautomorphism S of U with

S(F) = -FK,

S(E) = -K-1E,

S(K) = K-1.

(1)

One has

S2(u) = K-1uK

for all u E U.

(2)

PROOF. We have to show first that (-K-'E, -FK, K-', K) satisfy (Rl)-(R4) in U°rr. Let us denote the multiplication in U°pr by a so to distinguish it from that in U. Now (Rl) is obvious. For (R2) note that K(-K-'E)K-1 = q2(-K-'E)

K-1 (-K-1E) K =

since -K-'E E U1. One checks (R3) similarly. Finally, we get for (R4):

(-K-'E) (-FK) - (-FK) (-K-1E) = FKK-1E - K-1EFK

=FE-EF=

K

q-q-1

So there is indeed a homomorphism S : U --> U°rr (or an "antihomomorphism" S : U --> U) satisfying (1). Now S2 is an (ordinary) homomorphism from U to U. One checks easily that (2) is satisfied. This implies that S2 is bijective, hence that S is bijective. REMARK. One has obviously S(Kr) = K-r for all r E Z. Using induction on r one checks for all r > 0

S(E') =

(-1)''q''(''-1)K-rEr

and

S(Fr) =

(-1)rq-r(r-1)FrKr.

(3)

3.7. The map S from 3.6 is called the antipode of U. It is clear that also the inverse of S is an antiautomorphism. It is given by

S-1(E) = -EK-1,

S-1(F) = -KF,

S-1(K) = K-1

(1)

LEMMA. The following diagrams are commutative

U -A+ U®U t°et

U

11®s

U®U

U -A+ U®U t°et

U

is®1

U®U

where m : U ® U --> U is the multiplication map (m(u (9 u') = uu' for all u, u' E U) and where t : k --> U is the embedding t(a) = al (for all a E k).

3. TENSOR PRODUCTS OR: Uq(5I2) AS A HOPF ALGEBRA

35

PROOF. Let us restrict ourselves to the left diagram. The map f = m o (1 S) o A acts on our generators as follows:

--FK - FK=O,

F

K KK®K K®K-' KK-1 =1, K-' 4K-'®K-'-K-1 ®K,K-1K=1, as predicted by the diagram. To conclude the proof we have to check: If f (u) = t o e(u) and f (v) = t o E(v)

(where u, v E U), then also f (uv) = t o E(uv). (This is not obvious, since S and m are not ring homomorphisms.) Well suppose that A(u) = >i ui ® ui and A(v) = Ej vj ® v.' . Then f (uv) is given by

uivj ®uiv > uivj ®S

uv

ra EuivjS(vj)S(ui) _ Euif(v)S(u) 2.J

We assume that f (v) = t o E(v), so this element is a scalar multiple of I and thus central in U. Therefore f (uv) = E uiS(ui) f (v) = f (u) f (v) = G 0 '- (UV).

3.8. In general, a k-algebra A together with algebra homomorphisms A : A -> A ® A and E : A -> k and a linear map S : A -> A is called a Hopf algebra, if A is coassociative (i.e., 3.2(1) commutes) and if the diagrams in 3.4 and 3.7 (with U replaced by A) commute. One calls then A the comultiplication, E the counit, and S the antipode of the Hopf algebra. So the lemmas 3.2/4/7 say: (U, A, E, S) is a Hopf algebra. Other important examples are group algebras kG and enveloping algebras U(g) with comultiplication and antipode as described in the introductory

remarks to this chapter. The counit on kG takes any g E G to 1, on U(g) any

XEgto0.

In an arbitrary Hopf algebra the antipode S is uniquely determined by the condition that the diagrams in 3.7 commute. It is then automatically an algebra antihomomorphism and satisfies

EoS=E

and

AoS=Po(S®S)oA

(1)

where P is the map with P(a (9 b) = b ® a for all a and b. (One can find this in the books by Sweedler and Abe.) In our situation one can show (1) directly by checking it on the generators. Also, it is easy to see directly that there is only one choice for S(E), S(F), and S(K) that makes the diagram in 3.7 commute. In general, a Hopf algebra is called cocommutative, if P o A = A with P as in (1). Note that U in neither cocommutative nor commutative [= commutative as an algebra]. The group algebras and enveloping algebras, however, are always cocommutative.

36

3. TENSOR PRODUCTS OR: Uq(S12) AS A HOPF ALGEBRA

There can be more than one way to make an algebra into a Hopf algebra. For example, we could replace (A, E, s) on A by (PoA, E, S-'). If cp is an automorphism or an antiautomorphism of the algebra A, then we can define a new Hopf algebra structure ('PA, 5E, YS) where

y'A= Op®

)oAocp-1,

"'E=E000-1,

(2)

and

Ps =

if cp is an automorphism, { cp o s-1 o cp-1, if cp is an antiautomorphism. cp o s o cp-1,

(3)

(Note that m o (cp ®cp) = cp o m if co is an automorphism, and m o (cp ®cp) = cp o m o P

if cp is an antiautomorphism.) Take in particular for cp the antiautomorphism T from 1.2.b. One gets TE = E and

'A (E) = E ®1 + K-1 ®E, TA(F) = F ®K + 1®F, TO(K) = K ®K,

TS(E) = -KE, TS(F) = -FK-1, TS(K) = K-'.

(4)

3.9. If M is a U-module, then the dual space M* = Homk(M, k) can be made into a U-module such that

(uf)(m) = f (S(u)m)

for all f E M*, m E M, and u E U.

(1)

There is for each vector space M a natural homomorphism cp : M -> (M*)* that takes any m E M to the linear form f F-+ f (m) on M. (This is an isomorphism, if M is finite dimensional.) If M is a U-module, then this map is (in general) not a homomorphism of U-modules, since ucp(m) = cp(S2(u)m) for all u and m. (Observe that 3.6(2) shows that S2(u) # u in general.) However, 3.6(2) implies easily that cp'

:M

> (M*)*

with cp'(m)(f) = f(K-1m)

(2)

is a homomorphism of U-modules. The commutativity of the second diagram in Lemma 3.7 implies for each U-

module M that the map

M*®M--->k,

form f(m),

(3)

is a homomorphism of U-modules (where we regard k as the one dimensional triv-

ial module). The analogous map from M 0 M* to k (m ® f r--, f (m)) is not a homomorphism (in general). However, we can take (3) for M* and compose with the map from (2): we see thus that

Al ® Al -> k, is a homomorphism of U-modules.

m ®f f--1 f (K-1m),

(4)

3. TENSOR PRODUCTS OR: Uq(SI2) AS A HOPF ALGEBRA

37

3.10. If M and N are two U-modules, then there are a priori two reasonable ways of making Homk (M, N) into a (U ® U)-module: We can set for all u, u' E U, cp E Homk (M, N), and m E M ((u (9 u')cp)(m) = wp(S(u')m),

(1)

but we can also set

((u 0 u') cp)(m) =

(2)

For any f E M* and n E N let cp f.,, : M - N be the linear map with cp f,Z (m) _ f (m)n for all m E M. We have natural homomorphisms of vector spaces

No M* -> Homk(M, N),

(3)

and

M* ® N -> Homk(M,N),

f ®n'--' cpf,t.

(4)

(These maps are bijective, if M or N is finite dimensional.) Then (3) is a homomorphism of (U ® U)-modules, if we take the action from (1), and (4) is a homomorphism of (U ® U)-modules, if we take the action from (2).

We get now via 0 two possible structures as a U-module on Homk(M,N). They will differ (in general), since 0 is not cocommutative. We shall always take the structure arising from (1). So, if u e U and 0(u) = Ei ui ® ui, then (ucp)(m) _ >uzCp(S(uZ)m).

(5)

The natural map N 0 M* - Homk (M, N) from (3) is a homomorphism of Umodules, the map from (4) is not, in general.

Note: If we take for N the trivial module from 3.5, then the action of U on Homk(M, k) coincides with that on M* constructed in 3.9. (Combine (3) and 3.5(1).) For arbitrary N we get by (5) for any cp E Homy, (M, N) C Homk (M, N)

(w)(m) _ > uiS(ui)co(m) = (m o (1 ® S) o o)(u)vO(m) = i

by the first diagram in Lemma 3.7, i.e., satisfies w = then

p 0= E o

e(u)cp. Vice versa, if cp E Homk (117, N)

oK-', oK-'E=Eopp - po E, oF)K,

O=Fop= so p is U-linear. This says in the notation from 3.5(2) Homy; (M, N) = Homk(117, N) u

(6)

Note that this result depends on our choice of the U-action on the Hom space: It would not hold, if we had constructed it via (2).

38

3. TENSOR PRODUCTS OR: Uq(512) AS A HOPF ALGEBRA

For all vector spaces M, N, V over k there is an isomorphism of vector spaces Homk(M, Homk(N, V))

Homk(M ® N, V)

(7)

that take a linear map b : M Homk(N, V) to the linear map M®N V with z)(m ® n) = 0(m) (n) for all m E M and n E N. One can now check that (7) is an isomorphism of U-modules. It induces therefore an isomorphism of the spaces of U-invariants, i.e., by (6)

Homu(M, Homk(N, V)) => Homu(M ® N, V).

(8)

Let M be a finite dimensional U-module. The trace (op --> tr(op)) is in general not a homomorphism of U-modules from Endk(M) to k. If we identify Endk(M) with M ® M* via (3), then the trace corresponds to the map m ® f '--> f (m). Now 3.9(4) implies that

trq : Endk(M)

k,

op - tr(op o K-t),

(9)

is a homomorphism of U-modules, called the quantum trace.

We have now succeeded in carrying out the goals mentioned at the beginning of this chapter. But we have done it in a way that might appear to be defective. If M

and M' are U-modules, then the map P : M' o M M ® M' with P(m' (9 m) _ mom' is not an isomorphism of U-modules (in general). This is in contrast to the situation for groups and Lie algebras, and we have seen already in the last subsection

that it leads to unpleasant complications. Several natural constructions for vector spaces are no longer natural for U-modules. The truth, however, is that quantum groups were invented so as to have this defect. There is a way of correcting the non-cocommutativity of U and the correction terms have properties that people were looking for. The rest of this chapter is devoted to constructing these corrections and proving some of their properties. Suppose that q is not a root of unity and that the characteristic of k is not 2. If M and M' are finite dimensional U-modules, then they and their tensor products

are direct sums of their weight spaces. Since 0(K) = K 0 K is symmetric, one sees that the weight spaces (M 0 M') ,, and (M' 0 M)N have the same dimension for all p. However, one can easily deduce from 2.6 and 2.9 that these weight space dimensions determine the decomposition of a module into a direct sum of simple modules. So M ®M' and MO M are isomorphic as U-modules, even though the obvious map P does not work. But the existence of some isomorphism in itself is not satisfactory. One would like to have a functorial construction, i.e., isomorphisms RM,,M : M' ® M

M ® M' that are functorial in M and M'. Suppose for the moment that such a natural construction exists for all U-modules. Apply it in the case M = M' = U and set R = Ru,u(1 0 1) E U ® U. If we now consider arbitrary M and M', take M, a --> am arbitrary m E M and m' E M', and apply the functoriality to U and U M', b --> bm', then we see that necessarily Rmt,M (m' ®m) = R(m ®m'). We get in particular Ru,u(a ® b) = R R. (b ® a) for all a, b E U. The fact that Ru,u

39

3. TENSOR PRODUCTS OR: Uq($12) AS A HOPE ALGEBRA

is bijective implies easily that R is invertible in U ® U; the fact that Ru,u is a homomorphism of U-modules implies that

for all uE U.

R-1

(*)

On the other hand, if we have an invertible R E U ® U satisfying (*), then we get for all M and M' via RM',M(m' ® m) = R(m ® m') an isomorphism RM'.M as desired.

Drinfel'd has discovered/constructed an element R as above in the first version of a quantized enveloping algebra of .512. That first version is a completion of the

algebra we work with (in case k = C). He gets l,

R

-

(1 fn12)n q-n(n 1)/2Fn ®En) exp (4 H OH)

n=0

where h and H are related to our objects by

q = exp(-h/2)

K = exp(-hH/2).

and

Our algebra U does not contain his element. So we construct the RM',ni (for finite dimensional M and M') in a conceptually more complicated way. *

Assume for the rest of Chapter 3 that q is not a root of unity and that char(k)

2.

3.11. Set for all integers n > 0 where

On = an F® ® En E U ® U

(-1)nq-n(n-l)/2 (q

an =

[

j. 1)n

(1)

In particular, 60 = 1®1, and q 1 = - (q - q F ®E; set 6 _ 1 = 0. The coefficients an satisfy a n = q -(n-1) q - -1

-

[n]

an-1

(2)

LEMMA. We have for all integers n > 0

(E® 1)On + (K (&E)On-1 = On(E (91) + On_1(K-1 (&E),

(10F)On+(F(&K-1)6n_1 =On(1®F)+On-1(F®K), (K ®K)On = On(K ® K).

(3) (4) (5)

PROOF. The last formula is an immediate consequence of 3.1(2), since On is homogeneous of degree 0. The other two claims follow from elementary calculations using 1.3(5), (6).

40

3. TENSOR PRODUCTS OR: Uq(512) AS A HOPF ALGEBRA

3.12. Let M and M' be finite dimensional U-modules. By Proposition 2.1 both E and F act nilpotently on M and M'. We can therefore define a linear transformation

O=OM.M,:mom, -4Mom,

by

(1) n>0

The formulas from Lemma 3.11 imply (cf. 3.8(4))

0(u) o 0= O o T0(u)

for all u E U.

(2)

Since F ® E acts nilpotently on M ®117', we can find a basis of M ® M' such that the matrix of F®E with respect to this basis is strictly lower triangular. Each On is up to a scalar factor equal to (F ® E) 1L, so for n > 0 also its matrix is strictly lower triangular. Since Oo = 1 we see that Ohq,tit, is bijective.

(3)

More precisely, it is unipotent. Take for example M = M' equal to the module L(1, +) from 2.6 and take the basis mo, m1 from that theorem. Each O,, with n > 0 annihilates M ® M, and 01

annihilates all mi ® mj, except for e1(mo ® m1) = -(q - q-1)(ml 0 mo). So the matrix of Oht,ti1 with respect to the basis (mo ® mo, mo ® m1i m1 ® mo, m1 0 ml) is equal to 1

0

0

1

0 0

0

0

q-1-q

1

0 0

0

0

0

1

3.13. Recall from 2.3 that each finite dimensional U-module is the direct sum of its weight spaces (under our present assumptions on q) and that the weights are contained in

A=I±q'IaEZ}.

(1)

Suppose that we have a map f : A x A - k" such that f (A, l-t) = Af (A, I-tg2) = pf (\q2, IZ)

for all A, p E A.

(2)

(We shall discuss the existence of such f in 3.15.) Define for all finite dimensional

U-modules M and M' a bijective linear transformation f : M 0 M' -* M 0 M' by

f(m®m')= f(\, µ)m®m'

for all m e MA and m' E

(3)

(and all A, µ). Set

Of=Oof.

(4)

LEMMA. We have for all u E U

0(u) o Of = Of o (P o 0)(u)

(5)

where P : U ® U -* U ® U denotes the homomorphism with P(u ® u') = u' ® u.

3. TENSOR PRODUCTS OR: Uq(512) AS A HOPF ALGEBRA

41

PROOF. By 3.12(2) we have to show that TA (u) o f = f o (P o A) (u) for all u E U. It is enough to take for u the generators. So we have to show that

(E®1+K-1 ®E)o f = f o(E®K+1®E), (1(9 F+F® K) o f = f o(K-1 ®F+F®1), (K®K)o f = f o(K(9 K).

(5) (6) (7)

The last formula is obvious, since f stabilizes the weight spaces. The first two formulas require small calculations involving (2). We have for all m E Nf,\ and

m'EM. (and allA,y)

(E®1+K-1®E)o f(m®m')= f(A,u)Em®m'+A-1f(A,u)m®Em', fo(E®K+1®E)(m®m')= f(Ag2,µ)iEm®m'+f(A,,ug2)m®Em', and

(1®F+F®K)o f(m®m')= f(A,,u)m®Fm'+f(A,a)jFm®m', f o(K-1 ®F+F®1)(m®m') = f(A,µ4-2)A-1 m®Fm' +f(Aq-2, u)Fm®m'. 3.14. We extend the use of the notation P to all vector spaces M and M' over

kand define P:M'®M-+M®M' such that P(m'®m)=m®m' for all mE M

andm'EM'.

THEOREM. Let M and M' be finite dimensional U-modules. The map

of oP:M'®M-M®M'

(1)

is an isomorphism of U-modules.

PROOF. The map is clearly linear and bijective since Of and P are so. We

have for all uEUand vEM'®M P(uv) = P(0(u)v) = (P o L)(u)P(v). If we combine this with Lemma 3.13, we see that Of o P commutes with the action of U.

REMARK. Note that our construction above is functorial: If g : M - N and

-

g' : M' -+ N' are homomorphisms of U-modules, then we have a commutative diagram f

M'®M

tgog,

g'og t

N'®N

M®M'

HfoP

N®N'.

This is more or less clear, since g ® g' is a homomorphism of (U (& U)-modules.

3. TENSOR PRODUCTS OR: Uq(S(2) AS A HOPF ALGEBRA

42

3.15. If a map f

x A -> k" satisfies 3.13(2), then we have for all m, n E Z

and all El, E2 E {±1}

f

q-2mnf(el,

n

(elg2m, e2g2n)

f

= e1 e2

(elg2m+1 e2g2n)

n

e2),

q-2-+1)n f(e1q, e2),

= e1 e2

f(elg2m e2g2n+1) =le2 In q- (2n+1)mf(61' f(elg2m+1 e2g2n+1)

62q), q-(2nm+m+n) f(elq,e2q).

=e1e2

On the other hand, if we choose the sixteen possible f terms on the right hand sides of these formulas arbitrarily, then we can take the formulas as the definition of f and get a function that satisfies 3.13(2).

For example, if we choose f (q, q) = q-, then the formulas imply that f(q-, q-1) = q-' and f (q-1, q) = 1 = f (q, q-1). Then the matrix of Of (and of its inverse) on L(1, +) ® L(1, +) for the same basis as in 3.12 is equal to:

(q-' Of :

0

0 0

q-1-q

0

0

1

0 0 1

0 0 0

0

q-1

(Of)_1 '

q

0

0

0

0

1

0

0

1

0

0

q

0 q-q-1 0

0

3.16. The Of have other important properties. Before we prove them, we need a few more formulas. One checks easily that the coefficients an from 3.11(1) satisfy

anam = qnm r n + m 1 an+m n

for all n, m > 0.

-

(1)

Consider in U 0 U ® U the elements and

We claim that

O;=anFn®K-"®En.

(2)

n

(A ® 00n = E(1® On-i)O'i'.

(3)

i=0

Well, the left hand side is by 3.2(4) equal to n x

angi(n-i)

[n] F' ®F"-''K-i ®E",

=0

the right hand side is equal to n

0

Fn-i ® En-:)(F' ® K-' ® E').

i=0

So the equality follows from (1). One shows similarly (working with 3.2(3)) that n

(10 A)en =

1)e . i=0

(4)

3. TENSOR PRODUCTS OR: Uq(512) AS A HOPF ALGEBRA

43

The antiautomorphism T from 1.2.b satisfies clearly and

(T ®T)On = On

for all n > 0.

(TO T 0 T)O'n = On

Using this and the definition 3.8(2) one deduces from (3) and (4) that n

(TA

® 1) On =

Oi' (10 On-i),

(5)

i=0

n

(1 ®TA)On = E O.'y'(On-i ® 1).

(6)

i=0

3.17. Suppose that we have chosen an f as in 3.13(2). If M, M', and M" are finite dimensional U-modules, then we can use Of to construct three automorphisms of M ®M' ®M" (considered as a vector space) : We set O12 = Of ®1, 023 = 10 Of, and 013 equal to the composition of first 1 O P from M O MO M" to MOM" ®M',

then Of 0 1, and finally with 10 P back to M ® M' O M". THEOREM. One has for all finite dimensional U-modules M, M', and M"

012oe13oe23 =023 0 013 0 012.

(1)

PROOF. We can rewrite the two sides in the claim as

LHS=(001)0f120013of130(100)of23, RHS=(10 0)0 f2806130f130(001)of12-

(2)

Here f23 maps x = m®m'®m" with m E Ma, m' E Mµ, and m" E M,' to f (µ, v)x; similarly for the other fib. Furthermore, 013 is the analogue of Of3 just without f . We define endomorphisms of M ® M' ® M"

0'

On

0"

and

n>O

0n"-

(3)

n>O

This makes sense (as in 3.12(1)), since E and F act nilpotently on these modules. We claim first that

f12oe13=0'0f12

f23oe13=0"0f23

and

Well, we have for

anf

f12 o 013(x) _

(Aq-2n

iz) Fnm ® m' 0 Enmi"

n>0

_ E anUnf (A,,u) Fnm ® m' ® Enm"" n>O

_

anf (A, ,u) Fnm ® Knm' 0 n>O i

= 0 0 f12(x);

Enm"

(4)

44

3. TENSOR PRODUCTS OR: Uq(S[2) AS A 11OPF ALGEBRA

similarly for the second equality. We claim/ next that

f12of13o(1®O)=(100)of12of13

(5)

f23of13o(O®1)=(0®1)0f23of13

(6)

and

Indeed, we have for x as above f12 O f13 o (1 (90)(x) _

anf (A,,aq

2n)f (A, vg2n) m ® Fnm' ® Enm'

n>O

_ E anAnf (A, µ).-nf (A, v) m ® F"m ® Enm" n>O

=(1 (9 0)0f120f13i similarly for the second equality. We get now from (2)

LHS= (0®1)o0'o(1 ®0)ofl 3ofl 2of23, RHS=(1®0)o0"o(0(9 1)o f23of13of12.

(7) 1

Using 3.16(3), 3.12(2), 3.16(5) we get now for the theta terms from the last equation n

(1®19)00"o(19 0 1)E(1®On_i)0020(0(91) n>0 i=0

_ E(z ® 1)(On) 0 (0 (9 1) n>0

= E(0 0 1) 0 (TA (9 1)(On) n>O n E(O01)o02o(1®O.n-i)

n>O i=0

=(19 ®1)o0'o(1®19). Since the fig terms commute obviously, this shows that LHS = RHS as claimed. REMARK. In the special case where M = M' = M" one can express (1) as: Of is a solution to the quantum Yang Baxter equation. (Let V be a finite dimensional vector space. If R is an endomorphism of V ®V, then we can define endomorphisms R12, R23, and R13 of V ®V ®V by the same method that we used for the O. Then the equation R12R13R23 = R23R13R12 in Endk(V ® V ® V) is called the quantum Yang Baxter equation.)

3.18. In the diagrams of the next theorem we write R for several maps constructed using a suitable Of o P. More precisely, the R in the upper halves denote

maps of the form (Of o P) ® 1 and 10 (Of o P). The R in the lower halves are equal to a Of o P, where one of the two modules is a tensor product itself.

3. TENSOR PRODUCTS OR: Uq(512) AS A HOPF ALGEBRA

45

THEOREM. Let M, M', and M" be finite dimensional U-modules. Suppose that f(A,µv) = f(A,µ)f(A,v)

and

f(Ai,v) = f(A,v)f(µ,v)

(1)

for all weights A, it, and v of these modules. Then the following diagrams commute:

RI

M®(M"®M')

can

(M®M")(D M' \,R

(TM"®M)®M'

M®(M'(0 M") can

/can

(M(0 M')(0 M"

R

M"(0 (M®M')

(M' ® )Lf) (0 M"

can

M' ®(M (D M")

and

RI

\,R

(MOM')®M""

M'®(M"(D M).

cant'

M®(M'®M")

R

(M'®M")®M

/can

PROOF. The arguments for both diagrams are similar. Let us restrict to the first one. We are going to use freely notations from the proof of Theorem 3.17, e.g., the f,3. The two maps in the upper half of the diagram are (10 (9) 0 f23 0 P23 and 090 1) o f12 0 P12, where we set P12 = (P®1) and P23 = (10 P). One checks easily

that PL2o(1019)=013oP12 andP12of23=f130P12iwehave f12o013=O'of12 by 3.17(4). So the composition of the maps in the upper half is (0®1)o f120P120 (10 (9)0 f230P23=(0 0 1)0 01 0 f120 f130P120P23 The map in the lower half of the diagram is the composition of a permutation of the factors (equal to P12 o P23), a map f' that takes x E 117 O 1b1µ 0 M; to f (A, pv)x,

and finally (1 0 0)19. Since (1 0 0)O = (0 O 1) 0 0' by 3.16(4), we see that the maps are equal if and only if f' = f12 o f13. Here the right hand side takes an x as above to f (A, p) f (A, v)x. So the equality follows from the first assumption in (1). (The second one is used when dealing with the second diagram.)

3.19. If we have a function f satisfying 3.18(1) for all weights of the form q° with a E Z, then necessarily f (qa, qb) = f (q, q)ab for all a, b E Z. Furthermore, we get from 3.13(2) that f (q, q) f (q, q) = f (q, q2) = q-1 f (q, l) = q-', hence that f (q, q) is a square root of q- L. On the other hand: Suppose that k contains a square root of q; denote it by q1/2. Then we can define f on the q° by

f(q° qb) = (q1/2)-ab

for all a, b E Z,

and see that 3.13(2) and 3.18(1) are satisfied for all weights of this form.

(1)

3. TENSOR PRODUCTS OR: Uq(5(2) AS A HOPF ALGEBRA

46

However, we cannot extend f to all of A and still satisfy 3.18(1). We would

get from f (-1, 1) f (-1, 1) = f (-1, 1) that f (-1, 1) = 1. We can then compute f (-1, q) f (-1, q) in two ways. On the one hand it is equal to f (-1, q2) _ (-1) f (-1, 1) = -1, on the other hand it is equal to f (1, q) = 1, a contradiction. We say that a U-module M is of type 1, if all weights of M have the form qa with a E Z. (So we exclude weights of the form -qa.) The discussion above shows: If k contains a square root of q, then we can choose f such that the diagrams in Theorem 3.18 are commutative for all finite dimensional U-modules of type 1. The identities expressing the commutativity of these diagrams are usually called the hexagon identities.

There is a general notion of a tensor category, cf. [Saavedra] or [Deligne & Milne]. This is a category C with a functor C x C -+ C denoted by (A, B) ti A®B. One then would like to have an associativity constraint (= functorial isomorphisms a : (A ® B) ® C -> A ® (B ® C)) and a commutativity constraint (= functorial

isomorphisms c : A ® B -> B (9 A). To get nice properties one has additional axioms, e.g., the hexagon identities. On the category of finite dimensional U-modules we have an associativity constraint given by the canonical isomorphism, cf. 3.3(3), and a commutativity constraint given by Theorem 3.14. In contrast to the classical theory, our commutativity constraint does not satisfy c2 = 1 (in general), as we can read off the Of in 3.15.

Appendix 3.1(3),(4):

One checks first that the formulas say z(1) = 10 1 for r = 0, and that one gets for r = 1 the definitions from the lemma. Here are then the inductive steps: Er-iKi ®E2

A(Er+1) = (E ® 1 + K ®E) > qi(r-i) r

=E

L

i=O

i

J

q i(r-i) r l' 1 Er+1-zKi ®E'i +

_

r

i

qi(r-i) [ r ] Er+1-iKi

qi(r-

KEr-iK® ®Ei+1

rl

(9 E2

i=o

+ E qi(r- [r] q2(r-i)Er-iKi+1 ® Ei+1 r

i=o

_

r

qi(r-i) [ r ] Er+1- Kz®Ez

i=o r+1

+ E q(i+1)(r+1-0

j

r

i=1

_ Egi(r+l-i)(q-i [r] +qr-i+l T.- ti

i=0 r+1

_ E qi(r+1-i)

Er+1-iKi ® Ei

i-1 i

[r+1]

r

r IL

z

E liKz ®Ei

Er+1-iKi ®Ez

APPENDIX

KEr-i = q2(r-i)Er-iK by (R2), use 0.1(2) for the

(Use for the third equality that last step.) Similarly:

0(Fr+l) = (F

Fi ®Fr-iK Ii J Fi+l ®K-lFr-iK-i

+ 1®F) E

®K-1

47

qi(r-i)

i=o

r

L= qi(r-i) LJi L r' J

o

i

r

+ E q2(r-2) r I F2 ® L

i=O r

Fi+1 ®q2(r-i)Fr-iK-(i+1)

qi(r-i) i=o

Fr+1-iK-i

r

+ E qi(r-i rr l Fi ®Fr+l-iK-i Z

i=o

r+1

L q(i+1)(r-i+1) fL2-r I] F®®Fr-i+1K-i i=1

+E qi(r-2) Lrl

F2 ®

Fr+1-iK-i

i0 r+1

L= qi(r+1-i) (qr-i+1 Lz-1J + q-i [])Fi®Fr+1_iK_i i T

i

_

L

o

r+1 i=o

qi(r+I-i) Lr+ 11 Fi 0 Fr+l-iK-i L

2

J

3.2:

We have

(0®1)o0(E)=0(E)®1+0(K)®E =(E®1)®1+(K®E)®1+(K®K)®E, (1®0)o0(E)=E®0(1)+K®0(E) =E®(1®1)+K®(E®1)+K®(K®E), (Ao1)0 (F)=0(F)®K-1+0(1)®F =(F®K-1)®K-1+(1®F)®K-1+(1®1)®F, (10A)o0(F) =F®0(K-1)+1®0(F) =F®(K-1®K-1)+1®(F®K-1)+1®(1®F), (A 0 1) o 0(K) = 0(K) ®K = (K ®K) ®K,

(1®0)o0(K)=K®0(K)=K®(K®K).

3. TENSOR. PRODUCTS OR: Uq(5(2) AS A HOPF ALGEBRA

48

3.6(2): We have

S2(E) = S(-K-'E) _ -S(E)S(K-') = -(-K-'E)K = K-1EK, S2(F) = S(-FK) = -S(K)S(F) = -K-1(-FK) = K-1FK, S2(K) = S(K-1) = K = K-'KK. 3.6(3):

Since S is a antihomomorphism it suffices to show by induction on r that

(K-1E)r = qr(r-1)K-rEr

q-r(r-1)FrKr.

(FK)r =

and

This is obvious for r = 0, 1; induction yields

(K-lE)r+l = K-1Egr(r-1)K-rEr =

qr(r-1)K-(r+1)(KrEK-r)Er

= qr(r-1)K-(r+1)g2rEEr = qr(r+l)K-(r+i)Er+l and

(FK)r+l = FKq-r(r-1)FrKr _ qr(r-l)F(KFrK-1)Kr+l = q-r(r-1)Fq-2r Fr Kr+1 = q-r(r+1)Fr+lKr+1 3.10(7):

Let u E U; set 0(u) _ >i ui ® ui and 0(ui) _ >® xij ®3=ij and t,(ui) = Eh Yih ® yih We have for any linear map z4' : M --- Homk (N, V)

((u )(m))(n) _

(ui b(S(ua)m))(n) = Exi,( (S(ui)m')(S(x j)n))

for all m E M and n E N. We have

xij ®xij ®ui = (0 ®1)0(u) _ (10 A)A(U)

Ui ® yih ®yh i.h

i, j

If we apply 1 ® S ® S to this equation, we get

xij ®S(xij) ®S(ui) = E uh ® S(Yih) (9 S(yih)' i,h

i.,7

Therefore we can rewrite the formula above as uV)(m®n) = ((uV)) (m)) (n) _

ui(V)(S(yih)m)(S(yih)n)) i,h

On the other hand, we have for b : M 0 N -* V ui

(u,O) (m (9 n) _ i

®n)) _

Ui (;(S(yih)m (9 S(yih)n)) i.h

_ 1: Ui V)(S(yih)m)(S(yih)n)) = u7(m ® n) e.h

APPENDIX

49

where we use 3.8(1) for the second step. So we get u1/i = uw as desired. 3.11(3), (4): We may assume that n > 0. We have (E®1)®n-On(E®1)=an(EF'°-FEE)®E'°=anFn-1[n][K;1-n]®En

_

®n-1(-q-(n-1)(q - q-1)[K; 1 - n] 0 E)

®n-1(-q1-n(Kg1-n - K-1qn-1) ® E) _ Kg2(1-n)) ® E) = ®n-1(K-1 and

(K ® E)en-1 = =

an-1KFn-1

an-1q-2(n-1)F+n-1K

® En =

®n-1(82(1-n)K

® En

®E).

Now add and get 3.11(3). Similarly,

(1®F)®n,-on(1®F)=anFn0

(FEn-EnF)=an,Fn®(-[n]En-1[K;n-1])

= an-1q-(n-1) (q - q-1)Fn ® En'-1 [K; n - 1] = ®n-1(F (9 q1-n(Kgn-1 - K-1q1-n)) = ®n-1(F ® (K - K-182(1-n))) and

(F ® K-1)en-1 = an_1Fn (9 K-1En-1 = an_1Fn 0 En-1K-1q-2(n-1) = E)n-1 (F 0

K-182(1-n)).

CHAPTER 4

The Quantized Enveloping Algebra Uq(g) The definition of the quantized enveloping algebra for arbitrary semisimple Lie algebras is modeled on the s12 case (described in 1.1) and on Serre's presentation of the semisimple Lie algebras. Let me review the latter that you can find in Section 18 of Humphreys' book. Suppose that II is a basis of the root system (of the Lie algebra

with respect to a Cartan subalgebra); denote the entries of the Cartan matrix by aaQ = 2(a, 3)/(a, a). Then the Lie algebra has a presentation with 31111 generators xa, ya, ha where a E II and the following relations (for all a, i3) [ha, h,3] = 0,

[xa, Y,3] = ba3ha,

[ha, xa] = aaj xQ,

[ha, yo] _ -aa,Qy,Ci,

and (for all a03) (ad ya)1-a..sy,3 = 0.

(ad xa)1-a-13x/3 = 0,

The enveloping algebra of our Lie algebra is then the associative algebra with gen-

erators xa, ya, and ha and the same relations. However, one may then want to replace any [u, v] in the relations by uv - vu and express the last two relations as 1

na3

(-1)i i=o

(1 - aadl xlx;3xa =0 Ja

and

1-a,3

(-1)i

- a,,3 1 y11-a.3-i i i J yDya = 0.

i=0

(Note: If x and y are two elements in a Lie algebra, then one has in the enveloping algebra (ad x)r(y) = We give the definition of the quantized enveloping algebra Uq(g) in 4.3; given the s[2 formulas and Serre's presentation it should not look too unfamiliar. The main goals in the remainder of this chapter are the construction of a Hopf algebra structure on Uq(g) and the proof that Uq(g) has a triangular decomposition. E2-o(-1)2(2)x'-jyx,.)

4.1. Let g be a complex semisimple algebra. Let 1 be the root system of g with respect to a fixed Cartan subalgebra. Let W be the Weyl group of c. If c is irreducible, then there is a unique W-invariant scalar product ( , ) on the vector space generated by 4' over the reals (or over the rationals) such that (a, a) = 2 for 51

52

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

all short roots a in (F. So, if there is only one root length in (F, then (a, a) = 2 for all a E 4i. If D is of type B,,, C,,, or F4, then (a, a) = 2 for the short roots and (a, a) = 4 for the long ones. If (F is of type G2, then (a, a) = 2 for the short roots and (a, a) = 6 for the long ones. In general, 4i has a unique decomposition U 4 r into irreducible components. We then take on the span of 4i = (FI U 4 2 U each 4ii a scalar product as above (invariant under the Weyl group of (Fi), and we take then the orthogonal direct sum of these products. We get thus a W -invariant scalar product on the space spanned by all of (F such that the short roots a in each irreducible component satisfy (a, a) = 2. Set for each a E d« = (a, a) E {1,2,3}.

(1)

Fix a basis H of (F. Denote the fundamental weights by za,3, ,Q E H and the weight lattice of 4i by A = K3En Zza3. We have by definition for all a,.3 E (F:

a) =

(a, a)

2 (wo, )

= b«,3 d« E Z .

(2)

a)

This implies that (A, a) E Z for all A E A and all a E 4i. We shall use the notation (A

a v ) = 2(A' a) (a, a)

(3)

for all A E A and a E 4i, and we shall denote the reflection with respect to a by s«: s« (A) = A - (A, av)a.

(4)

4.2. Let k be a field with char(k) 0 2; assume in addition that char(k) 0 3 if 4i has an irreducible component of type G2. (We could admit fields of characteristic

2 as long as all roots have the same length, but I do not want to deal with the necessary modifications in that case.) Fix an element q E k, q 0 0 with g2da 0 1 for all a E 4i. So we always require q2 0 1: in some cases we additionally require q4 0 l and/or q6 0 1. Set now for all a E 11 q« =

qda = q(«,«)/2

(1)

and (for allaEZ) [a]« _ [Q]v=9a =

qa

- q-a «

gada

_1 = qd«

9« - q«

- q-ada

- q_da

Define similarly [n]a and [ a

(2)

n]« 4.3. The quantized enveloping algebra Uq(g) is defined as the k-algebra with generators E«, F«, K«, and K. I (for all a E II) and relations (for all a, E II) (R1)

K«K 1 = l = K- 1K«,

(R2)

K«E3Ka 1 = q(«,3)E3,

(R3)

K«F,3Ka I = q-(«,3)F3,

(R4)

E«FF - Fi3E0, = 6«3

K«K0 3 = K,3K«,

K«-Kg -1

q« - q«

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

where 6a,j is the Kronecker delta, and (for a 1-a ' s=U

L

1 -nn.i L

s

= 0,

Ck

a«;31

(R6) s=U

J

s

0)

Ea-nv3-sE3E'

aa3 ]

(R5)

53

Fa-no

F3 F(,, = 0,

Ja

where a,O = 2(a, 0) /(a, a) for all a, 3. Initially it will be convenient to work with the algebra Uq (g) defined by the same generators (Ea, Fa, K, Ka 1 with a E 11), but where we impose only the relations (Rl)-(R4). We have a canonical surjection Uq(g) -* Uq(g). We use the notations like Ea, Fn, etc., both for the corresponding elements in Uq(g) and in Uq(g); it will be clear from the context what is meant. 4.4. Denote by UU (g) (resp. Uy (g)) the subalgebra of Uq(g) generated by all Ea (resp. Fa) with a E H. Set Uq (g) equal to the subalgebra of Uq(g) generated by all Ka and K(1 with a E H. Define similarly Uq (g), Uq (g), and Uq (g). It is clear by (Rl) that U°(g) and U°(g) are commutative algebras. We can define for each A in the root lattice Z-D = EaE41 Za an element KA in U°(g) and U° (g) by

KA = f K3 `3

if A -

rn3a.

(1)

for all A, It E Z-D.

(2)

3E1I

/3E11

We have then obviously

KAKµ = KA+,

It is clear that the KA span U°(g) and U°(g) as a vector space. (In fact, they will turn out to be bases, cf. 4.17 and 4.21.) The relations (R2) and (R3) imply inductively

KAE3K, I = q(a.3)Eo

and

KAF;$KX 1 = q (a.3) F13

(3)

for allAEZ-D and/3E11. Note that (R2) and (R3) say for a = l3

KaEaK« 1 = q2E,

and

KaFaK« 1 = qa 2Fa.

(4)

This together with (R1) and (R4) shows that (Ea, Fa, Ka, Ka 1) satisfies the relations (RI)-(R4) from 1.1 with q replaced by qa. We have therefore for each a E II homomorphisms

Uq (s[2) _4 Uq(g)

and

Uqn (512) --> Uq(9)

(5)

that take E to Ea, F to Fa, and K to K. These homomorphisms will turn out to be isomorphisms onto their images, cf. 4.17 and 4.22. We can use these maps to

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

54

carry over commutator formulas from Uq, (5(2) to Uq(9). For example, we get from 1.3(5),(6) for all integers r > 0

EQFa = FLEa + [r]aFF-' [KQ; 1 - r],

(6)

FaEQ- aFa-[r]aEr-1[Ka;r-1],

(7)

and

where now for all a E Z [

K a;a] = Kaga qa - qa

K.-Iq-a

(8 ) 1

4.5. There are several ways to generalize the construction from 4.3. For example, consider a subgroup r of (m) ®Z Q containing the root lattice M such that E Z for all A E r and 0 E 4). Then define Uq(g, r) as the algebra with generators Ea, Fa (a E II) and KA (A E I') and relations (for all A, p E r, 3 E II)

Ko=1, KAK, = Ka+N., KAE,3Ka I = q(a.a)Ea,

KAFOKa l = q-(")F0,

(0)

(1) (2)

(3)

and the relations (R4)-(R6) from 4.3. Note that (2) and (3) make sense by our assumption on r. There is a natural homomorphism Uq(g) _* Uq(g, r) that takes any Ea E Uq(g)

to the Ea E Uq(g,F), similarly for the Fa and Ka. (Note that II C M c F.) In case r = M one gets an inverse map using the KA from 4.4 and the formulas 4.4(2),(3). So there is an isomorphism Uq(g) Uq(g, ZI). Another natural choice is to take r = A (the weight lattice); that is possible by the final remark in 4.1.

Some authors - for example Joseph and De Concini, Kac,

Procesi - call

Uq(9, M) the quantized enveloping algebra of adjoint type, whereas Uq(g, A) is called of simply connected type. This terminology seems to be inspired by that for semisim-

ple groups. However, it appears to me misleading. The multiplicative group of all KA with A E r plays in Uq(g, r) the role of the maximal torus in a semisimple group. The role of the group of characters of the maximal torus is then played by Homz(F, Z) and not by F itself. So we ought to talk of the simply connected type, when Homz(F, Z) = A, not when F = A. That is the point of view adopted by Lusztig in his book: He calls Uq(g,F) of simply connected (resp. adjoint) type, if F is the set of all A E (M) ®Z Q with (A, p) E Z for all p E A (resp. with (A, p) E Z

for all p E M). In order to define an algebra with generators and relations (R1)-(R6), we do not really need a root system. We can start with an index set II and an integral

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

55

(11 x 1I)-matrix with entries denoted by (a,,3). The matrix should be symmetric with diagonal entries in 12,4,6.... }. Finally each number aaR = 2(a, 0)/(a, a) should be an integer < 0. This type of generalization was first applied to Serre's presentation of semisimple Lie algebras in the late sixties. It led to the construction of infinite dimensional Lie algebras known as Kac-Moody algebras. If we apply this generalization to the construction in 4.3, then we get quantum analogues of these Kac-Moody algebras. For the purpose of these lectures, however, I shall restrict to the easiest (and "classical" case) of the algebra defined in 4.3.

4.6. From now on we simplify our notations and write only U = Uq(g) and U = Uq(g). Similarly, we write U+ = Uq+(g) and U+ = U+(q) and so on. LEMMA. a) There are unique automorphisms w of U and U with w(Ea) = Fa, w(Fa) = Ea, and w(Ka) = K,-,, I. One has w2 = 1 in both cases.

b) There are unique antiautomorphisms T of U and U with T(Ea) = E. T(Fa) = Fa, and T(Ka) = Ka I. One has T2 = 1 in both cases. PROOF. One has to check that the images of the generators satisfy the defining relations. That is more or less clear for (Rl)-(R3) and for (R4) in the case a 5-4,3. In (R4) for a = ,3 one uses the same calculation as for Uq(S(2). Finally w interchanges the left hand sides of (R5) and (R6), whereas T fixes both of them.

4.7. The algebras U and U are (M)-graded: We first define such a grading on the free algebra in our generators via deg Ea = a, deg Fa = -a, and deg Ka = 0 = deg Ka 1. Then all relations are homogeneous and we get thus a grading on U and U. The canonical homomorphism U -H U preserves the grading. We get from 4.4(3) for all A, µ E Z4

KAuK. I

= q(A"N)u

for all u E U.. resp. u E

U.

(1)

The subalgebrasU+, U-, and U° (resp. U+, U-, and U(') are graded subalgebras of U (resp. of U), since they are generated by homogeneous elements. 4.8. We want to make U into a Hopf algebra such that the maps from 4.4(5) are homomorphisms of Hopf algebras. This means that for all a E 11 the following diagrams commute, where we use the abbreviation Ua = Uqn (5(2) Ua -Ua ®Ua

Ua -L k

I

I

I

Iid

U -L k,

U. -- Ua I

I

(1)

-°> U®U, U U. This forces us to define (0, e, S) as in the lemma below. It turns out to be easier to construct the Hopf algebra structure first on U where we do not have to worry about the relations (R5) and (R6) that are unpleasant to handle. U

LEMMA. There is on U a unique Hopf algebra structure (D, e, S) such that for all c E11

A(Ea)=Ea®1+K.0Ea,

E(Ea)=0,

A(Fa) = F. ®KaI + 10 Fa,

E(Fa) = 0,

A(Ka)=Ka®Ka,

E(Ka)=1,

S(Ea)=-KQ.IEa, S(Fa) = -FaKK, S(Ko)=KaI.

56

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(Q)

PROOF. We have to show first that there are homomorphisms 0 : U U®U U°pp that map the generators as described above. So and e : U k and S : U we have to check that the images of the generators satisfy (Rl)-(R4). Now (Rl) is pretty obvious for all three maps, and (R2), (R3) follow from the fact that the gradings are preserved. For the case a = 0 in (R4) we can appeal to the calculations in Uq(512). For a # 0 we have to check that certain commutators are zero:

[0(E"), 0(Fo)] = [E" ®1, F0 ® KQ'] + [E" ®1,1 ® Fa]

+[K"®E",FF®Ka1]+[K"®E",1®Fp] q-(",0)F,3K" ® E"K 1 - F,3K" ®

q-(".a)EK31

= 0, [e(E"), e(Fo)] _ [0, 0] = 0,

[S(E"), S(F,3)] = K. 1E"F,3K,3 - F,3KOK,1E"

= K;1E"F,3K,3 - K" 1(K"F,3Ka')(K,3E"KQ 1)KR

= Ka 1E"F8K,3 - Kc 1q-(",R)F0q(",)3)E"K3

=Ka1[E.,F,3]K,3=0. We now have to show that the diagrams in 3.2, 3.4, and 3.7 (with U replaced by U) commute., It is enough to check this on the generators. There we can quote the results for Uq(5C2), since for each a E II the homomorphism Uqa (512) U is compatible with 0, e, and S.

4.9. The map S2 is an algebra endomorphism of U taking each K" to itself, each E" to -(-K; 1E")K" = q E", and each F" to -Ka 1(-F" K") _ The weight 1

P = 2 tz7)3 = 2 ,3E1I

'y 'yE4+,-y>O

(cf. [H], 10.2 and 13.3, where p is called S) satisfies (2p, a) = (a, a) for all a E II, and 2p is in the root lattice. The formulas above together with 4.4(3) imply that S2 (U)

= K-'uK2p

for all u E U.

(1)

This shows in particular that S2 and S are bijective. The inverse of S is an antiautomorphism of U given by

S-1(F") = -KF",

S-' (E") = -E"Ka 1,

S-1(K") =

(2)

It is easy to see that A(KA) = KA ®KA

S(KA) =

and

for all A E M.

(3)

Using induction on r one checks for all a E H and all r > 0 r

0(E'") _

qt(r-2)

i=0

r

A(Fr,) i=0

f rl Ea tKa ®E", 2

a (4)

q"(r-i) [ r ] FF ®Fa-%Ka 2, Li

01

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(q) and

S(E'r

(-1)Tq.(r-1)Kc

-

57

c'

S(Fa) _ (-1)rgar(r-I)Fr"Ka. These formu l as act ua lly follow from

4.10. For all a,/3EII with a =

S=0

(-1)s

f1 L

I-aa(i

_

- aa31 s

J

Ea-aa[i-sEEE5 E U,

L

S

J

(1)

a

1 - aa,31 Fa aaa-sFoFos

(-1)s S=0

3 . 6(3) .

,3 set

1-aa3 U+

3 . 1(3) , (4) an d

( 5)

E U,

(2)

a

where aa3 = 2(a, /3)/(a, a). So these terms are the left hand sides in (R5) and (R6).

LEMMA. Let a,,3 E II with a # 3; set r = 1 - aa3. Then 0(uQ0 ) = ua3 ® 1 + KrK3 0 u+Cto

0( ua3) = uQO 9 KaTK3 + 1 ®ua0,

S(ual3) = -KaT Kl3 1ua3,

-uaaKrKp

PROOF. Note that it is enough to prove the formulas for 0; those for S follow

then from the antipode property of S. Indeed, we have obviously e(ua3) = 0 = e(u-) for all a,,3. We get therefore in the notation from 3.7 0 = m(S ® 1)0(ua3) =

Ka rK3 1u43

and

0 = m(1 ® S)0(u-Q) = ua3Kc,K3 + S(ua3).

The proof of the formula for 0(u,+,,3) is hidden in the appendix. I am sure that

one can proceed in the same way for 0(ua3). However, it is easier to deduce it from the claim for u.3 using Lemma 4.12 (which is independent of 4.10/11 as long as we regard it as a statement in U).

4.11. PROPOSITION. There is on U a unique structure (0, E, S) of a Hopf algebra satisfying the same formulas as in 4.8. PROOF. Let I be the two-sided ideal in U generated by all and as in 4.10. So U = U/I by construction. We have clearly E(ua3) = 0 = E(ua3) for all a, , 3, hence E(I) = 0. So E factors through a homomorphisms U = U/I -* 0 that we denote again by E; it satisfies the formulas from Lemma 4.8.

The formulas in Lemma 4.10 imply that 0(I) C I ® U + U 9 I and S(I) C I. So 0 induces a homomorphism

U/I - (U ®U)/(I ®U + U 9 I)

(U/I) 9 (U/I),

i.e., a homomorphism U -+ U_9 U that we again denote by A. Similarly, S induces

an antihomomorphism U ^ U/I -* U/I ^ U again denoted by S. It is clear that these maps satisfy the formulas from Lemma 4.8. Finally, the diagrams in 3.2, 3.4, and 3.7 commute, since the corresponding diagrams for U commute.

4. THE QUANTIZED ENVELOPING ALGEBRA U. (g)

58

REMARK. It is now obvious that everything in 4.9 extends from U to U.

4.12. For each finite sequence I = (/31,/32, ... , /3,) of simple roots set and

El = E$1 E192 ... Ep,

FI = F$1 Fat ... F.

(1)

In particular, set E0 = 1 and F0 = 1. We use the notations from (1) both for the corresponding products in U and in U. The canonical homomorphism U -» U takes

the EI in U to the E1 in U; similarly for the FI. Set

wtI=01+02+...+0,-

ifI=(/31,/32,...,/3r)

(2)

We have then EI E U :t I resp. El E U wt I

and

F1 E U= wt I resp. F1 E U=Wt I

(3)

LEMMA. Let I be a sequence as above. We can find elements cA B E Z[v, v-1] (independent of q) indexed by finite sequences of simple roots A and B with wt I = wt A + wt B such that in U and in U

A(EI) = E CA,B(q)EAKwt B 0 EB A,B

and

A(FI)cA,B(q 1)FA®KwiAFB. A.B

We have CA 0 = SA,I and co B = 6B.1

PROOF. We use induction on the length of the sequence I. For I = 0 we have

E0 = 1 = F0 and 0(1) = 10 1; so the claim holds with c0 0 = 1. Suppose the formulas hold for some I, let a E II and consider I' = (a, I). We have A(EI) = (Ea ®1 + Ka ® Ea) E CA.B(q)EAKwt B 0 EB A.B

_

(CA,B(q)E(a,A)Kwt B ®EB + cA,B(q)KQEAKwt B (9 EaEB) A.B

E (cA,B(q)E(a.A)Kwt B ® EB +

CA,B(q)q(a.wt A)EAKa+wt

B ®E(a,B)

A.B

and

A(FI,) = (Fa0K+1®Fa)ECA.B(4-1)FA®KwtAFB A.B

E (cA.B (q-1)F(a.A) ®Ka+wt AFB + CA.B (q-' )FA ®FaKA' FB) A.B

_ j:(cA.B(q-1)F(a.A) ®Ka+wtAFB A.B

+CA.B(q-1)q-(a.wtA)FA

®KAIF(a.B)Jll

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

We see now that we can define

CIA

59

B satisfying the claim: If both A and B begin

with a (the first term in I'), then write A = (a, A') and B = (a, B') and set cA B = cA B + v(a'" t `')CA B,. If A begins with a and B does not, then set CA.B = cly, B where A = (a, A'). If B begins with a and A does not, then set cly B = v(a'" t `')cA B,

where B = (a, B'). If both A and B do not begin with a, then set cAB = 0. It is also clear that these elements satisfy the last claims on the cases

A=0 andB=0. REMARK. The proof shows that we have to take only sequences A and B that are subsequences of I. Proposition 4.16 below implies that the EAKt B ® EB and the FA ®KWt AFB are linearly independent in U ®U. So the CAB are uniquely determined if we work in U. (But we do not need this fact.) 4.13. Lemma 4.12 implies in particular for all µ E M, µ > 0 that

A( U+)C ®

(1)

0
A(Uµ) C (@ E C,

(2)

0
similarly for U. (Note that U+ is the direct sum of the U+, hence U ® U+ the direct sum of all U ® Uµ . This implies that the sum on the right hand side of (1) is direct. One argues similarly for (2) and the U analogues.) The formulas for S in 4.8 imply for all p > 0 that S(U+) = U+Kµ 1

and

S(U_µ) = U µKµ;

(3)

similarly for U. One can say more precisely: If x E U+ and y E U µ, then

S(x) _ (-1)ht (µ)gm(µ)KK'T(x) and S(y) _ (-1)ht (µ)q-m(V )T(y)Kµ where we set for µ = E.EII maa

ht(t)

mc.

and

m.(a,a)).

m(µ) = 2 ((µ,µ) -

(4)

(5)

cEH

cEH

It is enough to check (4) for x = El and y = FI with I = (a1, aa, ... ,13r) where µ = wt(I) = Then ht (A) = r and (µ, µ) _ Ei.j(Qj,aj) = 2Ejj(0ti,Oj) + Eif2.

Ey(,3i,Qi), hence m(µ) _ Ej<j(Ri,,3j). Well, we get

S(EI) _ (-KW'E,3,.) ... (-K, 1Ep2)(-KQ1'Ept) (-1)rgmK-1 ... K, _ 1K-'EQ,.... E,32E,31

_ (-1)rgmK, 1T(EI) where

r-1

m = E(i3 , Qi+1 + ... +,3r) = E (i3 , Qj) = m(µ). i=1 i<j The calculation for S(FI) is similar.

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

60

4.14. Consider for each extension field k' D k a vector space Mk' over k' with basis (v1)I indexed by all finite sequences of simple roots as in 4.12 (including 0).

LEMMA. Choose for each 0 E II an element cp E k' with ci3 # 0. There is on Mk' a structure as a U-module such that for all a E 11 and all finite sequences I = (01, 02, ... Or) of simple roots

=

F. vI = v(a.l),

K. vl

eaq(a.wt1)

(1) (2)

Vi,

caq-(a'µ') - eal -Aj)

Ea vl =

(3)

1

1<j
where µj =

Er

-j+1 /32.

PROOF. These formulas define endomorphisms f, ka, and ea of Mk,. It is clear that each ka is bijective with inverse given by

kalvl =

calq(a,wtI)vj

for all I.

(4)

We have to show that the (ea, fa, ka, ka 1) satisfy the relations (Rl)-(R4) from 4.3. Clearly (R1) is obvious. Denote for all p the span of all vj with wt(I) = p by (Mk')-u. Then (2) and (4) imply for all a and all v E ZID kav = eaq(a.v)v,

ka'v = ca 1q-(a,v)v

for all v E (Mk')v.

(5)

On the other hand (1) and (3) show for all v E Z4 and all /3 E 11 e,((Mk')v) C (Mk')v+o

and

ff((Mk')v) C

(Mk')µ-3.

(6)

Now (R2) and (R3) follow easily from (5) and (6). We are left with (R4). In case a 54,3 the definition of ea shows easily for all I eaf/3vI = eav(/3.I) = f3eavl,

whereas we get for a = /3 eaq-(a,wt 1) - ca 1q(a.wt I)

eafavl = eav(a.I) _

qa - qa_1

vi + faeavl

ka - ka 1 -1 vl + f«eavl qa - qa

So (R4) is satisfied in both cases.

4.15. Denote the module constructed in Lemma 4.14 by Mkw (c) where c = It is actually a (U ® k')-module. (We could identify U ® k' with the analogue of U constructed over k'; but we do not need this identification at this (ca)ac-n.

moment.)

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(B)

61

We can show similarly that there is a unique structure as a U-module on Mk, such that for all a E II and all sequences I

Ea vI = v(a,I),

Ka vI = c«q (a,wt I) Vi,

and -if I=(/31,I32,...,/3r)-

E 1<j
ca- lq-(a,ij) - c q(a.µ)) qa - qa-1

v(!3,.92....(3)

where µj = Ei=j+1133. We denote this module by Mk, (c). We have obviously F1v0 = vi in Mk' (c) and Eiv0 = vi in MM, (c). This implies

that the FI resp.the EI are linearly independent, hence a basis of U- resp.U+. This shows that U- is the free algebra in the generators F,,, a E II and that U+ is the free algebra in the generators E, a E II. However, that will follow also from the stronger result that we are going to prove next.

4.16. PROPOSITION. The elements F/KEJ with p E M and I, J finite sequences of simple roots are a basis of U. PROOF. Set V equal to the subspace of U spanned be these elements. We have

for all aEH F.F/KMEJ = F(a,I)KN.EJ and

KXFIKN.EJ =

q-(wtI,a)FIKµ+aEJ,

K,, I FIKµEJ = q(wtl.a)FIKP_aEJ.

This shows that FaV C V and KaV C V and K, ,-'V C V. Using (R4) one checks that also EaV C V; the fomula for EQFIKNEJ is a little bit more complicated and left to the eager reader. We get U V C V, and thus U = V (since 1 E V). So our elements span U. We now have to show their linear independence. Suppose that we have in U an equation (1) E al. J,,UFI KµEJ = 0 I, J.µ

with all ai.J.µ E k, almost all equal to 0, but at least one of them not equal to 0. Choose I0 such that there exist p and J with aio,J,µ 0 and such that wt(Io) is maximal for that. Consider now for all extension fields k' D k the tensor product Mk' (c') ®k' Mk',(c) with c'a = 1 for all a and arbitrary c = (ca). This is a tensor product of (U ®k k')-modules that we regard as a (U ®k k')-module via A. We get from (1)

_

E a,. j,,. FIK,. EJ(v0 (9 v0) = 0, I,J.µ

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(Q)

62

hence by Lemma 4.12

al,J,µFIKµ ECA,B(q)(EAKwt BVO ®vB) = 0. I.J,µ

A,B

Since Kt BVO = v0 and EAVO = 0 in Mk, (c') for all B and all A

0, and since

CO,B = SB.J, we get

E al,J.µFIKµ(v0 ®

0.

[,J,µ

Now A(Kµ) = Kµ ® Kµ yields E al,J,,Fr(v0 ® cµq(µ,wt

0,

I,J,µ

where we use the abbreviation

cµ= f ca(a)Ek'

m(a)a E Z4.

for each aEn

cEII

Using again Lemma 4.12 we get now al.J.µcAq(µ wt J) E CC.D(q-1)(vc ® Kw;CFDVJ) = 0. I,J,µ

C.D

For any I all Fc occurring in A(FI) satisfy wt C < wt I with equality only for C = I. So in the last equation the only terms in vlt, ® MM, (c) are E alo.J,µcAq(µ'wt J)VI® ® Kwt

Iovj

J,/a

_E

alu.J,µCµq(µ'wt J)C- wt loq-(wt lo,wt J)VI® ®

Vj.

J,µ

Therefore this sum has to be equal to 0. Since the vlo ®v are linearly independent, this implies alo.J,µCµq(µ.wt J)C- wt loq-(wt Io.wt J) = 0, Fi

hence

E

J)Cµ = 0

for all J.

We can regard the left hand side as a polynomial over k in III indeterminates and their inverses. It vanishes at all 11-tuples (ca) from any extension field k' of k where J) = 0 for all J and p, all ca 0; therefore it is 0 itself. This means hence alo,Jµ = 0 since the q power is nonzero. This contradicts the choice of Io. Therefore the FIKµE are linearly independent.

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

63

4.17. Proposition 4.16 implies in particular that the Ka with A E Z 1 are a basis of U°. It also implies that the EI (resp. FI) are a basis of U+ (resp. U-), and thus that U+ (resp. U-) is the free algebra in the generators Ea (resp. Fa) with a E II. (Of course, we have observed the last claim already in 4.15.) Furthermore, we see that the multiplication map

U ®U0®U+->U,

ul®U2 ®U3HU1U2U3

(1)

is an isomorphism of vector spaces. Using w one sees that also

U+ ® 0° ® U- -> U is an isomorphism of vector spaces.

U1 ® U2 ® U3 F-a U1U2U3

(2)

_

One checks also that the subalgebra of U generated by all E,,, K,,, and K. I with a E II has as a basis all KµEj. This subalgebra is denoted by U20. Similarly, the subalgebra U:0 of U generated by all Fa, Ka, and K,-,' with a E II has as a basis all FIKN,.

For each a E II the homomorphism Uqa (s(2) -> U from 4.4(5) takes the basis

of all FSKfE' from 1.5 to the linearly independent set of all F,',KaEa. So this homomorphism is an isomorphism onto its image. That image is the subalgebra of U generated by Ea, Fa, Ka, and K,-:,'. 4.18. If (A, A, E, S) is a Hopf algebra, then one defines the adjoint representation of A on itself as follows: First A is an (A(DA)-module via (a1®a2) - b = a,bS(a2)

for all al, a2, b E A. Then we make A into an A-module via A. We denote this action by ad. So we have for all a, b E A

ad(a)(b) = EaibS(ai),

if A (a) = Eat ®a'.

i

(1)

i

Note: We can consider for any A-module M the map A -> Endk(M) that takes any a E A to the linear map that. gives the operation of a on M. Now (1) and 3.10(5) say that this map is a homomorphism of A-modules if we take the adjoint representation on A. We apply this general construction in particular to A = U and to A = U. The formulas in 4.8 imply for all a E II and all u E U resp. u E U

ad(Ea)u = Eau - KauKQ 1Ea, ad(Fa)u = (Fau - uFa.)Ka, ad(Ka)u = KauK, I.

(2)

The last equation yields immediately

ad(K,)u = KuKµ I

for all p E M.

We get from 4.9(4), (5) for all a E II and all r > 0

ad(Ea)u =

Egar-i)

-o

r21 a

E.-'K.,

u(-1)'gat-1)KK'Ea,

(3)

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

64

hence

ad(E')u =

E(-1)'q'a(r-1)

ri

r-

K. a Ka iEa,

(4)

Ja

i=0 and r

ad(F )u = E q

r-t)

[r]

)(r-i- 1) F,r-iKr-

F,, u K a

i=0

hence

r

[r -(r-i)(r-1) Y

ad(Fa)u

u Fra -2Kar.

I.

(5)

a

i=0

LEMMA. We have in U for all a,,3 E II with a r/ ,3

ad(Ea)(E,o) = uaa

and

ad(F'«)(F,3K3) = u-,3 K,3 Kc,,

where r = 1 - aa,3. PROOF. Recall from 4.2(1) that aa3 = This implies g(a.3) hence

q(a.«)/2

qa

and from 4.3 that

- qa.0 = qa r

KaE3K« i = gi(a.13)E

(6)

3 = q'(1-r)

(7)

Plug this into (4) with u = Ep and we get the first claim. To get the second one, we apply (5) with u = F,3K3. In this case the claim follows from K,3F'«r-a = q-(r-i)(«,3)F'r-2K8

4.19. LEMMA. Let a, ,Q E II with a

[F-,, ua3] =0

and

(r-i)(1-r)Fr-zK3

= qct

,Q. Then

for all -y E H.

[E7, uaQ] = 0

(1)

PROOF. It is enough to deal with ua,3; the claim for u.3 follows then by applying w. For -y a, a we have [F.,, Ea] = 0 = [F-,, E3] by (R4), hence [F-,, ua3] = 0. Let us now look at -y = a. We have by 4.18(2) [Fa, ua3] = ad(Fa)(ua3) . K,-,

1.

So it is enough to show that ad(Fa)(ua3) = 0. We have by Lemma 4.18

ad(Fa E')(E3)

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(g)

65

where r = 1 - aaa. So 4.4(7) yields ad(F«)(uap) = ad(Ea) ad(Fa)E,3 - [r]a ad(EQ 1) ad([K«; r - 1])Ea.

(2)

Now (R4) and 4.18(2) imply ad (F.) E0 = 0.

(3)

On the other hand ad(KQ)E,3 = KaE,3Ka' = q'-'E,3, cf. 4.18(6),(7). So we have that ad(Ka 1)Ea = qQ ' E3 and

-a

ad(K.)qr 1 - ad(Ka')ql,-r E,3 = 0. I qa - qa-

ad([K«;r - 1])E3 =

(4)

Now (2), (3), and (4) yield ad(Fa)(uaa) = 0, hence our claim. Finally, look at ry =,3. We have (since [FO, Ea] = 0 by (R4)) r

[Fa,u«aE(-1)Z [r] E. '[Fa,E0]Ea a

i=O

with r as above. We have [Fa, E3] = (Ka' - K;3)/(Q3 - q_') and Ki3E,, qar)EaK3 and KQ'Ea = We get thus r (q

1

v+

- qp )[F'3, a)3 _

E(-1)i

[T]qT1ETK1 a i a

Z=0

-

E(-1)i

[r]qi(1r)ErK3a

Now these sums are equal to 0 by 0.2(3),(4) specialized at v = qa.

4.20. Denote by I+ (resp. I-) the two-sided ideal in U+ (resp. U-) generated by all u,+, (resp. all u..). LEMMA. The two-sided ideal in U generated by all uaa (resp. all

is equal

to the image of U- 0 Uo ® I+ (resp. of I- ® Uo ® U+) under the multiplication map 4.17(1).

PROOF. The arguments in the two cases are similar; let me restrict myself to the ideal generated by the uaa. (The claim in the other case follows then by applying Tow.)

Denote the image of U- ® Uo ® I+ by V. It is obviously contained in the two-sided ideal in U that we want to get, and it contains all the generators of the ideal. We have to show that V is a two-sided ideal. As a vector space V is spanned by all uuaaEj with a,,3 E 11, a # ,3, with u E U and with all sequences I as in 4.12. This shows that V is a left ideal in U. We have for all -y E IT and all µ E Z4 uua+aEjE, = uua+aE(l,.y)

and uuaaEjKµ = q-(µ.wt 1+ra+,3)UKNUQaE!

4. THE QUANTIZED ENVELOPING ALGEBRA Uq(9)

66

(where r = 1 - a,,,3). -This shows that V is stable under multiplication on the right with elements from U'-0. Finally, we have for all -y E II uuapEjFy = uF.yua+pEj - u[Fy, ua+p]EI - uua+p[F.y, EI].

(1)

Here the first summand is obviously again in V, the second one is equal to 0 by Lemma 4.19. The commutator [EI, F.y] is in U'0 by (R4). So our previous result implies that also the last summand in (1) is in V. This implies V Fy C V for all -y. Therefore V is a two-sided ideal, and the claim follows.

4.21. THEOREM. a) The multiplication map

U ®U®®U+ , U,

ul (9 U2 ®U3 I .' u1U2u3

(1)

is an isomorphism of vector spaces. b) The algebra U+ is isomorphic to the algebra with generators Ea, a E II and

relations (R5) for all a, 0 E II with a # 0. c) The algebra U- is isomorphic to the algebra with generators Fa, a E II and relations (R6) for all a, /3 E II with a # 0. d) The KN, with µ E Z4i are a basis of U°. PROOF. Let I be the kernel of the canonical map 7r : U - U. It is the two-sided

ideal generated by all uaa and all uap, hence equal to m(U- (& U° 0 I+ + I- ® UO ® U+) where m is the multiplication map m : U- ® U° ®U+ -a U. Now I n U° is equal to the image under m of

(U-®U°®I++i-®U°(D U+)nk®U°(D k=0, hence itself equal to 0. Therefore 7r maps U° injectively to U and induces an isomorphism

UO=>U°.

(2)

So part d) of the theorem follows from the corresponding statement for U° in 4.17. The intersection I n U+ is equal to the image under m of

(U-®U°(& I++i-®U°®U+)nk®k®U+=k®k®I+, hence itself equal to I+. So 7r induces an isomorphism

U+/I+ -' U+.

(3)

Since U+ is a free algebra in the Ea and since I+ is generated by the u+R, i.e., by the right hand sides of the relations (R5), the claim in b) follows. The proof of c) is symmetric; in particular, we have an isomorphism

U-/I- => U-.

(4)

Since

(U- (D 0-0 (D 0+)/(0- (D 00

P- + I- (D 00 (D U- +)

can be canonically identified with (U-/I-) ®U° ® (U+/I+), we see that m induces an isomorphism of vector spaces

(U-/I-) ® U° ® (U+/I+) => U/I,

(u1 + I-) ®u2 ® (u3 + I+) > UIU2U3 + I. Modulo the isomorphisms in (2), (3), and (4) this is the map in (1). So the claim in a) follows.

APPENDIX

67

4.22. Using w we see that also the multiplication map

U+®U°®U--,U

(1)

is an isomorphism of vector spaces.

The subalgebra of U generated by all K,,, K(-,', and all E,, (resp. and all Fa) with a E H is the image of U'0 (resp. under the canonical map. We denote this subalgebra by U'° (resp. U50). Theorem 4.21.a and the corresponding result in 4.17 show that the multiplication induces isomorphisms of vector spaces and U- 0 U0 - U<°. U° ® U+ Since each u,+,Q is homogeneous of degree (1-aap)a+,3, the ideal I+ is graded.

For each weight u of 1+ (i.e., with I, # 0) there are two distinct simple roots a and Q with u > a +,3, since this is true for the generators and since all weights of U+ are > 0. In particular, we have I = 0 for all a E H and all r > 0. So no E,r,, is contained in I+, and the images of the Er, in U are nonzero. Since they are homogeneous of distinct degrees, they are linearly independent. The same argument shows also that the F,', are linearly independent in U. Theorem 4.21 (parts a and d) implies now that the FQSKQE'r' are linearly independent. This shows as in 4.17 for each a E H that the homomorphism Uq,(512) -4 U from 4.4(5) is injective.

Appendix PROOF OF LEMMA 4.10. We have

= A(U+)

E(-1)i =o

[r] A(Ear a

a

(where A(EE) = Ep ®1 + Kp 0 Ep) and by 4.9(4) S-1

A(E«) = E« ®1 + Ka ®Ea + E j=1

q-(s-j)

E. jK.j®Ea.

[j] a

This implies easily that 0(ua+p) has the form r

A(ucxp)

= u«p ®1 +

®ucxp + E X. ®E. + EYmn ®E. E,3E.n +n.=1

(5)

rn.,n

with suitable X, and Y,,,n in U, and where the last sum is over the integers m, n > 0

with m + n < r. We have to show that all X,n and Y. are equal to 0. Well, for m and n as above the term Ynn is equal to r-n,

(-W [r]

i=n

a] E. i-m KT K. qn(i-n) [7i] (r-Z-"`) q E nKa Irm a a a

We can express the Gaussian binomial coefficients in terms of factorials and get then

[ri]a

[rmi]a [n] a

[r amn n]a [r nm] a [m]a

4. THE QUANTIZEI) ENVELOPING ALGEBRA Uq(8)

68

We have by (R2) and (4)

KaK3E -n = g2am(i-n)g(a.j3)(i-n)Eia-nKmKQ

Therefore

I,

mn =

b

=

g22m(i-n)+(1-r)(i-n)Eia--nKa

r-m

r Er-(m+n)Km+nK a a m a a

n

Kp

where

r-m b=

(_1)i I r - m - ] qm (r-i-m)+n(i-n)+m(i-n)+(1-r+m)(i-n)

r-m

i-n

L

i=n

Ja

(-1)i I r - m - n

i-n

L

i=n

qa (r-n-m)+(1-r+n+m)(i-n)

I

J

a

r-m-n

=qa

j

(r-n-m)(-1)n

[rfl]

-j(r-n-m-1) a

=0

[by 4.18(1)].

The term Xn,, (with 1 < m < r) is equal to r E(-1 )x

[r] i

i=0

a

j)

jJj

j

I

q(am-j)('-m+j)

Ia

m-j

Ea-m+jKm-j a

a

where we sum over all j with max(0, m - i) < j < min(m, r - i). We get now r-m amSEra-m-sEaKa,

Xm = E

S=0

where m+s

ams

()2 []a[mi]a[]a

= m+s

rr Sml rZmsl

i Lm] i=s

a

rr1

rr - ml

M

s

_ =0.

a

(-1)sga(s+1-m)

J

L

a

L

m+s

rr1

ma

J

q(m+s-i)(r-m+s+1-r)+(i-s)s a

a

m 1 gams+t-m)+(i-s)(m-t) i-s

rr - ml L

S

Ja

m 1)

LmJa

CHAPTER 5

Representations of Uq(g) Throughout this chapter we keep the assumptions and notations from the last chapter. In particular, we write U = Uq(g). We assume in addition that q is NOT a root of unity.

The theory of finite dimensional representations of Uq (g) is very similar to that of g as long as q is not a root of unity. We have seen that already in the case of s12i there the only real difference was that we got in 2.6 two simple modules for Uq(5(2) corresponding to each simple module for 512. In general there will be 21111 simple modules for Uq(g) corresponding to each simple module for g. These 21n1 modules arise from the choice of IIII signs. More or less the first thing we do in this chapter is to restrict ourselves to those modules where we choose all these signs equal to 1. (We first check that we do not lose any information that way.) After that the theory proceeds more or less as for g, for example as in the book by Humphreys [H], Sections 20 and 21. An essential tool are the subalgebras isomorphic to Uq (s[2) from 445) that play the role of the Lie subalgebras isomorphic to $12 in g. The arguments get occasionally more complicated, e.g., in the subsections 5.7/8

that replace the steps (3)-(5) in the proof of Theorem 21.2 in [H]. However, the technique of imitating the classical theory breaks down when we get to the description

of the simple modules by generators and relations, cf. Theorem 21.4 in [H]. The proof there uses Weyl's theorem on the complete reducibility of all finite dimensional

g-modules. Here we do not yet have the analogue of that theorem available. And when we prove it in 5.17 we use the description of the simple modules by generators and relations in a crucial way. The solution to this problem turns out to be cumbersome. We work for some time over the field Q(v) where v is an indeterminate over Q and take q = v. We consider Q[v, v-'] -lattices in our modules and reduce them modulo v-1. This leads to representations of our original Lie algebra g, and we can lift information from g to Uq(g). This applies not only to the generators and relations mentioned above, but also to Weyl's character formula. Once we have this over Q(v) we get it easily by base change for any field of characteristic 0 as long as q is transcendental over Q. The general case (q not a root of unity, k arbitrary) requires other techniques for which we are not yet ready. Let me mention that Lusztig proves the complete reducibility theorem early on in his book [L]. To do that he has to change the definition of Uq(g): He replaces the quantum Serre relations (R5) and (R6) by more abstract conditions that involve a bilinear form that we shall encounter in Chapter 6. 69

5. REPRESENTATIONS OF Uq(g)

70

5.1. Recall from 4.2 that we assume char(k) # 2 and from 4.1 that (A, p) E Z for all ) E A (the weight lattice) and all µ E Z4? (the root lattice). F o r a n y U-module M set f o r all A E A and all group homomorphisms o : Z4? -

{±1} MA,, = { m E M I Kµm = a(µ)q(X,A)m for all µ E Z4)}.

(1)

So each MA., is a subspace of M that we call a weight space of M. It suffices to take µ E II in the definition. If A = `/3Er1 m(3)zuo, then g(a,a)

= gm(a)(za ..a) = gm(a)(a,a)/2 = qa («)

for all a E H.

(2)

This shows (since q is not a root of unity) that ) is determined by the function q(A,A), hence that the sum of the MA,Q is direct.

PROPOSITION. Let M be a finite dimensional U-module. Then M is the direct sum of all MA,Q. We have for all ) and a

EcM,\ v C MA+a,,

and

FaMA,Q C M,\-a,a

(3)

for all a E II. Each Ea and each F0, acts nilpotently on M.

PROOF. The inclusions in (3) follow immediately from (R2),(R3), cf. also 4.4(3). We can regard M for each a E II as a Uq, (512)-module via the embedding from 4.4(5). Therefore Proposition 2.2 implies that Ea and Fa act nilpotently on M, and Proposition 2.3 implies that the action of Ka on M is diagonalizable with all eigenvalues of the form ±qa with a E Z. The KQ with 3 E II commute, so we can diagonalize them simultaneously on M. If m is a common eigenvector of all Ka,

then there are integers a(a) and signs as E { ±1 } such that Kern = aaq«(a)m for all a. Set A = EQEII a(3)zuo and define a group homomorphism a : Z4i - { ±1 } by a(a) = as for all a E II. Then (2) shows that m E MA,Q. So M is the sum of its weight spaces, hence their direct sum. 5.2. If M is a finite dimensional U-module, then Proposition 5.1 implies

M = ®M° or

where M° = (D M,,,.

(1)

AEA

(The first sum is over all group homomorphisms or : Z4i - { ±1 }.) By 5.1(3) each

M° is a U-submodule of M. We say that M is of type or, if M = M. We say that M is of type 1 if it is of type a with a(3) = 1 for all ,3. If M and N are finite dimensional U-modules with M of type or and N of type a' where or o'', then Homu (M, N) = 0. (Any homomorphism co : M - N of U-modules satisfies co(MA,,) C Na,o = 0 for all A.)

We can state the results from the last paragraph as follows: The category of all finite dimensional U-modules is the direct sum (over all a) of the categories of all finite dimensional U-modules of type or. For each a as above there is a unique automorphism v of U with

a(E.) = o'(a)E., a(F.) = Fa,

a(K0) = a(a)Ka,

for all a E H.

(1)

(We have necessarily a(K,-, I) = a(a)Ka 1. It is trivial to check that the defining relations of U are preserved.) It is clear that v2 = 1. We can twist U-modules with

5. REPRESENTATIONS OF Uq(g)

71

v, i.e., take the same vector space, but set the new action of u E U equal to the old action of v(u). It is easy to see that this twisting with v is a functor that takes finite dimensional U-modules of type 1 to finite dimensional U-modules of type 0', and vice versa. It induces isomorphisms on Hom spaces, takes simple modules to simple modules, direct sums to direct sums, and so on. In short: It is an equivalence of categories between the category of all finite dimensional U-modules of type 1 and those of type or.

5.3. Since we have on U a structure as a Hopf algebra, most of the remarks from Chapter 3 (up to 3.10) generalize from Uq(5(2) to U = Uq(g). In particular: We use 0 to define the tensor product of U-modules as in 3.3(1). The associativity statement 3.3(3) and its generalizations extend. Since 0(KN,) = KN, ®KN, for all µ, we have

Ma,o ® NA,,o, C (M ®N)a+a',oo'

(1)

for all U-modules M and N and all A, A', o , o'. Therefore a tensor product of two modules of type 1 is again of type 1. We have again a trivial one dimensional module given by e and usually denoted by k. More generally, each group homomorphism or : Z4 -> { ±1 } defines an algebra homomorphism Eo : U -> k with

Eo(Eq)=5,(F.)=0, E,(Kq)=o(a)

for all aEII.

(2)

This leads to a one dimensional U-module that we denote by L(0, v). One checks

that easily for any U-module M that M twisted by v as in 5.2 is isomorphic to L(0, o) ®M under m --> 10 m. We can use the antipode to make dual spaces of U-modules into U-modules as in 3.9(1). Now 3.9(3) extends immediately, whereas 3.9(2),(4) have to be modified: We have to use 4.9(1) instead of 3.6(2). So for each U-module M the map cp'

: M -, (M*)*

with

(m)(.f) = .f(K2plm)

(3)

is a homomorphism of U-modules. Similarly, replace K by K2p in 3.9(4). If M is a finite dimensional U-module, then M and M* are by Proposition 5.1 the direct sums of their weight spaces. One checks then easily for all A and v that

(M*)a.o = { f E M* I f (MA'.o') = 0 for all (A', o') # (-A, o) }.

(4)

Thus dim (M*)a,o = dim M_A,, for all A and o. We see in particular: If M is of type or, then so is M*. Finally, we can use 3.10(5) to make Hom spaces between U-modules into Umodules. The map in 3.10(3) is again a homomorphism of U-modules. We define fixed points again by 3.10(6), the formula 3.10(7) generalizes, and in the definition of the quantum trace 3.10(8) we have to replace K by K2p. 5.4. By 5.3 the category of finite dimensional U-modules of type 1 is closed under taking tensor products, dual modules, and Hom spaces. The discussion in 5.2 shows that we do not lose any information, if we restrict to finite dimensional U-modules of type 1. Since such a restriction will simplify many formulas and theorems, let's just do it:

5. REPRESENTATIONS OF Uq(g)

72

CONVENTION. From now on finite dimensional U-modules are supposed to be of type 1.

Furthermore, we use for weight spaces the abbreviated notation Ma = M).,1.

So any finite dimensional U-module M (under our convention) satisfies M = 0 is finite. So there is for M 0 always ).EA M. The set of A with Ma

0 and M)., = 0 for all A' > A. We have then in particular M).+, = 0 for all a E II, hence EaM). = 0. This proves the first part of: a A with M.

LEMMA. Let M be a finite dimensional U-module with Al

AEAand vEAT).,v

0. a) There exist

0with Eav=O for all a Ell.

b) Let A E A and v E Ma, v 0 with Eav = O for all a E [I. Then A is a dominant weight. We have F ('3)+l v = 0 for all,3 E II where m(,3) = 2(A, X3)/(13 3).

PROOF. Suppose that we have A and v as above. Take any a E II and consider M as a Uqa (5(2)-module via the embedding from 4.4(5). We have Eav = 0 and Kav = q().-")v = q'v where a = 2(A, a)/(a, a). Therefore the Uq (sC2)submodule generated by v is a homomorphic image of the module denoted by M(q,a:,) in 2.4. Since this image has finite dimension, Proposition 2.5 implies that a > 0, hence (A, a) > 0. Furthermore, we have F«+lv = 0, since otherwise the Uq (5(2)-submodule generated by Fa+'v is isomorphic to the simple infinite dimensional module M(q-a-2). 5.5. Any A E A defines a one dimensional U-°-module where each Ka acts as multiplication by and each Ea acts as 0. (Well, there is a linear map from U'-0 U° ® U+ that takes any Ku with u E U+ to Using E(KN,uKµ 1) = E(u) one checks easily that this map is an algebra homomorphism.) The kernel of this representation is an ideal Ja ° of codimension 1 equal to U>oEa

Ja o =

U''-o(Ka

+

- q().`))

aEII

aE17

We have U'-0 = k ® J, O. Let J,\ be the left ideal of U generated by Ja °. So we have

J). = E UEa + E U(Ka - q()..")), aEn

(1)

aEn

and 4.21(1) implies that U = U- ED J,\. Set

M(A) = U/J). = U/(E UEa + E U(Ka - q(). a))1l/ aEII

(2)

aEll

This is a U-module generated by the coset of 1; denote this coset by v).. Obviously

Eav). = 0

and

Kav). = q()..")v'\

for all a E II.

(3)

We call M(A) the universal highest weight module (or Verma module) of highest weight A. It has the following universal property: If M is any U-module and v E M,\ a vector with Eav = 0 for all a E II, then there is a unique homomorphism of Umodules cp : M(A) M with V(v).) = v. (Proof: The map U M, u --* uv takes J'\ to 0.)

5. REPRESENTATIONS OF Uq(9)

73

The direct sum decomposition U = U- ® Ja implies that the map U- ---> M(A),

u --> uvA

(4)

is bijective. (Of course, it is also an isomorphism of U--modules.) It takes each U, bijectively to M(A)A+µ. Each U, is finite dimensional, since it is spanned by the FI as in 4.12 with u = - wt I. So also each M(A)A+N. has finite dimension. One gets in particular M(A)A = kva

and

M(A)A_,,,a = kF,,vA

(5)

for all a E II and all integers n _> 0. We can now generalize to the present situation the standard results on standard cyclic modules for Lie algebras, e.g., in [H], 20.2: Each submodule N of M(A) is the direct sum of its weight spaces; if N is not equal to M(A), then N is contained in the direct sum of all M(A)A+µ with u 0; there is a unique maximal submodule in M(A). So M(A) has a unique simple factor module; denote this simple module by L(A). Any endomorphism ?/i of M(A) as a U-module has to preserve the weight spaces. So the first formula in (5) implies that there is an a E k with di(va) = a vA. Since vA generates the U-module M(A), this shows that b is scalar multiplication with a. The same argument applies to any homomorphic image of M(A), in particular to L(A). So we have EndUM(A) = k

and

EnduL(A) = k.

(6)

Lemma 5.4.a and the universal property of the M(A) imply that each finite dimensional simple U-module is a homomorphic image of some M(A), hence isomorphic to L(A). This A is unique, since it is the largest weight of the module. Lemma 5.4.b implies that A is dominant. In order to complete the classification of all finite dimensional simple U-modules, we have to show: If A is dominant, then dim L(A) < oc. This is our next goal.

5.6. LEMMA. Let A E A and a E II with (A, a) > 0; set n = 2(A, a)/(a, a). There is a homomorphism of U-modules V : M(A - (n + 1)a) -* M(A) with ,P(va-(n+I)a) = PROOF. We have Fn+'vA E M(A)A-(n+l).. Therefore the universal property of M(A - (n + 1)a) implies that it is enough to show that E,3F,,+IvA = 0 for all ,3 E II. This is obvious for ,Q 54 a, for example, since E,3 and Fa commute by (R4) and since E3vA = 0. (Alternatively: A - (n+ 1)a +,Q is not a weight of M(A), since A-(n+l)a+,Q $ A.) For ,3 = awe have (using 4.4(6) and KKvA = q(A,a)va = gnvA) EOFF+IvA = FF+1E,3vA + [n + 1]aFn, [Ka; -n]vA =0+[n+1],,F,-,gaga"-ganga

va

=0

.

qa - qa

5.7. An endomorphism f of a vector space V is called locally nilpotent, if there is for each v E V an integer n > 0 with f"L(v) = 0. (For finite dimensional V locally nilpotent means the same as nilpotent.)

5. REPRESENTATIONS OF U9(9)

74

LEMMA. Let A E A. Choose for all a E II integers m(a) > 0 and n(a) > 0. Then Let I be the left ideal in U generated by all E, (a), F,,n,(a), and Ka all Ea and F(,, act locally nilpotently on the U-module U/I. PROOF. Let us show first for all a,,3 E II with a # ,3 that r-1

FaFoE

kFaFpF,

forallm>rwherer=1- (a'a)

(1)

j=o

We use induction on m. For m = r this is just (R6). Induction shows for m > r r-I

r

j=o

j=1

Fa F, E Fa 1: kFQFQF. -1-j = EkFj, FfF. -j The terms with j < r on the right hand side are in the right hand of (1); using (R6) we see that the same is true for the summand with j = r. Pick now a E II and let us show that Fa acts locally nilpotently on U/I. We have to show that there is for each x E U an integer N > 0 with Fax E I. It is enough to take for x products of the form x = x1x2 . xr with all xi E { Eo, Fo, Kµ }. We want to use induction on r. For r = 0 we have x = 1 and get Fax = Fa E I for N = n(a). Consider for some r > 0 a product x = xoxlx2 xr with all xi as above. Suppose by induction that we have found for x' = x1x2

xr

an N with Fax' E I. We have now different cases according to what x0 is. If xo = K. for some u E ZP, then Fax = FaKµx' = K,,(K, 1 FaK,,)x' = KµgN(µ,(,,)Fax' E I.

(2)

If xo=E,owith ,3EH,/35La,then

Fax=FaEox' =EfFax' E I. If x0 = Ea, then Fa +1x = Fa +IEax = EaFa +1x' - [N + 1]aFa [Ka; -N]x' E I,

where we use (2) to get the last term into I. If xo = Fa, then

Fax =FaFax'=Fa+Ix'EI'. If x0 = Fp with 3 E II, 3 # a, then we apply (1) with r = 1 - 2(a 3)/(a, a) and get FaN+rx = FQN+rFax

r-1

E E kFaFpFa+r-ii C I. j=o

The proof for Ea is similar. Alternatively, we can deduce the claim for Ea from

that for Fa using w: If we twist U/I with w, we get U/w(I), and w(I) is of the same type as I.

5. REPRESENTATIONS OF Uq(g)

75

5.8. Let us return for the moment to the set up of Chapter 2 (with q not a root of unity).

LEMMA. Let V be a Uq(s12)-module such that V = EZ Vn where Vn = { v E V I Kv = qnv }. Suppose that E and F act locally nilpotently on V and that dim Vn < oo for all n. Then V is finite dimensional, and we have dim V = dim V_n for all ii Z. PROOF. We use induction on dim Vo + dim V1. We may assume that V 0. Pick an m with V,,,. # 0 and a w E V,,, with w 0. Then there is an integer r > 0

with E''w=0andE''-1w

0. Sety=E''-1w;then Ev=0andvE V,, where

n = m+2(r- 1). Therefore Uq(512)V is a homomorphic image of the module M(qn)

from 2.4. Since Fsv = 0 for s >> 0, this image cannot be isomorphic to M(qn). So 2.5 implies that n > 0 and that Uq(512)v is isomorphic to L(n, +). We have dim L(n, +)o = 1 for n even, and dim L(n, +) 1 = 1 for n odd. We can therefore apply induction to the module V' = V/Uq(s(2)v. So V' and hence V have finite dimension. The formula dim V = dimV_n holds, since it holds for every simple finite dimensional Uq(s(2)-module, cf. 2.6.

5.9. Let A E A be dominant, say A = E/3En n(,3)tzi . We have then n(a) = 2(A, a)/(a, a) for all a E II. Lemma 5.6 shows that there is for each a E II a homomorphism of U-modules V,, : M(A- (n(a)+1)a) -4M(A) with Va(va-(n(a)+1)a) _ Fan(a)+1 va. PROPOSITION. The U-module L(A) = M(A)/ Ea.En im(cpa) has finite dimension.

PROOF. If we identify M(A) with U- as in 5.5(1), then the image of cpa is This shows that

identified with Z(A)

U-F«(«)+1

U/(E UEa + E UFn(a)+I + E U(KK. - q(A,c ))). aEII

aEI7

aE17

Therefore Lemma 5.7 implies that all Ea and Fa act locally nilpotently on L(A). We want to argue as in the Lie algebra case, cf. [H], 21.2. So we want to show that the set of weights of L(A) is stable under the Weyl group of OP. This will imply dim L(A) < oo, since each Weyl group orbit contains a dominant weight, since there are only finitely many dominant weights less or equal to A, and since the weight spaces in M(A) and hence in L(A) are finite dimensional. So it is enough to show: If it is a weight of L(A) and a E II a simple root, then sa(µ) is a weight of L(A). Here sa is the reflection as in 4.1(4). Consider the direct sum

V = ®L(A)µ+na. nEZ

The subalgebra of U generated by E, F, and K 1 (isomorphic to Uq. (512)) stabilizes V, and the actions of E. and Fa on V are locally nilpotent. We can apply Lemma 5.8 to V, since we have for all n e Z L(A)µ+n,, = V,,,,

with

m = 2(µ'a) + 2n. (a, a)

5. REPRESENTATIONS OF Uq(9)

76

In particular, if we set r = 2(µ, a)/(a, a), then V,. = L(A)µ

and

V_,. =

So the equality dim V,. = dim V_, from 5.8 implies that sa(µ) is a weight of L(x). REMARKS. 1) The same proof shows also for each finite dimensional U-module

M that dim MN = dim M,,,(N,)

(1)

for all µ E A and all w in the Weyl group of P. 2) One can deduce from this proposition that dim U/I < oo for any left ideal I C U as in Lemma 5.7.

5.10. THEOREM. For each dominant A E A the simple U-module L(A) has finite dimension. Each finite dimensional simple U-module is isomorphic to exactly one L(\) with A E A dominant. PROOF. The first claim follows from Proposition 5.9, since L(7) is a homomorphic image of L(A). The second claim follows from 5.5.

5.11. PROPOSITION. Let u E U. If u annihilates all finite dimensional Umodules, then u = 0. PROOF. Consider for all dominant A E A the module L(A) as in Proposition 5.9. It is finite dimensional; it is generated by an element vA E L(A)A. The map y H yvA

takes any U7, surjectively onto L(A)A_,,. If v = EaEn maa and A = EaEII nawa with ma < na for all a E II, then this map U:, --f L(X)A_ is bijective, since the map U -4 M(.),\_ is bijective for all v, cf. 5.5(4), and since L(\)'\_ = M(A)A_ for these v by construction of L(.\). Consider for all dominant A E Aalso the module WL(7). This is L(A) twisted

by the automorphism w, i.e., it is L(a) as a vector space, but with any z E U acting as w(z) does on L(\). Denote vA by v' when considered as an element of WL(.\). We have then KN,v' = q-(",A)v' for all p. E M, i.e., v' E ("L(X))_\. Furthermore, we have Fav' = 0 for all a E II, and x --> xv, maps each U+ onto (WL(x))_A+,,. This map is bijective if ma < na for all a E II where v = EaEf maa and A = EaEII nawa. Each tensor product L(A) ®' 'L(A) with dominant A, A' E A is a finite dimen-

sional U-module. Suppose that u E U annihilates all these tensor products, in particular that u(vA (9 vi,) = 0 for all these A, A'. We want to show that this implies u = 0.

Choose bases (xi)i of U+ and (yj)j of U- consisting of weight vectors, say xi E U Vii) and yj E U :,,(j) with v(i) and v'(j) in M, > 0. Write

u = jEµEZ'' E

Eaj.,..,iyjKpxi

i

with aj,µ,i E k, almost all equal to 0. Suppose that u 54 0. Let vc E M be maximal among the weights v such that there exist i, µ, j with aj,µ,i 54 0 and v = v(i).

5. REPRESENTATIONS OF Ug(p)

77

Each 0(xi) is equal to Kv(i) ® xi plus a sum of terms in U> °U° 0 U+; these extra terms annihilate any vA ® v', as above. So we have xi(vA ® vi,) = q(v(i).a)vA ® xiva,.

Of course 0(Kµ) = Kµ ® K. implies Kµxi(vA (3 va,) =

q(v(i).A)+(µ.a-A'+v(i))va ®xiva,.

Each 0(yj) is equal to yj ®K-(j) plus a sum of terms in U- ®U°U
that xiV" + (*)

yjKµxi(va (D va,) =

(1)

where (*) is a sum of terms from certain L(A) ® (WL(A'))_A,+v with v < v(i). The maximality of v° implies that yj Kµxi (va ®va,) has a component in L(A) ® (' L(A'))-a,+vo only for v(i) = v°; if so, then the component is the right hand side in (1) without the (*) terms. So the projection of u(vA(Dva,) onto L(a)®(WL(A')) _a,+ is equal to

®xiva,.

(2)

j.µ,i:v(i)=v0

Since we assume that u(va (D v',) = 0, also this projection is equal to 0. We can find an integer N > 0 such that

v°<1: Na

and

v'(j)<1: Na aEn

aEII

for all a E II and all j such that there are a p and an i with aj,µ i

0 and v(i) = v°.

Set

all aEIII. aEn

The remarks in the second paragraph of the proof imply for all A' E A+ that

the map Ui -p (WL(A'))a,_vo, x , xv', is bijective. So the xiv', with v(i) = v° are linearly independent. Therefore the vanishing of the sum in (2) implies (for all A' E AN) Eaj.µ.iq(vo,A)+(µ,a-A'+vo)-(v'(j).-a'+vo)yjva

=0

(3)

for all i with v(i) = v°. The remarks in the first paragraph of the proof imply for all A E AN that all yjva that occur in (3) with a nonzero aj,µ,z are linearly independent. So we get from (3) E aj,µ.

(vo.a)+(µ.a-A'+vo)-(v'(j).-A'+vo) = 0

(4)

µEZ4)

for all i, j with v(i) = v°. We can cancel the (nonzero) factor in (4) that does not depend on u and get

E µEZ-f-

aj.µ.iq(µ,vo-a') q(,U,A) = 0

(5)

5. REPRESENTATIONS OF Uq(g)

78

for all i, j with v(i) = vo and all A, A' E A. Fix for the moment A' and regard the left hand side in (5) as a linear combination of the function A -4 q(µ°A) on A. These functions are distinct characters on the semigroup AN (since q is not a root of unity). Now Artin's theorem on the linear independence of characters (quoted already in 2.17) works equally well for semigroups. So the coefficients aj,N,iq(N'"0-A')

in (5) are all equal to 0. This implies that aj.µ.i = 0 for all i, j, and µ with v(i) = vo, contradicting the choice of vo. Therefore u = 0. REMARK. The proof of an analogous statement (in a more general setting) in [L],3.5.4 is rather sketchy. Probably Lusztig had in mind something along the lines of the proof given here (which is a variation of the argument used in 4.16).

5.12. Assume in the next subsections (until 5.14) that k = Q(v) and q = v, where v is an indeterminate over Q. Set A = Q[v, v-1]; this is a principal ideal domain (a localization of the polynomial ring Q[v]) with field of fractions Q(v) = k. Fix a dominant weight A E A, and set V equal to L(A) or to L(A). Denote the image of the generator va E M(A) again by VA. We have V = U-va, so V is spanned over k by all Fjva with FI as in 4.12. There are only finitely many I with Fjva 0, since the finite dimensional vector space V has only finitely many weights, since

Fjva E V\-wt 1, and since there are only finitely many I with wt(I) = µ for any given p. Set

VA = >2 AFIva

and

AF1va

Vµ,A =

I

(1)

Wt I=A-,u

(for all ,a E A). The remarks above show that VA and all Vµ,A are finitely generated

A-modules. They are obviously torsion free, hence free of finite rank over the principal ideal domain A. By construction VA = E,,, Vµ,A, hence VA = ® Vµ, A.

(2)

µEA

We claim now for each µ that the natural map

VµA®Ak-->V,u, m®c'-4cm

(3)

is an isomorphism. It is surjective, since Vµ is spanned by the Ftva with wt(I) _ A - µ. The injectivity is a general fact. (If A is any integral domain with field of fractions k, if V is a vector space over k and M an A-submodule of V, then the natural map M ®A k -4V is injective, cf. [Bourbaki 1], §7, n° 10, prop. 26.) The bijectivity of the map in (3) implies that a basis of Vµ,A over A is also a basis of V. over k, in particular rkAV,u,A = dimk Vµ

for all µ E A.

(4)

LEMMA. The A-module VA is stable under all E(,, F(,, Kim, Kg' , and [Ku; a]

withc EIIandaEZ.

5. REPRESENTATIONS OF Uq(g)

79

PROOF. This is clear for the Fa since

FFFlva = F(a,I)va.

(5)

It holds for Ka L, since we have for all µ E A and all in E V, ,A

K, Lm = q-(µ,«)m

and

Kim = q6µ-)M

(6)

and q, q-' E A. This yields also (for all a E Z and all µ, m as above) [K«; a]m =

K.q.a qa

- K- lq« 1

m, = [a + s],,m,

(7)

q«

where s = 2(µ,a)/(a, a). We have observed in 0.1 that [a + r] E Z[v, v-L]; this implies that (a+ r],,, E

Z[vd°,v-d°]

Q[v,v-L]

C

hence [K,,; a] m, E VA. Finally, if I = (01,32,... , /3n),

= A,

then

Fpl F02 ... Fp,-i [K«; 0]F,3,-, ... F0.va

EEFIvA =

(8)

By the cases dealt with already (Fp, [Ku; 0]) we know that the right hand side is in VA.

5.13. There is a unique homomorphism of Q-algebras cp : A = Q[v, v-I] -> C with W(v) = 1. Consider

V =VA®AC

and

Vµ =Vµ,AOA C

(1)

(for all y E A) where we regard C as an A-module via W. We have then

V=®Vµ

(2)

,UEA

and each V,,, is a vector space over C with dims VN, = rkAV,2,A = dimk V.

(3)

For each a E II the actions of Ea, Fa, Ka, and [Ku; 0] on VA yield endomorphisms

of V that we denote by e,,,, f, k, and h,,,. Recall from the introductory remarks to Chapter 4 that Serre has described the Lie algebra g by generators and relations, cf. [H], 18.3.

LEMMA. The endomorphisms ea, f, and ha of V satisfy Serre's relations. Each kc. is the identity map.

5. REPRESENTATIONS OF Uq(g)

80

PROOF. We get from 5.12(6) for all a and u

k«(m 0 1) = (K«m) 0 1=

(4(11,0)m)

0 1= m® p(4(µ'«)) = m® 1

for all m E Vµ,A. This shows that k« is the identity. We have cp([a]«) _ [a],;=I = a for all a E Z, cf. 0.1, so 5.12(7) implies

h«m =

a) m (a, a)

for all m E V. and all p, a.

(4)

Therefore the h« can be diagonalized simultaneously. We see in particular that [ha, h«] = 0 for all a and ,0; that is the first group of Serre's relations. The relation (R4) implies easily [e«, f81 = 6«,3h«; that is the second group of Serre's relations. We have for all a and p f«Vµ C Vµ_«,

and

e«Vµ C Vµ+«

(5)

since 5.1(2) implies E«Vµ C Vµ+«, hence E«VV,A C Vµ+«,A (similarly for F«). Now

(5) and (4) imply easily that and

ep

[h«, ep]

[h«, .fp] _ - ((a 'a)) fit

(6)

for all a,,3 E II; that is the third group of Serre's relations. Finally, we get from (R5) and (R6) for all a, 0 E II, a # /3 (setting a«o = 2(a, (3)/(a, a)) I-nay

S=o

I-aaA s

=0

a«3

l

s

ea

f« aaa-Sfi

a«p

s

aaa-Sepea

)

=0

(7)

J

/

fs

(8)

since we have for all a, n E Z with n > 0

{:]= [:]v1= (:). As pointed out in the introductory remarks to Chapter 4, the equations (7) and (8) are equivalent to the last two groups of Serre's relations.

5.14. The lemma implies by Serre's theorem that there is a homomorphism of Lie algebras from g to gl(V) that takes each x, to e«, and y« to f«, and h« (in g) to h« (the endomorphism of V). This makes V into a g-module.

_

LEMMA. Considered as a g-module, V is simple with highest weight A. Each Vµ is the p weight space of V.

5. REPRESENTATIONS OF Uq(g)

81

PROOF. Each it E A is determined by the ((a «)) , since it = LJaEn ALL «)) Therefore 5.13(4) implies

VN.={mEV I ham= 2(µ'a) m for allaEll}.

(1)

(a ,a)

This proves the second part of the lemma. We have clearly

ea(va®1)=0

for allaEll,

(2)

and

(Flea) ®1 = fa, ff2 ... far (va ® 1),

if I = ((31,(32, ...

, 00-

(3)

This shows (since va ®1 E VA) that the g-module V is standard cyclic in the sense of [H], 20.2. In particular, it has a unique simple quotient. On the other hand, it is finite dimensional, hence a semi-simple g-module by Weyl's theorem. So it has to be simple, hence the simple module with highest weight A. (Cf. also the proof of Theorem 21.4 in [H].)

REMARK. I have worked in the last two subsections with V = VA ®A C, so to apply Serre's theorem and the representation theory of complex semi-simple Lie algebras. It would be more natural to work instead with VA ®A Q and with split semi-simple Lie algebras over Q, cf. [Jacobson], chapter IV. Then one has to know that the representation theory of these Lie algebras is "the same" as for complex semi-simple Lie algebras (that can be found in [Jacobson], chapters VII and VIII) and that Serre's theorem extends (Exercise!). This stuff can also be found in [Bourbaki 3], chap. 8.

5.15. THEOREM. Suppose that char(k) = 0 and that q is transcendental over Q. Let A E A be a dominant weight. We have then L(A) = L(A), and the dimensions of the weight spaces L(A)N, are given by Weyl's character formula. PROOF. Let us consider first the case where k = Q(v) and q = v as in 5.12-14.

We can carry out the construction in 5.12 with V = L(A) and with V = L(A). We get by Lemma 5.14 in both cases for V the simple g-module with highest weight A. So 5.13(3) and Weyl's theorem for 9-modules ([H], 24.3) imply that the dimensions of the weight spaces in L(A) and in L(A) are given by Weyl's formula. We see in particular that L(A) and L(A) have the same dimension. Since L(A) is a homomorphic image of L(A), they have to be equal.

Return now to the general case. There is a unique homomorphism of Qalgebras Q[v, v-1] that takes v to q. Since q is transcendental over Q, it induces an embedding of Q(v) into k such that v is identified with q. I now want to consider objects such as U both over k and Q(v); so let me add an index (i.e., write Uk or UQ( )) to distinguish them. We can identify Uk with UQ(,,,) k; that is a general fact on algebras with generators and relations. (Since I could not find a suitable reference, I sketch the proof in the appendix.) This induces an identification of Uk with UQ(V)

k. (This follows also from 4.21.b and general

principles.) The universal property of M(µ)k yields for any it a homomorphism M(µ)k -> ®Q(,,.) k; this is an isomorphism, since the first module can

5. REPRESENTATIONS OF Uq(9)

82

be identified with U and the second one with UQ(v) ®Q(v) k. The definition of L(A)Q(v) says that there is an exact sequence

®M(\ - (n(a) + 1)a)Q(v) CEn

-f M(A)Q(v) --y L(A)Q(v) -> 0

(1)

with the n(a) as in 5.9. This sequence remains exact when we tensor over Q(v) with k. We get thus (modulo our identifications) an exact sequence

®M(A - (n(a) + 1)a)k -> M(\)k -> L(A)Q(v) ®Q(v) k -> 0.

(2)

cEn

On the other hand, there is an exact sequence

® M(A - (n(a) + 1)a)k -> M(A)k -> L(A)S -> 0

(3)

CEn

analogous to (1). The first map in (2) is the same as the first map in (3): both take each v,\-(a(r)+1)c, to Fn,(c')+1 v,. Therefore we get

L(A)k ^' L(A)Q(v) ®Q(v) k.

(4)

Now the theorem will follow, if we can show that also L(.)k ^' L(A)Q(v) ®Q(v) k.

(5)

We have to show that L(A)Q(v) ®Q(v) k is simple as a module for Uk - UQ(v) ®Q(v) k.

Since it is a homomorphic image of M(A)Q(v) ®Q(v) k ^J M(A)k, it then has to be isomorphic to L(\)k. This simplicity is a special case of the following general fact: Let A be an associative algebra over a field F, and let M be a finite dimensional

simple A-module with EndA(M) = F. Let k be an extension field of F. Then M OF k is a simple (A OF k)-module. [Proof: By Wedderburn's theorem the natural map A -> EndF(M) is surjective. It remains surjective after tensoring with k, i.e., A OF k -> Endk(M OF k) is surjective. The simplicity follows.] This fact can be applied by 5.5(6).

REMARK. This theorem can be generalized to arbitrary k and q as long as q is not a root of unity. This is proved in [Andersen, Polo, & Wen], Corollary 7.7. Alternatively, compare the discussion in 6.26. 5.16. Let wo be the unique element in the Weyl group of 4D with wo(II) = -II, cf. [H], exercise 10.9.

PROPOSITION. If A E A is dominant, then L(A)"` ^J L(-woA). PROOF. Do exercise 21.6 in [H]!

5.17. THEOREM. Suppose that char(k) = 0 and that q is transcendental over Q. Then every finite dimensional U-module is semisimple.

5. REPRESENTATIONS OF Uq(9)

83

PROOF. It is enough to show that an extension of two simple finite dimensional U-modules splits. (Cf. the argument at the beginning of the proof of Weyl's theorem in [H], 6.3.) So let A, µ E A be dominant weights and consider a short exact sequence

0->L(A)---'-4 M--7--4 L(µ)->0.

(1)

Pick a vector 'V E L(µ)µ, U# 0. Since M is the direct sum of its weight spaces, we have for all v a short exact sequence of vector spaces 0 -> L(A),.

My

L(µ)v -> 0.

(2)

In particular, we can find an element v E Mµ with 7r(v) = v. If we have E,,v = 0 for all a E II, then Uv is a homomorphic image of M(A). Since dim Uv < 00, the last part of Lemma 5.6.b implies that Uv is a homomorphic image of L(µ). So we get Uv a- L(µ) by Theorem 5.15. We have 7r(Uv) = Uir(v) = Uv = L(µ), hence Uv fl ker(ir) = 0 and M = Uv ® ker(7r). So (1) splits.

If there is an a E II with E,,v # 0, then µ + a is a weight of M. It is not a weight of L(µ), hence it is a weight of L(A). In particular, we get µ < A. Therefore the proof above shows: If µ 54 A, then (1) splits. Suppose now that µ < A. If we dualize (1), then we get an exact sequence

0->L(µ)*-M*-*L(A)*->0,

(3)

hence by 5.16 an exact sequence

0 -> L(-woµ)

M* -* L(-woA) -> 0.

(4)

Now µ < A implies -woµ < -woA (since -wo(II) = II), hence -woA A -woµ. Therefore we can apply the argument above to (4). We get

M* a, L(-woµ) ® L(-woA) ,., L(µ)* ® L(A)*, hence M a- (M*)* ^ L(µ) ® L(A).

5.18. LEMMA. Suppose that char(k) = 0 and that q is transcendental over Q. Let A = EaErl l(a)w,, E A be a dominant weight. Choose vA E L(A)A, vA # 0. Let

µ E DD, µ = E.En m,,a with 0 < ma < l(a) for all a. Then the map U_µ -* L(A)A_µ,

u --> uva

(1)

is bijective. For any basis u1i u2, ... , u,. of Uµ there is a basis u', u2. ... . u'r of

U=µ with uiujva = Sijva for all i and j.

PROOF. Let I- be the left ideal in U- generated by all F,1,(")+'. If -v is a weight of I- and if we write v = E.EI-I n,,a, then there is at least one a E II with nQ > l(a). This implies I- fl U=µ = O

for all a as in (1).

(2)

Since the map u --> uva from 5.5(4) induces a bijection from U-/1- to Z(A) = L(a), this implies that the map in (1) is bijective.

5. REPRESENTATIONS OF Uq(g)

84

To prove the second claim, let us begin with an arbitrary basis z1i z2, ... ,Zr of UT , for example with the w(ui). Each uizjv,\ has weight A, so there is an aij E k with uizjv,\ = aijva. We claim that the matrix of the aij is invertible. If not, then its nullspace is not equal to 0, so w e can find c1, c2, ... ,Cr E k, not all equal to 0,

with >j aijcj = 0 for all i. Then the element z = >j cjzj E U=µ is nonzero and satisfies

uizv,\ = E cjuizjv'\ = E cjaijv,\ = 0

for all i.

This implies

0 = Uµ zv,\ = (U+zv,\) fl L(.\), hence

U+zv,\ C ®L(.\),,,

Uzv,\ C ®L(.\),,.

hence

v
v
So Uzv,\ is a proper submodule of the simple module L(A), hence zva = 0. But z was nonzero, and the map in (1) is bijective. So we have reached a contradiction. So the matrix of the aij has an inverse matrix, say bid. Set now uj = >h bhjzh; we get then for all i

uiujvA _ E bhjuizhv,, _ h

6hja'ihv,\ = bijva h

as desired.

5.19. PROPOSITION. Suppose that char(k) = 0 and that q is transcendental over Q.

a) We have for allpEZID dim U+ = dim U-

P(p),

where P is Kostant's partition function. b) Let it E M, it > 0. If u E U=µ with [E,,, u] = 0 for all c E II, then u = 0. If u E Uµ with [F,,, u] = 0 for all c E H, then u = 0.

PROOF. Let µ E M. We may assume that µ _> 0. Choose a dominant weight .\ such that the assumption of 5.18(1) is satisfied. We get then dim U=µ = dim L(A)A_. By 5.15 this dimension is equal to the dimension of the (a-µ) weight space in the simple g-module with highest weight A. Under our assumption on µ and ), that dimension over C is equal to P(µ), e.g., by Kostant's formula. (Actually, it is more appropriate to note that Lemma 5.18 holds also for g-modules, if we replace U- by the enveloping algebra denoted by fl(N-) in [H]. Then use exercise 20.5 ibid.)

Suppose now that t > 0 and that we have u E UJ as in b). We have then with va as in 5.19(1)

UuvA = U-U°U+uvA = U-U°uU+vA = U-U°uv,\ = U uvA C

® L(A),.

(3)

v
So UuvA is a proper submodule of the simple module L(A), hence equal to 0. Therefore Lemma 5.18 implies u = 0. We get the results on U+ by applying w.

APPENDIX

85

REMARK. The results in 5.17-19 generalize to the general case (as long as q is not a root of unity), cf. the remark in 5.15 or 6.26.

Appendix Theorem 5.15:

I want to show here that extension of scalars is compatible with the construction of algebras by generators and relations Let R be a commutative ring and R' D R be an extension ring, again commutative and with the same 1. Let A be the R-algebra with generators xi, x2i ... , xn and relations zI, Z2.... Z. So the zj are elements in the free R-algebra Fn (R) with n generators. The inclusion of R into R' induces an inclusion of Fn(R) into the free R'-algebra Fn(R') with n generators. So we can regard the zj also as elements in F,, (R), and we can consider the R'-algebra A' with generators yi, y2, , yn and relations zI, Z2.... zm. The elements 1®xi in the R'-algebra R'®RA satisfy the relations zI, Z2.... zm. So there is a unique homomorphism of R'-algebras A' R' ®R A with yj -+ 10 xi for all i. On the other hand, we can regard A' by restriction of scalars as an RA' with algebra and have therefore a unique homomorphism of R-algebras A xi - yi for all i. Now the universal property of the tensor product yields a (unique) A' with 1 ® xi H yi for all i. Now it is homomorphism of R'-algebras R' OR A obvious that the two maps are inverse isomorphisms between R' ®R A and A'. This argument is similar to the proof in Bourbaki, Algebre III, (nouv. ed. 1971) §5, Prop. 5 that extension of scalars commutes with taking tensor algebras. In the special case of a free module this means that R' OR F,,(R) is isomorphic to F,, (R'). In the situation considered above, we could observe alternatively that A is given by an exact sequence F,,, (R) - Fn (R) - A - 0 and that this sequence remains exact under tensoring with R'.

CHAPTER 5A

Examples of Representations The goal of this supplement to Chapter 5 is to describe explicitly examples of finite dimensional representations of U = Uq(g). We keep the assumptions and notations from Chapters 4 and 5. We shall use the notations (A, av) and sa as in 4.1(3)-(4). 5A.1 (The Minuscule Case). Let A E A be a minuscule dominant weight, i.e., by definition (in [Bourbaki2], ch. VI, §1, exerc. 24) a dominant weight A # 0 with

(A, av) E {0, 1, -1} for all a E 4i. One has then also (µ, aV) E {0, 1, -1} for all µ E WA and a E 4i. Furthermore, there is no dominant weight v E A with v < A. So a minuscule dominant weight is the same thing as what Humphreys calls a nonzero minimal dominant weight in [H], §13, exerc. 13. That exercise contains also an explicit list of all minuscule dominant weights for 4i indecomposable. The weights of the simple g-module with highest weight A are precisely the conjugates of A under the Weyl group W each occurring with multiplicity 1. That follows from the basic properties of a minuscule weight (stated above) and with [H], Theorem 21.2. The same holds for the simple U-module with highest weight A: Use 5.9(1) and the definition of a highest weight module. However, we want to construct this module without any of the theory of Chapter 5. Let us consider a vector space L with a basis (xµ)µEWA indexed by the conjugates of A under the Weyl group W. We now define an action of the generators of U (from 4.3) on L as follows: We set for all a E II and all µ E WA

Kax, = q(µ'a)x,

and

K.-1Xµ = q-(A.a)x,

(1)

and

Eaxµ = 0,

Fax, = 0,

if (µ, av) = 0;

Eaxµ = 0,

Fax, = xµ-a,

if (µ, av) = 1;

Eax,=x,+a,

Fax,=0,

if (µ,av)=-1.

(2)

Note first that these definitions make sense: We have µ - a = saµ E WA if

(,u,av)=1,andp+a= saµEWAif(,u,av)=-1.

Next we claim that (1) and (2) make L into a U-module. So we have to check that the endomorphisms of L defined by these formulas satisfy the relations (RI)(R6) from 4.3. This is obvious for (R1) and trivial for (R2) and (R3). Actually, we could replace (1) by

for all tEWAandvEZ4i. 87

(3)

5A. EXAMPLES OF REPRESENTATIONS

88

We see then that each kxµ is equal to the weight space L. So (2) implies in particular EaLµ C L. and FaLµ C Lµ_a for all a E H and all p E WA; this makes (R2) and (R3) obvious. Note that (1) also says

Kaxµ = qa(µ.av) xµ.

(4)

Using this (R4) becomes obvious in the case a = 0. Consider now two simple roots a and 0 with a

and let us look at (R4) in this case as well as at (R5) and (R6). Suppose first that (a,,3) = 0. We have to show for all p E WA that EaFpxµ = FpEaxµ [R4], that EaE,3xµ = E,3Eaxµ [R5], and that FaFpxµ = F,3Faxµ [R6]. Well, we have EaFpxµ = xµ_a+a if (µ (3v) = 1 and (p - 0, av) _ (µ, av) = -1; otherwise EaFpxµ = 0. Similarly, we have FpEaxµ = xµ+(,_,3 if (,a, av) = -1 and (p + (3v) = 1; otherwise FpE(,,xµ = 0. So we have indeed EQ,Fpxµ = FpEaxµ. The argument for the other two cases is similar.

Suppose now that r = -((3,av) > 0. We claim first that EaFpxµ = 0 = FpEaxµ for all p. Indeed, if Fpxµ 0, then (p - 0, av) = (µ, av) + r > r - 1 > 0 implies 0 = Eaxµ_p = EQ,Fpxµ. Similarly, (,a+ /3v) _< 1 implies FpEaxµ = 0. So we get (R4) in this case. Finally look at (R5) and (R6). We claim that all summands in (R5) and (R6) act as 0 on L. By (2) obviously all E,2,, and Fa annihilate L. So the only summand in (R5) resp. in (R6) that could be nonzero on L is EQ,E,3EQ resp. FQ,FpFQ. It can occur only if r = 1. We have then for all

/EWA

(µ+a+,6,av)=(µ,av)+1>0, hence EaEpEaxµ = 0. Similarly (p - a - 0, av) = (µ, v) - 1 < 0 yields FaF,3Faxµ = 0. So indeed all summands in (R5) and (R6) annihilate L. This concludes the proof that (1) and (2) define a structure as a U-module on L. We have already observed that the weight spaces of L are the Lµ = kxµ. In particular, our A is the largest weight of L. Using 5.9(1) or a direct argument one checks easily that L is a simple U-module, hence isomorphic to the simple module L(A).

REMARK. One of the assumptions carried over from Chapter 5 is: "q is not a root of unity". You may have noticed that this assumption is not needed for the construction above of the U-module structure on L. (Actually, we mentioned weight

spaces that we did not define in general. But that can easily be modified.) Also the constructions in the next subsections can be generalized to general q. However, the modules constructed may very well be reducible for q a root of unity. (The L in this subsection is in fact simple for all q. This is not completely obvious: One has to show that distinct weights in WA still define distinct characters of U° even for q a root of unity.)

5A.2 (The Case a°). The modules constructed in 5A.1 include the quantum analogues of the "natural" representations for g of type A,, C,, and D,, as well as the quantum analogues of the "spin" or "half spin" representations for g of type B, or D,. What is missing is the analogue of the "natural" representation for g of type B,. That will be a special case of the construction in the present subsection. Assume in this subsection that ' is indecomposable. Denote by 4is C 4i the set of all short roots; if 4i has only one root length, then 4is = 4i. All roots of the

5A. EXAMPLES OF REPRESENTATIONS

89

same length are conjugate under the Weyl group. So (Ds is one orbit under W and contains exactly one dominant weight. Denote this dominant weight by ao; it is obviously the largest short root. The simple g-module with highest weight ao is the adjoint representation if there is only one root length. For g of type B,, it is the "natural" module, for g of type G2 resp. F4 is the the "smallest" module (of dimension 7 resp. 26). The weights of this g-module are the short roots (with multiplicity 1) and 0. The dimension of the 0 weight space is equal to the number I113I of short simple roots (setting [I = II n (P,,). Let us now construct a quantum analogue: Consider a vector space L over k with basis

(x7I'YE,Ps,

EE II,,).

h,3

We first turn L into a U°-module via q(,,v)x7

and

h;3

(1)

for all v E Z (D, y E 4 , and 0 E 11,. So L is a direct sum of weight spaces; the weight spaces are L° = E,3ErI, kh;3 and the L7 = kx7 with y E (D3.

We next want to define the action of all Ea and Fa on L. First of all, set for all aE11 and all -y c- 4(P, with -y 0 ±a

Eax-, = 0,

Fax, = 0,

Eax7 = 0, Eax7 = x7+a,

Fx,. = x7-a,

if (y, aa') = 0; if (-', aV) = 1; if ('Y, av) = -1.

Fax, = 0,

(2)

Note that the classification of rank 2 root system shows: If a and y are roots with -y short and -y 0 ±a, then (-y, av) E {0, 1, -1}. Therefore we have covered in (2) all possible cases. Note also that (as in 5A.1(2)) the x,fa that occur make sense since they are equal to X5.-Y,

It remains to deal with the case -y = ±a and the hp. For long (better: "not short") a set

Eah,3=0=Fah3 Fi na lly set for a ll a c-

11

for all 13E11sincase c

(3)

,

Eaxa = 0, Faxes = ha,

Eaha = [2]axa, Faha = [2]ax-a,

Eax-a = ha, Fax-a = 0,

(4)

and for a,/3E[I with a00 xa, 1 0,

if (0, av) _ otherwise;

1;

Faha

x-a, if (Q, a') _ -1; otherwise. = jl 0,

(5)

(Note that the condition (0, av) _ -1 in (5) is equivalent to (0, a) 0 since a and Q have the same length.) We now have to check the conditions (Rl)-(R6) from 4.3. Again (Rl) is obvious by (1). Furthermore, the definitions (2)-(5) show that EaL,,,, C L,,,,+a and FL,, c

5A. EXAMPLES OF REPRESENTATIONS

90

Lµ-« for all a E II and all weights µ of L. This implies easily that (R2) and (R3) are satisfied. The case a = /3 in (R4) is an easy exercise left to you. So let us now consider a, 0 E II with a : /3 and let us check (R4)-(R6) for these two roots. We deal first with (R4). We have to show that F«Epxy = EOF«xy

and

F«E,3hry = E,3F«h.y

(6)

for all -y E % resp. for all y E r l,. The second claim is very easy since F«Ephy and E3F«hy are contained in Lp_« and since Q - a U {0} for two distinct simple roots a and /3, hence L,3_« = 0. Let us now check the first equality in (6). We have v

F«E$xy =

F,,, h,3,

0,

if 'Y = -l3, otherwise,

xy+3-«, if (-Y" 3v) = -1 and (y + Q, av) = 1, X_«, if y = -Q, (/3, v) _ -1, and a short, 0,

otherwise,

Epxry-«,

if (y, av) = 1, if y = a,

and

E,3F«xy =

Ei3h«,

otherwise,

1 0,

-1,

Xy-«+p,

if ('Y, aV) = 1 and (-y -

x,3i

if y = a, (,,,3v) _ -1, and /3 short,

0,

otherwise.

,,3v)

So both F«E,3xy and E;3F«xy are either 0 or equal to the first equality in (6), we have to show F«E;3xy 54 0

E,3F«xy 54 0.

Well, the formulas above show for y = a E,3 Fxy

0

(a, Qv) _ -1 and /3 short (a,Nv) = = (Q,av) (-y, ,3v) _ -1, ('Y + Q, a") F«E3xy54 0

and for y = -/3 F«Ei3xy

0

(Q, aV) _ -1 and a short

(Q,av) _ -1 = (a,,3v)

('Y,aV) =1, ('Y-,/3v) _-1 E;3F«xy

0.

In order to prove

5A. EXAMPLES OF REPRESENTATIONS

Finally, for -y

91

a, -/3 we note that (-y, av) < 1 and (-y, /3) > -1, hence that

(-y+/3,av) = 1

(-y, a') = 1 and (/3,a) = 0

and

(-y - a, /3v) = -1

(-y,,3') _ -1 and (/3, a) = 0.

So we get from the earlier conditions E,3F«x-,

(-y, a') = 1, (Y, Qv) =

0.

1, (Q, a) = 0

FEax.y¢0. This concludes the proof of (6), i.e., of (R4).

Let us now turn to (R5). Consider first the case where (/3, a) = 0. We then have to show E,,E,3x-y = E,3E,,x-y

and

E.E,3h.y = EOEah-y

(7)

for all -y E % resp. for all -y E lls. The second claim is again very easy since E,,E,3h-y and E,3E,,h-, are contained in L,3+,, and since /3 + a 4D U {0} for two perpendicular simple roots a and /3, hence L,3+a = 0. Now for the first claim: We have

EQE,3x

Eax-,+,3,

if (-y,Qv) = -1,

E.h,3,

if -Y = -,3,

0,

otherwise,

= J x-,+o+a, if ('Y, /3v) = -1 = ('Y, av), 0,

otherwise,

since E,,,h,3 = 0 for (/3, a) = 0 and since -y +,3 -a. [Otherwise a +,3 = --y E D, which is impossible as observed above.] Now the formula for EQE$x.y is symmetric in a and /3, so we get indeed the first claim in (7).

Suppose now that r = -(/3,av) > 0. Each summand in (R5) has weight (r + 1)a + /3, hence maps any L,y to L.y+(r+l)a+a So, if the action is nonzero on L.y, than both -y and -y' = -y + (r + 1)a +,3 are weights of L. We have then

(-y' --y,v) = ((r + 1)a+/3,v) = r+2 > 3. On the other hand, we know that (-y', av) < 2 (with equality if and only if -y' _ a) and that (-y, av) > -2 (with equality if and only if -y = -a). This yields (-y' - -y, av) < 4 with equality only for -y' - -y = 2a (r + 1)a +,3. So the only remaining possibilities are r = 1 and y' = a [hence -y = -(a + /3)] or -y = -a. So we have to check for r = 1 and these two values of -y that E",E,3x, - [2].E«E,3EQx.y + E,3Eaxy = 0.

(8)

Note that we may assume a short since either -y' = a or -y = -a is a weight of L. Now (/3, av) = -r = -1 implies that also /3 has to be short, hence that (,,,3v) = -1. Now it is easy (and left to you) to check (8) for our two specific values of -y.

5A. EXAMPLES OF REPRESENTATIONS

92

We have now shown (R5) in all cases. We could proceed in the same way with (R6). However, it is easier to observe that the linear map o : L -* L with o (x.y) = x_.y for all -y E -Ds and with o-(hry) = h.y for all -y E II5 is an involution satisfying o o Fa. = Ea o o for all a E H. Using this equation it is easy to deduce (R6) from (R5).

So we have shown that (1)-(5) define a U-module structure on L. It is clear that ao is the highest weight of L. Under the assumptions of Theorem 5.15 we can deduce from that theorem that L ^ L(ao). However, it is not difficult to show for arbitrary q (not a root of unity) that L is simple, hence that L ^- L(ao). 5A.3 (The Adjoint Case for B2). Before we look at the quantum analogue of the adjoint representation in general, it turnes out to be useful to study first the rank 2 cases. Now the construction in 5A.2 covers the case where I is of type A2. The case AI x AI is a trivial consequence of rank 1 computations. We shall look now at B2 and we shall deal with G2 in the next subsection. Suppose that (D is of type B2. Set II = {a, ,3} with a long and 0 short. So we have (a, ,Q") = -2 and ((3, a") = -1 and (D+ = {a, ,3, a +,3, a + 20}. The largest root is a = a+ 2,3, and we want to describe L = L(a). We are going to use parts of the theory developed in Chapter 5 (something that we have avoided in the previous subsections). So we know by 5.10 that L has finite dimension and 5.9(1) implies that each weight of L is conjugate under W to a dominant weight < a, hence contained in (D U {0}. Furthermore, we have dim L. = 1 for all -y E WE = {±(a + 20), ±a}. Pick an arbitrary basis vector xa+23 of La+23. (In a moment we shall choose also basis vectors x, of L7 for the remaining long roots.) We have for all v E A Lv = U -(a+23)Lck+23 = FQLv+a + F,3L,+3

(1)

This yields in particular La+3 = F3La+23 = k(F3X.+23),

since La+3+a = 0. Now rank 1 calculations (as in Chapter 2) together with (a + 20,3v) = 2 show that E3F3xa+23 = [2],3xa+2,3, hence that F3xa+23 0 0. (Note that [2]3 = q3+qp' 0 0 since q is not a root of unity.) This shows that dim La+3 = 1, hence that also dim L. = 1 for all -y E W (a + L3) = {±(a + 3), ±8}. We shall choose a basis vector xry for each L7 also for each short root y. We begin by setting Xa+i3 = F,3xa+23,

xa =

1

F3xa+O-

(2)

E,3xa+3 = [2]3 xa+23

(3)

[2]3

Rank 1 calculations show

E3xa = xa+3,

(It is clear by 5.1 that the x.y defined in (2) are in Lry; it then follows from (3) that x_, 0, hence that x7 is a basis of L.y. The same argument will work also for the remaining x.y to be defined below; we shall not repeat it.) Set

x3 = Fax.+3

(4)

5A. EXAMPLES OF REPRESENTATIONS

93

A rank I calculation together with (a + (3, a') = 1 yields E«x;3 = X.+,3-

(5)

We have

Lo = FaLa + j33,3 = k(F«x«) + k(Fax3 by (1); set

ha=Fax.,,

(6)

h'3=F,3x;3.

So ha and h,3 span Lo; we shall see in a moment that they are in fact a basis of L0. Set also [2]3 F,3h,3 (7) x_p x_,, _ Faha, [2]a

Rank I calculations show

Ex_ ,, = ha,

Eaha = [2]a xa,

E3x-3= h;3,

E,3h3 = [2];3 x,3.

(8)

We see using (R4) that E,,h,3 = E,,Fi3x3 = FoEax3 = F3xa+o = [2]3 xa

(9)

E3ha = EOF,,xa = FaE3xa = Faxa+;3 = x3

(10)

and

Note (8)-(10) imply that ha and h3 are linearly independent. [If not, then E3h3 = [2],3E3h,, would imply h,3 = [2],3ha. If we then apply E,,, we would get [2];3x,, = [2]p[2],,x,,, hence I = [2],, = qa + qa 1, a contradiction.] In order to compute Fah3 and F3h,, we use the quantum Serre relations (R6): They say in particular F,2,F3x,,+3 - [2]aF«F,3Faxa+3 + F0F2 .xa+3

=0

and F,3Faxa+2a - [3]3FQF,,F3xa+2,3 + [3]0F0F«F02xa+23

- F«F33xa+23 = 0.

If we evaluate the different terms using the formulas above we get [2]a[2]3x_,, - [2],,F,,h,3 + 0 = 0 resp.

0 - [3]3[2]0x_3 + [3]a[2]$F3h,, - 0 = 0, hence

Faho = [2]3x-a,

F,3h,, = x-3

(11)

We set finally

x-(a+o) = F,3x-a,

x-(a+2p) =

[2]a

Fi

(12)

5A. EXAMPLES OF REPRESENTATIONS

94

Rank 1 calculations yield E0x-(a+0) = [2]0 x_a.

Eax-(a+2Q) = x-(a+p),

(13)

We see using (R4) that

Eax-(a+a) = EaFox_a = F,3Eax-a = Faha = x-a;

(14)

another rank 1 calculation yields now

Fax-a = x_(a+a)

(15)

The formulas (2)-(15) describe completely the action of E, F, E and Fp on L. Any Eaxry not mentioned explicitly is equal to 0, since a + ry 4i U {0}; similarly for the other generators. We know also the action of any K. on the basis vectors, since they are all weight vectors of a known weight. So we have a complete description of the U-module structure on L.

5A.4 (The Adjoint Case for G2). Suppose now that 4i is of type G2. Set II = {a,,31 with a long and 3 short. So we have (a,,3') = -3 and (,3, a') = -1 and 4i+ _ {a, /3, a + 0, a + 20, a + 33, 2a + 3,3 1. The largest root is a = 2a + 3,3. In order to describe L = L(a). we proceed more or less as in 5A.3. At first we note that the weights of L are contained in 4i U {0}. Then we see as before that dim L7 = 1 for all ry E 4i. We pick an arbitrary basis vector X2,+3,3 of L2,,+3,3, and we shall choose for all other ry E 4i a basis vector xry of L. More precisely, we first set

xa+3,3 = Fax2a+3a

(1)

and get (rank 1! = short for "by rank 1 calculations") Eaxa+3a = X2a+3a

(2)

Then set xa+2,3 = F Xa+3p,

xa+a = [2]a

Faxes+2a,

Xa = [3]0 Faxes+a

(3)

and get (rank 1!)

Eaxa - xa+a,

Epxa+2a = [3]O X.+33-

(4)

Next we set (5)

and get (rank 1!) (6)

Now set ha = Faxes,

x_a = [2]a Faha,

ha = Faxes,

X-a =

Faho. [2]a

(7)

5A. EXAMPLES OF REPRESENTATIONS

95

We get (rank 1!)

Ex_ a = ha,

E«h« _ [2]« x.,

Eax-Q = h,3,

E,3h0_ [2]0 x,3.

(8)

Using (R4) one then checks that Ei3ha = x3.

Eah$ = [3] 0 xa,

(9)

As in the B2 case we can use (8) and (9) to show that ha and hp are linearly independent, hence a basis of Lo. These formulas imply also that Ea (ha

[2]a ha) = 0,

Ep (ha - [2]Q ha) = 0,

hence (rank 1!) that

F. (ha - [2]0 ha) = 0,

F,3 (ha

- [2]a ha) = 0.

This yields using (7)

Faha = x-a.

Fah8 = [3]ax-a,

(10)

(Note that we could have used the same type of argument to deduce 5A.3(11) instead of using the quantum Serre relations.) Now set 1

x-(,,+3) = Fax-a, and x-(a+33) =

X_(a+2$) =

I

[2],3

Fax-(a+3),

F$X_(a+23)

[3]0

(11')

and get (rank 1!) Eax-(a+33) = x_(a+20),

Eax-(a+23) = [2]Q x_(a+3),

(12)

and

(12') Eax-(a+Q) = Using (R4) for the first equation and a rank 1 calculation for the second one we get [3]ax_a.

Eax-(a+3) = x-a,

Fax-,3 = x-(a+a).

(13)

Finally, we set x-(2a+33) = Fax-(a+3,3)

(14)

Eax-(2a+33) = x-(a+30).

(15)

and get (rank 1!) Again all Eax, etc. not mentioned explicitly in (1)-(15) are 0 since their weight is not in 4) U {0}. So these formulas yield a complete description of the U-module structure on L.

5A. EXAMPLES OF REPRESENTATIONS

96

5A.5 (The Adjoint Case in General). We now want to define for arbitrary the quantum analogue of the adjoint representation. Let L be a vector space with a basis h8I QEII). (1) (x7 I -yE4, Define first an action of U° on L such that each xry has weight ry and such that each h3 has weight 0.

Let a E H. We want to define the action of E. and F. on L. Consider first a $ and -y 0 a. The a-string of roots though ry consists of all -y - is with 0 < i < m where m = (ry, a'). We then set root -y E 4 with -y + a

[i + l]ax.y-(i+l)«, if 0 < i < m, 0

Eaxy-ia

(2)

if a ' = m,

- 1( [m+l-i]ax,y_(i_J)a, if0
(3)

ifi=0.

Futhermore, we set

Fax. = h., F.h. = [2]. x-a, Eax_a = h.,

Eaha = [2],, xa,

and (of course) Fax-. = 0 and E.x. = 0. Finally, we set for all -y E II, -y Eah7 = [-(a, Yv)]7 xa,

F'«h7 = [-(a,'Yv)]7 x-.

(4)

a (5)

(If you want to compare with the description of this type of module in [Lusztig 4], 2.1, note that one has to interpret his [h] as [h]i and his [-a3i] as [-aji]j.) In order to check that (2)-(5) define a structure as a U-module we again have to check the defining relations. It is easy to restrict to the case where is indecomposable. If there is only one root length (and 4 indecomposable), then one sees easily that the construction above coincides with that from 5A.2, and we are done. Similarly, if is of type B2 or G2i then we just compare with 5A.3/4. In general, it is easy to check (R1)-(R3). Also (R4) for a =,3 is easy. So the main problem is this: Take a, 0 E II with a 0 0 and check (R4)-(R6) for this pair. Here is a rough sketch of how to proceed. (I shall leave the details to you.) Set D.3 = $ f1 (Za + Z8). This is a root system (of rank 2) and we can construct a quantized enveloping algebra U.3 working with 1).,3 instead of 4. Now to check that (R4)-(R6) hold for a,,3 means to check that L is a U.p-module (since we know already that the other relations are satisfied). For each -y E 4 the sum L[-y] = ®i i L7+ia+j3 (sum over all i, j E Z) is stable under the generators of U. It is enough to show that each L[-y] is a U.3-module. Well, for -y E 41)a3 one checks that L[ry] is direct sum of the "adjoint" representation of Ua3 (described in the previous subsections) and of 1111 - 2 copies of the trivial one dimensional representation. If 4Da3 is of type AI x A 1, then U.3 ^ Uqo (5(2) ® Uq,j (512), and the L[ry] with -y 4D.3 are tensor products of representations for the rank 1 factors (as in Chapter 2). If 41)a3 is of type A2 or B2, then L[-y] is usually one of the representations decribed in 5A.1/2. The only exception occurs for 4 of type A2 where L[-y] may be the quantum analogue of the second symmetric power of the natural representation. We shall deal with that example in the next subsection.

5A. EXAMPLES OF REPRESENTATIONS

97

5A.6. For p = sCn+1 (C) the symmetric powers of the natural module (and of its dual) are simple. These modules can be thought of as representations on spaces of homogeneous polynomials. We now want to describe their quantum analogues, again realized as representations on polynomials. Consider a polynomial ring S = k[Xl, X2i ... , Xn, X,+l] in n + 1 indeterminates X1i X2, ... , Xn, Xn+l over k (for some integer n > 1). Define for all i with 1 < i < n + 1 an endomorphism Di of S (as a vector space) by [mi]XMIX2Mz...X"',-1 ...X"'77+'

n+1

2

1

1

(1)

n+1

i

2

for all monomials X i ' X2 2 . X +i ' . It is clear that the Di commute with each other. If k is infinite and if we interpret k as the ring of polynomial functions on kn+1, then Di is given by

(Dif)(x1,x2,... ,277+1) = f (x1 , ... , xi-1, qxi, xi+1, .

. .

,

x'77+l) - f (x1,

, xi-1, q-1xi, xi+1, . . .

, x77+l )

qxi - q-1xi for all f E S.

For all i with I < i < n + 1 define an algebra automorphism Mi of S by Mi(Xi) = qXi and Mi(Xj) = X3 for all j 0 i. We get then for all monomials z ... X +, ,) = q"',,X1 2 z ... X +i M2(X 1

2

(2)

The inverse of Mi satisfies

M1 1

(X1772X2 1 M2

77tH+1

X77+1

MI M2 ) = q-rn X1 X2

77277+' Xn+1

for all monomials. It is clear that the M, commute with each other. It is easy to check for all i and j that

DioM = and that

Xio.Mj =

MjoDi, q,Mj o Di,

1M3 0 Xi,

ifi0j, ifi = j,

ifi 0 j,

q-1.M;oXi, ifi=j,

()3 (4)

where we identify X. with the left multiplication by X. A look at (1) and (2) shows that each D2 can be expressed in terms of Mi as

DA(f)= M2(f)-Mi 1(f)

qXi-q-1Xi

for all f ES.

(5)

Now a simple calculation yields for all i and all f , g E S

Di(fg) = Dt(f)Mi(g)+.Mi 1(f)Di(g). Indeed:

- Mi 1(f)Mi 1(g) q-1X2 Di(fg) - M2(f)M2(g) qX2 (Mi(f) -.M2 1(f))Mi(g) +Mi 1(f)(Mi(g) - Mi 1(g)) gX2_q-1X2

(6)

5A. EXAMPLES OF REPRESENTATIONS

98

We get from (6) in particular Di(Xjf) = Di(Xj)Mi(f) +Mi 1(Xj)Di(f) for all f E S and all i, j . We have Di (X j) = 0 and M i (X j) = X j for j

Di o X3 = X j o Di

if j

i, hence

i.

(7)

On the other hand, we have Di(Xi) = 1 and Mi(Xi) = qXi, hence

Di o Xi = q-1 Xi o Di + M2

for all i.

(8)

Note also that we can express Xi o Di and Di o Xi using (5) as

Xi o Di =

Mi-M

1

and

g-g-1

Di o Xi =

gMi_q-1MZ' g-g_ 1

(9)

Define now for all i with 1 < i < n endomorphisms ei, fi, and ki of S as a vector space by

ei=XioDi+1,

fi=Xi+IoDi,

ki=MioMi+1

(10)

Each ki is an algebra automorphism of S with inverse k71 = M 1 o Mi+1 PROPOSITION. Suppose that 4 is o f type An and that 11 = {al, a2i ... , a } with the usual numbering. Then there exists a homomorphism from Uq (g) into the algebra of endomorphisms of S such that

F«. '-' fi,

E., '-' ei,

i

for alli(1 1, cf. [H], 12.1.)

PROOF. We have to show that the ei, fi, ki, kti 1 satisfy the relations (Ri)(R6) from 4.3. Since the M. commute, so do the ki; this yields (Rl). We get from (3) and (4)

if j = i, q-lei, if j = i + 1, qei,

MjeiMj 1 =

ei,

otherwise,

hence

if j = i, q-lei, if j = i + l or j = i - 1, g2ei,

kjeik.-1 =

ei,

otherwise.

This yields (R2); the proof of (R3) is similar. Let us now look at (R4). We get for j # i using (7)

fjei = Xj+1DjXiDi+1 = Xj+IXiDjDi+1 and (since j + 1

i + 1)

eifj = XiDi+1Xj+IDj = XiXj+tDi+IDj,

5A. EXAMPLES OF REPRESENTATIONS

99

hence f jei = ei fj. On the other hand, we have by (8) fiei = Xi+1DiXiDi+1 = q-1Xi+1XiDiDi+1 + Xi+1MiDi+l and

eifi = XiDi+1Xi+IDi = q-1XiXi+lDi+lDi + XiMi+1Di,

hence - using (9) [ei, fi] = XiMi+1Di - Xi+1MiDi+1 = XiDiMi+1 - Xi+1Di+1Mi 1)Mi+l

(Mi - Mi

- (Mi+1 - Mi+1)Mi q-q-1

MiMi+1 - Mi 1Mi+1

q-q-1

ki - k

1

q-q-1

Now for (R5). For i and j with Ij - ii > 2 we see that ei = XiDi+1 and ej = XjDj +1 commute since {i, i + 1 } fl {j, j + 1 } = 0. Consider now the case where j = i + 1. We have

eiei+l = XiDi+l XiDi+1Xi+1Di+2 = XiDi+l(q-1Xi+lDi+1 +Mi+1)Di+2 = q -1X?(q-1Xi+1Di+1 +Mi+1)Di+1Di+2 +X2(gMi+lDi+l)Di+2 = g 2X?Xi+1D?+1Di+2 + (q + q-1)X?Mi+1 Di+1Di+2, ei+lei = Xi+1Di+2XiDi+lXiDi+1 = Xi+lX2Di+2D2Z+1 = X?Xi+1D2i+1Di+2,

eiei+lei = XiDi+1Xi+1Di+2XiDi+1 =Xi(9 1Xi+1Di+1 +Mi+1)X2Di+2Di+l

= g-1 XiXi+1D2+1Di+2 + X?Mi+1Di+1Di+2, hence eiei+l + ei+lei = (q + q-1)eiei+lei

Similarly:

eiei+l = XiDi+1Xi+1Di+2Xi+1Di+2 = Xi(g 1Xi+1Di+1 +Mi+1)Xi+1D2i+2

= g-1XiXi+1(g-1Xi+1Di+1+Mi+1)Di+2 +Xi(gXi+1Mi+1)Di+2 = g-2XiX2 1Di+1D?+2 + (q + g-1)XXi+1Mi+1D?+2, ei+lei = Xi+1Di+2Xi+1Di+2XiDi+1 = XXz+1Di+1Di+2 ei+1 eiei+l = Xi+LDi+2XiDi+1Xi+1Di+2

= Xi+lXiDi+2(g 1Xi+1Di+1 +Mi+1)Di+2 = q 1XiXz+1Di+jDi+2 +XXi+1 Mi+1Di+2, hence 2

2

eiei+ l + ei+lei = (q + q-1)ei+leiei+l

5A. EXAMPLES OF REPRESENTATIONS

100

This shows that (R5) is satisfied. The proof of (R6) is similar. Alternatively, we can consider the automorphism p of S as a k-algebra with p(Xi) = Xn+2_i for all i. We have then ei

(p-1

= Xn+2-

Dn+l-i = fn+1-

So conjugating the formulas for (R5) we get the formulas for (R6).

5A.7. Keep the notations from 5A.6. For any integer m > 0 set Stm C S equal to the subspace of all homogeneous polynomials of degree m. It is clear that

the operators ei, fi, and ki from 5A.6(10) map each Sm to itself. Therefore the homomorphism from Proposition 5A.6 makes each Sm into a Uq(g)-module. Consider the basis of Snt that consists of all

Xm, Vm2 1

Xm1 m2.... .m,+

2

(1)

[ml]1 [m2]

+ mn+1 = m. (Recall that we with all m1i m2, ... , ,+l > 0 and m1 + m2 + assume that q is not a root of unity.) We have then for all i

-

m,-m,+1x K IX, a, ml.m2.....mn+1 - q m1.m2.....mn+1'

(2)

+1 is a weight vector of weight E'l I (mi -mi+1)zui. Distinct (n+1)tuples (ml, m2, ... , ,+1) with >i mi = n + ] lead to distinct weights Ei (mi mi+l )tz . So each nonzero weight space in Sm has dimension 1 and is spanned by

so x,,,1,,n2.,,.

one

The action of the remaining generators of Uq(g) on our basis is easily checked to be given by [m:+1 + 1] xm1... .mi-1 .m, -1.m,+1+1.m,+2.....mn+l F«,xnbt.m2.....mn+1

= 0,

E«Xm1.n'2.....mn+1

-

if m, > 0; if mi = 0;

[mi + 1] xm1.... ,mx_1.m,+l.m:+1 l.ni,+2.....m +1, 0,

if mi+1 > 0; if mi+1 = 0.

In particular, all E, annihilate xnl,().... ,o. Up to scalar multiples this is the only vector with this property. (It is enough to look at weight vectors. Since the nonzero weight spaces have dimension 1, it is enough to look at the xml.rn2.... }i.) This proves now that Sn1 is a simple Uq(g)-module. It highest weight is equal to the weight of xm.o.... ,o, i.e., to mza1. The construction in 5A.6 works of course equally well, when q is a root of unity.

Each Sm is a Uq (g)-module also in that case. However, in general it will not be simple.

5A.8. We want to add a discussion of tensor products. In particular, we want to understand the tensor product of any simple finite dimensional U-module with one of the modules in from 5A.1/2. We shall assume that all finite dimensional Umodules are semisimple and that the dimensions of the weight spaces of any L(.\)

5A. EXAMPLES OF REPRESENTATIONS

101

with A E A dominant are given by Weyl's character formula. We know by 5.15 and 5.17 that these assumptions hold if char(k) = 0 with q transcendental over Q. However, as pointed out in the remark in 5.15, one can show that the assumption is satisfied for arbitrary k if q is not a root of unity. In order to describe the decomposition of tensor products one can proceed in the same way as for 9-modules, say in [H], 24.4, and - under our assumptions -the answer will turn out to be "the same". One has to work with the group ring Z[A] and its standard basis (e(A) I A E A), cf. [H], 22.5. One defines then the formal character ch(M) of any finite dimensional U-module M by

ch (111) = E dim(A1/)e(µ).

(1)

µEA

Set for all dominant A E A X(A) = chL(A).

(2)

For an arbitrary finite dimensional U-module M there are integers m(X) > 0, almost all of them equal to 0, such that M is isomorphic to the direct sum of all L(A)m(l). One has then (3) ch (M) = E m(A)X(A), X

where we sum over all dominant weights X. This equation determines the m(A) uniquely, since the X(A) with A dominant are linearly independent, cf. [H], Proposition 22.5A. If Al and N are two finite dimensional U-modules, then

ch (M®N) = ch (111) ch (N).

(4)

This is proved in the same way as [H], Proposition 22.5B: One uses that 111/, ON, C

(M 0N)µ+ for all p, v E A, cf. 5.3(1). So, in order to decompose L(A') ® L(A") for two dominant weights A' and A", we have to express X(X')X(X") in the form EX n(X)X(X) where we sum over all dominant A; then L(A') ® L(A") is isomorphic

- by (3) and (4) - to the direct sum of all L(X)'(' . The same approach works for 9-modules, cf. [11], 24.4(1). Since the X(A) are given by Weyl's formula in both cases, the final result is the same. One possible formulation of this result is Steinberg's formula, cf. [11], Theorem 24.4. However, for our calculations another approach will be more useful. Set p equal

to the sum of the fundamental weights p = E.EII w ; so we have (p, a") = 1 for all a E H. We extend the definition of X(v) to all v E A as follows: If there is a root 0 E 1 with (v + p O") = 0, then set X(v) = 0. Otherwise there is a unique w E W with (w (v + p), a") > 0 for all a E 11, cf. [H], Lemma 13.2B. Then set X(v) = det(w)X(w(v + p) - p). For v dominant this is compatible with the old definition. Using the extended definition of X(A) one can show for all dominant A and all

finite dimensional U-modules M that X(A) ch (M) = E dim(11M/) X(A + p),

(5)

µEA

cf. [H], exerc. 24.9. (This formula is due to Brauer.) In order to get from (5) the decomposition of L(A) 0 M one has to rewrite the right hand side as a linear

5A. EXAMPLES OF REPRESENTATIONS

102

combination where only X(A') with A' dominant occur. This can be achieved using

the definition of X: We simply drop all X(A + p) that are 0; for the remaining non-dominant A + p we write X(A + p) in the form ±X()') with A' dominant. If 4) is of type Al then we have ch(L(n)) _ 0e(n - 2i) for all integers n > 0. In that case we get from (5) easily the Clebsch-Gordan formula: min(m.n)

L(m) ® L(n)

® L(m + n - 2i),

(6)

i=1

cf. [H], exerc. 22.7.

5A.9. LEMMA. Let A E A be a dominant weight. a) If )o E A is a minuscule dominant weight, then L(A) ® L(A0) is isomorphic to the direct sum of all L(A + p) with µ E W A0 and A + p dominant. b) Suppose that 4) is irreducible and that )o is the largest short root. Then

L(.\) ® L(eo) is isomorphic to the direct sum of all L(. + p) with p E W.0 and A + p dominant, and of m copies of L(A) where m is the number of short simple roots a E II3 with (A, a') > 0. PROOF. The subsections 5A.1/2 contain an explicit description of L(.\o), in particular of its weight spaces and their dimensions: Any p-weight space with p E W.o has (of course) dimension 1. In the minuscule case there are no other weights; in the case Ao = ao the only additional weight is 0, occurring with multiplicity Ill I. So 5A.8(5) yields

X( )X( o) =

/SAE Waa X(A + µ),

for Ao minuscule;

EI Ewa0 X(A + µ) + IIIsIX(A),

for Ao = ao.

1)

Consider a weight p. E W A0 with A + p not dominant. There is then an a E H with (A + aV) < 0. Since A is dominant, we have (A, aV) > 0, hence (µ, aV) < 0.

If (µ, aV) _ -1, then necessarily (A, aV) = 0; this yields (A + p + p, v) = 0 and thus X(A + µ) = 0. If (µ, a') < -1, then we have to be in the case A0 = ao and µ = -a and a E II,- We have then (µ, aV) = -2 and (A, aV) 1. If (A, av) = 1, then we get as above (A + µ + p, v) = 0 and thus X(A + µ) = 0. If (A, a') = 0,

then (A+µ+p,av) = -1; then sa()+µ+p) = )+µ+p+a = )+p and thus X(A + µ) = -X(A). This shows: If )o is minuscule, then in (1) all summands X(A + µ) with A + µ not dominant are equal to 0; so X(A)X(Ao) is equal to the sum of all X(A+µ) with p E W\0 and A + µ dominant. This yields the claim in a). Suppose now that A0 = ao. For each a E II3 with (A, aV) = 0 the term X(A+µ) for p. = -a is equal to -X(A); these summands cancel against part of III,IX(A) and leave exactly mX(A). All remaining X(A + µ) in (1) with A + µ not dominant are equal to 0 and can be dropped. This proves b). REMARK. It is occasionally useful not to work with A, but with other comultiplications on U, and then to define the U-module structure on a tensor product of U-modules using this new comultiplication. For example, we could take in the rank 1 case the TD from 3.8. (It will be generalized to arbitrary (D in 7.2.) One should then interpret 3.12(2) - and later on its generalization 7.2(4) - as saying

5A. EXAMPLES OF REPRESENTATIONS

103

that O is an isomorphism between M ® M' with one structure and M 0 M' with the other structure. The discussion in the last two subsections will still work with a new comultiplication A' on U as long as we have

0'(Kµ)=Kµ®Kµ

for all t

Z1.

(2)

This condition will imply 5.3(1), hence 5A.8(4) and thus all the other results in 5A.8/9 on decompositions of tensor products. 5A.10. PROPOSITION. Set A0 equal to the set of all A0 E A that are minuscule dominant weights or the largest short root of an indecomposable component of 1. For each dominant weight A E A there are A1, A2i ... , Ar E A0 such that L(A) is a composition factor of L(A1) ®L(A2) ® ®L(Ar).

PROOF. Since the decomposition of tensor products of simple U-modules is the same as that of g-modules, we might as well prove the result in that case. Equivalently, we can work with G-modules, where G is a connected, simply connected semisimple algebraic group over C with Lie algebra g. One can show: If V is a faithful G-module, then each irreducible representation of G occurs as a composition factor of some tensor product V ® V ®. . 0 V ®V * ® V * ®. . . ® V *, cf. [Demazure & Gabriel], II, §2, Proposition 2.9. We want to apply this with V equal to the direct sum of all L(A0) with A0 E A0. This module satisfies V* ^ V. (Exercise! Use Proposition 5.16 and the fact that -w0 maps A0 to itself.) Any tensor power of V is s direct sum of tensor products L(A1) ®L(A2) ®.. ®L(Ar) as in the proposition. So the claim follows as soon as we show that the direct sum of all L(A0) with A0 E A0 is a faithful G-module. In order to prove the faithfulness, we may assume that D is indecomposable. Then one can use case-by-case arguments. Alternatively, we can argue as follows: Since the representation is not trivial, its kernel is contained in the center of G. The center of G can be identified with the dual of the finite abelian group A/Z-D, i.e., with Hom(A/Z-D, C") where C" denote the multiplicative group of C. Under this identification any z E Z(G) acts on L(A) as the scalar z(A). So it is enough to show that the image of A0 in A/ZD generates this group. Actually, one has a more precise result: The elements in A0 are representatives for the classes in A/M, cf. [Bourbaki 2], Ch. VI, §2, exerc. 5a.

REMARK. Set Al equal to the set of all fundamental weights of It is easy to see that the proposition holds with Al instead of A0. So an alternative way of proving the proposition is to show first that each A E Al satsfies the claim. This can be done by case-by-case arguments. For your convenience, let me briefly sketch how one gets the result quoted from

[Demazure & Gabriel]. Suppose that G is a closed subgroup of SL(V) for some finite dimensional vector space V. Let L be a simple G-module. It is easy to show (using matrix coefficients) that L can be embedded into the regular representation of G on the ring C[G] of regular functions on G. Since G is contained in SL(V), the regular functions on G are the restrictions to G of the polynomial functions on End(V). The restriction of functions is a homomorphism of G-modules, so L is a composition factor of some symmetric power of End(V)*, hence of some tensor power of End(V)*. Since End(V)* ^ V 0 V*, the claim follows. Actually, one can

104

5A. EXAMPLES OF REPRESENTATIONS

now get rid of V*, since V* itself is isomorphic to the (n - 1)-st exterior power of V, hence occurs itself in some tensor power of V. (Here we use again G C SL(V). That is satisfied in our case, since G is equal to its own derived group.) If G is not contained in SL(V), then V* is necessary. In that general case, one needs besides the polynomial functions on End(V) also the inverse of the determinant function in order to generate C[G]. One gets it from an exterior power of V*, hence from a tensor power of V*.

CHAPTER 6

The Center and Bilinear Forms Throughout this chapter we keep the assumptions and notations from the last chapter. In particular, we assume that q is not a root of unity. In order to prove our main results we have to apply Theorem 5.15 and Proposition 5.17; therefore in those cases we have to assume that char(k) = 0 and that q is transcendental over Q. However, later on (in 8.30) we shall prove a result that will allow us to get rid of this additional assumption and to extend all results in this chapter to arbitrary k and q (not a root of unity), cf. 6.26.

One of the main goals of this chapter is the determination of the center Z(U) of U. Again there are strong similarities with the theory for the enveloping algebra U(g). Humphreys in his book emphasizes at this point the classification of central characters over the description of the center of U(g); so for a better comparison of the two theories you may want to look at the treatment of the center in (e.g.) /Dixmier].

In the Lie algebra case the center of U(g) is isomorphic to the algebra S(h)' of invariants of the Weyl group W in the symmetric algebra of a Cartan subalgebra fj of g. The isomorphism is given by the Harish-Chandra homomorphism, which is

the composition of a projection map U(g) -+ U(h) = S((,) with a "shift" by -p. This construction can obviously be generalized to our Uq(g) where we now get a homomorphism from Z(U) to U°. It is easy to show that this map is injective using the fact that the intersection of the annihilators of all finite dimensional modules is 0. (That argument could also be used for U(g) though usually one gets the injectivity there in a different way.) Using (as for g) nontrivial homomorphisms between Verma modules one sees that the image of the Harish-Chandra homomorphism is contained in (U°)W W. By analogy with the classical case you might expect equality here, but that is not quite true: The image turns out to be (U°) W where we denote by U° the span of all Kv, with p an "even" weight, i.e., in 2A. This deviation is caused by the Verma modules not of type 1 and the nontrivial homomorphisms between them that lead to extra conditions. (The proof below may somewhat obscure that fact.) Now the main problem is to prove the surjectivity of the Harish- Chandra homomorphism (now onto (U°,)'4'). There are several proofs available in the literature. I am going to follow Tanisaki's version of Rosso's approach that is closer to the

classical methods than the proofs by DeConcini £ Kac or by Joseph £ Letzter. In the classical situation one shows that U(g) and the symmetric algebra S(g) are isomorphic as g-modules under the adjoint representation, one then identifies S(g) and S(g*) via the Killing form, and constructs invariants in S(g*) via the trace map 105

106

6.

THE CENTER AND BILINEAR FORMS

for finite dimensional modules. These invariants correspond to central elements in U(g); going through the definitions and identifications one computes their images under the Harish- Chandra homomorphism and one checks that they yield generators for In the quantized situation we do not have a Lie algebra and the algebra of polynomial functions on the Lie algebra that we could use. However, we have for each finite dimensional module V the trace map that takes any u E U to the trace S(h)w.

of u acting on V. This is an element in the dual space U* of U. Now this does not produce a U-invariant in U', since the ordinary trace is not a homomorphism of U-modules. So we replace it by the quantum trace and map any u E U to the trace of uK-1 on V. This produces invariants in U* considered as the dual module of U for the adjoint representation. But we want elements in the center of U, i.e.,

invariants of U acting on U via the adjoint representation - and not in U*. To make this transition from U* to U Rosso has introduced a non-degenerate bilinear form on U that is invariant for the adjoint action. It induces an embedding of U into U*. The invariants in U* (constructed via the quantum trace) turn out to be in the image of this embedding, so their inverse images are central elements in U. We can then compute their image under the Harish-Chandra homomorphism and get generators for (U° )w Rosso's construction of the bilinear form on U is complicated and difficult to follow. In these notes I take another approach that was used in [Tanisaki] and more recently in [Joseph & Letzter 2]. (There are minor differences: Tanisaki works with a different notion of invariance, Joseph & Letzter work with a different version of U.) In Tanisaki's approach one gets the form from a bilinear pairing between and U'0 that is constructed first. It goes back to Drinfel'd, cf. Section 13 of [Drinfel'd 2]. It will play a crucial role in the construction of the universal Rmatrix in Chapter 7. Unfortunately the proofs of the main properties of this pairing and of the form on U involve lengthy computations.

6.1. Recall from 4.7 that U is a (M)-graded algebra. Since q is not a root of unity, 4.7(1) implies for all v E ZIP

U I KµuKµ1=q(,u, )u for allaEZ4}.

(1)

Since KµuKU 1 = ad(K, )u by 4.18(2), this shows that the notation U is compatible with the notation for weight spaces introduced in 5.1(2). for Each A E A defines an algebra homomorphism U° - k with Kµ H all p. By abuse of notation we denote also this homomorphism by A, i.e., we write A(K,.) = q('\*µ). We have obviously (A + A')(h) = A(h)A'(h) for all h and all A, A'. We can rewrite (1) as

{uEUIhu=v(h)uhforallhEU°},

(2)

and we can rewrite the definition of weight spaces in a U-module M as

Ma={meMI hm=A(h)mforallhEU°}.

(3)

6.2. The isomorphism u- ® U° ® U+ =-> U implies that we have a direct sum decomposition U0 = U° ® ® U_VU°U+. (1) v>0

6.

THE CENTER AND BILINEAR FORMS

107

Denote by 7r : Uo -> U° the projection with respect to this decomposition. One

checks easily that the direct sum of the U VU°U+ with v > 0 is a two-sided ideal in U0. This shows: 7r: U0 ---* U° is an algebra homomorphism.

(2)

6.3. Denote the center of U by Z(U); we have Z(U) C U°

(1)

by 6.1(1).

LEMMA. a) Let A E A. Any u E Z(U) acts on M(A) as scalar multiplication by A(ir(u)).

b) Suppose that char(k) = 0 and that q is transcendental over Q. The restriction of 7r to Z(U) is injective. PROOF. Let u E Z(U) and write u = >0 u, where each u E U U0 U,+; so 0 for all v > 0, 7r(u) = u0. If we take vA E M(A)A, vA 0 as in 5.5, then hence uvA = u°vA = A(uo)vA. Since u is central in U, this implies uv = A(uo)v for all v E M(A) = UvA. So u acts as scalar A(uo) = A(7r(u)) on M(A). If 7r(u) = 0, then we get now uM(A) = 0, hence uL(A) = 0 for all A. If we

make now the additional assumption as in b), then we can apply Theorem 5.17 and get that u annihilates each finite dimensional U-module, hence that u = 0 by Proposition 5.11.

6.4. Define for all A E A an algebra automorphism

rya: U°->U°

7A(h)=A(h)h

by

for allhEU°.

(1)

More explicitly, we set rya(Kµ) = q(A,µ)Kµ

for all A and it.

(2)

So we can rewrite 6.1(2) as

U= {uEUI

all hEU°}.

(3)

Obviously -yo is the identity, and we have for all A, A' E A '}'a 0''a' _ YA+A,

and

A'(-VA (h)) = (A+ A') (h)

for all h E U0.

(4)

We call ry-p o 7r

: Z(U) --* U°

(5)

a Harish-Chandra homomorphism for U. Here p is as in 4.9 the sum of the fundamental weights (or half the sum of the positive roots). So we have 2(p, a)/(a, a) = 1

for all aEII.

108

6.

THE CENTER AND BILINEAR FORMS

6.5. Denote the Weyl group of 4) by W. It is generated by the reflections sa with a E H, cf. 4.1(4). The Weyl group acts naturally on U° such that

for all wEWand µEZM.

w(KN.)=K,,,F,,

(1)

This is an action via algebra automorphisms. We have for all w E W, A E A, and h E U°.

(wA)(wh) = A(h)

(2)

(It is enough to check this for h = K,,; there it follows from the invariance of the scalar product under W.) The set of fixed points of the W action on U° (all h with w(h) = h for all w E W) is a subalgebra; we denote this subalgebra by (U°)w. LEMMA.

We have ^y_,, o 7r(Z(U)) c (U0)w

PROOF. Let u E Z(U) and set h = -y_,, o 7r(u). So it acts by Lemma 6.3.a on an M(A) as scalar multiplication by A(-y,,(h)) _ (A + p)(h). Lemma 5.6 says that there is for all A E A and a E II with (A, a) > 0 a nontrivial homomorphism M(A - ma) M(A) where

m=

2(A, a)

+ 1 = 2(A + p, a) (a, a) (a, a)

(3)

So it has to act by the same scalar on both modules, i.e., we get (A + p)(h) _ (A - ma + p)(h). Now (3) implies that A - ma + p = sa(A + p). So we get (A + p)(h) = sa.(A+ p)(h) = (A + p)(sh).

(4)

Now (4) holds also for A with (A, a) < 0: If 2(A, a)/(a, a) = -1, then sa(A + p) =

A + p and the claim is trivial; if 2(A, a)/(a, a) < -1, then we can apply (4) to A' = sa (A + p) - p and get then the claim also for A. Since (4) holds for all A and a, and since the sa generate W, we get now

(A+ p)(h) _ (A+ p)(wh)

for all w E W and A E A.

(5)

We can substitute A - p for A and get thus also

A(wh-h)=0

forallwEWandAEA.

(6)

Fix for the moment w and write wh - h = r-µ a..KN.. Then (6) says

E agi-\'i`i = 0

for all A E A.

(7)

µ

Each A H q(\.µ) is a character on A; distinct A yield distinct characters. So the linear independence of characters implies a.. = 0 for all A. We get thus wh - h = 0; since w E W was arbitrary, this shows h E (U°)W as claimed. 6.6. Set

U° = ® kK,,.

(1)

iEZ'n2A

This is a subalgebra of U°, since Zp f1 2A is a subgroup of M. The action of W on U° maps U. to itself, since Z4) fl 2A is W-stable.

6.

THE CENTER AND BILINEAR FORMS

109

PROPOSITION. The Harish-Chandra homomorphism -y_P o 7r maps Z(U) to (Ue°u)I l'.

PROOF. Take an arbitrary u E Z(U). We know by Lemma 6.5 that ry_polr(u) E (U°)Y1', so we can write aµ Kµ

'y_P o 7r (u) =

aµ for all w E W and µ E Z4i.

with a,,

(2)

PEZ4,

We have to show that aP 54 0 implies µ E 2A. Recall from 5.2(1) that we can associate to each group homomorphisms or Z4 -> {±1} an automorphism & of U. Obviously o maps the center Z(U) to itself. Each generator (Ea, Fa, K± 1) is mapped to a scalar multiple of itself. This implies

that o(U+) = U+ and o(U U) = U7,, for all v as well as o(U°) = U°. This shows that

ooTr=Trop. Furthermore, each rya commutes with o (restricted to U°) because o(,u)q(a.P)KP ='ya(o(,u)KP)

o(q(a.µ)K,.) = o('ya(KP)) = ='ya(o(KP))Therefore also the Harish-Chandra homomorphism ry_P o Tr commutes with o. We get now for our u from (2)

'y-P o 7r(a(u)) = o(1: a, K,.)

a. or (p)

Since Q(u) is central, this sum is in (U0)"'.; so we have aPO(µ) = au,PO(wp) = aPO(wµ)

for all w E W and µ E

where we use (2) for the second equality. So this means: If aP 54 0, then o(µ) _

o(wµ) for all w E W. This yields in particular I = o(µ - s,,µ) for all a E 11. (Actually, it is equivalent, given the W invariance.) We can choose or with o(a) _

-1 for all a. Then 2(µ, a) (a, a) Therefore aP 54 0 implies that 2(µ, a)/(a, a) is even for all a E II, i.e., that p E 2A.

o(µ - sp) = o(ra) = (-1)r

where r =

REMARK. We shall see in 6.25 that the image of ry_P o Tr is all of (U° )'" (at least if char(k) = 0 and q transcendental over Q). 6.7. Suppose that A is a k-algebra with a comultiplication A : A -> A ® A and a counit e : A ---> k such that the diagrams in 3.2(1) and Lemma 3.4 commute. We get then on the dual space A* a structure as an associative algebra as follows: We define for two linear maps f,g E A* their product fg E A* as the composition

AAA®Af k®k -2- k

(1)

where the last map is the isomorphism with x ® y -4 xy for all x, y E k. More explicitly, we have for all a E A:

If A(a) = Ei bi ® ci, then (fg) (a) = E.i f (bi)g(ci). (2) It is more or less clear that the map (f, g) --> fg is bilinear. The coassociativity of A (cf. 3.2) implies that the multiplication on A* is associative. The commutativity of the diagrams in Lemma 3.4 implies that e is a 1 for this multiplication.

110

6.

THE CENTER AND BILINEAR FORMS

6.8. We are going to apply the general construction from 6.7 to the subalgebra

U'0 of U. Recall from 4.22 that the multiplication induces an isomorphism (of vector spaces) U° 0 U+ => U'0. Since KN,u = (KN,uKµ 1)KN, and K uK, 1 E U+

for all u E U+ and all u E M, we see that the multiplication induces also an isomorphism U+ ® U° => U'0. Its restriction to each U® ® U° with v E M, v > 0 is an isomorphism U+ ® U° =+ UFO, and U'-0 is the direct sum of the U'-0.

Note that 4.13(1) and 4.9(3) imply that 0(U'-O) C U20 ® U'0. So we can apply the general construction of 6.7 to U'-O. If we combine Lemma 4.12 with 4.9(3), then we get more explicitly for all finite sequences I of simple roots and all

IEM

A(E1Kµ) = E CA,BEAKµ+wt B ® EBKN.,

(1)

A,B

where the sum is over all sequences A and B of simple roots with wt I = wt A+wt B, and where the cA.B are elements of k with cA 0 = 61,A and c0 B = 61.B. (Compared to 4.12 the cA B have been replaced by the cA,B(q).) Define now for each a E 11 a linear form fa on U'0 by

fa(EaK,) _ -(qa - qa 1)

1

for all µ E Z-P,

(2)

and fa(U>°) = 0

for all v E M with v # a.

(3)

This is well defined, since the EaKN, with µ E W are a basis of *0, and since U'0 is the direct sum of all U'-0. We have for all sequences I as above and for all

pEM

(qa - qa1)-1, if I = (a); otherwise. = { 0,

fa(E1Kµ) _

(4)

(The term -(qa - qa

1)-1 is a normalizing factor.) Set for each sequence I = (,3j,)32,... ,,3,) of simple roots

fi = fagfa2 ... far

(5)

where we take the product as described in 6.7. For I = 0 we interpret the empty product as 1, i.e., as the restriction of e to U'0:

f0(EjKµ) = 6j.0

for all J and u.

(6)

LEMMA. We have for all sequences I and J and all p E M

fj(E1Kµ) = fj(EI)

(7)

If wt I # wt J, then fj (Ej) = 0.

(8)

and:

6.

THE CENTER AND BILINEAR FORMS

11]

PROOF. We use induction on the length JJI of the sequence J. For JJI = 0, i.e., J = 0, both claims follow from (6). For JJI = 1, i.e., J = (a) for some a E II, they follow from (4). Suppose now that the claims hold for some J and consider J' = (a, J) with a E II. We have for all I and a

fj'(EIKµ) _ (fa 0 f,7)A(EIKN)

_ > CA,Bfa(EAKp+wt B)fJ(EBKz) A,B

_ > CIA. Bfa(EA)fJ(EB) A.B

Here we have used for the last step that fa and fj satisfy (7). We see that also fJ' satisfies (7). Furthermore, if this sum is nonzero, then there are A and B with cA B 0 fJ(EB). By assumption, the last condition 0 and fa(EA)

implies wt A = a and wt B = wt J. We sum in (2) only over A and B with wt I = wt A + wt B. So we get now wt I = a + wt J = wt J' as desired. 6.9. For all A E A let kA : U'° -> k be the algebra homomorphism with

ka(uK,) = e(u)q-(a,µ)

for all pEZ-(P anduEU+.

(1)

This is the homomorphism that decribes the action of U'0 on a highest weight vector in M(-A), cf. 5.5. We have for all finite sequences I of simple roots and all ,a as above

ka(EIK) µ

q-(a,µ)

-

1 0,

if I = 0;

(2)

otherwise.

LEMMA. We have for all A, A' E A and all sequences I of simple roots

kaka, = ka+a,

(3)

and

kafi = q-(a,WtJ)fJka

(4)

PROOF. We have for each linear form f on U'0 (and for all J, p) by 6.8(1)

(kaf)(EJKK) = EcA.Bka(EAKK+WtB)f(EBK,) A,B

On the right hand side each summand with A we get

0 is zero by (2). Since c0 B = bJ,B

(kaf)(EJK,..) = ka(K,+wtj)f(EJK,) =

q-(a,µ+wtJ)f(EJKµ).

Take in particular f = ka'; we get (kaka')(EJK,u) = q-(a,µ+Wt J)b j,mq-(a',µ) =

bJ,Oq-(\+a',µ)

= kA+A'(EJK )

(5)

112

6.

THE CENTER AND BILINEAR FORMS

hence our first claim. On the other hand, we have CA.Bf (EAKµ+wt B)ka(EBKµ).

(fk),)(EJKN.) = A.B

Now each term with B # 0 is 0 and we get

(fk),)(EJKl,) = q-(U,µ) f(EJKµ).

(6)

If we compare this with (5) we see that =q-(),,wtJ)(fka)(EJKN,) (kaf)(EJKµ) for all J and p. (7) For f = f, both sides in (7) are zero unless wt J = wt I (by Lemma 6.8). So we can replace the q factor in (7) for f = ft by q-(a.wt I) and get the second claim.

REMARK. Let us record that (6) together with 6.8(7) implies for all A, p, I,

and alluEU+

fJka(uK,.) = q-1".µ)fj(u).

(8)

6.10. The elements F1KN, with all finite sequences I of simple roots and all It E Z1 are a basis of U- ® U°, cf. 4.17. So there is a unique linear map U0)* with cp(FjKµ) = f1kµ (1) for all I and p. This map is in fact an algebra homomorphism, since we have for

,

all I, J, p, and v o(Flq-(µ.wt J)F.JK,.Kv) o(F1KN.FJKv) _ _ = q-(µ.wt J)f(l.J)kµ+v =

,P(q-(µ,wt J)F(l.J)K,+v) q-(µ.wt

where we use Lemma 6.9 for the equalities in the second line. Define a bilinear pairing

(, ) : U<° x U'0 -' k

(y,x) = o (y)(x) for all y E U-0 and x E U'-0. So we have for all I, J, p, and A by

(2)

(FJK),, EJKN.) = frka(EJK,) Note that 6.9(8) implies for all y E U- and x E U+ (yK),, xKµ) = q-0,-A)(y, x) Furthermore 6.8(8) implies

for all A, p E Z4'.

If p, v E Z4) with p # v, then (y, x) = 0 for all x E U+ and y E U We use the notation (,) also for the bilinear pairing

(3) (4)

(U<° ®U<0) x (U>-0 ® U>-0) -> k

with (y1 ®y2, xi ® x2) = (y1, x1)(y2, x2

for all x1i x2 E U'-O and y1, y2 E UFO. We have in this notation (Y1 Y2, x) = (y1 ®y2, 0(x)) for all x E U>--0 and all y1, y2 E

Indeed, the definition implies

(y1y2,x) = P(y1y2)(x) = (cP(y1)cP(y2))(x) = (cP(y1) 0 P(y2))(A(x)), hence (5).

(5)

6.

THE CENTER AND BILINEAR FORMS

113

LEMMA. We have for all x1,x2 E U'-0 and ally E U!50

(y, x1x2) _ (0(y), x2 ®x1).

(6)

PROOF. Consider first the case where y = KN with p E M. Since kN, is a homomorphism and since A(K,,,) = K® ® Kµ, we get (Kµ,x1x2) = kµ(x1x2) = kµ(x1)kµ(x2)

= (K®®Kµ, x2 ®x1) = (0(Kµ), x2 (9 x1).

Let us next look at y = Fa with a E II. Suppose that x1 = EIKI,, and X2 = EJK,,. Then

(0(y),x2®x1)=(Fa®K(, 1+1®Fa,x2®x1) = fa(x2)k-a(x1) + E(x2)fa(x1) _ -(qa - qa 1)-1

(7)

6J'06I.(a))

.

On the other hand (y,x1x2) _ (Fe,El (K,

I)-'q

1)Kµ+,) =

fa(q("'tJ)E([,J)) (8)

J)

There are exactly two cases where (I, J) = (a): Either I = (a) and J = 0 (and then J) = 1) or I = 0 and J = (a) (and then q(µ-Wt J) = So the right hand sides in (7) and (8) are equal. So far we have shown that (6) holds, if y is one of the generators Km and F,,, of Since (6) is clearly linear in y, our claim will follow if we can show: If (6) and Y2 E UFO, then it holds for y1y2. This is a straightforward holds for y1 E calculation better hidden in the appendix. 6.11. Recall the elements uQ, from 4.10(2).

LEMMA. We have (u-3, x) = 0 for all x E U'0 and all a, Q E II with a

Q.

PROOF. By 6.10(3),(4) it is enough to show (ua3, EI) = 0 for all I with wt(I) = ra + Q where r = 1 - 2(a, Q)/(a, a). We have I = (-y, I') with ry E {a, ,Q}

and where I' is sequence with wt(I') = wt(I) - 'y, hence wt(I) wt(I') # 0. We have now [using Lemma 6.10 and Lemma 4.101

(ua EI) = (ua3, E.,EI') = (0(ua,3), EI' ®E,) = (ua3 ®K 'K3 1 + 10 ua3, E[, ® E,) _ (ua EI')(KaT Ka I E7) + (1, E[')(uali, E.,). ,

Now 6.10(4) implies that all terms in the last sum are 0.

wt(I) and

6.

114

THE CENTER AND BILINEAR FORMS

6.12. PROPOSITION. There exists a unique bilinear pairing (, ) : UFO x

U'-0 -+ k such that for all x, x' E U'O, all y, y' E U5°, all a, v E M, and all (y, xx') _ (0(y), ®x), (Kµ Kv) = q-("-v), (Kµ, Ea) = 0,

(yy , x) = (y (9 y, o(x)),

(1)

(F.,Ea) = -b«a(q« - q«') (F, Kµ) = 0.

(2) (3)

[We use in (1) the notation (,) for the bilinear map (U-0(DU:5°) x (U'-°W!0) , k induced by (, ) as in 6.10.] PROOF. Lemma 6.11 says that the homomorphism cp from 6.10(1) maps each uaa to 0. Therefore the ideal generated by these terms is mapped to 0. Since Uis isomorphic to U- divided by the ideal generated by all uaa (by Theorem 4.21.c) and since U° = U° (by 4.21.d), this implies that cp factors through a homomorphism

and U'° via (y, x) = : U50 , (U'-O)". We get then a bilinear pairing of 5(y)(x) for all x E U'-0 and y E UFO. If y is the image of some u E UFO, then (y, x) = (u, x) where we have on the right hand side the pairing from 6.10(2). It is now clear, that this pairing inherits (1)-(3) from the analogous properties of the earlier one, i.e., from 6.10(6),(5), 6.8(4), and 6.9(2). The uniqueness of the pairing follows from the fact that (2) and (3) determine the form on the generators E,, and K,,, resp. Fa and KN, and that (1) shows that the values on the generators determines it on the whole algebras. REMARK. Note that (1, 1) = 1. [Take A = µ = 0 in (2)]

6.13. It is clear that the pairing in Proposition 6.12 inherits the properties of the pairing from 6.10(2). In particular, 6.10(3) and 6.10(4) extend. Let us state explicitly for future reference:

If y E U- and x E U+, then (yKA, xK,,,) = q- (a,1,,) (y, x) for all A, µ E Z( P,

(1)

and:

If t,v

Z4 with µ#v, then (y,x)=0for all xEU+and yEUC,.

(2)

for all x E U+, y E U-, and all) E Z4).

(3)

This implies

(KAyKa', KAxK1,1) _ (y, x)

Indeed, we may assume x E U+ and y E LC, with p, v E Z( P. Then

(KayKa 1,KAxKa I) =

q(",-v)(y,x),

(q-(a.v)y, gU.µ)x) =

which yields (3) for µ = v, whereas for µ # v all terms are 0 by (2). Here is a rank 1 example: We have for all a E I1 and all integers n > 0 (FFY E«) _ (-1),Lq«(- I)/2

[n ]'

(q0 - q«

I)7L

(4)

THE CENTER AND BILINEAR FORMS

6.

115

Indeed, this holds by definition for n = 1 and n = 0; we get for n > 4.9(4), (2) and (1) above, and 6.12(2)

0 using 6.12(1),

(Fn En) _ (Fr' ®Fq,O(Eq)) = = q.

(Fa-1 ®Fa,gq '[n]aEa-'K® ® Ea) 1

[n] Q(FF

E' -') (F., Ea)

_ -(qa -

qa')-'qa-I [n]a(FF-1,Ea-')

Now use induction.

6.14. We have by 4.13(1) for all xEU+ (with pEM,p>0)

0(x) E ® U± vKv ®U+. 0
In particular, there are for each a E II elements ra(x) and ra(x) in U±,, such that the component of 0(x) in U±vKv ®U+ is equal to ra(x)Ka ®Ea for v = a, and equal to EQKN,_q ®r'(x) for v = µ - a. So we have

(x) = x ® 1 + E ra(x)KQ ® Ea + (rest),

(1)

aE11

and

0(x) = Kµ ®x + E EaKp,-a ®r'n(x) + (rest),

(2)

aEII

where the "rest" involves terms in U± vK® ® U+ with v > 0, v V II in (1), resp. with µ - v > 0, µ - v V II in (2). For example, we have

ra(1) = r' (1) = 0

and

ra(E,3) = rq(Ep) = baa

for all /3 E H.

(3)

LEMMA. a) We have for all x E U+ and all x' E U±

r xx' = xr

x'+

a.

' r x x' and T xx'

a' xr'

+ r' x x'.

(4)

b) We have for all x E U+ and all y E U -

(Fay,x) _ (FF,EQ)(y,ra(x)) and (yFQ,x) _ (Fa,Ea)(y,ra(x))

(5)

c) We have r' (x) = TraT(x) for all x E L. PROOF. a) We have 0(xx') = 0(x)0(x'). If we write both factors as in (1),

then we see that the projection of 0(xx') to_aKa U+® ®U+ is equal to (x ® 1)(ra(x')K. (9 Ea) + (ra(x)K. 0 E.) (x' ® 1) _ = (xra(x')KK +ra(x)Kax) 0 Ea.

116

6.

THE CENTER AND BILINEAR FORMS

If we use (2) instead of (1), then we see that the projection of L(xx') to UKa U +µ, _ a is equal to (Kµ ® x)(EaKµ'_a ® r' (x')) + (EQKµ_a ® r'(x))(Kµ' ® x') = KNEaKN'_a ® xr'a(x) + EQKµ+µ'-a ® r'a(x)x'. Now the claim follows, since Kax' = q(a-µ')x'Ka and K.E. = b) We have (Fay, x) = (F®®y, A(x)) and (yFQ, x) = (y(9Fa, A(x)) by 6.12(1). Now the claims follow from (1),(2) and 6.13(2),(1).

c) The claim holds for x = 1 and for x = Ea with 0 E II by (3). Since both sides are linear in x, it is enough to show: If the equality holds for x E U+ and for x' E Uµ, , then it holds for xx'. Well, we have (using (4) and the induction hypothesis)

TraT(xx) = Tra(T(x') T(x)) = T (T(x') rar(x) + q(a,µ)rar(x') T(X)) = TraT (x) x' + q(a,µ)x TraT (x')

= rl(x) x' + q(a,µ)x r' (x') = r«(xx')

6.15. The construction of the last subsection can be carried out similarly in U-. We have by 4.13(2) for allyEUµ (with it E DD, it >0)

A(y) E ® U 1 7 ,0 0
So we can find for each a E II elements ra(y) and r' (y) in U (µ_a) such that

Kµia+(rest),

(1)

aEll and

(y) = 10 y +

Fa ®ra(y)K« 1 + (rest),

(2)

aE11

where the "rest" involves now terms in U , ®U (µ_V) KV 1 with p - v > 0, µ - v in (1), resp. with v > 0, v II in (2). For example, we have

ra(1)=r«(1)=0

and

ra(FF)=r'(Fp)=Sa3

forall,3EII.

II (3)

(Note that the definitions here and in 6.14 coincide on U = k = U+.) LEMMA. a) We have for all y E Uµ and all y' E U µ1

ra(yy) = gla,µ)yra(y) + ra(y)y and ra(yy) = yr«(y) + q(a,µ)r'(y)y

(4)

b) We have for all y E V i7- and all x E U+

(y,Eax) _ (FQ,Ea)(r(,,(y),x) c) We have for all y E Uµ ra(y) = wr' w(y)

and (y,xEa) _ (Fa,Ea)(ra(y),x).

(5)

and r«(y) = wraw(y)

(6)

THE CENTER AND BILINEAR FORMS

6.

117

PROOF. a), b) The argument is more or less the same as in 6.14. The only real difference is that by 6.12(1) now (y, Eax) = (A(y), x 0 Ea) and (y, xEa) = (A(y), E. ® x). This reversal of the order of the factors is the reason why ra and ra appear now in different places than in 6.14. c) The claim holds for y = 1 and for y = F0 with 33 E II by (3) and 6.14(4). Again it is enough to show: If the equalities hold for y E Uµ and for y' E U µ then they hold for yy'. Well, we have (using (4), 6.14(4) and the induction hypothesis) wraw(yy') = wr«(w(y) w(y')) = w(q(0',11w(y) r«w(y) + raw(y) w(y )) = q(a.µ) y wraw(y) + wraw(y) y = q(n.t,)y ra(y) +ra(y) y = ra(yy )

The proof of the second equality is similar. REMARK. If we combine (6) with 6.14.c, then we get for all y as above

ra(y) = (wT)ra(wT)(y) =Tr'T(y);

(7)

there is an analogous formula for ra. 6.16. LEMMA. We/ have for all x E U+ and y E U

(w(x),w(y)) = (y,x) = (T(y),T(x))

(1)

PROOF. We may assume y E UT, and x E U+ with p, v E M. We may even suppose that p = v, because otherwise all terms are 0 by 6.13(2). The claim is

obvious for v=IL =0,since U=kU+ and w(1)=l=r(1). In order to get the claim for all we want to use induction. We consider a weight p > 0 and suppose that the claim holds by induction for all weights less than p. By linearity, it is enough to consider y of the form y = Fay' with a E II and y' E U Then G_a).

(Fay , x) = = (Fn, E.) (y', r« (x))

= (Fa, Ea) (y', ra(x))

= (Fe, Ea)(wr'(x), w(y'))

= (Fa, Ea)(T(y'),Tra(x))

[by induction]

= (F.,Ea)(raw(x),w(y'))

= (Fe,Ea)(T(y),raT(x)) _ (T(y)F.,T(x)) _ (T(Fay),T(X)).

[by 6.15.c/6.14.c]

= (w(x), Eaw(y')) = (w(x),w(Fay'))

[by 6.14.b]

[by 6.15.b/6.14.b]

EXERCISE. Show that (S(y), S(x)) = (y, x) for all x E U'0 and all y E U:5° (Hint: Use (1), 4.13(4), and 6.10(3).)

118

6.

THE CENTER AND BILINEAR FORMS

6.17. LEMMA. Let a E H and µ E M. Then we have for all y E U --A

Eay - yEa = (qa - qa 1)-' (Kara(y) - r'(y)Ka 1),

(1)

and for all x E U+

(ra(x)Ka - Ka'r' (x))

xFa - Fax = (qa - qa')

(2)

PROOF. Let us look at (1) first. For y = 1 and for y = F$ with a E II the formula follows from 6.15(3) and (R4). Since both sides are linear in y, it is enough to show: If (1) holds for y E U µ and for y' E U µ then (1) holds for yy'. We have

Ea(yy) - (yy)Ea = [Ea, yy ] = (Eay - yEa)y + y(Eay - yEa), hence

(qa - qa 1) [E., yy] = Kara(y)y - r«(y)Ka'y + yKara(y)

= Kc(ra(y)y +

q(a.µ)yra(y))

= Kara(yy) - r' (yy)Ka

-

(q(a.µ )r'

- yrq(y)K-1

(y)y + yr' (y ))Ka 1

1.

(We use here the induction hypothesis, then K«'yKa = q(a°µ)y together with K-1y'Ka = finally 6.15.a.) We can use a similar induction argument in order to prove (2). We can also deduce it from (1) working with y = w(x):

xFa - Fax = w(yEa - Eay)

= (q. - qa 1)-'w(ra(y)K« 1 - Kara(y)) = (qa - qa 1)-'(ra(x)Ka - K- 1 r' (x)), using 6.15(6) for the last step. REMARK. The formulas in the lemma can be regarded as special cases of the following more general formulas, cf. [Tanisaki], Lemma 2.1.2(ii): Let x E U'-° and

y E UFO; write (1®0)o0(x) = Eixi®xi®xz' and (1(&0)o0(y) = Ej yj®yj®yj. Then

yx =

(S(yj ), xi) (yj , xj )xiyj

and

xy = E(yj , xi) (S(yj ), xj )yjx.

(3)

i.j

These formulas indicate the connection of U with the quantum double of U'-0, cf. [Drinfel'd 2], Section 13.

6.18. PROPOSITION. Assume that char(k) = 0 and that q is transcendental over Q. The restriction of ( ) to any Uµ x Uµ with µ E M, µ > 0 is a nondegenerate pairing.

6.

THE CENTER AND BILINEAR FORMS

119

PROOF. The claim says that the pairing induces an isomorphism between Uµ and the dual space (U+)". Since both spaces have the same dimension (e.g., because

of Uµ = w(U+)), it is enough to show: If y E U µ with (y, x) = 0 for all x E U+

then y = 0. We use induction on u for the usual ordering < on M. The claim holds for

p=0,since UC =k=U and (1, 1)=1. Consider now p > 0 and suppose that the claim holds for all v with 0 < v <

Let y E Uµ with (y, x) = 0 for all x E U+. We have then in particular

p.

(y, Eax) = 0 = (y, xEa) for all a E H and x E U+ a. Now Lemma 6.15.b implies that (ra(y), x) = 0 = (r' (y), x) for all these a and x. Since we assume that the claim holds for µ-a, this implies ra(y) = 0 = ra(y). Now 6.17(1) yields Eay = yEa for all a E H. But then Proposition 5.19.b says that y = 0.

REMARK. We can define a bilinear form on U+ via (x, x')' = (w(x), x') for all x, x'. The proposition says that this form is nondegenerate on each U+; by 6.10(4) distinct C µ+ are orthogonal for ( , )'. Lemma 6.16 implies that ( , )' is symmetric. We can lift the form to U+: If we denote the canonical map U+ - U+ by u H u, then set (u, v)' = (u, v)'. Then U+ is the quotient of U+ by the radical of this form. Lusztig uses this fact in his book to give a different definition of U = Uq(g). He constructs the form on U+ directly and then replaces the "quantum Serre relations" (R5) and (R6) in the definition of U by the condition that for all v in the radical of the form on U+ the image of v and w(v) in U should be 0. (Actually, Lusztig uses a different normalization of the form. He sets (Ea, Ea)' = qa/(qa - qQ 1), whereas

we get (Ea, Ea)' _ (-1)/(qa - qa 1). His form on a U+ is then ±q(p,µ) times our form. This different normalization is also responsible for the fact that the formulas in [L], 3.1.7/8, look somewhat different from those in 6.17(3) above.) 6.19. LEMMA. Let A, p, v E M. We have for all x E U- , y c- U yKA(q-('\`)Eax

ad(E,,)(yKAx) =

1

+

qa -

-1 qa

and a E H

- q(µ-v.a)xEa) (q-(v-a.a)ra(y)KA+a

-

\ ra(y)KA-a Jx

/

and

ad(Fa)(yKAx) =

q-(,u.a) (Fay

+ qa -1 qa-1

- q-(A'`Y)yFQ)KA+ax

y(q (µ-a.a)KAra(x) - q-2(µ-a,a)K.\+2ara(x)).

PROOF. We use the formulas for ad(Ea) and ad(Fa) from 4.18(2) together with Lemma 6.17 and 4.7(1):

ad(Ea)(yKAx) = EayKax - KayKaxKa IEa

_ (Eay - yEa)Kax + yEaKax - q(µ-V.a)yKaxEa _ (qa - qa 1)-1(Kra(y) - r' (y)K, 1)KAx + q-(,Va)yKAEax - q(µ-v.a)yKaxEa

_ (qa - qa 1)-1(q-("-a.a)ra(y)KA+a - ra(J)Ka-a)x yKa(q-(a.a)Eax - q(µ-v,a)xEa), +

120

THE CENTER AND BILINEAR FORMS

6.

and

ad(Fa)(yKAx) = (FayKax - yKaxFa)Ka = FayKaxKa - yKA(xFa - Fax)KQ. - yKaFaxKa yKAFF)Kax

= =

- yKA(ga - qa 1)-1(ra(x)Ka - K,-,, lra(x))KK - q-('\.a)yF'a)K,\+ax + (qa - q1 ) q-(µ.a)(Fey

lyK,\(q-(µ-a.a)r' (x)

KKq-2(µ-a.a)ra(x))

6.20. Suppose in this and the next subsections that q1/2 is a square root of q in k or in a quadratic extension k(q1/2) of k. The multiplication map U- ®U° ®U+ - U is by Theorem 4.21 an isomorphism of vector spaces. It induces for all µ and v an isomorphism LL-, ® U° ® Uµ U -,U°U,+. Since K is a unit in U°, the multiplication induces also isomorphisms (U:,&) ® U° ® U+ U :,U°U and (taking direct sums) µ,v>0

These considerations show that there is a well defined k-bilinear map

(, ):UxUk(q"2) with

((yKv)KAx, (y K,')KA'x) _ for all x E Uµ , x' E µ±, all y E U y' E U

(y',x)(y,x')q(2P,V)(gl/2)-(a.a')

(1)

and all A, A', µ, µ', v, v' E M.

Obviously 6.13(2) implies

(U vU°U+, U v,U('U) = 0

unless µ = v' and v = µ'.

(2)

PROPOSITION. We have (ad(u) v, v') = (v, ad(S(u)) v') for all u, v, v' E U.

PROOF. It is enough to take for u generators, i.e., the Ea, Fa, and K. Furthermore we may assume that

v=

and

v' =

with A, A' E Z4P and x E Uµ , x' E Uµ, y E Uv, and y' E U_v, with weights

µ,v,/L',v'EM,all>0. Consider first u = K.,7 with 'n E M. We have 0(K.) = K,7 ® K,7 and S(K,7) _ K, , hence ad(K.,7) a = K,7aK, I

and

ad(S(K.,7)) a = K,, 1 aK.,7

and

ad(S(K.,7))v' = q('7-"'-µ')v',

for all a E U. We get in particular ad(K,7) v = hence

)v

(3)

6.

THE CENTER AND BILINEAR FORMS

(ad (K.,) v, v') = q(77'µ-")(v, v')

and

121

(v, ad(S(K,7)) v') = q(7'"-µ')(v, v').

If the right hand side in one of these equations is nonzero, then v' = u and v =,u' by (2), hence It - v = v' - p'. Therefore the right hand sides are equal in any case, and our claim holds for u = K,7. Take now u = E. We have S(E«) = -Ka 1E«, hence

ad(S(E«)) = - ad (K.-) ad (E.)

-

Lemma 6.19 implies

l

ad(E«) v = q-(V'«)(yK,)K. (q-(A'«)Ex -

(g-("_«'a)(ra(y)K"-a)Ka+z« g«_1

ga

- (ra(y)K"-«)Ka)x

and - using (3) -

- x'E«)

-q-(«,a)(y K" )Ka (q-(u'+A',a)Eax'

ad(S(Ea)) v' =

q-(µ'.a)

- q« - qa ((ra(y )K"'-a)K.\'+2a

-

q')

(ra(y )K"'-a)Ka')x1

This implies using 6.12(1): If v'= it and it' = v - a, then (ad(EE) v, v') = (qa - qa') -'(y', x)g(2p,"-a)

(4)

(q-("-a,«)(ra(y),x)(gh/2)-(,\+2a,A') and

-

(v, ad(S(E«)) v') = q-(«.a)(y',x)q(2P.")(q1/2) ((y, x'E«)

(r'a(y),x')(gh/2)-(A.a')),

(A'A ) (5)

-

q-(µ'+a'.«)

(y, Eax'))

Since a is a simple root, we have (2p, a) = (a, a). So the product q-(«,«)g(2P.") in q(2P,"-a) in (4). From (4) and (5) pull out the common (5) is equal to the factor factors we are then left with

(r«(y) x'))

(g« - q« 1)-1

(4')

resp. (using v - a = u') (y, x'E«) - q-("-«+A',a) (y, E(,x')

(5')

Now 6.15(5) combined with (F«, Ea) = -(qa - q')', cf. 6.12(2), implies that the terms in (4') are equal to the terms in (5'), hence that the left hand side in (4) is equal to the left hand side in (5). If v' = p + a and u' = v, then we get (ad(Ea) v, v') = q-(",c') (y,x')q(2P'")

(q-(A'«) (y , Eax)

(q1/2)-(.\,)

)

- q(µ'a) (y xEE)), ,

(6)

122

6.

THE CENTER AND BILINEAR FORMS

and

(v, ad(S(E«)) v') = (q« -

qa')-1q-(µ .«)q(2Py)(y,X)

(q(r«(y),x)(91/2)-(a,A') _

(7) (r«(y),x)(g1/2)-(a,a'+2«)I

q-(",«)(y,x')q(21 )(q1/2)-(,\,a')(Recall that µ' = v). Pull out the common factors We are left with q-(a,«)(y, E«x) q(µ,«)(y,xE«), (6')

-

resp.

q-(a,«)(r«(y),x)

(q« -

(7')

Again 6.15(5) and 6.12(2) show that the terms in (6') and (7') are equal, hence so are the left hand sides in (6) and (7). If (ad(E«) v, v') 0 0 or (v, ad(S(E«)) v') 0 0, then (2) implies that we are in

one of the two cases (v' = µ, µ' = v - a or v' = µ + a, µ' = v) just considered. So we see that these inner products are always equal, i.e., that (ad(E«)v, v') _ (v,ad(S(E«))v') always. The argument for u = F« is similar; we leave it to the appendix. Alternatively, you can construct a proof using the following exercise. EXERCISE. Show that (w o S(v), w o S(v')) = (v', v) for all v, v' E U. (Hint: Use 4.13(3), 6.16(1), and the exercise in 6.16.) Show that w o S o ad(F«) = q2 ad(E«) o

woSand woSoad(S(F«))=q2 ad(-E«K(1)owoS. REMARK. If M is a U-module, then a bilinear form (,) : M x M -+ k is called U-invariant, if the induced map M®M -+ k, m®m' ' -. (m, m') is a homomorphism

of U-modules. (Here we regard k as a trivial module.) One can check that this condition is equivalent to (u in, m') = (m, S(u) m')

for all in, m' E M and u E U. So the form defined in (1) is invariant, if we regard U as a U-module under the adjoint representation.

6.21. PROPOSITION. Assume that char(k) = 0 and that q is transcendental over Q. Let u E U. If (v, u) = 0 for all v E U, then u = 0.

PROOF. Recall that U is the direct sum of all U U°U+ = U

By 6.19(2) it is enough to show: If u E II 7,U°U+ with (v, u) = 0 for all v E U µU°U+,

then u = 0. Choose for all µ E Z D an arbitrary basis ui , u2 , ... , u* (µ) of L. There is then by Proposition 6.18 a "dual" basis vi , v2 , ... , v*(µ) of U µ, i.e., a basis with (vµ,

Forallµandvthe(vz

6

for all i, j.

(1)

with 1
a basis of II 7,U°U+. The definition 6.20(1) implies ((vhKµ)KA'u(, (vi K1)KAu.;) = 6hi6li(q1

(2)

6.

THE CENTER AND BILINEAR FORMS

Consider now u E U vU°Uµ with (v, u) = 0 for all v E U

u = Ei

E aij

123

Write The assumption (v, u) = 0 for all v yields by (2) (g1/2)-(a.a') = 0 for all i and j and all A' E Z. (3) N,U°U+.

AEZ4'

is a group homomorphism from Zol) to the multiplicative group of k(ql/2). Distinct A lead to distinct homomorphisms, since q1/2 is not a root of unity. So by Artin's theorem on the linear independence of characters, (3) implies that all aija are 0, hence that u = 0 as claimed. Each A' -->

(q1/2)-(

REMARK. We have also: If u E U satisfies (u, v) = 0 for all v E U, then u = 0. We can argue as above. Alternatively, note that (4) for all u E U -,,U°U+ and v E U µU°U+. (v, u) = q(2P°µ-v) (u, v)

6.22. In the following subsections (6.22-6.25) we assume that char(k) = 0 and that q is transcendental over Q. Our lemmas are preceded by (TQ) to remind us of that assumption. The proof in 6.21 shows for all p and v: Let cp : Uµ x U+ -> k be a bilinear

map and let .\ E Z. Then there is an element u E

U 7,Ka+vUt+

such that for all xEU+,yEU,,A'EZ11) (1) ((yKv)KA'x, u) = o(y, (v%Kv)Kau [in the notations of that proof] will cp(v,, &z ')qx)(q1/2)-U,a')

Indeed, u = work.

We want to apply this to matrix coefficients: Let M be a finite dimensional U-module. For all m E M and f E M" let cf,m E U" be the linear form with for all v E U. (2) c f,m(v) = f(vm) LEMMA (TQ). Let M be a finite dimensional U-module such that all weights

.\ of M satisfy 2.\ E Z&. Then there is for each f E M' and m E M a unique element u E U with c f, (v) = (v, u) for all v E U. PROOF. The uniqueness follows from Proposition 6.21. To prove the existence, we may assume that f and m are weight vectors, since c f,,,,, depends linearly on

f and m. So suppose that there are weights ) and .\' of M with m E Ma and f E (M")_a,, i.e., with f (May,) = 0 for all .\" # A'. We have U+m C Ma+v for all p; since M has only finitely many weights, there are only finitely many v with for all p and v, we get c f,m(U µU°U+) = U+m # 0. Since U µU°U+m C 0 unless .\' = .\ + v - p. This shows (when combined with the previous statement) that there are only finitely many pairs (µ, v) with c f,,,,.(UµU°U+) # 0. We have for all xEU+, U µ,and77EZ11) , cf.m.((yKp)Knx) = f(yKµK,7xm) = q(n,a+v)f(yKµxm) = (q1/2)(n.2a+2v)f (yKµxm).

For all p and v the function (y, x) --> f (yKµxm) is bilinear. We now apply (1) to this bilinear form and to -2(.\ + v) instead of .\; here we use that 2(.\ + v) E Z4) by our assumption on the weights of M. W e get thus from (1) an element uvN, E U ,U°Uµ+ with (v, uvµ) = c f,m(v) for all v E U µU°U+. Then u = v µ uvN, will satisfy our claim (by the orthogonality 6.20(2)).

124

THE CENTER AND BILINEAR FORMS

6.

6.23. LEMMA (TQ). Let A E A be dominant such that 2A E M. Then there is a unique element zA E U such that (u, zA) is equal to the trace of uK2pI acting on L(A) for all u E U. The element zA is contained in the center Z(U) of U. PROOF. If MI, m2, ... , m,. is a basis of L(A) and if fl, f2,... , f, is the dual basis of L(A)*, then the trace of uK2pI acting on L(A) is equal to r

r 1

_ i=1

i=1

Cfi,K2plm,(U)

So the existence of zA follows from Lemma 6.22. The uniqueness follows from 6.21. Recall from 4.18 that the map U -> EndkL(A) that takes u E U to the action of u on L(A) is a homomorphism of U-modules, if we take the adjoint representation on U. Also the quantum trace is a homomorphism of U-modules (from EndkL(A)

to k with the trivial operation), cf. 5.3 and 3.10(8), hence so is the composition, which maps u E U to the trace of uK2pI acting on L(A). This means for all u, v E U

that e(u)(v,zA) = (ad(u)v,zA) = (v,ad(S(u))zA),

hence (by 6.21) ad(S(u))zA = e(u)z,\ for all u E U. We have e o S = e. (This holds in any Hopf algebra, cf. 3.8(1); it can also be checked directly from the formulas in 4.8.) So we have also ad(u)z,\ = e(u)z,\ for all u E U.Now 4.18(2) yields easily first that zA commutes with each K. with it E M, and then that zA commutes with all Eo and F,,, hence is central in U. 6.24. Keep the notations from the last subsection. Since zA is central, we have zA = µ>o zA,µ where each zA,µ E U µU°U+. Write zA.o a E k. We have then for all it E Z11) (KN., zA) _ (Kµ, zA,o) =

using the orthogonality 6.20(2) and the definition. On the other hand, this is the trace of K,K2p1 = K,,-2p acting on L(A), hence equal to dim

L(A),,g(\',,u-2p)

L(A),\,q-2(A'.p) (ql/2)(2A'.µ)

dim

=

A comparison of these two formulas shows that zA,o =

E dim

L(A)A,q(-2A',p)K-2A'

u

= Edim L(A)-(1/2) q(".p) K, V

We have zA.o = lr(zA) in the notation of 6.2, hence

'Y-p o 7r(zA) = E dim L(A)-(1/2),K,.

(1)

11

Q.

6.25. THEOREM. Assume that char(k) = 0 and that q is transcendental over The Harish-Chandra homomorphism is an isomorphism between Z(U) and

(Uoev)W

6.

THE CENTER AND BILINEAR FORMS

125

PROOF. By 6.3.b and 6.6 we only have to prove that the image of -y-p o 7r is E Z4

all of(U°)u'. Set for any

K..

av (t) =

(1)

vEWµ

We get a basis of (U° )W, if we take all av(p) with p running over a system of representatives for the W orbits in Zf? n 2A. Each orbit contains exactly one dominant weight, hence exactly one weight µ such that -µ is dominant. Therefore the av(-µ) with µ E Z4 n 2A dominant are a basis of (U)w. We want to show that all these av(-µ) are in the image of -y-p o 7r, and use induction on p to prove this. The case it = 0 is of course trivial. (We have av(0) = 1 = ry_Poir(1).) For any dominant u E Z4n2A the weight A = (1/2)µ E A is dominant with 2A E Z-1). So we can apply the construction from 6.23 to A and get an element zA E Z(U) with (by 6.24(1))

av(-µ) +

dim

ry-P o 7r(zA) _

dim L(A), av(-2v). v
v

Here we have used for the second equality that dim L(A),,,, = dim L(A), for all v E A and w E W, and that dim L(A)A = 1. All v occurring in the last sum satisfy 2v < µ and A - v E Z- I,, hence 2v E M. So by induction all av(-2v) in the sum are in the image of -y-p o 7r, hence so is av(-µ). 6.26. We shall see in 8.30 that we can drop in Proposition 6.18 the assumption that char(k) = 0 and that q is transcendental over Q and replace it by the weaker assumption that q is not a root of unity. I want to sketch here how we can use this fact to remove that stronger assumption from the other results (5.15, 5.17-5.19, 6.20-6.25) where it occurred. To start with, it is clear that with 6.18 also everything in 6.20-6.24 extends and that the proof in 6.25 shows that U,0,,,°

)w C 'Y-P 07r(Z(u))

(1)

We use this to show:

CLAIM. Let A, p E A with A dominant. If Z(U) acts on M(A) and M(µ) by the same character, then p + p E W (A + p). PROOF OF THE CLAIM. By (1) there is for each v E (Z-p) n 2A an element z E Z(U) with ry-P o E.EW K,,,,,,, hence with

1:

q

(wv.P)

Kwv

wEW

By Lemma 6.3.a each z acts on M(A) as scalar multiplication by

j

1` q(WV,A+P) = 1` q(v.w(A+P)). wEW

L

wEW

,

126

6.

THE CENTER AND BILINEAR FORMS

similarly on M(µ). If Z(U) acts on M(A) and M(p) by the same character, then we get

E q(v,w(a+P)) = E q(v,w(µ+P))

for all v E (Z4) n 2A.

(2)

wEW

WEW

Since (Z4) n 2A has finite index in Z- P, distinct µ' E A define distinct characters v ti on (Z4) n 2A. Because A is dominant, the w(A + p) with w E W are distinct. So (2) yields by the linear independence of characters that each w(A + p) has to occur among the w'(µ + p), hence the claim.

This claim implies now for all dominant A E A that L(A) = L(A). Indeed, if this were false, then L(A) had a composition factor L(µ) with µ < A dominant. Then Z(U) would act by the same character on L(p) and L(A), hence by the same character on M(p) and M(A). The claim yields now µ+p E W(A+p). Since p+p and A + p are dominant, this yields µ + p = A + p, hence µ = A, a contradiction. Since we have now the equality L(A) = L(A) in general (q not a root of unity), we see that also 5.17-5.19 extend to the general case. Having 5.17 we get now also Lemma 6.3.b and (hence) Theorem 6.25 in general. Finally, we can prove Weyl's character formula for L(A) using the approach in [H], 24.2.

Appendix Lemma 6.10:_ Let Y1, y2 E

such that 6.10(6) holds for yl and Y2 and arbitrary xi. We want

to prove it for Y1 Y2. Let x1, x2 E U'-0. Write for r = 1, 2

0(Y') =

yri ® yri

and

1 (xr) _ E xri

r

Then [using the assumptions and 6.10(5)] (yly2, x1x2) = (y1 ®y2, 0(xlx2)) = (y1 ®y2, 0(x1)i(x2))

E(y1 ®y2, xlix2j ®xlix2j ) i,j

= E(yl, xlix2j)(Y2, xiiX2j) i,j

= E(0(yl), x2j ®x1i)(i(y2), x2j ®xli) i,j

E (ylr ®ylr, x2j (9 x1i)(y2s (9 y2s, x2j ® xii) i,j,r,s

E (Ylr,x2j)(ylr,xli)(y2s,x2j)(y2s,xli)

i,j,r,s

E (Ylr ® y2s, x2j (9 x2j)(ylr (9 y2s, xii ® xii) i,j,r,s

APPENDIX

127

(yly2,xlx2) _ 1(y1r (9 y2s,1 (X2))(ylr ® Y2'., 0(xl)) r.s

=

(Y1ry2s, x2) (ylry2s, x1) r,s

E(Y1ry2s ® yirY s,X2 (9 x1) r,s

(O(y1)O(y2), x2 ®xl) = (O(y1y2), x2 ®xl)

Proposition 6.20, the case u = F«:

We have S(F«) = -F«K«, hence ad(S(F«))

ad(F«) ad(K«). Lemma 6.19

implies now

ad(F«) v = q-(µ,«) ((Fay

-

q-('+v,«)yF«)Kv+«)Kax

+ qqa - gQ- l (yKv)(KAr' (x) - 4 (li-a,a)Ka+2ara(x) and

ad(S(Fa)) v'=

-q-(v',a)

((Fay' - q-(\'+v'.a)y Fa)Kv'+a) Ka'x

qa - qa

(y'Kv') (KA'ra(x') - q-(µ -a,a)KA,+2ara(i )).

This shows: If v' = u and p' = v + a, then (ad(Fa)v, v') = q-(µ.a)(y ,x)q(2p'v+a)(q1/2)(1)

-(A+v . a) ((F'«y, x ' ) - q (yF'a, x ')) and x)q(2p,v)q-(v'-a.a) (v, ad(S(E«)) v') = (q« - q« 1)-1(y , (q-(µ'-a,a)(y,r«(x'))(q1/2)-(,\,a'+2a) - (y,r«(x'))(q1/2)

Pull out the common factors q(2p,v+a)(y' x)(q1/2from (Recall that v' = p and that q(2p>a) = q(«.a)) We are left with

(2)

(1) and (2).

(Fay, x') - q-(A+v'a) (yE«7 x'). resp.

(q,,,-q-')-' (q-(A+v.a)(y, ra(x')) - (y, r'«(xl)))

(2')

Now 6.14(5) combined with (Fa,Ea) _ (qt - qaimplies that the terms in (1') are equal to the terms in (2'), hence that the left hand side in (1) is equal to the left hand side in (2).

128

6.

THE CENTER AND BILINEAR FORMS

If v'=ti -aand µ'=v, then we get (ad(FF) v, v') = (qa - qQ

1)-1(y,

((y',

x')q(2p,v)q-(µ-a'Q)

ra(x))(g1/2)-(\,\')

- q-(µ-a,a) (y',

(3) ra(x))(g1/2)-('\+2a1'\')),

and (y,x')q(2p,v)(g1/2)

(v, ad(S(FF)) v') = q-(v 'a)

It

(q-(A'+v',a)

(y'FF, x) - (Fay , x)) .

that v' _

Pull out the common factors a). We are left with (qa - qa 1)-1 ((y', ra(x)) resp.

()

-

q-(u,-a,a) (y', ra(x))q-(a'\'))

q-(y Fa,x) - (Fay', x)

(3')

(4')

Again 6.14(5) shows that the terms in (3') and (4') are equal, hence so are the left hand sides in (3) and (4). For all other choices of µ' and v' we have

(ad(Ea) v, v') = 0 and (v, ad(S(Ea)) v') = 0 by 6.20(2). So we get (ad(Fa)v, v') = (v, ad(S(Fa))v') in all cases.

CHAPTER 7

R-matrices and kq [G] We keep the assumptions of the last chapter. Furthermore, we assume that char(k) = 0 and that q is transcendental over Q. (Everything can be extended to the case where q is not a root of unity using 8.30, cf. 6.26.) In this chapter we want to extend to the general case several constructions made in Chapter 3 in the case g = si2. This includes the construction of a "commutativity constraint", the relation with the quantum Yang-Baxter equations, and the hexagon identity. I briefly mention a relationship with the Hecke algebras in the case g = sin. Furthermore I describe how then U9(g) lead to quantum deformations of the rings of regular functions on semisimple algebraic groups.

7.1. Choose for each p E Z4, p > 0 a basis ui, u2 , ... ur(µ) of U+. By Proposition 6.18 we can find a basis vi , v2 , ... , vr(µ) of U µ such that (v , uµ) = big for all i and j, cf. the proof of 6.21. Set r(µ)

OµVA ®uµEU®U.

(1)

i=1

By linear algebra O, does not depend on the choice of the basis (uµ)2. For example,

we can replace the uµ by r(uµ); then 6.16(1) implies that we have to replace the vµ by r(vµ). This shows (T ®T)e = Oµ. (2) Alternatively, we can replace the uµ by w(vµ). Then 6.16(1) says that we have to replace the vµ by w(z ); we get thus

(w ®w)Oµ = P(e ),

(3)

where P is the operator with P(a ® b) = b ® a for all a, b.

We set Oµ = 0 for all u 0. We have e0 = 1 ®1 since (1,1) = 1. For each aeIIand each n>0we have enct=(FQ,Ea)-1F 0 En,, and we can find (F, EQ) in 6.13(4). In the case where g = si2 we see then that our present Ona is equal to the On from 3.11(1). So our next lemma generalizes Lemma 3.11. LEMMA. We have for all p E Z4, u > 0 and all a E II (E®(9 1)Oµ + (Ka E) E,)191,-,, = Oµ(&, ®1) + Oµ-a(Kct 1 ®E

(1®Fa)e

+(Fa®KK1)Oµ-a=Oµ(1®Fa)+Oµ-a(FQ®K.),

(Ka ® K.)O9 = Oµ(Ka ® Ka). 129

(4)

(5) (6)

7. R MATRICES AND kq[G]

130

PROOF. The last equation follows from the general fact that for all u E U and

(KA ®KA)(u ®u)(K- 1 (9 K-

1)

= (q(a`)u) ® (q(.,.')u) =

(u ®u').

The proof of (4) uses the fact that we have for all u E U+ (resp. all v E U µ)

u=

(vi , u)uµ

v=

resp.

(v, uµ)vN'.

We get (setting ca = (qa - qQ 1)-1)

(Ea®1)eµ-®µ(E.(9 1)(Eavi'-viEa)®ui i

= ca E(Kara(vµ) - r'(v!`)Ka

Ca E (Ka E(ra(vi ), j -a)vj Z

[by 6.17(1)]

1) ®uN'

-a

3

- (ra(v2 ),

-a)vj

-aKa

_ [ r-Ka (vµ' Eav7 -a)v7 -a 11

+1](v2 ,

®uN

[by 6.15(5)]

7

K,-

-

E(vl',ul-aE(,)up Eau -a)uN'

Kav"-a

f

/

_ 1:(v'-aKn 1

Kav7-a ® E(,uj-a)

j eµ-a(Ka 1 ® Ea) - (Ka 0 Ea)O,-a. Note that we have ra(vµ) = 0 = r' (vN') in case p - a the second line above. The calculation needed for (5) is similar.

0; we get then 0 already in

7.2. In the following subsections we are going to generalize the st2 results from 3.12-18. First we need the twisted comultiplication TA = (-r 0 T) o A o T, cf. 3.8. A look at the definitions (4.8 and 4.6) shows immediately for all a E 11

TA(Ea)=Ea®1+Knl®Ea,

TA(F.)F'a®Ka+1®Fa,

(1)

{A(K(,) = Ka ®Ka. Let 11 and til' be finite dimensional U -modules (as always: of type 1). We have

811(MA 0 M,1) C Ala_µ ®A1"+,,

for all A, A' E A, p E M.

(2)

7 R MATRICES AND kq[G)

131

Since the set of weights of M is finite, there are only finitely many i E Z4; that are differences of weights of M. Therefore we have Aµ (M ® M') = 0 for almost all u and we can define a linear transformation

©=E

9=eM,M':M®M'---,M®M,

(3)

.

µ>0

We can choose bases of M and M' consisting of weight vectors and then take their

tensor products as a basis of M ® M. If we order this basis suitably, then (2) shows that each Oµ with u > 0 acting on M ®1 i' has a strictly upper triangular matrix for this basis. Since ©o = 10 1 acts as the identity, this shows that ©M,M, is invertible, in fact unipotent. Lemma 7.1 implies now

0(u) 0 E)M,M' = ©M.M' o 'A(u)

for all u E U.

(4)

7.3. Consider a map f : A x A - k" with

f(A+v,,u) = q-("µ) f(A,i)

and

f(A,,a +v) =

(A,i)

(1)

for all A, ,u E A and v E Z. Such a map exists: Choose a system of representatives A1, A2, ... , A,. for A/Z4;, choose the f (Ai, Aj) arbitrarily, and set (v, v') f(Ai, A)

f (Ai + v, Aj + v') =

for all i, j and all v, v' E Z.

(2)

1

We define now (cf. 3.13(3)) for all finite dimensional U-modules M and M' a bijective linear map f : M ®117' -> M ®117' by

f(m®m') = f(A,p)m®in

for all m E M,, and m' E Mµ

(3)

(and all A, u). We denote again by P the map M' O M -> M ® M' with m' ® m m® m'. THEOREM. For all finite dimensional U-modules M and M' the map

©o foP:M'®M---*M®M' is an isomorphism of U-modules.

PROOF. More or less the same as in 3.13 and 3.14. REMARK. This construction has the same functorial properties as that for 512 in the remark in 3.14.

7.4. In order to generalize the formulas 3.16(3)-(6) we need something more sophisticated than 3.16(1). Consider an element x E U+ for some p E Z4) with p > 0. We know by Lemma 4.14 that 0(x) E U± K ®U,±. So there are c j E k such that (in the notations from 7.1) 0(x) c juµ-"K" ® u . We get then c" = (vµ-" ® v''. A(x)) = (v!'-`v",", x), where the last equality follows from 6.12(1). So we get

0(x)

E(viy-"v

0
,

x) ui "K,, ®u .

(1)

7. R-MATRICES AND kq[G]

1.32

One shows similarly for all y E (L-I"

0(y)

®v`-vKv 1.

_ 0
(2)

For any vector space V there are 6 natural embeddings V 0 V - V 0 V ® V; we denote them by z - zi j where i, j are two distinct numbers in 11, 2, 3}. More explicitly, if z = Ei ai 0 bi, then Z12 =

Z21 =

ai ® bi 0 1,

b® ® ai ® 1,

z31=bi®1®ai,

z13=Eai®1®bi,

(3)

i

Z23=E1®ai®bi,

1 ® bi 0 a,

Z32 = 1.

Of course, z12 = z 0 1 and Z23 = 10 z. LEMMA. We have for all

0

(eµ-,)23(1 ®K® 1 ® 1)(0,)13,

(4 ®1)Oµ =

(4)

0
(1 ®0)Oµ = E (6µ-,)12(1(9 K ®1)(0,)13

(5)

0
PROOF. We get from (2)

(0

0(vh) ®'ah

1)Oµ = h

=1: 1:

®vi-"K 1 ®uh

v.i.j

h

_ 1: vj' ®vz'-vKv v.i.j

_

1: (vh, uµ-vuJ) uh h

vj" ®vµ-vKvI

_ (eµ-,)23(1 ®K 1 ®1)(0,)13 v

This gives (4); the proof of (5) is similar.

7.5. We can use the two formulas from Lemma 7.4 together with 7.1(2) to generalize 3.16(5),(6): We get for all µ

(TA ®l)0µ =

(0,)13(1 ®K (9 1)(Oµ-,)23,

(1)

0
(1 ® T0)0µ = E (8013(10 K1-1 ®1)(eµ-,)12 0
(2)

7 R-?MATRICES AND kgIG]

133

Indeed, we have

(TA®l)19µ = (T®T(9 1)(A®1)(T(D 1)Oµ _ (T (D T ®T) (A ® 1) (T (D T)Oµ

=(T®T(D T)(A®1)Oµ. Now (1) follows from 7.4 (4), since TOT ® T is an antiautomorphism, hence reverses

the order of the factors, and since (in the notations from 7.4(3)) T ® T ® T(z,j) _ (T ® T(z))ij for all z E U ® U and all i, j. (The argument for (2) is similar.) Given three finite dimensional U-modules M, M', and M", we can define (as in 3.17) three automorphisms Oi2, 013, and 023 of M ®M' ®M" as a vector space. of). The description of (So we set 012 of) ®1 and O23 = 10 (Oh 19i3 is a bit more complicated.) THEOREM. We have for all finite dimensional U-modules M, M', and M" 1912°1913 -192f3 - 1923°1913°191'2'

(3)

PROOF. More or less as in 3.17. One introduces linear maps fig as in 3.17. One checks (using 7.2(2) and 7.3(1)) for all i E Z4> that f12 0 (1912)13 = (017)13 0 (1 0 K,1 0 1) 0 f12

(4)

f12 0 f13 0 (19023 = (19,q)23 ° f12 0 f13

(5)

and

In this way one gets that the left hand side in (3) is equal to 1912° (j:(0q)13(10 K,10 1)) 019230f120 f130 f23. 77

Similarly, one shows that the right hand side in (3) is equal to 1923 0 ((19+1)13(1 0 K,, 1 ® 1)) 0 1912 0 f23 ° f13 ° f12 n

The f j commute obviously. Furthermore, (1) resp. 7.4(4) implies

0120(E(On)13(1®K,1®1))0023=19120WA ®1)19) 11

resp.

e23°(

(e,7)13(10 K771®1))01912=((A0

Finally, 7.2(4) yields ((A ® 1)O) 0 1912 =1912 0 ((TA ® 1)9).

019)0012,

7. R-MATRICES AND kq[G]

134

7.6. Theorem 7.5 says that the Of yield solutions to the quantum Yang-Baxter equations, cf. the remark in 3.17. However, there is an additional way of looking at the result that I want to describe now. [n order to simplify notations let me work with just one finite dimensional U-

module V and take M = M' = M" = V in 7.5. Any permutation a of {1, 2, 3} defines a linear automorphism PQ of V ® V ® V via PP(v1 (9V2 ®v3) = VQ-1(1) ®V,-1(2) ®41,-1(3).

(1)

Set P12 resp. P23 equal to the PP where a is the transposition (1,2) resp. (2, 3). By 7.3 the maps R12 = OfI2 o P12 and R23 = 023 o P23 are automorphisms of the U-module V ® V ® V. Now I claim that 7.5(3) is equivalent (in the case

M=M'=M"=V)to

R12 ° R23 o R12 = R23 o R12 o R23

(2)

Well, note first that

PQ 0 O = Of

0 P°

(3)

for all permutations a and all i, j. (We define O analogously to zip in 7.4(3).) We get therefore R12 o R23 o R 12 = O 12 o P12 0 823 o P23 0 O 12 0 P12

=O120O130P120 O130P230P12 O120O130O230P(1.3) Similarly, R23oR120R23 =O230 P230 0120P120 O230 P23

= 023 0 013 o P23

° 013 0 P12 0 P23

=O230e13oO12oP(1.3).

Since P(1.3) is bijective, this shows that (2) and 7.5(3) are equivalent. Consider now an arbitrary tensor power

V®r = V ®V ®V ®... ®V

(r factors)

of V. We get then for 1 < i < r an automorphism Ri of the U-module V®r with Ri (v1 ® V2 ® (9 Vr) = VI ®... ® 41x_ 1 ® (Of o P(vi (9 vi+1)) ® vi+1 ®... ® V.

Now (2) implies for all i with 1 < i < r - 1 that Ri o Ri+i o Ri = Ri+1 o Ri o R,+1

(4)

On the other hand, we have obviously

Rio Rj =R; oRi

(5)

for all i and j with I j - il > 2. Now (4) and (5) say that the Ri satisfy the braid relations; so they define a representation of the braid group with values in Endu(V®r).

7. R-MATRICES AND kq[G]

135

7.7. Let me go back to the $C2 example in Chapter 3. In 3.15 1 wrote down the matrix of Of. The matrix of R = Of o P is then q-1

10

0

0

0

1

0

1

0

0

q-1

0 0

-q

0

(1)

0 q

The square of this matrix is equal to q-2

0

0

0

0

1

q-1 - q

0

0 0

q-1 - q 1+(q-1- q)2 0

hence to

(q_1 - q)

0

q-2

0

q-1

0

0

0

0 0 0

0

1

1

q-' -q

0 0

0

0

q-1

+

1.

This implies R2 = (q-1 - q)R + 1, which will look a bit more familiar to some, if we rewrite it as (qR-1)2 = (q2 - 1)(qR-1) + q2. (2) Consider now the constructions of the last subsection with g = 512 and V = L(1, +). Then the R above is equal to the Of o P involved in the definition of the qR7I, then the Ri still satisfy the braid relations, i.e., we Ri in 7.6. If we set Ri = have still

RiRi+I Ri = Rt+, R'R'+1

and

RfiRj = RJR; f o r Ij - iI > 2,

and we have additionally (R')2 = (q2 - 1)Ri + q2

for all i.

This implies that the subalgebra of Endu(V®r) generated by the Ri is a homomorphic image of a Hecke algebra of type A,_1i cf. [Humphreys 2], 7.4. This example has been generalized in [Jimbo 2] to g = stn, for any n with V the quantum analogue of the "natural representation" of g, i.e., with V = L(ZV1) for the usual numbering of the fundamental weights. His work shows not only that the R, in that case generate a homomorphic image of a Hecke algebra, but also that they generate the full endomorphism ring Endu (Vgr). That is the quantum analogue

of the classical result that the endomorphisms for the r-th tensor power of the natural are generated by the symmetric group Sr acting by permuting the factors of the tensor product. 7.8. THEOREM. Suppose that

f(A,,a +v) = f(A,,a)f(A,v)

and

f(A+µ, v) = f(A,v)f(µ,v)

(1)

for all A, µ, v E A. Then the Of o P satisfy the hexagon identities. PROOF. This theorem says that the diagrams in Theorem 3.18 commute for all finite dimensional U-modules M, M', and M". The proof is the same as in 3.18.

7. R-MATRICES AND kq[Gl

136

7.9. We can choose an integer d > 0 such that d(A, a) E Z

for all A, i

A.

(1)

For example, d = [A: M] will do, since then d\ E Z4 and (d),µ) E Z by 4.1(2). Suppose that k contains a d-th root of q, fix such a root and denote it by q1/d We can then set f(A ,u) = (q1/d)-d(A.I,)

for all A, µ E A.

(2)

This function satisfies 7.3(1) and 7.8(1). So it leads to maps ®f o P to which all results in 7.3-8 apply. 7.10. We can apply the construction from 6.7 in particular to A = U. Recall that we get thus a ring structure on U* where the product of two linear forms f,g E U* is given by:

If 0(u)

ui ® ui, then (fg)(u) = Lei f (ui)g(u')

(1)

Recall from 6.22 the definition of matrix coefficients: For a finite dimensional

U-module M and for all m E M and f E M' we denote by cf,,,,, E U* the linear form with cf.,(u) = f (um) for all u E U. LEMMA. Let M and M' be finite dimensional U-modules. Then we have Cf.mCf',m' = Cf5,;f'.m®m'

(2)

for all mEAl,m'EAV and all f EM*, f'EM'*. PROOF. We have for all u E U (writing A(u) _ Ei ui ® ui)

cf®f'.m®m'(u) _ (f ® f')(u(m ® m')) _ (f ® f')(Euim ® um') i

_

f (uim)f'(uxm) _

Cf,m(ui)Cf'.m' (Ui).

So the claim follows from (1).

7.11. Lemma 7.10 implies that the subspace of U* spanned by all cf,, (for all finite dimensional U-modules M and all f and m) is closed under multiplication. It contains the ring identity e of U*, since a is a matrix coefficient for the trivial one dimensional U-module. So this subspace is a subalgebra of U*. We denote this subalgebra by kq[G] where G is the simply connected, connected semisimple algebraic group with Lie algebra g. One should regard kq [G] as a quantum analogue of the algebra C[G] of regular functions on G.

Let me try to justify the last statement. The enveloping algebra U(g) is a Hopf algebra, so we have as in 6.7 an algebra structure on U(g)*. We can define matrix coefficients equally well for U(9)-modules and then generalize Lemma 7.10 to this situation. We get now that the subspace of U(g)* spanned by the matrix coefficients of all finite dimensional U(9)-modules is a subalgebra of U(g)*. This subalgebra is isomorphic to C[G]. In order to understand the last claim, let me approach it the other way. Let me start with C[G] and let m C C[G] be the maximal ideal of all regular functions

7 R-MATRICES AND kq[G]

137

on G that vanish at the identity of G. There is on C[G] a Hopf algebra structure, hence on C[G]* an algebra structure. One can check that {

E C[G]* I there is an integer r > 0 with O(mr) = 0 }

is a subalgebra of C[G]*. A theorem of Cartier says that this subalgebra is isomor-

phic to U(g). (Note that g - m/m2 is embedded into this subalgebra as the set of E C[G]* with o(1) = 0 and cp(m2) = 0.) We have thus an embedding of U(g) into C[G]*, hence dually a linear map t : C[G] -* U(g)*. This map is injective: all

If t(f) = 0, then f E m" for all r, since (C[G]/mr)* C U(g). But nr m" = 0 by Krull's intersection theorem, since G is connected, hence C[G] an integral domain. Furthermore one checks (chasing through the definitions) that t. is an algebra homomorphism.

So we have embedded C[G] as a subalgebra into U(g)*. It remains to check that the image of t. is exactly the space spanned by the matrix coefficients of finite dimensional U(g)-modules. Well, consider a finite dimensional rational G-module

M. Pick a basis e1, e2, ... , en of M. Then there are fji E C[G] with gei = E3 f3(g)e3 for all g E G. Then Al is made into a finite dimensional U (9) -module

via uei = Ef u(fji)ej for all u E U(g). By definition t(fji) is the function u H u(fji). So t(fj,) is a matrix coefficient of the U(g)-module M. Since C[G] is spanned by matrix coefficients of G-modules, this shows that the image of t. is the subspace of U(g)* spanned by the matrix coefficients of all finite dimensional U(g)modules that extend to G. So far everything works for any connected algebraic group over C. Now we need a special feature of G to finish the argument: All finite dimensional g-modules lift to (the simply connected) G. *

*

*

As in the classical case, there is a Hopf algebra structure on kq[G]: Consider a finite dimensional U-module M, choose a basis e I , e2, ... , en of Al, and denote the dual basis of M* by fl, f2,... , fn. Take for all m E M and f E M* the element n

1: cf,e, ®cf,.m E kq[G] ®kq[G] C U* ®U*; =1

we can regard it as a linear form on U ® U via the canonical embedding U* ® U* (U ®U)*. We get .e,

®cf,.,,,)(u ®u') _

.f(uei)fi(um)

= f (u

fi (u'm)e

= f(uu'm) = cf.m(uu') for all u, u' E U. By linearity it follows now that there is for each cp E kq [G] a unique A* (gyp) E kq [G] ® kq [G] with

A* (co) (u ®u) = o(uu)

for all u, u' E U.

The map A*: kq [G] _* kq [G] ®kq [G] can be checked to bean algebra homomorphism

and to be cocommutative: This is the comultiplication on kq[G]. The map e* kq[G] -* k with is the counit of kq[G]; it satisfies e(cf,m) = f(m). :

7. R-MATRICES AND kq[G]

138

Finally, we get an antipode S* on kq[G] via S* (cp) = cp o S. This maps takes kq [G] to itself since S* (c f.,,,,) = c,, f where we identify m with its image under the standard isomorphism (of vector spaces, not of U-modules) M (M*)*. It is not too hard to prove the Hopf algebra properties.

7.12. Let M be a finite dimensional U-module. Choose a basis e1, e2, ... , e,, of 111, and denote by cij the matrix coefficients with uej = Ei cij (u)ei for all j. (So c'j = where the fi are the dual basis of the ej.) Let M' be another finite dimensional U-module, with basis e', e2, ... , e' and corresponding matrix coefficients c' . Suppose that we have an isomorphism R : M'0 M -- Me M' of U-modules. Write R(ei (& ej)

R J eh ®el'

(1)

hJ

for all i and j. LEMMA. We have for all i, j, r, s rs / Rhl CliChj =

hJ

/ Rijhl CrhCsl

(2)

h.1

PROOF. Lemma 7.10 implies for all u E U

u(ei ® ej) = j:(CliChj)(u)el 0 eh, h,l hence

R(u(ei ® ej)) = j:(c chj)(u)R(ei ® eh) _ hJ

Rhl (CitChj)(u) er ® e'.. h.l.r,s

On the other hand

u(R(e' (9 ej)) _

R ? (CrhCsl)(u)(er 0 e').

R lu(eh ® el) _ hJ

h,l,r,s

Now compare coefficients.

7.13. Lemma 7.12 says: The R-matrices that yield the isomorphisms in 7.3 lead to relations between elements of kq[G]. Take as an example g = 512 and M = M' = L(1, +) with R given by the matrix in 7.7(1). We denote the basis of M now by e1, e2 as in 7.12 (instead of the previous mo, m1). We have in this case

R11 =R22 =q',

R12 =R2i = 1,

R2i -q-1 -q.

The remaining R2., are equal to 0. We get now from 7.12(1) the following relations C11C12 = gc12C11,

C12C22 = gc22c12,

C11C21 = gC21C11,

C21C22 = gc22c21,

C12C21 = C21C12,

C11C22 - C22C11

= (q - q-

(1)

)C12C21

In this case kq[G] = kq[SL2] is generated as an algebra by these four c,j. (Well, by 7.10(2) the algebra generated by the cij contains the matrix coefficients of all tensor powers of M = L(1, +). Each simple module L(n, +) is a direct summand of such a tensor power, therefore the matrix coefficients of each L(n, +) are in the algebra

APPENDIX

139

generated by the cit. We do not get any new matrix coefficients by taking direct sums. So the claim follows from the complete reducibility of all finite dimensional representations. Recall that we always look at modules of type 1 only.) The relations in (1) are not a complete set of relations defining kq[SL2]. The algebra generated by the ciJ and the relations (1) is called the algebra of quantum (2 x 2)-matrices kq[M2]. (If you want to compare with the version in Manin's Montreal notes, replace q by q-1; if you prefer Smith's survey, replace q by q2.) In order to get kq[SL2] one has to add the relation (2)

C11C22 - gc12C21 = 1.

This is usually expressed as: The quantum determinant of the ci7 is 1. It is not difficult to check that (2) holds in kq[SL2]; it is more work to show that (1) and (2) are a complete set of relations. For a discussion of the relation with other approaches to kq[G] in the case of Sln you may want to look at Polo's appendix.*

Appendix 7.1(5):

(1® F,,)©N - ®,,,,(1 ® F,,) _

vz' ® (F,,, - OF,,)

= CO > vµ ®(K; = cQ

vµ ® Kc 1 i

r0(uµ)K,,)

j

(vi

,

r.(uµ))

- E(v3

[by 6.17(2)] -a

K /

i

+

[by 6.14(5)]

(1: (vj-'Fc,U J

i

-

(F.vjl -',

uµ)v® K. 1uµ ®

®µ-Q(F. ® K.) - (F. ®Ka *to the Andersen-Polo-Wen paper in Invent. math. 104(1991), 1-59

7. R-MATRICES AND kq[G]

140

7.4(5):

(1 ®0)iµ =

LL}z

4 ®A(u}z) µ-vvv 2Lµ 7lµ-v Kv ®U v.i.j

EE( }z

It

%j ,u'h) 1µh® uL `Kv ®v'j

v

vµ'vvv

Evi_vv

®26i_vKv®v')

v.i,j 1:(o,,-v)12(1

® K® (9 1) (19 v)13

CHAPTER 8

Braid Group Actions and PBW Type Basis We suppose as in Chapter 5 that q is not a root of unity. *

*

*

Take a simple root a and suppose that we have chosen root vectors x. E g,, and y,, E g_,,, such that a([x,,, y,.,]) = 2. Then one defines for each finite dimensional g-module V an automorphism of the vector space V by i,, = eXp(x,,,.v) exp(-y,,v) exp(xa.v),

where x,,,v denotes the endomorphism of V given by the action of x"; similarly for ya,V. This automorphism has the property that saVN, = Vsnµ for ally E A; cf. [H], 21.2(5), (6). We have for all v E V a

b

a!

b!

x c

(-1)bXa y a.b,c>0

c!

v.

(*)

If v E Vµ, then xayaxav E Vµ+(a-b+c)c,; on the other hand

E Vsy = V,,-.,,,o

where m = 2(µ,a)/(a, a). Therefore we have i a.b,c>0.-a+b--c=m

i Nb

b xc a ya a

xa a!

b!

c!

all other summands in (*) together contribute 0. We want to generalize this construction to the quantum case. Lusztig has discovered two "natural" ways of doing it, cf. 8.2(3), (4) and 8.6(2), (3). There are two additional versions if one exchanges the roles of a and -a. However, all four maps turn out to be very closely related and we shall concentrate on one of them that we denote by T,,. In the Lie algebra case we can apply this construction in particular to the adjoint representation and get thus an automorphism of the Lie algebra g:

sa(X) = exp(ad xa)exp(-ad ya)exp(ad x,,,) (X)

for all X E Q.

This automorphism is then related to the action defined on finite dimensional gmodules via

SQ(Xv) -

for all X E g and all v in a module. We could work here equally well with the enveloping algebra U(g) instead of g, since the adjoint action of g on U(g) is locally finite. 141

142

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

Now in our situation the adjoint action of U on itself is not locally finite. However it turns out that we still have an automorphism Ta of U that has an analogous property: We have T (uv) = Ta(u)Ta(v)

for all u E U and all v in a finite dimensional U-module. The existence proof for the Ta requires long calculations. (At first these automorphisms of U were constructed by Lusztig by prescribing their values on the generators and then checking the relations. Independently, Levendorskii & Soibelman constructed them as conjugations by elements in a completion of U.) I shall add more comments on the contents of this chapter after 8.15.

8.1. We first need another property of the Gaussian binomial coefficients.

LEMMA. Let a, b, and m be integers with b, m > 0. Then

E ,Uai-b(m-i) IM a i=O

[a+ zJ Lbj

=

mb]

PROOF. We use induction on b. The claim is obvious for b = 0 since [°] = bio.

For b = 1 the claim is just 0.2(1). Suppose that we have the claim for some b and look at b + 1. We get using 0.2(1) and the induction hypothesis m

a

vai-(b+1)(m-z) r

1

r b+ 1 i

Lm-iJ L

LL

i=O

v ai-b(m-i)-m+i

_ i=o M

Evaz-b(m.-z)-m

a

(v

[rn_i]

-i [b] + i

vb-i+1

[ma-i] [b] zJ

i-o M

+

vai-b(m-i-1)-m+1 rma,

i] Iib1J

i=1

m-1

Lra+bl + E Vaj-b(m-j-1)+a+b-m+l I

-a

=v-,n

mJ

jI

ILI

b

Lm1-K

7=0

v-m ra+bl +,Ua+b-m+l r a+b 1 _ M

Lm-1J

IL

M

J

REMARK. If we apply the automorphism of Z[v,v-1] that interchanges v and v-1 (cf. 0.2), then we get from (1) M

Mvb(m-i)-air

a

m-i

1

rbl = [a+ bl MJ LzJ

.

(2)

i-o By symmetry we get then that (1) and (2) hold also if a > 0 and b E Z arbitrary. As pointed out in [L], 1.3.1(e), it can be extended to all a,b E Z.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

8.

143

8.2. Return for the moment (until 8.5) to the situation and notations from Chapter 2. So we take g = s(2 and write E, F, etc. instead of Ea, Fa, etc. and q = qa. However, we keep the convention from 5.4 and consider only finite dimensional U-modules V of type 1. So

V=®V,n

V,n={vEVI Kv=qm'v}.

where

(1)

mEZ

Let us introduce the divided powers

E(') =

E [r]

F

F(r) = [r]

and

(2)

for all r > 0. Define on each finite dimensional U-module V linear operators T, T', WT, and 'T' such that for all m E Z and all v E V,n

E a.b,c>0:-a+b-c=m E a,b,c>0:-a+b-c=m E (-1)bgb-acF(a)E(b)F(c)v.

(-1)bgb-acE(a)F(b)E(c)v,

T(v)

(-1)bgac-b E(a)F(b)E(c)v,

T'(v)

WT(v)

(3)

(4)

(5)

a,b,c>O;a-b+c=m

"'T'(v) =

(-1)bgac-bF(a)E(b)F(c)v E a,b,c>O;a-b+c=m

(6)

Note that these sums contain only finitely many nonzero terms since E and F act nilpotently on V. We get easily from 2.2(2) that all four operators map each Vm

to V,

We can twist each U-module V with the automorphisms w from 1.2, i.e., consider the same vector space and let u E U act via w(u) in the given module structure. Denote the twisted module by 'V; in order to distinguish the different module structures denote any v E V by "v when we regard it as an element of 'V. So we have by definition u Iv = w(w(u)v). Therefore (3)-(6) imply for all v and V

'T ('v) _ '(T(v))

and

'T' ("v) = W(T'(v)),

(7)

since w(E(a)F(b)E(c)) = F(a)E(b)F(c) and since v E V,, implies 'v E ('V)-m.

8.3. Consider in particular V = L(n, +) using the notation from 2.6. Let us replace the basis mo, ml,... , m,,, from 2.6 by the basis ve, v1, ... , v,,, where

vi =

li mi

(1)

[2]'

for all i. The formulas in 2.6 yield

Fvi =

[i + 1]vi+1, 0,

if i < n, if i = n,

Evi =

if i = 0, { [n+1 - i]vi_ 1, if i > 0. 0,

(2)

A trivial induction shows then for all i and r F(r)vi

=

rr+i l vi+r L

r

and

where we set vj = 0 for j < 0 and for j > n.

_ E (r)vi[n+r_i] r

(3)

144

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

LEMMA. We have for all i

WT(vi) _ (-1)igi(n+1-i)vn-z, WT'(vi) _ (-1)iq-i(n+l-i)Vn-i.

(-1)n-q(7L-i)(z+1)vn

T(vz.) =

e

(-1)n-zq-(n-z)(i+1)vn-

T'(vi) =

(4)

PROOF. Let us first look at the claims for T(vi) and T'(vi). We have vi E L(n, +)n_2x. So the definitions in 8.2(2),(3) show together with (3) that T(vi) resp. T'(vz) is equal to

b+i-C] [n+a-i-b+cvn

E(-,)b ge(b-ac) [n+c-i] LL

L

c

J

I

b

a

L

i

(5)

I

where we take e = 1 to get T(vi), and e = -1 to get T'(vi). Furthermore, the sum is over all a, b, c > 0 with -a + b - c = n - 2i. We get a zero summand unless c < i and b < n + c - i; since a = b - c - n + 2i we can thus assume also that a < i. Set j = n - i; we can rewrite (5) using b + i - c = n - i + a as E(-1)bge(b-ac)

rzl v n i+Cj+al z-C

(6)

I C J The claim is that the sum in (6) is equal to (-1)3gej(i+1)vn_z. So we have to show that La]

I

EE(-1)b+jqe(b-ac-j(i+l)) j+ c t [

a-0 c_0

C

j+a I

li-c] [a]

1'

(7)

where b = a + c + j - i. This is an identity for Gaussian binomial coefficients that we check in the appendix using Lemma 8.1. Let us now turn to the claims for WT and IT. The formulas in (2) and Kvi = qn-2ivz imply easily that the map co : W L(n, +) -* L(n, +) with V(Wv,) = vn_i for all i is an isomorphism of U-modules. Any homomorphism of U-modules commutes with the operators T, T', IT, and WT'. We get therefore using 8.2(7) WT (vi) = WT (,

Wvn-i))

= o ( IT (W'vn-z)) = o ( W (Tvn-i)) ((-1)igi(n+I-i) ;.'vi) = (-1)igi(n+l-i)vn-i,

similarly for WT'.

8.4. LEMMA. The operators T, T', IT, and WT' are bijective on each finite dimensional U-module V. We have T-1 = WT'

and

Ti-1 =WT.

and

WT'(v)

(1)

One has for all m E Z and all v e V,,,, WT(v) =

(-q)-m'T(v)

=

(-q)-T(v).

(2)

PROOF. It is enough to consider the case V = L(n, +), since every finite dimensional module is a direct sum of these simple modules. Then the bijectivity of the operators is clear by Lemma 8.3. In order to check (1) and (2) we may assume v = vi for some i. Again the claims follow easily from the formulas in Lemma 8.3.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

8.

145

8.5. LEMMA. We have for all finite dimensional U-modules V and all v E V

T (Ev) = (-FK)T(v), T (Fv) = (-K-1E)T(v),

ET(v) = T ((-K-'F)v), FT(v) = T ((-EK)v),

T (Kv) = K-1T(v),

KT(v) = T (K -'v).

PROOF. It is enough to consider V = L(n, +) and v = vi. We have Evi = [n+ 1 - i]vi_1, hence (-1)n-i+1gi(n-i+l)[n+

T (Ev) = On the other hand hence

T (v)

1 - i]vn-i+1

1)n-iq(i+ 1 )(n-i) ,U

FT (v) _ (-1)n iq(i+1)(n i)[n - i + 1]vn-i+1

Comparing with the first formula we get

T (Ev) = -q2i-nFT(v).

(1)

This yields the first formula in the lemma, since T(v) is a multiple of vn_i, hence has weight 2i - n. The proof of the remaining formulas is similar. (The formulas in the right column can easily deduced from those in the left one.)

8.6. Return now to the general case U = Uq(g) with g arbitrary. We extend the definition of divided powers from 8.2(2) and set for all simple roots a Ea E(r) = (1) and F(r) = F.' a

[r],

[r] a

a

for all r > 0. We define for all simple roots a and each finite dimensional U-module V linear operators T,,,, T,,,, WTa, and -T,,, on V such that for all A E A and all v E V,, (2)

(-1)bgb-acE(aa)Fab)E(c)v,

Ta(v) _ a,b,c>0;-a+b-c=m

(-1)bga

T., (v) _

bE(aa)Fa(b)E(ac)v

(3)

a,b,c>0;-a+b-c=m (-1)bgba-acF,(,a) Eab)Fa(c)v,

WTT(v) _

E

WTa(v) _

(4)

(- 1)bgac_bFaa)E(ctb)Fa`)v,

(5)

where* m = (A, aV ), If we consider V as a Uq, (sC2)-module via the embedding in 4.4(5), then Ta is just the operator T from 8.2(3); similarly for the other operators.

We can therefore apply the results from 8.2-5. We get in particular: All Ta and T; are bijective. Their inverses are given by (Ta)-1

= WTa

and

(T.,)-' = WTa

(6)

The formulas in 8.4(2) imply for all A E A and all v E VA WTa(v) =

(-qa)-("")

Ta(v)

and

'T«(v) =

*Recall the notation (a, a") = 2(A, a)/(a, a) from 4.1(3).

(-ga)(A.QV)

Ta(v)

(7)

146

BRAID GROUP ACTIONS AND PBW TYPE BASIS

8.

Each Ea°')F(c, )Ea°) maps any VA to V\+(a_6+c)«, and we have for each term in (2) and (3)

A + (a - b + c)a = A - (A, a') a = s,,A where s,, is the reflection corresponding to a. So we see (using the bijectivity) that for all simple roots a and all A E A T«(VA) = Vs \ = T«(VA) =

"T.(Va) = "T., (VA)

(8)

This implies easily for all it E M and all v E V T. (Kµv) = KsaµT. (v)

and

KT. (v) = T (Ksaµv).

(9)

Lemma 8.5 implies

T«(E. v) = (-F.K.)TT(v), Tc(Fov) _ (-K«'E.)Tc(v),

E«T. (v) =TT((-K«'FQ)v),

(10)

F. TQ(v) =T((-E.K«)v).

If 3 is a simple root with (a, (0) = 0, then EF commutes by (R5) with Ec. It always

commutes with F, hence with T. A similar argument works for F. We get thus

ET(v) = T (EFv) and FFT. (v) = T (FFv)

[if (a, (3) = 0].

(11)

REMARK. If you want to compare with the notation in Lusztig's book: The T0, here is his T"+1 and our T(, is his Ti','-,. Furthermore, our IT,, is his T'+1 and our "T,,, is his Ti'.-1. 8.7. We now want to have formulas for EFT (v) and T (EFv) for all ,3; similarly for FF instead of E. The formulas that we get will have analogues for T, IT, and IT,,, instead of Ta. They can usually be deduced from those for Ta using 8.6(6), 8.6(7), or a generalized version of 8.2(7). We are not going to write down all these formulas. But we now give one recipe explicitly.

LEMMA. Let a be a simple root. Suppose that we have u and u' in U such that

T (uv) = u' T0, (v)

(1)

for all v in all finite dimensional U-modules V. Then we have

IT. (w(u)v) = w(u) IT. (v)

(2)

for all v as above. If u E UN for some it E Z4, then we have also

T. (w(u)v) =

(-q.)-(µ.a") w(u)

T. (v)

(3)

for all v as above.

PROOF. In order to prove (2) we want to apply (1) to the U-module "V, cf. 8.2(7). We get for all v E V

IT. (w(u) "v) _ IT. ("(uv)) _ "(T. (uv)) _ "WT. (v)) = wW) "(T. (v)) = w(u') IT. ("v),

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

147

hence (2). In order to prove (3) we may assume v E VA for some A E A. We have then w(u)v E VA_N, since w(u) E U-µ. Now 8.6(7) yields T. (w(u)v) _

(-q.)(a-,,,v)

T. (w(u)v)

_ (-qQ)(a-µ,cv) w(u )WT. (v) _ (-q.)U-µ,av) (-q.)-(A,av) w(u) T. (v), hence (3).

8.8. The formula in 8.10 for TQ(Epv) will involve an element of the form ad(E,,(,m))EQ. We shall need several formulas for elements of this form that we prove in this and the next subsection. Let us first rewrite the formula for ad(E,,,) in 4.18 using divided powers: We

au

have for all u E U m

ad(E(m))u = i=O

K,,-,iE,(,,)

for all m > 0.

(1)

Fix ,3 E II, Q 54 a; set

r = -(Q, a").

(2)

We have then KQE8K;1 = q;TE,3. So (1) yields m

ad(E.m))E,3 =

for all m > 0.

(3)

i=0

(So ad(E,(,,m))Ea = x' ,(3;I.m;-1 in the notations of [L], 37.2.1.) The quantum Serre

relation (R5) implies that

for all m > r,

ad(E,(,,m))E,3 = 0

(4)

cf. Lemma 4.18.

We have ad(Fc,)EQ = 0 since F,,, and Ep commute by (R4). Therefore 4.4(7) implies easily - we could use the calculations in Chapter 2 instead - that

ad(F.)

[r + 1 -

m > 0, and more generally ad(FFs)) (ad(E(,m))E8) = l

J

[r + s- ml L

(6)

JCk

for all m, s > 0 with s < m; for s>mwe get ad(F,,(,s))ad(E(,m))EQ=0. 8.9. LEMMA. We have for all integers m, i > 0 (ad(E(m) )ER) E(,,)

=

(1)

[m+ j 1 qr-2m)-j(i-I)E -j) rad(Em+j))E) J

148

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

and

(ad(E(m))E3)FFZ)

(2)

=L.(-1)i [r-m+jl qj(i-L)F(i-j) i=a

3

L

J

(ad(E(--j))E3)K;

.

a

PROOF. We want to use induction on i; the case i = 0 is trivial for both formulas. Let us use the abbreviation

a(m) = ad(E(m))E3

for all integers m > 0.

(3)

We have ad(Eo)a(m) _ [m + 1]aa(m + 1); on the other hand

ad(Eo)a(m) = Eoa(m) -

Kaa(m)K,, LEa = Eaa(m) -

gain-ra(m)Ea,

hence

a(m)Ea = qa 2mEa(m) - qa 2m[m + 1]oa(m + 1) (4) which is the case i = 1 of (1). We have ad(F(,)a(m) = jr + 1 - m]aa(m - 1) by 8.8(5); on the other hand ad(Fo)a(m) = (Foa(m) - a(m)FQ)Ko by 4.18(2), hence

a(m)F0 = Foa(m) - [r + 1 - m]oa(m - 1)K,-,,1

(5)

which is the case i = 1 of (2). Now for the induction steps: We have (by induction)

a(m)E(i+1) = a(m) Eo')E.

[i + t],

\\

(-1)' i=o

[m+]

j

q r-2m)-i(i-1)E--?)a(m

+ j)E.

Ja

We get from (4)

r-2'- 21 [m+j+1]oa(m+j+1).

a(m+j)Eo=q« Plug this into the previous equation; we get i+1

ciEat+i-i)a(m+j)

a(m)E(t+L) _ i=o

with

i [,,n+j] aqa(i+I)(r-nij)-i(i+1) [i -j +

j

l.]o

[i+1]a

[m + j - i i q(i+1)(r-mj)-(i-L)( +I) [m+j]o

j-1

[i + 1]a

a

Using

[m

we get cj

1Ja [m+7]a = 1

= (- Lm+jl i Ja 1)

3(i+L)(r-rrtj)-ii

[m j h [j]a

q;i[i - j + 1]a +qni+i+L [j]o

[i+l]a

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

149

The last fraction is equal to a3(9a+1 -9az+j -l)+qa+li+I

9a

9a -

-i-I

= 1.

- qa

This yields (1).

Similarly, we have by induction

a(m)Fai+I)

m + j 1 Qa(i-1)Faz-7)a(rri

L. [il]a + L 7=II

7

J

- j)K;-jFa.

(6)

a

Now Ka Fa = qa-7 Fa Ka and

a(m - j)Fa = Faa(m - j) - [r + 1 - m + j]aa(m - j + 1)K,-,,' by (5), hence

a(m- j)Ka'F" = 4o3Faa(m -7)Ka3 - 4aj[r+ 1 - m+j]aa(m - j + c7F(a2+I-j)a(m Plug this into (6); we get a(m)Fai+1) = Ei+1 7=0

c = (-W

r-m+j 7

Qa(i+1)

1)Ka'-1.

j)K; with

[i + 1 - j]a [i + 1]a

a

IT - m + j - 11 qU-I)0+1) [r - m + j]a

r-m+j Ja

q az

4a[i+j]a+4«-i-1[j]a [i+1]a

The last fraction is easily checked to be equal to 1; so (2) follows.

REMARK. If we take m = r, then we have ad(E( +'))Ep = 0 for all 8.8(4). So in that case (1) says (ad(E(C,r))E3)Lai) =

gair

a2) (ad(Ear))E3).

j > 0 by (7)

8.10. Keep the notations a,,3, r from the last subsection. PROPOSITION. We have for all finite dimensional U-modules V and all v E V Ta(E5v) = (ad(Eor))E@)Ta(v).

(1)

PROOF. We may assume that v E VA with A E A. Set s = (A, a"); we have then KaaEac)v = gai(s+2c)Eac)v for all i, c > 0. (2)

150

BRAID GROUP ACTIONS AND PBW TYPE BASIS

S.

Fix n and use the abbreviation a(m) = ad(Ea'"))EE as in 8.9(3). We have (-1)bga-aca(r)Eaa)Fab)Eac)v

a(r)Ta(v) =

a,b,c>0;-a+b-c=s

_

_1)bgba-acgQarE(aa)a(r)FFb)F,(c)v

[by 8.9(7)]

a,b,c

min(b,r)

(-1)bq,ba-a(c+r)E(,a)(-1)iq'(b-1)Fab-z)a(r -

_

- a,b,c

i)K, ZEa°)v

i=0

a,b,c

r i

[by 8.9(2)] C

1:(-1)b+iq,ba-a(c+r)+i(b-I-s-2c)E(aa)F,(b-i).

j=0

(-1)j [r - i + j

j

q, c(-r+2i)-j(c-1) Eac-j)a(r - i + j)v Ia

[by (2) and 8.9(1)]

I 1` 1`(-1)b+i+jq,bb-a(c+r)+i(b-1-s)-cr-j(c-1) rr - i + j i

a.b.c

j

J

L E,

k

a

Fib-i)E(ac-j)a(r - i + j)v,

where I do not repeat the (unchanged) summation bounds. We have for all sum-

mands -a+(b-i)-(c-j) = s- (i-j). Note that a(r-i+j) is 0 for j > i by Proposition 8.8(4). So a(r)Ta(v) is a linear combination of terms Eaa)F, Eac)a(r - h)v with a, b, c, h > 0 and -a + b - c = s - h. The coefficient of Eaa)Fab)Eac)a(r - h)v is equal to

r-h j=U

r

( -1)b+jq,ba+i-a(c+j+r)+i(b+i-1-s)-(c+j)r-j(C+j-1)

L

11

I r - hl j

Ja

where i = j + h. If we plug this formula for i into the last expression, we get

r-h

(-1)

rr - h l

a

a

j=0

L

7

We rewrite the sum over j in (3) using -a + b - c = s - h and get

r-h

j:(-1)jq-j(r-h-1) rr - h] j=0

IL

j

a

= br-h,0

by 0.2(4). For h = r we get in (3)

(_1)bq,bb-ac+r(-a-c+b+r-s) _

(-1)bga-ac

since -a - c + b = s - h = s - r. So we get finally (-1)bga-acE(a)FQb)E(c)a(0)v.

a(r)Ta(v) _ a,b,c>0;-a+b-c=s-r

J a

(3)

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

151

Now a(0) = E3 and a(0)v E Va_;3 with (A - Q, a") = s - r; so we get a(r)TQ (v) _ Ta(E3v) as claimed. REMARK. We get now from 8.7(3) for all V and v

TT(Fpv) _ (-ga)rw(ad(E1r))Ep)Ta(v) r

_

(E(-1)i(-ga)rq,'F«r-Z)F,3F,',Z)T.(v)

i=o

1

r

_ (E(-1)r-iq« 'F(r-i)F0F'(i))Ta(v), i=O

r (E(-1)'q'F(i)Fi3F(r-'))TT(v)

Ta(F3v) =

i-o

(4)

8.11. We now want a formula for EoTa(v). It can be easily deduced from the formula for Ta (Epv), if we work with a twisted version of the adjoint action. Set for all x E U

Tad(x) = Toad(x)oT:U--->U

(1)

The map x F-> Tad(x) is obviously an algebra homomorphism from U to Endk(U), hence makes U into a U-module. The formulas in 4.18(2) imply for all a E II and

all uEU Tad(Ea)u = uEa - EaKauK, 1, Tad(Fa)u = K,-, 1(uFa - Fau), Tad(Ka)u = KauK,, 1

(2)

LEMMA. Let a be a simple root. Suppose that we have u and u' in U such that Ta (uv) = u'Ta (v) for all v in all finite dimensional U-modules V. Then we have TT((Tad(EQ)u)v) = (ad(FF)u)TQ(v) and

Ta( (Tad(FQ)u)v) = (ad(EE)u')TT(v)

for all v as above. PROOF. We have (using 8.6(9),(10))

Ta((Tad(Ea)u)v) = TT(uEav - EaKauK« 1v)

(-FaKa)Ta(v) - (-FaKa)K, 1u'KaTa(v) _ (Fau - u Fa)K«Ta(v) _ (ad(Fa)u')Ta(v),

152

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

and

Ta((Tad(Fa)u)v) = Ta(Ka luFav - Ka'Fauv) = Kau (-Ka'Ea)Ta(v) - Ka(-Ka'Ea)u'Ta(v) = (Eau' - Kau'EaKa 1)Ta(v)

= (ad(Ea)u')Ta(v) 8.12. Take ,Q as in 8.8 and set again r = -(Q, a'). We get from Proposition 8.10 together with Lemma 8.11 (by induction on m) Ta ((Tad(E(m.))E0)v) = (ad(F«m))ad(E,1,,r))E3)Ta(v)

(for all V and v as before) So 8.8(6) implies Ta((-ad(E,(,,m))ER)v) =

(ad(E,(,,r-m))EO)Ta(v)

(1)

In particular, we get for m = r a formula for E,3 Ta(v). We have T(EQ) = E3, hence Tad(E(m))EQ = T(ad(E. m))E8), and so by 8.8(3) m Tad(E(am.))E,3

(-1)Zqi(r+1-m)E,«i)E'3E,«'n-i)

=

(2)

i=0

(So Tad(Ea"'))E3 = fa $:1.m:-1 = xa,3;I.m:-1 in the notations of [L], 7.1.1 and 37.2.1.)

We can now rewrite (1) as m Ta(E(-1)igz(r+1-m)Eai)E. E(m-i)v)

i=0

r-rn

_

(3)

(E(-1)iq-i(m+1E(r-m-i)E,3E(,i))Ta(v). i=O

We get from (3) using 8.7(3) and a small calculation similar to the one leading to 8.10(4) m

1)iQar+1-m)F(m-i)F3Fa()v)

T. i=0

r-m

(4)

(?(-1)iga(m+1)Fa(i)F3Fa(r-m-

_

)TT(v)

i=o

8.13. PROPOSITION. Let a be a simple root. Then there is for all u E U a unique element u E U such that Ta(uv) = u'Ta(v) for all finite dimensional U-modules V and all v E V. The map u --f u' is an automorphism of U. PROOF. Let us first show the existence: If u1 and u2 are two elements of U, and if we have already found elements ui and u2 that satisfy the condition in the theorem, then we can take (u1u2)' = u' u' and (au1 + bu2)' = aui + bu'2 for all a, b E k. So it suffices to prove the existence for the generators of U. But for them the existence follows from 8.6(9), 8.6(10), Proposition 8.10, and 8.10(4).

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

153

Now for the uniqueness: If u' and u" both satisfy the condition for a given u, then (u' - u")TQ (v) = 0 for all V and v E V. Since TQ is bijective, this means that u' - u" annihilates each finite dimensional U-module. This implies u' - u" = 0 by Proposition 5.11. The formulas for (ulu2)' and (aul + bu2)' in the first part of this proof show

that the map u ra u' is an algebra endomorphism of U. The formulas in 8.6(9), 8.6(10), 8.12(1) [for m = r], and 8.12(4) [for m = r] show that all generators of U are in the image of this endomorphism; so it is surjective. If u is in the kernel of this homomorphism, then the definition implies TQ(uv) = 0 for all V and v E V. Since

TQ is bijective, this means that u annihilates each finite dimensional U-module, hence u = 0 by Proposition 5.11. So u --f u' is an automorphism. 8.14. By abuse of notation we denote also the automorphism from Proposition 8.13 by TQ. So we have by definition (1)

T.(UV) = TQ(u) TQ(v)

for all u E U and all v in a finite dimensional U-module. We get from 8.6(9)

TQ(Kµ)=K..N.=TQ1(Kµ)

for all /2 E Z,

(2)

and from 8.6(10)

TQ(EQ)-F.KQ, TQ(FQ)

-KQ IEQ,

T.-'(E.)

-K.-'F.,

T,1(FQ)

-EQKQ.

(3)

Finally, 8.12(3),(4) yield for all Q E II, Q # a (setting r = -(Q, cev)) and for

0<m
i(r+1-m)E(az)E3EQm-z)1

TQ

=

r-m

J

i=O M (/`(-1)igo(r+[-m)FQ("'-a)FpFQ''i

TQ

E(-I)agQ

m+l)E(,,r-m-i)E,3EQi), (4)

i=O

r-m

=E

(-1)agQ("'+1)FQ(i)F8FQr-m-i)

(5)

i=0

\i=OO

Occasionally it will be more convenient to work with the less explicit version TQ(Tad(E(,m))E3) =

ad(E(r-m))EE

(6)

of (4), cf. 8.12(1). Here are some frequently used special cases of (4) and (5): (-1)igiE(ar-i)E3E,,(,z)

TQ(E,3) _

(7)

i=0

r

T. -'(E3) _ (-1)Z4«ZEaZ)E3Ear-Z),

(7')

i=0

TQ(F) _

F,(i)FOFar-`),

(8)

Ta 1(F3) = E(-1)ZgaFar-')FNFnz).

(8')

i=O r

i=0

154

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

Note that 8.7(3) implies for all p E M

Ta(w(u)) _ (-4a)-4',av)w(7'a(u))

for all u E UN2.

(9)

The formulas above imply

Tom' =TOTaOT. (10) Indeed, both sides are automorphisms of U and they coincide on all generators of U.

Warning: Lusztig introduced such an automorphism first (in the simply laced case) in [Lusztig 1]. He worked there with a different version of Uq(g) and the printed version contains a few misprints that were corrected in [Lusztig 3]. In that paper he used the present version of Uq(g). He extended the construction to the non-simply laced case in [Lusztig 4]. However, the Ta here differ slightly from his construction there. Lusztig's original automorphisms are equal to Ta o cp,, where cp,, is the automorphism of U that is the identity on U° and satisfies cpa(E,3) = (-1)(3,av)E° and pa(F3.) _ (-1)(°,av)F8 for all /3 E II. REMARK. We get similar automorphisms of U corresponding to Tom, IT,,, and "Ta; they are again denoted by Tom, "Ta, and "T,,. Lemma 8.7 implies that "Ta = W o Ta o w. One deduces easily from 8.6(6) that "Ti = T,-,' and Ta = Here is an alternative approach to Ta: There is an isomorphism of k-algebras ("TQ)-I

o: U=Uq(0) --- U9-,(0) such that o(E3) = E,3 and or (FO) = F3 for all /3 E II and such that o(KN,) = Kµ' for all p E Z4). One has then Ta = 0-1 o Ta o or where Ta on the right hand side is constructed in Uq-i (g). The proof is very similar to that of 8.7(2).

8.15. Lusztig has shown that the T. (on modules and on U) satisfy the braid relations. This means for all simple roots a and ,3 with a 3: If sas3 E W has order rn, then TaT°

= T°Ta ...

(1)

where we have in factors on both sides. This generalizes the analogous property of the sa mentioned in the introductory remarks of this chapter, cf. Lemma 56 in [Steinberg].

It is easy to show that both sides of (1) coincide on U°: One has sas.3 . in the Weyl group, hence s°s,, TQT3 ... (Kµ) = Kg_s3...(l1) = K..,,,,...(µ) = Ti3Ta .

(Kµ)

_ (2)

for all p E Z4). The proof of the braid relations in general requires long calculations. Only the case in = 2 is trivial:

If (a, 3) = 0, then TaT3 = T3Ta on modules and on U.

(3)

Indeed, in this case both E,3 and F3 commute with both Ea and F, cf. 8.6(11). This yields clearly T,,T°(v) = T3Ta(v) for all v in a finite dimensional U-module, i.e., the claim for modules. We have then for all u E U and all v as before TT°(uv) = TT(T°(u)T°(v)) = TTT,3(u)T,T3(v) and similarly T3T,,(uv) = T;3Ta(u)T33Ta(v). This implies that T, T,,3 (u) - T3Ta (u) annihilates all finite dimensional U-modules, hence is 0 by 5.11. So we get the claim for U.

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

155

We are going to prove in the next subsections that (1) holds for m = 3 and for m = 4, if we consider the Ta as automorphisms of U. We shall refer to [L] for the case m = 6 as well as for the result on modules.

The braid group of type t is defined as the group with generators (o )aEn and = 0-0o-a for all a, 3 E II, a 54,3 with m(a,,<3) factors on both

relations QaQp

sides, where m(a,,3) is the order of sas3 in W. For I of type A, this yields the classical braid group B(n) introduced by Emil Artin. So the claim that the Ta on U satisfy the braid relations says: The braid group operates via automorphisms on U. The corresponding claim for the modules says: We have a representation of the braid group (of type on each finite dimensional U-module. Using 8.6(7) we can show: If the Ta satisfy the braid relations on modules, then so do the iITa. We can then extend that statement to the T", and the WT", using 8.6(6). If we had bothered to define automorphisms of U corresponding to these variations of T, then we could show equally well that also these automorphisms satisfy the braid relations. However, we shall concentrate on the Ta. The fact that the Ta satisfy the braid relations implies that one can define for each w in the Weyl group operators T,,, (on modules and on U) in a reasonable way, cf. 8.18. In the Lie algebra case the analogous operators permute (up to sign) the root vectors in a Chevalley basis of g. In our case we have (so far) analogues to root vectors only for simple roots and their negatives. We can now use the T,,, to construct more: For any positive root y there are a E II and w E W with w,3 = y.

One can show then that T,,,(E,3) E U+ and that Tu,(E,3) = Ey if y E II. So we would like to define E7 = T,,,(E,3) in general. However, it turns out that T,,,(E,3) depends (in general) on the choice of w and ,3. For example, if t is of type A2 and II = {c, ,(3} then 7 = a +,3 is equal to sa,3 and s8a. So we have two natural candidates for Ea+3: Ta(Eo) = E.E,3 - q-1Et3Ea

and

Ta(Ea) = Et3Ea - q-IEaE,3,

and we get two linearly independent vectors. In spite of this unpleasant phenomenon, it is still possible (in several ways) to make a consistent choice of "root vectors" E7 for all positive roots 7 so that we get the old E7 for -y simple and so that for some numbering {-yl, -y2i ... ,'yt} of the positive roots all monomials Eat Eat ... Ea, "Y1

,Y

7

are a basis of U+, cf. 8.24. This basis is analogous to a PBW basis of the enveloping algebra of the Lie subalgebra ®7>o p-, of g; therefore it is called a PBW type basis of U+. If we apply w to this basis, then we get a PBW type basis of U-. Using Theorem 4.21 we can then write down a basis of U. I shall discuss after 8.24 the contents of the remaining subsections.

8.16. Consider two simple roots a and ,3 with (,a, a") = -1 =

(1)

8.

156

BRAID GROUP ACTIONS AND PBW TYPE BASIS

(or, equivalently: such that sas3 has order 3). We have then (a, a) qa = q3. The formulas in 8.14 imply in this case Ta(E;3) =

E0E3-ga'E3Ea = TI(EQ),

Ta 1(E3) = E3E. - qa 'E.E3 = T3 (E.), Tn(F3) = F3Fa - gaFQF3 = T; '(F.),

T, 1(F,3) = F«F3 - q.F,3F«

= T3(F.).

hence

(2) (3) (4) (5)

We see in particular T«T33(Ea) = E3,

T«T3(Fa) = F3,

T3Ta(E3) = Ea, T3T.(F3) = Fa.

LEMMA. Let a and 0 be simple roots such that sas3 has order 3.

(6)

Then

TaT3Ta = T3TQT3 as automorphisms of U. PROOF. It is enough to show that TQT3Ta and T3TQT3 coincide on all generators of U. We have observed already in 8.15(2) that they are equal on U°. So we have to look only at the EE and Fy with ry E II. For -y = a and -y = ,3 we can apply the formulas above. Combining them with 8.14(2),(3) we get TTT3Ta(E3) = Tc(Ea) = -FaKa and

T,3T«T3(E3) = TTTT(-F3K3) _ hence TaT3Ta(E3) = T3TaT3(Ep). Similarly:

-FaKa,

TcT3Ta(F3) = Ta(Fa) = -Ka 1Ea and

T3TaT3(F,3) = T3Ta(-Ka 1F3) = -KSas,3Fc, = -Ka 1Fa, hence TaT3Ta(F3) = T3TaTQ(F3). Since the situation is symmetric in a and,3, we

get also TaT3Ta(Ea) = TTTaT3(Ea); similarly for F. Assume now that -y # a,,3. Since the Dynkin diagram contains no loops, we have (ry, a') = 0 or (-y, ,3") = 0. Since the situation is symmetric in a and ,3, we may assume that (y, 3") = 0; then [E-y, E3] = 0 = [Ey, F3].

(7)

This implies that T3(E,) = E. On the other hand, if we apply T3Ta to (7), then we get by (6) [T3Ta(Ey), Ea] = 0 = [ToTa(E-r), Fa].

This implies that T. (T3Ta(Ey)) = T3Ta(EY), hence

T«TI3T.(Ey) = T3Ta(Ey) = T3TaT3(E7) We can replace all Ey in this argument by FF, and get TaT,3Ta(Fy) = T3TaT3(FF).

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

157

8.17. Consider in this subsection simple roots a, ,Q with

(0,av)=-2

and

(1)

So a and 0 span a subsystem of type B2, the order of sas,3 is 4, and we have qa = q

and

(2)

q3 = q2

We can apply 8.14(7) with a and i3 interchanged and get

T3(Ea) = E3Ea - q-2EaE3,

(3)

T 1(Ea) = EaE3 - q-2E<3EQ.

(4)

On the other hand, if we take 8.14(4) with m = 1 for a and ,3 as above, then we get

Ta(E3Ea - q-2EaE,3) = EQE3 - q-2El3Ea,

(5)

i.e., Ta(T3(Ea)) = T1(Ea) and thus

T3TaT3(Ea) = Ea.

(6)

Furthermore, 8.14(7) yields q-1 EaE3Ea +

Ta(E,3) =

(7)

T,1(E3) = E.3E(2) - q-1E(E,3Ea If we multiply these equations with [2]a = qa +q-I =

(8)

q+q-I and

use (3),(4), then

we get

[2]QTa(E3) = EaE;3 - (1 + q-2)Ea.E,3Ea + q-2E3E'2

= Ea T; '(E,,)

- T3 1(Ea) Ea

and

[2]aT 1(E3) = E3Ea - (1 + q-2)EaE3Ea + q-2EaE3 = T3(Ea)Ea - E«T3(Ea), hence T3Ta (E3) = T, ,-'(E,3) and thus

TaT3T.(E,3) = E.

(9)

Now 8.14(9) together with (6) and (9) shows that T,3TaT3(Fa) = Fa

and

TaT3Ta(F3) = Fp.

(10)

(One checks easily that the (-q)-terms from 8.14(9) cancel.)

LEMMA. Let a and /3 be simple roots with sasp of order 4. Then we have TQT3T(T3 = TpTQT3Ta as automorphisms of U.

158

8.

BRAID GROUP ACTIOIJS AIJD PBW TYPE BASIS

PROOF. We may assume that (1) holds. We have to show that TQTpTQTp and T3TQT3TQ coincide on the generators of U. We know by 8.15(2) that they do so on all KN,. So we have to look only at the Ey and Fy with ry E H. We have TpTTT3(K«) = Ksos s0c. = Kim. Therefore (6) and (10) imply that

T«T3TTTa(E«) = TT(EE) = -FFK0, = -TpT«T3(F.K«) = T,3T.T,3T«(E.). Similar calculations show that TQTpTQTa and TpTQTpTQ coincide also on E3, FQ, and F3.

Consider now -y E II with ry # a, 0. Since the Dynkin diagram contains no loops, we have (y, a") = 0 or (y, Q") = 0. Suppose first that (y, /3") = 0; then [E7, Ea] = 0 = [E7, Fe]-

(11)

This implies that TO(EE) = E. On the other hand, if we apply TQTpTQ to (11), then we get by (9) and (10) [T'«T,3TQ(E-r), Ea] = 0 = [T TOT«(E-r), FR]

This implies that TO(T.ToTQ(EE)) = T. To T. (E,), hence

ToT.ToT.(E-r) = TQTaTQ(E1) = T. TO T. T3 (E-,).

We can replace all Ey in this argument by Fy, and get To T. T,3 T Q (F..,) = T Q T,3 T Q T,3 (F y) .

Finally, if we have (-y, a") = 0, then we interchange all /3 and a in the argument above and quote (6) instead of (9). The proof goes through as before. REMARK. If sQs,3 has order 6, then we have TQT3TQTj3TQTa = TpTQT3TQTJTQ

(as automorphisms of U). You can find a proof in [L], 39.2.2. (Lusztig assumes there II = {a, ,Q}, but the generalization to the general case is easy: Any ry E II with -y

a 0 is orthogonal to a and 0. So we have TQ(E1) = E, = TO(E.y);

similarly for Fy.) Lusztig proves the braid relations first for the TQ as automorphisms of U in all rank 2 cases, then on modules in the rank 2 cases, then on modules in general, and finally on U in general, cf. [L], 39.4. Besides giving the additional information on the module case, Lusztig's approach has the advantage to work in more general situations. For example, the Dynkin diagram may have loops. On the other hand, his proof relies on the "quantum Verma identities" ([L], 39.3.7) that require very unpleasant calculations.

8.18. Recall that W is generated by the sQ with a E H. If w E W is written sQt with t minimal, then this is called a reduced expression of w,

w = SQ18Q2

and t is called the length 1(w) of w. Let a and /3 be distinct simple roots. If sQsa has order 2 (hence S.So = sasQ) and if w has a reduced expression of the form w = SQ1 ... SQ.-ISQS3S

+2

... SQt,

then also w = SQ1 ... SQ.-, S3SQSQ:+2 ... SQt

BRAID GROUP ACTIONS AND PBW TYPE BASIS

8.

159

is a reduced expression of w. If sasp has order 3 (hence saspsa = spsasp) and if w has a reduced expression of the form w = $11 ... sa,-I SIS3Saga.+s ... Sal I

then also w = Sal...Sa,-ISQSaSQSai+3...Sal

is a reduced expression of w. Similarly, if sasp has order 4 (resp. 6), then we can replace a block saspsasp (resp. saspsaspsasp) in a reduced expression of w by spsaspsa (resp. spsaspsaspsa) and get again a reduced expression of w. Let me call the transition from one reduced expression to another one an elementary move,

if it has one of the forms described above. Now the following holds: Given two reduced expressions of an element w E W, one can get from one to the other by a sequence of elementary moves, cf. [Humphreys 2], 8.1 or [Steinberg], Lemma 83(a) or [Bourbaki 2], Ch. IV, §1, exerc. 13. The last result implies that we can now define for each w E W a linear operator T,,, on each finite dimensional U-module V and an automorphism T. of U as follows. For w = 1 set Ti = 1 (the identity). For w 1 choose a reduced expression W= and set

T.

Tai = Ta1Ta2

(1)

We have to show that the right hand side in (1) is independent of the reduced expression. By the result above we have to check only the case where we apply an elementary move. If it involves a and Q with sasp of order 2 (resp. 3, 4, 6), then the claim follows from 8.15(2) (resp. Lemma 8.16, Lemma 8.17, [L],39.2.2). It

is clear by definition that T,, = Ta for all a E II. If we have w, w' in W with l(ww') = l(w) + l(w'), then we get a reduced expression of ww' by joining reduced expressions of w and w'. This shows T,,,,,,' = T,T,,,,

if l(ww') = l(w) + l(w').

(2)

We get from 8.14(2) T,,,(KN)=K,,,N,

for all wE W and pE M.

(3)

We have (4) for all w E W and p E M. (This is clear for the generators by the formulas in 8.14; it then follows then in general.) If wu - µ = EaEII maa, then

T. (Up) = U,,,N

TT(w(u)) = (fl

(-4a)"in) w(T,,(u))

for all u E Uµ.

(5)

aE1I

This follows easily from 8.14(9). Finally, 8.14(10) yields T,;, 1= T o T,,,- i o T.

8.19. LEMMA. Let a, ,3 E 11 with a W generated by sa and s;3. If wa > 0, then generated by Ea and E. If wa E II, then

/3.

(6)

Let w E W be in the subgroup of is contained in the subalgebra E,,,a.

160

S.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

PROOF. The claim is trivial for w = 1 where T,,,(E,,) = E. So assume that

w01.

Denote the order of sas3 by m. If m = 2, then w = s3 and T,,,(E,,) = E. If m = 3, then w E {s3, sas3} and thus T. (E,,,) E {EEEa - qa 1EaE3, E3} by 8.16(3),(6). If m = 4, then w E {s3, sas3, s3sas3}. If (a, a) < (/3, /3), then 8.17(3)-(6) imply

Tw(Ea) E {E3Ea - q-2E.E(j, EE3 - q-2E,3Ea, Ea}. If (a, a) > (0, /3), then 8.17(7),(8),(9) (with a and 3 interchanged) imply that T,,;(E,,) is one of

E )Ea - q-1E3EaE3 + q-2E,,E(32), EaE32) - q-lEpEaEp + q-2E(2)Ea, Ea. For m = 6 the claim follows from [L], 39.2.2(a),(b).

8.20 PROPOSITION. Let w E W and a E H. If wa > 0, then Tw(Ea) E U+; if wa E H, then T,, ,(E,,) = E,,,a.

PROOF. We use induction on 1(w). If 1(w) = 0, then w = 1 and T,,, = 1; so the claim is obvious in that case. Suppose now that 1(w) > 0. Then there is /3 E II with w/3 < 0. We have obviously 3 0 a. There is now a decomposition w = w'w" with w" in the subgroup generated by s3 and sa such that w'0 > 0 and w'a > 0; we have then 1(w) = 1(w') + 1(w"), cf. [Humphreys 2], 1.10. Now wa > 0 and

w/3 < 0 implies that w"a > 0 and w"/3 < 0. (Both w"a and w"/3 are contained in D fl (Z/3 + Za), and w' maps positive resp. negative roots in this intersection to positive resp. negative roots in 4).) This implies in particular that w" 0 1 and thus 1(w') < 1(w). So we can apply induction to w' and get (e.g.) T,,,,(Ea) E U+ and T,,,' (E3) E U+. Lemma 8.19 implies that Tw,, (Ea) is contained in the subalgebra of U generated by E3 and E. Now the first claim follows since T. = T .'T"," by 8.18(2).

In order to get the claim in the case wa E H, we have to show that w"a E II in that case. Then induction applied to w' and w"a (together with the last part of Lemma 8.19) will yield the claim. Well, if w"a II, then w"a > 0 and w"a E 4) fl (Z/3 + Za) imply that w"a = 0 + ja with i > 0 and j > 0. Then wa = w'w"a = i(w'/3) + j(w'a) cannot be simple since w'/3 > 0 and w'a > 0. So w"a E H and the second claim follows. REMARK. W e have similarly: If wa > 0, then T,,,(Fa) E U if wa E II, then T,,,(Fa) = F. This follows easily from the lemma using 8.18(5). (It can also be

proved in the same way as the proposition.) 8.21. If W = Sal Sae

sa, is a reduced expression, then

a1, sala2, SaiSa2a3, ... , Salsa2...Sa,at

are t distinct positive roots, in fact, they are exactly the positive roots -y with w-1-y < 0, cf. [Humphreys 2], 5.6, Exerc. 1. All Sa1sa2 Sa, with i < t are reduced expressions, since they are intervals in a reduced expression. We have therefore Ts .l sn2 ...sn

T.

= T-1 T-2 ... Ta,

BRAID GROUP ACTIONS AND PBW TYPE BASIS

8.

161

Now Proposition 8.20 implies for all i < t Ta1Ta2 ...T, (Ea +i) E Ua1s«Z...sa

Ck.+l

(1)

hence also

Tai Tae ... Tas (Er,+1) E U+

for all r > 0. Therefore also all products Ta1Ta2...Ta1 1(E« t) ... TaiTa2(Ea3) Tai(Eaa2) Ea;

(2)

are in U+ (for all sequences a1, a2, ... , at of nonnegative integers). We want to show that the products in (2) are the basis of a subspace of U+ that depends only on w, not on the reduced expression. Let us show first the linear independence:

LEMMA. a) Let a E II and let uo, ul,... , u,,,, E U+ with T, 1(ui) E U+ for all i. If Emo uiEa = 0 or if Emo Eaui = 0, then ui = 0 for all i.

b) The products in (2) are linearly independent (for each w E W and each reduced expression of w).

PROOF. a) Suppose that Emo uiEa = 0. If we apply T, 1, then we get r

r

0 = T,, 1(I:uiE') = 1: T,-, 1(ui)Ta 1(Ea)x = Ta 1(ui)(-K, 1Fa)Z. i=0

i=0

Since the multiplication map U+ 0 U<0 - U is an isomorphism of vector spaces, and since the (-K,-,'F,,)' are linearly independent vectors in U50 while all Ta 1(ui) are in U+, we get Ta 1(ui) = 0 for all i, hence ui = 0. [The argument is similar if we assume that >m0 Eaui = 0.] b) We use induction on t. For t = 0 we have in (2) only the empty product, which is equal to 1 and linearly independent. Now suppose that t > 0. Then

the products in (2) have the form Ta(vj)Eax) where a = a1 and where the vj are the products as in (2) constructed for the reduced expression sae sat of saw. If we have scalars ai3 with E, ,j aijTa(vj)Ea = 0, then we can apply a) with ui = Ta (>j axe vj) and get Ej ai.9 vj = 0 for all i. By our induction on t the vj are linearly independent. So we get indeed ail = 0 for all i and j. 8.22. PROPOSITION. a) The subspace spanned by all products as in 8.21(2) depends only on w, not on the chosen reduced expression. b) Let a, ,3 be two distinct simple roots. If w is the longest element in the subgroup of W generated by sa and s3, then the products in 8.21(2) span the subalgebra of U generated by E, and E,3. PROOF. The proof of b) requires case-by-case considerations. We turn to them in the next subsection. At this point we show only that b) implies a): We use induction on l (w). For l (w) < 1 there is only one reduced expression, so the claim holds. Suppose now that l(w) > I and that the claim holds for all w' with l(w') < l(w). Denote by U+ [w'] the space spanned by the products as in (2) for w' instead of w. Let w = s;31 s,32 s,3, be another reduced expression. We may assume that we get it from the first one by an elementary move. Set a = al and ,a=/31

8.

L62

BRAID GROUP ACTIONS AND PBW TYPE BASIS

In case a =,3 set w' = saw. Then the subspace spanned by the elements as in (2) is (for both expressions) equal to

T.(U+[w']) (EkE'). a>O

This follows from our induction applied to w': The two reduced expressions lead just to two systems of generators for the subspace U+[w'].

Now suppose that a # 3. Then the elementary move from one reduced expression to the other changes the first factor of the reduced expression, hence has to take place at the beginning of the expression. Let w" be longest element in the subgroup of W generated by sa and sp. Then our two reduced expressions for w have to begin with the two reduced expressions of w" and to coincide after that. So we have a decomposition w = w"w' with 1(w) = 1(w") + 1(w'). Using b) we see that the span of the products in (2) is equal to T,,,,,(U+[w']) U+ [w"] for both expressions.

REMARK. A similar argument shows that also the subspace spanned by all products E.at TQ1

7.1T.2(E'3) ...

(3)

depends only on w, not on the reduced expression. [Actually, it is the same space, but we do not need that here.]

8.23. We now turn to the proof of Proposition 8.22.b. Let a, 3 E II with a # Q. Denote the order of s,,s,3 by m. Let w be the longest element in the subgroup of W generated by sc. and sp. It has length m and exactly two reduced expressions: w = sasp (m factors) and w = s,3s,, (m factors). Consider first the case m = 2, i.e., the case (a, Q) = 0. Since T'3 (E") = Ea and T,,(E;3) = E3, the two possible expressions w = sps, = saso yield the span of all E ) E(6) resp. of all E(ra) Ep i . Because of E,,Ep = E,3Ea we get in both cases the subalgebra generated by E,3 and E(,.

Suppose now that m = 3, i.e., that (0, a") _

1 = (a Q"). The situation is

symmetric in a and 3, so it suffices to work with the expression w = scsosa. The products in 8.21(2) are then all

T. T,3 (E.r) T.(E) Ea = EQ Ta(E) Ea,

(1)

cf. 8.16(6). Let us use the notation Ea+Q = TQ(Eo) = ad(Ea)E8 = EaER - qa I EREQ,

(2)

cf. 8.16(2). Note that this explicit formula for TQ(E,3) shows that all products as in (1) are contained in the subalgebra generated by EQ and E,3. In order to show that they span this subalgebra it will be enough to show that this span is stable under left multiplication by EQ and E,3. This stability is obvious in the case of E8; the proof for EQ uses a few commutator formulas that we are going to derive now. We have by 8.9(7)

E.+3E. = qaIEEE.+a, hence

EaEQ+Q = q'E,,+,3E,Y

for all n > 0.

(3)

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

163

We have by symmetry also T,3(E«)E,3 = q 1E,3T8(E«).

If we apply T« to this equation, then we get by 8.16(6)

E8Ta(Ea) = g3'T«(Ea)E,3, i.e., (since q13 = qa)

E«+0E8 = q«E,3E«+3

(4)

E. E,3 = q«' Ea Ea + E«+,3 .

(5)

We can rewrite (2) as

This yields (using induction) for all n > 1

E«E = q«"EEE« + [n]aE3"-'E«+0

(6)

Now we are ready to finish the proof in the case m = 3. We have for all a, b, c > 0:

E«EaE«+a E« = q« `EaE«E«+,3E« + [c]«Ea ' E«+aE«+a E. b-c

c

b

n+1

= q« EaE«+QE«

c-1 b+l

+ [c]«E3 E +Q

E.n

So the span of the products in (1) is indeed stable under left multiplication with E. As pointed out above, this yields the claim for m = 3. We shall deal with the case m = 4 in the appendix. For m = 6 let me refer you to [L], p. 312. (Cf. also the formulas in [Lusztig 4], 5.4.) 8.24. For any w E W denote the span of all products as in 8.21(2) by U+ [w] and set U-[w] = w(U+[w]). One can show that U+[w] is a subalgebra of U+, cf. [DeConcini, Kac & Procesi], 2.2. We shall see that later on in a special case (8.25); right now we need only the much weaker statement:

If w E W and a E II with w-1a < 0, then U+[w]Ea C U+[w].

(1)

Indeed, the assumption w-la < 0 implies that we can find a reduced expression W = sa,saZ . sat with a1 = a. Then the products in 8.21(2) end with Ea, so their span is clearly stable under right multiplication with E. THEOREM. Let wo be the longest element in W and let wo = sat saz a reduced expression. Then all products

Ta1Tc2 ...Tat_1(Ea) ... TaiTa2(E«3) T«i(E«Z) E;

sat be

(2)

with all ai E Z, ai > 0 are a basis of U4. PROOF. The products in (2) span U+[wo]. We have U+[wo]Ea C U+[wo] for all a E II by (1). Since the E. generate U+ and since 1 E U+[wo], this implies that U+ [wo] = U+, i.e., that the products in (2) span U+. They are linearly independent by Lemma 8.21.b, hence a basis.

164

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

REMARKS. 1) The same argument shows that also all products

Eai Ta,(Ea2 TaiTcr2(Ea3) ... Ta1Ta2...Tat(Eat)

(3)

are a basis of U+, cf. the remark in 8.22. 2) If we apply w to a basis of U+, then we get a basis of U-. Therefore the theorem together with 8.14(9) shows that all products Ta,Ta2 ... Tat-1 (Fat

... Tat Tae (Fa3) Ta, (F«22) F«1

(4)

are a basis of U-. 3) Set 1'i = sat sat_, ai for 1 < i < t. Then ryi, ^/2,. ..1't are exactly the positive roots in -P each occurring once, cf. 8.21. Set X7, = Ta1 Ta,_ Ea, for all i. Then X,, E U+ has weight ryi; we can write the basis in (3) in the form XaiXa2 ... Xat-1Xat (5) 71 7t-1 1

72

7t

This shows that the dimensions of the weight spaces in U+ (and U-) are given by Kostant's partition function. So we have extended Proposition 5.19.a to the general case where q is not a root of unity.

The X7i play a role similar to that of the root vectors x7 in 9. While there are nice commutator formulas for the x7, cf. [H], Proposition 8.4(d) and Theorem 25.2(d), things are more complicated in Uq(g). Levendorskii and Soibelman have shown for all i < j that X7jX,, - q(0-0>)X7iX7j is a linear combination of basis elements as in (3) that involve only X., with i < s < j, i.e., where as = 0 for all s with s < i and for all s with s > j. For a more accessible proof you may consult [De Concini & Procesi], Theorem 9.3.

The subspace U+ [w] introduced in 8.21-24 is a quantum analogue of the enveloping

algebra U(n,,,) where nw is the direct sum of all 97 with ry E I+ and w-try < 0. (This is a Lie subalgebra of 9.) However, there may be other reasonable analogues. For example, if w = sawn with a E 1I, then also Te (U+[sawo]) makes sense and is different from U+[sawo] unless a is orthogonal to all other simple roots. This is in contrast to the classical case where we have sa(ns,,,,0) = ns,,,,,a. We are going to look closer at U+[sawo] and Ta(U+[sawo]) and characterize them differently (8.25/26). These characterizations show that U+[se wo] and Ta(U+[sawo]) are subalgebras of U+. This investigation, however, is mainly a tool to achieve our main goal: We want to prove that the PBW type basis of U+ and the analogous basis of U- are (up to normalizing factors) dual bases with respect

to the pairing introduced in Chapter 6. We can then use this fact (by the results from Chapter 7) to get more explicit formulas for the universal R-matrices. We get thus also a new and more general proof for the non-degeneracy result in 6.18 that had been proved only under the assumption char(k) = 0 and q transcendental over Q. We are now able to get rid of that assumption also for the other results in Chapters 6 and 7. A crucial step is the invariance of the pairing between U+ and U- under each Te . Since U+ and U- are not stable under Ta, the last statement should really be formulated as: If x E U+ with Ta(x) E U+ and if y E U- with Ta(y) E U-, then (y, x) = (Ta(y), Ta(x)). The proof of this invariance is delegated to a special chapter 8A.

S.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

165

8.25. One can show that U has no zero divisors, cf. [DeConcini & Kac], 1.8 or [Joseph & Letzter 1], 4.10. Right now we need only a very special case that is the first claim in the next lemma.

LEMMA. Let a E II. Then Ea is not a zero divisor in U+. We have

U+ = ®T.(U+[s«wo])E«

(1)

i>O

and

Ta(U+[sawo]) = IU E U+ I T,,-,1(u) E U+}.

(2)

PROOF. Set V = { u E U+ I T,1(u) E U+ I. We can choose a reduced expression for wo that begins with sa. If we look at the corresponding basis of U+ as in 8.24(2), we see that (1) holds and that Ta(U+[sawo]) C V. Lemma 8.21.b implies now easily that we have here equality, i.e., that (2) holds. Since the elements in 8.24(2) are a basis, it is clear that Ea is not a right zero divisor in U+. Using the antiautomorphism T we see that Ea is also not a left zero divisor in U+. REMARKS. 1) Of course (2) is equivalent to

U+[sawo] = { u E U+ I Ta(u) E U+ }.

(3)

Since T,,-,' = T o Ta o T and since T (U+) = U+, we can deduce from (2) and (3) that Ta(U+[sawo]) = T(U+[sawo])

(4)

Note that (3) shows that U+[sawo] is a subalgebra of U+ (equal to Ta I(U+)flU+). 2) The lemma implies U+ = Ta(U+[sawo]) ® U+Ea = U+[sawo] ® EcU+.

(5)

Indeed, the first equality follows immediately from (1). We then apply T and (4) to get the second equality. If we apply w to (5) we get

U- =

Ta(U-[sawo])

® U-F,, = U-[sawo] ® FaU-.

(6)

8.26. We are now going to apply some of the results from 6.7-6.17, in particular

we shall work with the bilinear pairing from 6.12 and with the maps ra and r. introduced in 6.14/15. In 6.14/15 the maps ra and ra were defined only on weight spaces µ+ and EC,, We regard them now as linear maps on all of U+ and U-. We claim that we have for all integers m > 0 q 2m-1

E.-1

(1)

q2

Well, for m = 1 this is contained in 6.14(3). Then use induction on m: We get from 6.14(4) 2m

qa-1

2(m+I) qa q2a

-

-1

1

E.

166

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

To get the claim for r'a argue similarly or use Lemma 6.14.c instead. We have symmetrically 2m

Fa-1.

(2)

{xEU+ I ra(x) = 0 }.

(3)

qa

This follows from (1) using 6.15(6).

LEMMA. Let a E H. Then TaU+[sawo]

PROOF. Let us show first that ra(x) = 0 for all x E T«(U+[s«wo]). Well, by 8.25(1) there are ui and vj in U+[s«wo] with

r«(x) _ (q,,, - qa 1) I:T«(vj)Ea.

and

r. (x) _ (qa - qa') ETa(ui)Ea

j>0

i>O

Apply now Ta 1 to the equation

xFa - Fax = (q« - q« 1)-1(r«(x)K« - Ka 1r«(x)) from Lemma 6.17. We get

-T; 1(x)EaKa + E«KaT« 1(x)

_>

ui(-Ka

1F«)2Ka 1 - EKavj(-Ka

1Fa)j.

j>0

i>o

The left hand side is contained in U+Ka, the summands on the right hand side

in U+K«i-'Fi resp. U+K«-jFj,. Now the sum of all U+KaF« is direct. This yields ui = 0 for all i > 0 and vj = 0 for all j > 0, hence ra(x) = 0 and r' (x) E T«U+[saw0]

-

We now turn to the opposite direction: Let x E U+ with ra(x) = 0. We may assume that x E U+ for some µ. There are then xi E (T«U+[sawo])µ_ia with x = Ei>oxiEa, cf. 8.25(1). We have already seen that ra(xi) = 0 for all i. So we get now from (1) and 6.14(4) zi _

0=ra(x)q2-1 xiEa 1. i>O qa

Since the sum in 8.25(1) is direct, this implies xi = 0 for all i > 0, hence x = x0 E T«U+[sawo].

REMARK. The lemma implies by 8.25(4) and Lemma 6.14.c that

U+[s«wo]={xEU+ Ir'a(x)=0}.

(4)

U [sawo]={yEU- ra(y)=0}

(5)

Lemma 6.15.c yields now

and (using 8.14(9))

I

- I r«(y) = 0 }.

(6)

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

167

8.27. Combining 8,26(3)-(6) with 6.14(5), 6.15(5) we see that (F,,U

,

U+[sawo]) = 0 = (U-Fa,TcU+[sawo])

(1)

and

(U [sawo], EaU+) = 0 = (TaU [sawn], UEa)

(2)

We shall need a more precise statement.

LEMMA. Let a E II, let x E U+[sawo] and y E U- [sawn]. Then

(T.(y)Fa,T.(x)E,',) = (T. (y), T. (x)) (F., E.)

for all i E Z, i > 0.

(3)

If i and j are nonnegative integers with i 54 j, then (Ta(y) F.,T.(x)Ea) = 0.

(4)

PROOF. We may assume that x and y are weight vectors. We use induction on i + j where we set j = i in (3). The claim is trivial for i = j = 0. If i = 0 and j > 0, then (T.(y),T.(x)Ea) = 0 by (2). If i > 0, then 6.14(5) yields

(T.(y)Fa,T.(x)Ea) = (F.,E.)(T.(y)F'« ra(Ta(x)E«)) If j = 0, then we get 0, since ra(Ta(x)) = 0. If j > 0, then we get from 6.14(4) and 8.26(1)

(Fa,E.)(T.(y)Fa 1,Ta(x)Ea-1 We use now induction. In the case i 54 j we get (4) directly. For i = j we get inductively

2h-1

qa-1

h=1 q2

(Fa, Ea)z(Ta(y),T.(x))

If we take x = y = 1 we get 2h _

q2

h=1 qa

(F., E.)'.

We plug that into the last equation and get (3). 8.28. PROPOSITION. Let a E II, let x E U+[sawo] and y E U- [sawn]. Then (T. (y), T. (x)) = (y, x).

PROOF. This proof will be carried out in a special chapter 8A. 8.29. PROPOSITION. Let w = sal sae .

sa, be a reduced expression. Then Ta1Ta2...Ta,

is 0, if there is an i with bz 54 a21 otherwise it is equal to

(Fat)...7'ai(Ea2)E«i i

(Fa, E ).

168

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

PROOF. We use induction on 1(w). For 1(w) < 1 the claim is obvious. For t = 1(w) > 1 set ... Tcti2(E.3) 1!a2,

x=Ta2...T.,_i(E.t

y=

Ta2...TQt-i(F«t)

... Ta2(Fa3) Fate

and a = a1. We have then x E U+[sawo] by (e.g.) 8.25(3), and y E U-[sawo]. So the term in the proposition is equal to (Ta (y)Fa1 J. (x) E.'1) = ba, b, (Fa 1 , E« (Ta (Y), T. W) = baibi (Fa1, Ea1)(X,.Y)

by 8.27 and 8.28. Now we can apply induction to x and y, since l(saw) = 1(w) - 1 and since saw = sae sat is a reduced expression.

8.30. Fix a reduced expression wo = sat sae

sat .

sai...sa,_iai,

X.y, =Tat...TT._1(Ea.),

and

For 1 < i < t set yi =

Y.y. =Tai...T.._,(Fa,)

We have by 6.13(4)

(F.,, E'.,)

-

(-1)rga(r-1)/2 (ga

[r]ca:

- Qa,1

I )r'

So Theorem 8.24 and Proposition 8.29 imply for all u that the Xat ... XY2 Xay with

Ei aiyi = p are a basis of Uµ and that the t

- qa,l) a. yat...yry22yryi, 1)/2(ga, [ay]as

i=1

are a basis of U particular:

µ

J

that is the dual basis with respect to our pairing. We see in

COROLLARY. The restriction of (,) to any U x U+ with p E DD is nondegenerate.

REMARK. This corollary extends Proposition 6.18 to all q that are not a root of unity. As pointed out in 6.26 now everything in Chapter 6 extends to that case. The same is true for Chapter 7. Actually, we get now a more explicit formula for the E), from 7.1(1) and thus for the universal R-matrix. (This goes back to [Levendorskii & Soibelman] and independently to [Kirillov & Reshetikhin].) Consider for all i the

sum (in the direct product of all U L

)

o[i] _ E(_1)rgai (r-1)/2 (ga.

[Vi!

a,l)ry ®X7'

(1)

Then ®µ is the (U ® ® U+)-component of the product ®[t]o[t-1] ... ®[2]®[1].

(Note that the computation of that component involves only a finite sum.)

(2)

APPENDIX

169

Appendix 8.3(7):

If we plug b = a + c + j - i into the left hand side, we get

ccj]

Z

I

a=0 c=0

We have

IC+jI = (-1)C I-jc

c

11

+a[ull

It-cl

by 0.1(1), so this sum is equal to

E(-1)a+iqe a+c-ac-i-i ) (

l i

f

1J

a=oc=o

c Lj+aJ l i l LaJ

Since (-j - 1) (i - c) - (j + a)c = -ij - i + c - ac we can rewrite this sum as

(-1)a+igea r'i

a=0

qe(-(i+1)(i-c)-(j+a)c) [_i_ 11 r j

1

LJ

c

c=0

J

La-cJ

We can now use 8.1(1),(2) to evaluate the sum over c and get for the total sum

a+iqea [

The last factor is 0 for 0 < a - 1 < i; so only the term with a = 0 survives and yields (-1)Z

(cf. 0.1), as desired.

f of f ill = (-1)i

1

(-1)i = 1

170

BRAID GROUP ACTIONS AND PBW TYPE BASIS

8.

8.15(5):

Here is the induction step (where we write [r] = [r],, = [r],3 for all r E Z and q = qa) E(m)E(n+1)

a

[ -(

min(m,n)

=

1

)(n-Z)E(n-Z)E(Q

[n+1]i=o1 q

3

E(em-')EB

rnin(m.n)

l

(EaRE(,m-i-1) + qi-mE3E(am-i)1

q-(m-i)(n-i)E(n-i)E(i)

i

[n + 1]

as

3

i =o

min(m- l.n)

[n+ 1]] q-(rn-i)(n-i)E3n-z)Eaa

i=0

+

min(m,n)

jJ i=o

q-(rn-i)(n+1-i)E,(m-z)giE1E(a3E(am-i)

+1] a q-(m-i+1)(n+1-i) E(n+1-i)E(i) E(m-r)

[ZJ

min(m.n)

+

qi [n - i + 1] q-(mi)(n+1-i)E( +1-i)EQ?E(Qm-i)

E

ci

13

[n + 1]

i=O

rnin(m,n+1)

00

0

[n+1]

i=1

1)

1 [

min(m.n+1)

i

1)Eam

o

q-(mi)(n+1 i)E(n+1-i)E(i.)E(m-i) as a

a

i=O

where

[i]q-(n+1-z) + qi[n

ci =

- i + 1] -

[n + 1]

= 1,

8.23, Proof of 8.22.b for m = 4: Consider a, ,Q E II as in 8.17. So m = 4; the longest element in the subgroup generated by sa and s3 has two reduced expressions: sasas3sa and sasasas3. The first one leads to "root vectors" Ea+a = T1(Ea)

and

E2a+3 = T3Ta(Ea),

(1)

the second one to

E2a+3 = Ta(E3) = T(E2a+a)

and E«+3 = T«T3(Ea) = T(Ea+a).

(2)

We have Ea+3 = ad(E3)Ea and E2a+a = ad(Ea2))E3, so 8.9(7) implies

Ea+1E3 = q-2E3Ea+3 and E2a+aEa = q-2E«Eza+3

(3)

Apply Ta to the first equation in (3), apply T3 to the second one; we get Ea+,3E2a+3 = g-2Eza+3E.+a

and E2a+aEa+a = q-2Ea+3E2a+a

(4)

Apply T to both equation in (3); we get E,3 E,, +3 = q-2E«+3E,3

and EaE2o+I = q-2E2a+3Ea

(5)

The equations following 8.17(8) say that Ea+3Ea - EaEa+3 = [2]aE2a+a

and

EaE,,+3

- Ea+3Ea = [2]aE2a+0

(6)

APPENDIX

171

The definition in (1),(2) and the explicit formulas in 8.17(3),(4) imply q-2E3Ea

E,E3 = g2E3Ea - g2Ea+3 =

(7) - E«+,3 In order to get formulas for the commutators of E,3 with E2,,+,3 and E2',+,3, we notice first that

and

[Fp, Ea+,3] = ECK31

[F,3, E«+3] = -KpEa.

(8)

(Observe that the first equality says ad(Fp)ad(E,3)Ea = Ea, which follows from 8.8(5). Then apply r to get the second equality.) Now (8) and (6) together yield [F3, E2a+,3] = (q2 - 1)E.2)K31

and

[Fa, E2a+3]= (1 - g2)K,3Ex2).

Now K3 commutes with E2a+3 and E2a+,3 since (2a+,3,,3) = 0. So we can rewrite the last equations as [F3K3, E2a+a] = (q2 - 1)E.2) = [-K3 1F3, E2a+5J

Now apply T3 1 to the first equality and apply Tp to the second one; we get: [-E,3, E2a+3] = (q2 - 1)E'

and

[E3, E2a+,3] = (q2

- 1)E(2) Q+3*

We use here the convention E«+a

=

[n]

and

= T,3Ta (Ea ) ) =

E2a+a

E2a+,3 n [njQ

similarly for E«+,3 and E2a+,3. Our last equations say more explicitly E3E2a+3 = E2a+,3E,3 +

(g2

- 1)Ep+;3

(9)

and

E3E2a+3 = E2a+3E3 - (q2 - 1)E(2) «+a All these commutator formulas lead to commutator formulas for the powers of our root vectors. Those arising from (3)-(5) are quite simple; for example we get from (3) and (5) for all integers m, n > 0 2nmEm En En Em 3 a+;3 = q a+3 3

and Em Ena= q 2n"`E,,E2, 2a+,3 a 2a+3

() 10

The formulas to be derived from (6), (7), and (9) get more complicated. Let me state here only the simplest cases (proved by an easy induction): One has for all

n>0

- EaE«+3 + qn 1 [2]aE2a+0E«+3 E(3n)Ea - q-2n EQE() +

E(n) a+a Ea

(11)

Ea+aE(n-1),

E2a+3E ) = E(3n)E2a+l3 - (q2 -

(12) 1)q-2(n-1)E(n-1)E(2)

(13)

Let us now return to the proof of Proposition 8.22.b in the present case. If we take the reduced expression s,3sas3s,, then products in 8.21(2) are (up to scalar factors) all E«Eaa+3Ec+,(3E,

(14)

If we take sas;3sas,3, then we get all Ea(E«+3)b(E2a+3)`E.d.

(15)

172

8.

BRAID GROUP ACTIONS AND PBW TYPE BASIS

Let us look at (14) first. Denote the subalgebra of U+ generated by E« and E,3 by U« Both E«+Q and E2«+,3 are contained in U« cf. (1),(2) and 8.17(3),(8). So also all products in (14) are in U, p . In order to prove that these products span U+ it suffices to show that their span is stable under rightmultiplication with the generators E« and E,3 of Ua p . This is obvious in the case of E,3. For E« observe that (12),(11),(10) imply E2«+1E«+8EQE« = q-2dE2c +,3E«+19EEa

=q

2d

b

c

d

[d]AEz«+aE«+aE

E2«+f3E«E«+,3Ep + 9

c-1

b+1

-i

c-1

d

[2]«[c]«Ez«+pE«+pEa

+ [d],,E2«+3E«+r3Ea 1 = q2(b-d)E«E2«+f3E«+l3EE + qc-1 [2]«[c]«E2«+QE«+f3Ef3

+ [d]3E2«+/3E«+1 Fd 1

So indeed the products in (14) span U, Q,.

Instead of dealing directly with the products in (15), we look instead at all products ,3,+02.+q,. EaEb Ec (16) Ed Using the formulas (12),(10),(11),(4) one checks easily that E«E,3E.b+AE2c«+a

=

q2(a-c)EaE.b+,,Ez«+AE«

-

q2a-b+1[2] [b] E Ecb, 1 Ec+1 « « «+0 2«+/3

(17)

-

q2

Ec a Ea-lEb+1 «+Q 2«+3 [ ]Q p

This shows that the span of the products in (16) is stable under left multiplication with E«. It is obviously stable under left multiplication by E,3, hence all of Ua p] We get the products in (15) by applying r to the products in (16), cf. (2). Since -r induces an antiautomorphism of U+ 01, this implies that also the products in (15) span Ua 01 . This concludes the proof of 8.22.b in the case m = 4.

CHAPTER 8A

Proof of Proposition 8.28 We keep the assumptions of Chapter 8. We fix a E II. The claim in Proposition 8.28 is certainly true for y = 1: Since 1 E Uo is orthogonal to all U+ with µ :/- 0, we may assume that x = 1, and in that case the claim is obvious. Our strategy for the proof in general is this: We first find generators for the algebra U-[sawo]. Then we show: If the claim in Proposition 8.28 holds for a given y, then it holds for all yl y where yl runs through our generators of U- [s,,,wo]. This will inductively prove the proposition.

8A.1. LEMMA. The algebra TU+[s,,wo] is generated by all ad(E,(,,s))E1j with

,lEII, 0:/- a and 0<s<-(,(3,a"). PROOF. Let V be the subalgebra generated by all ad(E,(,))E,3 as above. Since

ad(E(s))Eo = T. (Tad(E(-(, Y)-s)) EE)

(1)

by 8.14(6), we get V C T,,,U+[s,,wo] by 8.25(2). We have by 4.18(2)

E. ad(E( ))E3 = ad(E.) ad(E( s))E0 + K. (ad(E( .s))Ep)K« 1E. = [s + 1].ad(Eas+1))E3 + q(sc+a,c) (ad(E(s))EO)E.

0 by 8.8(4). This shows that

and

ETV C VEn+V

(2)

by induction on the length of a monomial in the generators of V. Now (2) implies that Ei>0 VET is stable under left multiplication by all EE with ,(3 E II, since E0 is one of the generators of V in case ,(3 a. We have 1 E V, hence U+ = Since U+ is the direct sum of all by Ez>0VEc,.

8.25(1) and since V C T,,, U+[sawo], we get V = TQU+[s,,wo].

REMARK. Taking inverse images under T. we see that U+[sawo] is generated

by all Tad(Ea))E8 with 3 E II, 3

a and 0 < s < -(0, a'). So U-[s,,wo]

w(U+[s,,wo]) is generated by all S

(-1

)'q2(-%3,'v)+1-s)F.((s-i) F,3 Fa)

i=0 173

(3)

8A. PROOF OF PROPOSITION 8.28

174

with ,Q E II, ,3 # a and 0 < s < -(,(3, a"). (Note that the sum in (3) differs from w(Tad(Ea((3'«v)-S))E,3) by a scalar factor.)

As indicated above, we want to show that the claim in Proposition 8.28 holds for y, y, if it holds for y and if y1 is one of our generators of U-[s«w°] as in (3). We have then to evaluate (y1 y, x) for x E U+ [s«w°] and want to do this using the formula (yiy, x) = (y1 (D y, 0(x)) from 6.12(1). For this we need more information on 0(x) and also on 0(T«(x)) for these x. That is the object of the following subsections.

8A.2. Let ,3 E II, Q # a. Set r = - (,(3, a'). For O < i < r set xi = ad(E(i))E3.

and

xi = Tad(E,(,1))E3

(1)

So we have in particular x0 = x' = E3. Recall from 8.14(6) that T«(xi) = xr_i for all i. LEMMA. We have for 0 < i < r

ra(xi) = 0

and

g2(i-1)-r(grr+1-i - q- (r+I -i))xi

r«(x; =

1

(2)

(setting x-1 = 0) and i

r3(xi) _ (I(' - ga2(r+1-A)

E(z).

(3)

(j=] PROOF. We have xi E U+[s«w°] since T«(xi) E U+. So 8.26(4) yields rq(xi) 0. Now 6.17(2) implies

r«(xi) = (q« - qo 1)(xiF« - F«xi)KQ Because xi has weight is +)3 this formula is equivalent to

r« (xi) _ (q« -

q«1)q(«,(i-1)«+3)Kc 1(xiF«

- F«xi)

Using the definition of Tad - cf. 8.11(2) - we can rewrite this as r«(xi) _ (q« - q« -')q «(a

1)

r Tad(F«)(xi).

Rank 1 calculations as in 8.8(5) yield

'ad (F.) (xi) =T ad(F(,,) Tad(E«))(E3) = [r + 1 - i]« Tad(Ea 1))(E3)

= [r+1 -i]«xi-i. The definition of [r + 1 - i]« implies now the formula in (2). We now prove (3) by induction on i. For i = 0 we have xOl = E3, hence r'o3(x°) = 1. On the other hand, the product on the right hand side in (3) is empty for i = 0. Since E,(,°) = I we see thus that (3) holds for i = 0.

8A. PROOF OF PROPOSITION 8.28

175

We have for alli>0 [i + 1]« xj+1 = Tad(E)xi = x,E« - EK«xiKa 1

= xiE« -q(«.i«+3) Exi Now 6.14(4) and induction yield

[i + 1]« r'(xi+1) = r' (xi)E« - q2«i-rq(«.i3)EQr (xi) = ci I Ec(z)E« - g22i-rqa rE«F,(,,))

= ci(1 - q 2(r-i))[i + 1]« E(i+1), where I have denoted the product in (3) by ci. This proves (3) for i + 1. REMARK. Since x' = T(xi) we get from (2) and (3) using 6.14.c r«(xi) = 0

r ( x a)

and

= g22(i-1) r (q«r+1-i

-(r+1 -i)

- q«

i )xi-1

(4)

(setting x' 1 = 0) and i

r3(xi) =

fl(1 - q_2(r+1-j))) j=1

E+ («i).

(5)

/J

8A.3. Keep the notations 0, r, xi, x' from the last subsection. Set in addition (for 0 < i < r) (1)

yi = _ j=0

and

E(-1)igj(r+1-i)F(j)FaF«(i-j)

yi =

(2)

j=0

Recall that T«(y,) = y'-i by 8.14(5). LEMMA. We have for all i (0 < i < r)

I

(yi, xi) = (yi"xi) = L

i

(F'3, E3).

(3)

PROOF. We have (F«U-, xi) = 0 = (U-F«, xi) by 8.27(1). This implies

(yi,xi) = ((-1)iq'(r+1-i)FfF«(,i),xi) =

(-1)iga«(r+1-i)(F3,E3)(Faz),r'

(xi

8A. PROOF OF PROPOSITION 8.28

176

and I)

( yz+ xi

=

((_1)iq (r+l-i)Fpz)F3,xi) _

(xi))

Abbreviate the product in 8A.2(3) by ci. Since ra (xi) = ciEp2) = r,3 (xi) by 8A.2(3),(5) we get now (-1)iq'(r+l-i)ci(Fa(), E(

(yi, xi) = (yi, xi) _

EE).

(4)

We have by 6.13(4) 1)igpi-1)122

(F(t) EaZ))

-

(_

qa

)

(4ja

-

we can rewrite the product from 8A.2(3) as r

ci = g(i-I)/2qa

ar

j=r-i+l

The lemma follows as soon as we plug these two formulas into (4). REMARK. The lemma implies for all i

(Ta(yi),Ta(xz)) = (yr-i,xr-i) = I r r i (F,3, E3) L a (5)

= LrJ (F';3,E3)=(yi,Xi) So we get a special case of Proposition 8.28.

8A.4. In order to attack Proposition 8.28 in general, we shall need several generalizations of the maps ra and rn from 6.14. As before, 4.13(1) implies for all p E M, p > 0 and all x E µ+ that there are uniquely determined elements rna (x) E Uµ na and rna (x) E U± na (for each integer n > 0) such that

(x) = x 0 1 + E rn,a(x)Kna 0 Ea) + (rest)

(1)

n>O

and

0(x) = Kµ 0 x + E E( n)K,..-na 0 rna(x) + (rest)

(2)

n>O

where (rest) is a sum of terms in U+U° ®U± with v' Za in (1) resp. with v Za in (2). These definitions are (for n = 1) compatible with those in 6.14(1),(2). Obviously all rna and rna is linear on each U+; we extend them linearly to all of U+ . We have trivially for all n > 0 rna(1) = rna(1) = 0.

(3)

8A. PROOF OF PROPOSITION 8.28

177

If we want to evaluate rn« and r',, on products, then we get formulas that generalize 6.14(4), but are more complicated. For induction purposes we shall need only special cases. Using 0(E«) = E« ®1 + K« ® E« one gets easily for all x E U+

(and all n and u) rn«(Eax) = Earn«(x) +

[n]«q(«.µ-(n-1)a)r(n-1)a(x)

(4)

and

rn«(xE«) = rn«(x)E« +

[n]«q(«,µ-(n-0a)

1)«(x)

(5)

where we use the convention ro(x) = x = ro(x). Furthermore, we get for all 3 E II, ,(3

a rn«(E,3x) = EOrna(x)

and

rn«(x)E,3.

(6)

LEMMA. We have for all integers n > 0 and all x E U+ rn«(T(x)) = T(rn«(x))

(7)

and

(8)

and

(r«)n(x) = qa(n-1)/2rn«(x)

(r, )n(x) = q«(n-1)/2.1«(x)

PROOF. Since all terms are linear in x, it is enough to prove the claims for weight vectors. We then want to use induction on the weight. Since each claim holds trivially for x = I (where everything is 0), it is enough to show in each case (again by linearity): If the claim holds for a given x in some U+, then it holds for all E3x with /3 e II. In the case of (7) we get for 13 = a rn«(T(E«x)) =

rn(T(x))E« + 7(rn«(x))E« +

= T(E«rn«(x) + = T(rn«(E«x))

[n]«q(a.µ-(n-1)«)7-(r(n-1)«(x))

[n]«q(«,µ-(n-1)«)r(n-1)a(x))

where we use the definition of r, (5), induction, (4), and again the definition of T. The case 3 a is even easier using the two parts of (6) instead of (4) and (5). We get thus (7) for all x. Before we can proceed similarly with the first claim in (8), we need formulas for (r«)n(Eox). We claim that we have for all m > 0 and all x E U+ (ra)n2(E«x) = E.(r«)m(x) + [m]«q(a.µ)ga(m-1)(r«)m-1(x)

and for all 13EII, 3

(9)

a

(r«)n`(E8x) = E3(r«)'n(x).

(10)

8A. PROOF OF PROPOSITION 8.28

178

This can be checked using induction on m: For m = 1 these are just the formulas (4) and (6) for n = 1. The induction step is trivial for (10); in the case of (9) we get [m]«q(«.µ)q«(m-1)(r«)m-1(x))

r«(E«(rce)m(x) +

(rce)m+l(E«x) =

E.(ra)m+1(x)+q(a,µ-ma)(r«)m(x)

= +

[m]aq(«.µ)q«(m-1)

= Ea(r«)m+1(x)

(ra)m(x)

+q(«.µ)q«m(q«m

+ [m]«q«)(r«)m(x)

This yields (9) for m + 1 using the (easy) special case qa m + [m] «q« = [m + 1] « of 0.2(2).

We are now ready to show: If the first claim in (8) holds for some x in some U+, then it holds for each Epx with a E H. This is trivial for f a using (10) and (6). For a = a we have [n]«q(«.µ)q«(n-I)

(rc«)n(E«x) = E«(r«)'(x) +

(r(,,)n-1(x)

[n]«q(«,lt)q«(n

= E«q«(n-1)l2rn«(x) +

1)q(«n-L)(n-1)/2

r(n-1)«(x)

[n]«q(«,µ)q«2(n-1)r(n-1)«(x))

= n(n-l)/2 (E«rn«(x) + q0(n-I)/2rn«(Eax)

=

using (9), the claim for x, and (4). We get thus the claim for E«x. This argument proves the first claim in (8). The second one can be proved similarly. However, at this point it is easier to deduce it from (7) and the first claim in (8).

8A.5. Let a E H, /

a. Keep the notations R, xi, and x' from 8A.2.

PROPOSITION. We have for all i (0 < i < r)

A(xi) = Ki«+,3 ® xz + E aijxjK j ® Eat-j)

(1)

j=0 and

A(xi) = xi ®1 +

aijE(i-j) Kj«+,3 ®x'j

(2)

j=0=0

where

aij =

qa(i-j) fl(1

k=j

- q«2(r-k)) = q(i-j)(3i+i-2r-1)/2 rJ (qQ k=j

q«(r-k))

(3)

8A. PROOF OF PROPOSITION 8.28

179

PROOF. As usual, an empty product is to be interpreted as 1; so we have aii = 1 for all i. An easy calculation shows that the last equality in (3) holds. Since xi has weight is +/3, it is clear [e.g., by 4.13(1)] that A(xi) has the form

A(xi) =xi®1+Kia+,3®xi rna(xi)Kna ® L'anl + r E,")K(i_n)a+,0 ® ma(xi).

+ n=1

(4)

n=1=1

We have r'a(xi) = 0 by 8A.2(2), hence ma(xi) = 0 for all n > 0 by 8A.4(8). We have ra(xi) = bixi_1 with ga2(r+l-i)) bi = qa 1(1 by 8A.2(2). So 8A.4(8) implies n-1

ma(xi) =

gan(n-1)/2

(n-1

11 bi-mllxi-n = I ri bi_mga \ m=0 m=0 /

We have bi-m,qa (n-1-m)

qa

hence

n(1 - ga2(r+l-i++n)), i-1

n-1

fj bi_mga (n-1-m) =

qa(i-n)

fj

(1 - qa 2(r-s)) = ai,i_n.

s =i -n

m=0

So we get rna(xi) = ai,i-nxi-n If we plug this into (4) and recall the vanishing of the ma(xi) we get (1). Since also xi has weight is + /3, we can write also A(xe) in the form (4). We

have x' = T(xi), so 8A4.(7) implies now rna(xi) = 0 and ma(xi) = ai.i-nxi-n' This then yields (2). 8A.6. COROLLARY. We have A(U+[Sawo]) C

U+[Sawo]U°

® U+

(1)

and

A(TTU+[Sawo]) C U+U0 ® TU+[sawo].

(2)

PROOF. It is enough to check the claims for generators of these algebras. By the remark in 8A.1 the xi as in 8A.2(1) taken for all possible /3 generate U+[sawo];

they satisfy (1) by 8A.5(1). By Lemma 8A.1 the xi as in 8A.2(1) (again for all possible /3) generate TU+[sawo]; they satisfy (2) by 8A.5(2).

8A. PROOF OF PROPOSITION 8.28

180

8A.7. Let ,Q E II with 3 # a; set r = -(/3,av). None of the quantum Serre relations (R5) in 4.3 has a weight less than or equal to /3 + ra. Therefore all

with i, j > 0 and i + j < r are linearly independent in U+. This shows that there are for each x E U+ uniquely determined elements ri« a j( (x) E U -3_(i+j)« such that A(x) = Kµ ® x +

E( n)Kli-n« ® rn«(x) k

n>O

r

r-i

(1)

+ E E E(«i)EQE(j)Kµ-a-(i+j)« ® r'i«.3.j«(x) + (*) i=o j=0

where (*) is a sum of terms in U+Uo ® U+ with v Za, v ra + /3. Obviously ro«.a.o« is equal to the r' from 6.14(2). We shall abbreviate ri« 3 = r'.,,3,0,, and r'' j« = ro«.pj « Clearly each ri« Q j« is linear. We extend it linearly to all of U.

Let i be an integer with 0 < i < r. Use the notations xi, xi, yi, and y as in

8A.2(1) and 8A.3(1),(2).

LEMMA. We have for all y E U- and all x E U+

(yiy,x) = (yi,xi)(y,rR.i«(x))

(1)

(yiy, x) = (y', x') (y, ri«,3(x))

(2)

and

PROOF. We may assume x E U+ for some µ. We want to use that (yiy, x) = (yi ® y, A(x))

by 6.12(1). Since yj is orthogonal to all U+Uo with v :h is + /3 by 6.13(2), we can replace 0(x) in this equation by its weight component in U++aUo ®U± i«-,3, i.e., by

E(j)EpEai j)I{µ-i«-Q ® rji«,Q,(i-j)«(x). j=o

Now yi E U- [s«wo] implies (yi, E«U+) = 0 by 8.27(2). We get therefore (yiy, x) = (yi, E/3E«t))(y, rt«,,s(x)).

Now recall from 8.12(2) that

xi =

1`(_1)jq-j(r+l-i)E(j)E,3E(«i-j).

jj=o

So we have xi - EaEPP) (mod E«U+), hence (yi, E3E«t)) = (yi, xi). This yields (1).

The proof of (2) is similar. We use now that y' E T«U- [s«wo] satisfies (y', U+E«) = 0 by 8.27(2). Furthermore, observe that xi = T(xi) = T(EaE«)) = Eai)E,3

(mod T(E«U+) = U+E«).

8A. PROOF OF PROPOSITION 8.28

181

8A.8. Let us discuss the strategy for the proof of Proposition 8.28. The claim is for all y E U- [sawo] that for all x E U+[sawo].

(Ta(y), Ta(x)) = (y, x)

(1)

Since it holds for y = 1 and all x, it is enough to show: If (1) holds for some y and all x, then (1) holds for each yiy and all x, where we take all possible yi as in 8A.3(1) (for all 8 E H with /3 54 a). Fix now one of these yi (and thus one of these 8) and use the other notations from 8A.7. We have Ta(y) = yr_i and (yi, xi) = (y; -i, x'r-i) by 8A.3(5). Therefore Lemma 8A.7 shows that (1) (for yiy instead of y) is equivalent to

(y, rR.ia(x)) = (Ta(y), r(r-i)a.a(T.(x))) Corollary 8A.6 implies that y satisfies (1), we know therefore that

(2)

r(r-i)a.a(To.(x)) E TaU+[sawo]. Since we assume that

(T.(y)'r(r-i)a.,3(Ta(x))) = (y,Ta lr(r-i)a,a(Ta(x))) So (2) is equivalent to

(y,ra,ia(x)) = (y,Ta Ir(r-i)a.S(Ta(x)))

(3)

Now y E U- [sawo] satisfies (y, E, U+) = 0 by 8.27(2). We get thus: OBSERVATION. If we can show for all ,8 E H, 3 54 a and for all integers i with

0 < i < -(,Q, av) that (mod EaU+)

ra,ia(x) = Ta Ir(r-i)a,p(Ta(x))

for all x E U+[sawo],

then Proposition 8.28 follows.

8A.9. For the rest of this chapter we fix a simple root 8 54 a. We set r = -(8,a") and we use the notations xi and x? as in 8A.2(1). In order to prove the formula in the observation above we shall need induction formulas for the r'-functions involved. First, however, we record their values on the xi resp. x?. Since EpEa') occurs with coefficient 1 in xj = Tad(E,j))Ea, cf. 8.12(2), the formula 8A.5(1) implies (using the aij from 8A.5(3)) aijEa'-,

rai

On the other hand, So 8A.5(2) implies

(xi) _

,

if j
ifj>i.

10,

(1)

occurs with coefficient 1 in x'j = ad(Eaf))E,3, cf. 8.8(3).

rja,3(xi) _

11, if j = i; (2) 0,

otherwise.

8A. PROOF OP PROPOSITION 8.28

182

LEMMA. Let x c- U+ and x' E U±. We have for all i (0 < i < r) (µ.i«+13) xrpi«(x , ) rQ.i«(xx) = r',i«(x)x + q i-1 + [ji q(µ-(j«+Q),(i-j)«)ra,j«(x)r(i-j)«(x/) ,

J J«

and

ri« Q(xx') = ri«

/3(x)x'

q(la,i«+13)xri«,1

+

(x/) -j) (k Wri

J

PROOF. We expand 0(x) and A(x) as in 8A.7(1). The with i+j r are linearly independent; so the only contributions to rQ i«(xx') arise from the following terms in A(x)A(x')

',i«(x')) + (E/3E( )Kp-(i«+/3) (9 ',i«(x))(K, (9 x')

(Kµ (&

(9

i-1

+ 1:(E8E(j)K,,,-(j«+/3) ®r',j«(x))(E,,(,=-j)Kµ,-(i-j)« ®r'

(x/))

j=o

This leads easily to the claim on rQ.i«(xx'). Similarly, the contributions to ri« 3(xx') come from (Kµ 0 x)(E,=)E,3Km,-(i«+Q) ® ri«,R(x')) + (Ea)EgKm-(i«+Q) (9 ri«,o(x))(K, (9 i ) i-1

+ E(E(i-j)Kµ-(i-j)« ®r(i-j) ,(x))(E( i)EgKm,(j«+R) ® rj«Q(x')) j=o

Again it is easy to deduce the formula for ri« p(xx'). REMARK. If x' E U+[s«wo], then 0 for all n > 0 by 8A.4(8) and 8.26(4). So in this case the first formula above can be simplified: r(3.ice(x)x +q(li,z«+13)xrQi«(x')

if x' E U+[s«wo]

(3)

8A.10. We have by 8.25(5) a direct sum decomposition U+ = U+[s«wo] E«U+. Denote by 7r« : U+ -> U+[s«wo] the projection with kernel E«U+. This kernel is a right ideal in U+. That implies 1r«(yz) = 1r«(7r«(y)z)

for all y, z E U+,

(1)

in particular: 7r«(yz) = 7r«(y)z

for all y E U+ and z E U+[s«wo].

(2)

8A. PROOF OF PROPOSITION 8.28

183

LEMMA. We have for all x E U+[sawo]µ and all integers m > 0 q

-(µ,a) m(m+1)

a(xEa)

Ta 1(ra)mTa(x)

(qa - ga1)m

(3)

PROOF. We use induction on in. First m = 1: We have ra(Ta(x)) = 0 by Lemma 8.26; so 6.17(2) implies Ta (x)Fy - FaTy (x) _ -(qa - qa 1)- ' Ka lr' (Ta (x))

We apply T;' to this equation; since T, 1(Fa) = -EaKa by 8.14(2), we get 1)-1KyTa

xEaKa - E.Kax = (qa - qa

1(ra(Ta(x)))

Multiply by Ka' on the left and use 4.7(1): q-(Fi+a,a)xEa

- q-(a'a)Eax = (qa - qa 1)-1Ta 1raTa(x),

hence

xEa = q (11,a) Eax +

q

qa - qa-1

Ta 1raTa(x).

This yields (3) in the case m = 1. We use (1) for our induction step: 7ra(xE' +1) = 7fa(ira(xE' )Ea)

=

q

(µ+(m+1)a,a)

Ta 1rIkTa(ira(x2!a ))

-1

qa - qa

-1 T

q(µ,a)g2a(m+1)

rq

-(µ.a) qma(m+1)

qa - qa

1)

(qa - qa

(qa - qq- 1)'11+1

Ta

T-1 (r')"eTa(x))

m 1(ra)m+1Ta(x),

as desired. REMARK. Using Lemma 8A.4 we can rewrite (3) as q

a(xEa)

m(µ,a) m(3m+1)/2 9

( qa

qa1)m

8A.11. LEMMA. We have for all µ E Zt,

Ta lr"'aTa(x).

(4)

0, all u E U+[sawo]µ and for

all s with 0<s
(T. (u)) = T. o 7fa(rQ,sa(u))

(1)

8A. PROOF OF PROPOSITION 8.28

184

PROOF. Obviously (1) holds for u = 1 where both sides are 0. We want to show: If (1) holds for u = x, then it holds for u = xx' where x' runs over the generators of U+[sawo]. This will then prove (1) in general.

So suppose that x E U+[sawo]µ for some p such that (1) holds for u = x. Suppose further that x' = for some -y E II, -y a with 0 < i < -(1' av) CASE 1: We have -y : 3. The weight of x' is equal to -y + ia. Since 3 ¢ -y + is this implies rQ ja(x') = 0 for all j. So 8A.9(3) implies r3 sa(xx') = rp,sa(x)x' hence

7r, a rQ sa(xx') = 7ra(r3.sa(x) x') = ia(r3,sa(x)) x'.

Now Ta (x') has weight sa (-y + ia); so 3 ¢ sa (-y + ia) implies r, 3 (Ta (x')) = 0 for all j. Therefore the second formula in Lemma 8A.9 yields

r(r-s)a..Q(Ta(xx')) = r(r-s)a,p(Ta(x)) TT(x'). Now, since (1) is satisfied for u = x, we get

r(r-s)a.3(Ta(xx')) = Ta(7a(r,sa(x))) Ta(x) = Ta(ia ° r,sa(xx')) So (1) holds for u = xx'.

CASE 2: We have -y = 3. Then x' = xi; so 8A.9(3),(1) show (with ai.7 as in 8A.5(3))

rp sa(xes) = r3'sa(x)es + A where A = {

if s > i;

0,

i-s ), if S < i.

aisq(µ.sa+3)xEc

On the other hand, the second formula in Lemma 8A.9 yields now r(r-s)a.3 (Ta(x)xr-i) = r(r-s)a,3

(Ta(x))ir_x +

B

where

B=

0,

ifs > i;

q(snµ.(r-s)a+3) Ta(x),

if S = i;

q(s.µ-(i-s)a.(r-i)a+3) r - S 2-S

r('i-s)a(Ta(x))

if s < i.

a

We can rewrite the q terms in these formulas using the invariance of (,) under W

together with saa = -a and sa,0 =,3 + ra: 0,

B=

q(µ.sa+3) Ta(x)

2-r) I r - S I

r' 2 - S a (i-s)a

(Ta(x)), if S < i.

8A. PROOF OF PROPOSITION 8.28

(Actually, we use also that We have by assumption

q(a,a)

185

= qa and q(0'0) = qa' .)

r(r-s)a,Q (Ta(x)) = Tao 7ra(',a(x)), hence

r(r-s)a,Q (Ta(x)xr-i) = Ta(7ra(rQ,a(x)) xi) = Ta o 7ra(rp.a(x) xi)

So we get (1) for u = xx' if we can show that B = Ta7ra(A). Of course, the case s > i is trivial. For s = i we have ais = 1 and Eai-s) = 1, hence A = q(f`>sa+Q)x This yields Ira (A) = A and Ta o Ira (A) = q(µ,Sa+Q)To, (x) = B as desired.

Assume now that s < i. We get from 8A.10(4) Ta o Ira (A) = a r(z-s)a (Ta (x))

where

(t_s)(N a) (i-s)(3(i-s)+1)/2

a=aisq (li sa+Q) q-

qa

.

(qa - gal )Z-s[Z - s]a

Now 8A.5(3) yields

(i-s)(2i-r)

a = q(µ.ia+Q) qa

i

[Z - S] a

r-s 11

-i

qa - qa

j=r-i+1 ga

- qa

1

= q(Ii,ia+Q) q(ai-s)(2i-r)

r-s Z-S

Of

hence Tao Ira (A) = B. 8A.12. Note that Lemma 8A.11 implies the formula in the observation in 8A.8. Therefore we have now completed the proof of Proposition 8.28. We want now to use Proposition 8.28 to prove a formula that contains 8A.11(1)

as a special case and that looks nicer. We have a direct sum decomposition

U+ = U+Ea ® TaU+[sawo], and hence also a direct sum decomposition U+U° = (U+Ec)U° ® (TaU+[sawo])U°. Denote the corresponding projection by : U+U° -' Ta(U+[sawo])U°

7r'a

PROPOSITION. We have for all x E U+[sawo]

(Ira®id)oi oTa(x)=(Ta®Ta)o(id®7ra)oA(x)

(1)

PROOF. Set z = (7r' ®id) o i o Ta (x) - (T,,, ®Ta) o (id (9 7ra) o A(x). We want to show that z = 0. Corollary 8A.6 implies that z E ®(TaU+[Sawo])µKu ®

(TaU+[Sawo])v.

µ.v

Corollary 8.30 together with 6.13(1) implies that ( , ) induces a nondegenerate pairing between IJ µ®U and U+K®U+ for all p and v. Furthermore, U7., ®U is perpendicular to U+K ® U+ if (p', v') # (µ, v) by 6.13(2). So, in order to prove that z = 0, it is enough to show that (u ® u', z) = 0 for all u, u' E U-.

8A. PROOF OF PROPOSITION 8.28

186

We have (U-F,,, TaU+[sawo]) = 0 by 8.27(1), hence in our situation

(U-Fa 0 U- + U- 0 U-F,, z) = 0. Since U- = U-Fa®TaU- [sawo] by 8.25(6), it is enough to show that (u®u', z) = 0 for all u, u' E TaU+[sawo]. Well, consider y, y' E U- [sawo]. Also the product yy' is in U- [sawn], cf. Remark 1 in 8.25, so 8.28 implies (yy',x) = (Ta(y)Ta(y'),Ta(x)). So 6.12(1) yields (y ® y, A(x)) = (T. (Y) ® T. (Y,), A o T«(x))

(2)

We have (y', EaU+) = 0 by 8.27(2), hence (y ®y', U+ U° ®EaU+) = 0. This shows that the left hand side in (2) is equal to

LHS = (y®y', (id0ire) o A(x)). Applying Proposition 8.28 we get

LHS = (Ta(y) ®7'a(y ), (7'a ®Ta) o (id ®7ra) o A(x)).

(3)

On the other side, we have (Ta(y),U+Ea) = 0 by 8.27(2), hence (T,, (y) 0 T" (y), (U+Ea)U° ®U+) = 0. Therefore the right hand side in (2) is equal to

RHS = (Ta(y) 0T.(y), (7r' ®id) o A oTa(x))

(4)

If we now plug (3) and (4) back into (2), we get (Ta(y) 0 Ta(y'), z) = 0, i.e., (u ®u', z) = 0 for all u, u' E TaU+[sawo], hence the claim. REMARKS. 1) If you want to compare with Lusztig's approach in [L]: He proves first Proposition 8A.12, which is Lemma 38.1.8(b) in [L], and then uses it to prove Proposition 8.28, which is Proposition 38.2.1 in [L]. 2) Using the non-degeneracy result Corollary 8.30 we can make the formulas 8.27(1),(2) more precise. Let me mention only one example: We have U+[sawo] = { x E U+ I (FaU

,

x) = 0 }.

(5)

Indeed, it is enough to consider each weight space Uµ separately. We have by 8.25(5),(6)

dim U+[sawo]µ = dim U,, - dim U± a = dim U_µ - dimU (µ_a)

= dimU µ - dim(FaU Now Corollary 8.30 and 8.27(1) imply (5).

CHAPTER 9

Crystal Bases I Suppose for the moment that g is the Lie algebra S12 with the standard basis X,

Y, H (where [X, Y] = H, [H, X] = 2X, and [H, Y] = -2Y, cf. [H], § 7). If M is a finite dimensional g-module, then one can find a basis vi, V2, ... , vim, of M such

that each Yvti is either 0 or a nonzero multiple of another vj and such that also each XVh is either 0 or a nonzero multiple of another vI. Indeed, this is true for irreducible M by the explicit description, say in [H], 7.2. It follows for arbitrary M from the complete reducibility of M.

For general g we can consider for each simple root a nonzero root vectors Xc. E gc, and Yc. E g-,, that span together with Hc. = [Xe, Y] a Lie subalgebra isomorphic to 5(2i cf. [H], Proposition 8.3(f). So, if M is a finite dimensional gmodule, then one can find (for fixed a) a basis vi, V2, ... , vim, of M such that each Yvti is either 0 or a nonzero multiple of another vj and such that also each Xcvh is either 0 or a nonzero multiple of another vI. However, in general, there does not exist a basis that works simultaneously for all simple a. (There are exceptions, such as the adjoint representations or the "natural" modules for the classical Lie algebras. But those are very special cases.) The observations above carry over to quantized enveloping algebras for q not a root of unity. Here we use the explicit description of the simple Uq(5(2)-modules in 2.6. We get for arbitrary g and each finite dimensional U. (g) -module M: For each simple root a there is a basis v1 V2i ... , vim, of M such that each Favti is either 0 or a nonzero multiple of another vj and such that also each Ecvh is either 0 or a nonzero multiple of another vi. Again, there does not exist (in general) a basis that works simultaneously for all simple a. Now Kashiwara has shown that such a basis (working simultaneously for all a) exists at "q = 0". A priori this statement does not make sense, since we cannot take q = 0 in our constriction. What we mean is this: Suppose that k = Q(q) with q transcendental over Q. Then k is the fraction field of the polynomial ring Q[q] and it contains the local ring A of all fractions f /g with f, g E Q[q] and g(0) 0 0. Then Kashiwara shows that each finite dimensional Uq(g)-module M contains a "nice" A -lattice M such that there is a "nice" basis of the Q -vector space M/qM. We shall soon have to get more precise about the meaning of "nice"; at this point let me just say vaguely this: One important feature of the basis is the fact that it behaves well with respect to all Ec, and Fa at the same time. After giving the necessary definitions we shall prove the existence of these "nice"

lattices and bases. In Chapter 10 we shall consider bases "at q = 0" for U-. In Chapter 11 we shall then use these bases "at q = 0" to construct bases for the Uq(g)-modules and for U- over k. These bases turn out to have other remarkable

187

188

9.

CRYSTAL BASES I

properties.

Throughout this chapter we assume that k = Q(q) with q transcendental over Q. We set U = Uq(g). For each a E H denote by U' the subalgebra of U generated by Ea, Fa, K. and K, 1. It is isomorphic to U9.. (5(2), cf. 4.17.

9.1. Let M be a finite dimensional Ua-module (of type 1, as usual). Then M is a direct sum of simple modules M = L1 ®L2 ® ® L, with Li ^- L(n(i)) for some integer n(i) > 0. Choose for each i a highest weight vector mi of Li. Then

the F,)mi with I < i < r and 0 < j < n(i) are a basis of Al where we use the divided powers Fem.') = Fa/[j]! as in 8.6(2). If x E Al is a weight vector of weight 1 E Z (i.e., Kax = q1 x), then there are a, E k with

x= we get then

aiFQi)mi; j>O i,n(i)-2j=1

x = E F:j)xj

(1)

i>O.j>-l

where xj = Ei:n(i)-2j=1 aimi. (We get the inequality j > -l from j < n(i) = 1+2j.) Note that each x j is a weight vector of weight I+ 2j with Eax j = 0. We claim that this property determines the decomposition in (1) uniquely. More precisely: CLAIM. If X = F_j>o j>_1 Faj)x' with each x' E M1+2j and Eax'. = 0, then x'j = xJ for all j. Indeed, each Uax'j is either 0 or a simple submodule of M isomorphic to L(l + 2j). So FV)x'j is contained in the isotypic component of M of type L(l+2j). This shows that it is uniquely determined as the projection of x into this component,

and we get F,(,j)xf = Faj)xj. Now j > -l implies j < 1 + 2j. Therefore Faj) is injective on the 1 + 2j weight space of L(l + 2j), hence on the 1 + 2j weight space of a direct sum of modules isomorphic to L(l + 2j). This yields x'j = xj.

9.2. Let M be a finite dimensional U-module. Fix a E H. For each \ E A the (direct) sum of all Ma+ia with i E Z is a U'-submodule of M; any Ma+ja is the q(A+ja.a) eigenspace of Ka in this submodule. So the discussion in 9.1 shows (for each \ E A): Any x E Ma can be written uniquely

x=

FQj)xj

(1)

with xj E Ma+ja and Eaxj = 0 for all j. _ We now define (for each a E II) linear maps P. and Ea on M as follows: For any weight vector x E Ma write x as in (1) and set

Fax =

F'a(j+1)xj,

(2)

F«j-1)xj

(3)

and

Eax =

9.

CRYSTAL BASES 1

189

Then extend FQ and EQ linearly to M = ®A MA. Note: We have for a weight vector x E MA as above

FQEQx=x-x0

and

EQFQx=x-Fa''Mx,.

(4)

where r = -(\, a'). Furthermore, since + 1]- 1 F J)xj)

FQx = F.

and

+ 1]QFai)xj),

FQx = F« J

7

we see that FQMA = FQMA.

(5)

Here are a few obvious properties of these maps: (a) If M' C M is a U-submodule, then FQM' C M' and EQM' C M' for all a E II. (b) If W : M -* N is a homomorphism of U-modules, then W commutes with FQ

and k,,. Well, the main point in (b) is that W preserves decompositions as in (1). And in (a) we really ought to have said that the FQ for M' is the restriction of the FQ for M; similarly for E. Of course, that is just the special case of (b) where W is an inclusion.

Before I go on, let me return to the introductory remarks of this chapter. If V is a finite dimensional UQ (5L2) -module, then we can find a basis vi, v2, ... , vn of V such

that each Fvz (and each Evz) is either 0 or another vj. This is stronger than our original version ("is either 0 or a nonzero multiple of another vj ") and we really want to have this stronger property.

9.3. Recall that k is the fraction field of the polynomial ring Q[q]. Set I f, g E Q[q], g(0)

A

0 }.

(1)

g

This is a discrete valuation ring, i.e., a principal ideal domain that is also a local ring. The unique maximal ideal in A is generated by q. We shall identify the residue

class ring A/qA with Q. We denote for each a E A the coset a + Aq by a; so our identification equates any (f /g) with f (0)/g(0). We shall use, more generally, the

convention v = v + qM E M/qM for any A-module M and any v E M. (Of course, any given v might be considered as an element in different A-modules. So we use the notation v only when it should be clear which M we consider.) Let M be a finite dimensional U-module. DEFINITION. An admissible lattice in M is an A-submodule M of M such that

(Al) M is finitely generated over A and it generates M as a vector space over k,

(A2) M = ®AeA MA where MA = M fl MA, (A3) FQ.M C M and EQM C M for all a E II. Note: Any M as above (and any MA) is a free A-module of finite rank, since A is a principal ideal domain and since M is torsion free being contained in M. The condition (Al) says that the natural map

MOA k->M,

m®c- cm

(2)

190

9.

CRYSTAL BASES I

is surjective. In fact, it is an isomorphism. (Any x E M ®A k can be written x = y ® S-1 with y E M and s E A, s # 0: Find a common denominator! Then the map takes x to s-1y, which is 0 if and only if y = 0, i.e., if and only if x = 0.) The proof of the following lemma is left as an exercise:

LEMMA. a) Let W : M => M' be an isomorphism of finite dimensional Umodules, let M C M be an A-submodule of M. Then M is an admissible lattice in M if and only if W(M) is an admissible lattice in M'. b) Let M1i M2 be two finite dimensional U-modules, let M1 C Ml and M2 C M2 be A-submodules. Then M1 ® M2 is an admissible lattice in M1 ® M2 if and only if M1 C M1 and M2 C M2 are both admissible lattices. 9.4. Let M be a finite dimensional U-module. If M is an admissible lattice in M, then F., and f,,, induce maps on M/qM that are again denoted by F« and E«. DEFINITION. A crystal base of M is a pair (M, B) where M is an admissible lattice in M and B is a basis of the vector space M/qM over Q such that (Cl) B = UREA BA where Bx =13 n (Ma/qMA), (C2) F«13C13U{0}and E«13C13U{0} for allaEII, (C3)

foranyb,b'EBandaEll:

b=E«b' -# b'=F«b.

The condition (C3) means [given (C2)]: Let b E B and a E II. If F«b

0, then

b=EaF«b; if Eab#0, then b=FaEab. Again, it is easy to show (and left as an exercise) that:

LEMMA. a) Let W : M => M' be an isomorphism of finite dimensional Umodules. If (M, B) is a crystal base of M, then (cp(M), p(B)) is a crystal base of M' (where iP is the induced map M/gM -> cp(M)/gcp(M)). b) Let M1, M2 be two finite dimensional U-modules, let M1 C M1 and M2 C M2 be A-submodules, and let 131 C M1/qM1 and 132 C M2/qM2 be subsets. Set B = (131 X 0) U (0 X 132) C (M1 ® M2)/q(M1 (D M2).

Then (M1 ®M2, B) is a crystal base of M1®M2 if and only if (M1,131) is a crystal base of M1 and (M2,B2) is a crystal base of M2.

9.5. We want to show that each finite dimensional U-module admits a crystal base. By Lemma 9.4.b it is enough to look at simple modules. Let me describe now what the final result will be in that case. Choose for each dominant weight A E A a highest weight vector VA E L(A)a, va # 0. Set G(A) equal to the A-submodule of L(A) spanned by all Fq, Fa2 ... FcrvA for all integers r > 0 and all series a1 i a2, ... , a, of simple roots. We shall show (in 9.25) that G(A) is an admissible lattice in L(A). It is clear that G(A) is a finitely generated A-submodule of L(A) with FaL(A) C G(A) for all a E II. The generators of G(A) are weight vectors; this implies that G(A) is the direct sum of all £(A),.

Furthermore, each L(A), generates L(A), over k: This is trivial for µ = A; for µ < \ it follows inductively from L(A)j, =

FcL(A)m+. aEn

FcL(A)m+. OtEn

where we use 9.2(5) for the last equality. So the only condition from 9.3 that we have not yet checked is whether each Ea maps G(A) to itself.

9.

CRYSTAL BASES 1

191

Set B(A) C G(A)/qr(A) equal to the set of all nonzero Fa, Fa2 Fa .va (with r and the ai as above). We shall show (in 9.25) that (,C (A), B(A)) is a crystal base of L(A). At this point we know only that B(A) spans the vector space G(A)/qC(A) over Q, that it satisfies (Cl) and one part of (C2) [the inclusion FaB(A) C B(A) U {0}]. We do not yet know, whether B(A) is linearly independent and whether (C3) and the other part of (C2) hold. Note that different choices of va lead to pairs (G(A), B(A)) that differ by an automorphism of L(A) (as in Lemma 9.4.a). So it does not depend on that choice, whether we get a crystal base. 9.6. LEMMA. a) If 4i is of type A1, then (G(A), B(A)) is a crystal base of L(A) for all dominant weights A. b) If A is either 0 or a minuscule dominant weight or the largest short root in an indecomposable component of 4i, then (G(A), B(A)) is a crystal base of L(A).

PROOF. a) Denote the unique simple root by a. Let A be a dominant weight and set n = (A, a"). So L(A) is the simple module denoted by L(n) in 8.3. So the with 0 < i < n are a basis of L(A). We have Favi = vi+1 and Eavi = vi = vi_1 using the convention v_1 = 0 = vn+1 This shows that G(A) = Ei Avi and that B(A) = {vi 10 < i < n}. The formula Eavi = vi_1 shows that EaG(A) C G(A), hence that G(A) is an admissible lattice in L(A). It also shows that B(A) satisfies (C3) and both parts of (C2). Since B(A) is clearly linearly independent, the claim follows.

b) The case A = 0 is trivial. (We have G(A) = Ava and B(A) Consider next the case where A is minuscule. We have given in 5A.1 an explicit desription of L(A). Let us use the notation from that subsection. So L(A) has a basis xµ with

,u E WA. If µ E WA and a E l] with (µ, a') = 0, then EaXµ = 0 = Faxµ by 5A.1(2), hence

Eaxµ = 0 and FaXµ = 0. If (µ, a") = 1, then Eaxµ = 0 [cf. 5A.1(2)] implies EaXµ = 0

and

F.Xu = FFl)xµ = xµ-a

EaXN_a = Xµ

and

Fax, _a =

(1)

(2)

and

0.

(3)

Note that (3) describes all Eaxµ, and Fax,, with (µ', a") _ -1 since these µ' are exactly the µ - a = saµ with µ as above. Clearly (1)-(3) show that L = >J EWA Ax, is an admissible lattice in L(A). Set B = {-Xµ I µ E WA}; it is quite obvious by (1)-(3) that (G, B) is a crystal base of L(A).

We now have to compare (G, B) with (,C (A), B(A)). We may assume that va =

Xa. Since then va E G and since G satisfies (A3), we get G(A) C G. We claim that this inclusion is in fact an equality. We have to show that xµ E G(A) for all ,u E WA. This is done by induction on µ from above. For µ = A the claim holds by construction. If µ < A, then µ is not dominant, hence there is a E H with (µ, a") < 0. Then xµ = Faxu+a, hence xµ E G(A). The same argument shows also inductively Tµ E B(A) for all µ, hence B C B(A). On the other hand, since B satisfies (C2) and va E B, we get also the other inclusion, hence (G, B) = (G(A), B(A)).

192

9.

CRYSTAL BASES I

Suppose now that A is the largest short root in an indecomposable component of (P. If a E II belongs to _a different component of (b, then both Ea and Fa act as 0 on L(A). Hence so do Ea and Fa; therefore the conditions (A3), (C2), and (C3) are automatically satisfied for these a. In this way one sees that we can assume that (P is indecomposable and that A is the largest short root of C Under this assumption we have described L(A) in 5A.2. We are going to use the notations from that subsection. So L(A) has a basis (x. I 'y E (bs [= set of short roots = WA], hp Q E II, = II fl (s). (Recall that all roots are short, if there is only one root length.) Set

,C

Ax. + E Aho

C3={T., hp I -YEAS, 0 EIIs}.

and

(4)

/3Ens

We claim that (L, C3) is a crystal base of L(\). It is more or less obvious that the conditions (Al), (A2), and (Cl) are satisfied. In order to check the other conditions, we have to understand the actions of Ea and Fa on L(A). First of all, the arguments used in the minuscule case show for all y E (bs and

allc E II with -y 0 ±a Eax.y = 0,

Fax. = 0,

Eax7 = 0,

Fax. = x.-a) if ('Y, av) = 1; Fax. = 0, if (y, av) _ -1.

Eax7 = x.+a,

if (y, av) = 0; (5)

If a E II is not short, then Eahp = 0 = Fahp for all Q E IIS [cf. 5A.2(3)] implies

Eahp = 0 = Fahp.

(6)

Suppose now a E IIs. We have Eaxa = 0, hence

Faxa =

x-a,

Pcth, =

ha,

Fax-" =

0

(7)

and

Eaxa = 0,

Eaha = xa,

Finally, we have for all Q E IIs with Q

E'ax_a = hQ.

a

Ea (ha + (Q2] v) ha) = 0 l a

by 5A.2(4),(5), hence

Ea(ha+ (a2]a) ha) =0=Fa(hp+ (0,a) ha). This implies by (7),(8)

Eahp

(Q, av) Eaha [2].

(Q, av) xa [2].

F,ah

(Q, ay) F,,aha

(Q, ay) x_a

and

p

[2].

[2].

(8)

CRYSTAL BASES I

9.

193

Note that [2] = qQ. + q(k 1 = q, 1(q2 + 1), hence that

a")

(a, a")

q'+1

[2]

(9)

So we see that

E,hp E qG

and

Fhp E qC.

(10)

We have modulo qG

E'ho = F'hp = 0.

(11)

The formulas (5)-(11) imply easily that (A3), (C2), and (C3) are satisfied. So (G, B) is indeed a crystal base. Similar arguments as in the minuscule case show again that (G, B) = (G(A), B(A)) if we choose vA = xa. REMARK. More or less the arguments as above can be used to handle all examples studied in Chapter 5A. Details are left ...

We shall use the examples in this lemma in order to get crystal bases in general. Our strategy will be to proceed inductively: Suppose that we have a crystal base for some simple module M. Then we want to get one for all direct summands of the tensor product M ® L(A) where A is one of the weights in Lemma 9.6. b. So we have to develop a method to deal with tensor products as well as a method to descend from a module to its direct summands. Besides doing that, we shall also look at the question: How unique are crystal bases? 9.7. LEMMA. Let M be a finite dimensional U-module, let a E II and A E A. Consider a weight vector x E MA and write x =

E

(1)

i>o.i>-(A,Ck I)

with xi E MA+i, and Eaxi = 0 for all j. a) If M C M is an admissible lattice with x E M, then xi E M for all j. b) If M C M is an admissible lattice with E,,,x E qM, then xi E qM for all

j > 0. PROOF. a) We want to use induction on A from above; this makes sense, since the set of weights of M is finite. Now 9.2(4) says xo = x - FlElx, hence xo E M. And 9.2(3) says

E.x =

F.(j) xi

(2)

i>o,i>>-(A,&')

Since E,,x E M n MA+,, and since j > -(A, a") implies j > -(A + a, a") _ -(A, a") - 2, we get by induction xi E M for all j > 0. So the claim follows. b) Apply the claim in a) to E,,x and the admissible lattice gill of M. We have used already in the proof of a) that the expression in (2) is the analogue for E.x of the expression in (1). This implies the claim immediately.

194

9.

CRYSTAL BASES I

REMARK. Keep the notations of the lemma. Suppose that M is an admissible lattice in M. The lemma implies: If there is b E MA/qMA with Eab = 0 and V$ 0, then (A, a") > 0. _Indeed, we may assume b = Y with x as in (1). The assumption

Eab = 0 implies Eax E qM. If (A, a") < 0, then j = 0 does not occur in (1). So now b) yields x E qM, hence b = Y = 0, a contradiction.

9.8. Let M C M be an admissible lattice in a finite dimensional U-module M. We set then for each subset S C .M/qM

HW(S) = {xESIEax=O for all aEII},

(1)

the set of "highest weight vectors" in S. LEMMA. Let (M, 8) be a crystal base of a finite dimensional U-module M.

a) For each b E 13 there exist al, a2i ... , a,. E II and b' E HW(13) with b = Fa1Fa2...Fa,,b'. b) For each A E A the set HW(Ma/qM,\) is equal to the Q-subspace spanned by HW(13A).

c) If A is not dominant, then HW (13A) = 0.

PROOF. a) We use induction from above on the weight of b. If b E HW (13),

then we take r = 0 and are done. Otherwise pick an a with Eab $ 0 and set b1 = Eab. We have then b1 E 13 and b = Fab1 by (C2) and (C3), the claim holds for b1 by induction, hence for b. b) Any x E MA/qMA can be written uniquely as x = >abb with all ab E Q, where we sum over all b E 13a. We have then Eax = abEab for each a E II. The nonzero Eab with b E B,, are by (C2) in B and are distinct (hence linearly

independent) since b = FaEab by (C3). _So Eax = 0 (for a given a) implies ab = 0 for all b with Eab $ 0. Therefore Eax = 0 for all a implies ab = 0 for all b

HW(13a). This yields the claim. c) This follows from the remark in 9.7. 9.9. LEMMA. Let A E A be a dominant weight. a) If (1(A), 13(A)) is a crystal base of L(A), then HW(13(A)) = {va}

(1)

{ x E L(A)I, I Eax E L(A) for all a E II } = L(A)µ.

(2)

and for all µ $ A

b) Let M be a finite dimensional U-module isomorphic to L(\)r for some integer r > 0. Let (M, 8) be a crystal base of M. If HW (13) = 8,,, then there are nonzero homomorphisms of U-modules cci : L(A) -> M such that M = ®i 1 cpiL(A) and M = ®i=1 cc L(\) and such that 13 is the disjoint union of the (piB(A).

Fa v_a with r > 0 and all ai E II. PROOF. a) Let b E_B; write b = Fa, Fa2 Fa v_a. We have b = Fat b', so b $ 0 implies b' $ 0, hence b' E 13(A); now (C3) yields Ea, b = b' $ 0. On the other hand, it is trivial

If r > 0, then set b' = Fat

that Eava = 0 for all a E II. This proves (1). The definition of an admissible lattice implies that we have the inclusion "D" in (2). Consider now any x E L(A), with Eax E L(A) for all a E II. We may

9.

CRYSTAL BASES I

195

suppose that x # 0. Since L(A)µ spans L(A),, over k, there is an integer r > 0 with qrx E L(A). Choose r minimal with this property. We claim that r = 0; suppose

that this is false, i.e., that r > 0. In that case Ea(grx) = gr'Eax E qr(A) for all a E U, hence Ea(q''x_) = 0 in L(A)/qr(A). By Lemma 9.8.b this implies that q'x is in the span of HW(B(A)µ). We have HW(B(A)µ) = 0 by a) since p # A. So we get q''x- = 0 and thus q'x E qr(A) contradicting the minimality of r. Therefore r = 0 and x E L(A) as desired. b) We have r = dim MA, hence also r = I,CBA 1. Set 13, = {b1, b2, ... , b,.} and choose v1, v2, ... , v,. E MA with vi = bi for all i. The Nakayama lemma implies that v1, v2, ... vs is a basis of the A-module MA, hence also of the vector space MA over k. Since A is the largest weight of M ^ L(A)'', we have Eavi = 0 for all a E H and all i. So there are isomorphisms of U-modules oi : L(A) --> Uvi with (pi(vA) = vi. We have then M = ®i I oiL(A). Each cci commutes with all Fa. Since M satisfies (A3), the definition of L(A) shows that cpir(A) C M, hence C C M where C = ®i cir(A). We want to prove

for all weights p of M that Mµ = Gµ and that Bµ is the disjoint union of all ipiB(A)µ. We use induction on p from above. We have for p = A M

M

MA = ® Avi = i=1

oiL(A),\; i=1

so our claim holds holds in that case. So suppose that p < A. Consider any b E B. Our assumption that HW (13) = BA implies that b

HW(B); so there is then a E II with Eab # 0, hence - by (C2) and (C3) - with b = FaEab. This yields

MFi/qMµ = E Fa(Mµ+a/qMN.+a) =

E FPa ('CA+a l q'CA+a) ac-II

QC-II

where the last equality holds by induction. This implies

Mµ = qM, + E F.'C,'+a = qM, + Lµ. ac-II

(Recall that L(A) is stable under each Fa; hence so is each oir(A) and thus also L.) Now the Nakayama lemma implies that M = Gµ. We still have to prove the claim on B.. We have seen above that there is for each b E B. an a E H with b = FaEab. We have EabE L3,,,+a, so there is by induction an i with Eab E piB(A)N+a. But then also b = FaEab E piB(A), . So B, is contained in the union of all ipi,Ci(A)µ. On the other hand, given x E B(A),, there

is a simple root a and y E B(A)µ+a with x = Fay. By induction each ipi(y) is in B, hence so is pi (x) = since B satisfies (C2) and since 7i(x) # 0. So we see that indeed B. is the (necessarily disjoint) union of all PiB(A),,. REMARK. In the situation of b) each ipiB(A)) is by Lemma 9.4.b a crystal base of oiL(A). If r > 0 (i.e., if M # 0), then Lemma 9.4.a implies now that (L(A), B(A)) is a crystal base of L(A).

196

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CRYSTAL BASES I

9.10. For any finite dimensional U-module M and any dominant weight A we denote by M[A] the isotypic component of M of type L(A). So M is the direct sum of all M[A], we have M[A]A = { x E MA I Eax = 0 for all a E TI}, and each M[A] is isomorphic to L(A)m(A) where m(A) = dim M[A]A. PROPOSITION. Let M be a finite dimensional U-module, let (M,13) be a crystal

base of M. Suppose that M = ®AM[A] where M[A] = M f1 M[A]. Then the cardinality of each HW (13A) is greater than or equal to the multiplicity of L(A) in a direct sum decomposition of M. If we have here equality for all dominant weights A, then 13 is the disjoint union of all 13[A] = 13l (M[A]/qM[A]), and each (M[A],.B[A]) is a crystal base of M [A] with HW(13[A]) = 13[A]A.

PROOF. For each dominant weight A set m(A) = dim M[A]A as at the beginning

of this subsection. Then M[A]A = M f1 M[A]A is a free A-module of rank m(A). The assumption that M = ®M [A] implies that M [A] A is a direct summand of MA: therefore M[A]A/gM[A]a is a subspace of MA/qMA of dimension m(A).

All x E M[A]A satisfy Eax = 0 for all a E II, hence the same is true for all x E M [A] A /qM [A] A. Therefore Lemma 9.8.b implies that M [A] A/qM [A] A is

contained in the subspace spanned by HW (13A). So we get m(A) < IHW (13A) I , our first claim. Suppose now that we have equality for all dominant A. Then M [A],\ /qM [A],\ is equal to the subspace spanned by HW (13A). This implies that HW (13A) C 13[A] for

all dominant A. For an arbitrary b E 13 there exist by Lemma 9.8.a aweight A an element b' E HW (13A) and simple roots a1, a2, ... , a,. with b = P., Fa2 P., V. Lemma 9.8.c implies that A is dominant. So we get b' E HW(13a) C 13[A] from above, hence also b E 13[A]. (Note that FaM[A] C M[A] implies F,,13[A] C 13[A].)

This shows that 13 is the union of all 13[A]. The disjointness is clear since the sum of the M[A] is direct. Lemma 9.4.b implies that each (M[A],13[A]) is a crystal base of M[A]. Since A is the largest weight of M[A], each element in 13[A],\ is in HW(13[A]). On the other hand, consider any b E HW(13[A]). If b has weight

it, then b E HW(13,,) C l3[µ], where the inclusion follows from the proof in the last paragraph. Since the 13[A'] are disjoint for distinct A', we get p = A, hence HW(13[A]) =13[A]A.

REMARK. The last claim in the proposition says that we can apply Lemma 9.9.b to each (.M[A],13[A]). So we see by the remark in 9.9 that (G(A),13(A)) is a crystal base of L(A) for each composition factor L(A) of M. Of course the problem will be to find M that satisfy the assumptions of the proposition.

9.11. Each finite dimensional U-module M is isomorphic to a direct sum ®i=1 L(Ai) of simple modules with all Ai dominant. Choose nonzero homomorphisms of U-modules cpi L(Ai) -+ M such that M = ®i=1 cpiL(Ai). If each (G(Ai),13(Ai)) is a crystal base of L(Ai) for all i, then we get (by Lemma 9.4) a crystal base (M, 13) of M setting M = ®i=1 cpiC(A) and 13 equal to the disjoint union of the pi13(Ai). The following theorem says that we get thus all crystal bases of M provided that we know that the (G(A),13(A)) are crystal bases. :

9.

CRYSTAL BASES 1

197

THEOREM. Suppose that (L(A),13(A)) is a crystal base of L(A) for all dominant A E A. Let M be a finite dimensional U-module and let (M, 13) be a crystal base of M. Then there are dominant weights A1, A2, ... , A and nonzero ho-

momorphisms of U-modules cpi : L(Ai) - M such that M = ®i cpiL(Ai) and M = ®i=1 cp.. L(Ai) and such that 13 is the disjoint union of the pi13(Ai). 1

PROOF. We use induction on the dimension of M; the case M = 0 is trivial. So suppose that M 54 0; let A E A be maximal for MA 54 0. Then A is necessarily

dominant; set L equal to the isotypic component of type L(A) of M. We have L ^- L(A)m where m = dim MA; in particular, we have L 54 0. Set N equal to the direct sum of all isotypic components of M not of type L(A). We have then M = N ® L and (since L 54 0) dim N < dim M. [Our strategy will be to apply induction to N and to prove the theorem directly for L.] Set L = M n L and N = Nn N. We have by construction LA = MA, hence LA = MA. Choose v1, V2,... ,Vm E L such that BA = {1J1,4J2,... ,Vm}. The Nakayama lemma implies now that v1, V2, ... , v,,, is a basis of LA over A. There are isomorphisms cp.i : L(A) Uvi (for 1 G i G m) with cc(va) = vi; we have then L = ®i ` I cpiL(A). We want to show that L = ®i ` I cpiL(A) and that M = N®G. More precisely, we claim for all weights p of M m

M1, = Nµ ®®cc L(A)w

(1)

i=1

One inclusion is easy: We have cpi(va) E M for all i, each cpi commutes with all F«,

and we have k,M C M for all a. So the definition of G(A) yields cpiG(A) C M. Since also NC M, we get the inclusion "D" in (1). In order to prove the other inclusion, we use induction on µ from above. For A our claim holds by the choice of the cpi. Suppose now that p 54 A. Let x E M,,; since M is the direct sum of N and the cpiL(A) there are unique z E N, and yi E L(A),, with x = z + Ei cpi(yi). We have for all a E II m

E.x = E.z +

oi(Eayi) E Mµ+. = Nµ+a ®®0i1C(A)11+a i=1

where the last equality holds by induction. This implies EQyi E L(A)µ+a for all a (and all i), hence yi E L(A),, for all i by 9.9(2). This implies (as seen above) that cpi (yi) E M for all i, hence z = x - >i cpi (yj) E M n N = N. So we have now established (1). Note that it trivially implies that L = M n L is the direct sum of the cpiL(A). So we get also M = L. We want to show that this decomposition corresponds to a decomposition of 13. Set 13' = 13n (N/qAl). We claim: 13 is the disjoint union of 13' and the cpi13(A).

(2)

Since tpi(va) = vi and since 13 satisfies (C2), the definition of 13(A) together with the injectivity of Bpi implies that pi13(A) C 13 for all i. The union of the Pi13(A) is disjoint, since the sum L/qC = ®i ipi(L(A)/qL(A)) is direct. This union is a basis of L/qC contained in 13, hence equal to 13 n (L/qL). Consider now an arbitrary b E B. We have to show that b is in (N/q J)U(L/qL). There is a weight p with b E 13k; we want to use induction on p from above. We have BA C L/qC; so we may assume that p A. If there is an a E II with E«b 0,

198

9.

CRYSTAL BASES I

then we have b' = Eab E 13 and b = Feb'. Since b' has weight p + a we can apply induction to b'. If b' E N/qN, then also b = FQb' E N/qN; similarly for L instead of N. So consider now the case where Eab = 0 for all a E II. Decompose b = x + y with x E N/gNand y E G/qL. We have then also Eay = 0 for all a E II. Since we assume that a # A, 9.9(2) implies y = 0, hence b = x E N/qN. So we have shown (2).

Now Lemma 9.4.b implies that (N,13') is a crystal base of N. The induction on the dimension yields a decomposition as in the theorem for N and (N,13'). Together with the Wi they give the desired decomposition for M and (M,13). REMARK. Note that we need the assumption that (L(A),13(A)) is a crystal base of L(A) only for those A such that L(A) is a composition factor of M. Denote as (in 9.10) the isotypic components of M by M[A] and set M[A] _ M f M[A]. The theorem implies (under its assumptions) that M = ® M [A]. Here the inclusion "D" is obvious in general; the other one follows from WiL(Ai) C M[Ai] for all i.

Let me state explicitly what the theorem says in the particular case where M = L(A) for some dominant weight A. The only endomorphisms of the U-module M are scalar multiplications, so we get: Suppose that (L(A),13(A)) is a crystal base of L(A). For any crystal base (L,13) of L(A), there is an a E k, a # 0 such that L = aL(A) and such that the isomorphism L(A)/qL(A) --- L/qL [induced by x H ax] maps 13(A) to B.

9.12. Recall the algebras Ua with a E II introduced before 9.1. Any U-module M is by restriction also a Ua-module for each a E II. It then makes sense to talk of admissible lattices in M (or of crystal bases of M) for M considered as a Umodule since Ua Uqa (512). [Well, rigorously speaking our definitions apply only in the case qa = q. But the extension to the general case should be obvious.] We claim now: An A-submodule M of a finite dimensional U-module M is an admissible lattice in M considered as a U-module, if and only if it is an admissible lattice in M considered as a U"-module for each a E H. Indeed the condition (Al)

is the same for U as for each U. Furthermore, (A3) holds for U if and only if it holds for each U. Finally, the condition (A2) for U means that each projection map Pa : M --+ Ma (with respect to the weight space decomposition) satisfies PA(M) C M. Similarly, (A2) for Ua means that each projection map Pa,,. of M onto the qa eigenspace of M satisfies Pa,r(M) C M. Now the Pa,,. for distinct a commute, and each Pa is the product of all Therefore (A2) for all Ua yields (A2) for U. On the other hand, any Pa,,. is the sum of all Pa with (A, a') = r. So (A2) for U implies (A2) for all U. Given an admissible lattice M of M and a basis 13 of M/qM we get also: The pair (M, 13) is a crystal base of the U-module M if and only if it is a crystal base of M as a U"-module for each a E II. Indeed, (C2) and (C3) hold for U if and only if they hold for each U. For (Cl) one argues similarly as for (A2) above. Lemma 9.6.a shows that we can apply Theorem 9.11 to Uq(5(2), hence to U. [Again, this is rigorously true only for qa = q; as before one can extend to the general case.] Recall that there is for all n E Z, n > 0 a simple module L(n) of dimension n with a highest weight vector v such that Kav = q'v. We set then L(n) = Ei AFaz)v = >i AF." v and 13(n) = { Fw 0 < i < n }. Then (L(n),13(n)) is a crystal base of the U"-module L(n), cf. Lemma 9.6.a. Now Theorem 9.11

CRYSTAL BASES I

9.

199

applied to U° says: Let (M,G) be a crystal base of a finite dimensional Uamodule. Then there are integers ni > 0 and nonzero homomorphisms of U'-modules : L(ni) ---> M such that M = ®i=1 coiL(ni) and M = ®i=1 cpiG(ni) and such that 13 is the disjoint union of the (pi13(ni). It is this special case of Theorem 9.11 that we are going to use several times

coi

in the following subsections. And that is the reason for proving it at this time. (Actually, the general proof can be simplified in this rank 1 case.)

9.13. I have said before 9.7 that we want to use tensor products in order to construct crystal bases. However, the tensor product of two admissible lattices is, in general, not again an admissible lattice. Consider for example 4D of type Al and a tensor product C(n) ®G(m) C L(n) L(m). Denote by X E L(n) and y E L(m) the highest weight vectors. So we have G(n)n, = Ax and G(m).,,,, = Ay. Then z = x ®y E C(n) ®G(m) is a vector of weight n + m with Ez = 0; this implies

Fz=Fz=Fx®K-ly+x®Fy=q-mFx®y+x®Fy. On the other hand, we have £(n),i_2 = AFx and G(m)m,_2 = AFy, hence (G(n) (9 G(m)).,,,.+n_2 = A(Fx ® y) + A(x 0 Fy).

This shows: If m > 0, then Fz

£(n) ®G(m), so £(n) ®£(m) does not satisfy

(A3).

So, if we want to use tensor products in connection with crystal bases, then we have to come up with a new idea. The solution to this problem turns out to be this: We simply change the operation of U on the tensor product, i.e., we work with a different comultiplication.

Consider the antiautomorphism wT = Tw of U, cf. 4.6. We have WT(Ea) = Fa,

wT(FF) = Ea,

wr(KN.) = Kµ

(1)

for all a E II and all µ E W. It is clear that (wr)2 = 1.

(2)

As described in 3.8(2),(3) we can twist the Hopf algebra structure on U by WT. In particular, we get thus a new comultiplication A' given by 0' = ((wT) ® (wT))

O A O (wT).

(3)

A simple calculation yields for all a E II and all µ E Z4P

Ea®K,-,1+1®Ea, A'(F'a) = F. ®1 + K. 0 F.,

A'(E'a)

(4)

0'(Kµ) = Kµ (9 Kµ.

(5) (6)

We get !(EaT)) _ (wT 0 wr)A(FaT)) for all r > 0, hence by 4.9(4) and a simple calculation r A'(E( aT))

_ i=0

gi(r-i)E(i) 0 E(ar-i)Kai

(7)

200

CRYSTAL BASES I

9.

and (similarly) r

®Fr).

0'(F,I,'')) =

(8)

i=0

The new antipode is equal to S' = (wr) o S-1 o (wr), hence given by

S'(E«) = -E«K«,

S'(F«) _ -K« 1F«,

S'(Kµ) = Kµ

(9)

for all a E II and all i E M. The new counit e' = e o (wr) is equal to the old .

counit s.

CONVENTION. For the remainder of Chapter 9 we regard the tensor product of two U-modules as a U-module via 0, i.e., we let any u E U act as 0'(u).

Note: We can apply the remark in 5A.9 since we have 0'(Kµ) = K® ® Kµ for all a. So all the formulas on the decomposition of tensor products in 5A.8/9 are still valid for the structure that we are going to use now. In particular, Lemma 5A.9 and the Clebsch-Gordan formula 5A.8(6) hold. This means of course that a tensor product of two finite dimensional U-modules

(say Ml and M2) with the new structure is isomorphic to M1 ® M2 with the old one. However, such an isomorphism will not preserve tensor products of admissible lattices, in general.

9.14. Suppose that L and M are finite dimensional vector spaces over k and that G C L and M C M are A-submodules satisfying (Al). Then G® M is an A-submodule of L ® M and it satisfies (Al). We can identify (L ® .M)/q(G (9 M) with (G/qL) ®q (.M/q.M). If 131 is a basis of G/qG over Q and if 132 is a basis of .M/q.M over Q, then we can (and shall) identify 5® ®132 with a subset (in fact: a basis) of (G ® .M)/q(G ® M). We shall show in 9.17 for all U-modules L and M: If (L, 131) is a crystal base of L and if (.M, 52) is a crystal base of M, then (L ®.M, 51 (9 132) is a crystal base of L ® M. We first prove this result for U«-modules. The next lemma is the first step of an inductive procedure that we use in that case. LEMMA. Let a E H. Let x E L(1)1 be a highest weight vector such that G(1)1 = Ax and 13(1)1 = {-x}. Let (M, 8) be a crystal base of a finite dimensional U«-module M. Then (G(1) ®.M,13(1) (9 13) is a crystal base of L(1) (9 M. We have for all b E 13 F«

P,

®b)

F«-X ®b,

if E«b=0;

y ® F«b,

otherwise;

(1)

b) - F«Y®F«b, E« (Y ®b) _ 9 b)

(2)

0,

if E«b = 0;

Y 0 E«b,

otherwise.

Y ®b,

ifE«b=0;

F«Y 0 E«b,

otherwise.

(3)

(4)

9.

CRYSTAL BASES I

201

PROOF. We can reduce to the case where (M, M,13) = (L(m), G(m),13(m)) for some integer m > 0. (Use 9.12 and Lemma 9.4!) So let us assume that we are in this situation. The case m = 0 is trivial and is left as an exercise to you. Suppose

that m>0. Choose a highest weight vector y E L(m),,,, such that G(m),,,, = Ay and 13(m)m, = {y}. We have G(1) = Ax + AFax and G(m) = E''o AFat)y. The Clebsch-Gordan formula 5A.8(6) implies that L(1) ®L(m) a- L(m + 1) ®L(m - 1).

The tensor product zo = x 0 y of the highest weight vectors is again a highest weight vector (of weight m + 1). So one of the two simple summands of our tensor product is Uazo a- L(m + 1). Note that Azo = A(x ® y) = (,C(1) ®G(m))m+1

(5)

We have (recall: for the structure introduced in 9.13)

Ea(x(9 F«y) =x®EaFFy= [m].x®y and

Ea(Fax®y) = EaFax®Ka ly = gamx®y. Therefore

z1 =x®Fay - qa [m]cFax®y is a highest weight vector of weight m-1 in L(1)OL(m), we have Uazl ^- L(m-1) and L(1) 0 L(m) = Uazo ® Uazl. Set G = G(1) ® G(m). We want to show that m+1

m-I

G = E AFQi)zo + >2 AF(i)zl.

(6)

[That will show that L is an admissible lattice- isomorphic to G(m+ 1) ®G(m-1) - in Uazo ® Uazl = L(1) 0 L(m)).] Using 9.13(8) one checks easily that

F(r)z0 a

gax®Far)y+Fax®Far-l)y, for0
(7)

This implies clearly that F«)zo E £m+1_2r for 0 < r < m + 1 and AFam+1)zo = A(Fax (9 FQm)y) = G-(m+1).

(8)

Similarly, 9.13(8) yields also FaT)z1

= [r+1]aq«x®Far+1)y+[rj

=[r+1]a4ax®FQr+1)y-qa [m-r]aFax®FQr)y for 0 r < in. (Check that qa [m - r]a = formula as

qm-r [m]a

- [r]a.) Let us rewrite this

Far-l)zl = qQ I [r]a x ®Fr)y - q" [m - r + 1]a Fax ® Far-l)y

for0
(9)

202

9.

CRYSTAL BASES I

We have for all integers a, b > 0 g a[b]a - qa a

qa

- qal

= qaa ga

ga-b+1 qc

2-1 - ° ga (ga

4a - qa

)

2 qa

-1

-1

1)/(qa - 1) is a unit in A, congruent to 1 modulo Aq. So we see: The factor If a > b - 1, then ga[b]a E Aq; if a = b - 1, then ga[b],, is a unit in A, congruent to 1 modulo Aq.

Another look at (9) shows now that Fa)z1 E G for 0 < i < m - 1. Together with the earlier results on the Fat) zo this yields one inclusion in (10). Frthermore, Far)y, Fax Far-1)y

for 0 < r < m the transition matrix from the A-basis x ®

0

Far)zo has entries in A and determinant equal to of G,,,,+ 1_2r to the set qa +r [m - r + 1] a + qa 1 [r] a . This determinant is congruent to 1 modulo Aq, hence a unit in the local ring. Therefore also Far-1)z1i Far)zo together with (5) and (8) implies (6).

is an A-basis of £m+1-2r-This

We know now that G is an admissible lattice in L(1) ® L(m). It is clear that the basis 8(1) ® 8(m) of G/qC satisfies (Cl); we still have to prove (C2) and (C3). Well, (7) and (9) yield (in G/qC)

x®y,

ifr=0;

Fax®Faif0
F,zo

0,

ifr>m+1;

x®F4+1y

if0
0,

ifr > m.

and

Faz1

It is now easy to get (1)-(4): Any b E8(m) has the form b = Fay with 0 < i < m. Then x (9 b is equal to zo if i = 0, to FF 1z1 if i > 0. We get then

Fa(x® b) _ and

E®®b)

ifi=0; Faz1=x®Fa+1y=x(9 Fab, ifi>0; Fazo=Fax®y=Fax®b,

- jl(0,

ifi<1; F, 2z1=x®F 1y=x®Eab, ifi>1.

This yields (1) and (3) since Eab_= 0 (resp. Eab = 0) for b = Fay if and only if i = 0 (resp. i < 1). Frthermore, Fax ®b = Fa+1zo yields

Fa(Fax®b) =F'a+2zo= Fax®Fa+1Fax®Fab and

Ea(Fax (9 b) = F2zo =

ifi=0; x®y=x®b, Fax®Fa 1y=Fax®Eab, ifi>0.

This yields (2) and (4). Now (1)-(4) clearly show that 8(1) ® 8(m) satisfy (C2). A more careful look (exercise!) at these formulas shows that also (C3) is satisfied. This proves the lemma.

CRYSTAL BASES I

9.

203

9.15. LEMMA. Let a E II, let (M, 13) be a crystal base of a finite dimensional U'-module M. Then (1(n) ®M,,Gi(n) 013) is a crystal base of L(n) ®M (for any integer n > 0).

PROOF. We want to use induction on n. The claim is trivial for n = 0; for n = 1 it is contained in Lemma 9.14. Suppose that the claim holds for some n > 1 and let us prove it for n + 1. By induction (1 (n) OM, 5(n)05) is a crystal base of L(n) OM, so Lemma 9.14 implies that (G(1)01(n)®M,1(1)®3(n)013) is a a crystal base of L(1)OL(n)®M. Lemma 9.14 implies also that (G(1) 0.(n),13(1) 013(n)) is a a crystal base of L(1) OL(n). We have a direct sum decomposition L(1)OL(n) = N1®N2 with N1 L(n + 1) and N2 ^ L(n - 1). By 9.12 there are isomorphisms Cpl : L(n + 1) -- NI and 02 : L(n -1) --4 N2 such that .C(1) ®G(n) = NI ®N2 where Nl = o1G(n+ 1) and N2 = 021C (n-1), and such that 13(1) 013(n) is the disjoint union of p113(n+ 1) and (p213(n - 1). (We could refer to the proof of 9.14 instead of 9.12.) Now G(1) ®£(n) ® M is the direct sum of N® ® M and N2 ® M, and 13(1) ® 13(n) ®55 is the disjoint union of p1B(n + 1) ®13 and p2B(n - 1) ®13. So Lemma 9.4.b implies that (Ni ®M,1p113(n + 1) ®1) is a crystal base of N1 0 M. We apply now Lemma 9.4.a to cpl ®idM and get that (G(n+ 1) O M,13(n+ 1) ®13) is a crystal base of L(n + 1) 0 M. So our induction works. (Of course, Lemma 9.4.b can be applied also to N2 0 M; but there we do not get anything new.) 9.16. We shall need the generalizations of 9.14(1)-(4) from L(1) to all L(n). In our formulas below one should interpret F. Ix as 0. LEMMA. Let a E II, let (M, 13) be a crystal base of a finite dimensional Ua module M. Let n > 0 be an integer and denote the unique element of 13(n)n by x. Then we have for all b E 13 and all j with 0 < j < n

Fa(F«x®b) _

Fa+lx ®b, Fix ®Fab,

if EE-ib = 0;

E.(F.x ®b) =

Fa-1x ®b, FIx 0 Eab,

if E,n-i+Ib = 0;

(1)

otherwise; (2)

otherwise.

PROOF. We use again induction on n. The case n = 0 is trivial; for n = 1 (1) and (2) are equivalent to 9.14(1)-(4). Let us suppose that (1) and (2) hold for some n > 1 and prove them for n + 1. Note that (2) (taken for our given n) implies

forallbE13andallr>0 0 b) = 0

0,

(3)

and

E. (Fx 0 b)

for all j with 0 < j < n. (4) Consider the admissible lattice N = G(1) ®£(n) 0 M in L(1) ® L(n) ® M. 0

Denote the unique element in 5(1) 1 by v and set z = v ®x E 13(1) ® ,Gi(n). We have

then in (G(1) 0 £(n))/q(G(1) 0 £(n)) by 9.14(7)

Fai z=

ifi=0; P", v®F lx, ifi>0. v

®x,

(5)

9.

204

CRYSTAL BASES I

We get therefore in N/qN [using first (5), then 9.14(1),(2) - with v playing the role of - and finally (3), (1), and (5) again] F«(F«z(9 b) _

_

F«(v(9 x®b), if i = 0 F«(F«v(9 Fa 1x0 b), if i > 0

F«v®x®b, ifi=0, Ea(x(9 b) =0 v®F«(x®b), if i=0,Ea(x®b)0 F«v®F«(F« Ix®b), if i > 0 FFav®F«x®b=Fa+1z®b, «v®x®b=Faz®b, if i = 0, Ea+1b = 0 if i = 0,

a1b 0 ifi>0,E«-i+1b=0

v®x®Fab=z®Fab,

Fav®Fa 1x®Fab=Faz®Fab, ifi > 0,

Ea-i+lb

0

r Fi+1z ®b, Fa(Faz ®b) = { a

if E"-i+1b =0 « Faz ® Feb, if E--i+1b 0.

(6)

Similarly [working with (2) instead of (1), with 9.14(3),(4) instead of 9.14(1), (2), and later on with (4) in the case i > 1],

Ea(Fa z ® b) =

if i = 0 E«(Fav®F 1x(D b), if i > 0 Ea(v®x(D b),

if

ifi = 0, Ea(x ®b) = 0

if i>0,Ea(Fa 1x(D b)=0 Fav ®Ea(Fa-1x ®b), if i>0,EQ(F« v®Fa 1x®b,

ifi = 0,

1 0,

0

v®x®Eab=z®Eab, if i = O, E«+2b 0 v®x®b=z®b, if i=1,Ea+'b=0 Fav®Fa 2x®b=F 1z®b, if i > 1, En-i+2 b = 0 1x®Eab=Paz Fav®F ®Eab, if i > 0, Ea-i+2b # 0 Fc',,-'z (9 b,

if

E(--i+2 (7)

Faz ® Eab, if Ea-i+2 b 0. Denote by v E G(1)1 resp. by E G(n)u a representative of v resp. of x. Then v ®i is a representative of z. The proof of 9.14 shows that Uz is a submodule

of L(1) ®L(n) isomorphic to L(n_+ 1); we can find an isomorphism that takes

G':_EiAFaztoG(n+1)and{Faz1O
apply the isomorphism G' --- G(n + 1), then (6) and (7) turn into equations in (G(n + 1) ®M)/q(G(n + 1) 0 M), in fact, they turn into (1) and (2) for n + 1

9.

CRYSTAL BASES I

205

instead of n (with z playing the role of x). So our induction works and the lemma follows.

9.17. Given a crystal base (1,13) of a finite dimensional U-module L we set

for all bEBand all aEII

e«(b)=max{rJEab540} and f«(b)=max{rJFab540}.

(1)

By 9.12 we have a direct sum decomposition (depending on a) L = ®i Li into simple U«-modules such that G is the direct sum of all 1i = G n Li and such that S is the union of all Bi = S n (1x/qJ ). Furthermore, if Li a- L(n) as a U«-module, then (1i, Si) 2 (1(n), S(n)). We have then in Bi exactly one element (say bo) of weight n, and Bi_= {Ejbo 0 < j < n}. In this case our definition yields e«(Fabo) = j and f«(FFbo) = n - j. So the sum e«(Fabo) + f«(Fab0) = n determines the isotypic U«-component to which Fab_o belongs, and the difference

f«(Fabo) - e«(Fabo) = n - 2j yields the weight of F,7,bo under the action of K. We have thus in general (for all a E II and A E A)

f. (b) - e«(b) = (A, av)

for all b E BA.

(2)

We get also the following alternative description of the e« (b) and f« (b) (first for the Fbo and then in general)

e«(b)=max{rIbEFS} and f«(b)=max{rJbEEaS}.

(3)

THEOREM. Let L and M be finite dimensional U-modules. If (1,131) is a crystal base of L and if (M, 52) is a crystal base of M, then (1® M, Si 0 B2) is a crystal base of L ®1b7. We have then for all b E Si, b' E 132 and all a E II F'«(b(9 b') =

E«(b(9 b') -

Fb®b', if .f«(b) > e«(b'); b ®F«b',

if f« (b) < e« (b');

E«b®b',

iff«(b) ? e«(b');

b ®E«b',

if f,, (b) < e«(b').

(4)

(5)

PROOF. In order to prove (4) and (5) for a given a we may consider L and M just as U«-modules. Using a decomposition of (1,131) as above (i.e., using 9.12) we see that it is enough to prove (4) and (5) with (1,131) replaced by (1(n),S(n)) for each n > 0. But then (4) and (5) reduce to 9.16(1),(2): We have (in the notations

from 9.16) f«(Fax) = n - j, and the condition "E,-3 b = 0" in 9.16(1) can be rewritten as e«(b) < n - j, i.e., as e«(b) < f«(FJx) leading to the condition in (4). Similarly, the condition in 9.16(2) yields the condition in (5). This proves (4) and (5).

In order to show that (1® M, Si (9 S2) is a crystal base of L ® M it suffices that it is a crystal base of L ® M considered as a U«-module for all a E II (by 9.12). Therefore this claim follows from Lemma 9.16, since we can again reduce to the case where (1,131) = (1(n),S(n)).

9.18. Recall the notation HW(S) from 9.8(1).

206

9.

CRYSTAL BASES I

LEMMA. Let A E A such that (G(A),13(A)) is a crystal base of L(A). Suppose that (M, 13) is a crystal base of a finite dimensional U-module M. Then

HW(13(A) ®13) = {va ® b I b E 13, Ea= 0 for all a E lI }.

(1)

PROOF. We have k,,, (b (9 b') E {E«b ® b', b ® k,,, b} for all b E 13(A) and all

b ® E«b', thenf«(b) < e«(b'), hence a«(b') > 0, i.e., kb' # 0 and thus also E«(b (9 b') = b ®E«b' 0 0. So, if b' E 13 by 9.17(5). Furthermore, if E«(b (9 b')

E« (b ®b') = 0 for all a E II, then first of all k,, b ®b' = 0 for all a E II, hence b = U,\ by the preceding remark. Furthermore, we have to have k. (u,\ (& b') # va ® E« b', hence [by 9.17(5)] f«(va) >_ a«(b'), i.e, E(,nb = 0 where m = f«(va). However, it is easy to see that f«(va) = (A, a") + 1.

9.19. PROPOSITION. Let A E A such that (G(A),13(A)) is a crystal base of L(A). Let Ao be either a minuscule dominant weight or the largest short root in an irreducible component of 4). For each dominant weight v the the multiplicity of L(v) as a composition factor of L(A) ® L(Ao) is equal to IHW((13(A) (9 PROOF. Set

13={bE13(A0)

IEaa,«")+Ib=0

for all aEH}.

(1)

Lemma 9.18 says that HW(13(A) (9 13(A0)) _ {va} (9 13.

(2)

Therefore we want to determine 13. Let us assume that 4) is indecomposable; the general case can be easily reduced

to that one. We are going to use the notations from 5A.1/2. The proof of Lemma 9.6.b shows that 13(A0) is consists of all xµ with u E WA0 and (for A = ao) of the hQ with ,3 E II,. First of all, we claim for all p E WA0 and all a E II that E(«a,« )+1xµ = 0

(A + p, a") > 0.

(3)

Well, if (µ, a") > 0, then 9.6(1),(2) resp. 9.6(5),(8) say that E«xµ = 0, hence E(«a'«")+1xµ = 0; on the other hand, we have then obviously (A + u, a") > 0. If (µ, a") = -1, then E«xµ # 0 and E«xµ = 0 by 9.6(2)(3) resp. 9.6(5). So we E«a'« )+1xµ have in this case = 0 if and only if (A, a") >_ 1, i.e., if and only if (A + µ, a") > 0. The last case to look at is (µ, a") -2. This happens only for

µ= -a in the case A0=ao. We have then E«xµ=Oifandonly ifi>3by 9.6(8). Therefore now E(«a,«")+1xµ = 0 if and only if (A, a") > 2, i.e., if and only if (A +p, a") > 0. This proves (3) in each case. Now (3) taken for all a says for all

ILEW)0

-

(4) A +,u is dominant. Let us now look at the hp with ,(3 E 11, in the case Ao = ao. We have E«hp = 0 for TA E 13

E,a«")+l,Q all a = 3 by 9.6(6),(11), hence also = 0. On the other hand, 9.6(8) implies that Ephp = 0 if and only if i > 2, hence Ep hp = 0 if and only if

(A,3")> 1. So we get for all3EII, hp E 13

(A (3V) > 0.

(5)

9.

CRYSTAL BASES 1

207

So we see for all dominant weights v: The cardinality of HW ((13(A) 0 is equal to 1 if v has the form A + p with p E W A0 i it is equal to the number of all /3 E 11, with (,\,,3v) > 0 if v = A and A0 is a root; in all other cases the cardinality

is 0. Now a comparison with Lemma 5A.9 shows that this number is is equal to the multiplicity of L(v) as a composition factor of L(A) ® L(A0). But that is the claim of the proposition.

The proposition says that (G(A) ® £(A0), B(A) 0 B(A0)) satisfies the condition "If we have here equality ... " in Proposition 9.10. In order to be able to apply that lemma, we also need the other condition: We have to know that G(A) ®£(A0) is the direct sum of its intersections with the isotypic components of L(A) ® L(A0). The proof of that condition requires a new concept.

9.20. There is an automorphism v of U such that

a(E.) =

o,(F,) = -q.Fc,

v(KK) = Kµ

(1)

for all a E lI and p E Z-11. (It is trivial to check that the relations (Rl)-(R6) are preserved.) If we compose with the automorphism w from 4.6 and the antiautomorphism S' from 9.13(9), then we get an antiautomorphism

QowoS'

Ti

(2)

of U. A simple computation shows that

TI(E.)=q.F,K,',

TI(F.)=q_IKBE.,

TI(KN,)=Kµ

(3)

for all a E I1 and p E Z-11. Furthermore, one checks easily (by evaluating at the generators E,,, F,,, and KN.) that 2

=

1

(4)

and

A'OT1

=

(T1 (9 T1)OD'.

(5)

DEFINITION. We call a bilinear form (,) on a U-module M mod-rl -invariant if and only if (um, m') = (m, Ti (u)m')

for all m, m' E M and u E U.

(6)

LEMMA. a) If ( , ) is a mod-r1-invariant bilinear form on a U-module M, then distinct weight spaces in M are orthogonal with respect to ( , ). b) A tensor product of mod-rl-invariant bilinear forms is mod-rl-invariant. c) Let A E A be a dominant weight; choose a highest weight vector vA E L(A)A, vA 0. There is a unique mod-rl-invariant bilinear form ( , ) on L(A) with (vA, vA) = 1. This bilinear form is non-degenerate and symmetric. d) If (, ) is a mod-r1-invariant bilinear form on a finite dimensional U-module M, then distinct isotypic components of M are orthogonal with respect to (, ).

208

9.

CRYSTAL BASES I

PROOF. a) Suppose that A,,u E A and v E MA, v' E M. with (v, v') # 0. We have then for all v E DD

q(',') (v, v') = (Kv, v') = (v, Kv') = q(' ' (v, v'). So (v, v') # 0 yields q(\-µ,") = 1, hence (A - u, v) = 0 for all v (since q is not a root of unity). This implies A -,u = 0; the claim follows. b) Suppose that L and M are U-modules equipped with mod--ri -invariant bilinear forms both denoted by ( , ). The tensor product of these forms is then the bilinear form ( , ) on L ®M with (x ®y, x' ®y') = (x, x') (y, y') for all x, x' E L and y, y' E M. It is clear that the mod-7-1-invariance of the two forms implies ((u1 (9 u2) v, v') = (v, (7-1 (v'1) (9 7-1 (U2)) v')

for all u1, U2 E U and v, v' E L ® M. Since U operates on a tensor product via 0', we get for all u E U (v.v,v') = (0'(v.)v,v') = (v, (T1 ®7-1)oA'(u)v') = (v, A'(7-1 (U)) v') = (v, T1 (u) v')

(for all v, v' as above) using (5) for the third equality. c) For any U-module M we can make the dual space M* into a U-module such

that (u f) (m) = f (Tl (u)m) for all m E M, f E M*, and u E U. This is not the usual structure as in 5.3, which is constructed via S instead of 7-1. However, for the duration of this proof we work with the new structure. If M is finite dimensional, then the natural map M -> (M*)* is an isomorphism of U-modules since Ti = 1.

If N is a U-submodule of M*, then N1 ={ m E M I f (m) = 0 for all f E N} is a U-submodule of M; if dim M < 0 and N # 0, M*, then N1 # 0, M. We see thus for dim 11v1 < 0 that M is simple if and only M* is simple. In particular, this shows that L(A)* is simple.

For each v E A we can identify the dual space

with { f E M*

I

f (Mµ) = 0 for all u # v }. Then M* is the direct sum of all One checks easily that Km f = q("=µ) f for all f E and u E M. That shows C (M*),,. We have observed already that M = this yields now that (M*) for all v E A. So M and M* have the same weights. In particular, the simple modules L(A) and L(A)* have the same highest weights; so they are isomorphic. If Eli : L(A) L(A)* is an isomorphism, then (v, v') = O(v) (v') for all v, v' E L(A) defines a non-degenerate bilinear form on L(A). We have (for all v, v' E L(A) and U E U)

(UV, V') = W(uv)(v') = (ui,b(v))(v') = W(v)(v,T1 (u)v') = (v,Tl(u)v');

so (,) is mod-7-1 -invariant. By a) distinct weight spaces are orthogonal for (, ). Therefore its restriction to each weight space is non-degenerate. Applied to L(A)A = kva this implies (va, va) # 0. We can thus replace 0 by a scalar multiple and may assume that (VA, VA) = 1.

If (

,

)' is another mod-7-1 -invariant form on L(A) then the map cp : L(A) -->

L(A)* with cp(v)(v') = (v, v')' is a homomorphism of U-modules (by the same calculation as above). If we assume in addition that (v,\, v,\)' = 1, then we have p(va) = VG(va) by the orthogonality of the weight spaces. Since L(A) is simple, this implies cp = iL, hence ( , )' = ( , ). This proves the uniqueness of ( , ).

9.

CRYSTAL BASES I

209

The form ( , ) is non-degenerate by construction. If we define a new bilinear , )" by (v, v')" = (v', v) then this form is also mod-rr1-invariant (since rr, _ 1) and satisfies (va, va)" = 1. So our uniqueness result implies that hence that ( , ) is symmetric. d) Let A, µ E A be dominant weights, and let N1, N2 C M be submodules with N1 ^ L(A) and N2 ^ L(µ). The restriction of ( , ) to N1 x N2 defines a map co : N1 -> N2 by co(v)(v') _ (v, v'). This is a homomorphism of U-modules, form (

if we take on NN the same structure as in the proof of c). If A µ, then any homomorphism of U-modules from N1 ^ L(A) to NN = L(,u)* ^ L(µ) is 0. In particular, our co is 0, i.e., the restriction of (, ) to N1 x N2 is 0. So non-isomorphic simple submodules of M are orthogonal; this implies the claim.

9.21 EXAMPLE 1. Let A E A be a minuscule dominant weight. Recall from 5A.1 (and 9.6) that L(A) has a basis (xµ I p E WA), that each xµ spans the one dimensional weight space L(A)µ and that 5A.1(2) describes explicitly the action of all Ea and Fa on this basis. Lemma 9.20.c says that there is a unique mod-rr1-invariant bilinear form (, ) on L(A) with (xa, xa) = 1. CLAIM 1. We have (xµ, xµ) = 1 for all p E WA. PROOF OF CLAIM 1. We use induction from above on µ. For y = A the claim

holds by definition. Any p E WA with µ A is not dominant. So we can find a simple root a with (µ, a") < 0, hence with (µ, a") _ -1 since A is minuscule. We get then xµ = Faxµ+a, hence (xµ, xµ) = (Faxµ+a, xµ) = (xµ+a,Ti (Fc)xµ) = (xµ+a, q, 1 KaEaxµ) 1

(xµ+a,ga Kaxµ+a) = (xµ+a,xµ+a) using 5A.1(2). We have by induction (xµ+(,,,xµ+a) = 1; this yields the claim.

Note that Claim 1 determines (,) completely, since distinct xµ are orthogonal by Lemma 9.20.a. EXAMPLE 2. Suppose that D is indecomposable and that ao is the largest short root in D. We have described L(ao) explicitly in 5A.2; we keep the notations from

that chapter. So L(ao) has a basis (xµ I µ E -ts; h,Q 10 E fl,). (Here ops is the set of short roots and I= opS f111.) We have (again by Lemma 9.20.c) a unique mod-rr1-invariant bilinear form (, ) on L(ao) with (xa,,, xa0) = 1. By Lemma 9.20.a all (xµ, xµ,) with µ' 0. So ( , ) is completely described by:

µ as well as all (xµ, hp) are equal to

CLAIM 2. We have (xµ, xµ) = 1 for all µ E Wao; we have for all Q, ry E lls 1 + q3, (h3, h7) =

if ,Q = ry;

q3,

if (,3,'Y) < 0;

0,

if ('3' -Y) = 0.

210

9.

CRYSTAL BASES 1

PROOF OF CLAIM 2. We first show (xµ, xµ) = 1 for all it E Wao with it > 0 using the same arguments as in the proof of Claim 1 [using 5A.2(2) instead of 5A.1(2)]. Next we observe for all /3,y E IIs (h3,h-r) = (Fox,3,h-,-) = (xo,T1(F,3)h-r) = (x,3,ga1K0E0h7)

We have E,3h., = 0 if (/3, ry) = 0, hence (he, hy) = 0 in that case; if (a, ry) < 0 (or, equivalently, (ry, /3v) _ -1), then E,3h.y = x,3; since K,3x,3 = gax,3 we get in this case

(ho,h7) = gQ'gQ(xo,xa) = q,3 Finally, if ry =,3, then E,3h,3 = [2],3x,3, hence (ho, ho) = q' [2]agQ(xa, xa) = (qa + g51)ga = 1 + qQ.

We get now also for all a E IIy (x-a, x-a) = [2]01(Foho, x-a) = [2]a' (h,3, Tl (Fa)x-a) = [2] p 1(h,3,

K,3Eax-Q) = [2]q p' (h,3, Kaha)

= [2]13 'q 1(ho,h,3) = 1.

This yields the claim for xN. with u E -H5. The remaining cases (xi. with it < 0, it V -H5) are dealt with inductively as in Claim 1. 9.22. LEMMA. Let M be an admissible lattice in a finite dimensional U-module

M, let ( , ) be a mod-r1-invariant bilinear form on M. If (,) takes values in A on M x M, then we have (Eam, m') _ (m, Fam')

(mod Aq)

for all in, m' E M and a E 11.

(1)

PROOF. We may assume by (A2) that m and m' are weight vectors. Suppose

that in E Mµ; then Earn has weight u + a; so we may assume by 9.20.a that m' E M,,+a Write m = J:j Faj)mj with mj E Mµ+ja and Eamj = 0 where we sum over the j > 0 with j > -(p,av) (as in 9.2(2)). By Lemma 9.7 all Faj)mj are in Mµ; so it is enough to prove the result for these. We can make a similar reduction for m'.

So let us assume now that m = F,(j)x and m' = F,,(,)y with x E M,,+ja and y E Mµ+(i+1)a and with Eax = 0 = Eay. Set n = (p + ja, av); then m and Eam are contained in a simple Ua-module isomorphic to L(n). Similarly, m' and P.,'M/ are contained in a simple Ua-module with highest weight n' = (µ + (i + 1)a, a"). We can apply Lemma 9.20.d to Ua instead of U. It implies: If n n', then both sides in (1) are 0 and the claim holds. So let us assume that n = n', i.e., that

9.

CRYSTAL BASES I

211

i=j-1. We have then (Eam, m') = (EaF x m) = (F( -1)x, m') = [n+ 1 [n+ 1 -j]al(Faj)x,T1(Ea)m ) _ [n + 1 - j]a 1(m, gaF'aKa 1F« -1)y) qaq-(1'+a,a)

[n+1-j]a

1

(m, FaFaj- y)

qa

-

[n + 1 -

3.],q«-2j+2

W]a

[n + 1 - j]agan-2j+1

(m, [j]aFaj)y) Fam').

The fraction in the last line is equal to

[j]aq. [n + 1 - j] a qfl-.i+1 -

(gn+1-i

(q. - q 3)q« _ qa (n+1-a))qn+1-7

-

g« -1 g2a(n+1-j)

hence a unit in A, congruent to 1 modulo Aq. This implies the claim. REMARK. One shows similarly (.Pam, m') __ (m, Eam') for all m, m' E M and

a E II. Actually, we can apply (1) to the form (, )' with (x, y)' = (y, x) for all

x,yEM. 9.23. Let M be an admissible lattice in a finite dimensional U-module M. Any bilinear form (,) on M that takes values in A on M x M induces a bilinear form

(, )o :.M/q.M x .M/q.M -> Q

(1)

such that for all x, y E M (x + qM, y + qM)o = (x, y) + qA E A/qA

Q.

(2)

If (,) is mod-r1-invariant, then (, )o satisfies (Eam, m') = (m, Fam')

for all m, m' E .M/q.M and a E iI.

(3)

DEFINITION. Let (M, 13) be a crystal base of a finite dimensional U-module M. A polarization of (M, 13) is a symmetric, mod-rrl-invariant bilinear form (, )

on M that takes values in A on M x M such that (b, b')o = bb,b'

for all b, b' E 13.

(4)

Note that (4) says that ( , )o is positive definite on the vector space .M/q.M over Q, and that 13 is an orthonormal basis with respect to this form. EXAMPLE. The calculations in 9.21 show that the bilinear forms there are polarizations of (G(A),13(A)) for A minuscule dominant resp. for A the largest short root in case 4 indecomposable. (Recall from the proof of 9.6 that the xN, resp. the xN. and h$ span G(A) and that 13(A) consists of their classes modulo q.)

212

9.

CRYSTAL BASES I

PROPOSITION. Let (1,13) be a crystal base of a finite dimensional U-module

M, let (, ) be a polarization of (M, 8). a) The form (, ) is non-degenerate on M. We have

M={xEMI(x,y)EAforally EM}

(5)

M={xEMI(x,x)EA}.

(6)

and

b) Let (G,13') be a crystal base of a finite dimensional U-module L. Then the tensor product of a polarization of (G,13') with a polarization of (M, 13) is a polarization of (G 0 M, 13' ®13). c) The lattice M is the direct sum of its intersections with the isotypic components of the U-module M.

PROOF. a) Pick for each b E 13 C M/qM a representative b of b in M. The Nakayama lemma implies that the b with b E 13 are a basis of the A-module M, hence of the vector space M over k. Consider x E M with (x, y) E A for all y E M. Write x = > bEC3 abb with all ab E k; let r > 0 be minima] for qrab E A for all b E 13, i.e., for qrx E M. If r > 0, then we have (qrx, b) _ (x, b) qr E Aq for all b E 13, hence 0 = (grx, b)

E (grab' b', b) = grab + Aq,

_ b'E5

i.e., grab E Aq for all b E B. This contradicts the minimality of r. We get therefore r = 0 and thus x E M. This proves (5).

The radical of ( , ) is a k-subspace of M contained in the right hand side of (5). Any k-subspace of M contained in the lattice M is 0. Therefore (,) is non-degenerate.

Consider now x E M with (x, x) E A. Let r > 0 be minimal for qrx E M. Suppose that r > 0. Then (qrx, qrx) = g2r(x x) E Aq, hence (qTx, gTx)o = 0. Since

(, )o is positive definite on .M/qM, this implies g''y = 0, i.e., qrx E qM. This contradicts the minimality of r. So we have r = 0 and x E M. That proves (6). b) By 9.20.b the tensor product of the two forms is again mod--r1-invariant. It is obviously symmetric and takes values in A on G ® M. [Recall that it is given by (x (9 y, x' ® y') = (x, x') (y, y') for all x, x' E L and y, y' E M.] Since (b ®c, b' ®c')o = (b, b')o (c, c')o for all b, b' E 13' and c, c' E 13, it is clear that also (4) holds.

c) For each dominant weight A set M[A] equal to the isotypic component of M of type L(A) and set M[A] = M fl M[A]. We have M = ®a M[A] and thus M D ®A M [A] ; we have to prove the other inclusion "c". Let x E M; there are x A E M[A] (almost all equal to 0) with x = >a xA. Let r > 0 be minimal for grxA E M for all A. If r > 0, then (qrx, qrx) = q2r(X, x) E Aq

(since x E M). We have (qrx qrx) _ a(grxa,grxa) by Lemma 9.20.d. So r > 0 implies 0 = EA (grxa, grxa)o. Since (, )o is positive definite on M/qM, this yields grxa = 0 for all A, i.e., grxa E qM for all A. This contradicts the minimality of r. We get therefore r = 0, hence x E Ea M[A] for all x E M.

9.

CRYSTAL BASES]

213

9.24. PROPOSITION. Let A E A be dominant such that (G(A), 5(A)) is a crys-

tal base of L(A) that admits a polarization. Let A0 be either a minuscule dominant weight or the largest short root in an irreducible component of 4). Let v be a dominant weight such that L(v) is a composition factor of L(A) ® L(A0). Then (G(v), B(v)) is a crystal base of L(v) and it admits a polarization. PROOF. Let us suppose that 4i is indecomposable; the reduction to that case is easy. Furthermore, we may assume that v # A. Set Al = L(A) ®L(Ao), M = G(A) ®G(Ao), and 13 = 13(A) ®13(x0). So (M, 5) is a crystal base of M by Theorem 9.17. The claims in 9.21 show that (G(Ao),13(A0)) admits a polarization. We can take its tensor product with the given polarization of (,C (A), B(A)) and get by 9.23.b a polarization of (M,13). Denote the isotypic components of M by M [µ] (as in 9.10), set .M [µ] = M fl M[µ] and l3[µ] = 13(1(M[µ]/qM[µ]) for all µ. Proposition 9.23.c implies that M is the direct sum of all M[p]. Therefore Proposition 9.10 combined with Proposition

9.19 imply that each (.M[µ],13[µ]) is a crystal base of M[p] with HW(13[µ]) = B[p],. The restriction of the polarization of (M, 5) to M[µ] is a polarization of (.M[µ],13[µ]). (Indeed, the restriction is again symmetric and mod--T, -invariant, it takes values in A on .M[µ] C M, and B[p] C 13 is still an orthonormal system.)

We can apply this in particular to p = v. Since v # A the explicit formulas for the decomposition of L(A) ® L(A0) in the proof of 9.19 show that L(v) occurs with multiplicity 1 in this tensor product. So there is an isomorphism cp : L(v) M[v]. Lemma 9.9.b shows that we can choose cp such that coC(v) = M[v] and 75(v) = 5[v]). Therefore Lemma 9.4 implies that (G(v), B(v)) is a crystal base of L(v). Furthermore, if we pull back the polarization of (.M[µ],13[µ]) under cp, then we get a polarization of (G(v),13(v)).

9.25. THEOREM. For all A E A dominant (G(A), X3(A)) is a crystal base of L(A); it admits a polarization.

PROOF. Set A0 equal to the set of all A0 E A that are either a minuscule dominant weight or the largest short root in an irreducible component of P. For all dominant A set n(A) equal to the smallest integer r > 1 for which there are A1, A2, ... , Ar E A0 such that L(A) is a composition factor of L(A1)®L(A2)®.. ®L(A,.). This is well defined by Proposition 5A.10. We want to prove the theorem by induction on n(A). If n(A) = 1, then A E A0,

and the claim holds by Lemma 9.6 and the Example in 9.23. Now suppose that n(A) > 1. Then there exist a dominant weight A' with n(A') = n(A) - 1 and a weight A0 E A0 such that L(A) is a composition factor of L(A') ® L(A0). We can apply induction to A'. Then the claim follows from Proposition 9.24. REMARK. Let (, ) be a polarization of (G(A),13(A)); set a = (vA, vA). Now vA E £(A) implies a E A, and vA E 5(A) implies a = 1. In particular a is a unit. If we now multiply ( , ) by a-1, the we get a symmetric, mod-rr1-invariant bilinear form that still takes values in A on £(A). Furthermore, the reduction modulo q of the form has not changed. Therefore the new form is again a polarization. This shows that we can find a polarization of (,C (A), B(A)) with (vA, VA) = 1.

9.26. LEMMA. Let M be a finite dimensional U-module and let (M, 5) be a crystal base of M. For each dominant weight A E A the multiplicity of L(A) as a composition factor of M is equal to I HW (BA) I.

214

CRYSTAL BASES I

9.

PROOF. By Theorem 9.25 the assumption of Theorem 9.11 is satisfied. We can then apply 9.9(1) to each summand of a decomposition of M as in 9.11; this yields the claim.

9.27. PROPOSITION. Let A, A', A" E A be dominant weights. The multiplicity

of L(A) as a composition factor of L(A') ® L(A") is equal to the number of b E 8(A")a_, with +lb = 0 for all a E H. PROOF. This follows from Lemma 9.26 and Lemma 9.17, since (L (A') ®L (A"), 8(A') (9 8(A")) is a crystal base of L(A') ® L(A") by Theorem 9.17.

9.28. If we want to apply Proposition 9.27, then we have to understand B(All) and the action of each E. on that set. This information can be encoded in a graph that is associated to the crystal base. A graph consists of two sets, the set of vertices and the set of edges. Furthermore, there is a map that associates to each edge two vertices, the endpoints of the edge. (It is allowed that the two endpoints are equal.) A directed graph is (vaguely) a graph where the edges have a direction. More precisely, a directed graph consists of two sets, the set of vertices and the set of edges together with a map that associates to each edge an ordered pair of two vertices; we think of the edge as an arrow pointing from the first of its two vertices to the second one. A coloured directed graph is a directed graph together with a map from the set of edges (arrows) to an additional set, that associates to each arrow a type (or colour). Here is an example of a coloured directed graph where the set of types is {1, 2}, and where we write the type of each arrow close to the arrow: 1

2

(1)

1J

1

2

The crystal graph of a crystal base (M, B) is a coloured directed graph with set of vertices B and with set of types equal to H. If b and b' are vertices, then there is an arrow of type a from b to b' if and only if b' = F,,b. For example, take 4) of type A2 and write lI = {al, a2}. Set A equal to the largest root, i.e., A = a1 + a2. Then the crystal graph of (L (A), B(A)) can be identified with the graph in (1). One should interpret type i as type ai; a comparison with the calculations in 9.6 yields easily the graph. (The vertex in the upper left hand corner corresponds to xa.) It is clear by the definition of B(\) that any b E B(A) is linked by a chain of arrows to U. So the crystal graph of each (L (A), B(A)) is connected. If (M, B) is a crystal base of a finite dimensional U-module M and if M is isomorphic to ®L(Ai), then the crystal graph of (M, B) is isomorphic to the disjoint and disconnected union of the crystal graphs of all (L(Ai), 8(A )); this follows from Theorem 9.11. This implies in particular that the crystal graph of (M, B) depends on M only, not on the choice of the crystal base. Therefore we usually call it the crystal graph of M. It is clear that we can read off the crystal graph of (M, B) all ea(b) and fcx(b) - cf. 9.17(1) - with b E B and a E H. Therefore Theorem 9.17 tells us how to compute the crystal graph of a tensor product of finite dimensional U-modules, if

9.

CRYSTAL BASES[

215

we know the crystal graphs of both factors. Now the proof of Theorem 9.25 yields an algorithm for computing the crystal graphs of each L(A): For A E Ao (as in 9.25) one uses the explicit description in 9.6; in general the crystal graph of some L(A) will be a connected component of the crystal graph of some L(A') ® L(A') that is known by induction.

For the classical cases (i.e., for D of type A, B, C, or D,') one can find a description of the crystal graphs of the L(A) in [Kashiwara & Nakashima]. Based on those results there are formulas for tensor product decompositions in [Nakashima].

CHAPTER 10

Crystal Bases II We keep the assumptions and notations from Chapter 9. In particular, we work with k and A as in 9.3. We fix for all dominant \ E A a generator vA E L(A)A and define (G(A),13(A)) as in 9.5.

Consider a weight v > 0 in the root lattice; write v = EaEn maa with all ma > 0. The map u '-k uvA from LT, to L(A)A_ is surjective for all dominant weights \; if A satisfies (A, a' ) > ma for all a E II, then this map is bijective. Let me call these A `big' for the moment. (This notion depends of course on v.) We can now look at the inverse image of G(A)A_ in U, under this map. It will turn out to be independent of \ as long as \ is big. Denote this common inverse image by L(oo)-,. This is then a finitely generated A-submodule of U j:, that generates G(A)a_ given by u --> uva U:-, over k. For all big \ the isomorphism G(oo) induces an isomorphism G(oo)_,/gG(oo)_

We can pull back the basis 13(A)A_ of the right hand side and get a basis of the left hand side over Q. It turns out that also this basis is independent of (big) A; denote it by 13(oo)_,,. For arbitrary (not necessarily big) A the map u H uva will take G(oo)_ onto G(A)A_, ; the induced map modulo q will take 13(00)_ to U {0}.

Actually, we shall construct G(oo) and_ B(oo) intrinsically inside U. We introduce operators on U- analogous to the Ea and Fa from 9.2; then we can imitate the construction from 9.5. Afterwards we relate G(oo) and B(oo) to G(A) and 13(A) and prove other properties.

10.1. Let a E H. Recall the maps ro. and ra from 6.15. At that point they were introduced as maps Uv -* U v+a for all weights v; we now regard them as linear maps U- -* U- by taking direct sums. Recall from 6.17(1) that we have for

allyEUEay - yEa

K.ra(y) - ra(y)Ka'

(1)

qa - qa

and recall from 8.26(6) that

{ y E U- I ra (y) = 0 } = Ta U [sawo] 217

(2)

218

10.

CRYSTAL BASES 11

LEMMA. a) We have r« o rQ = rQ o r« for all a, ,Q E II.

b) Let a E II. For each y E U- there are uniquely determined elements yn E U-, almost all equal to 0, with r' (yn) = 0 for all n, such that y = En>0 Fanyn

PROOF. The claim r,,(ra(y)) = r'Q(r«(y)) is certainly true for y = 1 and for each y = F7 with -y E II; indeed, in these cases both sides are equal to 0 by 6.15(3). Since these elements generate U- as a k-algebra and since our maps are linear, it

is enough to show: If the claim holds for some y E Uµ and for some y' E U µ,, then it holds for yy'. Well, the product formulas 6.15(4) yield r«(rp(yy)) = rc (yrp(y) + q(Q,µ')ra(y)y )

=

+ ra(y)ra(y) + q(a>N')q(-,µ-a)ra(y)ra(y') +q(a.µ')r.(rp(y))y, q(ck,µ)yr.(ra(y))

rp(r. (yy)) = rQ (q(ck ,µ) yr. (y) + ra (y)y)

= q(a,µ)yr,'(r.(y')) + q(ck,µ)q(a,µ -Ck)rQ(y)r.(y') + ra(y)ra(y) + q(a,µ')ra(r«(y))y Now an easy comparison shows that the claim for y and y' yields the claim for yy'. b) By (2) it is equivalent to show that Fa is not a zero divisor in U- and that

U = (D Fn, Ta U-[s,,wo].

(3)

n>O

This claim is very similar to that of Lemma 8.25. We could imitate the proof given there: We would have to work with the basis in 8.24(3), or rather with its image under w. Here is an alternative approach: We first prove the uniqueness part of our claim. So we have to show: Given yn E U-, almost all equal to 0, with r'' (yn) = 0 for all n, such that En >0 Fn, yn = 0, then yn = 0 for all n. For that we could imitate the proof of Lemma 8.fl.a; on the other hand we can apply Lemma 8.21.a: Apply w to the equation; we get En>0 E,,w(yn) = 0. We have r,, (w(yn)) = w(r'(yn)) = 0 for all n, cf. 6.15(6), hence w(yn) E T«U+[sawo] by Lemma 8.26, hence T, 1(w(yn)) E U+. Now we can quote Lemma 8.21.a.

From the uniqueness result it can be seen that for all v the subspace En>0Fa of U , has dimension +na On the other hand, U+ has dimension equal to En>p by En>pdimTU-[scwo]

Lemma 8.25. Since U = w(U+) and TaU-[sawn] = w(T,,U+[sawo]) a comparison of dimensions yields that U-,, is the sum of all Fn, T ,,U-[sawn] +na REMARK. If you want to compare with the approach in [Kashiwara 2], 3.4.2: Suppose that the index i there corresponds to the simple root a. Then the e?' there

is the present r, the e' there is the map y --> K,,r'(y)K« 1. In particular, the e' there and the r' here have the same kernel. Therefore the definitions in the next subsection coincide with Kashiwara's.

CRYSTAL BASES II

10.

219

10.2. Let a E H. We can replace each F, in Lemma 10.1.b by Fan), since q is not a root of unity. So there are for each y E U- uniquely determined elements yn E U-, almost all equal to 0, with r(',(yn) = 0 for all n, such that F,n)yn.

y =

(1)

n>0

If we have y E LL-, for some v, then yn E II v+na for all n. We define linear maps k(, and F(, from U- to itself by

Fay = >

and

Eay = E F'',n-1)yn

(2)

n>0

n>>O

for all y as in (1). We have obviously

FLU v c U v_

and

E.U_,, C it

v+

(3)

for all v, and

EaFoy = y

for all y E U-.

(4)

The proof of 9.2(5) shows now that

ForUi,=F,,Ui,

for all v.

(5)

Note that (4) implies

Each F,,, with a E II is injective on U-.

(6)

10.3. We can now imitate the construction in 9.5 with 1 E U0 replacing the highest weight vector vA. We set G(oo) equal to the A-submodule of U- spanned

with r>0anda1ia2,...,aTEII. ForallvEZ4,v>_0

by

1 with >i ai = v. Then

set G(oo) _ equal to the span of all those F., F.2

r(oo)_ C U7, and G(oo) = v r(oo)_,,, hence G(oo) = (@,C(00)-,

(1)

v>o

and

G(oo)_ = G(oo) fl t c,

for all v.

(2)

Each G(oo)_ is a finitely generated A-module that generates U!:, as a vector space over k (as in 9.5). We have by construction F r(oo) C C(oo) for all a E II. with r and Set 8(oo) C G(oo)/gr(oo) be the set of all the a, as before. Set 13(oo)_ equal to the set of those P., P.1, 1 + gr(oo)

with >t ai = v. Then 13(00)_ = 13(o0) n (G(oo)_,/gr(oo)_,)

(3)

and 8(oo) is the union of all 13(oo)_,,. This union is disjoint unless 0 occurs in more than one 13(oo)_,,. Note: In the definition of 8(A) in 9.5 we took only nonzero

220

10.

CRYSTAL BASES 11

F«, va. We did not make this restriction here. However, we shall prove in 10.11 that all elements in 13(oo) are nonzero; this will then also imply that the union of all B(oo)_ is disjoint. We have clearly G(oo)o = Al and 8(oo)o = {1}. More generally, we have for F«, F«2

allaEIIand allnEZ,n>0

G(oo)_n« = AFa 1 = AF(') and (4) In the next subsections we want to do two things: On the one hand, we want to show that (G(oo),13(oo)) is something like a crystal base. This means that (Al)(A3) and (C1)-(C3) hold, where we now take the f,, and F« defined above, and not those from 9.2. However, we have to modify the finite generation condition in (Al) and require now that each G(oo),, is finitely generated. In particular, we want to show that E«G(oo) C G(oo) for all a E II; the other conditions in (Al)-(A3) are satisfied by our remarks above. After that, we shall have to prove that B(00) is a basis and that (C2) and (C3) hold. The other goal will be to look at the maps cpa : U- -+ L(A) with oa(u) = uvA for all u E U- and all dominant A E A. We want to show that cpAG(oo) = G(A) and ipa8(oo) = 8(A) U {0} (where ?a is the induced map modulo q). The proof of the equality cpaC(oo) = G(A) would be easy, if the VA would commute with the F«. However, that is not true in general. For example, consider two simple roots a, 3 E II that span a subsystem of type A2. Set F«+,3 = T«(F3) =

F3F« - q«F«F8, cf. 8.16(4). We have T, 1F«+,3 E U-, hence r'(F«+p) = 0 and thus P. F«+3 = F«F«+a So we have on one side cpa(F«F«+8) = F«F«+ava. Let us look on the other side at F«cpa(F«+,3) = F«(F«+8va). We have to write F«+,3va in the form of 9.2(1). A simple calculation yields m+]

F'«+,ava = (FaF« -

[

[ + 1]« F«F3)v - [m + :l]«

F«F,3va

where m = (A, a'). One checks that

E. (FaF« -

[m]«

[m + 1]«

F«F,3)vA = 0

and

E F,QvA = 0.

So the definition in 9.2 yields m+1

F«F8)v,\

F«(F«+ava) = F. (FFF« - [M + 1]« hence (with a little additional calculation) F,,(F«+ava) = F«F'«+ava +

qc,

F«2)F,3 v,\,

[m + 1]«

qa (q« - q« + 1) Fa2)F,3va

_ (F«F'«+3)va +

[m + 1] a

q' (q2 - q« + 1) [m + 1]«

(5) Fa2) F,3va.

So we see that F«(F«+ava) # in general. This example shows what our problem is: If we have an element y E U- with r' (y) = 0, then in general we do not get E«yv , = 0 (which would then imply F«(yva) = F«yva = (F«y)va). The next lemma looks more closely at the failure of this equality.

CRYSTAL BASES II

10.

221

10.4. LEMMA. Let a E II be a simple root. Let M be a finite dimensional U-module and let v e MA be a weight vector for some A E A with Eav = 0. Let y E U for some v E ZIP, v > 0 be an element with r'«(y) = 0. Then we have for all integers n > 0 n((A-v,«v)+n+l)

E« yv = q«

1

(q« - q« )n

rn(y) v.

(1)

PROOF. We use induction on n. The claim is trivial for n = 0. So suppose that we have the claim for some n and prove it for n + 1. Let us abbreviate the exponent of q« in (1) by m(n). We get by induction Ea+lyv = (q« - q« 1)-nq

(n)Er(y)v.

(2)

We have ra(rn(y)) = rn(r'«(y)) = 0 by Lemma 10.1.a, hence

E«r«(y) - ra(y)E« = (q« - q«1)-1Kra+l(y) by 10.1(1). If we plug this into (2), we get using E«v = 0

En+lyv = (q. - qa I)-(n+1)q« (n)K«rn+l (y )v.

(3)

Now rn+1(y)v has weight A - v + (n + 1) a, so we get

&r n+1 (y)v =

q(A-v.«")+2(n+1)rn+1

(y)v.

'Ale plug this into (3) and see that the claim follows from

m(n+1) = m(n) + (A - v, a') + 2(n + 1) which one checks easily.

10.5. LetaEll. We have for allaEZ a

9« - q«1

= q« (a-1)

qa

- I E qa (a-1) (1 + Aq), q« - 1

(1)

hence for all n e Z, n > 0

[n]« E q(')2(1 + Aq),

(2)

and or all a,nEZwith n>[a] 0 n

E q_ (a-n)n(1 + Aq).

(3)

a

The factors in I + Aq are units in the local ring A. Using these formulas it is easy to show that the coefficient of Fat)Fpva in 10.3(5) belongs to Aq. One can now deduce in that situation that Fa(Fa+3va) (FaF, )va (mod qG(A)). We want such a formula in general, for all u E G(oo) instead of just u = Fa+p. The next lemma will allow us to prove it in certain cases:

222

10.

CRYSTAL BASES II

LEMMA. Let a E H. Let A E A be a dominant weight, let x E L(A). Suppose that m > 1 is an integer such that

gai(i+m)KaF'ai)x E qC(A)

(4)

for all i with 1 < i < m. Then we have modulo qr(A) FaFa(m-1)x

= Fam)x

EaFa(m-1)x = F(m-2)x.

and

(5)

PROOF. We may assume that x is a weight vector, say x E L(A),, for some ji E A; set a = (ja, a'). Write x as in 9.2 (1) F(3)x3

X=

(6)

j>_0,7>-a

with all xj E L(A)N,+ja and Eaxj = 0. We have by 8.3(3) E«i) F«j)xj a

for all i and j with i < j; we have EaZ)Faj)xj = 0 for i > j. Since Fay-Z)xj has weight p + is we get for all i > 0

[a++il

Eai)x _

j >i j >- a

KaF.(j-i)xj

i

[a++i] i

j>i.j>-a

J

qi

a+2i)F(j-

a

So (4) says for I < i < m

E

[a++i] qa+i-m)Fj-i)xj

E qr(A),

a

j>_%>7>-a

hence by Lemma 9.7.b

[a+i+il qa+i-m)xj i

E qr(A)

Ja

for all j (j > i, j > -a) and all i (1 < i < m). The Gaussian binomial coefficient in this formula is by (3) equal to gaa(a+j) times a unit in A; so we get q«Z-j-"`)xj E q,C(A) for all i and j as above. If we take here i = m, then we get gamjxj E qG(A) for all j > m; if we take i = j, then we get q;jmxj E qr(A) for all j with 1 < j < m. So we have shown that gam3 xj E qr(A)

for all j > 0.

(7)

We get from (6)

m - i +j

Fa(m-i)x _ j>O, j>-a

F(am-i+j)x a

j

CRYSTAL BASES II

10.

223

for 0 < i < m. The Gaussian binomial coefficient here is by (3) equal to q;2(- i) _ qa-mJ q, times a unit in A. So (7) implies that FFm_i)x - F.1m-i)xo

for 0 < i < m.

(mod qG(\))

(8)

Since f,,, and Fa preserve qC(A) we get now (always modulo qG(A)) ExiFF(m-i)x = E.F(m-i)x0 = F.(??,-22)xo

F(,n,-2:)x

and

m_i)x =

F(m)xo -

for 0 < i < m. For i = 1 we get the claim in the lemma. REMARK. The proof yields actually a stronger result: We have for all i with 0 < i < m (always modulo qG(\))

F.'F(--i)x

- F(-)x

E.F.((m-i)x = F(--2i)x.

and

(9)

Here (as well as in the lemma) one has to interpret an F.(--22)x with m - 2i < 0 as 0.

Note: If we have (µ, av) < 0, then j = 0 does not occur in (6). So then (8) says Fpm-2)x E qC(A) for 0 < i < m, in particular x E qG(A).

10.6. LEMMA. Let u E U-. a) There is an integer n > 0 such that q"uva E G(A) for all dominant weights AE A.

b) Let a E II. There exists an integer r > 0 such that (F,,,u) va - F. (uva)

and

(E,,,u) va - E,,, (uva)

(1)

modulo qC(.\) for all dominant weights A E A with (A, av) > r. PROOF. We may assume that u E Q7, for some v E ZID, v > 0.

a) We use induction on v. For v = 0 there is an n > 0 with q"u E A 1 C 1 = U0. Then we get also q"uva E Ava = G(\)a for all dominant A. Suppose now that v > 0. Since LC, = EOEII FQU +Q, it suffices to look at the case where u = Fpy with some ,Q E II and some y E U U+p. By induction there is an m > 0 with gmyva E G(A) for all dominant A. We write then gmyva as in k

9.2(1) gmyv,\

r

_

(2)

F(2)xi,a i=0

with each xi,a E

and E$xi,a = 0. Furthermore, we may assume

that xi,a = 0 if i < -(\ - v

Now Lemma 9.7.a implies that xi,a E G(A)

for all i. Note that the upper bound r for i in (2) can be chosen independently of A: We can take r as the largest integer with (r + 1),3 < v. So we get for all A r

q gm'uv,\ = g5q'F,3yva = gQFQ

r

qa[i + 1],3F(i+1)xi"\.

F(i)xi,a = i=0

i=0

224

CRYSTAL BASES II

10.

Fat+I)xi,,\ = Fp+1xi Now xi,a E L(A) implies also a E L(\). On the other hand, 10.5(1) implies that qp[i + 11p E A for 0 < i < r. So we get gpq'uva E L(.\) for all A. We choose n such that qn = gpgl; the claim follows.

b) We can write u = n>o Fan)un with all un E U- and ra(un) = 0. It is enough to prove the result for each summand._ So let us assume that u = Fan)y with y E U +na and r' (y) = 0. We have now Fau = Fan+1)y and Eau = Fan-1)y. So the claim in (1) says Fan+1)yva

and

= Fa (Fan)yvA)

Fan-1)yva = Ea (F n)yva)

(3)

modulo qL(A) for suitable A. Lemma 10.4 implies for all i and all dominant A Ea) yva = x

qa

_ 1)t [zlix(ga - qa

ria( y)va.

Since Eat)yva has weight A - v + (i + 1)a this implies i(2(a-v.av)+2i+4-n)

ga i(i+n+1)K.' E(ai)yvA = ga

_ [Z]a(ga - qa 1)t f

ria(yv

(4)

Denote the fraction in this formula by ci. By a) we can find an integer s > 0 such that q,',r,',(y)va E L(A) for all dominant A and all i with 0 < i < n + 1. If we have ci,gas E Aq for all i with 1 < i < n+l, then (4) yields ggt(i+n+l)K,,Eat)yva E qL(.\) for 1 < i < n + 1. This means that the condition 10.5(4) is satisfied for m = n + 1 and x = yva. Then Lemma 10.5 yields the congruences in 10.5(5) that are exactly the congruences in (3), hence those in (1). So we have to find a condition on A that assures that ciga s E Aq for all i with 1 < i < n + 1. By 10.5(2) ci is equal to a unit in A times gti(2(a-v. a')+2i+4-n)qa t-1)/2q

= gi (a.a")-(v.a")-n)Qa((A.av)-(v,av)+2i+5)ga

So if A satisfies (A, a') > (v, a') + n and (A, a') > (v, v) + s, then indeed ciq_ s E Aq for 1 < i < n + 1. So we have found a bound r as in the claim of the lemma.

10.7. We want to have the congruence (Fau) va - Fa (uva) from 10.6(1) for all u E G(oo) without any restriction on A. We shall make the transition from the special A as in the lemma to all A using tensor products. Also in this chapter all tensor products of U-modules will be regarded as Umodules via the comultiplication A' from 9.13. So the convention made in at the end of 9.13 remains in force in Chapter 10. Let A, p be dominant weights. From Proposition 9.17 we have that (L(A) ® L(p),C3(A) (9 13(p)) is a crystal base of L(\) ® L(µ). So we get from Theorem 9.11 dominant weights Al i )'2, ... , A, and nonzero homomorphisms Vi : L(,\i) --

10.

CRYSTAL BASES 11

225

L(A)®L(p) such that L(A)®L(p) = ®z-0 viL(A) and G(A)®G(µ) = ®'_o V (A,) and such that B(A) ® 5(µ) is the disjoint union of the cpi13(A ). The highest weight of L(A) ® L(µ) is A + µ; the corresponding weight space (L(A) ® L(µ))A+µ = k(vA ®vµ) has dimension 1. This implies that L(A+µ) occurs with multiplicity 1 as a composition factor in L(A) ® L(µ). So there is exactly one index i with A, = A + µ; let us assume that Al = A + µ. We have A(vA ® vµ) = ('C(A) ®'C(p))A+Fi = c,1 G(A + ll)A+µ = AV, (vA+µ).

So there is a unit a E A with cp1(vA+µ) = a(vA ® vµ). We have then ?1(va+µ) _ a(va ®vµ). Since p1 maps B(A + µ) into B(A) ® 5(µ) and since va ® vµ is the only element of weight A + µ in B(A) ® l3(µ), we have to have ip1(va+µ) = va ®vµ,

hence a = 1. If we replace VI by a-lcpl, then we do not change VIC(A +,u) and ipl. Therefore we may assume a = 1. We now denote VI by VA,,,,. So

Va,µ : L(A + µ) -> L(A) ® L(µ)

(1)

is the unique homomorphism of U-modules with VA.A(va+µ) = VA ® V.

(2)

vA,, C(A+ µ) c G(A) ®£(p)

(3)

cpa,µ8(A + µ) C B(A) ® 13(µ)

(4)

We have

and

since cpl has these properties.

10.8. Let A, µ be dominant weights. We shall need another map involving the tensor product L(A) ® L(µ). Since L(µ) is the direct sum of its weight spaces and since L(µ)µ = kvµ, there is a unique linear map

0A, i : L(A) ®L(µ) -; L(A)

(1)

with

o,A,µ(v®v,,)=v

for allvEL(A)

(2)

and

aA,µ(L(A) (9 L(µ)µ,) = 0

for all µ' < µ.

(3)

LEMMA. Let A, µ E A be dominant weights. The map 0A,µ is a homomorphism of U- -modules. We have ora,µ(G(A) ®G(µ)) = G(A)

(4)

and

0a,µ(F'ax) = F, oA,µ(x) for all x EC(A) ®G(µ).

(mod qr(A))

(5)

226

CRYSTAL BASES II

10.

PROOF. We want to prove first that a,\,, is a homomorphism of U--modules. We have to show that Q,\,µ(Fax) = FaCA,µ(x) for all x E L(A) ® L(µ) and a E II.

Well, if x = v ® vµ with v E L(A), then Fax = (Fav) ® vµ + (Kav) ® (Favµ). (Recall that we work with the comultiplication A' as in 9.13(4)-(6).) Since Favµ E we get Q,\,µ(Fax) = Fav = FaCA,µ(x). If x = v ® v' with v E L(A) and

v' E L(µ)µ' for some µ' < µ, then QA,µ(x) = 0, hence FaCA,µ(x) = 0. On the other hand, Fax = (Fav) ® v' + (Kav) ® (Fav') and Fav' E L(µ)µ'_a, hence also Qa,µ(Fax) = 0.

Since G(µ) is the direct sum of its weight spaces, and since G(µ)µ = Avµ, hence C(A) ®C(µ),, = C(A) 0 vµ, the claim (4) is obvious. Let us abbreviate G = G(A) ®G(µ). The claim in (5) says that the induced map modulo q :

G/qC -> C(A)1gC(A)

commutes with each Fa. It is enough to check this on the basis B (A) ® B(µ) of G/q,C. Let b E 5(A) and b' E B(µ). We have Fa(b ® b') E {(Fab) ® b', b ® (Fab')} by 9.17(4). If b' E B(µ)µ- for some µ' < µ, then also Fab' has a weight less than µ, hence we get in both cases 7a,µ(Fa(b ®b')) = 0 = FaQa,µ(b ®b').

(6)

If b' has weight µ, then b' = vµ. We have ea(vµ) = 0, hence by 9.17(4)

(Fab) ®vµ, ifFab # 0;

Fa(b®vµ)

b ® (Favµ), if Fab = 0.

So oa,µ(Fa(b®vµ)) is in the second case equal to 0, in the first case equal to Fab. But in the second case Fab = 0, so we get in both cases (Fa(b ®vµ)) = Fab = Fa 7A,µ(b ®vµ).

(7)

Now (5) follows from (6) and (7).

REMARK. There is no general analogue to (5) involving the E. However, one

has still a partial result. We have Ea(b ®vµ) _ (Eab) ®vµ for all b E 5(A) by 9.17(5), hence

Qa>µ(Ea(b 0 vµ)) = Eab = Ea7A,µ(b ®vµ) This implies for all v E G(A)

Qa µ(Ea(v ®vµ)) - Eav (mod qC(A)).

(8)

10.9. PROPOSITION. We have for all dominant weights A E A Vai.(oo) C 'C (A)

(1)

and for all uE C(oo) and all a EH (Fau)va

Fa(uva)

(mod qC(A)).

(2)

10.

CRYSTAL BASES II

227

PROOF. We want to show for all v E Z4'i, v _> 0 that (2) holds for all u E

G(oo)- and that c,AG(oo)

C G(a)y

(3)

This will imply the claim.

We want to use induction on v. For v = 0 we have G(oo)o = A 1; so (3)

is obvious. In (2) it is enough to look at the case u = 1. Then Fau = Fa and F0(uva) = Fava = Fava. So (2) holds for v = 0.

Suppose v > 0. We have G(oo)- = 1:,3EI1 For any )3 E II we have by induction V,\(y) = yva E and any y E and (Fpy)va F$(yva) (mod qG(A)). The first fact implies that F$(yva) E G(A)a-,,, the second one yields then that also cpa(F$y) = (F$y)va E C(,\),\-,,. Therefore (3) holds.

Fix now u E G(oo)- and a E H. Take an integer r as in Lemma 10.6.b. Choose a dominant weight p such that (A + p, av) > r. Then Lemma 10.6.b implies that

(Fau)va+µ - F0(uva+µ)

(mod qC(A+ p)).

Furthermore, (3) applied to A +,u instead of A yields uva+µ E G(A +,u). The homomorphism cpa,N, commutes with Fa, cf. 9.2(b), so we get using 10.7(2),(3)

(Fau) (va (9v,,) - F0(u(va ®vu,))

(mod qG)

and

u(va 0 vj,,) E G

where G = G(A) ®G(p). Now apply v,\,,; we get using 10.8(4),(5) that modulo qG(A)

F.

(u(va 0 vu)) = or,,,. (F.(u(va ® vµ))) = 6a,µ ((Fau) (VA 0 vµ)).

(4)

Now v,\,,, commutes with U-, so we get 6a,µ ((Fau) (VA ®vµ)) = (Fau)aa,la(va ®vµ) = (Fau)na

and similarly va,,(u(va®vµ)) = uva. If we plug this into (4), then we get as desired

(FFu)va - FFuva

(mod qG(A)).

10.10. THEOREM. a) We have for all dominant A E A c,AG(oo) = G(A)

and

ipA8(oo) = B(A) U {0}.

b) We have EaG(oo) C G(oo) for all a E H.

PROOF. a) We know already that cpAG(oo) C G(A). So there is an induced map PA : G(oo)/gG(oo) -> G(A)/qG(A). Furthermore, SPA commutes with each Fa by 10.9(2). So we have

a(FaiFa2 ...Farm) =

Fa,Fa2...Farv

228

10.

CRYSTAL BASES II

for all finite sequences al, a2.... , a,,,, of simple roots. This shows that paB(oo) _ F« va = 0 for large m.) 13(A) U {0}. (Note that 0 has to occur since 'P-1 'P-2 Since 13(A) is a basis of G(A)/qL(A), we see that 5A is surjective, hence that G(A) = cpaG(oo) +qL(A). Now the Nakayama lemma implies that G(A) = cpaG(oo). E b) It suffices to show that av v = EQEn mp13. Consider any u E L(oo)-,. Let r be an

integer as in Lemma 10.6.b. Choose a dominant weight such that (A, /(3v) > mo for

all 3 E II and such that also (A, cc') > r. We have uva E G(A) by Lemma 10.9, hence Ea(uva) E G(A). The condition (A, a') > r implies by Lemma 10.6.b that

E,,(uva) - (Eau)va (mod qG(A)). We get therefore that also (Eau)va E G(A) and so that Eau E U +a n co 1G(A).

(2)

The first condition on A ensures that cpa induces an isomorphism of vector spaces L(A)A

morphism claimed.

+a By part a) of this theorem it induces now also an iso-k G(A)a-v+a So (2) implies that Eau E

as

REMARK. We can now extend Lemma 9.7.a to G(oo). Take any y E U- and

write y = >o F,(," y as in 10.2(1), i.e., with r'(yn) = 0 for all n. We claim: If y E G(oo), then yn E G(oo) for all n.

(3)

Indeed, we have yo = y - FaEay and Eay = En>o F.(i-1)yn The second part of the theorem implies that both terms are in G(oo); we can now apply induction to Eay to get the claim. 10.11. By construction each Fa maps G(oo) to itself, by Theorem 10.10.b so does each Ea. They induce therefore maps on G(oo)/gL(oo). We denote the induced maps again by Fa and Ea. They inherit the property Ea o Fa = id from 10.2(4).

PROPOSITION. a) We have 0 V B(oo).

b) For each v E VD, v > 0 the set B(oo)_ is a basis of PROOF. a) Each Fa is injective on G(oo)/gL(oo) since E_a o Fa = id. Therefore Fa, 1 is not Fa is injective. In particular, each Fa, F_a2 also each Fa, Fa2 F,,,,, 1. So we have equal to 0. By definition 13(oo) consists of all these Fa, Fa2 indeed 0 V B(oo). b) Given v, we can find a dominant weight A such that cpa induces an isomorphism G(oo)_ - 2 4 > G(A)a-,,, hence also an isomorphism

G(oo)-,lgG(oo)

-- G(A)a- /gL(A)A

(1)

Since 0 V 13(oo), we have also 0 V ipaB(oo)_, . So Theorem 10.10.a implies ip,B(oo)_ = CB(A)a-,,. So B(oo)_ is mapped under the isomorphism in (1) to a basis, hence it is a basis itself.

10.

CRYSTAL BASES II

229

10.12. PROPOSITION. Let a E H. Then Fa13(oo) C 13(oo)

and

Ea13(oo) C 5(oo) U {0}.

(1)

We have EaFa6 = b for all b E 13(oo) and

FaEab = b

for all b E 13(oo) with Eab ¢ 0.

(2)

_PROOF. The definition of 13(oo) implies that F 13(oo) C 13(oo). The equality

ECF b = b follows from the more general identity Ea o Fa = id stated as the _ beginning of 10.11. Consider now any b E 13(oo) with Eab

_ 0. We want to show that Eab E 13(oo)

and FaEa6 = b: this will prove the proposition. There is a weight v E Z4, v > 0 with b E 5(oc)_,,. Choose a representative u E C(oo)_ of b. Pick a dominant weight A as in the proof of 10.10.b. So we can apply Lemma 10.6.b and get (Eau)va = Ea(uva)

(mod qr(A)),

hence

ipa(Ea6) = Ea;pa(b) E Ea13(A) C 13(A) U {0}.

(3)

(We have used here 10.10.a and the fact that B(A) satisfies (C2).) Our choice of A implies that ;5a induces an isomorphism .C(oo)-v+cjq,C(oo)-,+,,

that maps the basis first that pa(Eab) E

,C(A)a-v+a/gC(A)a-v+a

to the basis hence that Eab E

So Eab # 0 implies by (3) Furthermore, since

B(A) satisfies (C3), we get also 7a(6) =

Since ;5a is bijective also on C(oo)_,/gC(oo)_,,, we see that b = FaEab. So we get the proposition.

REMARK. We said in 10.3 that (G(oo), G(oo)) is a crystal base in a suitably generalized sense. The proposition above is the last step in proving this claim. In fact, they show that the analogues of (C2) and (C3) are satisfied. The `basis condition' from the definition in Proposition 10.11; the result that 0 V 5(oo) was also the last ingredient to prove (Cl). The conditions (A1)-(A3) [with (Al) suitably modified] were clear from the beginning, except for the condition Ear(oo) C G(oo) proved in 10.10.b. 10.13. LEMMA. Let A E A be a dominant weight. If b E 5(oo) with ;5a (b)

then Ea?pa(b) = pa(Eab) for all a E II.

0,

230

10.

CRYSTAL BASES 11

PROOF. By definition of B(oo) there is asequence al, a2i ... , a, of simple roots such that b = U where u = Fat Fa2 Fam 1. The claim says for all a E II that E«(uva) _ (E«u)vA (mod qL(A)). (1) Fix a E II. Using Lemma 10.6.b and arguing as in the proof of Proposition 10.9 we find a dominant weight µ such that

E«(uva+µ) - (E.u)va+µ

(mod qL(A+µ)).

(2)

We have

uva+µ = F'E, & ... Famva+µ (mod qC(A +,u)) by 10.9(2). Combine (2) and (3) and then apply spa µ; we get E. F., Fa2

Fam (va ®vµ.) _ (Eau)(va ®vµ)

(3)

(mod qL)

(4)

where L=L(A)®G(µ). The assumption pa(b) # 0 says uva

qL(A), hence F_a1Fa2

This implies for all i (1 < i < m) that & (Fa;+,

P,-v.\

qC(A).

qL(A), in other Famva) # 0 in G(A)/qL(A). Therefore 9.17(4) implies

words, that Fa; (Fa;}1

P,-v,\)

F'a,+...Famva) ®vµ

F«; ((Fa.}1 ... Famva) (&vµ) = (F'a,

for all i. This yields inductively Pa,

Fa2... F«m (v.\ ®vµ) _ (Fa1 Fa2 ... F«,n va) ®vµ

Fat Fa2

Fam (va ®vµ) - (uva) ®vµ

(mod qL).

Apply Ea and compare with (4). We get (mod qL).

Ea(uva ® vµ.) - (Eau)(va 0 vµ.) Now (1) follows, if we apply Q,\,,, and use 10.8(8).

10.14. PROPOSITION. Let \ E A be a dominant weight. Then pa (b) induces a bijection (1) { b E B(oo) I pa (b) 310 } _ B(A). PROOF. By Theorem 10.10.a we have to show only: If we have b, b' E B(oo) 0 p.\ (Y), then p.\ (b) p.\ (Y). So consider b, b' E B(oo) with pa (b) = pa (b') # 0. Obviously b and b' have the

with b 31 b' and p.\ (b)

same weight, say -v, where v E M, v > 0. We want to prove by induction on v that b = Y. For v = 0, we have B(oo)o = {1}; so the claim is trivial in that case. Suppose now that v > 0. By 9.9(1) there is a simple root a with Eapa(b) # 0. Lemma 10.13 and the assumptions on b and b' imply pa(Eab') = Eacpa(b') =

pa(Eab) 310.

by 10.12(1). So we can apply induction and get We have Eab, Eab' E Eab = Eab'. Now b = b' follows from 10.12(2).

10.

CRYSTAL BASES II

231

10.15. Set for all aEHandforallbEB(oo)

ea(b) = max{rEZIr>0, Erb#0}.

(1)

This generalizes the definition in 9.17(1). Each e0,(b) is a non-negative integer and we have as in 9.17(3)

e,, (b) = max { r E Z I r > 0, b E Fr,I3(oo) }.

(2)

(IfEab#0, then b=FrErb;if b=Fib'with b'6B(oo),then Erb=b'#0.) Occasionally it will be easier to work with a somewhat different formulation:

e,, (b) = max { r E Z I r> 0, b E Fr, (G(oo)/gr(oo)) }.

(3)

Well, if b = Frx for some x E G(oo)/gC(oo), then write x = EcE,3(.) a,c with a, E Q. We get then b = E. a,Frc. By 10.12 the Fac are distinct and in B(oo). Since B(oo) is a basis with b E B(oo), there has to be a c with b = FFc and a. = 1 whereas a,, = 0 for all c' # c. This shows that b E FF(G(oo)/gr(oo)) if and only if b E Fr,B(oo). Therefore (3) is equivalent to (2). PROPOSITION. Let A E A be a dominant weight. Then we have for all b E B(oo): If ipA(b) # 0, then ea(?a(b)) = ea(b) for all a E H. PROOF. Let b E B(oo) with TA(b) #_ 0. Pick a simple root a E II; set n = e,, (b)

and b' = Enb. Then b' E B(oo) with Eab' = 0 and b =Fib. We have ipA(b) = FF ?,\(b'). So ipa(b) # 0 implies ipA(b') # 0. This yields pA(b), 7,\ (Y) E B(A), hence

E.n?a(b) = ?a(b'). Furthermore 10.13 implies that

E,,ipa(b') _

1pa(E,,.,b) = 0. This shows that e,, (cpA(b)) = n = ea(b) as claimed.

REMARK. Note that there is no point in generalizing the definition of fa(b) from 9.17, since all b E B(oo) satisfy F,,b # 0. We can compute f,,(ipA(b)) using 9.17(2) and get:

If b E B(oo)_ with pa(b) # 0, then .ff(?,,(b)) _ (A - v, av) + e«(b)

(4)

for all a E II.

10.16. The `crystal base' (G(oo), B(oo)) admits something like a 'polarization' generalizing that for (,C (A), B(A)) defined in 9.23. This will be a bilinear form (, ) on U- with certain properties similar to those from 9.23. The construction of this form makes use of the antiautomorphism Ti from 9.20. Recall that Tl (F.) = qa 1 K. E. = ga E. K.. This implies for all v E DD, v > 0

Ti(U

= U tKv.

The second ingredient of our construction is the bilinear pairing between

(1)

and

U'O from 6.12. In order to distinguish it from the bilinear form ( , ) on U- to be constructed, let us denote the old pairing by ( , `the pairing from section

232

CRYSTAL BASES II

10.

VI'. We could now define a bilinear form on U- by mapping each pair (x, y) to (x, r1(y))vi; this makes sense by (1). However, we have to introduce a normalizing factor in order to get the `right' properties of a polarization.

For all v E M, v > 0 set ma

qa

cv = q-(v,v) aED

(Fa, Ea)vi

(2)

'

where v = EaEn maa. Each c, is a nonzero element in k; we have co = 1 and ca = q-1(Fa, Ea)vi1 for all a E U. We set now for

all v, v'EZtI)

(y,x) = cv(y,Ti(x))vi

(3)

We extend this (taking direct sums) to a bilinear form on all of U-. Note that (1) and 6.13(2) imply

If yEU7,and xEwith v0v',then (y,x)=0.

(4)

We get from 6.13(1) that

(y,x) =

for all x,y E U_-,.

1) vi

(5)

Since x --> Tl (x)Kv 1 is a bijection from U17, to U+, Proposition 6.18 implies

The restriction of (

,

) to each U 7, is non-degenerate.

(6)

We have (using 6.12(2) and Ti(1) = 1 and co = 1)

(1,1) = 1.

(7)

Similarly, the formulas for T1(Fa) and ca yield

(Fa, F,,) = 1

for all a E U.

(8)

10.17. Before we can prove more about of the bilinear form (,) defined in 10.16, we need more properties of the automorphism Ti. Note that 10.16(1) shows for each y E U U that we can apply all ra and ra with a E TI to Ti (y) K,- 1. One can now show (for all y E EC, and all a E II) r,,,(7-j(y)K,-,1)

-1(v'a) = qa q T1 (ra (y) ) Kv-a 1

(1)

and

ra(T1(y)Kv

1) = q-1q(v.a)T1(ra(y))Kv

la

(2)

One proves these formulas more or less in the same way as (e.g.) those in 6.16(6). One first checks them for y = 1 and for y = Fp with ,Q E H. (One gets 0 = 0 in

both equations for y = 1 and for y = Fp with 0 0 a; for 0 = a one gets qa = qa.)

10.

CRYSTAL BASES II

233

Then one shows: If the equations hold for y = yl and for y = y2i then they hold for y = y1y2. That is a straightforward calculation using the product formulas 6.14(4) and 6.15(4).

We can now translate the formulas 6.14(5) and 6.15(5) into statements on our new bilinear form; we get for all x, y E U- and all a E II

(Fay,x) = (y,Kar'(x)Kal), (yFa,x) = (y,Kara(x)Kal),

(3)

(x, Fay) = (Kara(x)Ka 1,y), (x,yFF) = (Kara(x)Ka 1, y),

(5)

(4)

(6)

This is again just a straightforward calculation; it uses (1) resp. (2) in order to get (3) resp. (4). One works with weight vectors and one observes first that 10.16(2) implies Cv+a = cv4

2(v.a)

4a

1

(FE a, a)vx

( 7)

LEMMA. We have for all x, y E U(x, y) = (y, x) = (T (y), 7- (X)).

(8)

PROOF. The bilinear form (, ) is completely determined by (3) and by 10.16(4) and 10.16(7). Indeed, the last two properties determine (1, x) for all x E U-; now (3) yields an inductive formula to calculate (y, x) for all y. We can define a bilinear form ( , )' on U- by (y, x)' = (x, y). It is clear that also (, )' satisfies 10.16(4) and 10.16(7). Furthermore (5) implies that (3) holds for (, )'. So the uniqueness statement above says that i.e., that (y, x) _ (x, y) for all x, y.

Define another bilinear form (, )" on U- by (y,x)" _ (T(y),T(x)). Since 7-(1) = 1 and T(U,-) = U,-, this new form satisfies 10.16(4) and 10.16(7). Finally, we have for all x, y (using (4) and 6.15(7)) (Fay, x)" _ (T(Fay),T(x)) = (T(y)Fa,T(x))

_ (T(y),Kara(T(x))K, ') = (T(y),Kar(r(x))K, ') _ (y, Kara(x)Ka 1)" So (, )" satisfies also (3) and we get as before

i.e., (T(y),T(x)) _

(y, x)

10.18. We have for all a E II and all integers m > 0

ra(F(m)) = r' (F()) = 4m

(1)

-1F(m-1),

cf. 8.26(2).

LEMMA. Let x, y E U- with r' (x) = 0 = r' (y). Let r, s E Z, r > s > 0 with r > 0. Then if r > s; 0, (Far)x, Fas)y) = rr1ga(r-1)(F,ar-1)x Far-1) (2) l

a

y),

if r = S.

234

10.

CRYSTAL BASES II

PROOF. We may assume that x and y are weight vectors, say x E 1C, and yEU

We have by 10.17(3) (F,,(,r) x, FFs)y) _ ([r]

F,(5)y)

_ If s = 0, then the last term is 0 since r' (y) = 0; this is compatible with our claim since then r > 0 = s. Suppose now that s > 0. We have r' (F,(') y) _ q(a,v')r'(FFs))y by 6.15(4). So (1) yields

(FFr)x, F,(s)y) _ [r]alq(a'v )q

1(F,(T-1)x,Ka(FFs_1)y)Ka

lq(a.v')q _ [r]a

l)

Fas-1)y)

If r > s, then r - 1 > s - 1; in that case we can use induction on r to show that (FFr-l)x, F.(`1)y) = 0, hence the claim. If r = s, then the claim follows from q(a,v')q lq(a,-v'-(s-1)a) = q lq-(r-1)(a,a) = qc (r-1) 10.19. PROPOSITION. We have (y, x) E A for all y, x E G(oo).

PROOF. We may assume that y, x E G(oo)- for some v E

v > 0. We

want to prove the claim by induction on v. For v = 0 we have x, y E A A. 1, so the claim follows from (1, 1) = 1.

Suppose now that v > 0. Then G(oo)-v = EaEn FaG(oo)-v+,. So we may assume that there are an a E II and a y' E C(oo)_v+a with y = Fay'. Write and x = E5>oFa8)xs as in 10.2(1) with r' (yr) = 0 = ra(xs) Y' = for all r, s. We have yr, xs E G(oo) for all r, s by 10.10(3). Since y = Fay' _ Er>o Fa(r+1)yr, it is enough to show that (F,(,r+1)yr, F«s)xs) E A for all r, s. If Fa(s)xs) r + 1 # s, then = 0 by 10.18(2) and the symmetry of the form. If r + 1 = s, then 10.18(2) yields (F,(,r+l)yr,

(F,(r+l)yr, FF(s)xs) = [r + 1] a lgar(F(r)yr, Fps-11 x51

Since Fa( )yr,

F«s-11x3

the other hand, 10.5(1) implies that [r +

(F,(,r)yr, F,a(s-l)xs)

E A. On is a unit in A congruent to 1

E G(oo)-v+a, induction yields

modulo Aq. The claim follows.

REMARK. This proof actually shows more. The last steps prove that 1

(FF(r+l)yr,

Fas)xs) =

Fps-1)x3)

(mod Aq)

in case s = r + 1. However, this congruence is true also for r + 1 # s, since then both sides are 0. (For s = 0 one has to interpret Fa(s-1)x3 as 0.) If we sum over all r and s, then we get now

(Fay', x) - (y', Eax) (mod Aq).

(1)

This congruence holds for all x, y' E G(oo): As in the proof above we can reduce to the case x E U1, and y' E U +a. Then the proof above deals with arbitrary x and y of this form.

10.

CRYSTAL BASES 11

235

10.20. Proposition 10.19 implies that (, ) induces a (symmetric) bilinear form (with values in Q = A/Aq) on G(oo)/gG(oo). We denote this form by (, )o. PROPOSITION. a) We have (Fay, x)o = (y, Eax)o for all x, y E G(oo)/gC(oo). b) 13(oo) is an orthonormal basis of G(oo)/gC(oo) with respect to (, )o. c) G(oo) _ { y c- U- I (y, x) E A for all x c- G(oo) }. d) G(oo) _ { y c- U- I (y, y) E A}.

PROOF. The claim in a) follows immediately from 10.19(1). Let us look at b).

By induction on v, 13(oo)_ is an orthonormal basis of This is obvious for v = 0 since 13(oo)o = {1}. Suppose that v > 0, let b,b' E 13(oo) There is a E H and b1 E with b = Fab1. Then (b, b)o = (Fab1, b)o = (bl, Eab)o = (b1, E.FFb1)o = (b1, b1)o = 1

by induction. If b' # b, then (b, b')o = (Fab1, b')o = (bl, Eab')o = 0

since either Eab' = 0 or Eab' E Eab'# b1.

with b' = F«E(,b' # F(,b1, hence

In order to prove c) and d) we can more or less copy the proof of 9.23(5) and 9.23(6). I shall leave the details to you. 10.21. COROLLARY. We have -G(oo) = G(oo).

PROOF. This follows from 10.20.d, since (r(y), r(y)) = (y, y) by 10.17(8).

CHAPTER 11

Crystal Bases III We keep the assumptions and notations from Chapters 9 and 10.

So far we have constructed bases of G(oo)/gL(oo) and of all L(A)/qL(A). Now we want to find bases of U- and of all L(\) that have similar properties. In particular, we want to have an analogue of Proposition 10.14: For each dominant A the map u --> uva takes the u in our basis of U- with uva # 0 bijectively to our basis of L(A).

We can construct a basis of U- as follows: Choose for all v representatives in L(oo)_ for the elements in B(oo)-,. By the Nakayama lemma these representatives are a basis of L(oo)_ over A, hence a basis of U over k. However, an arbitrary choice of the representatives will not have the special property mentioned above.

In order to get this property, we have to impose two conditions on our representatives. The first one is that they should belong to a certain `integral' form of U. More precisely, this is a form of U over Z[q, q-1 ] introduced by Lusztig, cf. [Lusztig 1J, 4.1. In 11.1-8 we define this form and prove some of its properties. The second condition on our representatives is that they should be fixed by a certain automorphism of U regarded as a Q-algebra. This automorphism maps q

to q-' and fixes all Ea and F. with a E H. This automorphism is constructed in 11.9. Then we state the main results in 11.10 and prove them in 11.11-15.

F,"

11.1. Set UZ equal to the Z[q, q-']-subalgebra of U generated by all E,), , and K1 1 with a E H and n E Z, n > 0. This construction (due to Lusztig)

should be regarded as a quantum version of the Z-form U(g)z of the enveloping algebra U(g), cf. [H], 26.4. (If you compare, you will notice that one takes in the enveloping algebra case divided powers x« /n! for all roots a, whereas we take here only a E ±H. However, one can show in the enveloping algebra case that it is enough to take the a E ±H.) One word on the notation: It might be more appropriate to replace the index Z in Uz by Z [q, q-1]. However, that would be too complicated as an index. We could get around this problem by setting say A' = Z[q, q-1] and then writing UA,. However, I do not want to introduce another notation; so I follow more or less Kashiwara's example by writing Uz. In the enveloping algebra case an important role is played by the elements (n ) with a E H and n E Z, n > 0 where h« E [g«, g-«] with a(h«) = 2. In the present 237

238

CRYSTAL BASES III

11.

quantum case their role is taken over by the elements

Kaga i+1 _ K-1q- (a-i+l)

[Ka;aJ =

11

J

qa _ qa-i

i=1

(1)

with a E II and a, n E Z, n _> 0. So we get for n = 0 the empty product [ KO' a 1

= 1 and for n = 1 we get I K1' a ] = [Ka; a] in the notation from 4.4(8).

Note that one can rewrite Lemma 1.7 using this new notation as min(r,s)

E

E all E II. (Actually, this is not Lemma 1.7, but Lemma 1.7 applied to all Ua C U using Ua = Uqa (s C2) as before 9.1.) We claim that

[Kn;a

E Uz

for all a,n E Z, n > 0 and all a E 11.

(3)

We prove this by induction on n; the claim is obvious for n = 0. For n > 0 we get the claim for a = 0 by induction from (2) applied to r = s = n. We get then the claim for all other a (inductively) using the formula

q-n[Ka;a]-[Ka;a-11 n n

-K,lgaa[Ka;a-11

n-1

(4)

that can be checked by an elementary calculation. The definition of a weight space implies for any finite dimensional U-module V and all a, a, n as above. [K.'a1

v = [a+ (li,av)]v n

for all vEVm

(5)

(and all µ E A). In the case of the adjoint action, we have for all u E U (where v E Z4') and

Kau = u (q((,A'av)K(,)

K( lu = u (q. (W,«

)Kq 1),

(6)

hence

[K'a1 u

=

u

[Ka'a +(v'av)n

(7)

Set Uz resp. UZ equal to the subalgebra over Z[q,q-1] generated by all E((,,,) resp. all F(n) with a E H and n E Z. Set Uz° equal to the subalgebra over Z[q,q-1]

generated by all K 1 and all [Kay a as above. Then Uz , UZ , and Uz° are subJ

algebras of L. (In the case of Uz° this requires (3); in the other two cases it is obvious by definition.) We claim now that UZ Uz° Uz = Uz.

(8)

CRYSTAL BASES III

11.

239

Well, it is clear by our last remark that the left hand side is contained in the right hand side. On the other hand, the left hand side is stable under left multiplication 1

with all F,,(,) [contained in the subalgebra UZ ], with all K 1 and

[by

Ka; n aJ (6) and (7), since UZ is generated by weight vectors], and with all E(n) [use (2) together with (6) and (7) and also E(n)F(m) for all a 0 Q]. So the left hand side is stable under multiplication with all generators of Uz, hence equal to L

U. We have just appealed to the fact that UZ is generated by weight vectors. This implies that

UZ = ® (Uz)-v

(9)

vEZ'',v>O

where (UZ )_v = UZ fl U=,,. There are similar weight space decompositions for UZ and Uz. Furthermore, there are only finitely many products in the F(,n) (with a E 11 and n E Z, n > 0) that have a given weight. This implies that the weight spaces (UZ) _ v are finitely generated modules over Z [q, q-1 ] We have rr(E(n)) = E(,n) and r(F,,(,n)) = F(,n) for all a and n, hence

T(Uz) = Uz,

T(UZ) = Uz

,

T(Uz) = Uz .

(10)

We get also rr(Uz°) = UZ since UZ = U° n Z.z by (8). As an alternative, you may want to evaluate T on the generators of Uz°. 11.2. LEMMA. Let a E H. a) We have

ra(UZ)CUZ

and

ra(UZ)CUZ

(1)

EQ(UZ)CUZ

and

FF(UZ)CUZ.

(2)

and

b) If U = En>0 FQn)un E U- where un E U- with ra(u) = 0 for all n, then u is in UZ if and only if un is in UZ for all n. PROOF. al) We have ra(1) = r0'' (1) = 0 and ra(FFm)) = ra(FFm)) = 0 for all ,Q E II, Q 0 a and all m > 0, since a resp. -m,Q + a is not a weight of U- . We have already seen before that ra(F m)) = qr for all in > 0, cf. 10.18(1). So ra and ra map all generators of UZ into UZ . Therefore the product formulas 6.15(4) yield (1). b) Now consider u = >n>0 as in b). If each un is in UZ , then clearly so is u. Suppose now that u E UZ . We may assume by 11.1(9) that u is a weight

vector, say u E (UZ )_ with v E M, v > 0. We want to use induction on v to show that un E UZ for all n. For v = 0 we have u = u0 and un = 0 for all n > 0; so the claim is clear in that case. For v > 0 we have by 6.15(4) and 10.18(1) q(a'v-na)ra(FQn))un = E FcYn-1)(q(a.v-na)ga-1 un)

ra(u) n>0

n>O

240

CRYSTAL BASES III

11.

We have r(u) E U-

by (1). Since r'

(«,v-n«) n-lu

= 0 for all n > 0,

un E UZ for all n > 0. we can apply our induction hypothesis and get Because UZ is a Z[q, q-I ]-module, this implies also un E UZ for all n > 0. Finally

we see that also uo = u -En >OF.(n)un

E UZ .

a2) Now (2) follows immediately from b) and from the definitions of E« and F«.

11.3. LEMMA. Let a E 11 and v E Z4, v > 0. We have for all integers r > 0 and for all u E (UZ) _,

ku - F«(r)u

(mod

(1)

We have for all r > 0

_

Fc(UZ

F(r+s) (Uz)

+s

(2)

s>o

and

= FLU V n UZ.

F«(UZ)

(3)

PROOF. Let u E U. and write u = En>o Fa)un as in 11.2.b, i.e., with un E U +n« and rer(un) = 0 for all n. We have then F«ru

F«(r+n)un

=

(4)

n>O

and

[r±n]

F,«(r+n)un=F «

>0

n>o

(5)

r

Suppose first that u E U. Then all un are in UZ by 11.2.b, so (4) and (5) imply

Far)(UZ)- c FF(UZ)

F«(r+n)(Uz)

C

+n«

(6)

n>O

(In order to get the first inclusion we have to recall that the Gaussian binomial coefficients are in Z[q, q-1], cf. 0.2.) We can apply (6) to any v - sa instead of v and get thus F«(r+s)(UZ

)_v+s« C Fr+S(UZ )-L"+s« C j''a(U7 )_V

(7)

for all s > 0 using 11.2(2) to get the last inclusion. Now (6) and (7) together yield (2).

A comparison of (4) and (5) shows that Far)u - F.r u =

1: (f r r +nJ «

n>0

1)F(,r+n)un

L

= 1 «r+,

(I

L

n>O

r+n r J l

«

-

1)F(n-')u « n.

11.

CRYSTAL BASES III

241

This implies (1). It is clear by (4) that Fau E FaU V for all u E U this yields the inclusion "C" in (3). Now drop the assumption that u E UZ and assume instead that Fau E UZ . We have Far+n) Fau FaF(n)un (anUn) n>O

n>OO

where an = [n+r]l([n]')-1. Now Lemma 11.2.b applied to Fau yields anun E UZ for all n, hence Fau = Fa Fan)(a.nun) E Fa(UZ )-v. n>O

That proves the other inclusion in (3). REMARK. The definition of UZ implies that 1: Fan)(TlZ )-v+na

(UZ )-v =

aEII n>O

for all v > 0. Now (2) applied to each v - a shows that

for allv>0.

(8)

aEII

11.4. For all dominant weights A E A set Lz(A) = Uzva = UZ vA.

(1)

Here the second equality follows from 11.1(8) and 11.1(5). It follows from 11.1(9) that Lz(A) is the direct sum of its weight spaces; we have LZ(A)A = Z[q,q-']va. Let us look at a few examples. If P is of type A1i then

UZ = 1: Z[q, q- 1]Fan) r>O

where II = {a}. This implies (in the notation from 8.3) that the vi = Faz)v0 with 0 < i < n are a basis of Lz(n) over Z[q, q-1] Consider now the case where P is arbitrary and where A E A is a minuscule dominant weight. Recall the basis (x, I µ E WA) of L(A) from 5A.1. I claim now that these xN. are also a basis of Lz(A) over Z[q, q-1] if we set va = xa. Indeed, first induction on µ from above shows that x,, E Lz(A) for all µ: For µ = A this holds by the choice of vA. For p < A there exists a E II with (µ, a") < 0; then x,, = Fax +a E Lz(A) by induction. Next, the formulas in 5A.1(2) show that EMEw.a Z[q, q-1]x,.. is stable under all Fa. They imply also that all Fa (and thus all Fan) with n > 2) annihilate L(A). So the span over Z[q, q-1] is stable under UZ, hence equal to Lz(A). Suppose now that is indecomposable and that A = aO is the dominant short root. Recall the notations -tos and TI, from 5A.2 and the basis (x7 ry E -10s, hp /3 E H8) described in that subsection. Again I claim that these elements are also a I

242

CRYSTAL BASES III

11.

basis of Lz (A) over Z[q, q-1] if we set va = xA. As before one checks first that all x.y and hp are in Lz(A). (One gets first the x-1 with y > 0 arguing as in the minuscule

case. Then one gets h« = F«x« and x_« = F' ( 2)x« for all a E H. Finally, for the x.y with -y < 0, -y V -II one argues again as in the minuscule case.) We now have to show that the span of all x.y and hp over Z[q, q-1] is stable under all F«n). This can be read off 5A.2(2)-(5) for n = 1. All F«n) with n > 3 annihilate L(A) while F,(2) annihilates all basis vectors except for x« where we have F,(, 2)x« = x_«. Now the claim follows.

Similar arguments show also in the cases from 5A.5-7 that the bases of L(A) constructed there are also bases of Lz(A) over Z[q, q`1] if we choose va suitably. Let me however return to the case A = ao. If we have two short simple roots a and 3 with (,(3, a") = -1, then the calculations in the proof of 9.6 [immediately preceding 9.6(9)] show that

_

1

1

[2]« '-c' = q« + q. 1 x

F«ha

Ct

This shows that in general F«Lz(A) 0 Lz(A) even though F«UZ C UZ .

11.5. Let a E II. Recall the operators T«, T«, etc. from 8.6. We have by definition for all weight vectors v E L(A), (with u E A)

E

T« (v) =

(-1)bg6«-ac F, (a)F«(b)E («c)V

(1)

a,b, c> 0; - a+b-c=m

where m = (µ, a"). So T« maps Lz (A) to itself. The other formulas in 8.6 show that the same is true for the other operators (T«, -T,,,, and -T,,). Since WT« = T; 1, this implies for all µ E A

T«Lz(A)µ = Lz(A)Saµ

(2)

LEMMA. Let a E II, let A E A be dominant. Let µ E A with (µ, ") > 0. Then

Lz(A)sRµ = E

(3)

i>0

PROOF. Set m = (µ,a"). By (2) and (1) it suffices to show for all v E Lz(A),,

that E«a)F.(b)E«°)v E

(4) i>O

for all nonnegative integers a, b, c with b = a + c + m. Since we have then b > a, the commutator formula 11.1(2) yields a

F(b-i) i=0 Since

fL K«; 2i

-a - b]J

i

for all i, we get (4), hence (3).

LK«; 2i

i a - bl Era-i)E«c) J

E(«a_i

)E.(c)v E Lz(,\)µ+(a-i+c)«

V.

11.

CRYSTAL BASES III

243

11.6. Set

Gz(oo) = UZ ngoo).

(1)

This is a Z[q]-module since Z[q] = Z[q, q-11 n A. Therefore 11.2(2) implies for all

r>0andalla1ia2,...,a,Ell

Fa1F-z ... Fer I E GZ(0O)

(2)

Both UZ and G(oo) are direct sums of their weight spaces, hence so is their intersection (@ Gz(00) (3) Gz(OO) _ vEZ4),v>0

Note that Cz(oo)o = A 1 n Z[q, q-1] I = Z[q] 1. We have q,Cz(oo) = UZ n gG(oo),

(4)

since q is a unit in Z[q, q-1]. This implies that Cz(oo)/gCz(oo) can be identified with the image of Cz(oo) in G(oo)/gr(oo). By (2) we have under this identification

B(oo) c Lz(oo)/gCz(oo) c G(oo)/gL(oo)

(5)

For all dominant weights A E A set

Lz(A) = Lz(A) nG(A).

(6)

This is a Z[q]-submodule of L(A). Since both Lz(A) and G(A) are the direct sum of their weight spaces, so is their intersection:

Gz(A) _ ® Gz(A),,.

(7)

µEA,A
We have Gz(-\)a = Z[q]VAThe map co U- - L(A) takes UZ onto Lz(A). So Theorem 10.10 implies

co L z(oo) C Gz(A)

(8)

gGz(A) = Lz(A) n qG(A)

(9)

We have

as in (4). So we can identify Gz(A)/gGz(A) with the image of Gz(A) in G(A)/qG(A). Now (8) and (5) together with Theorem 10.10 imply that (under this identification)

13(A) c Cz(A)/gCZ(A) C C(A)/gC(A).

(10)

11.7. PROPOSITION. The Z-module Gz(A)/gGz(A) is free with basis l3(/) for each dominant weight A E A. The Z-module Gz(oo)/gCz(oo) is free with basis l3(oc).

244

11.

CRYSTAL BASES III

PROOF. Let A E A be dominant. There is by Proposition 9.25 (and the remark in 9.25) a polarization of (L(A),13(A)) with (va, va) = 1. We claim first that for all v, w E Lz(A).

(v, w) E Z[q, q-1]

(1)

We may assume that v, w E Lz(A),,, for some weight µ < A; we then want to prove

(1) by induction on µ from above. For µ = A the claim follows from Lz(A)A = Z[q, q-1]va and (v,\, v,\) = 1. Suppose now that µ < A. We have then

Lz(A)µ = E E F,,(,n)Lz(A)µ+na. aEn n>O

Therefore it is enough to look at the case where v = FFn)v' for some v' E Lz(A)µ+na

(with a E II and n > 0). We have then (v, w) = (F,(,n)v', w) = (v', 7-1(F,(,n))w)

(2)

Now T1(F.) = q.E.KK implies that 7.1(F(( n)) has the form q1 E(('n)K( for a suitable integer m. Therefore Tl (F,(,n)) E Uz i hence ri (F,(,n) )w E Lz (A)µ+n« So we can

apply induction to the last term in (2) and get that it is in Z[q, q-1]. Now (1) follows from (2).

Since Z[q] = A fl Z[q, q-1] the definition of a polarization together with (1) implies

for all v, w E Lz(A).

(3)

for all v,w E Lz(A)/gLz(A).

(4)

(v, w) E Z[q]

Working modulo q then yields

(v,w)o E Z

Since 13(A) is an orthonormal basis of L(A)/qL(A), any v E L(A)/qL(A) satisfies V = > bEB(A)(v,b)b. Now, if v E Lz(A)/qLz(A), then (4) and 11.6(10) imply that (v, b) E Z for all b, hence

Lz(A)/gLz(A) C E Zb. bEB(A)

The reverse inclusion is clear by 11.6(10). So we have here equality. Since 13(A) is linearly independent over Q, this implies that it is a basis of Lz(A)/gLz(A) over Z.

Now for the claim on B(oo). It is enough to show for all v E M, v > 0 Choose a domithat 13(oo)_ is a basis of the Z-module nant weight A such that cpa maps U7, bijectively to L(A)A_,,. Then (UZ )_ goes bijectively to Lz(A)A_,,, so 10.10 implies that Lz(oo)_, is mapped bijectively to Lz(A)A_,,. Therefore cpa induces an isomorphism

Lz(oo)-,./gLz(oo)

-- Lz(A)a-,./gLz(A)A

(5)

Since it maps 8(oo)_ bijectively to 13(A)A_ and since the latter set is a basis of the right hand side in (5) by the first part of this proposition, it follows that L3(oo)_ is a basis of the left hand side. But that is our claim.

It. CRYSTAL BASES 111

245

11.8. The only vectors of length 1 in the Z-span of an orthonormal basis are the basis vectors themselves and their negatives. Applied to the bilinear form in the first part of the proof in 11.7 this shows

B(A) U (-B(A)) = { x c Lz(\)/q/.z(A) I (x, x)o = 1 }.

(1)

Applied to the form from 10.16 we get using 10.20.b

B(oo) U (-B(oo)) = { x E Lz(oo)/gCz(oo) I (x, x) o = 1 }.

(2)

The antiautomorphism T maps both G(oo) and Uz° to itself, cf. 10.21 and 11.1(10), hence also their intersection Lz(oo). It therefore induces an automorphism (also denoted by T) of the Z-module £z(oo)/gGz(oo). This automorphism preserves the

bilinear form (, )0i since the original T preserves (,) by 10.17(8). Therefore (2) implies

T(B(oo) U (-13(oo))) = B(oo) U (-B(oo)).

(3)

(We shall see in 11.18 that actually T!3(oo) = B(oo).)

11.9. There is a unique isomorphism z/' : Q[q] - Q[q-1] of Q-algebras with V)(q) = q-1. It can be extended uniquely to an automorphism of the fraction field k = Q(q). We have for all integers a and n with n > 0 and all a E II ([a]te) _ [a].,

')([n]«)

= [n]om,

[

([a],) = n]

,

(1)

cf. 0.2. If V is a vector space over k, then we denote by V1101 the vector space over

k that is equal to V as an abelian group, and where the scalar operation of any a E k is equal to the scalar operation of V)(a) on V. In the next proof I shall use a to denote the new operation; so we have [in V«]]

=

O(a)v

[in V]

for all aEkandvEV. PROPOSITION. a) There is a unique automorphism algebra with V)(E.)

= E.,

V) (F«) = F.,

'(Kµ) = Kµ 1,

of U regarded as a QV)(q)

= q-1

(2)

for all aEII andallµEM. We have V)2 = 1. b) There is for all dominant weights A E A a unique Q-linear map Va : L(A) L(A) with z/ia(va) = va and z/ia(uv) = Vi(u)Via(v)

for all u E U and v E L(A),

where Vi is the map from a). We have Va = 1, and V),\ is bijective.

(3)

246

11.

CRYSTAL BASES 111

PROOF. a) The functor V --* V101 takes k-algebras to k-algebras. In particular

we can regard U1'' as a k-algebra. One now checks easily [using (1)] that the (E., F., Ka 1, K.) satisfy the defining relations of U. So the universal property of U yields a homomorphism U -+ UE''1 of k-algebras that takes each Ea to Ea, each Fa to F, and each K0, to Ka 1, hence each KN, to Kµ 1. We regard this map as a homomorphism of Q-algebras. It maps q = q1 to =(q)1 = q-1, hence satisfies (2). We denote it by 0, since it restricts to the earlier V) on k kl C U. The uniqueness of Eli satisfying (2) is clear, since b(q) = q-1 determines its restriction to k ^ kl as the old i,b, and since the other elements in k generate U over k. We have 02 = 1 since V)2 fixes the generators and induces the identity on k. The bijectivity of ili follows. b) Let A E A be dominant. We can make L(A)[ '1 into a U-module via [in L(A)[ ']] = i,h(u)v

[in L(a)]

for all u E U and v E L(7). (Since i,) on U restricts to the old i,b on k, this is compatible with the vector space structure on L(A)[ 1.) The bijectivity of Eli on U implies that L(A)[? ] is a simple U-module. We have E,,-v,\ = Eava for all a E II, and q-(µ,-\)va K,.-v,\ = K, lva = = q(µ+-\).v'\

for all µ E Z. So vA regarded as an element of L(A)101 is a highest weight vector of weight A. Therefore the simple module L(A)[ '] has to be isomorphic to L(A); we can find an isomorphism Via : L(A) - L(A)[ '1 of U-modules with Via(va) = vA. If we regard -+/)a as a map from L(A) to itself, then it is Q-linear, and the `isomorphism

of U-modules' translates into (3) and the claim on bijectivity. The uniqueness of -+/)a satisfying Via(va) = vA and (3) follows from L(A) = UvA. Finally V)2(uva) _ V)a(V)(u)va) ='1V12(u)vA = uva shows that 02 = 1. REMARKS. 1) We get from (2) and (1) that b(F(n)) = F,(,") for all a E II and

n E Z, n > 0. Since 0 induces an automorphism of Z[q, q-1] this implies that Vi(UZ)=UZ. We have similarly Vi(UZ)=UZ,b(UZ)=lz,and b(UU°)=UU°. 2) We have for all dominant weights A E A and for all u E U ba (uva) = 0(u) v.\ .

(4)

Since Lz(A) = UZ va the first remark implies now 0a(LZ(A)) = Lz(\) for all A. 3) It is easy to see that Vi commutes with the antiautomorphism T of U. More precisely, one can check that T o Eli = V) o T is the antiautomorphism of U (regarded

as an algebra over Q) that fixes all Ea, Fa (with a E II) and all K. (with µ E Zf), and that maps q to q-1.

11.10. We are now ready to state the main theorem on the basis of U- that lifts the basis B(oo) and has similar properties. We formulate it as a result for each weight space. It will then be proved (in 11.11-15) by induction on the weight. THEOREM. Let v E Zt, v _> 0.

a) There is for each b E 13(oo)- a unique element G(b) E £z(oo)- with i,LG(b) = G(b) and G(b) = b.

11.

CRYSTAL BASES III

247

b) The G(b) with b E B(oo)- are a basis of Gz(oo)_ over Z[q], of (UZ )_ over Z[q, q 1], and of {x E Lz(oo)_ I V) (x) = x} over Z. c) Let b E 13(oo)_,,, a E H, and n E Z, n > 0. Then G(b) E FnU- if and only

ifn<e,, (b). d) Let A E A be dominant. The G(b)vA with b E B(oo)-,, and pa(b) # 0 are a basis of Lz(A)A_ over Z[q,q-1], of Lz(A)a_ over Z[q], and of {x E Lz(A)A-v Via(x) = x} over Z. We have G(b)vA = 0 for all b E B(oo)_ with pa(b) = 0.

Observe that the theorem holds for v = 0: We have 13(oo)o = {1} and Lz(oo)o = Z[q] 1. The fixed points of z/i in this module are Z 1; so G(1) = 1 is the only element satisfying a). The description of Lz(oo)o and its ib fixed points yield two of the claims in b); the last one follows from (U)0 = Z[q, q-1] 1. Part c) is obvious for v = 0 since ea(1) = 0 and 1 FQU-. Finally d) holds since vA is a basis for all modules considered, cf. 11.4, 11.6, and since pa(1) is never equal to 0.

Note that one direction in c) is clear for all v: If G(b) E FnU-, then G(b) E FQ

by 10.10(3). This yields b E F(n,, (L(00)_v+nQ/gL(oo)-v+nQ7

hence n < e,,(b) by 10.15(3). 11.11. Before we start proving Theorem 11.10 we need an auxiliary lemma: LEMMA. Let L be a vector space over k, let L C L be an A-submodule of L, and let S = {vi, v2, ... , vr} C L be a finite subset of L. Suppose that the classes

.. , Ur in L/qL are linearly independent over Z. Then v1, v2, ... , yr are linearly independent over k, we have

V1, V2i

r

r

L n > Z[q, q-1]vi = >Z[q]vi, i=1

i=1

and the restriction of the canonical map L -> L/qL to

I:r

(1)

1 Zvi is injective.

PROOF. Set E = E 1 Zvi. The canonical map 7r : L -> L/qL maps the generators of E to a linearly independent set. Therefore the vi themselves are linearly independent over Z, and E is a free Z-module with basis v1, v2i ... , vr. Furthermore the restriction of 7r to E is injective.

Let us show now that qjE C L is direct. Suppose that we have eo, e 1 , ... , en E E with > 0 gjej = 0. Then 0 = 7r(> o gjej) = 7r(eo), hence eo = 0 by the injectivity of 7r on E. We divide by q and get En=-o gjej+1 = 0, hence ej = 0 for all j by induction on n. Now we can check the linear independence of the vi over k. Let a1, a2i ... , ar E k with >i aivi = 0. We can multiply the ai with a (nonzero) common denominator and assume that ai E Z[q] for all i, say ai = Ej aijgj with all aij E Z. We get then

Ej qj (Ei aijvi) = 0. The directness of Ej qjE yields now >i aijvi = 0 for all i, and the linear independence of the vi over Z implies then aij = 0 for all i and j hence ai = 0 for all i. It is left to prove (1). Here the inclusion "D" is obvious. On the other hand, any x in the left hand side can be written x = Ej'=-r gjej with all ej E E and 0 < s, r. If r > 0, then Er+s qrx E qL, hence 7r(e-r) = 7r(grx - Er+s J=1 qjej_r) = 0. This implies a-r = 0, since 7r is injective on E. Now induction yields ej = 0 for all j < 0, i.e., x E Z[q]E.

248

11.

CRYSTAL BASES III

REMARK. Keep the notation E = Ei 1 Zvi from the proof. The lemma implies in particular that S = {v1, V2.... , v,.} is a basis of Z[q, q-1]E over Z[q, q-1], and a basis of L n Z[q, q-1]E over Z[q].

In our applications of this lemma there will be an extra feature. There will be a map i/i' : L -p L with z/i'(x + y) = V'(x) + V'(y) and z/i'(ax) = z/i(a)z/i'(x) for all x, y E L and a E k. (More explicitly, L will be a subspace of U or of an L(a), and V' will be the restriction of the V resp. of the i' from 11.9.) We have then in addition: If t,b'(vi) = vi for all i, then E = {x E Z[q]E I V,'(x) = x}.

(2)

Indeed, any x E Z[q]E can be written uniquely x = Ei aivi with all ai E Z[q]. The assumption in (2) implies then that V'(x) = >i V1(ai)vi. Therefore we have V'(x) = x if and only if V1(ai) = ai for all i. However, we have Vi(a) = a for a E Z[q] if and only if a E Z. This yields (2).

11.12. Let v E Z4P, v > 0. Assume by induction that Theorem 11.10 hold for all weights v' E Z4P with 0 < v' < v. Denote the canonical map G(oo) -p G(oo)/gL(oo) by ir.

for all n > 0 and L = L(oo)-,. a) For each integer n > 0 and for each b E B(oo)_ with ea(b) > n there is a unique element Ga(b) E Lz(oo)_ n m,, with 'Ga(b) = Ga(b) and irGa(b) = b. b) For each integer n > 0 the set { G- (b) I b E B(oo)-,, ea (b) > n } is a basis o f M n over Z[q, q-1], o f M ,, , n L over Z[q], and o f { x E M .n n L I b(x) = x } over LEMMA. Let a E H. Set Mn = FQ (UZ

Z.

PROOF. Before we start, note that the Mn form a descending chain M1 D M2 D M3 D since Fa(UZ C (UZ)-1+na Furthermore we have ikMn) = Mn for all n; this follows from 11.3(2). We want to prove the lemma by induction on n from above. For large n we have v - na 0, hence Mn = 0 and ea(b) < n for all b E B(oo)-,. Therefore the claim holds for large n. Fix now an n > 0 and suppose that we have the lemma for

n+1. Let me first prove the existence part in a). Our induction on n yields the GO, (b) E

.n+1 C Mn for all bE B(oo)_ with ea(b) > n. Consider now b E 13(00)_

with Eab' = 0. By our

with ea(b) = n. Set b' = Enb; then b' E

induction on v Theorem 11.10 holds for v - na, so we have G(b') E (UZ

n

L(oo) as in 11.10.a. We have F,(n)G(b') = FnG(b') (mod MM+1) by 11.3(1), so there

is y E MM+1 with FQ')G(b') = FFG(b') + y. By induction on n the Ga(c) with c E B(oo)-,, and ea(c) > n are a basis of M.n+1 over Z[q,q-1]. Any a E Z[q,q-1] can be decomposed uniquely

a = a+ + ao + a_

with a+ E qZ[q], a_ E q-1Z[q-1], and ao E Z.

Therefore y can be decomposed uniquely y = y+ + yo + y_ with

y+ E> gZ[q]Ga(c), C

y- E E q-'Z [q-1]Ga(c), C

yo E E ZGa(c) C

So we have

y+, V)(y-) E q(M

i n L)

and

yo E M -,-1 n L with ''(yo) = yo.

11.

CRYSTAL BASES III

249

Set now

+y+ -'b(y-)

G"(b) = F«n)G(b') - (y- +'b(y-) +yo) =

(1)

The first expression implies ?PG' (b) = G" (b), the second one shows G" (b) E Mn n C

and G"(b) - FnG(b') (mod q(Mn+l fl G)), hence 7rG" (b) = ir(FF G(b')) = Fa irG(b') = Fa b' = b.

So we have found an element as desired in a). We want to show next that

E

Mn =

Z[q,q-1]G"(b)

(2)

bEB(oc)_,,,e, (b)>n

Well, we can apply Theorem 11.10 to v - na and get from 11.10.b that

E

Mn =

Z[q, q-1]F« G(d).

for all d with e((d) > 0,

Furthermore 11.10.b implies G(d) E hence Fa G(d) E Mn+1. This shows that

E

Mn = Mn+1 +

Z[q,q-1]FFG(d).

dE13(-)- +na,ea (d)=o

The map d -4 Fnd is a bijection between Id E

I

a"(d) = 0} and

e" (b) = n } with inverse_ b - En, b. The G" (b) constructed above have by (1) the property that G"(b) - FnG(EEb) (mod Mn+1). We get therefore { b E 13(oo)

Mn=Mn+1+

E

Z[q, q-I]G"(b),

bE23(oc)-,,,ea (b)=n

and the induction on n yields (2). We now want to apply Lemma 11.11 with C = G(oo)- and S = G" (b) I b E 13(oo)_,,, a (b) > n}. Since 7rG"(b) = b, we get the linear independence condition on 7r(S) in 11.11. Therefore (2) and Lemma 11.11 together with 11.11(2) yield the claims in b). We get now also the uniqueness part in a): If x E Mn n C is another choice for some G" (Y), then ?P (x) = x implies by b) that x E E ZG"(b). Since 7r is injective on this Z-module and since 7r(x) = b' = 7rG"(b') we get x = G"(b'). REMARK. Part a) of the lemma says for n = 1: There is for each b E 13(oo)_ with e" (b) > 0 a unique element G" (b) E GZ (oo) nF" UZ with ?PG" (b) = G" (b) and 7rG"(b) = b. If we then apply this part of the lemma to n = e,, (b), we get that G"(b) E F"e°(b)Uz

for all these b.

(3)

250

11.

CRYSTAL BASES III

11.13. Keep the assumptions on v from 11.12 until 11.15. Let a E H and let A E A be a dominant weight. Set for all integers n > 0

Mn =

F«r)Lz(A)A-v+ra C Lz(A)A

(1)

r>n

The MM are a descending chain of Z[q, q- 1 ]-submodules of Lz(A)A_ with 0),(M' ) = Mn for all n, cf. 11.9(4). We have

E(F«r)(UZ )-v+ra)va = Mnva r>n

with Mn as in Lemma 11.12, cf. 11.3(2). So Lemma 11.12.b implies

Mn =

Z[q,q-1]Ga(b)va

(2)

bEB(oo)_,,,ea (b)>n

We want to show more precisely: LEMMA. Let n > 0.

a) The Ga(b)va with b E 13(oo)-,,, ea(b) > n, and ,\(b) 0 0 are a basis of Mn over Z[q, q-1], of Mn f1 G(A) over Z[q], and of {x E Mn n G(A) I 0,\(x) = x} over Z.

b) We have G°(b)va = 0 for all b E 13(00)- with ea(b) > n and ;5A (b) = 0.

PROOF. We use again induction on n from above. For large n there is no b E 13(oo)- with ea(b) _> n, and we have Mn = 0. Suppose now that the lemma holds for n + 1 and prove it for n. We have Mn = M,M+1 + F«n)Lz(A)a-v+na, hence

Mn = Mn+1 +

Z[q,

q-1]F«n)G(d)va

(3)

by Theorem 11.10.d applied to v - na. If ea(d) > 0, then Theorem 11.10.c implies (as in the proof of 11.12) that FQn)G(d) E Mn+1, hence Fan)G(d)va E Mn+1. If ea(d) > 0, then we have Fan)G(d) = G'(F(-,d) (mod Mn+1), cf. 11.12(1), hence Fan)G(d)va - G'(F(n,,d)v, (mod Mn+1). So (3) yields Mn = M,+1 +

Z[q,q-1]Ga(FQd)va

(4)

ipa (d)#0,e. (d)=0

We now have to distinguish two cases:

CASE 1. We have (A - v, a") > -n. For all d occurring in (4) we have 0. fa(ipa(d)) = (A-v+na,a")+0 > n by 10.15(4), hence ;pa(Fad) = Therefore (4) yields Mn = Mn+1 +

Z[q, q-1]Ga(b)va bEB(oo)__ ip, (b)#0,ea (b)=n

(5)

11.

CRYSTAL BASES III

251

CASE 2. We have (A - v, a') < -n. We claim that (5) holds also in this case. First of all, 10.15(4) implies now for all b E 13(oo)

with ;p,\ (b) # 0 and ea (b) = n

that fa(;p,\(d)) = (A - v, av) + n < 0, which is impossible. So there is no b with this property, and the sum over b in (5) is equal to 0 in this case. On the other hand, Lemma 11.5 implies Lz(A)a

= i>O

where m = (sa(A - v), a") _ -(A - v,av) > n. This shows that Lz(A)a_, Mm = Mn+1 = Mn. (Recall that the MiA are a descending chain.) Therefore (5) holds also in this case. We get (in both cases) from (5) and the induction hypothesis on Mn+1 that

M, =

Z[q,q-1]Ga(b)vA.

(6)

bE.l3(oo) _,,,

We now apply Lemma 11.11 with C = E(A) and S = {b E 13(oo)_ y5,,(b) 0, ea(b) > n}. The canonical map 7r : E(A) -> E(A)/qE(A) maps any Ga(b)v,, to a (Ga (b)) = ,\(b). Now Proposition 10.14 implies that 7r maps the elements of S injectively to a subset of the basis 13(A). Therefore the linear independence assumption in 11.11 is satisfied. Now (2) shows that Lemma 11.11 and 11.11(2) imply the claims in a). We see also that 7r is injective on E = {x E Mn fl E(A)

fa(x) = x}. with a (b) = 0. We have Ga (b)va E M,, flE(A) Consider finally any b E 13(oo) with OGa(b)va = Ga(b)va, hence Ga(b)va E E. Furthermore, we have 7rGa(b)va = P,\(b) = 0. Since 7r is injective on E, this implies Ga(b)va = 0.

11.14. LEMMA. Let b E 8(oo)_,,. If a, a' E II are simple roots with ea(b) > 0 and ea' (b) > 0, then Ga (b) = Ga' (b).

PROOF. We have b = Fal Fat Fa,.T for some sequence al, a2, ... , a, of simple roots; set ,3 = a,. Write v = E-YEII m yy and choose a dominant weight A with (A,,3') = 0 and (A, y") > m.y for all y E II, -y # 3. Then Theorem 5.15 implies that pa induces a bijection

(U / U Fa)

L(A)a

We have Fpva = Fpv), = 0, hence y,\(b) = Fal

(1)

Fa,._lP,,(Fp) = 0. By Lemma

11.13 this yields Ga (b) va = 0 and Ga' (b) va = 0, hence G" (b), Gay (b) E U-FQ by (1).

There is by 11.8(3) an element b' E 13(oo)_ with T(b) = ±b'. We have now ±TG'(b) E Ez(oo)- [since TEz(oo) = Ez(oo), cf. 11.8] with ±7-G' (b) = ±T(b) = b'. Since T is an antiautomorphism, the result Ga(b) E U-F,3 from above implies ±TG'(b) E FQU-, hence b' E FF(E(oo)/gC(oo)) and thus ep(b') > 0 by 10.15(3). We have 0(±7-Ga(b)) = ±TG'(b) since 0 and T commute by Remark 3 in 11.9. Now the uniqueness part in Lemma 11.12.a (applied to 3 instead of a) implies that G'(b') = ±7-Ga(b). The same argument yields also GO (b') = ±TG''(b), hence G' (b) = Gay (b) and the claim.

252

CRYSTAL BASES III

11.

11.15 (Proof of Theorem 11.10). For each b E B(oo)_ there is an a E II with b E F 13(oo) [since v # 0], hence with ea(b) > 0. Set then G(b) = GI(b); this definition is by Lemma 11.14 independent of the choice of a. Then G(b) satisfies 11.10.c by 11.12(3).

We want to apply Lemma 11.11 to S = We have G(b) = b for all b, so the natural map 7r : G(oo) - G(oo)/gC(oc) maps the G(b) to a basis. Therefore the linear independence condition in 11.11 is satisfied. Now 11.3(8) and 11.12 imply

1:

F'«(Uz )-"+a. = 1:

(Uz)-v = aEII

Z[q,q-']Ga(b).

QE[I bE13(oo)_,,.ea (b)>0

So the definition of the G(b) implies

_

(Uz)

Z[q, q-1]G(b). bEB(oc)_

Now Lemma 11.11 together with 11.11(2) yields 11.10. We see also that 7r is injective

on {x E Gz(oo)_ I *(x) = x}. This then implies the uniqueness part in 11.10.a Consider now a dominant weight \ E A and let us prove 11.10.d. For any b E B(oo)- with ip),(b) = 0, pick a E II with e ,,(b) > 0. We get then G(b)v), = G'(b)v), = 0 by 11.13. So we get the last claim in 11.10.d. Furthermore, we see now that

E

(Uz )-ova =

Z[q,q-']G(b)vA.

bE 13(oo) _,, ,ipA (b) 00

Now the claims on bases in 11.10.d follow from Lemma 11.11 and 11.11(2) applied to ,C = G(A) and S = {b E B(oo)_ I ilp),(b) # 0}, cf. the argument in 11.13. REMARK. Note that 11.10.b implies { x E Gz(oo)

I

V) (X) = x } = Gz(oo)

nv)Gz(oo)_,,.

(1)

We can therefore replace in 11.10.a the condition "7PG(b) = G(b)" by "G(b) E That is the way how these elements are introduced in Gz (oo) - n i,Gz (oo) [Kashiwara 2].

11.16. Since we have G(b) = b and G(b)v), = ilpa(b) for all b E B(oo), the Nakayama lemma [applied to each G(oo)- and to G(,\)] shows that the G(b) with b E B(oo) are a basis of G(oo) over A, and that (for all dominant \) the G(b)v,\ with b E 8(oo) and G(b)v), # 0 are a basis of G(,\) over A. Any basis of UZ over Z[q, q-1] is also a basis of U- over k (since that is the field of fractions of Z[q,q-1]). The same holds for Lz(A) and L(.A). So Theorem 11.10 implies:

THEOREM. The G(b) with b E B(oo) are a basis of U- over k. For each dominant weight \ E A the map u H uvA, takes {G(b) b E B(oo), G(b)v), 0 0} I

bijectively to a basis of L(.A) over k.

A basis with this property was first discovered by Lusztig in the simply laced case, cf. [Lusztig 5]. We have described here Kashiwara's later work in [Kashiwara

11.

CRYSTAL BASES III

253

2]. It has several features in common with Lusztig's earlier work: Lusztig started

with a lattice G in U+ over Z[q-I]; he then got a basis of G/q-',C; these basis elements then have a unique representative in G that is fixed by V). These representatives are then a basis of U+ that satisfies the theorem above. Lusztig called this

basis the canonical basis of U+. He showed later that it coincides with the basis found by Kashiwara (after identifying U- and U+ via 0 o w). The construction by Kashiwara and his proofs work equally well for quantized enveloping algebras associated to arbitrary symmetrizable Cartan matrices, not just for those corresponding to finite dimensional Lie algebras. Independently of Kashiwara (and more or less at the same time) Lusztig extended in [Lusztig 6] his construction to arbitrary symmetric Cartan matrices and sketched in the final section of that paper an extension to the symmetrizable case. Grojnowski and Lusztig showed later that also in this case Kashiwara's and Lusztig's bases coincide.

In Chapter II of [L] you can find Lusztig's general construction. In Chapter III of [L] Lusztig describes his approach to Kashiwara's construction; it implies the equality of the two bases. Lusztig's approach (involving perverse sheaves) yields additional results that nobody has been able to prove using Kashiwara's elementary approach: Since UZ

is an algebra and since the G(b) are a basis of UZ over Z[q, q-I], each product G(b1)G(b2) with b1, b2 E 13(oo) can be written >2bEB(oo) abG(b) with ab E Z[q, q-I]

Now Lusztig's approach yields the stronger result that each ab is a polynomial in q and q-1 with nonnegative coefficients.

11.17. Let me describe as an example the case where 4D is of type A2. Set We have qq = qp. Set F.+p = TT(F) = F,3F. - q.FgFQ, cf.

II =

8.16(4), and F p = F+,/[n]- for all n > 0. Using the quantum Serre relations an induction proof shows for all r, s > 0 that min(r.s)

F(S)F'(T)

F(5-')l 1)(s-i)F,(r-i) ,(i) «+Q a

_ i=0

(1)

cf. [Lusztig 2], Lemma 1.6, or [L], 42.1.2(b). We have F1 = FjT) for all r > 0, 0 implies since r'(1) = 0. Now F(5)F(T)

=

E G(oo).

(2)

We can now apply 10.10(3) to (1), since r' (F(+) F(j)) = 0 for all i, j _> 0 [by F(s-a) with 0 < i < min(r, s) are in 8.26(6)]; it implies that all q(,(r-i)(s-i)F(x) a+a a G(oo). Given integers m, n > 0 we can apply this to r = m, s = m + n, and i = m; we get F(+)) F(n) E G(oo)

for all m, n > 0.

(3)

Using again r' (F(') a+3F(n)) 0 = 0 we get then also FFT)F(-) F(n) We claim next:

- FrF(-) F(n) E G(oo)

for all m, n, r > 0.

_ If r < s, then E,,Fps)F(T) E gL(oo).

(4)

(5)

254

CRYSTAL BASES III

11.

Well, (1) yields r-1

EaF(s)F(r) = j:q(r-i)(s-i)F(r-1-

F(s-i)

,(i)

a+Q 0

i=O

Here all exponents of qa are positive. By (4) all Far-1-Z)F(Z) F(s-t) are in C(oo); this implies (5). Of course (5) says that EaFQs)F(c,,r) = EaFQFcr, 1 = 0 in G(oo)/gr(oo). Since F« and F$ map B(oo) to itself, this implies If r < s, then FFFF 1 E 13(00)_(ra+s0) with ea(FQF« 1) = 0.

(6)

Now we can show:

F«-T3 Fai 1 I 0 < i < 0-

If r < s, then B(00)_(ra+sp)

(7)

Indeed, to start with all these elements are in 13(00)_(ra+s$) since they have the right weight and since Fa and F$ map B(oo) to itself. We get from (6) for each i that

ea(Fr-ZFQF 1) = r - i + ea(FQF 1) = r - i. This shows that the 1 are r + 1 distinct elements. Finally the Kostant partition function of ra + s,3 is r + 1 for r < s. So we have r + 1 = dim U (ra+s$) _ I13(00)-(ra+s3)I. This implies (7). We get by symmetry Ifs < r, then 13(0o)_(ra+sl3)

1

I 0 < i < s }.

(8)

So we have a description of each 13(00)_,,.

Now we can lift these basis elements to U. We claim:

If 0 < i < r < s, then G(F«-iFQF 1) = Fir-z)F(s)Fa(z) Well, we have FaF 1 =

F(s)F«Z)

-

9)

(mod gr(oo)) by (2), (1), and (3).

This implies Z>

F.r-'F'!3F'. 1 =

=

Fir-t)Fa(+aFas

Z)

(mod gr(oo)).

(10)

On the other hand,

F,(r-t)F(s)F(') - F(r-i) 1: q(ai-j)(s-j)F,(i-j)Fj)

F(s-j)

j=0

i-1

F,s-')F(') Far Z) +

q(i-j)(s-j)

j=0

rr -

j

1 F«r j)F«+aFas-j)

The Gaussian binomial coefficient in the sum is by 10.5(3) equal to qa (r-i)(i-j) q(i-j)(s-r+i-j) times a unit in A. When multiplied by q(i-j)(s-j) we get times a

CRYSTAL BASES III

11.

255

unit in A, hence an element in Aq since i > j and s > r. So the sum over j in our last equation is contained in gC(oo). This implies Fir-Z)Fas)FFZ) = FF5

_2)

(mod gC(0O)),

hence by (10) F.r

FAF. 1 - F.S

Z)Fe+/3Far-Z)

(mod gC(oo)).

So we see that F«s-Z)FQ+aFaT-2) E G(oo) and is a representative of Fa-iFaFa 1 E G(oo)/q,C(oo). This product is also in UZ [hence in £z(oo)], and it is fixed by z/i since each factor is fixed. So it satisfies the condition in 11.10.a, and we get (9). Now (9) and its analogue (with a and ,Q interchanged) imply G(13(o0) -(re+sp)

{

10 < i < r }, if r < s;

(s-i) (r) (i) {FQ Fa FQ

10
11.18. PROPOSITION. We have r13(oo) = 13(00) and TG(b) = G(b) for all b E 13(0o).

PROOF. We shall prove for all v E Z-P, v > 0 using induction on v that r13(oo)_v = 13(oo)_v and rG(b) = G(b) for all b E 13(0o)_v. For v = 0 this is obvious since r(1) = 1. So suppose that v > 0 and that the claim holds for all

v'
Let b E8(0o)_v. Since v # 0 there is a E _II with ea(b) > 0. Set n = ea(b) and b' = Ecnb. Then b' E 13(oo)_v+na with Eab' = 0 and b = Fib'. Write G(b') _ E U v+(n+j)a and r'(ub) ,-E 0 for all j. All u' are in £z(oo) by Lemma 11.2.b and 10.10(3). Now Eab' = 0 implies EaG(b') = Ej>oF(( j-')u' E gr(oo), hence u' E qC(oo) for all j > 0. So we have G(b') uo (mod qC(oo) fl F(,U-), in particular b' = uo. Set u = r(uo). We have then b = F b' = Fn U0 = hence r(b) = uFc(n). Choose a dominant weight A with (A, a") = 0 and with (A, Q") large enough for all ,Q E II, 0 ,6 a such that cpa induces an isomorphism

(U-/U-F.)-,+n.

L(A)a-v+na

We have G(b) V FaU- by 11.10.c since ea(b') > 0, hence G(rb') = rG(b') V U-F,,. This implies by our choice of A that G(Tb')va ,-E 0, hence by 11.10.d that spa (T Y) = G(T b')va ,-E 0, in fact G(T b')va E 13(A) by 10.14. Furthermore, we have

it - G(T b') (mod gr(oo) fl U-F,,), hence uva = G(T b')vA and uva E 13(A). Choose a second dominant weight µ such that (µ, a") > n. This condition implies F(,(n)vµ E 13(p). We have for all w E L(µ)

Fa(va ® w) _ (Fava) ® w + (Keva) 0 (Few)

=

A,av)0 0 w+ q

va 0 (Few) = va 0 (Fw),

11.

256

CRYSTAL BASES III

hence by induction F n)(va(&vµ) = vA®F(n)v,,. On the other hand, 9.13(5) implies So we get now L(A) ® L(µ)µ'.

uF(n)(va ®vµ) E uva ®F(n)vµ. ®

(1)

µ' <µ-na

By 11.8(3) there is e E {±1} with T(b) E e13(oo), hence with CPA+µ(ET(b)) = euF.( )va+µ. E B(A +,u) U {O}.

This implies by 10.7(4) [using the notation from 10.7] that euF(,n)(va ® v,4)

= iPa,µ(euFan)vA+N) E cPa µB(. +,u) U {0} C 13(A) ®13(µ) U {0}.

By (1) the projection (with respect to the weight space decompositions) of ®(G(µ)lgG(µ))µ_na is an element of euF(,n) (va (& vµ) to eB(A) ®13(µ) since vv, E B(A) and F,nivµ E 13(µ). This implies in particular that euF(( ,n) (v,\ ® vµ) 0 0, hence euF('('n)(va

® vµ) E 13(A) ® B(µ).

Since the elements in B(A) and 13(µ) are weight vectors, this implies in particular

that euF(n)(va (9 vµ) has only one nonzero projection into a tensor product of weight spaces and that this projection is in B(A) ® 8(µ). On the other hand we have found above a nonzero projection that is in eB(A) ®13(µ). Since there is only one nonzero projection, this projection is in (e13(A) (& B(µ)) fl (B(A) (& B(µ)). Since B(A) ® B(µ) is a basis this implies e = 1, hence T(b) E B(oo). The equality TG(b) = G(T(b)) follows now from TG(b) E Lz(oo) and TG(b) _ T(b) and ib TG(b) = TG(b), cf. the proof in 11.14.

11.19. We finally want to use the bases for quantized enveloping algebras and their representations to get bases for the original enveloping algebras and their representations. These bases should have the property from Theorem 11.16. Set n- equal to the direct sum of all root spaces go with !3 E 4?, /3 < 0. This is a Lie subalgebra of g, and we want to consider its enveloping algebra U(n-), which was the model for U-. Fix for all dominant weights A E A a (nonzero) highest weight vector v' in a simple 9-module with highest weight A.

Theorem: There is a basis B of U(n-) such that for all dominant weights A the uva with u E B and uva

0 are a basis of the simple g-module with highest weight

A.

Proof : Recall the description of U(g) by generators and relations in the introductory

remarks to Section 4. So we consider in g root vectors x, E g, and y,, E g-,,, for all a E II such that each hQ _ [xe, y.] satisfies a(h(,,) = 2. Then U(g) is generated by all xQ, yam, h(,, with the relations mentioned in those remarks.

The algebra U(n-) is now the algebra with generators yc. (a E II) and relations (ad y(,')1-a°' yo = 0 (with a,,3 E II, a : /3) where aao = (,3,a'). This

11.

CRYSTAL BASES III

257

can be proved in the same way as the corresponding results in Theorem 4.21.b/c. Alternatively, see [Bourbaki 3], Ch. VIII, §4, n° 4, Prop. 4 (ii). Regard C as a Z[q,q-1]-algebra via the ring homomorphism Z[q,q-1] --+ C that takes q to 1. (Recall a similar construction in 5.13 with Q instead of Z.) Set now

Uc = UZ

C.

(1)

This is an associative algebra over C with basis {G(b) ®1 b E 13(oo)} (by 11.10).

The description of U(n-) by generators and relations implies that there is a unique algebra homomorphism

K: U(n-) -' Uj

(2)

with K(y0,) = F,, ® 1 for all a E II. Here the Fc. ® 1 satisfy the required relations since the F0, satisfy (R6), cf. the proof of 5.13(8). We want to show first that i is an isomorphism. We have ic(ya/n!) = F,(,n) ®1 (for all a and all n > 0) since F ®1 = ([n]aFini) ®1 = F,( n) ®n!, cf. 0.3. Because UZ is generated as an algebra by all Fin), this implies that i is surjective. It is clear that i maps any weight space U(n-)_µ (with p E ZIP, p > 0) onto (UZ) _N, ® Z. Both spaces have the same dimension (given by Kostant's partition function at p). Therefore is is injective on each weight space, hence on all of U(n-). So r. is an isomorphism. Set now 13={ub=K;-1(G(b)(9 1) bEC3(oo)}. (3)

Our results so far tell us that 13 is a basis of U(n-). We want to show now that it has the property claimed in the theorem. Consider for each dominant weight \ E A the tensor product Lc(A) = LZ(A) ®Z[q,q-1] C.

(4)

Since each Ka acts on the weight spaces of LZ(A) as multiplication by a q-power, the action of K0, ® 1 on Lc(A) is trivial. Arguing as in 5.13 we see that we can make Lc(A) into a g-module such that each x0, acts as E0, ® 1, each y , as F,,, ® 1, and each h0, as [K,,;0] ® 1. The induced action of U(n-) can also be constructed

via , and the obvious action of Uc = Uj ® C on Lc (,\) = LZ(A) ® C. I claim that Lc(A) considered as a g-module is simple with highest weight A. Well, one sees easily that any Lz (,\),, ® C is contained in the p weight space of the

g-module Lc(A). In fact, we get here equality, since LZ(A) is the direct sum of all LZ(A)µ. We see in particular that v), ® 1 is a weight vector of weight A. Since E,,v), = 0 implies x(v), ®1) = 0 this shows that v), ®1 generates a simple g-module of highest weight A. Now this g-module is all of Lc(A) since LZ(A) = UZ v) hence Lc (,\) = (UZ v),) ® C = Uj (v), (3 1) = U(n )(v), (9 1). So Lc(A) is simple. The G(b)v.>, with b E 13(oo) and G(b)v), # 0 are a basis of LZ(A) over Z[q,q-1] Therefore the corresponding (G(b)v),) ® 1 are a basis of Lc(A) over C. Our construction and (3) imply that ub(v), ® 1) = (G(b)v),) (9 1 for all b. This shows that ub(v), ® 1) = 0 whenever G(b)v), = 0 and that the ub(v), ® 1) with G(b)v), # 0 are a basis of Lc(A). So the basis 13 from (3) has indeed the desired property.

References I refer to the books and papers by the names of the author(s) plus (where appropriate) a number. There are two exceptions: [H] = [Humphreys 1] and [L] _ [Lusztig 7].

E. Abe Hopf Algebras (Cambridge Tracts in Mathematics), Cambridge etc. 1980 (Cambridge Univ.) H. H. Andersen, P. Polo, Wen K. Representations of quantum algebras, Invent. math. 104 (1991), 1-59 N. Bourbaki 1) Algebre, Chap. 2, Paris (3e ed.) 1967 (Hermann) 2) Groupes et algebres de Lie, Chap. 4, 5 et 6, Paris 1968 (Hermann) 3) Groupes et algebres de Lie, Chap. 7 et 8, Paris 1975 (Hermann)

V. Chari, A. N. Pressley A Guide to Quantum Groups, Cambridge etc. 1994 (Cambridge Univ.) C. De Concini, C. Procesi

Quantum Groups, pp. 31-140 in: L. Boutet de Monvel et al., D-modules, Representation Theory, and Quantum Groups, Proc. Venezia 1992 (Lecture Notes in Mathematics 1565), Berlin etc. 1993 (Springer) C. De Concini, V. G. Kac Representations of quantum groups at roots of 1, pp. 471-506 in: A. Connes et al. (eds.), Operator Algebras, Unitary Representations, Enveloping Algebras, and Invariant Theory (Colloque Dixmier), Proc. Paris 1989 (Progress in Mathematics 92), Boston etc. 1990 (Birkhauser) C. De Concini, V. G. Kac, C. Procesi Some quantum analogues of solvable Lie groups, preprint P. Deligne, J. S. Milne Tannakian categories, pp. 101-228 in: P. Deligne, J. S. Milne, A. Ogus, K.y. Shih: Hodge Qvcles, Motives, and Shimura Varieties (Lecture Notes in Mathematics 900), Berlin etc. 1981 (Springer) M. Demazure, P. Gabriel Groupes Algebriques. Tome I, Paris & Amsterdam 1970 (Masson & NorthHolland)

259

REFERENCES

260

J. Dixmier Algebres Enveloppantes, Paris etc. 1974 (Gauthier-Villars) V. G. Drinfel'd 1) Hopf algebras and the quantum Yang-Baxter equation, Soviet Math. Doklady 32 (1985), 254-258, (russ. orig.:) A.nre6pu H KBaHTOBOe ypaBHeHIIe Slara-BaxcTepa, .Zlox.n. AxaR. HayK CCCP 283 (1985), 1060-1064

2) Quantum groups, pp. 798-820 in: A. M. Gleason (ed.), Proceedings of the International Congress of Mathematicians 1986, vol. 1, Proc. Berkeley 1986, [Providence, R. 1., 1987 (Amer. Math. Soc.)] J. E. Humphreys 1) Introduction to Lie Algebras and Representation Theory (Graduate Texts in Mathematics 9), New York etc. (3rd printing) 1980 (Springer) 2) Reflection Groups and Coxeter Groups (Cambridge Studies in Advanced Mathematics 29), Cambridge 1990 (Cambridge Univ.) N. Jacobson Lie Algebras (Interscience Tracts in Pure and Applied Mathematics 10), New York etc. (3rd printing) 1966 (Interscience/Wiley) M. Jimbo 1) A q-difference analogue of U(g) and the Yang-Baxter equation, Lett. Math. Phys. 10 (1985), 63-69 2) A q-difference analogue of U(gr(N + 1)), Hecke algebra and the Yang-Baxter equation, Lett. Math. Phys. 11 (1986), 247-252

A. Joseph Quantum Groups and Their Primitive Ideals, Ergebnisse der Mathematik (3) 29, Berlin etc. 1995 (Springer)

A. Joseph, G. Letzter 1) Local finiteness of the adjoint action for quantized enveloping algebras, J. Algebra 153 (1992), 289-318 2) Rosso's form and quantized Kac-Moody algebras, preprint M. Kashiwara 1) Crystalizing the q-analogue of universal enveloping algebras, Comm. Math. Phys. 133 (1990), 249-260 2) On crystal bases of the q-analogue of universal enveloping algebras, Duke Math. J. 63 (1991), 465-516 3) The crystal base and Littelmann's refined Demazure character formula, Duke Math. J. 71 (1993), 839-858

M. Kashiwara, T. Nakashima Crystal graphs for representations of the q-analogue of classical Lie algebras, J. Algebra 165 (1994), 295-345 C. Kassel

Quantum Groups, Graduate Texts in Mathematics 155, New York etc. 1995 (Springer)

A. N. Kirillov, N. Reshetikhin q-Weyl group and a multiplicative formula for universal R-matrices, Comm. Math. Phys. 134 (1990), 421-431

REFERENCES

261

S. Z. Levendorskii, Ya. S. Soibelman Some applications of the quantum Weyl groups, J. Geom. Phys. 7 (1990), 241254

G. Lusztig 1) Quantum deformations of certain simple modules over enveloping algebras, Adv. in Math. 70 (1988), 237-249 2) Finite dimensional Hopf algebras arising from quantized universal enveloping algebras, J. Amer. Math. Soc. 3 (1990), 257-296 3) On quantum groups, J. Algebra 131 (1990), 466-475 4) Quantum groups at roots of 1, Geom. Dedicata 35 (1990), 89-114 5) Canonical bases arising from quantized enveloping algebras, J. Amer. Math. Soc. 3 (1990), 447-498 6) Quivers, perverse sheaves, and quantized enveloping algebras, J. Amer. Math. Soc. 4 (1991), 365-421 7) Introduction to Quantum Groups (Progress in Mathematics 110), Boston etc., 1993 (Birkhauser)

I. G. Macdonald Symmetric Functions and Hall Polynomials, Oxford 1979 (Clarendon, Oxford Univ.)

Yu. I. Manin Quantum Groups and Non-commutative Geometry, Montreal 1988 (CRM, Univ. de Montreal) T. Nakashima Crystal base and a generalization of the Littlewood-Richardson rule for the classical Lie algebras, Comm. Math. Phys. 154 (1993), 215-243 M. Rosso

Analogues de la forme de Killing et du theoreme d'Harish-Chandra pour les groupes quantiques, Ann. scient. Ec. Norm. Sup. (4) 23 (1990), 445-467 A. N. Rudakov, I. R. Shafarevich Irreducible representations of a simple three-dimensional Lie algebra over a field of finite characteristic, Math. Notes Acad. Sci. USSR 2 (1967), 760-767, (russ. orig.:) HenpHBO,z Mble npe,ucTasneHHB npocTOH TpexMepxoH anre6pbl JIH Haz rloJneM KOHe11HOiiI xapa}TepHCTHKH, MaTeM. 3aMeTKH 2 (1967), 439-454,

N. Saavedra Rivano Categories Tannakiens (Lecture Notes in Mathematics 265), Berlin etc. 1972 (Springer)

S. P. Smith Quantum groups. An introduction and survey for ring theorists, pp. 131-178 in: S. Montgomery, L. Small (eds.), Noncommutative rings, Proc. Berkeley 1989 (MSRI Publications 24), New York etc. 1992 (Springer) R. Steinberg

Lectures on Chevalley Groups, New Haven, Conn. 1968 (Yale Univ.) M. E. Sweedler Hopf Algebras, New York 1969 (Benjamin)

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T. Tanisaki Killing forms, Harish-Chandra isomorphisms, and universal R-matrices for quantum algebras, pp. 941-961 in: A. Tsuchiya, T. Eguchi, M. Jimbo (eds.), Infinite Analysis, Part B, Proc. Kyoto 1991 (Advanced Series in Mathematical Physics 16), River Edge, N. J., 1992 (World Scientific)

Notations We use the standard notations Z for the ring of integers, Q for the field of rational numbers, and C for the field of complex numbers without extra explanation

roman letters

T , T' , wT wTr

8.2

T.

8.6, 8.14

1 1

Ta, WT"" wTa

8.6

8 2

T.

8.18

U

1.2, 4.6

U+, U-, U° U°

6.6

Ua

4.8 9.0

0-

4 6

C E

2 7

E(T)

E.

.

.

.

4 3 .

Ear) Ea

8.6

El

4.12

F

1.1

9.2, 9.4, 10.2

F(*)

8 2

F.

4.3

F(r)

F. F1

G(b) Ga (b)

Homk(M, N)

HW(S) K K. Ka

.

,

,

.

U+ [W], U [w]

8.24

Uq

4 3

(

9)

8.6

.

Uq (9), Uq (9), U°(9)

4.4

Uq(9)

4.3

9.2, 9.4 , 10.2

4.12 11.10, 11.15 11.12 3.10 9.8 1.1

4.3

L(n, +), L(n, -)

2.6 5.5

L(a)

5.9

Lz(.\) M (A)

_

11.4 2.4, 5.5

Mk' (c)

4.15

P

3.8. 7.1

R

3.18. 7.7

R13

7.6

S S*

36 4 8 .

.

7.11 9.13

.

Uq(5(2)

1.1

Uqa (s (2)

4.4

Uz i Uz , Uz, UZ

11.1

W

4

Z

2.19

Z(U)

6.3

.

1

Zb(A)

2.11

aaQ

4.3

ad(a)

4.18

'ad (x)

8.11

ch (M)

5A.8

da

4.1

ea (b)

fa (b)

9.17 9.17

ht (µ)

4.13

4.4

L(A)

S'

1.6 4.6

k

1.1

kq [G]

7.11

q

1.1

4a

4.2

ra, ra

6.14, 6.15, 8.26

263

NOTATIONS

264

rna

ri,,P, r/ s, trq

type 1 type or

uqa, u; v, WO

wt I

8A.4 8A.7

3.4, 4.8

4.1

3.10 3.19, 5.2 5.2 4.10 5.5,9.5 8.24 4.12

fraktur letters 4.1

g

6.4

rya

7.11

eo

5.3

2.16, 6.2, 11.12

8A.10

lrQ

u

4.1

p

4.9, 5A.8

T TI

1.2, 4.6

X(A)

5A.8

Wa

10.3 11 . 9

9.20

calligraphic letters

1.2, 4.6

9.5

13(A) 1i(oo)

10.3

G(A) G(oo) GZ(oo), GZ(A)

9.5

P(µ)

10.3 11.6 5.19

brackets, etc.

I

0.1

[n] 1.

4.2

[a]

[n]

greek letters

4.2 0.1

4 2 .

0.1

!

3.1, 4.8

[K; a]

1.3

7.11 9.13

[K.; a]

4.4

E) = OM,M, On

3.12, 7.2 3.11

( (A, av)

4.1, 6.10

Oµ

7.1

(u, v)

Of

6.20

3.13

O

3.1 7, 7.5

My

A

4.1

M*

A

3.13

ma

II

4.1

IIs

5A.2

W0, °s, WE

K

)

;a]

11.1 4.1

exponents 3.5

3.9 5.3 ,

5.2 3.8, 7.2

indices

4)

4.1

4is

5A.2 5A.2

MA

2.2, 5.4

ao

MA

9.3

1'i

1.6

MA,o

5.1

Index This index contains mainly references to definitions and main results. In some cases (such as "semisimplicity of U-modules") you will be directed to subsections containing results on the topic, but where the phrase from the index does not occur explicitly in that form. adjoint representation 4.18 admissible lattice 9.3 antipode 3.6/7, 3.8, 9.13 augmentation 3.5 braid relations

7.6, 8.15 7.6, 8.15

canonical basis

11.16

center of U

2.17, 2.10, 6.25

character formula Clebsch-Gordan formula coassociative cocommutative comultiplication counit crystal base crystal graph divided powers dominant short root dual space o f a module elementary move fixed points formal caharcters fundamental weights Gaussian binomial

5.15

braid group

coefficient

hexagon identities Hom space of modules Hopf algebra

invariant form on U isotypic component Kostant's partition function largest short root

3.8

3.1/2, 3.8, 9.13 3.4/5. 3.8 9.4, 10.16 9.28

modules of type 1 mod-,rl -invariant pairing of UFO and

8.2, 8.6, 11.1

5A.2, 5A.9, 11.4

U?O 3.9, 5.3, 5.16

8.18

5A.8 4.1

0.1. 0.4. 8.1. 10.5

6.4. 6.6. 6.25

6.20 9.10 5.19

5A.2, 5A.9, 9.6, 5.7

6.22, 7.10

5A.1, 5A.9, 9.6, 9.21, 11.4 3.19, 5.2 9.20 6.12, 6.18, 8.28-30 1.5, 8.24 9.23, 10.16, 11.7

PBW type bases polarization quantized enveloping algebra 1.1, 4.3/4 quantum determinant 7.13 quantum trace 3.10, 5.3 quantum Yang-Baxter equation 3.17, 7.5/6 reduced expression 8.18 R-matrices 7.13, 8.30

3.5

Harish-Chandra homomorphism Hecke algebra

3.8, 4.8, 4.11,

9.21, 11.4

locally nilpotent matrix coefficients minuscule dominant weights

3.2

1.9. 4.7. 6.1

3.10, 5.3 7.11

5A.8, 9.14

grading of U

3.18/19, 7.8

7.7

265

INDEX

266

scalar product of

roots semisimplicity of quad U-modules simple U-modules tensor categories tensor products of U-modules

4.1

2.9, 5.17 2.6, 2.13, 5.10 3.19 3.3, 3.14, 5.3,

5A.8/9, 7.3, 9.13

trivial module twisted comultiplication twisted modules

3.5, 5.3

3.8, 7.2

2.13, 5.2, 5.11,

8.2, 11.9

type of a U-module universal highest weight module weight lattice

5.2

weight space weights of a module

2.2/3, 5.1 2.2/3

zero divisors

1.8, 8.25

Z-form

11.1

2.4, 5.5 4.1