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Infinite-dimensional Lie algebrasEdinburgh, 2008/9

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

16 Root spaces 32

1

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

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.

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

Der(A) = {

dz : j ∈Z

[ Lm ,Ln

= ( (n +1)zm+n+1 f ′+ zm+n+2 f ′′)− (

(m +1)zm+n+1 f ′+ zm+n+2 f ′′) = (n −m)zm+n+1 df

dz = (m −n)Lm+n f

This is called the Witt algebra over C, denoted Witt; its basis is { L j : j ∈Z}

and its structure is[ Lm ,Ln

]= (m −n)Lm+n .

Infinite-dimensional Lie algebras 3

:

} ,

i.e. we impose that each element is in the finite span of cos(nθ) d dθ and sin(nθ) d

dθ , or equivalently

in the finite span of { e i nθ d

dθ : n ∈Z} . Without the finiteness assumption, we would have to deal

with the Fourier analysis and convergence issues. Thus we have a basis{ Ln := i e i nθ d

dθ : n ∈Z

} .

Set z := e iθ, then Ln =−zn+1 d dz , and we recover the Witt algebra.

Example (Diffeomorphisms of S1). Let G := Diff+ ( S1

) be the group of (orientation-preserving)

diffeomorphisms of S1 under composition. It acts on C∞( S1,C

) via (g . f )(z) := f

( g−1(z)

) for

g ∈G . An element of the tangent space at 1 ∈G has the form

g (z) = z ( 1+ε(z)

)= z + ∞∑

)= z − ∞∑

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) =−x } .

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)

such that z⊆ Z ( g )= {

x ∈ g : [x, g] = 0 } .

As vector spaces the sequence (2.1) splits as g= g′⊕ z, say via a section σ : g→ g′. Let

β : g×g→ z , (x, y) 7→ [ σ(x),σ(y)

]−σ( [x, y]

3

) := {

] := (

.

Then gβ is a central extension. However, there exist isomorphisms of such gβ:

z −−−−→ gβ = g⊕ z −−−−→ g θ

y∼=

z −−−−→ gγ = g⊕ z −−−−→ g

We find quickly that θ : gβ→ gγ 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 θ2 ( [g , g ′]

)= γ(g , g ′)−β(g , g ′). So we find that central extensions are classified by

H 2(g;z )= C 2

} , and dθ(g , g ′) = θ

) .

Exercise 2.2. Show that if g is a finite-dimensional simple Lie algebra, then H 2 ( g;C

)= 0.

) := 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, (Lm ,Ln) 7→β(m,n) satisfy the two conditions from above, namely

• 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 , Lm 7→ θ(m) .

Writing θβ(m) := β(0,m) /

m for m 6= 0, replace β by β+dθβ: We see that w.l.o.g. β(0,m) = 0 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,−nβ(m), where β(m) =−β(−m).

With l +m +n = 0, we produce a relation which for n = 1 gives

(1−m)β(m +1)+ (m +2)β(m)− (2m +1)β(1) = 0 .

This recurrence relation has a two-dimensional solution space:

β(m) =α1m +α2m3 , with α1,α2 ∈C .

Now replace β with β+dθβ, where θβ(m) = δm,0 α1 2 , to get β(m,n) = δm,−nα2m3.

3 The Virasoro algebra

Definition 3.1. The Virasoro algebra (denoted by Vir) is the central extension of the Witt algebra with the choice β(m) = 1

12

) in the notation of Proposition 2.3.

The Virasoro Algebra is spanned by { Lm : m ∈Z}∪ {c} such that c is central, i.e. [c,Lm] = 0

for all m ∈Z, and [ Lm ,Ln

]= (m −n)Lm+n +δm,−n 1

12

) c .

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

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)

and ρ : h 7→ ρ(h) = (−×h) . (4.1)

Exercise 4.2. If h 6= 0, then B(µ,h) is irreducible, while if h = 0 it is not (e.g. the subspace of constants C< B(µ,h) is invariant).

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

Link to Virasoro algebras. Define

Lk := 1

: a− j a j+k : for k ∈Z .

The notation “: a− j a j+k :” means “normal ordering”, i.e.

: a− j a j+k : := {

a− j a j+k if − j ≤ j +k,

a j+k a− j if − j ≥ j +k.

Explicitly, we have

an− j an+ j if k = 2n,∑ j>0

an+1− j an+ j if k = 2n +1.

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) an acts as zero for n À 0, and

3. an acts locally nilpotently for n > 0.

Theorem 4.3. With Lk defined as above,

[ Lm ,Ln

12

( m3 −m

) as operators on B(µ,1), i.e. as a representation of the Virasoro algebra. (But not of the Witt algebra!)

5

1 if |x| ≤ 1,

0 if |x| > 1,

and for ε > 0, define Lk (ε) := 1 2

∑ j∈Z : a− j a j+k : ψ(ε j ). This has only finitely many non-zero

terms, and Lk (ε) → Lk as ε→ 0. Now calculate: [ an ,Lk

]= nak+n (using Lk (ε)), and hence

[ Lm(ε),Ln

]= 1

2

( j +m) : a− j a j+m+n :ψ(ε j )+

1

2

: an− j a j+m :ψ(ε j )−δm,−n 1

2

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).v1, v2⟩ = ⟨v1,ρ(ω(x)).v2⟩ for all x ∈ Vir, v1, v2 ∈V , i.e. x† =ω(x).

5.1 Family 1: Chargeless representations

Let

vn : n ∈Z} , (5.1)

where α,β ∈C and P ∈C[z, z−1]. Define a representation of the Virasoro algebra by the following action:

c 7→ 0 and Ln(vk ) =−( k +α+β(n +1)

) vn+k . (5.2)

This is the representation one would discover by using the definition Ln := −zn+1 d dz on the

elements vk := zk+α(dz)β. Now we start to use the structure of Vir. Set

h :=Cc ⊕CL0 (abelian); Vir± := span { Lk : k ∈Z±

} .

Vir = Vir−⊕ h⊕Vir+ . (5.3)

Exercise 5.2. Check that h=Cc ⊕CL0 is a Cartan subalgebra of Vir, i.e. that it is nilpotent and self-normalising.

6

Infinite-dimensional Lie algebras 7

Remark 5.3. In the representation Vα,β, h acts by scalars on each vk : L0vk =−(k +α+β)vk . For a general representation ρ : Vir → gl(V ), the decomposition

V = ⊕ λ∈h∗

Vλ

is called a weight space decomposition, where Vλ := { v ∈ V : ρ(z).v = λ(z)v for all z ∈ h

} . Obvi-

ously, the existence of a weight space decomposition is a restriction on the representation.

Lemma 5.4. Let V be a representation of an abelian Lie algebra A, such that V = ⊕ λ∈A∗ Vλ

and such that each Vλ is finite-dimensional. Then any subrepresentation U ≤V decomposes as U =⊕

λ

) .

Proof. Any v ∈V can be written as v =∑m j=1 v j with v j ∈Vλ j for j = 1, . . . ,m, i.e. a.v j =λ j (a)v j

for all a ∈ A. Suppose v ∈U . We need to show that v j ∈U for each j . Since λ j 6=λk for j 6= k, we can find a ∈ A such that λ j (a) 6=λk (a) if j 6= k. Then

v = v1 +·· ·+ vm

a.v = λ1(a)v1 +·· ·+λm(a)vm

am−1.v = λ1(a)m−1v1 +·· ·+λm(a)m−1vm .

Since each a j .v ∈U , we see that [ λi (a) j

] (v1, . . . , vm)T ∈U m , where 1 ≤ i ≤ m and 0 ≤ j ≤ m −1.

But the matrix [ λi (a) j

] is a Vandermonde matrix and thus, since the λi (a) are pairwise distinct,

it has non-zero determinant, and hence (v1, . . . , vm)T ∈U m , as required.

Theorem 5.5.

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

α,β of codimension one in Vα,β. (As vector spaces we may

consider these to be span {

v−(α+β) }

(If Vα,β is irreducible, then set V ′ α,β =Vα,β.)

2. V ′ α,β is unitary if and only if β+β= 1 and 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 vk ∈U . Now given vk ∈U , we have Ln(vk ) ∈U for all n, and so we get scalar conditions from Equation (5.2), i.e.

vn+k 6∈U ⇒ k +α+β(n +1) = 0 .

If vn+k , vm+k 6∈ U , then β = 0 and α = −k, and hence Cvk is a subrepresentation, and Vα,β

/ Cvk is irreducible (because Ln(vt ) = (t −k)vt+n for t 6= k).

Now assume that only one vector, say vm , does not belong to U . Then for l 6= k,m we must have Ln−k (vk ) = 0 = Lm−l (vl ), whence β= 1, a =−(m +1) and U = span

{ vt : t 6= m

} .

Part (2): Unitarity implies ⟨vk , vk⟩ > 0 for all relevant k. Hence

• ⟨L0(vk ), vk⟩ = ⟨vk ,L0(vk )⟩, whence α+β ∈R, and

7

8 Iain Gordon

• ⟨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)

whence β+β= 1.

Conversely, ⟨vk , vm⟩ = δkm is a well-defined positive-definite hermitian form on V ′ α,β because of

the conditions on α, β.

Exercise 5.6. Is Vα,β completely reducible? That is, if it is not irreducible, is Vα,β = C⊕V ′ α,β a

decomposition of Lie algebra representations?

5.2 Family 2: Highest-weight representations

The Virasoro algebra has a representation on B(µ,1) =C[x1, x2, . . . ] as in Equation (4.1) above. Recall that

L0 := 1

2 a2

2 + ∑

On a monomial ∏

i xni

i , the term a− j a j acts by multiplication by j n j , and so we see that∑ j>0 a− j a j picks out the “degree” of a homogeneous polynomial, where deg(x j ) = j , i.e.

B(µ,1) = ∑ λ∈h∗

homogeneous polynomials of degree N λ(c) = 1, λ(L0) =µ2/2+N ,

0 otherwise.

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

⟨ xk1

j=1 j 2k 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 ak , for k < 0.)

Proposition 5.7. Let V be a unitary representation of the Virasoro algebra (with c acting as a scalar) such that

V = ⊕ λ∈h∗

Vλ with dimVλ <∞ .

Then V is completely reducible.

Proof. Let U <V be an invariant subspace. Observe that ω(L0) = L0. Unitarity forces λ ∈ h∗ R

by considering the effect of L0 on ⟨v, v⟩ > 0. It then follows that ⟨Vλ,Vµ⟩ = 0 ifλ 6=µ. Set Uλ :=U∩Vλ, and define U⊥

λ ⊂ Vλ as the orthogonal complement to Uλ in the form restricted to Vλ. Then

Vλ =Uλ⊕U⊥ λ

, and ⊕

λU⊥ λ =: U⊥ is invariant under the Virasoro algebra. So V =U ⊕U⊥.

8

Infinite-dimensional Lie algebras 9

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

Uµ2/2 ⊕U⊥ µ2/2 = B ′(µ,1)µ2/2 =C.1

to see that 1 ∈Uµ2/2 ⇒ B ′(µ,1) =U , i.e. B ′(µ,1) is irreducible.

At this point we do not know whether B ′(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. cVir.v = cv (here c is a complex number on the right hand side),

2. L0(v) = hv for h ∈C,

} .

Remark 5.9.

• cVir acts as the complex number c on all of V .

• L0 ( L−ik · · ·L−i1 (v)

)= ( h + ik +·· ·+ i1

]= i L−i . So we get a weight space decomposition

V = ⊕ λ∈h∗

Vλ ,

where Vλ 6= 0 only if λ(cVir) = c and λ(L0) = h +Z≥0.

• Lk (v) = 0 for k > 0 (because Vλ = 0 for λ(L0) = h −k).

The (c,h) are called the highest weights of V .

) .

Definition 5.11. If V =⊕ λ∈h∗ Vλ, then the formal character of V is

chV (q, t ) := ∑ λ∈h∗

dim(Vλ)qλ(L0)tλ(c) .

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 − y x. We write Aad for this Lie algebra structure.

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 : g→U (g)ad satisfying the following universal property:

For every arbitrary associative algebra with unit A over k and a Lie algebra homomorphism j : g→ Aad , there exists a unique homomorphism of associative algebras φ : U (g ) → A such that j =φ .

9

10 Iain Gordon

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)

where T (g) denotes the tensor algebra of (the vector space) g, i.e. T (g) :=⊕ i≥0 g⊗i . 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 {

xb : b ∈ B }

be a basis for g, where B is ordered. Then the set of ordered monomials{

xi1

} 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 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:

Lk .λ= 0 for k > 0, L0.λ= hλ , cVir.λ= cλ , for all λ ∈C.

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.

1. M(c,h) = ⊕ s∈Z≥0

} ,

j=1 i j = s.

(This has dimension P (s) = the number of partitions of s; thus

chM(c,h)(q, t ) = t h

j≥1 (1−q j ) .

Note that L0.(1⊗1) := L0 ⊗1 = 1⊗ (L0.1) = h(1⊗1), and c.(1⊗1) = c(1⊗1), which proves that this is a highest-weight representation.)

10

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−ik · · ·L−i1 :

0 < i1 ≤ ·· · ≤ ik } .

Part (2). If M(c,h) = V ⊕W , this respects the weight space decomposition by Lemma 5.4. Hence M(c,h)h =C, so without loss of generality, Vh =C and Wh = 0. Then 1⊗1 ∈V . But 1⊗1 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 , L0.v = hv and Lk .v = 0 for all k > 0. Hence Cv =C(c,h). The evaluation map ev: M(c,h) :=U (Vir)⊗C(c,h) N , ρ⊗1 7→ ρ.v thus factors through the tensor product over 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 Ph = 0. (Note that all proper sub- 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 Virasoro algebra with V =⊕

λ∈h∗ Vλ is precisely one of the following three:

• V ′ α,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

V (c,h)∗ :=⊕ 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−ik · · ·L−i1 and Q = L− jl · · ·L− j1 , and for v = 1⊗1, let

⟨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−ik · · ·L−i1 L− jl · · ·L− j1 v .

Exercise 8.1. Check that this form is contravariant and sesquilinear (though not necessarily non-degenerate). Check that M(c,h)λ and M(c,h)µ are pairwise orthogonal if λ 6=µ.

Now let M ′(c,h) ≤ M(c,h) be the maximal proper submodule, which we know to exist.

Lemma 8.2. M ′(c,h) = ker (⟨−,−⟩) := {

k : ⟨k,−⟩= 0 } .

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⟩ 6= 0, so the kernel is indeed a proper submodule. Now suppose P.v ∈ M ′(c,h), and thus ω(Q)P.v ∈ M ′(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 det1(c,h) = 2h. Next,

.

) coeff v , working in the Verma module,

= ((

) coeff v

) .v +0

) coeff v

= 4h + c 2 .

With similar calculations we get that det2(c,h) = 2h(16h2 +2hc −10h + c).

Theorem 8.3 (Kac determinant formula).

det n(c,h) = K ∏

( h −hr,s(c)

where P (m) is the number of partitions of m,

K = ∏ r,s∈N

1≤r,s≤n

( (2r )s s!

and

(c −1)(c −25)(r 2 − s2)−24r s −2+2c )

.

• V (1,h) = M(1,h) if and only if h 6= m2

4 , m ∈Z.

• V (0,h) = M(0,h) if and only if h 6= m2−1 24 , m ∈Z.

Exercise 8.4 (Unitarity). Show that detn(c,h) > 0 for all n if c > 1 and h > 0.

Note that ⟨L−n .v,L−n .v⟩ = 2nh+c n3−n 12 , so considering large enough n we see that unitarity

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 detn(c,h) > 0, the form is positive semi-definite in this region (i.e. it is positive-definite for V (c,h)) if it is so at least once in this region. But we already found one unitary irreducible

highest-weight representation of Vir with h = µ2

4 , c = 1, namely B ′(µ,1).

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:

c(m) := 1− 6

Then

)2 −1

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

Let V = ⊕ n∈ZCvn . Then gl(V ) =: gl∞ is the Lie algebra of matrices with a finite number of

non-zero entries. It is spanned by the Ei , j : v j 7→ vi . Let

F (0) :=Λ∞ (0)V = span

} .

A.vi = A.vi0 ∧ vi2 ∧·· ·+ vi1 ∧ A.vi2 ∧·· ·+ · · · ,

(note that this action preserves tails). In detail,

if i 6= j , then Ei , j .vi = {

0 if j does not appear in i or if i appears in i,

replace v j with vi in vi otherwise,

and E j , j .vi = {

vi if j appears in i,

0 otherwise.

The space F (0) has a basis labelled by partitions λ ` n, where λ = (λ1 ≥ λ2 ≥ . . .) ∈ Z∞ ≥0 with∑

i λi = n. To see this, map

i = ( i0, . . . , im , . . .

) ,

and this map is clearly reversible. It follows that F (0) = span {

vλ } .

Define deg(vλ) := |λ| = n if λ ` n, so that |λ| is the number of which λ is a partition; i.e. 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 contains a big abelian subalgebra spanned by Λk for k ∈ Z, where Λk := ∑

i∈ZEi ,i+k , so that Λk .v j = v j−k for all j .

Exercise 9.2. Show that the elementsΛk 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.

A typical element of a∞ is a finite linear combination of terms ∑

i∈Zλi Ei ,i+k . If k 6= 0, then∑ i∈Zλi Ei ,i+k acts on F (0), because for i À 0 or i ¿ 0 we have an action by 0 for Ei ,i+k on vi.

However, if k = 0, then we would have ∑

i∈Zλi Ei ,i .vj =∑ s≥0λ js vj, which is not well defined!

To remedy the situation, we adjust the “action” as follows:

• Ei , j acts as Ei , j for i 6= j .

• Ei ,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.

0 otherwise.

Exercise 9.4. Show that the Ei , j differ from Ei , j only by a multiple of I , so commutators are unchanged. Show further that[

Ei , j , Ek,l ]= 0 for j 6= k, i 6= l ,

[ Ei , j , E j ,l

]= Ei ,l for i 6= l ,[ Ei , j , Ek,i

]=−Ek, j for k 6= j , [ Ei , j , E j ,i

]= Ei ,i − E j , j +α ( Ei ,i , E j , j

) I ,

where

) :=

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 ,

a central extension with α as the defining 2-cocycle. As before, we define the diagonal elements Λk :=∑

i∈Z Ei ,i+k , and we get an action on F (n) via

Λ0.vj = nvj and [ Λk ,Λl

]= 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

) .vn ∧ vn−1 ∧·· · , and check that

• this is an H -map (using that B(n,1) is irreducible to see injectivity), and that

• P ( Λ−1,Λ−2, . . .

) .vn ∧vn−1 ∧·· · spans F (n). (Count dimensions of the homogeneous com-

ponents on both sides.)

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?

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[x1, x2, . . . ] and F (0).

Exercise 9.7. Let k ∈Z≥0 and fk = vk ∧vk−1∧·· ·∧v1∧v−k ∧v−k−1∧·· · . Show that L0. fk = k2 fk

and L j . fk = 0 for all j > 0. Deduce that B(0,1) is not irreducible as a Virasoro module, and that it 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⊗CR 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.)

In the case R =C[ t , t−1

] , the Laurent polynomials in one variable, we get L g := g⊗R, the

loop algebra of g. This has a description as Calg ( C×,g

) or as Csf

) , where “Csf” stands for

the space of smooth maps with finite Fourier expansion, and the correspondence is between ψ : S1 → g and its Fourier coefficients in ψ = ∑

n∈Z e2πi nθXn . Either way, a basis {

Xa }

determines a basis { Xa(n) := Xa ⊗ t n}

for L g, and the Lie algebra structure is[ Xa(n), Xb(m)

]= [ Xa , Xb

] (n +m) .

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 6= I ( g such that

[ g, I

] ⊆ I ). Recall that g acts on itself via the adjoint 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 := {

.

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

B ∈ ( (g⊗g)∗

0 otherwise.

K (x, y) = tr ( ad x 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). H 2(L g;C

)=C .

Exercise 10.3. Prove the theorem.

Let d ∈ Der(R), i.e. d(r s) = d(r )s + r d(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

(g⊗R)oCd , where [

X ,Y ]⊗ r s +µY ⊗ds −λX ⊗dr .

By applying this to the loop algebra, we get the new algebra

L g :=L goCd ,

where d := t d dt . (The derivation tells us about degrees.)

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 needβ : L g⊗L g→Cwhich 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

L g=⊕ i∈Z

X ∈L g : [d , X ] = i X }

, and hence

i .

Claim. H 2 i = 0 if i 6= 0. (In fact, generalising what we’re about to do, it is simply better to see that

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 )

) .

)= 1 i β

i = 0. //

Now (L g)0 = g⊗1⊕Cd , hence H 2 ( (L g)0;C

)= 0 by (a generalisation of) Exercise 2.2. So we can ensure that β

( (L g)0, (L g)0

)= 0. Define βi : g⊗g→C by βi (x, y) =β

( x(i ), y(−i )

) . Then g⊗ t n is a (g⊗1)-representation (via

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 )

)=−λi K (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- stituent algebraC

[ t±1

] . 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

.

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 , [hi ,h j ] = 0 , [h j ,ei ] =αi (h j )ei , and [h j , fi ] =−αi (h j ) fi ,

and [ei , f j ] = 0 for i 6= j , [ei , fi ] = hi for all i , (11.1)

where

]= 0 if i − j

]= 0 = [ ei+1, [ei+1,ei ]

] for all i .

We write this concisely as ( ad(ei )

)1−αi (h j )(e j ) = 0 (for i 6= j ). There are completely similar rela- tions for the fi .

17

18 Iain Gordon

Exercise 11.1. Find similar generators and relations for the Lie algebras

g(M) := {

, where

0 0 In

0 In 0

Definition 11.2. The matrix A := ( αi (h j )

) i j is called the Cartan matrix, and it determines finite-

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 ∈Matn(Z) and A is indecomposable (i.e. not block-diagonal with more than one block);

• Ai i = 2 for all i ;

• Ai j ∈ { 0,−1,−2,−3

} for all i 6= j ;

• Ai j = 0 if and only if A j i = 0;

• if Ai j =−2 or −3, then A j i =−1.

These conditions imply that A is “non-degenerate” in appropriate sense which we will meet soon; in particular this includes det A 6= 0.

12 Kac-Moody Lie algebras

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

• Ai i = 2,

• Ai j ≤ 0 if i 6= j , and

• Ai j = 0 if and only if A j i = 0.

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 ∈Matn(C). A realisation of A is a triple (h,Π,Π∨), where

• h is a finite-dimensional vector space over C,

• Π∨ = { h1, . . . ,hn

• Π= { α1, . . . ,αn

18

such that α j (hi ) = Ai j .

Lemma 12.3. If (h,Π,Π∨) is a realisation of A, then dimh≥ 2n − rk A.

Proof. Let r = rk A and dimh = m. Extend Π and Π∨ to bases { α1, . . . ,αm

} and

( α j (hi )

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.

Example. Let A = ( 2 −2−2 2

) . Then rk A = 1, so a minimal realisation must have dimh= 3. So let

h=C3 and h∗ =C3, with respective bases { h0,h1,d

} and

{ α0,α1,γ

2 ,0 )

(This is associated to the Lie algebra sl2, 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 isΦ : (h,Π,Π∨) → (h′,Π′,Π′∨), whereΦ : h→ h′ is an isomorphism such thatΦ(hi ) = h′

i andΦ∗(α′ i ) =αi for all i .

Proof of 12.5. Let r = rk A, so that possibly after re-ordering

A = (

)

with A11 non-singular. Extend this to a (2n − r )× (2n − r )-matrix

C = A11 A12 0

A21 A22 In−r

0 In−r 0

.

This is non-singular, e.g. detC =±det A11. Let α1, . . . ,αn be the first n coordinate functions, and 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

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

,

The free Lie algebra L(X ) generated by X is defined as follows: Let

C⟨X ⟩ :=C⟨e1, . . . ,en , f1, . . . , fn , x⟩ be the free associative algebra generated by X . Then C⟨X ⟩ad is a Lie algebra (under commuta- tors), and L(X ) is the Lie subalgebra of C⟨X ⟩ad generated by the elements of X . Its elements are called Lie words. (More generally, any Lie algebra is spanned by Lie words in a set of generators.)

Definition 12.8. Let L(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,

• [ x, y

• [ x,ei

]+αi (x) fi for all i = 1, . . . ,n.

Remark 12.9. The Lie algebra L(A) is independent of the choice of minimal realisation of A thanks to Proposition 12.5.

Theorem 12.10 (Structure of L(A)).

1. L(A) = n−⊕ h⊕ n+, where h ∼= h is abelian and n−, n+ are freely generated by f1, . . . , fn , e1, . . . ,en , respectively.

2. L(A) =⊕ α∈Q Lα, where

Lα := {

} and Q =Zα1 +·· ·+Zαn ⊂ h∗.

3. L0 = h.

4. Lα = 0 unless α ∈Q+ or α ∈Q−, where Q+ =Z≥0α1 +·· ·+Z≥0αn , and analogously for Q−.

5. [ Lα, Lβ

]⊆ Lα+β.

Proof. Define ω : ei ↔− fi and x ↔−x. Then ω induces a Lie algebra involution on L(A) (just check it preserves the relations). Let n± be the space generated by respectively ei and fi for i = 1, . . . ,n, and h the space spanned by elements x for all x ∈ h. The involution swaps n+ and n−.

For λ ∈ h∗, let θλ : X → End

( T (V )

) ,

e j 7→

1 7→ 0

vi 7→ δi jλ(h j ).1

vi1 · · ·vis 7→ δi1 j (λ−αi2 −·· ·−αis )(h j )vi2 · · ·vis + vi1

( θλ(e j )(vi2 · · ·vis )

Infinite-dimensional Lie algebras 21

It is an exercise to check that this induces a Lie algebra representation of L(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 : L(A) → End

( T (V )

We can now prove part (1) of the theorem.

• h∼= h: There is a natural surjective Lie algebra homomorphism x 7→ x. If x = 0 in L(A), we would have θλ(x) = 0 ∈ End

( T (V )

) for all λ ∈ h∗; but θλ(x)(1) =λ(x). This can be 0 for all λ

only if x = 0.

• n− is free: Let φ(w) = θλ(w).1 (by construction this is independent of λ), i.e.

φ ( w( f1, . . . , fn)

)= w(v1 · · ·vn) .

So we get a surjective Lie algebra homomorphism φ : n− → L(V ) onto the free Lie algebra 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.

• n−+ h+ n+ is a direct sum: Let w−+ x +w+ = 0. Then θλ(w−)+θλ(x)+θλ(w+) = 0 for 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.

• So n−⊕ h⊕ n+ ⊆ L(A). To see that it is all of L(A), it is enough to check that it is closed under taking brackets since it contains the generating set X . In other words, we need

ad(ei ) ( n−⊕ h⊕ n+) ⊂ n−⊕ h⊕ n+ ,

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 h and n+. For n− we have

ad(ei )( f j ) = δi j h j ∈ h⊕ n− .

By induction,

]= [ ad(ei )(w1), w2

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 on

[ ei1 , [ei2 , . . . ]

Lα = L(A) .

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= L0.

Lemma 12.11. L(A) contains a unique ideal I that is maximal with respect to the condition I ∩h= 0.

21

Proof. Let J = { I C L(A) : I ∩h = 0

} . Then for I ′, I ′′ ∈ J , I ′ and I ′′ inherit the weight space

decomposition by Lemma 5.4, so I (′,′′) = ⊕

0 6=α∈Q I (′,′′) α .

Therefore (I ′+ I ′′)0 = 0. So ∑

I∈J I is the desired ideal.

Definition 12.12. Let L(A) := L(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 ω induces an involution ω on L(A) which exchanges n+ and n−. Finally, L(A) inherits the weight space decomposition

L(A) = ⊕ α∈Q

Lα ,

where Q =Zα1 +·· ·+Zαn is the weight lattice. We collect a few obvious results:

Proposition 12.13.

2. dimLαi = 1.

3. kαi is a root if and only if k =±1.

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 Sn , where A ∼ A′ if and only if Ai j = A′

π(i ),π( j ) for π ∈ Sn . Furthermore we assume that A is indecomposable, i.e. if A = A1 ⊕ A2, then at least one of A1 and A2 is zero.

Definition 13.1. Let v = (v1, . . . , vn) ∈Rn . We say v ≥ 0 if vi ≥ 0 for all i , and we say v > 0 if v ≥ 0 and v 6= 0. We define v ≤ 0 and v < 0 similarly.

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 6= 0,

• if Av ≥ 0, then v ≥ 0.

2. A has affine type if

• corank(A) = 1,

• 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 ∈Matn(R) such that Ai i ≥ 2 for all i . For J ⊆ {1, . . . ,n} let A J be the submatrix of A whose entries are labelled by J × J .

Lemma 13.4. Assume that A is indecomposable. Then

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

• if A is affine, then A J 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

A = (

) .

If A is finite, then by definition there exists a v > 0 such that Av > 0, so

( A | P

v1 ...

vn

> 0 ,

and so A J 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 ) 6= 0, and so λ is not an eigenvalue of A. So A is a real, symmetric matrix all of whose eigenvalues are strictly positive, and hence A is positive definite.

Conversely, if A is positive definite, then det A 6= 0, and so A is not affine. If A were indef- 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

( d1, . . . ,dn

) and a symmetric matrix B such that A = DB .

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 i1, . . . , ik ∈ {1, . . . ,n}.

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 6= 0 for all k = 1, . . . , t −1. Let 0 6= d1 ∈R and define

d j = A jt jt−1 · · · A j2 j1

A j1 j2 · · · A jt+1 jt

d1 .

(Exercise: Check that this is well-defined, i.e. independent of the route from 1 to j .) Let D := diag

( d1, . . . ,dn

) and Bi j := d−1

i Ai j . We need Bi j = B j i , i.e.

Ai j

d j .

which is obvious if Ai j = 0. Assume thus that Ai j 6= 0, choose a sequence from 1 to i , i.e. 1 = j1, . . . , jt = i , and augment it by jt+1 = j . Then

d j = A j jt

( A jt jt−1 · · · A j2 j1

)( A j1 j2 · · · A jt−1 jt

) A jt j

A j jt

A jt j di .

the indexes seem wrong...

Remark 13.9. The proof of Lemma 13.8 shows that without loss of generality, di > 0 for all i , and that B ∈Matn(Q).

Proposition 13.10. If A is indecomposable and v > 0 is such that Av > 0, then vi > 0 for all 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∑n j=1 Ai j v j ≥ 0 for all i (recall that Ai j < 0 if i 6= j ).

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

} .

A J =

,

where ri , si > 0, and this matrix is either finite or affine, i.e. there exists v = (v1, . . . , vk ) > 0 such that A J v ≥ 0. Let

M := diag(v)−1 A J diag(v) =

2 −r ′

. . . 0 ...

−r ′ k . . . −s′k−1 2

where

s′i = v−1 i+1si vi > 0

r ′ i s′i = ri si ∈Z

and ∑ j

j=1 v−1

i (A J )i j v j = v−1 i A J v ≥ 0 .

So ∑

k + s′k ) .

i + s′i = 2

. . . . . .

. . . ...

0 0 0 −1 2 −1 0 0 0 0 −1 2 −1 −1 0 0 . . . 0 −1 2

(13.1)

This is symmetric; v = (1, . . . ,1) > 0, and A J v = 0, so A J is affine, and so A J = A.

25

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

1. A is finite if and only if det A J > 0 for all J ,

2. A is affine if and only if det A = 0 and det A J > 0 for all proper J , and

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 di > 0 and B symmetric, and B is of the same type as A. The conclusions 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

A J =

0 ??? 2 0 0 0 ...

. . . . . .

. . . ...

0 0 0 2 ??? 0 0 0 0 ??? 2 −bs−1

−b′ s 0 0 . . . 0 −b′

s−1 2

i ∈N.

Kac claims that det A J < 0 unless bi = b′ i = 1 for all i .

We want to show that if either of the conditions in the lemma on principal determinants (Which lemma?) holds, then bi = b′

i = 1 for all i : Suppose there exists a unique bt or b′

t > 1. Then change the basis to1 ... 1

,e1, . . . ,et , . . . ,es ,

from which det A J < 0 follows by an easy calculation (det A J = s.(−x), where −x is the unique entry in column t ). So there exist at least two non-zero bt ,b′

t > 1. Taking the shortest route from t to t ′ produces a submatrix, which is proper unless s ≤ 4.

If s > 4, we get a tridiagonal matrix

2 −b1

. . . . . .

. . .

−b′ t 2

.

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

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 Ai j = 0,

• there is one edge between i and j if Ai j =−1 = A j i ,

• there is a directed double edge between i and j if Ai j A j i = 2, pointing towards j if j < i ,

• there are three edges between i and j if Ai j A j i = 3, two of which form a directed double edge pointing towards j if j < i ,

• there are four (undirected) between i and j if Ai j A j i = 4 and Ai j =−1,

• there is a crossed double edge between i and j if Ai j A j i = 4 and if Ai j =−2 = A j i , and

• for Ai j A j i ≥ 5, there is one edge between i and j labelled “|Ai j | |A j i |”.

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 : . . .

• 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

• A1:

• Cn (n ≥ 3):

• Dn (n ≥?):

• GT 2 :

These diagrams are called affine Dynkin diagrams, and they correspond precisely to affine generalised Cartan matrices. Note that the matrix determined by An 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 ).

...

Then we get back to the finite case. In the affine case, check what happens in rank 2:

( 2 −b−c 2

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

L (A) =⊕ α

} .

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.

Proof. Let A = DB , where D = diag ( d1, . . . ,dn

) with di ∈ N and B T = B . Let

( h,Π,Π∨)

h1, . . . ,hn )

( Π∨) ⊆ h,

and let h′′ be some complement such that h′⊕h′′ = h. Define a bilinear form (−,−) on h as follows:

(hi ,h) := diαi (h) for all h ∈ h, (h′′ 1 ,h′′

2 ) := 0 for all h′′ 1 ,h′′

2 ∈ h′′. (15.1)

We claim that this is symmetric:

(hi ,h j ) := diαi (h j ) = di A j i = di d j B j i = d j Ai j =: (h j ,hi )

It is also non-degenerate, which we leave as an Exercise. In particular,

ker(−,−)|h′ = { x ∈ h′ :αi (x) = 0 for all i = 1, . . . ,r

} is the null space of A, which (Exercise:) equals the centre of L (A).

Now consider the principal grading on L (A),

L (A) = ⊕ k∈Z

L (A)k ,

where deg(ei ) = 1 =−deg( fi ) for i = 1, . . . ,n and deg(x) = 0 for x ∈ h. Let

L (N ) = N⊕

k=−N L (A)k , so L (0) =L (A)0 = h .

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( [ei , f j ],h

) .

Now for N > 1, ( Li ,L j

)= 0 if i + j 6= 0, and we need to define ( L±N ,LN

) . Take y ∈LN and

express it as y =∑ i [ui , vi ] for some ui , vi ∈L(N−1). For x ∈L±N , define

(x, y) :=∑ i

) ,

and note that [x,ui ] ∈L±(N−1) and vi ∈L(N−1), so this is indeed an inductive definition. This is well-defined: Suppose x =∑

j [u′ j , v ′

(x, y) = ∑ i

)+ ( u′

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)α,L (A)β

)= 0 unless α+β= 0. So(−,−) : L (A)α×L (A)−α→C

is a non-degenerate pairing, and [x, y] = (x, y)h′ α, where x ∈ L (A)α, y ∈ L (A)−α, and h′

α is defined by (

h′ α,−)

h =α(−) .

Proof. Let α+β 6= 0. Then there exists h ∈ h such that (α+β)(h) 6= 0. So

−α(h) ( x, y

) ,

and hence (x, y) = 0. Now we have [x, y]− (x, y)h′ α ∈ h if x ∈L (A)α and y ∈L (A)−α. It follows

that ( [x, y]− (x, y)h′

α,h )= 0 from the definition for all h ∈ h.

Reminder: representation theory for sl2.

sl(2;C) = {(

)} ,

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)),

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

Theorem 15.3.

1. If V is a representation of sl2, then there exists v ∈V such that H v =λv. Setting v j = F j v, we have H v j = (λ−2 j )v j .

2. There exists a unique (n +1)-dimensional irreducible representation of sl2 (where n ≥ 0). It has a basis (v0, . . . , vn) and satisfies

H v j = (n −2 j )v j ,F v j = v j+1,Ev j = j (n − j +1)v j−1 ,

• F E •

F E •→ . . . →•

F E • .

Proof. (1) follows from the relations above (REFERENCE!), (2) is a fun Exercise.

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

[hi ,ei ] = 2ei , [hi , fi ] =−2 fi , [ei , fi ] = hi .

Lemma 15.4. In L (A),

(adei )1−Ai j e j = 0 and (ad fi )1−Ai j f j = 0 for all i , j = 1, . . . ,n, i 6= j .

30

Infinite-dimensional Lie algebras 31

Proof. Let us show this for the fi ’s, the proof for the ei ’s is analogous. Let x = (ad fi )1−Ai j f j . We prove that [ek , x] = 0 for all k. First, set

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 6= i , j , then this is clear (check!), since [ek , fi ] = 0 = [ek , f j ].

• If k = j , then

[e j , (ad( fi ))1−Ai j ( f j )] = (ad( fi ))1−Ai j [e j , f j ] = ad f 1−Ai j i h j ,

and if Ai j ≤−1, we get zero. If Ai j = 0, then [ fi ,h j ] =αi (h j ) fi = A j i fi = 0.

• If k = i , then we use sl(2;C)-representation theory: Let v = f j . Then [ei , f j ] = 0 and [hi , f j ] =−α j (hi ) f j =−Ai j f j . Now (ei ,hi , fi ) act on L (A) via the adjoint action, and we consider the orbit through v . Then we use a result from representation theory:

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

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).

Proof. We have (adei )1−Ai j e j = 0, (adei )2h = 0, (adei )ei = 0, (adei ) f j = 0, (adei )3 fi = 0. So 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:

(ad x)k[ y, z

] ,

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

exp(x) := ∞∑

xn

n!

on V whose inverse is exp(−x), and x 7→ exp(x) is a Lie algebra homomorphism by the Leibniz rule.

Definition 15.7. We define the elements

ni := exp ( ad(ei )

32 Iain Gordon

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

e = ( 0 1 0 0

) exp(e) =

)

Proposition 15.8. The element si = ni |h : h→ h satisfies si (x) = x −αi (x)hi .

Proof. This is a fascinating calculation, left as an Exercise.

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

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

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

So we have an action of W on h∗ via

si (λ) =λ−λ(hi )αi for all λ ∈ h∗.

Proposition 15.11. The map ni : Lα 7→Lsi (α) is an identification.

Proof. We compute directly:[ h,ni (x)

]= [ ni (s−1

i h)ni (x) = si (α)(h)ni (x)

Theorem 15.12 (Properties of the Weyl group). If i 6= j , then the element si s j ∈ W has the following order:

order(si s j ) =

∞ Ai j A j i ≥ 4 .

Proof. Let

} ,

so dimK = dimh∗−2. Writing V =Cαi ⊕Cα j , we get a decomposition h∗ =V ⊕K , and si s j acts trivially on K . On the other hand, we have

(si s j )|V 7→ (−1+ Ai j A j i Ai j

−A j i −1

) .

The result follows from a calculation of the eigenvalues and the Jordan Normal Form of this matrix.

32

16 Root spaces

Let L :=L (A). We have a decomposition (of vector spaces)

L = L0 ⊕ ⊕ α∈R

Lα ,

where R ⊂Q =Q+tQ−. We have Lα 6= 0. Here Q+ =Z≥0α1 +·· ·+Z≥0αn is the positive weight lattice. The elements of R are called roots, and the decomposition of Q induces a decomposition R = R+tR− into positive and negative roots.

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.

}⊂ R+.

• R is W -stable, i.e. mwα = mα for all w ∈W .

Definition 16.1. An element α ∈ R is called real if there exists αi ∈ Π and w ∈ W such that w(αi ) =α. Otherwise we call α imaginary.

Note. α ∈ Rre if and only if −α ∈ Rre, since w(siαi ) =−w(αi ). Furthermore, R+ = R+ re tR+

im.

)= R+ im.

Proof. Positivity is the key here. We have α = ∑n j=1 k jα j ∈ R+

im, where k j ≥ 0, and at least two k j ’s are non-zero. So si (α) = α−α(hi )αi has at least one positive k j , so all k j must be non- negative.

Let C := { λ ∈ h∗

}⊂C := { . . . ≥ 0

{ ht(β) : β = w(α), w ∈ W

} . Pick an element β achieving

minimal height. Then si (β) = β−β(hi )αi . Since β has minimal height, β(hi ) ≤ 0 for all i , and thus β ∈−C .

Theorem 16.4 (Kac). Let

} .

} , where α = ∑

j k jα j ; supp(α) corresponds to the vertices in the Dynkin diagram.

Exercise 16.6. Show that for [DIAGRAMME], −C = { k(α1 +α2 +α3) : k ∈ N}

in any minimal realisation, and w(α1 +α2 +α3) =α1 +α2 +α3.

Corollary 16.7. α ∈ R+ im ⇒ kα ∈ R+

im for all k ∈N.

Corollary 16.8. If A is an indecomposable generalised Cartan matrix, then

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

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 factors. Let u = (a1, . . . , an) and δ=∑n

i=1 aiαi . Then R+ im = {

kδ : k ∈Z\ {0} } .

3. if A is indefinite, then there exists α ∈ R+ im such that α=∑n

i=1 kiαi , ki > 0 and α(hi ) < 0 for all i .

Proof. Exercise.

Corollary 16.9. Since A is symmetrisable by (REFERENCE), if α ∈ Rre, then (α,α) > 0, and if α ∈ Rim, then (α,α) ≤ 0.

Proof. (hi , x) := diαi (x) where hi ↔ diαi , so

( αi ,α j

> 0, which proves the first statement. If α=∑n

i=1 kiαi is imaginary with ki ≥ 0 and α(hi ) ≤ 0 for all i , then

( α,α

We prove Theorem 16.4 in several steps.

Step 1: α ∈ R ⇒ supp(α) is connected.

Proof. Without loss of generality, letα ∈ R+. Let supp(α) = J ⊆ { 1,2, . . . ,n

} with J = J1∪ J2 discon-

nected (i.e. A j1 j2 = 0 for all j1 ∈ J1 and j2 ∈ J2). Let i ∈ J1, j ∈ J2. Then [ei ,e j ] = (adei )1−Ai j e j = 0. So Lα ⊂ n+, and we claim that Lα = 0.

...

Step 2: α ∈ R, α 6= ±αi with α±αi 6∈ R ⇒α(hi ) = 0. α ∈ R, α 6= −αi with α+αi 6∈ R ⇒α(hi ) ≥ 0.

Proof. Pick x ∈ Lα. Then ni (x) ∈ Lsiα. Now calculate ni (x):

adei (x) = 0 ⇒ exp(adei )x = x

ad fi (x) = 0 ⇒ exp(ad(− fi ))x = x

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

Step 3: α = ∑ i kiαi ∈ K , Ψ := {

β ∈ R+ : β = ∑ i = 1nmiαi , mi ≤ ki for all i

} . Let β have the

largest height amongst the β ∈Ψ. Then supp(β) = supp(α).

Proof. First note that supp(β) ⊆ supp(α). If equality did not hold, then there would exist j ∈ supp(α) \ supp(β) and j ′ ∈ supp(β) such that A j j ′ 6= 0. So in β, m j = 0, and so β−α j 6∈ R+

and β+α j 6∈ R+. So β(h j ) = 0, and for a contradiction note that β(h j ) =∑ i∈supp(β) miαi (h j ) =∑

i∈supp(β) mi A j i < 0.

34

}= supp(α). (This implies K ⊆ R+.)

Proof. Let i ∈ supp(α) \ J . Then β+αi 6∈ R+. So by β(hi ) ≥ 0 by A PREVIOUS Lemma. Let M ⊆ supp(α) \ J be the connected component of i . So β(h j ) ≥ 0 for all j ∈ M . Let

β′ :=∑ j∈M m jα j , so

β′(hi ) = β(hi )− ∑ j∈supp(α)\M

m jα j (hi ) .

Recall that α( j )(hi ) = Ai j . If j ∈ M , then β(h j ) ≥ 0, and since Ai j ≤ 0, β′(hi ) ≥ 0. By choosing i carefully (there exist i ′ ∈ M and j ′ ∈ supp(α) \ M such that Ai ′ j ′ < 0), we have β′(hi ) > 0 for some i ∈ M .

Let AM be the principal matrix (Ai j )i , j∈M , and u = (m j )T j∈M . So

β′(hi ) = ∑ j∈M

Ai j m j for all i ∈ M ,

i.e. AM u ≥ 0 (and 6= 0). So AM has finite type BY (REFERENCE). Let γ :=∑ i∈M (ki −mi )αi , where

we have ki −mi > 0 for all i ∈ M . Then

α− β= ∑ t∈supp(α)\J

(kt −mt )αt .

So (α− β)(hi ) = ∑

j∈M (k j −m j )Ai j ,

since M is the connected component of i in supp(α) \ J . But α(hi ) ≤ 0 since α ∈ K , we have β(hi ) ≥ 0 from above. Therefore, γ(hi ) ≤ 0 for all i ∈ M .

Set u = (ki −mi )T i∈M , and so AM u ≤ 0. Since AM has finite type, u = 0. (That is, AM (−u) ≥ 0,

and since AM has finite type, either −u > 0 or u = 0.)

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 θ =∑

i aiαi ∈ h∗.

Example. The Lie algebra associated to An has ai = 1 for all i = 1, . . . ,n, and the root spaces are Ei j , i 6= j .

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 ⟨αi ,αi ⟩ = 2/αi .)

Let hθ be the coroot corresponding under h ∼−−→ h∗ to the element 2θ

⟨θ,θ⟩ = θ.

36 Iain Gordon

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

Ai j = Ai j for i , j ∈ {1, . . . ,n},

A00 = 2 ,

j=1 a j Ai j for i ∈ {1, . . . ,n},

A0 j = − n∑

i=1 ci Ai j for j ∈ {1, . . . ,n}.

This process produces the so-called untwisted affine Dynkin diagrams.

Exercise 17.1. Show that the matrix A is an affine generalised Cartan matrix.

Example. Starting with G2, we the Weyl group is W (G2) = ⟨s1, s2 : s2 1 = s2

2 = e,(s1s2)6 = e⟩ ∼= D6

2 −1−3 2

R+ = R+ re =

} .

To find the coroot hθ, we write hθ = c1h1 + c2h2 and compute:

⟨θ,α1⟩ = 2⟨α1,α1⟩+3⟨α2,α1⟩ = 4−3 = 1 =α1(hθ) = 2c1 −3c2

⟨θ,α2⟩ = 0 =α2(hθ) =−c1 +2c2

This implies that hθ = 2h1 +h2. So the affine Cartan matrix A is

.

L g= g [ t±1]⊕Cd ⊕Cc

with structure[ X ⊗ f +λd +µc,Y ⊗ gλ′+µ′c

]= [X ,Y ]⊗ f g +λY ⊗ t dg

dt −λ′X ⊗ t

Rest=0 f dg )⟨X ,Y ⟩c .

Theorem 17.2. If g is simple with Dynkin diagram of finite type, then L g 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

ei := Ei ⊗1 , fi := Fi ⊗1 , hi := H1 ⊗1 ∈ g [ t±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(Ei ) =−Fi . There exists a non-degenerate pairing

⟨−,−⟩ : gθ×g−θ →C .

Choose Eθ,Fθ ∈ g±θ with ⟨Eθ,Fθ⟩ = 1 and ω(Eθ) = −Fθ. (For example, pick Fθ ∈ g−θ, set Eθ := −ω(Fθ) ∈ gθ, then ⟨Eθ, Fθ⟩ = ξ 6= 0, so set Fθ := Fθ

/√ ξ etc.) Then let

36

Infinite-dimensional Lie algebras 37

So we have H := (h⊗1)⊕Cd ⊕Cc 3 h0 =: −hθ⊗1+ c .

It follows that [e0, f0] = [Fθ,Eθ]⊗1+⟨Fθ,Eθ⟩c = h0 ,

where we used [Fθ,Eθ] = ⟨Fθ,Eθ⟩h′ −θ =−hθ.

We need a realisation of A given by α0,α1, . . . ,αn ∈ H∗, where α1, . . . ,αn are extended from h∗ by αi (d) =αi (c) = 0 for all i = 1, . . . ,n. Do the same for θ ∈ h∗, which gives θ ∈ H∗. Let δ be the dual of d . Then α0 :=−θ+δ ∈ H∗, and(

H ,Π= {α0,α1, . . . ,αn},Π∨ = {h0,h1, . . . ,hn} )

is a minimal realisation of A. (Clearly the αi and hi are all linearly independent.) So we get

α j (hi ) = Ai j for all i , j = 1, . . . ,n,

α0(hi ) = −θ(hi )+δ(hi ) =− n∑

n∑ j=1

a j Ai j = Ai 0 for i 6= 0,

α j (h0) = similarly for j 6= 0.

The relations are as follows:

[ei , fi ] = hi for all i , [ei , f j ] = 0 for i 6= j ,

[x, ei ] =αi (x)ei for x ∈ H , all i , [x, fi ] =−αi (x) fi for x ∈ H , all i .

To see the relation [ei , f j ] = 0 when j 6= i = 0, note that [e0, f j ] = [Fθ,F j ]⊗t = 0. To see the relation [x,ei ] = 0 when i = 0, note that (with x0 ∈ h)

[x0 +λd +µc,e0] = [x0,Fθ]⊗ t +λFθ⊗ t =−θ(x0)Fθ⊗ t +λFθ⊗ t = (δ−θ)(x)e0 =α0(x)e0 .

} :

• S 6= 0 because S contains Fθ.

• g acts on S, i.e. S is an ideal in g, [y ⊗1, x ⊗ t ] = [y, x]⊗ t .

• By simplicity, S = g.

Since [g,g] = g, we have [g⊗ t ,g⊗ t k−1] = g⊗ t k for k ≥ 2, and this g [ t±1

] ⊆ M , where c,d ∈ H . Hence L g= M . For x ∈ H ,[

x, vα⊗ t i ]= [ x0 +µd +λc, vα⊗ t i ]=α(x0)vα⊗ t i +µi vα⊗ t i = (α+ iδ)(x)vα⊗ t i ,

i.e. L g= L0 ⊕∑ (i ,α) 6=(0,0)

( L g

( L∩ (L g)α+iδ

) ,

so J ∩Lα+iδ 6= 0 for some (i ,α) if J 6= 0. But x ⊗ t i ∈ J for 0 6= x ∈ gα. Pick y ∈ g−α such that ⟨x, y⟩ 6= 0. But

0 =[ x ⊗ t i , y ⊗ t−i ]= [x, y]⊗1− i ⟨x, y⟩c ∈ H ,

so i = 0, so [x, y] = 0 = ⟨x, y⟩h′ α 6= 0, contradiction.

37

38 Iain Gordon

Remark 17.3. Observe that the real roots are {α+ iδ :α 6= 0}, the imaginary roots are {iδ : i 6= 0}, 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 (g) correspond precisely to affine Lie algebras g[t±]⊕Cd ⊕Cc =: L g.

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.

A2n−1 → B T n = 2 A2n−1

A2n (n ≥ 2) → C ′ n = 2 A2n

Dn+1 → C ′ n = 2Dn+1

A2 → ?

D4 → GT 2 = 3D4

The diagram automorphism σ extends to an automorphism, also denoted by σ, of g and H via

σ(c) = c , σ(d) = d , ei 7→ eσ(i ) , fi 7→ fσ(i ) , hi 7→ hσ(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(L g) by

τ(x ⊗ t i ) = ξ−iσ(x)⊗ t i , 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 (

L g )T =

{ x ∈ L g : τ(x) = x

} is isomorphic to the Kac-Moody Lie algebra corresponding to the twisted affine Dynkin diagram .

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:

• V =⊕ λ∈H∗ Vλ, where Vλ =

{ v ∈V : h.v =λ(h)v for all h ∈ H

} .

• dimCVλ <∞ for all λ ∈ H∗.

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

j=1 n jα j for n j ∈Z≥0.)

38

ch: O → Fun(H∗) , V 7→ ch(V ) := ∑ λ∈H∗

dimVλeλ ,

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

} ,

.

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 vλ := 1⊗1λ that satisfies

ei .vλ = 0 and h.vλ =λ(h)vλ .

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

M(λ)µ = {

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. ch M(λ) = eλ∏

α∈Φ+

)mα ,

whereΦ+ is the set of positive roots and mα the multiplicity of the root α.

Proof. We need ∑ ν∈Q+

P (ν)e−ν := ∑ ν∈Q+

dimU (n−)−νe−ν = ( ∏ α∈Φ+

( 1−e−α

) .

By Poincaré-Birkhoff-Witt, a basis for U (n−)−ν is∏ α∈Φ+

( f (i ) α

)aα,i , where ∑ α,i

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) /

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

16 Root spaces 32

1

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

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.

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

Der(A) = {

dz : j ∈Z

[ Lm ,Ln

= ( (n +1)zm+n+1 f ′+ zm+n+2 f ′′)− (

(m +1)zm+n+1 f ′+ zm+n+2 f ′′) = (n −m)zm+n+1 df

dz = (m −n)Lm+n f

This is called the Witt algebra over C, denoted Witt; its basis is { L j : j ∈Z}

and its structure is[ Lm ,Ln

]= (m −n)Lm+n .

Infinite-dimensional Lie algebras 3

:

} ,

i.e. we impose that each element is in the finite span of cos(nθ) d dθ and sin(nθ) d

dθ , or equivalently

in the finite span of { e i nθ d

dθ : n ∈Z} . Without the finiteness assumption, we would have to deal

with the Fourier analysis and convergence issues. Thus we have a basis{ Ln := i e i nθ d

dθ : n ∈Z

} .

Set z := e iθ, then Ln =−zn+1 d dz , and we recover the Witt algebra.

Example (Diffeomorphisms of S1). Let G := Diff+ ( S1

) be the group of (orientation-preserving)

diffeomorphisms of S1 under composition. It acts on C∞( S1,C

) via (g . f )(z) := f

( g−1(z)

) for

g ∈G . An element of the tangent space at 1 ∈G has the form

g (z) = z ( 1+ε(z)

)= z + ∞∑

)= z − ∞∑

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) =−x } .

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)

such that z⊆ Z ( g )= {

x ∈ g : [x, g] = 0 } .

As vector spaces the sequence (2.1) splits as g= g′⊕ z, say via a section σ : g→ g′. Let

β : g×g→ z , (x, y) 7→ [ σ(x),σ(y)

]−σ( [x, y]

3

) := {

] := (

.

Then gβ is a central extension. However, there exist isomorphisms of such gβ:

z −−−−→ gβ = g⊕ z −−−−→ g θ

y∼=

z −−−−→ gγ = g⊕ z −−−−→ g

We find quickly that θ : gβ→ gγ 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 θ2 ( [g , g ′]

)= γ(g , g ′)−β(g , g ′). So we find that central extensions are classified by

H 2(g;z )= C 2

} , and dθ(g , g ′) = θ

) .

Exercise 2.2. Show that if g is a finite-dimensional simple Lie algebra, then H 2 ( g;C

)= 0.

) := 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, (Lm ,Ln) 7→β(m,n) satisfy the two conditions from above, namely

• 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 , Lm 7→ θ(m) .

Writing θβ(m) := β(0,m) /

m for m 6= 0, replace β by β+dθβ: We see that w.l.o.g. β(0,m) = 0 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,−nβ(m), where β(m) =−β(−m).

With l +m +n = 0, we produce a relation which for n = 1 gives

(1−m)β(m +1)+ (m +2)β(m)− (2m +1)β(1) = 0 .

This recurrence relation has a two-dimensional solution space:

β(m) =α1m +α2m3 , with α1,α2 ∈C .

Now replace β with β+dθβ, where θβ(m) = δm,0 α1 2 , to get β(m,n) = δm,−nα2m3.

3 The Virasoro algebra

Definition 3.1. The Virasoro algebra (denoted by Vir) is the central extension of the Witt algebra with the choice β(m) = 1

12

) in the notation of Proposition 2.3.

The Virasoro Algebra is spanned by { Lm : m ∈Z}∪ {c} such that c is central, i.e. [c,Lm] = 0

for all m ∈Z, and [ Lm ,Ln

]= (m −n)Lm+n +δm,−n 1

12

) c .

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

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)

and ρ : h 7→ ρ(h) = (−×h) . (4.1)

Exercise 4.2. If h 6= 0, then B(µ,h) is irreducible, while if h = 0 it is not (e.g. the subspace of constants C< B(µ,h) is invariant).

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

Link to Virasoro algebras. Define

Lk := 1

: a− j a j+k : for k ∈Z .

The notation “: a− j a j+k :” means “normal ordering”, i.e.

: a− j a j+k : := {

a− j a j+k if − j ≤ j +k,

a j+k a− j if − j ≥ j +k.

Explicitly, we have

an− j an+ j if k = 2n,∑ j>0

an+1− j an+ j if k = 2n +1.

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) an acts as zero for n À 0, and

3. an acts locally nilpotently for n > 0.

Theorem 4.3. With Lk defined as above,

[ Lm ,Ln

12

( m3 −m

) as operators on B(µ,1), i.e. as a representation of the Virasoro algebra. (But not of the Witt algebra!)

5

1 if |x| ≤ 1,

0 if |x| > 1,

and for ε > 0, define Lk (ε) := 1 2

∑ j∈Z : a− j a j+k : ψ(ε j ). This has only finitely many non-zero

terms, and Lk (ε) → Lk as ε→ 0. Now calculate: [ an ,Lk

]= nak+n (using Lk (ε)), and hence

[ Lm(ε),Ln

]= 1

2

( j +m) : a− j a j+m+n :ψ(ε j )+

1

2

: an− j a j+m :ψ(ε j )−δm,−n 1

2

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).v1, v2⟩ = ⟨v1,ρ(ω(x)).v2⟩ for all x ∈ Vir, v1, v2 ∈V , i.e. x† =ω(x).

5.1 Family 1: Chargeless representations

Let

vn : n ∈Z} , (5.1)

where α,β ∈C and P ∈C[z, z−1]. Define a representation of the Virasoro algebra by the following action:

c 7→ 0 and Ln(vk ) =−( k +α+β(n +1)

) vn+k . (5.2)

This is the representation one would discover by using the definition Ln := −zn+1 d dz on the

elements vk := zk+α(dz)β. Now we start to use the structure of Vir. Set

h :=Cc ⊕CL0 (abelian); Vir± := span { Lk : k ∈Z±

} .

Vir = Vir−⊕ h⊕Vir+ . (5.3)

Exercise 5.2. Check that h=Cc ⊕CL0 is a Cartan subalgebra of Vir, i.e. that it is nilpotent and self-normalising.

6

Infinite-dimensional Lie algebras 7

Remark 5.3. In the representation Vα,β, h acts by scalars on each vk : L0vk =−(k +α+β)vk . For a general representation ρ : Vir → gl(V ), the decomposition

V = ⊕ λ∈h∗

Vλ

is called a weight space decomposition, where Vλ := { v ∈ V : ρ(z).v = λ(z)v for all z ∈ h

} . Obvi-

ously, the existence of a weight space decomposition is a restriction on the representation.

Lemma 5.4. Let V be a representation of an abelian Lie algebra A, such that V = ⊕ λ∈A∗ Vλ

and such that each Vλ is finite-dimensional. Then any subrepresentation U ≤V decomposes as U =⊕

λ

) .

Proof. Any v ∈V can be written as v =∑m j=1 v j with v j ∈Vλ j for j = 1, . . . ,m, i.e. a.v j =λ j (a)v j

for all a ∈ A. Suppose v ∈U . We need to show that v j ∈U for each j . Since λ j 6=λk for j 6= k, we can find a ∈ A such that λ j (a) 6=λk (a) if j 6= k. Then

v = v1 +·· ·+ vm

a.v = λ1(a)v1 +·· ·+λm(a)vm

am−1.v = λ1(a)m−1v1 +·· ·+λm(a)m−1vm .

Since each a j .v ∈U , we see that [ λi (a) j

] (v1, . . . , vm)T ∈U m , where 1 ≤ i ≤ m and 0 ≤ j ≤ m −1.

But the matrix [ λi (a) j

] is a Vandermonde matrix and thus, since the λi (a) are pairwise distinct,

it has non-zero determinant, and hence (v1, . . . , vm)T ∈U m , as required.

Theorem 5.5.

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

α,β of codimension one in Vα,β. (As vector spaces we may

consider these to be span {

v−(α+β) }

(If Vα,β is irreducible, then set V ′ α,β =Vα,β.)

2. V ′ α,β is unitary if and only if β+β= 1 and 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 vk ∈U . Now given vk ∈U , we have Ln(vk ) ∈U for all n, and so we get scalar conditions from Equation (5.2), i.e.

vn+k 6∈U ⇒ k +α+β(n +1) = 0 .

If vn+k , vm+k 6∈ U , then β = 0 and α = −k, and hence Cvk is a subrepresentation, and Vα,β

/ Cvk is irreducible (because Ln(vt ) = (t −k)vt+n for t 6= k).

Now assume that only one vector, say vm , does not belong to U . Then for l 6= k,m we must have Ln−k (vk ) = 0 = Lm−l (vl ), whence β= 1, a =−(m +1) and U = span

{ vt : t 6= m

} .

Part (2): Unitarity implies ⟨vk , vk⟩ > 0 for all relevant k. Hence

• ⟨L0(vk ), vk⟩ = ⟨vk ,L0(vk )⟩, whence α+β ∈R, and

7

8 Iain Gordon

• ⟨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)

whence β+β= 1.

Conversely, ⟨vk , vm⟩ = δkm is a well-defined positive-definite hermitian form on V ′ α,β because of

the conditions on α, β.

Exercise 5.6. Is Vα,β completely reducible? That is, if it is not irreducible, is Vα,β = C⊕V ′ α,β a

decomposition of Lie algebra representations?

5.2 Family 2: Highest-weight representations

The Virasoro algebra has a representation on B(µ,1) =C[x1, x2, . . . ] as in Equation (4.1) above. Recall that

L0 := 1

2 a2

2 + ∑

On a monomial ∏

i xni

i , the term a− j a j acts by multiplication by j n j , and so we see that∑ j>0 a− j a j picks out the “degree” of a homogeneous polynomial, where deg(x j ) = j , i.e.

B(µ,1) = ∑ λ∈h∗

homogeneous polynomials of degree N λ(c) = 1, λ(L0) =µ2/2+N ,

0 otherwise.

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

⟨ xk1

j=1 j 2k 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 ak , for k < 0.)

Proposition 5.7. Let V be a unitary representation of the Virasoro algebra (with c acting as a scalar) such that

V = ⊕ λ∈h∗

Vλ with dimVλ <∞ .

Then V is completely reducible.

Proof. Let U <V be an invariant subspace. Observe that ω(L0) = L0. Unitarity forces λ ∈ h∗ R

by considering the effect of L0 on ⟨v, v⟩ > 0. It then follows that ⟨Vλ,Vµ⟩ = 0 ifλ 6=µ. Set Uλ :=U∩Vλ, and define U⊥

λ ⊂ Vλ as the orthogonal complement to Uλ in the form restricted to Vλ. Then

Vλ =Uλ⊕U⊥ λ

, and ⊕

λU⊥ λ =: U⊥ is invariant under the Virasoro algebra. So V =U ⊕U⊥.

8

Infinite-dimensional Lie algebras 9

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

Uµ2/2 ⊕U⊥ µ2/2 = B ′(µ,1)µ2/2 =C.1

to see that 1 ∈Uµ2/2 ⇒ B ′(µ,1) =U , i.e. B ′(µ,1) is irreducible.

At this point we do not know whether B ′(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. cVir.v = cv (here c is a complex number on the right hand side),

2. L0(v) = hv for h ∈C,

} .

Remark 5.9.

• cVir acts as the complex number c on all of V .

• L0 ( L−ik · · ·L−i1 (v)

)= ( h + ik +·· ·+ i1

]= i L−i . So we get a weight space decomposition

V = ⊕ λ∈h∗

Vλ ,

where Vλ 6= 0 only if λ(cVir) = c and λ(L0) = h +Z≥0.

• Lk (v) = 0 for k > 0 (because Vλ = 0 for λ(L0) = h −k).

The (c,h) are called the highest weights of V .

) .

Definition 5.11. If V =⊕ λ∈h∗ Vλ, then the formal character of V is

chV (q, t ) := ∑ λ∈h∗

dim(Vλ)qλ(L0)tλ(c) .

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 − y x. We write Aad for this Lie algebra structure.

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 : g→U (g)ad satisfying the following universal property:

For every arbitrary associative algebra with unit A over k and a Lie algebra homomorphism j : g→ Aad , there exists a unique homomorphism of associative algebras φ : U (g ) → A such that j =φ .

9

10 Iain Gordon

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)

where T (g) denotes the tensor algebra of (the vector space) g, i.e. T (g) :=⊕ i≥0 g⊗i . 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 {

xb : b ∈ B }

be a basis for g, where B is ordered. Then the set of ordered monomials{

xi1

} 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 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:

Lk .λ= 0 for k > 0, L0.λ= hλ , cVir.λ= cλ , for all λ ∈C.

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.

1. M(c,h) = ⊕ s∈Z≥0

} ,

j=1 i j = s.

(This has dimension P (s) = the number of partitions of s; thus

chM(c,h)(q, t ) = t h

j≥1 (1−q j ) .

Note that L0.(1⊗1) := L0 ⊗1 = 1⊗ (L0.1) = h(1⊗1), and c.(1⊗1) = c(1⊗1), which proves that this is a highest-weight representation.)

10

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−ik · · ·L−i1 :

0 < i1 ≤ ·· · ≤ ik } .

Part (2). If M(c,h) = V ⊕W , this respects the weight space decomposition by Lemma 5.4. Hence M(c,h)h =C, so without loss of generality, Vh =C and Wh = 0. Then 1⊗1 ∈V . But 1⊗1 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 , L0.v = hv and Lk .v = 0 for all k > 0. Hence Cv =C(c,h). The evaluation map ev: M(c,h) :=U (Vir)⊗C(c,h) N , ρ⊗1 7→ ρ.v thus factors through the tensor product over 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 Ph = 0. (Note that all proper sub- 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 Virasoro algebra with V =⊕

λ∈h∗ Vλ is precisely one of the following three:

• V ′ α,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

V (c,h)∗ :=⊕ 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−ik · · ·L−i1 and Q = L− jl · · ·L− j1 , and for v = 1⊗1, let

⟨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−ik · · ·L−i1 L− jl · · ·L− j1 v .

Exercise 8.1. Check that this form is contravariant and sesquilinear (though not necessarily non-degenerate). Check that M(c,h)λ and M(c,h)µ are pairwise orthogonal if λ 6=µ.

Now let M ′(c,h) ≤ M(c,h) be the maximal proper submodule, which we know to exist.

Lemma 8.2. M ′(c,h) = ker (⟨−,−⟩) := {

k : ⟨k,−⟩= 0 } .

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⟩ 6= 0, so the kernel is indeed a proper submodule. Now suppose P.v ∈ M ′(c,h), and thus ω(Q)P.v ∈ M ′(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 det1(c,h) = 2h. Next,

.

) coeff v , working in the Verma module,

= ((

) coeff v

) .v +0

) coeff v

= 4h + c 2 .

With similar calculations we get that det2(c,h) = 2h(16h2 +2hc −10h + c).

Theorem 8.3 (Kac determinant formula).

det n(c,h) = K ∏

( h −hr,s(c)

where P (m) is the number of partitions of m,

K = ∏ r,s∈N

1≤r,s≤n

( (2r )s s!

and

(c −1)(c −25)(r 2 − s2)−24r s −2+2c )

.

• V (1,h) = M(1,h) if and only if h 6= m2

4 , m ∈Z.

• V (0,h) = M(0,h) if and only if h 6= m2−1 24 , m ∈Z.

Exercise 8.4 (Unitarity). Show that detn(c,h) > 0 for all n if c > 1 and h > 0.

Note that ⟨L−n .v,L−n .v⟩ = 2nh+c n3−n 12 , so considering large enough n we see that unitarity

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 detn(c,h) > 0, the form is positive semi-definite in this region (i.e. it is positive-definite for V (c,h)) if it is so at least once in this region. But we already found one unitary irreducible

highest-weight representation of Vir with h = µ2

4 , c = 1, namely B ′(µ,1).

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:

c(m) := 1− 6

Then

)2 −1

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

Let V = ⊕ n∈ZCvn . Then gl(V ) =: gl∞ is the Lie algebra of matrices with a finite number of

non-zero entries. It is spanned by the Ei , j : v j 7→ vi . Let

F (0) :=Λ∞ (0)V = span

} .

A.vi = A.vi0 ∧ vi2 ∧·· ·+ vi1 ∧ A.vi2 ∧·· ·+ · · · ,

(note that this action preserves tails). In detail,

if i 6= j , then Ei , j .vi = {

0 if j does not appear in i or if i appears in i,

replace v j with vi in vi otherwise,

and E j , j .vi = {

vi if j appears in i,

0 otherwise.

The space F (0) has a basis labelled by partitions λ ` n, where λ = (λ1 ≥ λ2 ≥ . . .) ∈ Z∞ ≥0 with∑

i λi = n. To see this, map

i = ( i0, . . . , im , . . .

) ,

and this map is clearly reversible. It follows that F (0) = span {

vλ } .

Define deg(vλ) := |λ| = n if λ ` n, so that |λ| is the number of which λ is a partition; i.e. 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 contains a big abelian subalgebra spanned by Λk for k ∈ Z, where Λk := ∑

i∈ZEi ,i+k , so that Λk .v j = v j−k for all j .

Exercise 9.2. Show that the elementsΛk 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.

A typical element of a∞ is a finite linear combination of terms ∑

i∈Zλi Ei ,i+k . If k 6= 0, then∑ i∈Zλi Ei ,i+k acts on F (0), because for i À 0 or i ¿ 0 we have an action by 0 for Ei ,i+k on vi.

However, if k = 0, then we would have ∑

i∈Zλi Ei ,i .vj =∑ s≥0λ js vj, which is not well defined!

To remedy the situation, we adjust the “action” as follows:

• Ei , j acts as Ei , j for i 6= j .

• Ei ,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.

0 otherwise.

Exercise 9.4. Show that the Ei , j differ from Ei , j only by a multiple of I , so commutators are unchanged. Show further that[

Ei , j , Ek,l ]= 0 for j 6= k, i 6= l ,

[ Ei , j , E j ,l

]= Ei ,l for i 6= l ,[ Ei , j , Ek,i

]=−Ek, j for k 6= j , [ Ei , j , E j ,i

]= Ei ,i − E j , j +α ( Ei ,i , E j , j

) I ,

where

) :=

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 ,

a central extension with α as the defining 2-cocycle. As before, we define the diagonal elements Λk :=∑

i∈Z Ei ,i+k , and we get an action on F (n) via

Λ0.vj = nvj and [ Λk ,Λl

]= 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

) .vn ∧ vn−1 ∧·· · , and check that

• this is an H -map (using that B(n,1) is irreducible to see injectivity), and that

• P ( Λ−1,Λ−2, . . .

) .vn ∧vn−1 ∧·· · spans F (n). (Count dimensions of the homogeneous com-

ponents on both sides.)

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?

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[x1, x2, . . . ] and F (0).

Exercise 9.7. Let k ∈Z≥0 and fk = vk ∧vk−1∧·· ·∧v1∧v−k ∧v−k−1∧·· · . Show that L0. fk = k2 fk

and L j . fk = 0 for all j > 0. Deduce that B(0,1) is not irreducible as a Virasoro module, and that it 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⊗CR 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.)

In the case R =C[ t , t−1

] , the Laurent polynomials in one variable, we get L g := g⊗R, the

loop algebra of g. This has a description as Calg ( C×,g

) or as Csf

) , where “Csf” stands for

the space of smooth maps with finite Fourier expansion, and the correspondence is between ψ : S1 → g and its Fourier coefficients in ψ = ∑

n∈Z e2πi nθXn . Either way, a basis {

Xa }

determines a basis { Xa(n) := Xa ⊗ t n}

for L g, and the Lie algebra structure is[ Xa(n), Xb(m)

]= [ Xa , Xb

] (n +m) .

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 6= I ( g such that

[ g, I

] ⊆ I ). Recall that g acts on itself via the adjoint 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 := {

.

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

B ∈ ( (g⊗g)∗

0 otherwise.

K (x, y) = tr ( ad x 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). H 2(L g;C

)=C .

Exercise 10.3. Prove the theorem.

Let d ∈ Der(R), i.e. d(r s) = d(r )s + r d(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

(g⊗R)oCd , where [

X ,Y ]⊗ r s +µY ⊗ds −λX ⊗dr .

By applying this to the loop algebra, we get the new algebra

L g :=L goCd ,

where d := t d dt . (The derivation tells us about degrees.)

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 needβ : L g⊗L g→Cwhich 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

L g=⊕ i∈Z

X ∈L g : [d , X ] = i X }

, and hence

i .

Claim. H 2 i = 0 if i 6= 0. (In fact, generalising what we’re about to do, it is simply better to see that

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 )

) .

)= 1 i β

i = 0. //

Now (L g)0 = g⊗1⊕Cd , hence H 2 ( (L g)0;C

)= 0 by (a generalisation of) Exercise 2.2. So we can ensure that β

( (L g)0, (L g)0

)= 0. Define βi : g⊗g→C by βi (x, y) =β

( x(i ), y(−i )

) . Then g⊗ t n is a (g⊗1)-representation (via

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 )

)=−λi K (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- stituent algebraC

[ t±1

] . 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

.

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 , [hi ,h j ] = 0 , [h j ,ei ] =αi (h j )ei , and [h j , fi ] =−αi (h j ) fi ,

and [ei , f j ] = 0 for i 6= j , [ei , fi ] = hi for all i , (11.1)

where

]= 0 if i − j

]= 0 = [ ei+1, [ei+1,ei ]

] for all i .

We write this concisely as ( ad(ei )

)1−αi (h j )(e j ) = 0 (for i 6= j ). There are completely similar rela- tions for the fi .

17

18 Iain Gordon

Exercise 11.1. Find similar generators and relations for the Lie algebras

g(M) := {

, where

0 0 In

0 In 0

Definition 11.2. The matrix A := ( αi (h j )

) i j is called the Cartan matrix, and it determines finite-

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 ∈Matn(Z) and A is indecomposable (i.e. not block-diagonal with more than one block);

• Ai i = 2 for all i ;

• Ai j ∈ { 0,−1,−2,−3

} for all i 6= j ;

• Ai j = 0 if and only if A j i = 0;

• if Ai j =−2 or −3, then A j i =−1.

These conditions imply that A is “non-degenerate” in appropriate sense which we will meet soon; in particular this includes det A 6= 0.

12 Kac-Moody Lie algebras

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

• Ai i = 2,

• Ai j ≤ 0 if i 6= j , and

• Ai j = 0 if and only if A j i = 0.

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 ∈Matn(C). A realisation of A is a triple (h,Π,Π∨), where

• h is a finite-dimensional vector space over C,

• Π∨ = { h1, . . . ,hn

• Π= { α1, . . . ,αn

18

such that α j (hi ) = Ai j .

Lemma 12.3. If (h,Π,Π∨) is a realisation of A, then dimh≥ 2n − rk A.

Proof. Let r = rk A and dimh = m. Extend Π and Π∨ to bases { α1, . . . ,αm

} and

( α j (hi )

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.

Example. Let A = ( 2 −2−2 2

) . Then rk A = 1, so a minimal realisation must have dimh= 3. So let

h=C3 and h∗ =C3, with respective bases { h0,h1,d

} and

{ α0,α1,γ

2 ,0 )

(This is associated to the Lie algebra sl2, 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 isΦ : (h,Π,Π∨) → (h′,Π′,Π′∨), whereΦ : h→ h′ is an isomorphism such thatΦ(hi ) = h′

i andΦ∗(α′ i ) =αi for all i .

Proof of 12.5. Let r = rk A, so that possibly after re-ordering

A = (

)

with A11 non-singular. Extend this to a (2n − r )× (2n − r )-matrix

C = A11 A12 0

A21 A22 In−r

0 In−r 0

.

This is non-singular, e.g. detC =±det A11. Let α1, . . . ,αn be the first n coordinate functions, and 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

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

,

The free Lie algebra L(X ) generated by X is defined as follows: Let

C⟨X ⟩ :=C⟨e1, . . . ,en , f1, . . . , fn , x⟩ be the free associative algebra generated by X . Then C⟨X ⟩ad is a Lie algebra (under commuta- tors), and L(X ) is the Lie subalgebra of C⟨X ⟩ad generated by the elements of X . Its elements are called Lie words. (More generally, any Lie algebra is spanned by Lie words in a set of generators.)

Definition 12.8. Let L(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,

• [ x, y

• [ x,ei

]+αi (x) fi for all i = 1, . . . ,n.

Remark 12.9. The Lie algebra L(A) is independent of the choice of minimal realisation of A thanks to Proposition 12.5.

Theorem 12.10 (Structure of L(A)).

1. L(A) = n−⊕ h⊕ n+, where h ∼= h is abelian and n−, n+ are freely generated by f1, . . . , fn , e1, . . . ,en , respectively.

2. L(A) =⊕ α∈Q Lα, where

Lα := {

} and Q =Zα1 +·· ·+Zαn ⊂ h∗.

3. L0 = h.

4. Lα = 0 unless α ∈Q+ or α ∈Q−, where Q+ =Z≥0α1 +·· ·+Z≥0αn , and analogously for Q−.

5. [ Lα, Lβ

]⊆ Lα+β.

Proof. Define ω : ei ↔− fi and x ↔−x. Then ω induces a Lie algebra involution on L(A) (just check it preserves the relations). Let n± be the space generated by respectively ei and fi for i = 1, . . . ,n, and h the space spanned by elements x for all x ∈ h. The involution swaps n+ and n−.

For λ ∈ h∗, let θλ : X → End

( T (V )

) ,

e j 7→

1 7→ 0

vi 7→ δi jλ(h j ).1

vi1 · · ·vis 7→ δi1 j (λ−αi2 −·· ·−αis )(h j )vi2 · · ·vis + vi1

( θλ(e j )(vi2 · · ·vis )

Infinite-dimensional Lie algebras 21

It is an exercise to check that this induces a Lie algebra representation of L(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 : L(A) → End

( T (V )

We can now prove part (1) of the theorem.

• h∼= h: There is a natural surjective Lie algebra homomorphism x 7→ x. If x = 0 in L(A), we would have θλ(x) = 0 ∈ End

( T (V )

) for all λ ∈ h∗; but θλ(x)(1) =λ(x). This can be 0 for all λ

only if x = 0.

• n− is free: Let φ(w) = θλ(w).1 (by construction this is independent of λ), i.e.

φ ( w( f1, . . . , fn)

)= w(v1 · · ·vn) .

So we get a surjective Lie algebra homomorphism φ : n− → L(V ) onto the free Lie algebra 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.

• n−+ h+ n+ is a direct sum: Let w−+ x +w+ = 0. Then θλ(w−)+θλ(x)+θλ(w+) = 0 for 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.

• So n−⊕ h⊕ n+ ⊆ L(A). To see that it is all of L(A), it is enough to check that it is closed under taking brackets since it contains the generating set X . In other words, we need

ad(ei ) ( n−⊕ h⊕ n+) ⊂ n−⊕ h⊕ n+ ,

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 h and n+. For n− we have

ad(ei )( f j ) = δi j h j ∈ h⊕ n− .

By induction,

]= [ ad(ei )(w1), w2

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 on

[ ei1 , [ei2 , . . . ]

Lα = L(A) .

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= L0.

Lemma 12.11. L(A) contains a unique ideal I that is maximal with respect to the condition I ∩h= 0.

21

Proof. Let J = { I C L(A) : I ∩h = 0

} . Then for I ′, I ′′ ∈ J , I ′ and I ′′ inherit the weight space

decomposition by Lemma 5.4, so I (′,′′) = ⊕

0 6=α∈Q I (′,′′) α .

Therefore (I ′+ I ′′)0 = 0. So ∑

I∈J I is the desired ideal.

Definition 12.12. Let L(A) := L(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 ω induces an involution ω on L(A) which exchanges n+ and n−. Finally, L(A) inherits the weight space decomposition

L(A) = ⊕ α∈Q

Lα ,

where Q =Zα1 +·· ·+Zαn is the weight lattice. We collect a few obvious results:

Proposition 12.13.

2. dimLαi = 1.

3. kαi is a root if and only if k =±1.

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 Sn , where A ∼ A′ if and only if Ai j = A′

π(i ),π( j ) for π ∈ Sn . Furthermore we assume that A is indecomposable, i.e. if A = A1 ⊕ A2, then at least one of A1 and A2 is zero.

Definition 13.1. Let v = (v1, . . . , vn) ∈Rn . We say v ≥ 0 if vi ≥ 0 for all i , and we say v > 0 if v ≥ 0 and v 6= 0. We define v ≤ 0 and v < 0 similarly.

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 6= 0,

• if Av ≥ 0, then v ≥ 0.

2. A has affine type if

• corank(A) = 1,

• 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 ∈Matn(R) such that Ai i ≥ 2 for all i . For J ⊆ {1, . . . ,n} let A J be the submatrix of A whose entries are labelled by J × J .

Lemma 13.4. Assume that A is indecomposable. Then

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

• if A is affine, then A J 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

A = (

) .

If A is finite, then by definition there exists a v > 0 such that Av > 0, so

( A | P

v1 ...

vn

> 0 ,

and so A J 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 ) 6= 0, and so λ is not an eigenvalue of A. So A is a real, symmetric matrix all of whose eigenvalues are strictly positive, and hence A is positive definite.

Conversely, if A is positive definite, then det A 6= 0, and so A is not affine. If A were indef- 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

( d1, . . . ,dn

) and a symmetric matrix B such that A = DB .

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 i1, . . . , ik ∈ {1, . . . ,n}.

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 6= 0 for all k = 1, . . . , t −1. Let 0 6= d1 ∈R and define

d j = A jt jt−1 · · · A j2 j1

A j1 j2 · · · A jt+1 jt

d1 .

(Exercise: Check that this is well-defined, i.e. independent of the route from 1 to j .) Let D := diag

( d1, . . . ,dn

) and Bi j := d−1

i Ai j . We need Bi j = B j i , i.e.

Ai j

d j .

which is obvious if Ai j = 0. Assume thus that Ai j 6= 0, choose a sequence from 1 to i , i.e. 1 = j1, . . . , jt = i , and augment it by jt+1 = j . Then

d j = A j jt

( A jt jt−1 · · · A j2 j1

)( A j1 j2 · · · A jt−1 jt

) A jt j

A j jt

A jt j di .

the indexes seem wrong...

Remark 13.9. The proof of Lemma 13.8 shows that without loss of generality, di > 0 for all i , and that B ∈Matn(Q).

Proposition 13.10. If A is indecomposable and v > 0 is such that Av > 0, then vi > 0 for all 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∑n j=1 Ai j v j ≥ 0 for all i (recall that Ai j < 0 if i 6= j ).

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

} .

A J =

,

where ri , si > 0, and this matrix is either finite or affine, i.e. there exists v = (v1, . . . , vk ) > 0 such that A J v ≥ 0. Let

M := diag(v)−1 A J diag(v) =

2 −r ′

. . . 0 ...

−r ′ k . . . −s′k−1 2

where

s′i = v−1 i+1si vi > 0

r ′ i s′i = ri si ∈Z

and ∑ j

j=1 v−1

i (A J )i j v j = v−1 i A J v ≥ 0 .

So ∑

k + s′k ) .

i + s′i = 2

. . . . . .

. . . ...

0 0 0 −1 2 −1 0 0 0 0 −1 2 −1 −1 0 0 . . . 0 −1 2

(13.1)

This is symmetric; v = (1, . . . ,1) > 0, and A J v = 0, so A J is affine, and so A J = A.

25

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

1. A is finite if and only if det A J > 0 for all J ,

2. A is affine if and only if det A = 0 and det A J > 0 for all proper J , and

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 di > 0 and B symmetric, and B is of the same type as A. The conclusions 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

A J =

0 ??? 2 0 0 0 ...

. . . . . .

. . . ...

0 0 0 2 ??? 0 0 0 0 ??? 2 −bs−1

−b′ s 0 0 . . . 0 −b′

s−1 2

i ∈N.

Kac claims that det A J < 0 unless bi = b′ i = 1 for all i .

We want to show that if either of the conditions in the lemma on principal determinants (Which lemma?) holds, then bi = b′

i = 1 for all i : Suppose there exists a unique bt or b′

t > 1. Then change the basis to1 ... 1

,e1, . . . ,et , . . . ,es ,

from which det A J < 0 follows by an easy calculation (det A J = s.(−x), where −x is the unique entry in column t ). So there exist at least two non-zero bt ,b′

t > 1. Taking the shortest route from t to t ′ produces a submatrix, which is proper unless s ≤ 4.

If s > 4, we get a tridiagonal matrix

2 −b1

. . . . . .

. . .

−b′ t 2

.

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

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 Ai j = 0,

• there is one edge between i and j if Ai j =−1 = A j i ,

• there is a directed double edge between i and j if Ai j A j i = 2, pointing towards j if j < i ,

• there are three edges between i and j if Ai j A j i = 3, two of which form a directed double edge pointing towards j if j < i ,

• there are four (undirected) between i and j if Ai j A j i = 4 and Ai j =−1,

• there is a crossed double edge between i and j if Ai j A j i = 4 and if Ai j =−2 = A j i , and

• for Ai j A j i ≥ 5, there is one edge between i and j labelled “|Ai j | |A j i |”.

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 : . . .

• 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

• A1:

• Cn (n ≥ 3):

• Dn (n ≥?):

• GT 2 :

These diagrams are called affine Dynkin diagrams, and they correspond precisely to affine generalised Cartan matrices. Note that the matrix determined by An 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 ).

...

Then we get back to the finite case. In the affine case, check what happens in rank 2:

( 2 −b−c 2

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

L (A) =⊕ α

} .

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.

Proof. Let A = DB , where D = diag ( d1, . . . ,dn

) with di ∈ N and B T = B . Let

( h,Π,Π∨)

h1, . . . ,hn )

( Π∨) ⊆ h,

and let h′′ be some complement such that h′⊕h′′ = h. Define a bilinear form (−,−) on h as follows:

(hi ,h) := diαi (h) for all h ∈ h, (h′′ 1 ,h′′

2 ) := 0 for all h′′ 1 ,h′′

2 ∈ h′′. (15.1)

We claim that this is symmetric:

(hi ,h j ) := diαi (h j ) = di A j i = di d j B j i = d j Ai j =: (h j ,hi )

It is also non-degenerate, which we leave as an Exercise. In particular,

ker(−,−)|h′ = { x ∈ h′ :αi (x) = 0 for all i = 1, . . . ,r

} is the null space of A, which (Exercise:) equals the centre of L (A).

Now consider the principal grading on L (A),

L (A) = ⊕ k∈Z

L (A)k ,

where deg(ei ) = 1 =−deg( fi ) for i = 1, . . . ,n and deg(x) = 0 for x ∈ h. Let

L (N ) = N⊕

k=−N L (A)k , so L (0) =L (A)0 = h .

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( [ei , f j ],h

) .

Now for N > 1, ( Li ,L j

)= 0 if i + j 6= 0, and we need to define ( L±N ,LN

) . Take y ∈LN and

express it as y =∑ i [ui , vi ] for some ui , vi ∈L(N−1). For x ∈L±N , define

(x, y) :=∑ i

) ,

and note that [x,ui ] ∈L±(N−1) and vi ∈L(N−1), so this is indeed an inductive definition. This is well-defined: Suppose x =∑

j [u′ j , v ′

(x, y) = ∑ i

)+ ( u′

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)α,L (A)β

)= 0 unless α+β= 0. So(−,−) : L (A)α×L (A)−α→C

is a non-degenerate pairing, and [x, y] = (x, y)h′ α, where x ∈ L (A)α, y ∈ L (A)−α, and h′

α is defined by (

h′ α,−)

h =α(−) .

Proof. Let α+β 6= 0. Then there exists h ∈ h such that (α+β)(h) 6= 0. So

−α(h) ( x, y

) ,

and hence (x, y) = 0. Now we have [x, y]− (x, y)h′ α ∈ h if x ∈L (A)α and y ∈L (A)−α. It follows

that ( [x, y]− (x, y)h′

α,h )= 0 from the definition for all h ∈ h.

Reminder: representation theory for sl2.

sl(2;C) = {(

)} ,

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)),

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

Theorem 15.3.

1. If V is a representation of sl2, then there exists v ∈V such that H v =λv. Setting v j = F j v, we have H v j = (λ−2 j )v j .

2. There exists a unique (n +1)-dimensional irreducible representation of sl2 (where n ≥ 0). It has a basis (v0, . . . , vn) and satisfies

H v j = (n −2 j )v j ,F v j = v j+1,Ev j = j (n − j +1)v j−1 ,

• F E •

F E •→ . . . →•

F E • .

Proof. (1) follows from the relations above (REFERENCE!), (2) is a fun Exercise.

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

[hi ,ei ] = 2ei , [hi , fi ] =−2 fi , [ei , fi ] = hi .

Lemma 15.4. In L (A),

(adei )1−Ai j e j = 0 and (ad fi )1−Ai j f j = 0 for all i , j = 1, . . . ,n, i 6= j .

30

Infinite-dimensional Lie algebras 31

Proof. Let us show this for the fi ’s, the proof for the ei ’s is analogous. Let x = (ad fi )1−Ai j f j . We prove that [ek , x] = 0 for all k. First, set

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 6= i , j , then this is clear (check!), since [ek , fi ] = 0 = [ek , f j ].

• If k = j , then

[e j , (ad( fi ))1−Ai j ( f j )] = (ad( fi ))1−Ai j [e j , f j ] = ad f 1−Ai j i h j ,

and if Ai j ≤−1, we get zero. If Ai j = 0, then [ fi ,h j ] =αi (h j ) fi = A j i fi = 0.

• If k = i , then we use sl(2;C)-representation theory: Let v = f j . Then [ei , f j ] = 0 and [hi , f j ] =−α j (hi ) f j =−Ai j f j . Now (ei ,hi , fi ) act on L (A) via the adjoint action, and we consider the orbit through v . Then we use a result from representation theory:

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

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).

Proof. We have (adei )1−Ai j e j = 0, (adei )2h = 0, (adei )ei = 0, (adei ) f j = 0, (adei )3 fi = 0. So 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:

(ad x)k[ y, z

] ,

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

exp(x) := ∞∑

xn

n!

on V whose inverse is exp(−x), and x 7→ exp(x) is a Lie algebra homomorphism by the Leibniz rule.

Definition 15.7. We define the elements

ni := exp ( ad(ei )

32 Iain Gordon

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

e = ( 0 1 0 0

) exp(e) =

)

Proposition 15.8. The element si = ni |h : h→ h satisfies si (x) = x −αi (x)hi .

Proof. This is a fascinating calculation, left as an Exercise.

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

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

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

So we have an action of W on h∗ via

si (λ) =λ−λ(hi )αi for all λ ∈ h∗.

Proposition 15.11. The map ni : Lα 7→Lsi (α) is an identification.

Proof. We compute directly:[ h,ni (x)

]= [ ni (s−1

i h)ni (x) = si (α)(h)ni (x)

Theorem 15.12 (Properties of the Weyl group). If i 6= j , then the element si s j ∈ W has the following order:

order(si s j ) =

∞ Ai j A j i ≥ 4 .

Proof. Let

} ,

so dimK = dimh∗−2. Writing V =Cαi ⊕Cα j , we get a decomposition h∗ =V ⊕K , and si s j acts trivially on K . On the other hand, we have

(si s j )|V 7→ (−1+ Ai j A j i Ai j

−A j i −1

) .

The result follows from a calculation of the eigenvalues and the Jordan Normal Form of this matrix.

32

16 Root spaces

Let L :=L (A). We have a decomposition (of vector spaces)

L = L0 ⊕ ⊕ α∈R

Lα ,

where R ⊂Q =Q+tQ−. We have Lα 6= 0. Here Q+ =Z≥0α1 +·· ·+Z≥0αn is the positive weight lattice. The elements of R are called roots, and the decomposition of Q induces a decomposition R = R+tR− into positive and negative roots.

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.

}⊂ R+.

• R is W -stable, i.e. mwα = mα for all w ∈W .

Definition 16.1. An element α ∈ R is called real if there exists αi ∈ Π and w ∈ W such that w(αi ) =α. Otherwise we call α imaginary.

Note. α ∈ Rre if and only if −α ∈ Rre, since w(siαi ) =−w(αi ). Furthermore, R+ = R+ re tR+

im.

)= R+ im.

Proof. Positivity is the key here. We have α = ∑n j=1 k jα j ∈ R+

im, where k j ≥ 0, and at least two k j ’s are non-zero. So si (α) = α−α(hi )αi has at least one positive k j , so all k j must be non- negative.

Let C := { λ ∈ h∗

}⊂C := { . . . ≥ 0

{ ht(β) : β = w(α), w ∈ W

} . Pick an element β achieving

minimal height. Then si (β) = β−β(hi )αi . Since β has minimal height, β(hi ) ≤ 0 for all i , and thus β ∈−C .

Theorem 16.4 (Kac). Let

} .

} , where α = ∑

j k jα j ; supp(α) corresponds to the vertices in the Dynkin diagram.

Exercise 16.6. Show that for [DIAGRAMME], −C = { k(α1 +α2 +α3) : k ∈ N}

in any minimal realisation, and w(α1 +α2 +α3) =α1 +α2 +α3.

Corollary 16.7. α ∈ R+ im ⇒ kα ∈ R+

im for all k ∈N.

Corollary 16.8. If A is an indecomposable generalised Cartan matrix, then

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

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 factors. Let u = (a1, . . . , an) and δ=∑n

i=1 aiαi . Then R+ im = {

kδ : k ∈Z\ {0} } .

3. if A is indefinite, then there exists α ∈ R+ im such that α=∑n

i=1 kiαi , ki > 0 and α(hi ) < 0 for all i .

Proof. Exercise.

Corollary 16.9. Since A is symmetrisable by (REFERENCE), if α ∈ Rre, then (α,α) > 0, and if α ∈ Rim, then (α,α) ≤ 0.

Proof. (hi , x) := diαi (x) where hi ↔ diαi , so

( αi ,α j

> 0, which proves the first statement. If α=∑n

i=1 kiαi is imaginary with ki ≥ 0 and α(hi ) ≤ 0 for all i , then

( α,α

We prove Theorem 16.4 in several steps.

Step 1: α ∈ R ⇒ supp(α) is connected.

Proof. Without loss of generality, letα ∈ R+. Let supp(α) = J ⊆ { 1,2, . . . ,n

} with J = J1∪ J2 discon-

nected (i.e. A j1 j2 = 0 for all j1 ∈ J1 and j2 ∈ J2). Let i ∈ J1, j ∈ J2. Then [ei ,e j ] = (adei )1−Ai j e j = 0. So Lα ⊂ n+, and we claim that Lα = 0.

...

Step 2: α ∈ R, α 6= ±αi with α±αi 6∈ R ⇒α(hi ) = 0. α ∈ R, α 6= −αi with α+αi 6∈ R ⇒α(hi ) ≥ 0.

Proof. Pick x ∈ Lα. Then ni (x) ∈ Lsiα. Now calculate ni (x):

adei (x) = 0 ⇒ exp(adei )x = x

ad fi (x) = 0 ⇒ exp(ad(− fi ))x = x

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

Step 3: α = ∑ i kiαi ∈ K , Ψ := {

β ∈ R+ : β = ∑ i = 1nmiαi , mi ≤ ki for all i

} . Let β have the

largest height amongst the β ∈Ψ. Then supp(β) = supp(α).

Proof. First note that supp(β) ⊆ supp(α). If equality did not hold, then there would exist j ∈ supp(α) \ supp(β) and j ′ ∈ supp(β) such that A j j ′ 6= 0. So in β, m j = 0, and so β−α j 6∈ R+

and β+α j 6∈ R+. So β(h j ) = 0, and for a contradiction note that β(h j ) =∑ i∈supp(β) miαi (h j ) =∑

i∈supp(β) mi A j i < 0.

34

}= supp(α). (This implies K ⊆ R+.)

Proof. Let i ∈ supp(α) \ J . Then β+αi 6∈ R+. So by β(hi ) ≥ 0 by A PREVIOUS Lemma. Let M ⊆ supp(α) \ J be the connected component of i . So β(h j ) ≥ 0 for all j ∈ M . Let

β′ :=∑ j∈M m jα j , so

β′(hi ) = β(hi )− ∑ j∈supp(α)\M

m jα j (hi ) .

Recall that α( j )(hi ) = Ai j . If j ∈ M , then β(h j ) ≥ 0, and since Ai j ≤ 0, β′(hi ) ≥ 0. By choosing i carefully (there exist i ′ ∈ M and j ′ ∈ supp(α) \ M such that Ai ′ j ′ < 0), we have β′(hi ) > 0 for some i ∈ M .

Let AM be the principal matrix (Ai j )i , j∈M , and u = (m j )T j∈M . So

β′(hi ) = ∑ j∈M

Ai j m j for all i ∈ M ,

i.e. AM u ≥ 0 (and 6= 0). So AM has finite type BY (REFERENCE). Let γ :=∑ i∈M (ki −mi )αi , where

we have ki −mi > 0 for all i ∈ M . Then

α− β= ∑ t∈supp(α)\J

(kt −mt )αt .

So (α− β)(hi ) = ∑

j∈M (k j −m j )Ai j ,

since M is the connected component of i in supp(α) \ J . But α(hi ) ≤ 0 since α ∈ K , we have β(hi ) ≥ 0 from above. Therefore, γ(hi ) ≤ 0 for all i ∈ M .

Set u = (ki −mi )T i∈M , and so AM u ≤ 0. Since AM has finite type, u = 0. (That is, AM (−u) ≥ 0,

and since AM has finite type, either −u > 0 or u = 0.)

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 θ =∑

i aiαi ∈ h∗.

Example. The Lie algebra associated to An has ai = 1 for all i = 1, . . . ,n, and the root spaces are Ei j , i 6= j .

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 ⟨αi ,αi ⟩ = 2/αi .)

Let hθ be the coroot corresponding under h ∼−−→ h∗ to the element 2θ

⟨θ,θ⟩ = θ.

36 Iain Gordon

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

Ai j = Ai j for i , j ∈ {1, . . . ,n},

A00 = 2 ,

j=1 a j Ai j for i ∈ {1, . . . ,n},

A0 j = − n∑

i=1 ci Ai j for j ∈ {1, . . . ,n}.

This process produces the so-called untwisted affine Dynkin diagrams.

Exercise 17.1. Show that the matrix A is an affine generalised Cartan matrix.

Example. Starting with G2, we the Weyl group is W (G2) = ⟨s1, s2 : s2 1 = s2

2 = e,(s1s2)6 = e⟩ ∼= D6

2 −1−3 2

R+ = R+ re =

} .

To find the coroot hθ, we write hθ = c1h1 + c2h2 and compute:

⟨θ,α1⟩ = 2⟨α1,α1⟩+3⟨α2,α1⟩ = 4−3 = 1 =α1(hθ) = 2c1 −3c2

⟨θ,α2⟩ = 0 =α2(hθ) =−c1 +2c2

This implies that hθ = 2h1 +h2. So the affine Cartan matrix A is

.

L g= g [ t±1]⊕Cd ⊕Cc

with structure[ X ⊗ f +λd +µc,Y ⊗ gλ′+µ′c

]= [X ,Y ]⊗ f g +λY ⊗ t dg

dt −λ′X ⊗ t

Rest=0 f dg )⟨X ,Y ⟩c .

Theorem 17.2. If g is simple with Dynkin diagram of finite type, then L g 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

ei := Ei ⊗1 , fi := Fi ⊗1 , hi := H1 ⊗1 ∈ g [ t±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(Ei ) =−Fi . There exists a non-degenerate pairing

⟨−,−⟩ : gθ×g−θ →C .

Choose Eθ,Fθ ∈ g±θ with ⟨Eθ,Fθ⟩ = 1 and ω(Eθ) = −Fθ. (For example, pick Fθ ∈ g−θ, set Eθ := −ω(Fθ) ∈ gθ, then ⟨Eθ, Fθ⟩ = ξ 6= 0, so set Fθ := Fθ

/√ ξ etc.) Then let

36

Infinite-dimensional Lie algebras 37

So we have H := (h⊗1)⊕Cd ⊕Cc 3 h0 =: −hθ⊗1+ c .

It follows that [e0, f0] = [Fθ,Eθ]⊗1+⟨Fθ,Eθ⟩c = h0 ,

where we used [Fθ,Eθ] = ⟨Fθ,Eθ⟩h′ −θ =−hθ.

We need a realisation of A given by α0,α1, . . . ,αn ∈ H∗, where α1, . . . ,αn are extended from h∗ by αi (d) =αi (c) = 0 for all i = 1, . . . ,n. Do the same for θ ∈ h∗, which gives θ ∈ H∗. Let δ be the dual of d . Then α0 :=−θ+δ ∈ H∗, and(

H ,Π= {α0,α1, . . . ,αn},Π∨ = {h0,h1, . . . ,hn} )

is a minimal realisation of A. (Clearly the αi and hi are all linearly independent.) So we get

α j (hi ) = Ai j for all i , j = 1, . . . ,n,

α0(hi ) = −θ(hi )+δ(hi ) =− n∑

n∑ j=1

a j Ai j = Ai 0 for i 6= 0,

α j (h0) = similarly for j 6= 0.

The relations are as follows:

[ei , fi ] = hi for all i , [ei , f j ] = 0 for i 6= j ,

[x, ei ] =αi (x)ei for x ∈ H , all i , [x, fi ] =−αi (x) fi for x ∈ H , all i .

To see the relation [ei , f j ] = 0 when j 6= i = 0, note that [e0, f j ] = [Fθ,F j ]⊗t = 0. To see the relation [x,ei ] = 0 when i = 0, note that (with x0 ∈ h)

[x0 +λd +µc,e0] = [x0,Fθ]⊗ t +λFθ⊗ t =−θ(x0)Fθ⊗ t +λFθ⊗ t = (δ−θ)(x)e0 =α0(x)e0 .

} :

• S 6= 0 because S contains Fθ.

• g acts on S, i.e. S is an ideal in g, [y ⊗1, x ⊗ t ] = [y, x]⊗ t .

• By simplicity, S = g.

Since [g,g] = g, we have [g⊗ t ,g⊗ t k−1] = g⊗ t k for k ≥ 2, and this g [ t±1

] ⊆ M , where c,d ∈ H . Hence L g= M . For x ∈ H ,[

x, vα⊗ t i ]= [ x0 +µd +λc, vα⊗ t i ]=α(x0)vα⊗ t i +µi vα⊗ t i = (α+ iδ)(x)vα⊗ t i ,

i.e. L g= L0 ⊕∑ (i ,α) 6=(0,0)

( L g

( L∩ (L g)α+iδ

) ,

so J ∩Lα+iδ 6= 0 for some (i ,α) if J 6= 0. But x ⊗ t i ∈ J for 0 6= x ∈ gα. Pick y ∈ g−α such that ⟨x, y⟩ 6= 0. But

0 =[ x ⊗ t i , y ⊗ t−i ]= [x, y]⊗1− i ⟨x, y⟩c ∈ H ,

so i = 0, so [x, y] = 0 = ⟨x, y⟩h′ α 6= 0, contradiction.

37

38 Iain Gordon

Remark 17.3. Observe that the real roots are {α+ iδ :α 6= 0}, the imaginary roots are {iδ : i 6= 0}, 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 (g) correspond precisely to affine Lie algebras g[t±]⊕Cd ⊕Cc =: L g.

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.

A2n−1 → B T n = 2 A2n−1

A2n (n ≥ 2) → C ′ n = 2 A2n

Dn+1 → C ′ n = 2Dn+1

A2 → ?

D4 → GT 2 = 3D4

The diagram automorphism σ extends to an automorphism, also denoted by σ, of g and H via

σ(c) = c , σ(d) = d , ei 7→ eσ(i ) , fi 7→ fσ(i ) , hi 7→ hσ(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(L g) by

τ(x ⊗ t i ) = ξ−iσ(x)⊗ t i , 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 (

L g )T =

{ x ∈ L g : τ(x) = x

} is isomorphic to the Kac-Moody Lie algebra corresponding to the twisted affine Dynkin diagram .

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:

• V =⊕ λ∈H∗ Vλ, where Vλ =

{ v ∈V : h.v =λ(h)v for all h ∈ H

} .

• dimCVλ <∞ for all λ ∈ H∗.

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

j=1 n jα j for n j ∈Z≥0.)

38

ch: O → Fun(H∗) , V 7→ ch(V ) := ∑ λ∈H∗

dimVλeλ ,

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

} ,

.

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 vλ := 1⊗1λ that satisfies

ei .vλ = 0 and h.vλ =λ(h)vλ .

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

M(λ)µ = {

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. ch M(λ) = eλ∏

α∈Φ+

)mα ,

whereΦ+ is the set of positive roots and mα the multiplicity of the root α.

Proof. We need ∑ ν∈Q+

P (ν)e−ν := ∑ ν∈Q+

dimU (n−)−νe−ν = ( ∏ α∈Φ+

( 1−e−α

) .

By Poincaré-Birkhoff-Witt, a basis for U (n−)−ν is∏ α∈Φ+

( f (i ) α

)aα,i , where ∑ α,i

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) /

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