Infinite-dimensional Lie algebras

Lecture course by Iain Gordon

Edinburgh, 2008/9

Contents

1 Introduction 1

2 Central extensions 2

3 The Virasoro algebra 3

4 The Heisenberg algebra 4

5 Representations of the Virasoro algebra 5 5.1 Family 1: Chargeless representations ...... 5 5.2 Family 2: Highest-weight representations ...... 7

6 Enveloping algebras 8

7 The universal highest-weight representations of Vir 9

8 Irreducibilty and unitarity of Virasoro representations 10

9 A little infinite-dimensional surprise 12 9.1 Fermionic Fock ...... 12 9.2 Central extensions of gl ...... 13 ∞ 10 Hands-on loop and affine algebras 14

11 Simple Lie algebras 16

12 Kac-Moody Lie algebras 17 12.1 Step I: Realisations ...... 17 12.2 Step II: A big Lie algebra ...... 19 12.3 Step III: A smaller Lie algebra ...... 20

13 Classification of generalised Cartan matrices 21 13.1 Types of generalised Cartan matrices ...... 21 13.2 Symmetric generalised Cartan matrices ...... 22

14 Dynkin diagrams 26

15 Forms, Weyl groups and roots 27

16 Root spaces 32

17 Affine Lie algebras and Kac-Moody Lie algebras 34

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18 The Weyl-Kac formula 37 18.1 Category O ...... 37 18.2 Kac’s Casimir ...... 39 18.3 The Weyl-Kac formula ...... 40

19 W − K + A = M 43 19.1 Affine Weyl groups ...... 43 19.2 Formulae ...... 44

20 KP hierarchy and Lie theory 45

21 The Kazhdan-Lusztig Conjecture 48 21.1 The Shapovalov form and Kac-Kazhdan formulas ...... 48 21.2 Bringing in the Weyl group ...... 51 21.3 The Kazhdan-Lusztig Conjecture ...... 52

A References 53

B Notational reference 53

C Generation of Lie algebras 54

1 Introduction

Lie algebras may arise in the following ways in the wild:

• Derivations of an associative algebra.

• Vector fields on a smooth manifold.

• Tangent spaces at the identity of Lie groups.

We should often think of a Lie algebra of being one of those (intimately related) concepts.

1 Example (Witt algebra over C). Let A C[z,z− ] be the algebra of Laurent polynomials in one = variable and consider

n d £ 1¤o n j 1 d o Der(A) f (z) : f C z,z− span L j : z + : j Z . = dz ∈ = = − dz ∈

We compute the structure of Der(A):

£ ¤ h m 1 d n 1 d i Lm,Ln f z + , z + f = − dz − dz ¡ m n 1 m n 2 ¢ ¡ m n 1 m n 2 ¢ (n 1)z + + f 0 z + + f 00 (m 1)z + + f 0 z + + f 00 = + + − + + m n 1 df (n m)z + + (m n)Lm n f = − dz = − +

This is called the Witt algebra over C, denoted Witt; its basis is ©L : j Zª and its structure is j ∈ £ ¤ Lm,Ln (m n)Lm n . = − +

2 Infinite-dimensional Lie algebras 3

Example (Vector fields on S1). Consider the complex vector fields on S1 ©eiθ : θ R± ª: = ∈ ∼

n d 1 o Vectfin : f : f : S C smooth with finite Fourier expansion , = dθ →

d d i.e. we impose that each element is in the finite span of cos(nθ) dθ and sin(nθ) dθ , or equivalently in the finite span of ©einθ d : n Zª. Without the finiteness assumption, we would have to deal dθ ∈ with the Fourier analysis and convergence issues. Thus we have a basis

n inθ d o Ln : ie : n Z . = dθ ∈

iθ n 1 d Set z : e , then L z + , and we recover the Witt algebra. = n = − dz Example (Diffeomorphisms of S1). Let G : Diff ¡S1¢ be the group of (orientation-preserving) 1 = + ¡ 1 ¢ ¡ 1 ¢ diffeomorphisms of S under composition. It acts on C ∞ S ,C via (g.f )(z) : f g − (z) for = g G. An element of the tangent space at 1 G has the form ∈ ∈

¡ ¢ X∞ n 1 g(z) z 1 ²(z) z λn²z + . = + = + n =−∞ Then 1 ¡ ¢ X∞ n 1 g − (z) z 1 ²(z) z λn²z + . = − = − n =−∞ Thus

¡ ¢ ¡ ¢ ¡ X∞ n 1¢ g.f (z) f z(1 ²(z)) f z λn²z + = − = − n =−∞ ³ X∞ n 1 d ´ ¡ X∞ ¢ 1 λn²z + f (z) 1 λn²Ln f (z). = − n dz = + n =−∞ =−∞

The Ln for a topological basis for Lie(G).

The three preceding examples all give the same Lie algebra structure. The Witt algebra has two further properties:

1. There is an anti-linear Lie algebra involution ω(Ln) : L n. = − 2. Thence we can build a real form of the Witt algebra as ©x Witt : ω(x) xª. ∈ = −

2 Central extensions

Definition 2.1. A central extension of a Lie algebra g is a short exact sequence of Lie algebras

0 z g g 0 (2.1) −→ −→ b −→ −→ such that z Z ¡g¢ ©x g :[x,g] 0ª. ⊆ b = ∈ b b =

As vector spaces the sequence (2.1) splits as g g0 z, say via a section σ: g g0. Let b = ⊕ → β: g g z ,(x, y) £σ(x),σ(y)¤ σ¡[x, y]¢ . × → 7→ − We check that β satisfies

• β(x, y) β(y,x), and = −

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• β¡x,[y,z]¢ β¡y,[z,x]¢ β¡z,[x, y]¢ 0. + + = Conversely, given β C 2¡g;z¢ : ©β: g g z : satisfying the aboveª, we can construct ∈ = × → £ ¤ ¡ ¢ g : g z as vector spaces, (g,z),(g 0,z0) : [g,g 0],β(g,g 0) . bβ = ⊕ = Then bgβ is a central extension. However, there exist isomorphisms of such bgβ: z g g z g −−−−→ bβ = ⊕ −−−−→ °  ° ° θ ° ° y=∼ ° z g g z g −−−−→ bγ = ⊕ −−−−→ ¡ ¢ ¡ ¢ We find quickly that θ : bgβ bgγ is of the form θ (g,z) g,θ2(g) z , where θ2 : g z; and for θ → =¡ ¢ + → to be a Lie algebra homomorphism, we require that θ [g,g 0] γ(g,g 0) β(g,g 0). 2 = − So we find that central extensions are classified by C 2¡g;z¢ H 2¡g;z¢ , = dB 1¡g;z¢ 1 © ª ¡ ¢ where B θ : g z , and dθ(g,g 0) θ [g,g 0] . = → = Exercise 2.2. Show that if g is a finite-dimensional simple Lie algebra, then H 2¡g;C¢ 0. = Proposition 2.3. H 2¡Witt¢ : H 2¡Witt;C¢ C. In other words, there is a one-dimensional space = =∼ of central extensions of the Witt algebra. Proof. Let β: Witt Witt C,(L ,L ) β(m,n) satisfy the two conditions from above, namely × → m n 7→ • β(m,n) β(n,m), and = − • (m n)β(l,m n) (n l)β(m,l n) (l m)β(n,l m) 0. − + + − + + − + = Freedom of choice is provided by dθ for some θ : Witt C , L θ(m). → m 7→ Writing θ (m) : β(0,m)±m for m 0, replace β by β dθ : We see that w.l.o.g. β(0,m) 0 β = 6= + β = for all m Z. Now with l 0, we get nβ(m,n) mβ(n,m) 0, i.e. (m n)β(m,n) 0. Thus ∈ = − = + = β(m,n) δm, nβb(m), where βb(m) βb( m). = − = − − With l m n 0, we produce a relation which for n 1 gives + + = = (1 m)βb(m 1) (m 2)βb(m) (2m 1)βb(1) 0 . − + + + − + = This recurrence relation has a two-dimensional solution space: 3 βb(m) α1m α2m , with α1,α2 C . = + ∈ α1 3 Now replace β with β dθβ, where θβ(m) δm,0 , to get β(m,n) δm, nα2m . + = 2 = −

3 The Virasoro algebra

Definition 3.1. The Virasoro algebra (denoted by Vir) is the central extension of the Witt algebra 1 ¡ ¢ with the choice βb(m) m3 m in the notation of Proposition 2.3. = 12 − The Virasoro Algebra is spanned by ©L : m Zª {c} such that c is central, i.e. [c,L ] 0 m ∈ ∪ m = for all m Z, and ∈ £ ¤ 1 ¡ 3 ¢ Lm,Ln (m n)Lm n δm, n m m c . = − + + − 12 − The number 1/12 is chosen so that c acts as the scalar 1 on a natural irreducible representation that we are about to see.

4 Infinite-dimensional Lie algebras 5

4 The Heisenberg algebra

The Witt algebra was a simple point of departure for infinite dimensional Lie algebras. Another simple case is given by constructing loops on a (1-dimensional) abelian Lie algebra, and again making a central extension. This leads to © ª Definition 4.1. The Heisenberg algebra H is spanned by an : n Z {h}, where h is central £ ¤ ∈ ∪ and am,an mδm, nh. (We could even rescale to get rid of m.) = −

This algebra has a representation as an algebra of operators on B(µ,h) : C[x1,x2,...], on ¡ ¢ = which the H -module structure is given by ρ : H gl B(µ,h) as follows: →

 ∂  n 0 ,  ∂xn > ρ : an ρ(an) hnx n n 0 , and ρ : h ρ(h) ( h) . (4.1) 7→ = − − < 7→ = − × µ n 0 , = Exercise 4.2. If h 0, then B(µ,h) is irreducible, while if h 0 it is not (e.g. the subspace of 6= = constants C B(µ,h) is invariant). <

There exists an involution ω(an) a n, ω(h) h. = − =

Link to Virasoro algebras. Define

1 X Lk : : a j a j k : for k Z . = 2 j Z − + ∈ ∈

The notation “: a j a j k :” means “normal ordering”, i.e. − + ( a j a j k if j j k, : a j a j k : : − + − ≤ + − + = a j k a j if j j k. + − − ≥ + Explicitly, we have  X 1 a2 a a if k 2n,  2 n n j n j  + j 0 − + = L > k = X  an 1 j an j if k 2n 1. j 0 + − + = + >

The operator Lk is an infinite sum, but it is well-defined when acting on any element of B(µ,h) because of Property (2) below: On B(µ,h) we have that

1. a0 and h act diagonally,

2. on a given element of B(µ,h) a acts as zero for n 0, and n À 3. a acts locally nilpotently for n 0. n >

Theorem 4.3. With Lk defined as above,

£ ¤ 1 ¡ 3 ¢ Lm,Ln (m n)Lm n δm, n m m = − + + − 12 − as operators on B(µ,1), i.e. as a representation of the Virasoro algebra. (But not of the Witt algebra!)

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Sketch of proof. Define a function

( 1 if x 1, ψ: R R , x ψ(x) : | | ≤ → 7→ = 0 if x 1, | | > 1 P and for ² 0, define Lk (²) : 2 j Z : a j a j k : ψ(²j). This has only finitely many non-zero > = ∈ − + £ ¤ terms, and Lk (²) Lk as ² 0. Now calculate: an,Lk nak n (using Lk (²)), and hence → → = +

£ ¤ 1 X Lm(²),Ln (j m): a j a j m n : ψ(²j) = 2 j Z + − + + + ∈ m 1 X 1 −X : an j a j m : ψ(²j) δm, n j(j m)ψ(²j), 2 j Z − + − − 2 j 1 + ∈ =− and let ² 0. →

Exercise 4.4. Is B(µ,1) irreducible as a Virasoro module?

5 Representations of the Virasoro algebra

Recall that we have an anti-linear involution ω(Ln) L n, ω(c) c. = − = Definition 5.1 (Unitarity). A representation ρ : Vir gl(V ) of the Virasoro algebra is unitary if → there exists a non-degenerate hermitian form , on V such that 〈− −〉 1. v,v 0 for all v V , with equality if and only if v 0, and 〈 〉 ≥ ∈ = 2. ρ(x).v ,v v ,ρ(ω(x)).v for all x Vir, v ,v V , i.e. x† ω(x). 〈 1 2〉 = 〈 1 2〉 ∈ 1 2 ∈ =

5.1 Family 1: Chargeless representations

Let α β 1 © ª V Vα,β P(z)z (dz) C[z,z− ] span vn : n Z , (5.1) = = =∼ = ∈ 1 where α,β C and P C[z,z− ]. Define a representation of the Virasoro algebra by the following ∈ ∈ action: ¡ ¢ c 0 and Ln(vk ) k α β(n 1) vn k . (5.2) 7→ = − + + + + n 1 d This is the representation one would discover by using the definition Ln : z + dz on the k α β = − elements v : z + (dz) . k = Now we start to use the structure of Vir. Set

© ª h : Cc CL0 (abelian); Vir : span Lk : k Z . = ⊕ ± = ∈ ± We have the so-called triangular decomposition

Vir Vir h Vir . (5.3) = − ⊕ ⊕ + Exercise 5.2. Check that h Cc CL is a Cartan subalgebra of Vir, i.e. that it is nilpotent and = ⊕ 0 self-normalising.

6 Infinite-dimensional Lie algebras 7

Remark 5.3. In the representation V , h acts by scalars on each v : L v (k α β)v . α,β k 0 k = − + + k For a general representation ρ : Vir gl(V ), the decomposition → M V Vλ = λ h ∈ ∗ is called a weight space decomposition, where V : ©v V : ρ(z).v λ(z)v for all z hª. Obvi- λ = ∈ = ∈ ously, the existence of a weight space decomposition is a restriction on the representation.

L Lemma 5.4. Let V be a representation of an abelian Lie algebra A, such that V λ A Vλ = ∈ ∗ and such that each V is finite-dimensional. Then any subrepresentation U V decomposes as λ ≤ U L ¡U V ¢. = λ ∩ λ Proof. Any v V can be written as v Pm v with v V for j 1,...,m, i.e. a.v λ (a)v ∈ = j 1 j j ∈ λj = j = j j for all a A. Suppose v U. We need to show= that v U for each j. Since λ λ for j k, we ∈ ∈ j ∈ j 6= k 6= can find a A such that λ (a) λ (a) if j k. Then ∈ j 6= k 6= v v v = 1 + ··· + m a.v λ (a)v λ (a)v = 1 1 + ··· + m m . . m 1 m 1 m 1 a − .v λ (a) − v λ (a) − v . = 1 1 + ··· + m m j £ j ¤ T m Since each a .v U, we see that λi (a) (v1,...,vm) U , where 1 i m and 0 j m 1. £ ∈ j ¤ ∈ ≤ ≤ ≤ ≤ − But the matrix λi (a) is a Vandermonde matrix and thus, since the λi (a) are pairwise distinct, it has non-zero determinant, and hence (v ,...,v )T U m, as required. 1 m ∈ Theorem 5.5.

1. V is reducible if and only if either α Z and β 0, or α Z and β 1. In those cases, α,β ∈ = ∈ = Vα,β has a composition series whose irreducible sections are a trivial representation and

an irreducible representation Vα0 ,β of codimension one in Vα,β. (As vector spaces we may © ª © ª consider these to be span v (α β) and span vk : k α β , respectively.) − + 6= − − (If V is irreducible, then set V 0 V .) α,β α,β = α,β

2. V 0 is unitary if and only if β β 1 and α β α β. α,β + = + = + Proof. Part (1): By Lemma 5.4, any subrepresentation U decomposes as a direct sum of its weight spaces, so if U is a non-zero subrepresentation, then there exists k with v U. Now k ∈ given v U, we have L (v ) U for all n, and so we get scalar conditions from Equation (5.2), k ∈ n k ∈ i.e.

vn k U k α β(n 1) 0 . + 6∈ ⇒ + + + =

If vn k ,vm k U, then β 0 and α k, and hence Cvk is a subrepresentation, and ± + + 6∈ = = − Vα,β Cvk is irreducible (because Ln(vt ) (t k)vt n for t k). = − + 6= Now assume that only one vector, say vm, does not belong to U. Then for l k,m we must © 6= ª have Ln k (vk ) 0 Lm l (vl ), whence β 1, a (m 1) and U span vt : t m . − = = − = = − + = 6=

Part (2): Unitarity implies v ,v 0 for all relevant k. Hence 〈 k k 〉 > • L (v ),v v ,L (v ) , whence α β R, and 〈 0 k k 〉 = 〈 k 0 k 〉 + ∈

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• L 1(vk ),vk 1 vk ,L1(vk 1) and L 2(vk ),vk 2 vk ,L2(vk 2) . Using this to produce 〈 − − 〉 = 〈 − 〉 〈 − − 〉 = 〈 − 〉 two different looking relations between vk ,vk and vk 2,vk 2 for infinitely many differ- 〈 〉 〈 − − 〉 ent integers k, we find an equality of polynomials

¡X β¢¡X (1 β)¢¡X 2(1 β)¢ ¡X (1 β)¢¡X (2 β)¢¡X 2β¢ , − − + − + = − − − − − whence β β 1. + =

Conversely, v ,v δ is a well-defined positive-definite hermitian form on V 0 because of 〈 k m〉 = km α,β the conditions on α, β.

Exercise 5.6. Is V completely reducible? That is, if it is not irreducible, is V C V 0 a α,β α,β = ⊕ α,β decomposition of Lie algebra representations?

5.2 Family 2: Highest-weight representations

The Virasoro algebra has a representation on B(µ,1) C[x ,x ,...] as in Equation (4.1) above. = 1 2 Recall that 2 1 2 X µ X L0 : a0 a j a j a j a j . = 2 + j 0 − = 2 + j 0 − > > Q ni On a monomial i xi , the term a j a j acts by multiplication by jn j , and so we see that P − j 0 a j a j picks out the “degree” of a homogeneous polynomial, where deg(x j ) j, i.e. > − = X B(µ,1) B(µ,1)λ , = λ h ∈ ∗ where

( 2 homogeneous polynomials of degree N λ(c) 1, λ(L0) µ /2 N , B(µ,1)λ = = + = 0 otherwise.

Observe that B(µ,1) for µ R is unitary, and in fact it is induced from a unitary representation ∈ of the Heisenberg algebra H . The monomials form an orthonormal basis:

Qn k ! D k k k k E j 1 j 1 n 1 n = x1 xn , x1 xn . ··· ··· = Qn 2k j j 1 j = (This just says that P,Q is the constant coefficient of ω(P)Q.1, where P and Q are polynomials 〈 〉 in the (commuting) variables a , for k 0.) k < Proposition 5.7. Let V be a unitary representation of the Virasoro algebra (with c acting as a scalar) such that M V Vλ with dimVλ . = λ h < ∞ ∈ ∗ Then V is completely reducible.

Proof. Let U V be an invariant subspace. Observe that ω(L ) L . Unitarity forces λ h∗ by < 0 = 0 ∈ R considering the effect of L on v,v 0. It then follows that V ,V 0 if λ µ. Set U : U V , 0 〈 〉 > 〈 λ µ〉 = 6= λ = ∩ λ and define Uλ⊥ Vλ as the orthogonal complement to Uλ in the form restricted to Vλ. Then ⊂ L V U U ⊥, and U ⊥ : U ⊥ is invariant under the Virasoro algebra. So V U U ⊥. λ = λ ⊕ λ λ λ = = ⊕

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By our previous observation we conclude that B(µ,1) is a completely reducible Virasoro module. Write B 0(µ,1) B(µ,1) for the submodule generated by 1. It is unitary, hence completely ≤ reducible, and if B 0(µ,1) U U ⊥, then look at = ⊕

U 2 U ⊥ B 0(µ,1) 2 C.1 µ /2 ⊕ µ2/2 = µ /2 = to see that 1 U 2 B 0(µ,1) U , i.e. B 0(µ,1) is irreducible. ∈ µ /2 ⇒ = At this point we do not know whether B 0(µ,1) B(µ,1). = Definition 5.8 (Highest-weight representations). A highest-weight representation of the Virasoro algebra is a representation V such that there exists a v V satisfying ∈ 1. c .v cv (here c is a complex number on the right hand side), Vir = 2. L (v) hv for h C, 0 = ∈ © ª 3. V span (L i L i )(v) : 0 i1 ik . = − k ··· − 1 ≤ ≤ ··· ≤ Remark 5.9.

• cVir acts as the complex number c on all of V . ¡ ¢ ¡ ¢¡ ¢ £ ¤ • L0 L i L i (v) h ik i1 L i L i (v) thanks to L0,L i iL i . So we get − k ··· − 1 = + + ··· + − k ··· − 1 − = − a weight space decomposition M V Vλ , = λ h ∈ ∗ where Vλ 0 only if λ(cVir) c and λ(L0) h Z 0. 6= = = + ≥ • L (v) 0 for k 0 (because V 0 for λ(L ) h k). k = > λ = 0 = − The (c,h) are called the highest weights of V .

¡ 2 ¢ Remark 5.10. B 0(µ,1) is a highest-weight representation with weight 1,µ /2 . L Definition 5.11. If V λ h Vλ, then the formal character of V is = ∈ ∗

X λ(L0) λ(c) chV (q,t) : dim(Vλ)q t . = λ h ∈ ∗

Key idea: There exists a universal highest-weight representation for any (c,h) C2. To define ∈ this we introduce first the universal enveloping algebra of a Lie algebra.

6 Enveloping algebras

Note that any associative algebra A is a Lie algebra under the commutator bracket [x, y] : = x y yx. We write A for this Lie algebra structure. − ad Definition 6.1. Let g be an arbitrary Lie algebra over some field k. The universal enveloping algebra of g is an associative algebra U(g) over k with unit and a Lie algebra homomorphism ı : g U(g) satisfying the following universal property: → ad For every arbitrary associative algebra with unit A over k and a Lie algebra homomorphism j : g A , there exists a unique homomorphism of associative algebras φ: U(g) A such that → ad → j φ ı. = ◦

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Exercise 6.2. Show that if U(g) exists, then it is unique.

Exercise 6.3. Show that any Lie algebra representation of g is a (left) U(g)-module and vice-versa. There is an equivalence of categories between representations of g and left U(g)-modules.

The universal enveloping algebra U(g) exists since we can construct it explicitly as

U(g) : T (g)± x y y x [x, y]: x, y g , (6.1) = 〈 ⊗ − ⊗ − ∈ 〉 L i © ª where T (g) denotes the tensor algebra of (the vector space) g, i.e. T (g) : i 0 g⊗ . If xb : b B = ≥ ∈ is a basis for g, the T (g) is just the free algebra on the variables xb.

Exercise 6.4. Show that U(g) as defined in Equation (6.1) satisfies the universal property from Definition 6.1.

Theorem 6.5 (Poincaré, Birkhoff, Witt). Let ©x : b Bª be a basis for g, where B is ordered. Then b ∈ the set of ordered monomials n o i1 i2 i3 x x x : b1 b2 b3 ,i j Z 0 b1 b2 b3 ··· < < < ··· ∈ ≥ is a basis for U(g).

7 The universal highest-weight representations of Vir

Recall from Remark 5.9 that a highest-weight representation V of the Virasoro algebra has a L weight space decomposition V λ h Vλ. The Virasoro algebra has a sub-Lie-algebra Vir : = ∈ ∗ ≥ = Vir h (cf. Equation (5.3); but be careful, this is only a vector space direct sum, not a Lie algebra + ⊕ direct sum). We will now construct a universal highest-weight representation of the Virasoro algebra. To this end, define

M(c,h) : U(Vir) U(Vir ) C(c,h) , (7.1) = ⊗ ≥ where the action of U(Vir ) on U(Vir) is by right multiplication, and on the one-dimensional ≥ space C(c,h) on the right-hand side it is as follows:

L .λ 0 for k 0, L .λ hλ , c .λ cλ , for all λ C. k = > 0 = Vir = ∈ We call this M(c,h) the Verma module.

Exercise 7.1. Check that the Verma module M(c,h) is a well-defined representation of the Virasoro algebra, and moreover that it is a highest-weight representation with highest weight (c,h). (See also Proposition 7.2 (1).)

Proposition 7.2. L © ª 1. M(c,h) s Z 0 M(c,h)h s, where M(c,h)h s has a basis of vectors (L ik L i1 ) 1 , = ∈ ≥ + Pk + − ··· − ⊗ where 0 i1 i2 ik and j 1 i j s. < ≤ ≤ ··· ≤ = = (This has dimension P(s) the number of partitions of s; thus = h t Y j chM(c,h)(q,t) , where φ(q) (1 q ) . = φ(q) = j 1 − ≥ Note that L .(1 1) : L 1 1 (L .1) h(1 1), and c.(1 1) c(1 1), which proves 0 ⊗ = 0 ⊗ = ⊗ 0 = ⊗ ⊗ = ⊗ that this is a highest-weight representation.)

10 Infinite-dimensional Lie algebras 11

2. (Universality.) M(c,h) is indecomposable, and any highest-weight representation of highest weight (c,h) is a quotient of M(c,h).

3. M(c,h) has a unique irreducible quotient V (c,h). © Proof. Part (1) is clear, since by Theorem 6.5 U(Vir) is free over U(Vir ) with basis L i L i : ≥ − k ··· − 1 0 i i ª. < 1 ≤ ··· ≤ k Part (2). If M(c,h) V W , this respects the weight space decomposition by Lemma 5.4. = ⊕ Hence M(c,h) C, so without loss of generality, V C and W 0. Then 1 1 V . But 1 1 h = h = h = ⊗ ∈ ⊗ generates M(c,h) as an algebra, so M(c,h) V . = If N is a highest-weight representation with highest weight (c,h), then there exists v N ∈ such that c.v cv, L .v hv and L .v 0 for all k 0. Hence Cv C(c,h). The evaluation map = 0 = k = > = ev: M(c,h) : U(Vir) C(c,h) N, ρ 1 ρ.v thus factors through the tensor product over = ⊗  ⊗ 7→ U(Vir ) and so is the desired quotient. ≥ Part (3). This is equivalent to a unique maximal proper submodule. Any submodule inherits a weight space decomposition, so any proper submodule P has P 0. (Note that all proper sub- h = modules lie in the complement of the top weight space.) Thus the sum of all proper submodules is again proper, and clearly maximal.

Theorem 7.3 (Mathieu; Chari, Pressley (unitary case)). An irreducible representation V of the L Virasoro algebra with V λ h Vλ is precisely one of the following three: = ∈ ∗

• Vα0 ,β from Equation (5.1) and Theorem 5.5,

• V (c,h), the irreducible quotient of M(c,h) from Proposition 7.2 (3), or

• V (c,h)∗, the restricted dual representation of V (c,h), which is defined as M£ ¤ V (c,h)∗ : V (c,h)t ∗ . = t

It is acted on in the usual way by (X .θ)(v) : θ( X .v) for X Vir, θ V (c,h)∗ and v = − ∈ ∈ ∈ V (c,h); it is a lowest-weight representation.

The proof of Theorem 7.3 is tricky; it proceeds by an analysis in positive characteristic.

8 Irreducibilty and unitarity of Virasoro representations

First note that the involution ω on Vir extends to U(Vir). Define a form on M(c,h) as follows: For any P L i L i and Q L j L j , and for v 1 1, let = − k ··· − 1 = − l ··· − 1 = ⊗ P(v),Q(v) : ¡(ω(P)Q).v¢ 〈 〉 = coeff v be the coefficient of v in the weight space decomposition, i.e. the coefficient of v in the weight decomposition of L i L i L j L j v. − k ··· − 1 − l ··· − 1 Exercise 8.1. Check that this form is contravariant and sesquilinear (though not necessarily non-degenerate). Check that M(c,h) and M(c,h) are pairwise orthogonal if λ µ. λ µ 6=

Now let M 0(c,h) M(c,h) be the maximal proper submodule, which we know to exist. ≤ ¡ ¢ © ª Lemma 8.2. M 0(c,h) ker , : k : k, 0 . = 〈− −〉 = 〈 −〉 =

11 12 Iain Gordon

Proof. The kernel is a submodule since x.k, k,ω(x). 0 for all k ker¡ , ¢ and x Vir. 〈 −〉 = 〈 −〉 = ∈ 〈− −〉 ∈ Note that v is not in the kernel since v,v 0, so the kernel is indeed a proper submodule. 〈 〉 6= Now suppose P.v M 0(c,h), and thus ω(Q)P.v M 0(c,h) for any Q. Hence Q(v),P(v) 0, ∈ ∈ 〈 〉 = and thus P.v is in the kernel.

To understand irreducibility or unitarity properties of M(c,h) we need to understand the form , . Define , N to be the restriction of , to M(c,h)h N . This is a form on a 〈− −〉 〈− −〉 〈− −〉 + finite-dimensional vector space; let detN (c,h) be its determinant. We compute the first two values: det1(c,h) L 1.v,L 1.v , which is the (L1L 1.v)-coeffi- = 〈 − − 〉 − cient of v, so det (c,h) 2h. Next, 1 = ¯ ¯ ¯ L 2.v,L 2.v L 2.v,L 1L 1.v ¯ det 2(c,h) ¯ 〈 − − 〉〈 − − − 〉 ¯ . = ¯ L 1L 1.v,L 2.v L 1L 1.v,L 1L 1.v ¯ 〈 − − − 〉〈 − − − − 〉 But

¡ ¢ L 2.v,L 2.v L2L 2.v , working in the Verma module, 〈 − − 〉 = − coeff v ³¡ ¢ ´ [L2,L 2] L 2L2 .v = − + − coeff v ³¡ 1 ¢ ´ 4L0 (8 2)c .v 0 = + 12 − + coeff v 4h c . = + 2 With similar calculations we get that det (c,h) 2h(16h2 2hc 10h c). 2 = + − + Theorem 8.3 (Kac determinant formula).

Y ¡ ¢P(n r s) det n(c,h) K h hr,s(c) − , = r,s N − 1 r,∈s n ≤ ≤ where P(m) is the number of partitions of m,

Y ¡ s ¢P(n r s) P(n r (s 1)) K (2r ) s! − − − + , = r,s N 1 r,∈s n ≤ ≤ and h (c) 1 ¡(13 c)(r 2 s2) p(c 1)(c 25)(r 2 s2) 24r s 2 2c¢ . r,s = 48 − + + − − − − − + Examples.

2 • V (1,h) M(1,h) if and only if h m , m Z. = 6= 4 ∈ m2 1 • V (0,h) M(0,h) if and only if h − , m Z. = 6= 24 ∈ Exercise 8.4 (Unitarity). Show that det (c,h) 0 for all n if c 1 and h 0. n > > > n3 n Note that L n.v,L n.v 2nh c − , so considering large enough n we see that unitarity 〈 − − 〉 = + 12 implies that c 0. ≥ We saw that unitarity implies c 0 and h 0. If we restrict to the region c 1 and h 0, then, ≥ ≥ ≥ ≥ since det (c,h) 0, the form is positive semi-definite in this region (i.e. it is positive-definite n > for V (c,h)) if it is so at least once in this region. But we already found one unitary irreducible µ2 highest-weight representation of Vir with h , c 1, namely B 0(µ,1). = 4 =

12 Infinite-dimensional Lie algebras 13

For c 0, it can be shown that the only unitary representation is V (0,0) C. We want to = =∼ understand what happens in the range 0 c 1. Reparametrise as follows: ≤ < 6 c(m) : 1 for m 0. = − (m 2)(m 3) ≥ + + Then ¡ ¢2 ¡ ¢ (m 3)r (m 2)s 1 hr,s c(m) hr,s(m) + − + − . ≡ = 4(m 2)(m 3) + + Theorem 8.5 (Friedan, Qiu, Shenker; Goddard, Kent, Olive). In the region 0 c 1, unitary ≤ < representations occur precisely at

¡ ¢ (c(m),hr,s(m) for m,r,s Z 0 such that 1 s r m 1 . ∈ ≥ ≤ ≤ ≤ +

9 A little infinite-dimensional surprise

9.1 Fermionic Fock L Let V n Z Cvn. Then gl(V ) : gl is the Lie algebra of matrices with a finite number of = ∈ = non-zero entries. It is spanned by the∞E : v v . Let i,j j 7→ i

(0) n o F : Λ(0)∞ V span vi : vi0 vi1 vim 1 v m v m 1 : i0 i1 im 1 m . = = = ∧ ∧ ··· ∧ − ∧ − ∧ − − ∧ ··· > > ··· − > −

Then gl acts on F (0) via ∞ A.v A.v v v A.v , i = i0 ∧ i2 ∧ ··· + i1 ∧ i2 ∧ ··· + ··· (note that this action preserves tails). In detail,

( 0 if j does not appear in i or if i appears in i, if i j, then Ei,j .vi 6= = replace v j with vi in vi otherwise, ( vi if j appears in i, and E j,j .vi = 0 otherwise.

(0) The space F has a basis labelled by partitions λ n, where λ (λ λ ...) Z∞ with ` = 1 ≥ 2 ≥ ∈ 0 P λ n. To see this, map ≥ i i = i ¡i ,...,i ,...¢ λ : ¡i ,i 1,i 2,...,i m,...¢ , = 0 m 7→ i = 0 1 + 2 + m + and this map is clearly reversible. It follows that F (0) span©v ª. = λ Define deg(vλ) : λ n if λ n, so that λ is the number of which λ is a partition; i.e. P ¡ =¢ | | = ` | | deg(vi) s 0 is s . = ≥ + More generally, we have F (n), which is defined as for F (0), but with the basic vector being vn vn 1 instead of v0 v 1 . The Lie algebra gl acts naturally on it. ∧ − ∧ ··· ∧ − ∧ ··· ∞ Remark 9.1 (Etymology). The notation F (n) comes from the term “Fermionic Fock”. The earlier notation B(µ,h) for Heisenberg representations stands for “Bosonic”.

13 14 Iain Gordon

9.2 Central extensions of gl ∞ Let a gl be the algebra of matrices with a finite number of non-zero “diagonals”. This ∞ ⊃ ∞ P contains a big abelian subalgebra spanned by Λk for k Z, where Λk : i Z Ei,i k , so that ∈ = ∈ + Λk .v j v j k for all j. = −

Exercise 9.2. Show that the elements Λk commute with each other.

Exercise 9.3. Consider the representation Vα,β of the Witt algebra. Show that the Witt algebra embeds in a using this. ∞ P A typical element of a is a finite linear combination of terms i Z λi Ei,i k . If k 0, then P (0) ∞ ∈ + 6= i Z λi Ei,i k acts on F , because for i 0 or i 0 we have an action by 0 for Ei,i k on vi. ∈ + ÀP ¿ P + However, if k 0, then we would have i Z λi Ei,i .vj s 0 λjs vj, which is not well defined! = ∈ = ≥ To remedy the situation, we adjust the “action” as follows:

• Ebi,j acts as Ei,j for i j. 6=

• Ebi,i acts as Ei,i for i 0, and as Ei,i 1 for i 0. ≥ − < For example, ( vj if i 0 and js i for some s. Ebi,i .vj ≥ = = 0 otherwise.

Exercise 9.4. Show that the Ebi,j differ from Ei,j only by a multiple of I, so commutators are unchanged. Show further that £ ¤ £ ¤ Ebi,j ,Ebk,l 0 for j k, i l, Ebi,j ,Ebj,l Ebi,l for i l, £ ¤ = 6= 6= £ ¤ = 6= ¡ ¢ Ebi,j ,Ebk,i Ebk,j for k j, Ebi,j ,Ebj,i Ebi,i Ebj,j α Ebi,i ,Ebj,j I , = − 6= = − + where   1 if i 0, j 0, ¡ ¢ − ≥ < α Ebi,i ,Ebj,j : 1 if i 0, j 0, = + < ≥ 0 otherwise.

We can say this in two ways: We either get a projective representation of gl and a , or a ∞ genuine representation for a central extension of gl and a : ∞ ∞ ∞ a : a Cc , c∞ = ∞ ⊕ a central extension with α as the defining 2-cocycle. As before, we define the diagonal elements P (n) Λb k : i Z Ebi,i k , and we get an action on F via = ∈ + £ ¤ Λb 0.vj nvj and Λb k ,Λb l δk, l kI , = = − i.e. we get a representation of the Heisenberg algebra with a0 acting as n and h acting as 1.

Lemma 9.5. The representation B(n,1) of the Heisenberg algebra is isomorphic to F (n) as graded H -modules (but obviously not as algebras). ¡ ¢ ¡ ¢ Proof. Let φ P(x1,x2,...) : P Λb 1,Λb 2,... .vn vn 1 , and check that = − − ∧ − ∧ ··· • this is an H -map (using that B(n,1) is irreducible to see injectivity), and that

¡ ¢ (n) • P Λb 1,Λb 2,... .vn vn 1 spans F . (Count dimensions of the homogeneous com- − − ∧ − ∧··· ponents on both sides.)

14 Infinite-dimensional Lie algebras 15

Exercise 9.6. Recall that Witt , a . Show that this embedding extends to an embedding → ∞ Vir , a . This depends on α and β; what is the central charge? → c∞ The isomorphism from Lemma 9.5 allows us to consider the module structure of one side acting on the other side. For instance, we see that the Virasoro algebra acts on F (n) as in Theorem 4.3. In particular, we have a distinguished isomorphism between B(0,1) C[x ,x ,...] and F (0). = 1 2 2 Exercise 9.7. Let k Z 0 and fk vk vk 1 v1 v k v k 1 . Show that L0.fk k fk ∈ ≥ = ∧ − ∧···∧ ∧ − ∧ − − ∧··· = and L .f 0 for all j 0. Deduce that B(0,1) is not irreducible as a Virasoro module, and that it j k = > is in fact a direct sum of infinitely many unitary irreducible representations.

10 Hands-on loop and affine algebras

Let g be any complex Lie algebra and and R any commutative algebra over C. Then the vector space g C R is a Lie algebra via ⊗ £X r,Y s¤ : £X ,Y ¤ r s . ⊗ ⊗ = ⊗ Thus we have a rich source of infinite-dimensional Lie algebras. (We could also try to consider a sheafified version by replacing R with some OX -algebra.) £ 1¤ In the case R C t,t − , the Laurent polynomials in one variable, we get L g : g R, the = ¡ ¢ ¡ 1 ¢ = ⊗ loop algebra of g. This has a description as Calg C×,g or as Csf S ,g , where “Csf” stands for the space of smooth maps with finite Fourier expansion, and the correspondence is between 1 P 2πinθ © ª ψ: S g and its Fourier coefficients in ψ n Z e Xn. Either way, a basis Xa for g → = ∈ determines a basis ©X (n) : X t nª a = a ⊗ for L g, and the Lie algebra structure is

£X (n), X (m)¤ £X , X ¤(n m). a b = a b +

Review. A finite-dimensional Lie algebra g is simple if g is not abelian and g has no proper £ ¤ non-zero ideals (i.e. 0 I ( g such that g, I I). Recall that g acts on itself via the adjoint 6= ⊆ action ad(x).y : [x, y] for all x, y g. Then g is simple if and only if g is itself irreducible under = ∈ the adjoint representation.

Lemma 10.1. If g is simple, then there exists an invariant bilinear form on g, unique up to scaling. Here invariance means

B : g g C , such that B¡[x, y] z¢ B¡x [y,z]¢ , (10.1) ⊗ → ⊗ = ⊗ which is just the infinitesimal (i.e. Lie algebra) version of group invariance.

Proof. Let M, N be representations of g; then M N is a representation via ⊗ x.(m n) : x.m n m x.n , ⊗ = ⊗ + ⊗ and M ∗ HomC(M,C) is a representation via =

(x.f )(m) : f ( x.m) for f M ∗ , m M . = − ∈ ∈ Define Mg : ©m M : x.m 0 for all x gª . = ∈ = ∈

15 16 Iain Gordon

The adjoint representation induces via tensor product a representation on g g, and then by ⊗ ¡ taking duals on (g g)∗, the space of bilinear maps on g. Then Equation (10.1) states B [y,x] ⊗ ⊗ z¢ B¡x [y,z]¢ 0 for all x, y,z g, which is equivalent to y.B 0 for all y. Hence + ⊗ = ∈ = ( ¡ ¢g ¡ ¢ ¡ ¢ C if g ∼ g∗ by Schur’s Lemma, B (g g)∗ Homg g g,Ctriv. Homg g,g∗ = ∈ ⊗ =∼ ⊗ =∼ =∼ 0 otherwise.

Now here is an explicit form:

K (x, y) tr¡adx ad y¢, the Killing form. = ◦ The Killing form in invariant by the Jacobi identity, symmetric by definition, and non-zero (by some theory which boils down to the fact that simple Lie algebras are not nilpotent).

Theorem 10.2 (Garland – I think). 2¡ ¢ H L g;C C . = Exercise 10.3. Prove the theorem.

Let d Der(R), i.e. d(r s) d(r )s rd(s). This extends to g R via d(X r ) : X d(r ), and thus ∈ = + ⊗ ⊗ = ⊗ d¡[X r,Y S]¢ £d(X r ),Y s¤ £X r,d(Y s)¤ £X dr,Y s¤ £X r,Y ds¤ . ⊗ ⊗ = ⊗ ⊗ + ⊗ ⊗ = ⊗ ⊗ + ⊗ ⊗ Hence we get a new Lie algebra £ ¤ £ ¤ (g R) o Cd , where X r µd,Y s λd X ,Y r s µY ds λX dr . ⊗ ⊗ + ⊗ + = ⊗ + ⊗ − ⊗ By applying this to the loop algebra, we get the new algebra

L g : L go Cd , = where d : t d . (The derivation tells us about degrees.) = dt ¡ ¢ Lemma 10.4. H 2 L g;C C, so there exists a unique (up to scalar) central extension of L g. It is = given by vanishing on d and ¡ ¢ β(X f ,Y g) Rest 0 f dg K (X ,Y ) . ⊗ ⊗ = = Remark 10.5. This produces a central extension which restricts to the central extension arising from Theorem 10.2.

Proof. We need β: L g L g C which satisfies anti-symmetry and the Jacobi relations, modulo ⊗ → linear functionals. We know that L g acts on (L g L g)∗, and a check shows that this descends to an action of ¡ ¢ ⊗ L g on H 2 L g;C . We get decompositions into d-eigenspaces M © ª L g L gi , where L gi X L g :[d, X ] i X , and hence = i Z = ∈ = ∈ 2¡ ¢ M 2 H L g;C Hi . = i Z ∈

Claim. H 2 0 if i 0. (In fact, generalising what we’re about to do, it is simply better to see that i = 6= L g acts trivially on H 2.) To see this, note that

β¡d,[x(i), y(j)]¢ β¡y(j),[d,x(i)]¢ β¡x(i),[y(j),d]¢ (i j)β¡x(i), y(j)¢ . = − − = +

16 Infinite-dimensional Lie algebras 17

So for i 0, setting F ¡a(i)¢ 1 β¡d,a(i)¢ ensures that H 2 0. ±± 6= = i i = 2¡ ¢ Now (L g)0 g 1 Cd, hence H (L g)0;C 0 by (a generalisation of) Exercise 2.2. So we can =¡ ⊗ ⊕ ¢ = ensure that β (L g) ,(L g) 0. 0 0 = Define β : g g C by β (x, y) β¡x(i), y( i)¢. Then g t n is a (g 1)-representation (via i ⊗ → i = − ⊗ ⊗ the adjoint representation). The Jacobi identity for β implies that βi is invariant. Therefore βi λi K , and by anti-symmetry λi λ i . The Jacobi identity again implies that λi 1 λi λ1. = = − − + = + Therefore β¡x(i), y( i)¢ λiK (x, y) − = − for some scalar λ, which gives the desired unique central extension.

Definition 10.6. We call L g Cc the affine Lie algebra, and L g Cd Cc the affine Kac-Moody ⊕ ⊕ ⊕ Lie algebra.

Remark 10.7 (Relation to the Virasoro algebra). In the central extension L g we have the con- £ 1¤ stituent algebra C t ± . However, natural constructions should be independent of the parameter t, so we expect an action of the Virasoro algebra on L g.

11 Simple Lie algebras

Consider the algebra n o sln 1 sl(n 1;C) : X Matn 1(C) : tr X 0 . + ≡ + = ∈ + =

In the classification of simple Lie algebras, this is of type An. This Lie algebra has Lie subalge- bras h, n and n of diagonal, strictly upper-triangular and strictly lower-triangular matrices, + − respectively, and there is a decomposition sln 1 n h n as a direct sum of vector spaces, + =∼ − ⊕ ⊕ + where each summand is actually a subalgebra. Note that for i 1,...,n there are elements =

ei Ei,i 1 , = + hi Ei,i Ei 1,i 1 , = − + + fi Ei 1,i , = + which generate the Lie algebra (e.g. Ei,j [ [[ei ,ei 1],ei 2] ,e j 1] for i j). Observe that = ··· + + ··· − < the following relations are obviously satisfied:

For all i, j,[h ,h ] 0 , [h ,e ] α (h )e , and [h , f ] α (h )f , i j = j i = i j i j i = − i j i and [e , f ] 0 for i j,[e , f ] h for all i, (11.1) i j = 6= i i = i where  2 i j ,  ¯ = ¯ αi (h j ) 1 ¯i j¯ 1, = − − = 0 otherwise.

Less obviously, we have £ ¤ ¯ ¯ £ ¤ £ ¤ ei ,e j 0 if ¯i j¯ 1 , ei ,[ei ,ei 1] 0 ei 1,[ei 1,ei ] for all i. = − 6= + = = + +

¡ ¢1 αi (h j ) We write this concisely as ad(e ) − (e ) 0 (for i j). There are completely similar rela- i j = 6= tions for the fi .

17 18 Iain Gordon

Exercise 11.1. Find similar generators and relations for the Lie algebras n o g(M) : X Mat (C): X T M MX 0 , where = ∈ k + = 2 0 0  1. M 0 0 In for k 2n 1, = = + 0 In 0

µ 0 I ¶ 2. M n for k 2n, and = I 0 = − n µ 0 I ¶ 3. M n for k 2n. = In 0 =

Definition 11.2. The matrix A : ¡α (h )¢ is called the Cartan matrix, and it determines finite- = i j i j dimensional simple Lie algebras completely.

It is known from the theory of finite-dimensional simple complex Lie algebras that the Cartan matrix satisfies the following properties:

• A Mat (Z) and A is indecomposable (i.e. not block-diagonal with more than one block); ∈ n • A 2 for all i; ii = • A ©0, 1, 2, 3ª for all i j; i j ∈ − − − 6= • A 0 if and only if A 0; i j = ji = • if A 2 or 3, then A 1. i j = − − ji = − These conditions imply that A is “non-degenerate” in appropriate sense which we will meet soon; in particular this includes det A 0. 6=

12 Kac-Moody Lie algebras

Definition 12.1. We relax the conditions on A slightly and say that A Mat (Z) is a generalised ∈ n Cartan matrix (GCM) if

• A 2, ii = • A 0 if i j, and i j ≤ 6= • A 0 if and only if A 0. i j = ji = From the data of a GCM we now construct a complex Lie algebra in three steps. If the GCM is actually a Cartan matrix, this will produce the associated finite-dimensional simple Lie algebra.

12.1 Step I: Realisations

Definition 12.2. Let A Mat (C). A realisation of A is a triple (h,Π,Π∨), where ∈ n • h is a finite-dimensional vector space over C, © ª • Π∨ h ,...,h h is a linearly independent subset, and = 1 n ⊂ © ª • Π α ,...,α h∗ is a linearly independent subset, = 1 n ⊂

18 Infinite-dimensional Lie algebras 19 such that α (h ) A . j i = i j

Lemma 12.3. If (h,Π,Π∨) is a realisation of A, then dimh 2n rk A. ≥ − © ª © ª Proof. Let r rk A and dimh m. Extend Π and Π∨ to bases α ,...,α and h ,...,h = = 1 m 1 m respectively, producing an invertible matrix

µAB ¶ ¡α (h )¢ . j i = CD

The submatrix of the first n rows, ¡A B¢, has rank n since the whole matrix is invertible, and | A already has r linearly independent columns. So rkB n r . But B has m n columns, so ≥ − − m n n r . − ≥ −

Definition 12.4. A minimal realisation of A is a realisation of A such that dimh 2n rk A. = −

¡ 2 2 ¢ Example. Let A 2− 2 . Then rk A 1, so a minimal realisation must have dimh 3. So let 3 =3 − = © ª © ª = h C and h∗ C , with respective bases h ,h ,d and α ,α ,γ : = = 0 1 0 1

h ¡0,2,0¢ α ¡ 1,1,1¢ 0 = 0 = − h ¡1, 1,0¢ α ¡1, 1,0¢ 1 = − 1 = − d ¡0,0,1¢ γ ¡ 1 , 1 ,0¢ = = 2 2

(This is associated to the Lie algebra slb 2, to which we will return later.)

Proposition 12.5. Any square matrix has a minimal realisation, and any two minimal realisa- tions are isomorphic.

Remark 12.6. An isomorphism of realisations is Φ: (h,Π,Π∨) (h0,Π0,Π0∨), where Φ: h h0 is → → an isomorphism such that Φ(h ) h0 and Φ∗(α0 ) α for all i. i = i i = i Proof of 12.5. Let r rk A, so that possibly after re-ordering =

µA A ¶ A 11 12 = A21 A22 with A non-singular. Extend this to a (2n r ) (2n r )-matrix 11 − × −   A11 A12 0 C A21 A22 In r  . = − 0 In r 0 −

This is non-singular, e.g. detC det A . Let α ,...,α be the first n coordinate functions, and = ± 11 1 n let h1,...,hn be the first n rows of C. This is a minimal realisation. The proof of uniqueness is left as an exercise.

Exercise 12.7. Complete the proof of Proposition 12.5, i.e. show that any two minimal realisa- tions of a square matrix A are isomorphic.

19 20 Iain Gordon

12.2 Step II: A big Lie algebra

Let A Matn(Z) be a generalised Cartan matrix and (h,Π,Π∨) a minimal realisation of A, where ¡∈ ¢ Π∨ α1,...,αn . Let = n o X : e ,...,e , f ,..., f ,x : x h , = 1 n 1 n e ∈ The free Lie algebra L(X ) generated by X is defined as follows: Let

C X : C e ,...,e , f ,..., f ,x 〈 〉 = 〈 1 n 1 n e〉 be the free associative algebra generated by X . Then C X is a Lie algebra (under commuta- 〈 〉ad tors), and L(X ) is the Lie subalgebra of C X generated by the elements of X . Its elements are 〈 〉ad called Lie words. (More generally, any Lie algebra is spanned by Lie words in a set of generators.) ± Definition 12.8. Let Le(A) : L(X ) R , where R is the set of relations modelled on those in (11.1). = 〈 〉 Specifically, R consists of

• x λy µz whenever x λy µz in h, for x, y,z h, e− e− e = + ∈ • £x, y¤ whenever x, y h, e e ∈ £ ¤ £ ¤ • ei , f j for i j, ei , fi hei for all i 1,...,n, 6= − = • £x,e ¤ α (x)e and £x, f ¤ α (x)f for all i 1,...,n. e i − i i e i + i i = Remark 12.9. The Lie algebra Le(A) is independent of the choice of minimal realisation of A thanks to Proposition 12.5.

Theorem 12.10 (Structure of Le(A)).

1. Le(A) n he n , where he h is abelian and n , n are freely generated by f1,..., fn, = e− ⊕ ⊕ e+ =∼ e− e+ e1,...,en, respectively. L 2. Le(A) α Q Leα, where = ∈ n £ ¤ ¡ ¢ o Leα : x Le(A): he,x α he x for all he he = ∈ = ∈

and Q Zα Zα h∗. = 1 + ··· + n ⊂ 3. Le0 he. = 4. Leα 0 unless α Q or α Q , where Q Z 0α1 Z 0αn, and analogously for Q . = ∈ + ∈ − + = ≥ + ··· + ≥ − £ ¤ 5. Leα,Leβ Leα β. ⊆ + Proof. Define ω: ei fi and x x. Then ω induces a Lie algebra involution on Le(A) (just e ↔ − e ↔ −e e check it preserves the relations). Let n be the space generated by respectively ei and fi for e± i 1,...,n, and he the space spanned by elements x for all x h. The involution swaps n and n . = e ∈ e+ e− For λ h∗, let ∈ θ : X End¡T (V )¢ , where V span©v ,...,v ª , λ → = 1 n be given by

x ¡v v (λ α α )(x)v v ¢ , e 7→ i1 ··· is 7→ − i1 − ··· − is i1 ··· is f j multiplication by v j , 7→  1 0  7→ e j vi δi j λ(h j ).1 7→  7→ v v δ (λ α α )(h )v v v ¡θ (e )(v v )¢ i1 ··· is 7→ i1 j − i2 − ··· − is j i2 ··· is + i1 λ j i2 ··· is

20 Infinite-dimensional Lie algebras 21

It is an exercise to check that this induces a Lie algebra representation of Le(A) (again, it induces a representation of the free Lie algebra generated by X ; now check the relations). So in fact we ¡ ¢ get an (abusively denoted) map θλ : Le(A) End T (V ) . → We can now prove part (1) of the theorem.

• h he: There is a natural surjective Lie algebra homomorphism x xe. If xe 0 in Le(A), we =∼ ¡ ¢ 7→ = would have θ (x) 0 End T (V ) for all λ h∗; but θ (x)(1) λ(x). This can be 0 for all λ λ e = ∈ ∈ λ e = only if x 0. =

• n is free: Let φ(w) θλ(w).1 (by construction this is independent of λ), i.e. e− = φ¡w(f ,..., f )¢ w(v v ). 1 n = 1 ··· n So we get a surjective Lie algebra homomorphism φ: n L(V ) onto the free Lie algebra e− → generated by V ; it has an inverse which sends vi to fi .

• n is free: Apply ω to n to see that n is free. e+ e e− e+

• n he n is a direct sum: Let w x w 0. Then θλ(w ) θλ(x) θλ(w ) 0 for e− + + e+ − + e+ + = − + e + + = all λ h∗. Evaluating this at 1 T (V ) gives φ(w ) λ(x) 0, where φ(w ) T (V ) 0 and ∈ ∈ − + = − ∈ > λ(x) T (V )0. Thus λ(x) 0 and φ(w ) 0, which by the arguments above implies that ∈ = − = x 0 and w 0. It then follows that w 0, too. e = − = + = • So n he n Le(A). To see that it is all of Le(A), it is enough to check that it is closed e− ⊕ ⊕ e+ ⊆ under taking brackets since it contains the generating set X . In other words, we need ¡ ad(ei ) n he n ) n he n , e− ⊕ ⊕ e+ ⊂ e− ⊕ ⊕ e+

and similarly for fi and x˜. We deal only with the ei claim, the rest are similar. It is easy to see for the direct summands he and n . For n we have e+ e−

ad(ei )(f j ) δi j hej he n . = ∈ ⊕ e− By induction,

ad(e )£w (f ,..., f ),w (f ,..., f )¤ £ad(e )(w ),w ¤ £w ,ad(e )w ¤ , i 1 1 n 2 1 n = i 1 2 + 1 i 2 which completes part (1).

The further parts (2)–(5) are straight-forward: n consists of Lie words in the ei ’s, and so x acts e+ e on £e ,[e ,...]¤ by multiplication by ¡α α ¢(x). Thus i1 i2 i1 + i2 + ··· X X ne ,ne ,he Leα , and so Leα Le(A). + − ⊆ α Q α Q = ∈ ∈ Parts (3)–(5) follow immediately.

12.3 Step III: A smaller Lie algebra

We want to get close to a simple without losing the information carried in A. This means that we would like to factor out ideals, but preserve h Le0. = Lemma 12.11. Le(A) contains a unique ideal I that is maximal with respect to the condition I h 0. ∩ =

21 22 Iain Gordon

© ª Proof. Let J I Le(A): I h 0 . Then for I 0, I 00 J , I 0 and I 00 inherit the weight space = C ∩ = ∈ decomposition by Lemma 5.4, so ( , ) M ( , ) I 0 00 Iα0 00 . = 0 α Q 6= ∈ P Therefore (I 0 I 00)0 0. So I J I is the desired ideal. + = ∈ ± Definition 12.12. Let L(A) : Le(A) I, where I is the ideal from Lemma 12.11. We call L(A) the = Kac-Moody Lie algebra associated to A.

It is immediate that we have a decomposition (the so-called triangular decomposition

L(A) n h n . = + ⊕ ⊕ − We call h the Cartan subalgebra. It is also clear that ωe induces an involution ω on L(A) which exchanges n and n . Finally, L(A) inherits the weight space decomposition + − M L(A) Lα , = α Q ∈ where Q Zα Zα is the weight lattice. We collect a few obvious results: = 1 + ··· + n Proposition 12.13.

1. α1,...,αn are positive roots.

2. dimL 1. αi = 3. kα is a root if and only if k 1. i = ±

13 Classification of generalised Cartan matrices

For the duration of this section, let A denote a generalised Cartan matrix of size n n. To simplify × matters, we shall only consider matrices up to permutation by S , where A A0 if and only if n ∼ A A0 for π S . Furthermore we assume that A is indecomposable, i.e. if A A A , i j = π(i),π(j) ∈ n = 1 ⊕ 2 then at least one of A1 and A2 is zero.

Definition 13.1. Let v (v ,...,v ) Rn. We say v 0 if v 0 for all i, and we say v 0 if v 0 = 1 n ∈ ≥ i ≥ > ≥ and v 0. We define v 0 and v 0 similarly. 6= ≤ <

13.1 Types of generalised Cartan matrices

Definition 13.2. Let A be a generalised Cartan matrix.

1. A has finite type if

• det A 0, 6= • there exists v 0 such that Av 0, and ≥ > • if Av 0, then v 0. ≥ ≥ 2. A has affine type if

• corank(A) 1, = • there exists v 0 such that Av 0, and > = • if Aw 0, then w λv for some λ C. ≥ = ∈

22 Infinite-dimensional Lie algebras 23

(Note that this says that there does not exist any w such that Aw 0.) > 3. A has indefinite type if

• there exists v 0 such that Av 0, and > < • if Av 0 and v 0, then v 0. ≥ ≥ = Theorem 13.3 (see Kac or Carter). An indecomposable generalised Cartan matrix A is either finite, affine or indefinite. Moreover,

• A is finite if and only if there exists v 0 such that Av 0, > > • A is affine if and only if there exists v 0 such that Av 0, and > = • A is indefinite if and only if there exists v 0 such that Av 0. > < The conclusions of Theorem 13.3 hold under weaker assumptions. Let A Mat (R) such ∈ n that A 2 for all i. For J {1,...,n} let A be the submatrix of A whose entries are labelled by ii ≥ ⊆ J J J. × Lemma 13.4. Assume that A is indecomposable. Then

• if A is finite, then AJ is finite, and

• if A is affine, then AJ is finite (for proper J).

Proof. Let J {1,...,n}, so J (1,...,m) after reordering. Let P,Q Mat(Z 0) such that we can ⊆ = ∈ ≤ write in block form µA P¶ A J . = QR If A is finite, then by definition there exists a v 0 such that Av 0, so > >   v1 . ¡A P¢ .  0 . |  .  ≥ vn

But     v1 v1 . . P  .  0 , so A  .  0 ,  .  ≤ J  .  > vn vn and so AJ is finite. The proof for the affine case uses the same strategy.

13.2 Symmetric generalised Cartan matrices

We continue to assume that all generalised Cartan matrices are indecomposable. In this subsec- tion we consider the special case in which A is a symmetric matrix, i.e. AT A. = Proposition 13.5. Let A be a symmetric generalised Cartan matrix.

• A is finite if and only if A is positive definite.

• A is affine if and only if A is positive semi-definite and of corank 1.

• A is indefinite precisely if it is neither finite nor affine.

23 24 Iain Gordon

Proof. If A is finite, then by definition there exists v 0 such that Av 0, and so for all λ 0, > > ≤ we have (A λI)v 0. By the remark in Definition 13.2 (2),(A λI) has finite type. shouldn’t it − > − be affine type? So det(A λI) 0, and so λ is not an eigenvalue of A. So A is a real, symmetric − 6= matrix all of whose eigenvalues are strictly positive, and hence A is positive definite. Conversely, if A is positive definite, then det A 0, and so A is not affine. If A were indef- 6= inite, then there would exist v 0 such that Av 0, so that vT Av 0, contradicting positive > < ≤ definiteness. Hence A is finite. For affinity, we can use a similar argument (exercise) to deduce positive semi-definiteness. The converse proof that a positive semi-definite matrix is affine is essentially the same as above.

Definition 13.6. An (n n)-matrix A is symmetrisable if there exists a non-singular matrix × D diag¡d ,...,d ¢ and a symmetric matrix B such that A DB. = 1 n = Exercise 13.7. Find the smallest possible non-symmetric generalised Cartan matrix.

Lemma 13.8. A generalised Cartan matrix A is symmetrisable if and only if

Ai1i2 Ai2i3 Aik 1ik Aik i1 Ai2i1 Ai3i2 Aik ik 1 Ai1ik ··· − = ··· − for all i ,...,i {1,...,n}. 1 k ∈ Proof. The “only if” direction is trivial. For the “if” direction, recall that A is assumed indecom- posable. For each j, choose 1 j1, j2,..., jt j such that A jk jk 1 0 for all k 1,...,t 1. Let = = + 6= = − 0 d R and define 6= 1 ∈ A jt jt 1 A j2 j1 d j − ··· d1 . = A j1 j2 A jt 1 jt ··· + (Exercise: Check that this is well-defined, i.e. independent of the route from 1 to j.) Let D : ¡ ¢ 1 = diag d ,...,d and B : d − A . We need B B , i.e. 1 n i j = i i j i j = ji

Ai j A ji . di = d j which is obvious if A 0. Assume thus that A 0, choose a sequence from 1 to i, i.e. i j = i j 6= 1 j1,..., jt i, and augment it by jt 1 j. Then = = + = ¡ ¢ A j jt A jt jt 1 A j2 j1 A ji A j jt d j ¡ − ··· ¢ d1 d1 and d1 di . = A j1 j2 A jt 1 jt A jt j = Ai j = A jt j ··· − the indexes seem wrong...

Remark 13.9. The proof of Lemma 13.8 shows that without loss of generality, d 0 for all i, i > and that B Mat (Q). ∈ n Proposition 13.10. If A is indecomposable and v 0 is such that Av 0, then v 0 for all i. > > i > This proposition should move up, but where?

Proof. After reordering, v (v1,...,vm,0,...,0) and vi 0 for i 1,...,m. So if Av 0, then Pn = > = > j 1 Ai j v j 0 for all i (recall that Ai j 0 if i j). = ≥ < 6= Proposition 13.11. If A is a finite or affine generalised Cartan matrix, then A is symmetrisable.

24 Infinite-dimensional Lie algebras 25

Proof. Suppose there exist Ai1i2 0, Ai2i3 0, . . . , Aik 1ik 0, Aik i1 0 (where k 3 for non- 6= 6= − 6= 6= ≥ triviality). Minimality means that

A 0 for all (s,t) ©(1,2),(2,3),...,(k,1),(2,1),(3,2),...,(1,k)ª . is it = 6∈ For J {i ,...,i }, A has the form = 1 k J   2 r ... s − 1 − k  . .   .. .   s1 0 .  AJ −.  , =  .  . 0 2 rk 1  − −  rk ... sk 1 2 − − − where r ,s 0, and this matrix is either finite or affine, i.e. there exists v (v ,...,v ) 0 such i i > = 1 k > that A v 0. Let J ≥   2 r 0 ... s0 − 1 − k  . .   s .. 0 .  1  10  M : diag(v)− AJ diag(v) −.  = =  .   . 0 2 rk0 1 − − rk0 ... sk0 1 2 − − − where

1 ri0 vi− ri vi 1 0 = + > 1 si0 vi− 1si vi 0 = + > r 0s0 r s Z i i = i i ∈ and n X X 1 1 Mi j vi− (AJ )i j v j vi− AJ v 0 . j = j 1 = ≥ = P So i,j Mi j 0, but ≥ X Mi j 2k (r10 s10 ) (rk0 sk0 ). i,j = − + − ··· − + So ri0 si0 q + r 0s0 pri si 1 . 2 ≥ i i = ≥ So

r 0 s0 2 r 0 s0 2 i + i ≥ ⇒ i + i = r s 1 , ⇒ i i = and so  2 1 0 ... 0 0 1 − −  1 2 1 0 0 0  − −   0 1 2 1 0 0 0   − −   . . . . .  AJ  ......  (13.1) =  . .     0 0 0 1 2 1 0   − −   0 0 0 1 2 1 − − 1 0 0 ... 0 1 2 − − This is symmetric; v (1,...,1) 0, and A v 0, so A is affine, and so A A. = > J = J J =

25 26 Iain Gordon

Corollary 13.12. If A is an indecomposable, symmetrisable generalised Cartan matrix, then

1. A is finite if and only if det A 0 for all J, J > 2. A is affine if and only if det A 0 and det A 0 for all proper J, and = J > 3. A is indefinite precisely if it is neither finite nor affine.

Proof. For (1) and (2), note that if A is finite or affine, then by Proposition 13.11 it is symmetris- able. So A DB with d 0 and B symmetric, and B is of the same type as A. The conclusions = i > of (1) and (2) follow by applying Proposition 13.5 to B. Conversely, we first show that A is symmetrisable. We have a precise condition on Ai j to ensure symmetrisability, namely Ai1i2 Ai2i3 Aik 1ik Aik i1 Ai2i1 Ai3i2 Aik ik 1 Ai1ik . This is ··· − = ··· − obviously satisfied unless there is a cycle in the matrix. Pick a minimal cycle

 2 b 0 ... 0 0 b  − 1 − s  b0 2 ??? 0 0 0  − 1   0 ??? 2 0 0 0     . . . . .  AJ  ......  , an (s s)-matrix with bi ,b0 N. =  . .  × i ∈    0 0 0 2 ??? 0     0 0 0 ??? 2 bs 1 − − bs0 0 0 ... 0 bs0 1 2 − − −

Kac claims that det A 0 unless b b0 1 for all i. J < i = i = We want to show that if either of the conditions in the lemma on principal determinants (Which lemma?) holds, then b b0 1 for all i: i = i = Suppose there exists a unique b or b0 1. Then change the basis to t t > 1  .   . ,e1,...,et ,...,es , 1 from which det A 0 follows by an easy calculation (det A s.( x), where x is the unique J < J = − − entry in column t). So there exist at least two non-zero b ,b0 1. Taking the shortest route from t t > t to t 0 produces a submatrix, which is proper unless s 4. ≤ If s 4, we get a tridiagonal matrix >  2 b  − 1  1 2 1  − −   1 2 1   − −   . . .   ......  .      1 2 1   − −   1 2 bt  − − b0 2 − t This is a determinant that can be calculated (by expanding along row 1 and then along row t), and it is always 0. So A is symmetrisable, A DB, and the principal determinants of B have ≤ = the same property as A. So we can use the results about quadratic forms again. (REFERENCE – is this Proposition 13.5?)

Exercise 13.13. Our proof of Kac’s claim was very inelegant. Find a nicer proof.

26 Infinite-dimensional Lie algebras 27

14 Dynkin diagrams

Attach a diagram ∆(A) to an (n n)-matrix A as follows: ∆(A) has vertices 1,...,n, and if i, j are × distinct vertices, then

• there is no edge between i and j if A 0, i j = • there is one edge between i and j if A 1 A , i j = − = ji • there is a directed double edge between i and j if A A 2, pointing towards j if j i, i j ji = < • there are three edges between i and j if A A 3, two of which form a directed double i j ji = edge pointing towards j if j i, < • there are four (undirected) between i and j if A A 4 and A 1, i j ji = i j = − • there is a crossed double edge between i and j if A A 4 and if A 2 A , and i j ji = i j = − = ji • for A A 5, there is one edge between i and j labelled “ A A ”. i j ji ≥ | i j | | ji |

Tautological observation. Generalised Cartan matrices correspond precisely to Dynkin dia- grams, and A is indecomposable if and only if ∆(A) is connected.

List of finite Dynkin diagrams

• An: ...

• B (n 2): ... n ≥ • C (n 3): ... n ≥ • D (n 4): ... n ≥

• E6:

• E7:

• E8:

• F4:

• G2:

These diagrams are called finite Dynkin diagrams, and they correspond precisely to finite generalised Cartan matrices. Note that the list is closed under transposition (which corresponds to reversing the arrows in the diagrams) and taking sub-diagrams.

Exercise 14.1. Show that the determinant of any of the associated matrices is positive.

27 28 Iain Gordon

List of affine Dynkin diagrams

• Ae1:

• Ae10 :

• Aen (n 2): ≥ T • Ben (n 3) and Be : ≥ n

• Cen (n 3): ≥ • CeT (n 3): n ≥

• Den (n ?): ≥

• Ee6:

• Ee7:

• Ee8:

• Fe4:

T • Fe4 :

• Ge2:

T • Ge2 : These diagrams are called affine Dynkin diagrams, and they correspond precisely to affine generalised Cartan matrices. Note that the matrix determined by Aen is the one in (13.1).

Exercise 14.2. Show that the determinant of any of the associated matrices vanishes.

Theorem 14.3. Finite (affine) Dynkin diagrams are in one-to-one correspondence with finite (affine) generalised Cartan matrices.

Proof. One direction follows from Exercises 14.1 and 14.2. To go the other way, in the finite case use the fact that if a Dynkin diagram produces a finite generalised Cartan matrix, then so does any sub-diagram (where you are allowed to remove edges, e.g. is a sub-diagram of ). Now suppose we had the following parts in a given diagram:

...

Then we get back to the finite case. In the affine case, check what happens in rank 2: ¡ 2 b ¢. For rank 3, we need to check that c −2 ≥ all sub-diagrams (by removing vertices only) are finite.−

15 Forms, Weyl groups and roots

Let A be a generalised Cartan matrix and L (A) its associated Kac-Moody Lie algebra. Recall that we have a weight space decomposition M © ª L (A) L (A)α for α Z α1,...,αn . = α ∈

28 Infinite-dimensional Lie algebras 29

Theorem 15.1. Let A be a symmetrisable generalised Cartan matrix. Then L (A) has a non- degenerate invariant symmetric bilinear form. ¡ ¢ T ¡ ¢ Proof. Let A DB, where D diag d1,...,dn with di N and B B. Let h,Π,Π∨ be a = = ¡ ¢ ∈ ¡ ¢= ¡ ¢ minimal realisation of A, where Π∨ h ,...,h and Π α ,...,α . Let h0 span Π∨ h, = 1 n = 1 n = ⊆ and let h00 be some complement such that h0 h00 h. Define a bilinear form ( , ) on h as ⊕ = − − follows: (h ,h) : d α (h) for all h h,(h00,h00) : 0 for all h00,h00 h00. (15.1) i = i i ∈ 1 2 = 1 2 ∈ We claim that this is symmetric:

(h ,h ) : d α (h ) d A d d B d A : (h ,h ) i j = i i j = i ji = i j ji = j i j = j i It is also non-degenerate, which we leave as an Exercise. In particular, © ª ker( , ) h x h0 : α (x) 0 for all i 1,...,r − − | 0 = ∈ i = = is the null space of A, which (Exercise:) equals the centre of L (A). Now consider the principal grading on L (A), M L (A) L (A)k , = k Z ∈ where deg(e ) 1 deg(f ) for i 1,...,n and deg(x) 0 for x h. Let i = = − i = = ∈ N M L (N) L (A)k , so L (0) L (A)0 h . = k N = = =− Define a form ( , ) on L (N) by induction: The case N 0 is given by Equation (15.1). For − − = N 1, define = ¡ ¢ ¡ ¢ (ei , f j ) : δi j di , and L 1,L0 0 L 1,L 1 . = ± = = ± ± Invariance holds since

¡[e , f ],h¢ δ (h ,h) δ d α (h) (e , f )α (h) ¡e ,[f ,h]¢ . i j = i j i = i j i j = i j j = i j ¡ ¢ ¡ ¢ Now for N 1, Li ,L j 0 if i j 0, and we need to define L N ,L N . Take y L N and > P = + 6= ± ∓ ∈ ∓ express it as y i [ui ,vi ] for some ui ,vi L (N 1). For x L N , define = ∈ ∓ − ∈ ± X¡ ¢ (x, y) : [x,ui ],vi , = i and note that [x,ui ] L (N 1) and vi L (N 1), so this is indeed an inductive definition. ∈ ± − P∈ ∓ − This is well-defined: Suppose x j [u0 ,v0 ], where u0 ,v0 L (N 1). Then = j j j j ∈ ± − X¡ ¢ (x, y) [x,ui ],vi = i X¡£ ¤ ¢ [u0j ,v0j ],ui ,vi = i,j X¡£ ¤ ¢ ¡£ ¤ ¢ [u0j ,ui ],v0j ,vi [v0j ,ui ],u0j ,vi = i,j − X¡ ¢ ¡ £ ¤¢ [u0j ,ui ],[v0j ,vi ] u0j , [v0j ,ui ],vi = i,j + X¡ £ ¤ £ ¤¢ u0j , ui ,[v0j ,vi ] [v0j ,ui ],vi = i,j + X¡ £ ¤¢ u0j , v0j ,[ui ,vi ] = i,j X¡ ¢ u0j ,[v0j , y] , = j

29 30 Iain Gordon where we used induction to conclude the second and the penultimate line. This calculation shows that the form ( , ) is well-defined, and moreover that it is symmetric. The proof that it is − − invariant is similar and we omit it. The form is non-degenerate: Let I : ker( , ). By invariance, I is an ideal. But we know that = − − ( , ) h is non-degenerate. Therefore I h 0, and by construction of L (A), I 0. − − | ∩ = = ¡ ¢ Corollary 15.2. L (A) ,L (A) 0 unless α β 0. So α β = + = ¡ ¢ , : L (A)α L (A) α C − − × − → is a non-degenerate pairing, and [x, y] (x, y)hα0 , where x L (A)α, y L (A) α, and hα0 is = ∈ ∈ − defined by ¡ ¢¯ h0 , ¯ α( ) . α − ¯h = − Proof. Let α β 0. Then there exists h h such that (α β)(h) 0. So + 6= ∈ + 6= α(h)¡x, y¢ ¡[x,h], y¢ ¡x,[h, y]¢ β(h)¡x, y¢ , − = = = and hence (x, y) 0. Now we have [x, y] (x, y)hα0 h if x L (A)α and y L (A) α. It follows ¡ = ¢ − ∈ ∈ ∈ − that [x, y] (x, y)h0 ,h 0 from the definition for all h h. − α = ∈

Reminder: representation theory for sl2.

nµa b ¶o sl(2;C) , = c a − so sl2 is of type A1. It has a basis E, H,F , where

[H,E] 2E ,[H,F ] 2F ,[E,F ] H . = = − =

By induction (inside the universal enveloping algebra U(sl2)),

k k k k k k 1 k 1 [H,F ] 2kF ,[H,E ] 2kE ,[E,F ] k(k 1)F − kF − H . = − = = − − + Theorem 15.3.

1. If V is a representation of sl , then there exists v V such that Hv λv. Setting v F j v, 2 ∈ = j = we have Hv (λ 2j)v . j = − j 2. There exists a unique (n 1)-dimensional irreducible representation of sl (where n 0). It + 2 ≥ has a basis (v0,...,vn) and satisfies

Hv j (n 2j)v j ,F v j v j 1,Ev j j(n j 1)v j 1 , = − = + = − + − F F F   ...  . • E • E • → → • E • Proof. (1) follows from the relations above (REFERENCE!), (2) is a fun Exercise.

In L (A), we have ei ,hi , fi with

[h ,e ] 2e ,[h , f ] 2f ,[e , f ] h . i i = i i i = − i i i = i Lemma 15.4. In L (A),

1 A 1 A (ade ) − i j e 0 and (ad f ) − i j f 0 for all i, j 1,...,n, i j. i j = i j = = 6=

30 Infinite-dimensional Lie algebras 31

1 A Proof. Let us show this for the f ’s, the proof for the e ’s is analogous. Let x (ad f ) − i j f . We i i = i j prove that [e ,x] 0 for all k. First, set k = © ª c(x) : y L (A):[y,x] 0 L (A), = ∈ = ⊆ this is a Lie subalgebra. Therefore, [n ,x] 0. (Please check, I think I’m missing a statement.) + = If A is symmetrisable, then (n ,x) 0 implies x 0. Otherwise, if (n ,x) 0, then U(n )x + = = + = + = Cx, and U(n )x U(n )U(h)U(n )x U(n )U(h)x U(n ) U(n ). + = − + = − = − ⊆ − So x n generates an ideal in L (A) wholly contained in n , so x 0 by construction of ∈ − − = L (A).

Another proof. Let x be as in the first proof. Then x n L (A). We show that [ek ,x] 0 for all ∈ − ⊂ = k, which implies that x 0. We distinguish three cases: = • If k i, j, then this is clear (check!), since [e , f ] 0 [e , f ]. 6= k i = = k j • If k j, then =

1 A 1 A 1 Ai j [e ,(ad(f )) − i j (f )] (ad(f )) − i j [e , f ] ad f − h , j i j = i j j = i j and if A 1, we get zero. If A 0, then [f ,h ] α (h )f A f 0. i j ≤ − i j = i j = i j i = ji i = • If k i, then we use sl(2;C)-representation theory: Let v f . Then [e , f ] 0 and = = j i j = [h , f ] α (h )f A f . Now (e ,h , f ) act on L (A) via the adjoint action, and we i j = − j i j = − i j j i i i consider the orbit through v. Then we use a result from representation theory:

1 Ai j T ei .(v1 A ) : [ei ,(ad(fi )) − f j ] (1 Ai j )( Ai j 1 Ai j 1)v1 A (ad(f j )) v 0 − i j = = − − − + + − i j = =

Definition 15.5. Let V be a representation of a Lie algebra g. An element x g is called locally ∈ nilpotent if for all v V there exists N(v) such that xN(v).v 0. ∈ =

Lemma 15.6. The elements ad(ei ) and ad(fi ) act locally nilpotently on L (A).

1 A 2 3 Proof. We have (ade ) − i j e 0, (ade ) h 0, (ade )e 0, (ade )f 0, (ade ) f 0. So i j = i = i i = i j = i i = ETS that since these are a generating set, we can act locally nilpotently on L (A) (i.e. what is generated). This is due to Leibniz:

k k £ ¤ X ¡k¢£ i k i ¤ (adx) y,z i (adx) (y),(adz) − (z) , = i 0 = which implies the statement.

If V is again a representation of a Lie algebra g, then a locally nilpotent action x g produces ∈ an automorphism n X∞ x exp(x) : = n 0 n! = on V whose inverse is exp( x), and x exp(x) is a Lie algebra homomorphism by the Leibniz − 7→ rule.

Definition 15.7. We define the elements ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ n : exp ad(e ) exp ad( f ) exp ad(e ) Aut L (A) . i = i − i i ∈

31 32 Iain Gordon

Example (sl(2;C)). The Lie algebra sl(2;C) has a basis {e, f ,h}, and so we have:

µ0 1¶ µ1 1¶ e exp(e) = 0 0 = 0 1 µ0 0¶ µ 1 0¶ f exp( f ) = 1 0 − = 1 1 −

Proposition 15.8. The element s n h : h h satisfies s (x) x α (x)h . i = i | → i = − i i Proof. This is a fascinating calculation, left as an Exercise.

Definition 15.9. The map s : h h, x x α (x)h is called a fundamental simple reflection. i → 7→ − i i The subgroup of Aut(h) generated by the elemets si is called the Weyl group of L (A), denoted by W .

We have s (h ) h , s2, s (x) x if (h ,x) 0. An easy calculation shows: i i = − i i i = i =

Lemma 15.10. The bilinear form ( , ) h is invariant under W . − − |

So we have an action of W on h∗ via

s (λ) λ λ(h )α for all λ h∗. i = − i i ∈

Proposition 15.11. The map n : L L is an identification. i α 7→ si (α) Proof. We compute directly:

£ ¤ £ 1 ¤ £ 1 ¤ ¡ 1 ¢ 1 h,n (x) n (s− h),n (x) n s− h,x n α(s− h)x α(s− h)n (x) s (α)(h)n (x) i = i i i = i i = i i = i i = i i

Theorem 15.12 (Properties of the Weyl group). If i j, then the element s s W has the 6= i j ∈ following order:  2 A A 0 ,  i j ji  = 3 Ai j A ji 1 ,  = order(si s j ) 4 A A 2 , = i j ji =  6 Ai j A ji 3 ,  =  A A 4 . ∞ i j ji ≥ Proof. Let © ª K : λ h∗ : λ(h ) λ(h ) 0 , = ∈ i = j = so dimK dimh∗ 2. Writing V Cα Cα , we get a decomposition h∗ V K , and s s acts = − = i ⊕ j = ⊕ i j trivially on K . On the other hand, we have

µ 1 A A A ¶ (s s ) − + i j ji i j . i j |V 7→ A 1 − ji − The result follows from a calculation of the eigenvalues and the Jordan Normal Form of this matrix.

32 Infinite-dimensional Lie algebras 33

16 Root spaces

Let L : L (A). We have a decomposition (of vector spaces) = M L L0 Lα , = ⊕ α R ∈ where R Q Q Q . We have Lα 0. Here Q Z 0α1 Z 0αn is the positive weight ⊂ = + t − 6= + = ≥ + ··· + ≥ lattice. The elements of R are called roots, and the decomposition of Q induces a decomposition R R R into positive and negative roots. = + t − We are lead to ask the following questions: What is R, and what is dimLα? Writing mα for the multiplicity of α, we know:

• dimLα dimL α. = − © ª • Π α1,...,αn : mα 1 R . = i = ⊂ + • R is W -stable, i.e. m m for all w W . wα = α ∈ Definition 16.1. An element α R is called real if there exists α Π and w W such that ∈ i ∈ ∈ w(α ) α. Otherwise we call α imaginary. i =

Note. α R if and only if α R , since w(s α ) w(α ). Furthermore, R+ R+ R+ . ∈ re − ∈ re i i = − i = re t im ¡ ¢ Lemma 16.2. w R+ R+ . im = im Pn Proof. Positivity is the key here. We have α k α R+ , where k 0, and at least two = j 1 j j ∈ im j ≥ k ’s are non-zero. So s (α) α α(h )α has at least= one positive k , so all k must be non- j i = − i i j j negative. © ª © ª Let C : λ h∗ : λ(h ) 0 for all i 1,...,n C : ... 0 . = ∈ R i > = ⊂ = ≥ S ¡ ¢ Proposition 16.3. R+ w W w C . im ⊆ ∈ − © ª Proof. For α R+ , we have a set ht(β): β w(α),w W . Pick an element β achieving ∈ im = ∈ minimal height. Then s (β) β β(h )α . Since β has minimal height, β(h ) 0 for all i, and i = − i i i ≤ thus β C. ∈ − Theorem 16.4 (Kac). Let

K : ©α Q : α 0, supp(α) connected, α C ª . = ∈ + 6= ∈ − S Then R+ w W w(K ). im = ∈ Remark 16.5. Here supp(α) : ©j : k 0ª, where α P k α ; supp(α) corresponds to the = j 6= = j j j vertices in the Dynkin diagram.

Exercise 16.6. Show that for [DIAGRAMME], C ©k(α α α ): k Nª in any minimal − = 1 + 2 + 3 ∈ realisation, and w(α α α ) α α α . 1 + 2 + 3 = 1 + 2 + 3

Corollary 16.7. α R+ kα R+ for all k N. ∈ im ⇒ ∈ im ∈ Corollary 16.8. If A is an indecomposable generalised Cartan matrix, then

1. if A is of finite type, then R+ ∅, i.e. there are only real roots, im =

33 34 Iain Gordon

2. if A is affine, then there exists an integral u 0 such that Au 0, and u has no common Pn > © = ª factors. Let u (a1,...,an) and δ i 1 ai αi . Then Rim+ kδ : k Z \ {0} . = = = = ∈ Pn 3. if A is indefinite, then there exists α R+ such that α k α , k 0 and α(h ) 0 for ∈ im = i 1 i i i > i < all i. =

Proof. Exercise.

Corollary 16.9. Since A is symmetrisable by (REFERENCE), if α R , then (α,α) 0, and if ∈ re > α R , then (α,α) 0. ∈ im ≤ Proof. (h ,x) : d α (x) where h d α , so i = i i i ↔ i i ³ ´ ¡ ¢ hi h j 1 αi ,αj , d − A ji . = di d j = j

2 So (αi ,αj ) 0, which proves the first statement. di P=n > If α i 1 ki αi is imaginary with ki 0 and α(hi ) 0 for all i, then = = ≥ ≤ n n ¡ ¢ X ¡ ¢ X ki α,α ki αi ,α α(hi ) 0 . = i 1 = i 1 di ≤ = = We prove Theorem 16.4 in several steps.

Step 1: α R supp(α) is connected. ∈ ⇒ © ª Proof. Without loss of generality, let α R+. Let supp(α) J 1,2,...,n with J J1 J2 discon- ∈ = ⊆ = ∪1 A nected (i.e. A 0 for all j J and j J ). Let i J , j J . Then [e ,e ] (ade ) − i j e 0. j1 j2 = 1 ∈ 1 2 ∈ 2 ∈ 1 ∈ 2 i j = i j = So Lα n , and we claim that Lα 0. ⊂ + = But Lα is spanned by Lie monomials in the ek ’s. We show by induction on degree that any Lie monomial involving ei and e j (with i, j chosen as above) is zero. ...

Step 2: α R, α α with α α R α(h ) 0. α R, α α with α α R α(h ) 0. ∈ 6= ± i ± i 6∈ ⇒ i = ∈ 6= − i + i 6∈ ⇒ i ≥ Proof. Pick x L . Then n (x) L . Now calculate n (x): ∈ α i ∈ si α i ade (x) 0 exp(ade )x x i = ⇒ i = ad f (x) 0 exp(ad( f ))x x i = ⇒ − i =

Since adei (x) Lα α and ad fi (x) Lα α , we conclude that ni (x) x. So si α α, and so ∈ + i ∈ − i = = α(h ) 0. The second claim is a variation on this theme. i =

P © P n ª Step 3: α ki αi K , Ψ : β R+ : β 1 mi αi , mi ki for all i . Let βb have the = i ∈ = ∈ = i = ≤ largest height amongst the β Ψ. Then supp(βb) supp(α). ∈ = Proof. First note that supp(βb) supp(α). If equality did not hold, then there would exist j ⊆ ∈ supp(α) \ supp(βb) and j 0 supp(βb) such that A j j 0 0. So in βb, m j 0, and so βb αj R+ ∈ 6= = P − 6∈ and βb αj R+. So β(h j ) 0, and for a contradiction note that β(h j ) mi αi (h j ) + 6∈ = = i supp(βb) = P m A 0. ∈ i supp(βb) i ji ∈ <

34 Infinite-dimensional Lie algebras 35

© ª Step 4: J : i supp(α): k m supp(α). (This implies K R+.) = ∈ i = i = ⊆

Proof. Let i supp(α)\ J. Then βb αi R+. So by βb(hi ) 0 by A PREVIOUS Lemma. ∈ + 6∈ ≥ Let M supp(α) \ J be the connected component of i. So βb(h j ) 0 for all j M. Let P ⊆ ≥ ∈ βb0 : j M m j αj , so = ∈ X βb0(hi ) βb(hi ) m j αj (hi ). = − j supp(α)\M ∈

Recall that α(j)(hi ) Ai j . If j M, then βb(h j ) 0, and since Ai j 0, βb0(hi ) 0. By choosing i = ∈ ≥ ≤ ≥ carefully (there exist i 0 M and j 0 supp(α) \ M such that Ai j 0), we have βb0(hi ) 0 for some ∈ ∈ 0 0 < > i M. ∈ T Let AM be the principal matrix (Ai j )i,j M , and u (m j )j M . So ∈ = ∈ X βb0(hi ) Ai j m j for all i M, = j M ∈ ∈ P i.e. AM u 0 (and 0). So AM has finite type BY (REFERENCE). Let γ : i M (ki mi )αi , where ≥ 6= = ∈ − we have k m 0 for all i M. Then i − i > ∈ X α βb (kt mt )αt . − = t supp(α)\J − ∈ So X X (α βb)(hi ) (kt mt )αt (hi ) (k j m j )Ai j , − = t supp(α)\J − = j M − ∈ ∈ since M is the connected component of i in supp(α) \ J. But α(h ) 0 since α K , we have i ≤ ∈ βb(hi ) 0 from above. Therefore, γ(hi ) 0 for all i M. ≥ ≤ ∈ Set u (k m )T , and so A u 0. Since A has finite type, u 0. (That is, A ( u) 0, = i − i i M M ≤ M = M − ≥ and since A has finite∈ type, either u 0 or u 0.) M − > =

Step 5: K R+ . ⊆ im

Proof. We already know that K R+. If α K , then also 2α K . Hence 2α R+, and so α is ⊆ ∈ ∈ ∈ imaginary.

This concludes the proof of Theorem 16.4.

17 Affine Lie algebras and Kac-Moody Lie algebras

Simple Lie algebras of finite type are determined by the Dynkin diagram of their Cartan matrix, which is one of An, Bn, Cn, Dn, E6, E7 E8, F4 or G2. Write g for the corresponding simple Lie algebra. It acts on itself by the adjoint representation. There exists a unique highest root P θ a α h∗. = i i i ∈ Example. The Lie algebra associated to A has a 1 for all i 1,...,n, and the root spaces are n i = = E , i j. i j 6= We had constructed for all Kac-Moody Lie algebras a “standard form” , . We know that 〈− −〉 simple Lie algebras have a unique invariant inner product up to scale, the Killing form, so that the standard form must be a multiple of the Killing form. For the “long root” θ, we have θ,θ 2. 〈 〉 = (This follows from W -invariance of the form and α ,α 2/α .) 〈 i i 〉 = i ∼ 2θ Let hθ be the coroot corresponding under h h∗ to the element θ,θ θ. −−→ 〈 〉 =

35 36 Iain Gordon

Observation. Let A be the Cartan matrix for a Dynkin diagram of finite type. Let Ab be the matrix given by

Abi j Ai j for i, j {1,...,n}, = ∈ Ab00 2 , = n X Abi0 a j Ai j for i {1,...,n}, = − j 1 ∈ = n X Ab0j ci Ai j for j {1,...,n}. = − i 1 ∈ = This process produces the so-called untwisted affine Dynkin diagrams.

Exercise 17.1. Show that the matrix Ab is an affine generalised Cartan matrix. 2 2 6 Example. Starting with G2, we the Weyl group is W (G2) s1,s2 : s1 s2 e,(s1s2) e ∼ D6 ¡ 2 1 ¢ = 〈 = = = 〉 = (the dihedral group). The Cartan matrix is 3− 2 , and d diag(1,3). The positive roots are − = © ª R+ R+ α ,α ,α α ,α 2α ,α 3α ,2α 3α . = re = 1 2 1 + 2 1 + 2 1 + 2 1 + 2 To find the coroot h , we write h c h c h and compute: θ θ = 1 1 + 2 2 θ,α 2 α ,α 3 α ,α 4 3 1 α (h ) 2c 3c 〈 1〉 = 〈 1 1〉 + 〈 2 1〉 = − = = 1 θ = 1 − 2 θ,α 0 α (h ) c 2c 〈 2〉 = = 2 θ = − 1 + 2

This implies that hθ 2h1 h2. So the affine Cartan matrix Ab is = +  2 1 0  − Ab  1 2 1 . = − − 0 3 2 − Recall that we had the affine Kac-Moody Lie algebra

£ 1¤ Ldg g t ± Cd Cc = ⊕ ⊕ with structure

£ ¤ dg df ¡ ¢ X f λd µc,Y gλ0 µ0c [X ,Y ] f g λY t λ0X t Rest 0 f dg X ,Y c . ⊗ + + ⊗ + = ⊗ + ⊗ dt − ⊗ dt − = 〈 〉

Theorem 17.2. If g is simple with Dynkin diagram of finite type, then Ldg is isomorphic to the Kac-Moody Lie algebra of the associated untwisted affine Dynkin diagram.

Proof. We have generators and relations for the Kac-Moody Lie algebra: e0,...,en, f0,..., fn, h0,...,hn. Since g is simple, we already have generators E1,...,En, F1,...,Fn and H1,..., Hn. Set

£ 1¤ e : E 1 , f : F 1 , h : H 1 g t ± . i = i ⊗ i = i ⊗ i = 1 ⊗ ∈

We need to define the generators e0, f0, h0. We know that dimgθ dimg θ 1, and we have an = − = involution ω: g g such that ω(E ) F . There exists a non-degenerate pairing → i = − i

, : gθ g θ C . 〈− −〉 × − →

Choose Eθ,Fθ g θ with Eθ,Fθ 1 and ω(Eθ) Fθ. (For example, pick Feθ g θ, set Eeθ : ∈ ± 〈 〉 = ±=p − ∈ − = ω(Fθ) gθ, then Eeθ,Feθ ξ 0, so set Fθ : Feθ ξ etc.) Then let − ∈ 〈 〉 = 6= = 1 e : F t , f : E t − . 0 = θ ⊗ 0 = θ ⊗

36 Infinite-dimensional Lie algebras 37

So we have H : (h 1) Cd Cc h : h 1 c . = ⊗ ⊕ ⊕ 3 0 = − θ ⊗ + It follows that [e , f ] [F ,E ] 1 F ,E c h , 0 0 = θ θ ⊗ + 〈 θ θ〉 = 0 where we used [Fθ,Eθ] Fθ,Eθ h0 θ hθ. = 〈 〉 − = − We need a realisation of Ab given by α0,α1,...,αn H ∗, where α1,...,αn are extended from ∈ h∗ by α (d) α (c) 0 for all i 1,...,n. Do the same for θ h∗, which gives θ H ∗. Let δ be i = i = = ∈ ∈ the dual of d. Then α : θ δ H ∗, and 0 = − + ∈ ¡ ¢ H,Π {α ,α ,...,α },Π∨ {h ,h ,...,h } = 0 1 n = 0 1 n is a minimal realisation of Ab. (Clearly the αi and hi are all linearly independent.) So we get

α (h ) A for all i, j 1,...,n, j i = i j = n n X X α0(hi ) θ(hi ) δ(hi ) a j αj (hi ) a j Ai j Abi0 for i 0, = − + = − j 1 = − j 1 = 6= = = α (h ) similarly for j 0. j 0 = 6= The relations are as follows:

[e , f ] h for all i,[e , f ] 0 for i j, i i = i i j = 6= [x, e ] α (x)e for x H, all i,[x, f ] α (x)f for x H, all i. i = i i ∈ i = − i i ∈ To see the relation [e , f ] 0 when j i 0, note that [e , f ] [F ,F ] t 0. To see the relation i j = 6= = 0 j = θ j ⊗ = [x,e ] 0 when i 0, note that (with x h) i = = 0 ∈ [x λd µc,e ] [x ,F ] t λF t θ(x )F t λF t (δ θ)(x)e α (x)e . 0 + + 0 = 0 θ ⊗ + θ ⊗ = − 0 θ ⊗ + θ ⊗ = − 0 = 0 0 What about the map Le(Ab) Ldg – is it surjective, what is its kernel? It is surjective, i.e. the Lie → ª algebra M generated by e0,...,en, f0,..., fn and H is Ldg. Consider the set S : {x g : x t M : = ∈ ⊗ ∈ • S 0 because S contains F . 6= θ • g acts on S, i.e. S is an ideal in g,[y 1,x t] [y,x] t. ⊗ ⊗ = ⊗ • By simplicity, S g. = k 1 k £ 1¤ Since [g,g] g, we have [g t,g t − ] g t for k 2, and this g t ± M, where c,d H. = ⊗ ⊗ = ⊗ ≥ ⊆ ∈ Hence Ldg M. For x H, = ∈ £x,v t i ¤ £x µd λc,v t i ¤ α(x )v t i µi v t i (α iδ)(x)v t i , α ⊗ = 0 + + α ⊗ = 0 α ⊗ + α ⊗ = + α ⊗ P ¡ ¢ i.e. Ldg L0 (i,α) (0,0) Ldg α iδ. Next, = ⊕ 6= + X ¡ ¢ J (L0 J) L (Ldg)α iδ , = ∩ ⊕ (i,α) (0,0) ∩ + 6= i so J Lα iδ 0 for some (i,α) if J 0. But x t J for 0 x gα. Pick y g α such that ∩ + 6= 6= ⊗ ∈ 6= ∈ ∈ − x, y 0. But 〈 〉 6= £ i i ¤ 0 x t , y t − [x, y] 1 i x, y c H , = ⊗ ⊗ = ⊗ − 〈 〉 ∈ so i 0, so [x, y] 0 x, y h0 0, contradiction. = = = 〈 〉 α 6=

37 38 Iain Gordon

Remark 17.3. Observe that the real roots are {α iδ : α 0}, the imaginary roots are {iδ : i 0}, + 6= 6= and the multiplicity of the set of imaginary roots is the rank of g.

The Kac-Moody Lie algebras associated to affine Dynkin diagrams L (eg) correspond precisely to affine Lie algebras g[t ±] Cd Cc : Ldg. ⊕ ⊕ = Dynkin diagrams have automorphisms: [Explain the procedure that gets from the orbit space to the affine Lie algebra.] [something] produces the so-called twisted affine Dynkin diagrams from the classical Dynkin diagrams.

T 2 A2n 1 Ben Ae2n 1 − → = − 2 A2n (n 2) Ce0 Ae2n ≥ → n = 2 Dn 1 Cen0 Den 1 + → = + A ? 2 → T 2 E6 Fe Ee6 → 4 = T 3 D4 Ge De4 → 2 = The diagram automorphism σ extends to an automorphism, also denoted by σ, of g and H via

σ(c) c , σ(d) d , e e , f f , h h . = = i 7→ σ(i) i 7→ σ(i) i 7→ σ(i) All that remains is to define the action of σ on t. If σ has order m, let ξ e2πi/m, and let τ act on = Aut(Ldg) by i i i τ(x t ) ξ− σ(x) t , τ(c) c , τ(d) d . ⊗ = ⊗ = = Theorem 17.4. If ∆ is the (classical) Dynkin diagram of the simple Lie algebra g, then the twisted affine Kac-Moody Lie algebra

¡ ¢T n o Ldg x Ldg : τ(x) x = ∈ = is isomorphic to the Kac-Moody Lie algebra corresponding to the twisted affine Dynkin diagram ∆e.

18 The Weyl-Kac formula

Throughout this section, let L : L (A) denote a symmetrizable Kac-Moody Lie algebra. We = want to study L through its representations.

18.1 Category O

The idea is to study representations of L by their combinatorics.

Definition 18.1. Let V be a representation of L . We say that V belongs to category O, or V O, ∈ if the following hold: L © ª • V λ H Vλ, where Vλ v V : h.v λ(h)v for all h H . = ∈ ∗ = ∈ = ∈

• dimC V for all λ H ∗. λ < ∞ ∈

• There exist λ1,...,λs H ∗ such that if Vλ 0, then λ λi for some i. (This is to say that ∈ Pn 6= < λi λ Q , i.e. λi λ j 1 n j αj for n j Z 0.) − ∈ + − = = ∈ ≥

38 Infinite-dimensional Lie algebras 39

Definition 18.2. There is a character function X ch: O Fun(H ∗), V ch(V ) : dimVλeλ , → 7→ = λ H ∈ ∗ where eλ is the character function for λ H ∗, and eλeµ eλ µ. Explain what eλ is. Is it the ∈ = + same as e(λ) below? We define © ª R : f Fun(H ∗): λ ,...,λ s.t. supp(f ) S(λ ) S(λ ) , = ∈ ∃ 1 s ⊆ 1 ∪ ··· ∪ s where © ª S(λ) : λ Q µ H ∗ : µ λ . = − + = ∈ ≤ Observe that the set R is a ring and that the character function is in fact ch: O R. → Remark 18.3. Suppose 0 A B C 0 is a short exact sequence of L -representations. → → → → Then if B O, it follows that A,C O, but the converse does not hold. However, if D,E O, then ∈ ∈ ∈ D E O. (Here x.(d e) x.d e d x.e.) ⊗ ∈ ⊗ = ⊗ + = ⊗ The category O has universal objects (cf. the Virasoro algebra, in particular Equation (7.1)): Recall that L n H n . For λ H ∗, we define the Verma module = − ⊕ ⊕ + ∈

M(λ) : U(L ) U(n H) C(λ) , (18.1) = ⊗ +⊕ where the action of U(n H) is as follows: + ⊕

1λ C(λ), n .C(λ) 0 , and h.1λ λ(h)1λfor all h H. ∈ + = = ∈ There is a special element v : 1 1 that satisfies λ = ⊗ λ e .v 0 and h.v λ(h)v . i λ = λ = λ

By the Poincaré-Birkhoff-Witt Theorem (Theorem 6.5), M(λ) H U(n ) Cλ. explain this nota- =∼ − ⊗ tion; is it “isomorphism as H-algbras”? We have further ( 0 if µ λ, M(λ)µ 6≤ = P (λ µ) if µ λ, − ≤ where P is the Kostant partition function: P (ν) : dimU(n ) ν, which is by the Poincaré- = − − Birkhoff-Witt Theorem the number of ways to write ν Q as a sum of positive roots. ∈ + Lemma 18.4. e chM(λ) λ , Y ¡ ¢mα = 1 e α α Φ − − ∈ + where Φ is the set of positive roots and mα the multiplicity of the root α. + Proof. We need

X X ³ Y ¡ ¢mα ´ P (ν)e ν : dimU(n ) νe ν 1 e α . ν Q − = ν Q − − − = α Φ − − ∈ + ∈ + ∈ +

By Poincaré-Birkhoff-Witt, a basis for U(n ) ν is − −

Y ¡ (i)¢aα,i X fα , where aα,i α ν . α Φ α,i = ∈ +

39 40 Iain Gordon

A very similar argument shows that the representation M(λ) has a unique maximal submod- ule J(λ), and hence a unique maximal (irreducible) quotient:

Definition 18.5. If M(λ) is the Verma module (as in (18.1)) for the Kac-Moody Lie algebra L and J(λ) M(λ) is the unique maximal submodule, let ⊂ L(λ) : M(λ)±J(λ) = be the unique maximal quotient of M(λ).

Observe that every irreducible object in O is of the form L(λ) for some λ. (We did something similar for the Virasoro algebra.) The goal is to compute chL(λ), which is unkonwn in general.

Proposition 18.6. Let V O and λ H ∗. Then there exists a filtration ∈ ∈ V V V V V 0 = 0 ⊃ 1 ⊃ 2 ⊃ ··· ⊃ t = ± ± such that either Vi Vi 1 L(µ) for some µ λ or (Vi Vi 1)µ 0 for all µ λ. − =∼ ≥ − = ≥ If µ λ, then the number of factors L(µ) is independent of the choice of the filtration and of ≥ the choice of λ.

Definition 18.7. We define £V : L(µ)¤ to be the multiplicity of L(µ) in a filtration V from Proposi- tion 18.6. P Proof of Proposition 18.6. Existence: V has spectrum bounded above, a(V,λ) µ λ dimVµ is = ≥ finite. We use induction on a(V,λ). If a(V,λ) 0, then we have a filtration V V V 0. = = 0 ⊃ 1 = If a(V,λ) 0, then there exists a weight µ with µ λ in V . Choose a maximal weight and let > ≥ 0 v Vµ. By maximality, n .v 0, so N : U(L ).v V , so N is a quotient M(µ)  N, vµ v. 6= ∈ + = = ⊂ 7→ Let N 0 be the image of J(µ) M(µ) under the quotient map. Then V N N 0 0, and ⊂ ⊇ ⊃ ⊇

a(N 0,λ) a(V,λ), a(V /N,λ) a(V,λ) < < ± because (N N 0) 0. So induction works. µ 6= The proof of uniqueness is tricker, but not very illuminating. P £ ¤ Corollary 18.8. If V O, then chV λ H V : L(λ) chL(λ). ∈ = ∈ ∗

18.2 Kac’s Casimir

If L is a finite-dimensional simple Lie algebra, then there is a special element, the Casimir operator fd X Ω : xi yi U(L ), = i ∈ where {xi } and {yi } are dual bases with respect to the non-degenerate invariant form on g. It is easy to see that in fact Ωfd lies in the centre of U(L ).

There exists an explicit version (via the Killing form). If {h10 ,...,hn0 } is a basis for h∗ and {h00,...,hn00} is the dual basis, and if eα Lα, fα L α such that [eα, fα] hα0 for all α Φ , then 1 ∈ ∈ − = ∈ + n fd X X ¡ ¢ Ω hi0 hi00 eα fα fαeα = i 1 + α Φ + = ∈ + n X X ¡ ¢ hi0 hi00 2fαeα hα0 = i 1 + α Φ + = ∈ + n X X ¡ ¢ hi0 hi00 2 fαeα 2hρ0 , = i 1 + α Φ + = ∈ +

40 Infinite-dimensional Lie algebras 41

P where 2ρ α Φ α satisfies ρ(hi0 ) 1, hρ0 ,h ρ(h). = ∈ + = 〈 〉 = For infinite-dimensional Lie algebras, we define

X X X (i) (i) Ω : hi0 hi00 2hρ0 2 fα eα , = i + + α Φ i ∈ + (i) (i) D (i) (j)E where the fα form a basis for L α and the eα form a basis for Lα such that fα ,eα δi j ; − = the h0 and h00 are mutually dual bases of H, and h0 ,h ρ(h) defined by ρ(h ) 1 for all i i 〈 ρ 〉 = i = i 1,...,n. (Note that ρ is not unique, as the h do not necessarily form a basis!) = i Then Ω acts on V for V O, denoted by Ω . ∈ V Lemma 18.9 (Technical manipulation of elements). ¡ ¢ Ω End id , u : V V, Ω .u u.Ω for u U(L ) ∈ O O → V = V ∈ explain this more verbosely

Proposition 18.10. If M(λ) is the Verma module for the Kac-Moody Lie algebra L as in (18.1), then Ω ¡ λ ρ,λ ρ ρ,ρ ¢id , M(λ) = 〈 + + 〉 − 〈 〉 M(λ) where the inner product on H ∗ is induced from inner product on H.

Proof. With v M(λ), we have λ ∈ ¨ ¡X X X (¨i) ¨(i)¢ Ω.vλ hi0 hi00 2hρ0 2 ¨fα eα .vλ = i + + α Φ i ¡X ∈ + ¢ λ(hi0 )λ(hi00) 2λ(hρ0 ) vλ = i + ¡ λ,λ 2 λ,ρ ¢v . = 〈 〉 − 〈 〉 λ Now use Lemma 18.9 an the fact that M(λ) U(L ).v . = λ Corollary 18.11. Ω ¡ λ ρ,λ ρ ρ,ρ ¢id . L(λ) = 〈 + + 〉 − 〈 〉 L(λ)

18.3 The Weyl-Kac formula L Definition 18.12. A representation V of L is integrable if V λ H Vλ and ei , fi : V V act = ∈ ∗ → locally nilpotently for i 1,...,n. = Example. The adjoint representation L is integrable. Any finite-dimensional representation is integrable.

Lemma 18.13. If V is integrable, then dimV dimV for all w W. λ = w(λ) ∈ Sketch of proof. It is easy to show that dimV dimV for each simple reflection s . An sl - λ = si (λ) i 2 calculation for e ,h , f acting on V shows that integrability implies that V is locally finite. 〈 i i i 〉 Finally, sl -representation theory shows that dimV dimV . 2 λ = si (λ)

Proposition 18.14. L(λ) is integrable if and only if λ(hi ) Z 0 for all i 1,...,n. In this case we ∈ ≥ = say that λ is dominant and integral.

Remark 18.15. Since integrability says that elements of the Lie algebra act locally nilpotently, the exponential is well defined and we may pass from the Lie algebra to a Lie group. In some sense integrability is an analogue of finite-dimensionality (of the representation).

41 42 Iain Gordon

Proof. We know the sl2-representation theory. Let vλ be the highest weight of L(λ). Then r {fi vλ}r 0 span a sl2-representation (where sl2 (ei , fi ,hi )). This representation is finite-dimen- ≥ r = sional if and only if f vλ 0 for r 0, and only if λ(hi ) Z 0 i = À ∈ ≥ Conversely, if λ(hi ) Z 0 for all i, then take v L(λ) such that v u.vλ for some u U(L ) ∈ ≥ ∈ = ∈ and apply the Leibniz rule:

N Ã ! N X N ¡ k ¢ N k fi (u.vλ) (ad fi ) (u) .(fi − .vλ) 0 , = k 0 k = = k N k N where (ad f ) 0 for k 0 and f − 0 for N k 0. So for N 0, f (u.v ) 0. i = À i = − À À i λ =

Theorem 18.16 (Weyl-Kac character formula). Let λ X +, where ∈ © ª X + : λ H ∗ : λ(h ) Z, λ(h ) 0 for all i 1,...,n . = ∈ i ∈ i ≥ = Then X ²(w)e¡w(λ ρ) ρ¢ w W + − chL(λ) ∈ , Y mα = ¡1 e( α)¢ α Φ − − ∈ + where ²: W { 1} is defined as a group homomorphism with ²(s ) 1. → ± i = − Clarify whether e( α) is the same as e α before, and maybe recall what it is. − − Proof. We have X chM(λ) £M(λ): L(µ)¤chL(µ), = µ H ∈ ∗ and µ λ if £M(λ): L(µ)¤ 0. The Casimir operator Ω acts on ≤ 6= ° °2 ° °2 M(λ) via °λ ρ° °ρ° , + − ° °2 ° °2 and on L(µ) via °µ ρ° °ρ° . + − ° °2 ° °2 Therefore °λ ρ° °µ ρ° [check; condition missing?], and + = + X chM(λ) £M(λ): L(µ)¤chL(µ) . (18.2) = µ λ ≤ λ ρ 2 µ ρ 2 k + k =k + k Now define n ° °2 ° °2o C : µ λ : °λ ρ° °µ ρ° . λ = ≤ + = + Then µ C C C . Also note that £M(λ): L(λ)¤ 1. Running through the entire set C and ∈ λ ⇒ µ ≤ λ = λ looking at the formula for chM(µ) in Equation (18.2) we find that for all µ C ∈ λ X chM(µ) aτµ chL(τ), = τ C ∈ λ where a 1 and a 0 unless τ µ ( i.e. with a total ordering on C that refines the ordering µµ = τµ = ≤ λ ). The matrix (aτµ)τ,µ C is triangular with 1 on the diagonal. Hence ≤ ∈ λ X chL(λ) bλµ chM(µ) = µ C ∈ λ with b Z. For example, λµ ∈ Y mα X e(ρ) ¡1 e( α)¢ chL(λ) b e(µ ρ). − − = λµ + α Φ µ Cλ ∈ + ∈

42 Infinite-dimensional Lie algebras 43

Since L(λ) is integrable, chL(λ) is W -invariant. So

³ Y mα ´ s e(ρ) ¡1 e( α)¢ ¡1 e( α )¢chL(λ) i − − − − i α Φ \{αi } ∈ + Y mα e(ρ α ) ¡1 e( α)¢ ¡1 e( α )¢chL(λ) = − i − − − − i α Φ \{αi } ∈ + Y mα e(ρ) ¡1 e( α)¢ chL(λ), = − α Φ − − |∈ + {z } ∆ whence we deduce

w¡e(ρ)∆chL(λ)¢ ²(w)e(ρ)∆chL(λ) ⇒ ¡X ¢= X w bλµe(µ ρ) ²(w) bλµe(µ ρ) ⇒ µ + = µ + b ²(w)b if w(µ ρ) ν ρ . ⇒ λµ = λν + = + Suppose that b 0. Then b 0 for all ν such that w(µ ρ) ν ρ and ν λ. Pick ν such λµ 6= λµ 6= + = + ≤ that λ ν has minimal height. Then ν ρ X + since − + ∈ s w(µ ρ) s (ν ρ) ν ρ (ν ρ)(h )α . i + = i + = + − + i i We have ν ρ,ν ρ λ ρ,λ ρ since ν C , and so (λ ρ) (ν ρ) P k α with k 0 〈 + + 〉 = 〈 + + 〉 ∈ λ + − + = i i i i ≥ for all i. So ° °2 ° °2 0 °λ ρ° °ν ρ° λ ν 2ρ,λ ν = + − + = 〈X+ + − 〉 ki λ ν 2ρ,αi = i 〈 + + 〉 X αi ,αi ¡ ¢ ki 〈 〉 λ ν 2ρ (hi ) = i 2 + + But note that each summand is non-negative: k 0, α ,α 0 and (λ ρ)(h ),(µ ρ)(h ) 0. i ≥ 〈 i i 〉 ≥ + i + i ≥ Therefore λ ν. So = b 0 µ ρ w(λ ρ) for some w W λµ 6= ⇒ + = + ∈ bλµ ²(w)bλλ ²(w) ⇒ = =X e(ρ)∆chL(λ) ²(w)e¡w(λ ρ)¢ ⇒ = w W + P ∈ ¡ ¢ w W ²(w)e w(λ ρ) ρ chL(λ) ∈ + − . ⇒ = ∆

Corollary 18.17 (Kac denominator formula).

Y mα X e(ρ) ¡1 e( α)¢ ²(w)e¡w(ρ)¢ α Φ − − = w W ∈ + ∈ Proof. Take λ 0, i.e. the trivial representation L(0) C. = =

There exists a formula for dimL(λ) for λ X + (the Kostant partition function). µ ∈ Exercise 18.18. Using this style of representation theory (and more work), one can show the presentation for a symmetrisable Kac-Moody Lie algebra is what we used plus

1 A 1 A (ade ) − i j e 0 , (ad f ) − i j f 0 i j = i j = (see Carter §19.4, Kac Theorem 9.11).

43 44 Iain Gordon

19 W − K + A = M

Let L g be an untwisted Lie algebra (i.e. g L (Ab) for some untwisted, affine generalised = b b = Cartan matrix Ab), where g L 0 is a simple, finite-dimensional Lie algebra with root system Φ0 = and Weyl group W 0. We know the roots of L :

Φ ©α nδreal : α Φ0,n Zª ©nδim : n 0ª , = + ∈ ∈ ∪ 6= 0 where the imaginary roots have multiplicity rkL : l. The root δ a0α0 al αl is dual to 1 = = +···+ a− d (but a 1 in the untwisted case). The positive roots are 0 0 = Φ ©α Φ0 ª ©α nδ : n 0,α Φ0ª ©nδ : n 0ª . + = ∈ + ∪ + ≥ ∈ ∪ > 19.1 Affine Weyl groups

The Weyl group of L is W s ,s ,...,s , and W 0 s ,...,s . Let θ : a α a α Φ0, = 〈 0 1 l 〉 = 〈 1 l 〉 = 1 1 + ··· + l l ∈ so θ,θ 2. We can write 〈 〉 = l X αi ,αi 2αi θ ai 〈 〉 , = i 1 2 αi ,αi = 〈 〉 and the associated reflection is s W 0. Writing s (h) h θ(h)h , we get θ ∈ θ = − θ l l X αi ,αi X 2ci hθ ai 〈 〉hi hi c c0hh c h0 . = i 1 2 = i 1 2 = − = − = = The elements {s0sθ,s1,...,sl } generate W . We have

s s (h) s ¡h θ(h)h ¢ 0 θ = 0 − θ h α (h)h θ(h)¡h α (h )h ¢ = − 0 0 − θ − 0 θ 0 h δ(h)h ¡θ(h) δ(h)¢c = + θ − + h δ(h)h ¡ h ,h 1 h ,h δ(h)¢c . = + θ − 〈 θ 〉 + 2 〈 θ θ〉 For x H 0, there is a map ∈ t : H H , h h δ(h)x ¡ x,h 1 x,x δ(h)¢c . x → 7→ + − 〈 〉 + 2 〈 〉 1 0 Lemma 19.1. tx ty tx y and wtx w − tw(x) for w W . = + = ∈ Proof. Easy calculation.

0 © ª 0 Proposition 19.2. W t(M)oW , where t(M) tm : m M and M H is the space generated = = ∈ ⊂ as an abelian group by w(h ) for all w W 0. θ ∈ Ll Remark 19.3. In the untwisted case, an elementary computation shows that M i 1 Zhi . = = 0 ¡ ¢ 1 Proof. We have H H Cc Cd, and δ w(h) (w − δ)(h) δ(h). The set = ⊕ ⊕ = = H : ©h H : δ(h) 1ª ©h0 λc dª 1 = ∈ = = + + has an action of W . Since w(c) c for all w W , this action descends to an action of W on ± 0 0 0= ∈ 0 0 0 0 0 H1 Cc H (and W acts on H as usual, and tm(h ) h m); h λc d h d h . =∼ = + + + 7→ + 7→ ∼ 0 Since H = H ∗, we need the action of W on H ∗. We have W t(M ∗)oW , where M ∗ is the −→ = lattice spanned by long roots and

¡ 1 ¢ tα(λ) λ λ(c)α (λ,α) (α,α)λ(c) δ . = + − + 2

44 Infinite-dimensional Lie algebras 45

19.2 Formulae

Recall that ρ(hi ) 1 and ρ(d) 0. The number ρ(c) h∨ c0 cl is the dual Coxeter number. 0 0 = 0 = = = +···+ With ρ (H )∗ such that ρ (h ) 1 for i 1,...,l we have ∈ i = = 0 w(ρ) ρ w tα(ρ) ρ − = ³ − ´ w 0 ρ ρ(c)α ¡ ρ,α 1 α,α ρ(c)¢δ ρ = + − 〈 〉 + 2 〈 〉 − ¡ 0 0 0 0 ¢ 0 0 0 ρ h∨α,ρ h∨α ρ ,ρ δ w (h∨α ρ ) ρ 〈 + + 〉 − 〈 〉 . = + − − 2h∨ We define ° °2 ° °2 c(λ) : °λ ρ0° °ρ0° , = + − 0 so that the finite-dimensional Casimir operator on L (λ) is Ω 0 c(λ)id (cf. Corollary 18.11). L (λ) = Then we have a character formula X . X χ0(λ) : chL0(λ) ²(w)e¡w(λ ρ0) ρ0¢ ²(w)e¡w(ρ0) ρ0¢ . = = w W 0 + − w W 0 − ∈ ∈ So

X ¡ ¢ X X 0 ¡ 0 0 0¢ ³ c(h∨α) ´ ²(w)e w(ρ) ρ ²(w )e w (h∨α ρ ) ρ e − δ − = 0 + − 2h w W α M ∗ w W ∨ ∈ ∈ ∈ X 0 0 0 0 X 0 ³ c(h∨α) ´ ²(w )e(w ρ ρ ) χ (h∨α)e − δ = 0 − 2h w W α M ∗ ∨ ∈ ∈ Y ¡ ¢ X 0 ³ c(h∨α) ´ 1 e( α) χ (h∨α)e − δ . = 0 − − 2h α Φ α M ∗ ∨ ∈ + ∈ Theorem 19.4 (MacDonald). Set q : e( δ). Then = −

Y n l Y n X 0 c(h∨α)/2h∨ (1 q ) (1 q e α) χ (h∨α) q , (19.1) − 0 − − = n 0 α Φ α M ∗ > ∈ ∈ where q e( δ). = − 0 0 1 Example. Consider Ae1. We have l 1, Φ { α1}, h∨ 2, M ∗ Zα1 and ρ α1. Set z : e α ; = = ± = = = 2 = − 1 then n n 1 0 e(nα1) e( (n 1)α1) z− z + χ (nα1) − − + − . = 1 e( α ) = 1 z − − 1 − The Casimir operator is multiplication by

c(2nα ) ­¡2n 1 ¢α ,¡2n 1 ¢α ® ­ 1 α , 1 α ® 4n(2n 1) . 1 = + 2 1 + 2 1 − 2 1 2 1 = + Hence the right-hand side of Equation (19.1) is

2n 2n 1 X z− z + n(2n 1) 1 X m m m(m 1)/2 − q + ( 1) z q − . n Z 1 z = 1 z m Z − ∈ − − ∈ Plugging in the left-hand side, we get

Y n n 1 n 1 X m m m(m 1)/2 (1 q )(1 q − z)(1 q z− ) ( 1) z q − , n 0 − − − = m Z − > ∈ which is the Jacobi triple product identity.

45 46 Iain Gordon

Exercise 19.5. Do this for Ae10 :

Y n n 1 n 1 2n 1 2 2n 1 2 X ¡ 3n 3n 1¢ n(3n 1)/2 (1 q )(1 q z− )(1 q − z)(1 q − z− )(1 q − z ) z z− + q − n 0 − − − − − = n Z − > ∈ This is the quintuple product identity, known already in antiquity.

If we set e 1 for all α Φ0, this produces other identities. (For instance, dimL0(λ) d 0(λ).) α = ∈ = Theorem 19.6. Let φ(q) : (1 q)(1 q2)(1 q3) . What about convergence? Then = − − − ··· 0 dimL X 0 c(h∨α)/2h∨ φ(q) d (h∨α) q . = α M ∈ ∗ 1 3 P n(2n 1) Example. For Ae , φ(q) n Z(4n 1)q + . = ∈ + Exercise 19.7. Repeat the entire section for the twisted version.

20 KP hierarchy and Lie theory

In 1834 John Scott Russel created the first soliton in history in historic Edinburgh. The basic governing equation of a soliton (a solitary wave) is the Korteveg-de Vries (KdV) equation:

1 3 ut uxxx u ux 0 (20.1) − 4 − 2 =

1 1 It is a non-linear partial differential equation in R + that describes the behaviour of a (one- dimensional) wave in shallow water. There is a two-dimensional version, called the KP equation:

³ 1 3 ´ 3 ut uxxx u ux uy y (20.2) − 4 − 2 x = 4

Greek mythology. An integrable system is a Procrustean bed. Integrable systems have

• exact solutions (in this case solitons),

• remarkable symmetries,

• Lax-pair formalism.

The KP equation fints into an infinite family of PDEs, all of which are governed simulta- neously. We will see (a) solutions of Equation (20.2), and (b) a family of solutions, which is an inifinite-dimensional homogeneous manifold. (“If you have one solution, you have every solution.”) L (0) Recall from Section 9.1 that from a space V n Z Cvn we can form the space F , on which = ∈ we have an action of gl End(V ), which is spanned by elements E of “infinite matrices with ⊂ i j only finitely many non-zero∞ entries”. The Lie group attached to this algebra is n o GL : A (ai j )i,j Z : A is invertible and almost all ai j δi j 0 ∞ = = ∈ − = We define n o Ω : GL .v0, 1, 2,... u0 u 1 u 2 : u m v m for m 0 , = ∞ − − = ∧ − ∧ − ∧ ··· − = − À

46 Infinite-dimensional Lie algebras 47

where v0, 1, 2,... is the vacuum vector in degree 0. There is a natural scaling action on Ω, so if we − − define a Graßmannian-like object

n k M ¡ ± k ¢ o Gr : U V : there is k such that U V ≤ Cvi and dim U V ≤ k , = < ≥ = i k = ≤ then we have a bijection (Exercise: Check!)

∼ M PΩ = Gr , u0 u 1 u 2 Cui . −−→ ∧ − ∧ − ∧ ··· 7→ i 0 ≤ (0) (1) (0) ( 1) There are two operators, F F and F F − , respectively via V andV ∗ (the restricted → → dual, cf. Theorem 7.3), defined as follows. For v V and f V ∗, we have operations ∈ ∈ v.v v v v v , and b i0 ∧ i1 ∧ ··· = ∧ i0 ∧ i1 X j fb.vi0 vi1 ( 1) vi0 vi1 vi j 1 f (vi j ) vi j 1 . ∧ ∧ ··· = j 0 − ∧ ∧ ··· ∧ − ∧ + ∧ ··· ≥

(0) The element Ei j acts on F via vbi vb∗j .

Lemma 20.1. If τ Ω, then ∈ X vbj (τ) vb∗j (τ) 0 , (20.3) j Z ⊗ = ∈ and if τ 0 and τ satisfies Equation (20.3), then τ Ω. 6= ∈

Proof. For τ v0, 1, 2,... Equation (20.3) is obviously satisfied. We prove that the left-hand side = − − of Equation (20.3) is GL -invariant: ∞ Ã ! X 1 X A vbj (τ) vb∗j (τ) A− vbj (τ) vb∗j (τ). j Z ⊗ = j Z ⊗ ∈ ∈

Then it follows that Equation (20.3) holds in general, since τ Av0, 1, 2,..., and so = − − Ã ! X 1 A vbj (τ) vb∗j (τ) A− A v0, 1, 2,... 0 . j Z ⊗ − − = ∈

Conversely, assume that τ satisfies Equation (20.3) and that τ 0. Write τ PN c τ , where 6= = k 1 k k τ are semi-infinite monomials and all c 0. Without loss of generality, τ has maximal= degree k k 6= 1 among the τ . If τ is of the form τ E .τ with i j, then E 2 .τ 0, and so exp¡ c E ¢.τ k 2 2 = i j 1 < i j 1 = − 2 i j removes the τ2-term. P ¡ ¢ (Note that the degree is deg(vi) m 0(im m) and that exp c2Ei j GL .) = ≥ + − ∈ ∞ Now by GL -invariance, the new expression also satisfies Equation (20.3), and by iterating ∞ we obtain τ τ φ, where φ has no terms of the form E .τ . Then Equation (20.3) implies = 1 + i j 1 X ¡ ¢ vbj .τ1 vb∗j .φ vbj .φ vb∗j .τ1 vbj .φ vb∗j .φ 0 . j Z ⊗ + ⊗ + ⊗ = ∈ So for each jwe must have, up tp sign, that

v .τ v .(semi-infinite wedge in φ), bj 1 = bk which forces E .τ , contradiction. Clarify this sentence. We conclude that φ 0. k j 1 =

47 48 Iain Gordon

(0) Recall further that we had shown that F C[x1,x2,...] B(0,1), the bosonic representation. =∼ = Since gl acts on F (0), it also acts on C[x ,x ,...]. How does it act, and where do the v ( v ) 1 2 λ = i go? We introduce∞ the vertex operators

(0) (1) X j (0) (1) X j X : F Fd , X (u) : u vbj , and X ∗ : F Fd , X ∗(u) : u− vb∗j . → = j Z → = j Z ∈ ∈ These become operators on B(0,1), too, by the following fact:

j j ³X j ´ ³ X u− ∂ ´ ³ X j ´ ³X u− ∂ ´ X (u) u exp u x j exp , X ∗(u) exp u x j exp 7→ j 1 − j 1 j ∂x j 7→ − j 1 j 1 j ∂x j ≥ ≥ ≥ ≥ This answers the question how gl acts. For the image of vλ, note that vλ Sλ(x), the Schur ¡∞ ¢ 7→ polynomial, given by Sλ(x) det Sλ j 1(x) , where the elementary Schur polynomials Sk are = i + − such that X k ³ X∞ k ´ Sk (x)z exp xk z . k Z = k 1 ∈ = We translate Ω to B(0,1) C[x ,x ,...] by identifying = 1 2 (0) (0) F F with C[x0 ,x0 ,...] C[x00,x00,...]. ⊗ 1 2 ⊗ 1 2 0 Then X (u).τ X ∗(u).τ has vanishing u -term, i.e. ⊗ j ³X j ¡ ¢´ ³X u− ¡ ¢´ u exp u x0 x00 exp ∂ ∂ .τ(x0)τ(x00) j j x00j x0j j 1 − j 1 j − ≥ ≥ 0 has vanishing u -term. Rewrite x0 X Y , x00 X Y . So ( ).τ C[x ,x ,...] belongs to Ω if = − = + − ∈ 1 2 and only if the u0-term vanishes in the expression

j ³ X j ´ ³X u− ∂ ´ u exp 2u Y j exp .τ(X Y )τ(X Y ) . (20.4) − j 1 j 1 j ∂Y j − + ≥ ≥

The Hirota form. Let P C[x ,x ,...] with a finite number of x ’s, and let f and g be functions ∈ 1 2 i in those variables. Define an operation

³ ∂ ∂ ´¡ ¢¯ (P f .g)(x) : P , ,... f (x1 u1,x2 u2,...)g(x1 u1,x2 u2,...) ¯ . = ∂u ∂u − − + + ¯u 0 1 2 = So Equation (20.4) becomes

³ X j ´³ X j ´ u u S j ( 2Y ) u− S j (e∂Y ) .τ(X Y )τ(X Y ), j Z − j Z − + ≥ ≥ ¡ ∂ 1 ∂ 1 ∂ ¢ where Y (Y1,Y2,...) and e∂Y , , ,... . It follows that Equation (20.3) is equivalent = = ∂Y1 2 ∂Y2 3 ∂Y3 to X S j ( 2Y )S j 1(e∂Y )τ(X Y )τ(X Y ) 0 . (20.5) j 0 − + − + = ≥ We compute: ¯ S j 1(e∂Y )τ(X Y )τ(X Y ) S j 1(e∂u)τ(X Y u)τ(X Y u)¯ + − + = + − − + + ¯u 0 = ³X ∂ ´ ¯ S j 1(e∂u)exp Ys τ(X u)τ(X u)¯ = + ∂u − + ¯u 0 s 1 s = ³ ≥ ´ ¯ X ¯ S j 1(Xe)exp Ys Xs τ(X )τ(X )¯ , = + s 1 u 0 ≥ =

¡ 1 1 ¢ where Xe X1, X2, X3,... . = 2 3

48 Infinite-dimensional Lie algebras 49

Exercise 20.2. Show that P f .f 0 if P(x) P( x). = = − − Theorem 20.3. τ Ω if and only if τ is a solution of ∈ ³X∞ ¡P ¢´ S j ( 2Y )S j 1(Xe)exp s 1 Ys Xs τ(X )τ(X ) 0 , (20.6) j 0 − + ≥ = = where Y1,Y2,... are parameters, i.e. we have an infinite family of differential equations. ¡P ¢ Now expand Equation (20.6) in the Yi . The variable Yr appears in exp s 1 Ys Xs once with ≥ coefficient X . In the expression S ( 2Y ) the variable Y appears when j r , with coefficient r j − r = 2. So Equation (20.6) has the Y -term − r ¡ ¢ ¡ ¢ S1(Xe)Xr 2Sr 1(Xe) ττ 0 X1Xr 2Sr 1(Xe) ττ 0 . − + = ⇒ − + = If r 1 or r 2, this is odd and there are no conditions. If r 3, the term in parentheses is = = = x4 3x2 4x x . Setting x : x , y : x and t : x , we obtain 1 + 2 − 1 3 = i = 2 = 3 ³ ∂4 ∂2 ∂2 ´ ¯ 3 4 τ(X u)τ(X u)¯ 0 . 4 + 2 − ∂u ∂u + − ¯u 0 = ∂u1 ∂u2 1 3 =

2 A long exercise shows that u(x, y,t) 2 ∂ ¡logτ¢ is a solution of the KP equation. = ∂x2

21 The Kazhdan-Lusztig Conjecture

Let A be a symmetrisable generalised Cartan matrix.

21.1 The Shapovalov form and Kac-Kazhdan formulas

The character forlumals for integrable irreducible reprepresentations L(µ) involves the Weyl group. Recall that we had the Verma modules M(λ), and for M(λ µ) the character formula has + the form L¡w(λ ρ)¢. The Kac-Kazhdan formula is interesting because it shows that the Weyl + group W does play a rôle in general.

Definition 21.1. Let g be a Kac-Moody Lie algebra. We have a formula for the universal envelop- ing algebra of g:

U(g) U(n ) U(H) U(n ) U(H) ¡n .U(g) U(g.n ¢ . = − ⊗ ⊗ + = ⊕ − + + Denote by π: U(g U(H) the projection onto the first direct summand. → Recall that U(H) is a polynomial ring. There exists an anti-linear involution σ: g g, → σ(e ) f , and σ: h h. Let i = i → F : U(g) U(g) U(H), F (x, y) π¡σ(x)y¢ . ⊗ → =

This is a family of symmetric, bilinear forms over H ∗.

Definition 21.2. The Shapovalov form F (λ): M(λ) M(λ) C is defined by ⊗ → F (λ)(xv , yv ) F (x, y)(λ). λ λ = Exercise 21.3. Check that the Shapovalov form is well defined.

Lemma 21.4. The Shapovalov form has the following properties:

49 50 Iain Gordon

• F (λ)¡xv,w¢ F (λ)¡v,σ(v)w¢ for x U(g), v,w M(λ). = ∈ ∈ • F (λ)¡M(λ) ,M(λ) ¢ 0 if µ µ . µ1 µ2 = 1 6= 2 • radF (λ) is the maximal submodule of M(λ).

We would like to find out about F (λ), or at least about detF (λ)η, the restriction of F (λ) to M(λ)λ η, to help us understand simplicity or composition factors of the M(λ)’s. − Definition 21.5. Let Fη be the restriction of F to U(n ) η. − − Theorem 21.6 (Theorem A). ³ ´mult(α)P (η nα) Y Y∞ ­ nα ® − detFη hα ρ 2 ,α , = α ∆ n 1 + − ∈ + = where P is the Kostant partition function, and detF U(H). η ∈ Theorem 21.7 (Theorem B). L(µ) is an irreducible subquotient of M(λ) if and only if there exists β1,...,βn ∆ and n1,...,nk N such that ∈ + ∈ ­ ® ni ­ ® λ ρ n1β1 n2β2 ni 1βi 1,βi βi ,βi for all i 1,...,k (21.1) + − − − ··· − − − = 2 = Pk and λ µ i 1 ni βi . − = =

Observation. Pick λ H ∗ such that λ,δ h∨, the dual Coxeter number, and λ ρ,α N ∈ 〈 〉 = − 〈 + 〉 6∈ for each positive real root α. (Such weights exist, e.g. λ ρ.) This weight is called a KK weight = − (for “Kac-Kazhdan”). Then ­λ ρ,δ, ® n δ,δ 0 for all n N, and so L(λ nδ) is a factor of + = 2 〈 〉 = ∈ − M(λ) for all n N. But for α ∆re, ∈ ∈ + ­λ ρ nδ,α® ­λ ρ,α® N + − = + 6∈ and n ­ ® α,α ndi , where α w(αi ), 2 = = and so there are no subquotients of the form L(λ nδ mα). In particular, the Weyl group plays − − no rôle. Conjecture 21.8 (Kac-Kazhdan, Theorem: Hayashi; Goodman, Wallach). Let λ be a KK weight on an affine Lie algebra. Then

λ Y ¡ α¢ 1 chL(λ) e 1 e− − . = α ∆re − ∈ + What are te ingredients of the proof?

• detFη is a polynomial in U(H), so it can be factored. First calculate the “leading term” and get Y Y∞ mult(α)P (η nα) hα − . (21.2) α ∆re n 1 ∈ + =

• Use Kac’s Casimir to observe that if v M(λ)λ β is a highest-weight vector, then ∈ − ° °2 ° °2 ° °2 ° °2 °λ ρ° °ρ° °λ β ρ° °ρ° , + − = − + − i.e. ­λ ρ,β® 1 β,β . + = 2 〈 〉 So if this cannot be satisfied, then M(λ) is irreducible and so F (λ) is non-degenerate; hence ¡ 1 ¢ detFη is a product of linear factors hβ ρ 2 β,β , and the formula for leading terms shows +〈 − 〉 ¡ 1 ¢ that β nα for some α ∆ , which we call a quasi-root. So we get products of hα ρ α,α . = ∈ + +〈 − 2 〉

50 Infinite-dimensional Lie algebras 51

The trick. We had already shown that the leading term of detFη (given by Equation (21.6)) is the expression (21.2), and F is a product of finitely many linear terms of the form h ­ρ nα ,α®. η α + − 2 (Note that the Kostant partition function P (η nα) vanishes for large n or large α.) Now extend − everything to C[t]. For example, if we have the universal enveloping algebra Ue(g) extended over C[t], then M( ) U(g) C[t] , f λ e Ue(n H) λe = ⊗ +⊕ where λe H ∗ C[t] is given by ∈ ⊗ λe(n ) 0 and λe(h) λ(h) t z(h), + = = + where z H ∗ is some generic element in the sense that that z(hα) 0 for all α Q \{0}. This ∈ 6= ∈ + produces a form F t (λ) taking values in C[t], and when we specialise t 0, we recover F (λ), η → η and Mf(λ) M(λ). Filter → 0 1 2 Mf(λ) Mf Mf Mf , = ⊇ ⊇ ⊇ ··· where i © t i ª Mf v Mf(λ): F (λ)(v,w) (t ) for all v,w Mf(λ) , = ∈ ∈ ∈ and specialise t 0: → M(λ) M 0 M 1 M 2 = ⊃ ⊇ ⊇ ··· This is called Jantzen’s filtration; here M 1 is a maximal submodule. Pick λ such that λ(h ) ρ 1 β,β vanishes for some quasi-root but does not vanish for all β + 〈 − 2 〉 other quasi-roots. Then we see that

1. M(λ β) is irreducible, and − 2. M(λ) has a submodule U which is a direct sum of modules of the form M(λ β). − t Now by construction of λe we see that the highest power of t dividing detFη(λ) equals the oerder of the linear factor h ­ρ nα ,α® in detF (λ). But M 1 LM(λ β), so if M(λ β) appears in α + − 2 η = − − M i with multiplicity s (β), we see that the order of t dividing detF t (λ) is P s (β)P (η β). i η i i − Example. Suppose M(λ β) contains three highest-weight modules of highest weight β, with − 1 3 2 2 3 4 multiplicities respectively 1, 2 and 5. Then M M(λ β)⊕ , M M(λ β)⊕ , and M M = − = − = = M 5 M(λ β). = −

The functions φβ : Q R, η P (η β) are all linearly independent. But comparing the + → 7→ − leading term of Fη(λ) (cf. (21.2)) with the multiplicities in Jantzen’s filtration for the quasi-root β nα, we get = P Y Y∞ mult(α)P (η nα) Y Y∞ P (η nα) i si (nα) hα − hα − . α ∆re n 1 = ∆ n 1 ∈ + = + = This proves Theorem A. The proof of Theorem B is an inductive variation on the proof of Theorem A, producing the Jantzen Sum Formula X X chM(λ)i chM(λ nα) , (21.3) i 1 = (α,n) D − ≥ ∈ λ where the right-hand-side sum is over the set

n ­ 1 ® o Dλ : (α,n) ∆e N : λ ρ α,α 0 . = ∈ + × + − 2 =

(Here ∆e is the set of positive roots with multiplicities.) +

51 52 Iain Gordon

21.2 Bringing in the Weyl group

Definition 21.9. We define an equivalence relation on H ∗ by λ µ if there exists a sequence ∼ ∼ λ λ0,λ1,...,λn µ such that for each 0 i k, either {λi ,λi 1} or {λi 1,λi } satisfy (21.1) for = = ≤ < + + some β1,...,βn ∆ and n1,...,nk N. ∈ + ∈ ± Definition 21.10. Given equivalence class [λ] C (λ) : Λ in H ∗ , we say that a module M O ≡ = ∼ ∈ has type Λ if all its “composition factors” belong to Λ. (Cf. Section 18.1 for the category O.) We write OΛ for the subcategory of modules of type Λ.

Example. It follows from Theorem B of Kac-Kazhdan (Theorem 21.7) that M(λ) has type Λ.

Theorem 21.11 (Deodhar, Gabher, Kac). Let M O. Then there exists a unique set {MΛ}Λ such L 1 ∈ that M M . Furthermore, Ext (F,E) 0 if E O ,F O and Λ Λ0. = Λ Λ O = ∈ Λ ∈ Λ0 6=

Theorem 21.11 says that to understand category O, we only need to understand OΛ. The moral will be that some choices of Λ are better than others. For w W , we define an action w on H ∗ by w λ w(λ ρ) ρ. We set ∈ ∗ ∗ = + − © ­ ® ­ ® isotropicª C : λ H ∗ : λ ρ,αi 0 and λ ρ,α 0 if α ∆ , = ∈ + ≥ + 6= ∈ + and define Tits’ cone of “good” elements [ K g : w C . = w W ∗ ∈ Define

n 1 ­ ®o S : λ H ∗ : λ ρ,α α,α , α = ∈ 〈 − 〉 = 2 [ S : S , and = α α ∆im ∈ + w.g. K : H ∗ \ S . = Elements of S are “critical hyperplanes”, and W acts on the set K w.g. of “weakly good” elements.

g w.g. Exercise 21.12. Compare H ∗, K and K in the cases

1. finite generalised Cartan matrices,

2. affine generalised Cartan matrices, and

µ 2 a¶ 3. − for a 3. a 2 ≥ − (It is clear that K g K w.g..) ⊆ g Definition 21.13. Define an equivalence relation as for , using K instead of H ∗, and an w.g. ≈ ∼ equivalence relation ◦ as for , using K instead of H ∗. ≈ ∼ w.g. Proposition 21.14. If λ,µ K , then λ ◦ µ if and only if there exists w W (λ), the integral ∈ ≈ ∈ Weyl group generated by ½ ¾ re 2 λ ρ,β sβ : β ∆ such that 〈 + 〉 Z , ∈ + β,β ∈ 〈 〉 such that w λ µ. ∗ =

52 Infinite-dimensional Lie algebras 53

Proof. We prove a better result. Let λ K w.g.. Then L(µ) is a subquotient of M(λ) if and only if re ∈ there exist β1,...,βt ∆ with λ (sβ ) λ (sβ sβ ) λ µ. ∈ > 1 ∗ > ··· > t ··· 1 ∗ = We pick λ µ Q and+ do induction on ht(λ µ): If ht(λ µ) 0, then λ µ and there is − ∈ + − − = = nothing to do. Now assume ht(λ µ) 0. Then L(λ) M(λ)±M 1 from Jantzen’s filtration. Since − > = λ µ, L(µ) is a subquitient of M 1, and hence by Jantzen’s Sum Formula (21.3) also a subquotient 6= of M(λ nα) for some (α,n) D . − ∈ λ Now if we assume that α ∆im, we have ­λ ρ,α® n ­α,α®. Set β : nα ∆im and ∈ + = 2 = ∈ ­λ ρ,β® 1 ­β,β®; and so λ S +, contradiction. So α ∆re, and then ­λ ρ,α® n+­α,α® + = 2 ∈ β ∈ + = 2 with ­α,α® 0, and so n n(α) is uniquely determined. + 6= ≡ w.g. Finally we note that λ n(α)α (sα) λ K , and by induction we conclude the “if” direc- − = ∗ ∈ tion. The converse is just Jantzen’s Sum Formula (21.3).

Remark 21.15. The two equivalence relations from Definiton 21.13 are related by ◦ g , ≈=≈ |K since both have the same description in terms of W . In the affine (and finite) case, ◦ w.g. , ≈ =∼|K but this is false in general.

Definition 21.16. Let O g be the full subcategory of O whose objects have composition factors belonging to K g. We say that M O g has type Λ if each composition factor of M belongs to Λ. ∈ Example. If λ K g, then M(λ) O g, and M(λ) has type Λ. ∈ ∈ There is a decomposition g M g O OΛ . = Λ S ± Moreover, the equivalence classes are labelled by λ C W W (λ) thanks to Proposition 21.14. ∈

Exercise 21.17. Check that if λ λ0, then W (λ) W (λ0). ≈ =

21.3 The Kazhdan-Lusztig Conjecture

53 54 Iain Gordon

A References

• Kac, Raina: Infinite-dimensional Lie algebras and highest-weight representations.

• Kac: Infinite-dimensional Lie algebras.

• Carter: Lie algebras of finite and affine type.

• Goddard, Olive (eds.): Kac-Moody and Virasoro algebras.

• Kumar: Kac-Moody groups, their flag carieties and representation theory.

• E. Frenkel: Langlands correspondence for loop groups.

• Pressley, Segal: Loop Groups.

• Humphreys: Lie algebras.

• Segal, Wilson: ? (probably IHES)

• Deodhar, Gabher, Kac: Adv. Math. 45 (1982).

• Kumar: J. Algebra 108 (1987)

B Notational reference

Witt the Witt algebra over C

Vir the Virasoro algebra over C, a central extension of Witt

Lie(G) the Lie algebra of the Lie group G, g T G = e g either a general Lie algebra or a finite-dimensional, simple one

g a central extension of a Lie algebra g; 0 z g g 0 b → → b → → 1 L g the loop algebra g C[t,t − ] ⊗ d ¡ 1 ¢ L g L go Cd, where d t Der C[t,t − ] = = dt ∈ ¡ ¢ ¡ ¢ Note: H 2 L g;C C H 2 L g;C =∼ =∼ Ldg L g Cd Cc, the central extension of L g, affine Kac-Moody Lie algebra = ⊕ ⊕ L (A) the Kac-Moody Lie algebra corresponding to the generalised Cartan matrix A U(g) the universal enveloping algebra of the Lie algebra g; U(g) T (g)±(x y y x [x, y]) = ⊗ − ⊗ − , a positive-definite sesquilinear form on a representation V , 〈− −〉 non-degenerate if V is unitary

Ab the affine, untwisted gen. Cartan matrix derived from the Cartan matrix A

Ae the affine, twisted generalised Cartan matrix derived from A

54 Infinite-dimensional Lie algebras 55

C Generation of Lie algebras

Lie algebra Dynkin diagram Generalised Cartan Matrix Name Notation

simple g ∆, classical A, finite type

£ 1¤ loop g t ± n/a n/a £ 1¤ affine g t ± Cc n/a n/a ⊕ £ 1¤ ¡ d ? g t ± o C dt ) n/a n/a £ 1¤ affine Kac-Moody g t ± Cd Cc ∆b, untwisted affine Ab, untwisted affine ⊕ ⊕ ¡ £ 1¤ ¢T twisted affine Kac-Moody g t ± Cd Cc ∆e, twisted affine Ae, twisted affine ⊕ ⊕

55