Infinite-dimensional Lie algebras - School of Mathematicsigordon/LA1.pdfInﬁnite-dimensional Lie algebras 5 4 TheHeisenbergalgebra TheWittalgebrawasasimplepointofdepartureforinﬁnitedimensionalLiealgebras

Infinite-dimensional Lie algebras

Lecture course by Iain Gordon

Edinburgh, 2008/9

Contents

1 Introduction 1

2 Central extensions 2

3 The Virasoro algebra 3

4 The Heisenberg algebra 4

5 Representations of the Virasoro algebra 55.1 Family 1: Chargeless representations . . . . . . . . . . . . . . . . . . . . . . . . . . 55.2 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 129.1 Fermionic Fock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129.2 Central extensions of gl∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

10 Hands-on loop and affine algebras 14

11 Simple Lie algebras 16

12 Kac-Moody Lie algebras 1712.1 Step I: Realisations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1712.2 Step II: A big Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1912.3 Step III: A smaller Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

13 Classification of generalised Cartan matrices 2113.1 Types of generalised Cartan matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 2113.2 Symmetric generalised Cartan matrices . . . . . . . . . . . . . . . . . . . . . . . . . 22

14 Dynkin diagrams 26

15 Forms, Weyl groups and roots 27

16 Root spaces 32

17 Affine Lie algebras and Kac-Moody Lie algebras 34

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2 Iain Gordon

18 The Weyl-Kac formula 3718.1 Category O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

18.2 Kac’s Casimir . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

18.3 The Weyl-Kac formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

19 W − K + A = M 4319.1 Affine Weyl groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

19.2 Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

20 KP hierarchy and Lie theory 45

21 The Kazhdan-Lusztig Conjecture 4821.1 The Shapovalov form and Kac-Kazhdan formulas . . . . . . . . . . . . . . . . . . . 48

21.2 Bringing in the Weyl group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

21.3 The Kazhdan-Lusztig Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

A References 53

B Notational reference 53

C Generation of Lie algebras 54

1 Introduction

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

• Derivations of an associative algebra.

• Vector fields on a smooth manifold.

• Tangent spaces at the identity of Lie groups.

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

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

Der(A) =

f (z)d

dz: f ∈C[

z, z−1]= span

L j :=−z j+1 d

dz: j ∈Z

.

We compute the structure of Der(A):

[Lm ,Ln

]f =

[−zm+1 d

dz,−zn+1 d

dz

]f

= ((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 isL j : j ∈Z

and its structure is[Lm ,Ln

]= (m −n)Lm+n .

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Infinite-dimensional Lie algebras 3

Example (Vector fields on S1). Consider the complex vector fields on S1 = e iθ : θ ∈R/∼

:

Vectfin :=

fd

dθ: f : S1 →C smooth with finite Fourier expansion

,

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

dθ , or equivalently

in the finite span ofe 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 basisLn := i e i nθ d

dθ: n ∈Z

.

Set z := e iθ, then Ln =−zn+1 ddz , 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 +∞∑

n=−∞λnεzn+1 .

Then

g−1(z) = z(1−ε(z)

)= z −∞∑

n=−∞λnεzn+1 .

Thus (g . f

)(z) = f

(z(1−ε(z))

) = f(z −

∞∑n=−∞

λnεzn+1)=

(1−

∞∑n=−∞

λnεzn+1 d

dz

)f (z) = (

1+∞∑

n=−∞λnεLn

)f (z) .

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

The three preceding examples all give the same Lie algebra structure. The Witt algebra hastwo 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]

).

We check that β satisfies

• β(x, y) =−β(y, x), and

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4 Iain Gordon

• β(x, [y, z]

)+β(y, [z, x]

)+β(z, [x, y]

)= 0.

Conversely, given β ∈C 2(g;z

):=

β : g×g→ z : satisfying the above, we can construct

gβ := g⊕ z as vector spaces,[(g , z), (g ′, z ′)

]:= (

[g , g ′],β(g , g ′))

.

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

(g;z

)dB 1

(g;z

) ,

where B 1 = θ : g→ z

, and dθ(g , g ′) = θ

([g , g ′]

).

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

)= 0.

Proposition 2.3. H 2(Witt

):= H 2

(Witt;C

)∼=C. In other words, there is a one-dimensional spaceof 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) = 0for all m ∈ Z. Now with l = 0, we get nβ(m,n)−mβ(n,m) = 0, i.e. (m +n)β(m,n) = 0. Thusβ(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,0α12 , 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 algebrawith the choice β(m) = 1

12

(m3 −m

)in the notation of Proposition 2.3.

The Virasoro Algebra is spanned byLm : 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,−n1

12

(m3 −m

)c .

The number 1/12 is chosen so that c acts as the scalar 1 on a natural irreducible representationthat we are about to see.

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4 The Heisenberg algebra

The Witt algebra was a simple point of departure for infinite dimensional Lie algebras. Anothersimple case is given by constructing loops on a (1-dimensional) abelian Lie algebra, and againmaking a central extension. This leads to

Definition 4.1. The Heisenberg algebra H is spanned by

an : n ∈Z∪ h, where h is centraland

[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, . . . ], onwhich the H -module structure is given by ρ : H → gl

(B(µ,h)

)as follows:

ρ : an 7→ ρ(an) =

∂∂xn

n > 0 ,

−hnx−n n < 0 ,

µ n = 0 ,

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 ofconstants C< B(µ,h) is invariant).

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

Link to Virasoro algebras. Define

Lk := 1

2

∑j∈Z

: 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

Lk =

12 a2

n + ∑j>0

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

]= (m −n)Lm+n +δm,−n1

12

(m3 −m

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

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6 Iain Gordon

Sketch of proof. Define a function

ψ : R→R , x 7→ψ(x) :=

1 if |x| ≤ 1,

0 if |x| > 1,

and for ε > 0, define Lk (ε) := 12

∑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∈Z

( j +m) : a− j a j+m+n :ψ(ε j )+

1

2

∑j∈Z

: an− j a j+m :ψ(ε j )−δm,−n1

2

−m∑j=−1

j ( j +m)ψ(ε j ) ,

and let ε→ 0.

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

5 Representations of the Virasoro algebra

Recall that we have an anti-linear involution ω(Ln) = L−n , ω(c) = c.

Definition 5.1 (Unitarity). A representation ρ : Vir → gl(V ) of the Virasoro algebra is unitary ifthere 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

V =Vα,β = P (z)zα(dz)β ∼=C[z, z−1] = span

vn : n ∈Z, (5.1)

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

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 ddz on the

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

h :=Cc ⊕CL0 (abelian); Vir± := spanLk : k ∈Z±

.

We have the so-called triangular decomposition

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 andself-normalising.

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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 asU =⊕

λ

(U ∩Vλ

).

Proof. Any v ∈V can be written as v =∑mj=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, wecan 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 andan irreducible representation V ′

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

consider these to be span

v−(α+β)

and span

vk : k 6= −α−β, respectively.)

(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 itsweight spaces, so if U is a non-zero subrepresentation, then there exists k with vk ∈U . Nowgiven 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, andVα,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 musthave 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

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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 producetwo different looking relations between ⟨vk , vk⟩ and ⟨vk−2, vk−2⟩ for infinitely many differ-ent integers k, we find an equality of polynomials(

X −β)(X − (1+β)

)(X −2(1+β)

)= (X − (1−β)

)(X − (2−β)

)(X −2β

),

whence β+β= 1.

Conversely, ⟨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

2a2

0 +∑j>0

a− j a j = µ2

2+ ∑

j>0a− j a j .

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∗

B(µ,1)λ ,

where

B(µ,1)λ =

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 representationof the Heisenberg algebra H . The monomials form an orthonormal basis:

⟨xk1

1 · · ·xknn , xk1

1 · · ·xknn

⟩=

∏nj=1 k j !∏n

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 polynomialsin the (commuting) variables ak , for k < 0.)

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

byconsidering 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⊥.

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

Uµ2/2 ⊕U⊥µ2/2 = B ′(µ,1)µ2/2 =C.1

to see that1 ∈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 Virasoroalgebra 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,

3. V = span(L−ik · · ·L−i1 )(v) : 0 ≤ i1 ≤ ·· · ≤ ik

.

Remark 5.9.

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

• L0(L−ik · · ·L−i1 (v)

)= (h + ik +·· ·+ i1

)(L−ik · · ·L−i1

)(v) thanks to

[L0,L−i

]= i L−i . So we geta 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 .

Remark 5.10. B ′(µ,1) is a highest-weight representation with weight(1,µ2/2

).

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 definethis 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 envelopingalgebra of g is an associative algebra U (g) over k with unit and a Lie algebra homomorphismı : g→U (g)ad satisfying the following universal property:

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

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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 fromDefinition 6.1.

Theorem 6.5 (Poincaré, Birkhoff, Witt). Let

xb : b ∈ B

be a basis for g, where B is ordered. Thenthe set of ordered monomials

xi1

b1xi2

b2xi3

b3· · · : b1 < b2 < b3 < ·· · , i j ∈Z≥0

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 aweight 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 algebradirect sum). We will now construct a universal highest-weight representation of the Virasoroalgebra. 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-dimensionalspace 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 theVirasoro 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

M(c,h)h+s , where M(c,h)h+s has a basis of vectors(L−ik · · ·L−i1 )⊗ 1

,

where 0 < i1 ≤ i2 ≤ ·· · ≤ ik and∑k

j=1 i j = s.

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

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

φ(q), where φ(q) = ∏

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 provesthat this is a highest-weight representation.)

10


2. (Universality.) M (c,h) is indecomposable, and any highest-weight representation of highestweight (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 basisL−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⊗1generates 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 ∈ Nsuch that c.v = cv , L0.v = hv and Lk .v = 0 for all k > 0. Hence Cv =C(c,h). The evaluation mapev: M(c,h) :=U (Vir)⊗C(c,h) N , ρ⊗1 7→ ρ.v thus factors through the tensor product overU (Vir≥) and so is the desired quotient.

Part (3). This is equivalent to a unique maximal proper submodule. Any submodule inheritsa 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 submodulesis again proper, and clearly maximal.

Theorem 7.3 (Mathieu; Chari, Pressley (unitary case)). An irreducible representation V of theVirasoro 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

[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 weightdecomposition of L−ik · · ·L−i1 L− jl · · ·L− j1 v .

Exercise 8.1. Check that this form is contravariant and sesquilinear (though not necessarilynon-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.

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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 theform ⟨−,−⟩. Define ⟨−,−⟩N to be the restriction of ⟨−,−⟩ to M(c,h)h+N . This is a form on afinite-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,

det 2(c,h) =∣∣∣∣ ⟨L−2.v,L−2.v⟩ ⟨L−2.v,L−1L−1.v⟩⟨L−1L−1.v,L−2.v⟩ ⟨L−1L−1.v,L−1L−1.v⟩

∣∣∣∣ .

But

⟨L−2.v,L−2.v⟩ = (L2L−2.v

)coeff v , working in the Verma module,

=((

[L2,L−2]+L−2L2).v

)coeff v

=((

4L0 + 112 (8−2)c

).v +0

)coeff v

= 4h + c2 .

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∏

r,s∈N1≤r,s≤n

(h −hr,s(c)

)P (n−r s) ,

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

K = ∏r,s∈N

1≤r,s≤n

((2r )s s!

)P (n−r s)−P (n−r (s+1)) ,

and

hr,s(c) = 148

((13− c)(r 2 + s2)+p

(c −1)(c −25)(r 2 − s2)−24r s −2+2c)

.

Examples.

• 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= m2−124 , 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−n12 , 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-definitefor 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


For c = 0, it can be shown that the only unitary representation is V (0,0) ∼= C. We want tounderstand what happens in the range 0 ≤ c < 1. Reparametrise as follows:

c(m) := 1− 6

(m +2)(m +3)for m ≥ 0.

Then

hr,s(c(m)

)≡ hr,s(m) =((m +3)r − (m +2)s

)2 −1

4(m +2)(m +3).

Theorem 8.5 (Friedan, Qiu, Shenker; Goddard, Kent, Olive). In the region 0 ≤ c < 1, unitaryrepresentations 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

vi := vi0 ∧ vi1 ∧·· ·∧ vim−1 ∧ v−m ∧ v−m−1 ∧·· · : i0 > i1 > ·· · im−1 >−m

.

Then gl∞ acts on F (0) via

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 , . . .

) 7→λi := (i0, i1 +1, i2 +2, . . . , im +m, . . .

),

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 beingvn ∧ 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 earliernotation B(µ,h) for Heisenberg representations stands for “Bosonic”.

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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”. Thiscontains a big abelian subalgebra spanned by Λk for k ∈ Z, where Λk := ∑

i∈ZEi ,i+k , so thatΛk .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 algebraembeds 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,

Ei ,i .vj =

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 areunchanged. 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

α(Ei ,i , E j , j

):=

−1 if i ≥ 0, j < 0,

+1 if i < 0, j ≥ 0,

0 otherwise.

We can say this in two ways: We either get a projective representation of gl∞ and a∞, or agenuine 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 elementsΛk :=∑

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 gradedH -modules (but obviously not as algebras).

Proof. Let φ(P (x1, x2, . . . )

):= P

(Λ−1,Λ−2, . . .

).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.)

14


Exercise 9.6. Recall that Witt ,→ a∞. Show that this embedding extends to an embeddingVir ,→ 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 sideacting on the other side. For instance, we see that the Virasoro algebra acts on F (n) as in Theorem4.3. 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 itis 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 vectorspace 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 asheafified 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

(S1,g

), where “Csf” stands for

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

n∈Z e2πi nθXn . Either way, a basis

Xa

for g

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 propernon-zero ideals (i.e. 0 6= I ( g such that

[g, I

] ⊆ I ). Recall that g acts on itself via the adjointaction ad(x).y := [x, y] for all x, y ∈ g. Then g is simple if and only if g is itself irreducible underthe 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 .

DefineMg :=

m ∈ M : x.m = 0 for all x ∈ g

.

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16 Iain Gordon

The adjoint representation induces via tensor product a representation on g⊗g, and then bytaking 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)∗

)g ∼= Homg

(g⊗g,Ctriv.

)∼= Homg

(g,g∗

)∼=C if g∼= g∗ by Schur’s Lemma,

0 otherwise.

Now here is an explicit form:

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 (bysome 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

]+ [X ⊗ r,d(Y ⊗ s)

]= [X ⊗dr,Y ⊗ s

]+ [X ⊗ r,Y ⊗ds

].

Hence we get a new Lie algebra

(g⊗R)oCd , where[

X ⊗ r +µd ,Y ⊗ s +λd]= [

X ,Y]⊗ r s +µY ⊗ds −λX ⊗dr .

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

L g :=L goCd ,

where d := t ddt . (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 isgiven 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 arisingfrom Theorem 10.2.

Proof. We needβ : L g⊗L g→Cwhich satisfies anti-symmetry and the Jacobi relations, modulolinear functionals.

We know that L g acts on (L g⊗L g)∗, and a check shows that this descends to an action ofL g on H 2

(L g;C

). We get decompositions into d-eigenspaces

L g=⊕i∈Z

L gi , where L gi =

X ∈L g : [d , X ] = i X

, and hence

H 2(L g;C)=⊕

i∈ZH 2

i .

Claim. H 2i = 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 )

).

16


So for i 6= 0, setting F(a(i )

)= 1i β

(d , a(i )

)ensures that H 2

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 canensure 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. Thereforeβi =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-MoodyLie 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

Consider the algebra

sln+1 ≡ sl(n +1;C) :=

X ∈Matn+1(C) : tr X = 0

.

In the classification of simple Lie algebras, this is of type An . This Lie algebra has Lie subalge-bras h, n+ and n− of diagonal, strictly upper-triangular and strictly lower-triangular matrices,respectively, and there is a decomposition sln+1

∼= n−⊕h⊕n+ as a direct sum of vector spaces,where each summand is actually a subalgebra.

Note that for i = 1, . . . ,n there are elements

ei = Ei ,i+1 ,hi = Ei ,i −Ei+1,i+1 ,fi = Ei+1,i ,

which generate the Lie algebra (e.g. Ei , j = [· · · [[ei ,ei+1],ei+2] · · · ,e j−1] for i < j ). Observe thatthe 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

αi (h j ) =

2 i = j ,

−1∣∣i − j

∣∣= 1,

0 otherwise.

Less obviously, we have[ei ,e j

]= 0 if∣∣i − j

∣∣ 6= 1 ,[ei , [ei ,ei+1]

]= 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 .

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18 Iain Gordon

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

g(M) :=

X ∈Matk (C) : X T M +M X = 0

, where

1. M =2 0 0

0 0 In

0 In 0

for k = 2n +1,

2. M =(

0 In

−In 0

)for k = 2n, and

3. M =(

0 In

In 0

)for k = 2n.

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 theCartan 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 meetsoon; 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 generalisedCartan 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 isactually 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

⊂ h is a linearly independent subset, and

• Π= α1, . . . ,αn

⊂ h∗ is a linearly independent subset,

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 basesα1, . . . ,αm

and

h1, . . . ,hm

respectively, producing an invertible matrix

(α j (hi )

)= (A BC D

).

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, som −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 basesh0,h1,d

and

α0,α1,γ

:

h0 =(0,2,0

)α0 =

(−1,1,1)

h1 =(1,−1,0

)α1 =

(1,−1,0

)d =(

0,0,1)

γ=(12 , 1

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′ isan 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 =(

A11 A12

A21 A22

)

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, andlet h1, . . . ,hn be the first n rows of C . This is a minimal realisation. The proof of uniqueness isleft 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.

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20 Iain Gordon

12.2 Step II: A big Lie algebra

Let A ∈Matn(Z) be a generalised Cartan matrix and (h,Π,Π∨) a minimal realisation of A, whereΠ∨ = (

α1, . . . ,αn). Let

X :=

e1, . . . ,en , f1, . . . , fn , x : x ∈ h

,

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 arecalled 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

]whenever x, y ∈ h,

•[ei , f j

]for i 6= j ,

[ei , fi

]− hi for all i = 1, . . . ,n,

•[x,ei

]−αi (x)ei and[x, fi

]+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 Athanks 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 :=

x ∈ L(A) :[h, x

]=α(h)x for all h ∈ h

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) (justcheck it preserves the relations). Let n± be the space generated by respectively ei and fi fori = 1, . . . ,n, and h the space spanned by elements x for all x ∈ h. The involution swaps n+ and n−.

For λ ∈ h∗, letθλ : X → End

(T (V )

), where V = span

v1, . . . , vn

,

be given by

x 7→ (vi1 · · ·vis 7→ (λ−αi1 −·· ·−αis )(x)vi1 · · ·vis

),

f j 7→ multiplication by v j ,

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 )

)

20


It is an exercise to check that this induces a Lie algebra representation of L(A) (again, it inducesa representation of the free Lie algebra generated by X ; now check the relations). So in fact weget 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), wewould 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 algebragenerated 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 forall λ ∈ h∗. Evaluating this at 1 ∈ T (V ) gives φ(w−)+λ(x) = 0, where φ(w−) ∈ T (V )>0 andλ(x) ∈ T (V )0. Thus λ(x) = 0 and φ(w−) = 0, which by the arguments above implies thatx = 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 closedunder 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 tosee 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( f1, . . . , fn), w2( f1, . . . , fn)

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

]+ [w1,ad(ei )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 actson

[ei1 , [ei2 , . . . ]

]by multiplication by

(αi1 +αi2 +·· ·)(x). Thus

n+, n−, h⊆ ∑α∈Q

Lα , and so∑α∈Q

Lα = L(A) .

Parts (3)–(5) follow immediately.

12.3 Step III: A smaller Lie algebra

We want to get close to a simple without losing the information carried in A. This means that wewould 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 conditionI ∩h= 0.

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22 Iain Gordon

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, soI (′,′′) = ⊕

0 6=α∈QI (′,′′)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) theKac-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) whichexchanges 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.

1. α1, . . . ,αn are positive roots.

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 simplifymatters, we shall only consider matrices up to permutation by Sn , where A ∼ A′ if and only ifAi 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 ≥ 0and 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,

• there exists v ≥ 0 such that Av > 0, and

• if Av ≥ 0, then v ≥ 0.

2. A has affine type if

• corank(A) = 1,

• there exists v > 0 such that Av = 0, and

• if Aw ≥ 0, then w =λv for some λ ∈C.

22


(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 eitherfinite, 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) suchthat Ai i ≥ 2 for all i . For J ⊆ 1, . . . ,n let A J be the submatrix of A whose entries are labelled byJ × 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 canwrite in block form

A =(

A J PQ R

).

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

(A | P

)v1...

vn

≥ 0 .

But

P

v1...

vn

≤ 0 , so A J

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.

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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 itbe affine type? So det(A−λI ) 6= 0, and so λ is not an eigenvalue of A. So A is a real, symmetricmatrix 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 positivedefiniteness. 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 asabove.

Definition 13.6. An (n ×n)-matrix A is symmetrisable if there exists a non-singular matrixD = 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. Let0 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

di= A j i

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

d1 =A j i

Ai jd1 and d1 =

A j jt

A jt jdi .

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∑nj=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


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

Ais i t = 0 for all (s, t ) 6∈ (1,2), (2,3), . . . , (k,1), (2,1), (3,2), . . . , (1,k)

.

For J = i1, . . . , ik , A J has the form

A J =

2 −r1 . . . −sk

−s1. . . 0

...... 0 2 −rk−1

−rk . . . −sk−1 2

,

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

M := diag(v)−1 A J diag(v) =

2 −r ′

1 . . . −s′k−s′1

. . . 0...

... 0 2 −r ′k−1

−r ′k . . . −s′k−1 2

where

r ′i = v−1

i ri vi+1 > 0

s′i = v−1i+1si vi > 0

r ′i s′i = ri si ∈Z

and ∑j

Mi j =n∑

j=1v−1

i (A J )i j v j = v−1i A J v ≥ 0 .

So∑

i , j Mi j ≥ 0, but ∑i , j

Mi j = 2k − (r ′1 + s′1)−·· ·− (r ′

k + s′k ) .

Sor ′

i + s′i2

≥√

r ′i s′i =

pri si ≥ 1 .

So

r ′i + s′i ≥ 2 ⇒ r ′

i + s′i = 2

⇒ ri si = 1 ,

and so

A J =

2 −1 0 . . . 0 0 −1−1 2 −1 0 0 00 −1 2 −1 0 0 0...

. . .. . .

. . ....

0 0 0 −1 2 −1 00 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.

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26 Iain Gordon

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

1. A is finite if and only if det A 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 conclusionsof (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 isobviously satisfied unless there is a cycle in the matrix. Pick a minimal cycle

A J =

2 −b1 0 . . . 0 0 −bs

−b′1 2 ??? 0 0 0

0 ??? 2 0 0 0...

. . .. . .

. . ....

0 0 0 2 ??? 00 0 0 ??? 2 −bs−1

−b′s 0 0 . . . 0 −b′

s−1 2

, an (s × s)-matrix with bi ,b′

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 uniqueentry in column t ). So there exist at least two non-zero bt ,b′

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

If s > 4, we get a tridiagonal matrix

2 −b1

−1 2 −1−1 2 −1

. . .. . .

. . .

−1 2 −1−1 2 −bt

−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 havethe 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 aredistinct 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 doubleedge 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 : . . .

• Bn (n ≥ 2): . . .

• Cn (n ≥ 3): . . .

• Dn (n ≥ 4): . . .

• E6:

• E7:

• E8:

• F4:

• G2:

These diagrams are called finite Dynkin diagrams, and they correspond precisely to finitegeneralised Cartan matrices. Note that the list is closed under transposition (which correspondsto 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.

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28 Iain Gordon

List of affine Dynkin diagrams

• A1:

• A′1:

• An (n ≥ 2):

• Bn (n ≥ 3) and B Tn :

• Cn (n ≥ 3):

• C Tn (n ≥ 3):

• Dn (n ≥?):

• E6:

• E7:

• E8:

• F4:

• F T4 :

• G2:

• GT2 :

These diagrams are called affine Dynkin diagrams, and they correspond precisely to affinegeneralised 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 caseuse the fact that if a Dynkin diagram produces a finite generalised Cartan matrix, then so doesany sub-diagram (where you are allowed to remove edges, e.g. is a sub-diagram of ).

Now suppose we had the following parts in a given diagram:

...

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

(2 −b−c 2

). For rank ≥ 3, we need to check that

all sub-diagrams (by removing vertices only) are finite.

15 Forms, Weyl groups and roots

Let A be a generalised Cartan matrix and L (A) its associated Kac-Moody Lie algebra. Recall thatwe have a weight space decomposition

L (A) =⊕α

L (A)α for α ∈Zα1, . . . ,αn

.

28


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∨)

be aminimal realisation of A, where Π∨ = (

h1, . . . ,hn)

and Π = (α1, . . . ,αn

). Let h′ = span

(Π∨) ⊆ h,

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

(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=−NL (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). ForN = 1, define

(ei , f j ) := δi j di , and(L±1,L0

)= 0 = (L±1,L±1

).

Invariance holds since([ei , f j ],h

)= δi j (hi ,h) = δi j diα j (h) = (ei , f j )α j (h) = (ei , [ f j ,h]

).

Now for N > 1,(Li ,L j

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

). Take y ∈L∓N and

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

(x, y) :=∑i

([x,ui ], vi

),

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 ′

j ], where u′j , v ′

j ∈L±(N−1). Then

(x, y) = ∑i

([x,ui ], vi

)= ∑

i , j

([[u′

j , v ′j ],ui

], vi

)= ∑

i , j

([[u′

j ,ui ], v ′j

], vi

)− ([[v ′

j ,ui ],u′j

], vi

)= ∑

i , j

([u′

j ,ui ], [v ′j , vi ]

)+ (u′

j ,[[v ′

j ,ui ], vi])

= ∑i , j

(u′

j ,[ui , [v ′

j , vi ]]+ [

[v ′j ,ui ], vi

])= ∑

i , j

(u′

j ,[v ′

j , [ui , vi ]])

= ∑j

(u′

j , [v ′j , y]

),

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30 Iain Gordon

where we used induction to conclude the second and the penultimate line. This calculationshows that the form (−,−) is well-defined, and moreover that it is symmetric. The proof that it isinvariant 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 isdefined 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

)= ([x,h], y

)= (x, [h, y]

)=β(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) =(

a bc −a

),

so sl2 is of type A1. It has a basis E , H ,F , where

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

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

[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). Ithas 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 ,

•FE•

FE•→ . . . →•

FE• .

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


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 . Weprove 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, andU (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 ofL (A).

Another proof. Let x be as in the first proof. Then x ∈ n− ⊂L (A). We show that [ek , x] = 0 for allk, 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 ji 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 weconsider 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 locallynilpotent 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. SoETS that since these are a generating set, we can act locally nilpotently on L (A) (i.e. what isgenerated). This is due to Leibniz:

(ad x)k[y, z

]= k∑i=0

(ki

)[(ad x)i (y), (ad z)k−i (z)

],

which implies the statement.

If V is again a representation of a Lie algebra g, then a locally nilpotent action x ∈ g producesan automorphism

exp(x) :=∞∑

n=0

xn

n!

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

Definition 15.7. We define the elements

ni := exp(ad(ei )

)exp

(ad(− fi )

)exp

(ad(ei )

) ∈ Aut(L (A)

).

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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 10 0

)exp(e) =

(1 10 1

)f =

(0 01 0

)exp(− f ) =

(1 0−1 1

)

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), denotedby W .

We have si (hi ) =−hi , s2i , 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)]= ni

[s−1

i h, x]= ni

(α(s−1

i h)x)=α(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 thefollowing order:

order(si s j ) =

2 Ai j A j i = 0 ,

3 Ai j A j i = 1 ,

4 Ai j A j i = 2 ,

6 Ai j A j i = 3 ,

∞ Ai j A j i ≥ 4 .

Proof. Let

K := λ ∈ h∗ :λ(hi ) =λ(h j ) = 0

,

so dimK = dimh∗−2. Writing V =Cαi ⊕Cα j , we get a decomposition h∗ =V ⊕K , and si s j actstrivially 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 thismatrix.

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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 weightlattice. The elements of R are called roots, and the decomposition of Q induces a decompositionR = R+tR− into positive and negative roots.

We are lead to ask the following questions: What is R, and what is dimLα? Writing mα forthe multiplicity of α, we know:

• dimLα = dimL−α.

• Π= α1, . . . ,αn : mαi = 1

⊂ 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 thatw(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.

Lemma 16.2. w(R+

im

)= R+im.

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

im, where k j ≥ 0, and at least twok 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∗

R:λ(hi ) > 0 for all i = 1, . . . ,n

⊂C := . . . ≥ 0

.

Proposition 16.3. R+im ⊆⋃

w∈W w(−C

).

Proof. For α ∈ R+im, we have a set

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 , andthus β ∈−C .

Theorem 16.4 (Kac). Let

K := α ∈Q+ :α 6= 0, supp(α) connected, α ∈−C

.

Then R+im =⋃

w∈W w(K ).

Remark 16.5. Here supp(α) := j : k j 6= 0

, where α = ∑

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

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

in any minimalrealisation, 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,

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34 Iain Gordon

2. if A is affine, then there exists an integral u > 0 such that Au = 0, and u has no commonfactors. 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 forall i .

Proof. Exercise.

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

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

(αi ,α j

)= (hidi

,h j

d j

)= d−1

j A j i .

So (αi ,α j ) = 2di

> 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

)= n∑i=1

ki(αi ,α

)= n∑i=1

ki

diα(hi ) ≤ 0 .

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.

But Lα is spanned by Lie monomials in the ek ’s. We show by induction on degree that anyLie monomial involving ei and e j (with i , j chosen as above) is zero.

...

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 soα(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.

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Step 4: J := i ∈ supp(α) : ki = mi

= 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 icarefully (there exist i ′ ∈ M and j ′ ∈ supp(α) \ M such that Ai ′ j ′ < 0), we have β′(hi ) > 0 for somei ∈ M .

Let AM be the principal matrix (Ai j )i , j∈M , and u = (m j )Tj∈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 ) = ∑

t∈supp(α)\J(kt −mt )αt (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 haveβ(hi ) ≥ 0 from above. Therefore, γ(hi ) ≤ 0 for all i ∈ M .

Set u = (ki −mi )Ti∈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 isimaginary.

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 simpleLie algebra. It acts on itself by the adjoint representation. There exists a unique highest rootθ =∑

i aiαi ∈ h∗.

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

We had constructed for all Kac-Moody Lie algebras a “standard form” ⟨−,−⟩. We know thatsimple Lie algebras have a unique invariant inner product up to scale, the Killing form, so thatthe 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.

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36 Iain Gordon

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

Ai j = Ai j for i , j ∈ 1, . . . ,n,

A00 = 2 ,

Ai 0 = −n∑

j=1a j Ai j for i ∈ 1, . . . ,n,

A0 j = −n∑

i=1ci 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 : s21 = s2

2 = e,(s1s2)6 = e⟩ ∼= D6

(the dihedral group). The Cartan matrix is(

2 −1−3 2

), and d = diag(1,3). The positive roots are

R+ = R+re =

α1,α2,α1 +α2,α1 +2α2,α1 +3α2,2α1 +3α2

.

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

A = 2 −1 0−1 2 −10 −3 2

.

Recall that we had the affine Kac-Moody Lie algebra

L g= g[t±1]⊕Cd ⊕Cc

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

]= [X ,Y ]⊗ f g +λY ⊗ tdg

dt−λ′X ⊗ t

df

dt− (

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 theKac-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 aninvolution 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

e0 := Fθ⊗ t , f0 := Eθ⊗ t−1 .

36


So we haveH := (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 fromh∗ by αi (d) =αi (c) = 0 for all i = 1, . . . ,n. Do the same for θ ∈ h∗, which gives θ ∈ H∗. Let δ bethe 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∑

j=1a jα j (hi ) =−

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 .

What about the map L(A) → L g – is it surjective, what is its kernel? It is surjective, i.e. the Liealgebra M generated by e0, . . . ,en , f0, . . . , fn and H is L g. Consider the set S := x ∈ g : x ⊗ t ∈ M

:

• 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

)α+iδ. Next,

J = (L0 ∩ J )⊕ ∑(i ,α)6=(0,0)

(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.

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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 preciselyto affine Lie algebras g[t±]⊕Cd ⊕Cc =: L g.

Dynkin diagrams have automorphisms: [Explain the procedure that gets from the orbit spaceto the affine Lie algebra.] [something] produces the so-called twisted affine Dynkin diagramsfrom the classical Dynkin diagrams.

A2n−1 → B Tn = 2 A2n−1

A2n (n ≥ 2) → C ′n = 2 A2n

Dn+1 → C ′n = 2Dn+1

A2 → ?

E6 → F T4 = 2E6

D4 → GT2 = 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 onAut(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 twistedaffine 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. Wewant 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 thatλi −λ ∈Q+, i.e. λi −λ=∑n

j=1 n jα j for n j ∈Z≥0.)

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Definition 18.2. There is a character function

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 thesame as e(λ) below? We define

R := f ∈ Fun(H∗) : ∃λ1, . . . ,λs s.t. supp( f ) ⊆ S(λ1)∪·· ·∪S(λs)

,

whereS(λ) :=λ−Q+ =

µ ∈ H∗ :µ≤λ

.

Observe that the set R is a ring and that the character function is in fact ch: O →R.

Remark 18.3. Suppose 0 → A → B → C → 0 is a short exact sequence of L -representations.Then if B ∈O , it follows that A,C ∈O , but the converse does not hold. However, if D,E ∈O , thenD ⊗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 =

0 if µ 6≤λ,

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+

(1−e−α

)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

)mα

).

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

(f (i )α

)aα,i , where∑α,i

aα,iα= ν .

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40 Iain Gordon

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

Definition 18.5. If M(λ) is the Verma module (as in (18.1)) for the Kac-Moody Lie algebra L

and J (λ) ⊂ M(λ) is the unique maximal submodule, let

L(λ) := M(λ)/

J (λ)

be the unique maximal quotient of M(λ).

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

Proposition 18.6. Let V ∈O and λ ∈ H∗. Then there exists a filtration

V =V0 ⊃V1 ⊃V2 ⊃ ·· · ⊃Vt = 0

such that either Vi/

Vi−1∼= L(µ) for some µ≥λ or (Vi

/Vi−1)µ = 0 for all µ≥λ.

If µ≥λ, then the number of factors L(µ) is independent of the choice of the filtration and ofthe choice of λ.

Definition 18.7. We define[V : L(µ)

]to be the multiplicity of L(µ) in a filtration V from Proposi-

tion 18.6.

Proof of Proposition 18.6. Existence: V has spectrum bounded above, a(V ,λ) =∑µ≥λdimVµ is

finite. We use induction on a(V ,λ). If a(V ,λ) = 0, then we have a filtration V =V0 ⊃V1 = 0.If a(V ,λ) > 0, then there exists a weight µ with µ≥λ in V . Choose a maximal weight and let

0 6= v ∈Vµ. By maximality, n+.v = 0, so N :=U (L ).v ⊂V , so N is a quotient M(µ) N , vµ 7→ v .Let N ′ be the image of J (µ) ⊂ M(µ) under the quotient map. Then V ⊇ N ⊃ N ′ ⊇ 0, and

a(N ′,λ) < a(V ,λ) , a(V /N ,λ) < a(V ,λ)

because (N/

N ′)µ 6= 0. So induction works.The proof of uniqueness is tricker, but not very illuminating.

Corollary 18.8. If V ∈O , then chV =∑λ∈H∗

[V : L(λ)

]chL(λ).

18.2 Kac’s Casimir

If L is a finite-dimensional simple Lie algebra, then there is a special element, the Casimiroperator

Ωfd :=∑i

xi yi ∈U (L ) ,

where xi and yi are dual bases with respect to the non-degenerate invariant form on g. It iseasy to see that in factΩfd lies in the centre of U (L ).

There exists an explicit version (via the Killing form). If h′1, . . . ,h′

n is a basis for h∗ andh′′

1 , . . . ,h′′n is the dual basis, and if eα ∈Lα, fα ∈L−α such that [eα, fα] = h′

α for all α ∈Φ+, then

Ωfd =n∑

i=1h′

i h′′i +

∑α∈Φ+

(eα fα+ fαeα

)=

n∑i=1

h′i h′′

i +∑α∈Φ+

(2 fαeα+h′

α

)=

n∑i=1

h′i h′′

i +2∑α∈Φ+

(fαeα

)+2h′ρ ,

40


where 2ρ =∑α∈Φ+α satisfies ρ(h′

i ) = 1, ⟨h′ρ ,h⟩ = ρ(h).

For infinite-dimensional Lie algebras, we define

Ω :=∑i

h′i h′′

i +2h′ρ+2

∑α∈Φ+

∑i

f (i )α e(i )

α ,

where the f (i )α form a basis for L−α and the e(i )

α form a basis for Lα such that⟨

f (i )α ,e( j )

α

⟩= δi j ;

the h′i and h′′

i are mutually dual bases of H , and ⟨h′ρ ,h⟩ = ρ(h) defined by ρ(hi ) = 1 for all

i = 1, . . . ,n. (Note that ρ is not unique, as the hi do not necessarily form a basis!)ThenΩ acts on V for V ∈O , denoted byΩV .

Lemma 18.9 (Technical manipulation of elements).

Ω ∈ EndO

(idO

), u : V →V ,ΩV .u = u.ΩV for u ∈U (L )

explain this more verbosely

Proposition 18.10. If M(λ) is the Verma module for the Kac-Moody Lie algebra L as in (18.1),then

ΩM(λ) =(⟨λ+ρ,λ+ρ⟩−⟨ρ,ρ⟩) idM(λ) ,

where the inner product on H∗ is induced from inner product on H.

Proof. With vλ ∈ M(λ), we have

Ω.vλ = (∑i

h′i h′′

i +2h′ρ+2

∑α∈Φ+

∑i

f (i )α e(i )

α

).vλ

= (∑iλ(h′

i )λ(h′′i )+2λ(h′

ρ))vλ

= (⟨λ,λ⟩−2⟨λ,ρ⟩)vλ .

Now use Lemma 18.9 an the fact that M(λ) =U (L ).vλ.

Corollary 18.11.ΩL(λ) =

(⟨λ+ρ,λ+ρ⟩−⟨ρ,ρ⟩) idL(λ) .

18.3 The Weyl-Kac formula

Definition 18.12. A representation V of L is integrable if V =⊕λ∈H∗ Vλ and ei , fi : V →V act

locally nilpotently for i = 1, . . . ,n.

Example. The adjoint representation L is integrable. Any finite-dimensional representation isintegrable.

Lemma 18.13. If V is integrable, then dimVλ = dimVw(λ) for all w ∈W .

Sketch of proof. It is easy to show that dimVλ = dimVsi (λ) for each simple reflection si . An sl2-calculation for ⟨ei ,hi , fi ⟩ acting on V shows that integrability implies that V is locally finite.Finally, sl2-representation theory shows that dimVλ = dimVsi (λ).

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

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

41

42 Iain Gordon

Proof. We know the sl2-representation theory. Let vλ be the highest weight of L(λ). Then f r

i vλr≥0 span a sl2-representation (where sl2 = (ei , fi ,hi )). This representation is finite-dimen-sional if and only if f r

i vλ = 0 for r À 0, and only if λ(hi ) ∈Z≥0

Conversely, if λ(hi ) ∈Z≥0 for all i , then take v ∈ L(λ) such that v = u.vλ for some u ∈U (L )and apply the Leibniz rule:

f Ni (u.vλ) =

N∑k=0

(N

k

)((ad fi )k (u)

).( f N−k

i .vλ) = 0 ,

where (ad fi )k = 0 for k À 0 and f N−ki = 0 for N −k À 0. So for N À 0, f N

i (u.vλ) = 0.

Theorem 18.16 (Weyl-Kac character formula). Let λ ∈ X +, where

X + := λ ∈ H∗ :λ(hi ) ∈Z, λ(hi ) ≥ 0 for all i = 1, . . . ,n

.

Then

chL(λ) =

∑w∈W

ε(w)e(w(λ+ρ)−ρ)

∏α∈Φ+

(1−e(−α)

)mα,

where ε : W → ±1 is defined as a group homomorphism with ε(si ) =−1.

Clarify whether e(−α) is the same as e−α before, and maybe recall what it is.

Proof. We havech M(λ) = ∑

µ∈H∗

[M(λ) : L(µ)

]chL(µ) ,

and µ≤λ if[M(λ) : L(µ)

] 6= 0. The Casimir operatorΩ acts on

M(λ) via∥∥λ+ρ∥∥2 −∥∥ρ∥∥2 ,

and on L(µ) via∥∥µ+ρ∥∥2 −∥∥ρ∥∥2 .

Therefore∥∥λ+ρ∥∥2 = ∥∥µ+ρ∥∥2 [check; condition missing?], and

ch M(λ) = ∑µ≤λ

‖λ+ρ‖2=‖µ+ρ‖2

[M(λ) : L(µ)

]chL(µ) . (18.2)

Now defineCλ :=

µ≤λ :

∥∥λ+ρ∥∥2 = ∥∥µ+ρ∥∥2

.

Then µ ∈Cλ⇒Cµ ≤Cλ. Also note that[M(λ) : L(λ)

]= 1. Running through the entire set Cλ andlooking at the formula for ch M(µ) in Equation (18.2) we find that for all µ ∈Cλ

ch M(µ) = ∑τ∈Cλ

aτµ chL(τ) ,

where aµµ = 1 and aτµ = 0 unless τ≤µ ( i.e. with a total ordering on Cλ that refines the ordering≤). The matrix (aτµ)τ,µ∈Cλ

is triangular with 1 on the diagonal. Hence

chL(λ) = ∑µ∈Cλ

bλµ ch M(µ)

with bλµ ∈Z. For example,

e(ρ)∏α∈Φ+

(1−e(−α)

)mα chL(λ) = ∑µ∈Cλ

bλµe(µ+ρ) .

42


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

si

(e(ρ)

∏α∈Φ+\αi

(1−e(−α)

)mα(1−e(−αi )

)chL(λ)

)= e(ρ−αi )

∏α∈Φ+\αi

(1−e(−α)

)mα(1−e(−αi )

)chL(λ)

= −e(ρ)∏α∈Φ+

(1−e(−α)

)mα

︸︷︷︸∆

chL(λ) ,

whence we deduce

⇒ w(e(ρ)∆chL(λ)

)= ε(w)e(ρ)∆chL(λ)

⇒ w(∑µ

bλµe(µ+ρ))= ε(w)

∑µ

bλµe(µ+ρ)

⇒ bλµ = ε(w)bλν if w(µ+ρ) = ν+ρ .

Suppose that bλµ 6= 0. Then bλµ 6= 0 for all ν such that w(µ+ρ) = ν+ρ and ν≤ λ. Pick ν suchthat λ−ν has minimal height. Then ν+ρ ∈ X + since

si w(µ+ρ) = si (ν+ρ) = ν+ρ− (ν+ρ)(hi )αi .

We have ⟨ν+ρ,ν+ρ⟩ = ⟨λ+ρ,λ+ρ⟩ since ν ∈Cλ, and so (λ+ρ)− (ν+ρ) =∑i kiαi with ki ≥ 0

for all i . So

0 = ∥∥λ+ρ∥∥2 −∥∥ν+ρ∥∥2 = ⟨λ+ν+2ρ,λ−ν⟩= ∑

iki ⟨λ+ν+2ρ,αi ⟩

= ∑i

ki⟨αi ,αi ⟩

2

(λ+ν+2ρ

)(hi )

But note that each summand is non-negative: ki ≥ 0, ⟨αi ,αi ⟩ ≥ 0 and (λ+ρ)(hi ), (µ+ρ)(hi ) ≥ 0.Therefore λ= ν. So

bλµ 6= 0 ⇒ µ+ρ = w(λ+ρ) for some w ∈W

⇒ bλµ = ε(w)bλλ = ε(w)

⇒ e(ρ)∆chL(λ) = ∑w∈W

ε(w)e(w(λ+ρ)

)⇒ chL(λ) =

∑w∈W ε(w)e

(w(λ+ρ)−ρ)∆

.

Corollary 18.17 (Kac denominator formula).

e(ρ)∏α∈Φ+

(1−e(−α)

)mα = ∑w∈W

ε(w)e(w(ρ)

)Proof. Take λ= 0, i.e. the trivial representation L(0) =C.

There exists a formula for dimL(λ)µ for λ ∈ X + (the Kostant partition function).

Exercise 18.18. Using this style of representation theory (and more work), one can show thepresentation for a symmetrisable Kac-Moody Lie algebra is what we used plus

(adei )1−Ai j e j = 0 , (ad fi )1−Ai j f j = 0

(see Carter §19.4, Kac Theorem 9.11).

43

44 Iain Gordon

19 W − K + A = M

Let L = g be an untwisted Lie algebra (i.e. g = L (A) for some untwisted, affine generalisedCartan matrix A), where g=L 0 is a simple, finite-dimensional Lie algebra with root systemΦ0

and Weyl group W 0. We know the roots of L :

Φ= α+nδreal :α ∈Φ0,n ∈Z∪

nδim : n 6= 0

,

where the imaginary roots have multiplicity rkL 0 =: l . The root δ= a0α0 +·· ·+alαl is dual toa−1

0 d (but a0 = 1 in the untwisted case). The positive roots are

Φ+ = α ∈Φ0

+∪

α+nδ : n ≥ 0,α ∈Φ0∪nδ : n > 0

.

19.1 Affine Weyl groups

The Weyl group of L is W = ⟨s0, s1, . . . , sl ⟩, and W 0 = ⟨s1, . . . , sl ⟩. Let θ := a1α1 +·· ·+alαl ∈Φ0,so ⟨θ,θ⟩ = 2. We can write

θ =l∑

i=1ai

⟨αi ,αi ⟩2

2αi

⟨αi ,αi ⟩,

and the associated reflection is sθ ∈W 0. Writing sθ(h) = h −θ(h)hθ, we get

hθ =l∑

i=1ai

⟨αi ,αi ⟩2

hi =l∑

i=1

2ci

2hi = c − c0hh = c −h0 .

The elements s0sθ, s1, . . . , sl generate W . We have

s0sθ(h) = s0(h −θ(h)hθ

)= h −α0(h)h0 −θ(h)

(hθ−α0(hθ)h0

)= h +δ(h)hθ−

(θ(h)+δ(h)

)c

= h +δ(h)hθ−(⟨hθ,h⟩+ 1

2 ⟨hθ,hθ⟩δ(h))c .

For x ∈ H 0, there is a map

tx : H → H , h 7→ h +δ(h)x − (⟨x,h⟩+ 12 ⟨x, x⟩δ(h)

)c .

Lemma 19.1. tx ty = tx+y and w tx w−1 = tw(x) for w ∈W 0.

Proof. Easy calculation.

Proposition 19.2. W = t (M )oW 0, where t (M) = tm : m ∈ M

and M ⊂ H 0 is the space generated

as an abelian group by w(hθ) for all w ∈W 0.

Remark 19.3. In the untwisted case, an elementary computation shows that M =⊕li=1Zhi .

Proof. We have H = H 0 ⊕Cc ⊕Cd , and δ(w(h)

)= (w−1δ)(h) = δ(h). The set

H1 := h ∈ H : δ(h) = 1

= h0 +λc +d

has an action of W . Since w(c) = c for all w ∈ W , this action descends to an action of W onH1

/Cc ∼= H 0 (and W 0 acts on H 0 as usual, and tm(h0) = h0 +m); h0 +λc +d 7→ h0 +d 7→ h0.

Since H∼=−→ H∗, we need the action of W on H∗. We have W = t (M∗)oW 0, where M∗ is the

lattice spanned by long roots and

tα(λ) =λ+λ(c)α− ((λ,α)+ 1

2(α,α)λ(c)

)δ .

44


19.2 Formulae

Recall that ρ(hi ) = 1 and ρ(d) = 0. The number ρ(c) = h∨ = c0+·· ·+cl is the dual Coxeter number.With ρ0 ∈ (H 0)∗ such that ρ0(hi ) = 1 for i = 1, . . . , l we have

w(ρ)−ρ = w0tα(ρ)−ρ= w0

(ρ+ρ(c)α− (⟨ρ,α⟩+ 1

2 ⟨α,α⟩ρ(c))δ)−ρ

= w0(h∨α+ρ0)−ρ0 −(⟨ρ0 +h∨α,ρ0 +h∨α⟩−⟨ρ0,ρ0⟩)δ

2h∨ .

We definec(λ) := ∥∥λ+ρ0

∥∥2 −∥∥ρ0∥∥2

,

so that the finite-dimensional Casimir operator on L0(λ) isΩL0(λ) = c(λ) id (cf. Corollary 18.11).Then we have a character formula

χ0(λ) := chL0(λ) = ∑w∈W 0

ε(w)e(w(λ+ρ0)−ρ0) / ∑

w∈W 0

ε(w)e(w(ρ0)−ρ0) .

So ∑w∈W

ε(w)e(w(ρ)−ρ) = ∑

α∈M∗

∑w∈W 0

ε(w0)e(w0(h∨α+ρ0)−ρ0)e

(−c(h∨α)

2h∨ δ)

= ∑w∈W 0

ε(w0)e(w0ρ0 −ρ0)∑

α∈M∗χ0(h∨α)e

(−c(h∨α)

2h∨ δ)

= ∏α∈Φ0+

(1−e(−α)

) ∑α∈M∗

χ0(h∨α)e(−c(h∨α)

2h∨ δ)

.

Theorem 19.4 (MacDonald). Set q := e(−δ). Then∏n>0

(1−qn)l∏α∈Φ0

(1−qne−α) = ∑α∈M∗

χ0(h∨α) qc(h∨α)/2h∨, (19.1)

where q = e(−δ).

Example. Consider A1. We have l = 1,Φ0 = ±α1, h∨ = 2, M∗ =Zα1 and ρ0 = 12α1. Set z := e−α1 ;

then

χ0(nα1) = e(nα1)−e(−(n +1)α1)

1−e(−α1)= z−n − zn+1

1− z.

The Casimir operator is multiplication by

c(2nα1) = ⟨(2n + 1

2

)α1,

(2n + 1

2

)α1

⟩−⟨12α1, 1

2α1⟩= 4n(2n +1) .

Hence the right-hand side of Equation (19.1) is

∑n∈Z

z−2n − z2n+1

1− zqn(2n+1) = 1

1− z

∑m∈Z

(−1)m zm qm(m−1)/2 .

Plugging in the left-hand side, we get∏n>0

(1−qn)(1−qn−1z)(1−qn z−1) = ∑m∈Z

(−1)m zm qm(m−1)/2 ,

which is the Jacobi triple product identity.

45

46 Iain Gordon

Exercise 19.5. Do this for A′1:∏

n>0(1−qn)(1−qn z−1)(1−qn−1z)(1−q2n−1z−2)(1−q2n−1z2) = ∑

n∈Z

(z3n − z−3n+1)qn(3n−1)/2

This is the quintuple product identity, known already in antiquity.

If we set eα = 1 for allα ∈Φ0, this produces other identities. (For instance, dimL0(λ) = d 0(λ).)

Theorem 19.6. Let φ(q) := (1−q)(1−q2)(1−q3) · · · . What about convergence? Then

φ(q)dimL 0 = ∑α∈M∗

d 0(h∨α) qc(h∨α)/2h∨.

Example. For A1, φ(q)3 =∑n∈Z(4n +1)qn(2n+1).

Exercise 19.7. Repeat the entire section for the twisted version.

20 KP hierarchy and Lie theory

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

ut − 1

4uxxx − 3

2u ux = 0 (20.1)

It is a non-linear partial differential equation in R1+1 that describes the behaviour of a (one-dimensional) wave in shallow water. There is a two-dimensional version, called the KP equation:(

ut − 1

4uxxx − 3

2u ux

)x= 3

4uy y (20.2)

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

• exact solutions (in this case solitons),

• remarkable symmetries,

• Lax-pair formalism.

The KP equation fints into an infinite family of PDEs, all of which are governed simulta-neously. We will see (a) solutions of Equation (20.2), and (b) a family of solutions, which isan inifinite-dimensional homogeneous manifold. (“If you have one solution, you have everysolution.”)

Recall from Section 9.1 that from a space V =⊕n∈ZCvn we can form the space F (0), on which

we have an action of gl∞ ⊂ End(V ), which is spanned by elements Ei j of “infinite matrices withonly finitely many non-zero entries”. The Lie group attached to this algebra is

GL∞ :=

A = (ai j )i , j∈Z : A is invertible and almost all ai j −δi j = 0

We define

Ω :=GL∞.v0,−1,−2,... =

u0 ∧u−1 ∧u−2 ∧·· · : u−m = v−m for m À 0

,

46


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

Gr :=

U <V : there is k such that U ≥V ≤k =⊕i≤k

Cvi and dim(U

/V ≤k)= k

,

then we have a bijection (Exercise: Check!)

PΩ∼=−−→ Gr , u0 ∧u−1 ∧u−2 ∧·· · 7→⊕

i≤0Cui .

There are two operators, F (0) → F (1) and F (0) → F (−1), respectively via V andV ∗ (the restricteddual, cf. Theorem 7.3), defined as follows. For v ∈V and f ∈V ∗, we have operations

v .vi0 ∧ vi1 ∧·· · = v ∧ vi0 ∧ vi1 , and

f .vi0 ∧ vi1 ∧·· · = ∑j≥0

(−1) j vi0 ∧ vi1 ∧·· ·∧ vi j−1 f (vi j )∧ vi j+1 ∧·· · .

The element Ei j acts on F (0) via vi v∗j .

Lemma 20.1. If τ ∈Ω, then ∑j∈Z

v j (τ)⊗ v∗j (τ) = 0 , (20.3)

and if τ 6= 0 and τ satisfies Equation (20.3), then τ ∈Ω.

Proof. For τ= v0,−1,−2,... Equation (20.3) is obviously satisfied. We prove that the left-hand sideof Equation (20.3) is GL∞-invariant:

A

(∑j∈Z

v j (τ)⊗ v∗j (τ)

)A−1 = ∑

j∈Zv j (τ)⊗ v∗

j (τ) .

Then it follows that Equation (20.3) holds in general, since τ= Av0,−1,−2,..., and so

A

(∑j∈Z

v j (τ)⊗ v∗j (τ)

)A−1 A v0,−1,−2,... = 0 .

Conversely, assume that τ satisfies Equation (20.3) and that τ 6= 0. Write τ=∑Nk=1 ckτk , where

τk are semi-infinite monomials and all ck 6= 0. Without loss of generality, τ1 has maximal degreeamong the τk . If τ2 is of the form τ2 = Ei j .τ1 with i < j , then E 2

i j .τ1 = 0, and so exp(−c2Ei j

).τ

removes the τ2-term.(Note that the degree is deg(vi) =∑

m≥0(im +m) and that exp(−c2Ei j

) ∈GL∞.)Now by GL∞-invariance, the new expression also satisfies Equation (20.3), and by iterating

we obtain τ= τ1 +φ, where φ has no terms of the form Ei j .τ1. Then Equation (20.3) implies∑j∈Z

(v j .τ1 ⊗ v∗

j .φ+ v j .φ⊗ v∗j .τ1 + v j .φ⊗ v∗

j .φ)= 0 .

So for each j we must have, up tp sign, that

v j .τ1 = vk .(semi-infinite wedge in φ) ,

which forces Ek j .τ1, contradiction. Clarify this sentence. We conclude that φ= 0.

47

48 Iain Gordon

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

X : F (0) → F (1) , X (u) := ∑j∈Z

u j v j , and X ∗ : F (0) → F (1) , X ∗(u) := ∑j∈Z

u− j v∗j .

These become operators on B(0,1), too, by the following fact:

X (u) 7→ u exp(∑

j≥1u j x j

)exp

(− ∑

j≥1

u− j

j

∂

∂x j

), X ∗(u) 7→ exp

(− ∑

j≥1u j x j

)exp

(∑j≥1

u− j

j

∂

∂x j

)This answers the question how gl∞ acts. For the image of vλ, note that vλ 7→ Sλ(x), the Schurpolynomial, given by Sλ(x) = det

(Sλi+ j−1(x)

), where the elementary Schur polynomials Sk are

such that ∑k∈Z

Sk (x) zk = exp( ∞∑

k=1xk zk

).

We translateΩ to B(0,1) =C[x1, x2, . . . ] by identifying

F (0) ⊗F (0) with C[x ′1, x ′

2, . . . ]⊗C[x ′′1 , x ′′

2 , . . . ] .

Then X (u).τ⊗X ∗(u).τ has vanishing u0-term, i.e.

u exp(∑

j≥1u j (x ′

j −x ′′j

))exp

(∑j≥1

u− j

j

(∂x ′′

j−∂x ′

j

)).τ(x ′)τ(x ′′)

has vanishing u0-term. Rewrite x ′ = X −Y , x ′′ = X +Y . So (−).τ ∈ C[x1, x2, . . . ] belongs to Ω ifand only if the u0-term vanishes in the expression

u exp(− ∑

j≥12u j Y j

)exp

(∑j≥1

u− j

j

∂

∂Y j

).τ(X −Y )τ(X +Y ) . (20.4)

The Hirota form. Let P ∈C[x1, x2, . . . ] with a finite number of xi ’s, and let f and g be functionsin those variables. Define an operation

(P f .g )(x) := P( ∂

∂u1,∂

∂u2, . . .

)(f (x1 −u1, x2 −u2, . . . ) g (x1 +u1, x2 +u2, . . . )

)∣∣∣u=0

.

So Equation (20.4) becomes

u( ∑

j≥Zu j S j (−2Y )

)( ∑j≥Z

u− j S j (∂Y )).τ(X −Y )τ(X +Y ) ,

where Y = (Y1,Y2, . . . ) and ∂Y = (∂∂Y1

, 12

∂∂Y2

, 13

∂∂Y3

, . . .). It follows that Equation (20.3) is equivalent

to ∑j≥0

S j (−2Y )S j+1(∂Y )τ(X −Y )τ(X +Y ) = 0 . (20.5)

We compute:

S j+1(∂Y )τ(X −Y )τ(X +Y ) = S j+1(∂u)τ(X −Y −u)τ(X +Y +u)∣∣∣u=0

= S j+1(∂u)exp(∑

s≥1Ys

∂

∂us

)τ(X −u)τ(X +u)

∣∣∣u=0

= S j+1(X )exp(∑

s≥1Ys Xs

)τ(X )τ(X )

∣∣∣u=0

,

where X = (X1, 1

2 X2, 13 X3, . . .

).

48


Exercise 20.2. Show that P f . f = 0 if P (x) =−P (−x).

Theorem 20.3. τ ∈Ω if and only if τ is a solution of( ∞∑j=0

S j (−2Y )S j+1(X )exp(∑

s≥1 Ys Xs))τ(X )τ(X ) = 0 , (20.6)

where Y1,Y2, . . . are parameters, i.e. we have an infinite family of differential equations.

Now expand Equation (20.6) in the Yi . The variable Yr appears in exp(∑

s≥1 Ys Xs)

once withcoefficient Xr . In the expression S j (−2Y ) the variable Yr appears when j = r , with coefficient−2. So Equation (20.6) has the Yr -term(

S1(X )Xr −2Sr+1(X ))ττ= 0 ⇒ (

X1Xr −2Sr+1(X ))ττ= 0 .

If r = 1 or r = 2, this is odd and there are no conditions. If r = 3, the term in parentheses isx4

1 +3x22 −4x1x3. Setting x := xi , y := x2 and t := x3, we obtain

( ∂4

∂u41

+3∂2

∂u22

−4∂2

∂u1∂u3

)τ(X +u)τ(X −u)

∣∣∣u=0

= 0 .

A long exercise shows that u(x, y, t ) = 2 ∂2

∂x2

(logτ

)is a solution of the KP equation.

21 The Kazhdan-Lusztig Conjecture

Let A be a symmetrisable generalised Cartan matrix.

21.1 The Shapovalov form and Kac-Kazhdan formulas

The character forlumals for integrable irreducible reprepresentations L(µ) involves the Weylgroup. Recall that we had the Verma modules M(λ), and for M(λ+µ) the character formula hasthe form L

(w(λ+ρ)

). The Kac-Kazhdan formula is interesting because it shows that the Weyl

group W does play a rôle in general.

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

U (g) =U (n−)⊗U (H)⊗U (n+) =U (H)⊕ (n−.U (g)+U (g.n+

).

Denote by π : U (g→U (H) the projection onto the first direct summand.

Recall that U (H) is a polynomial ring. There exists an anti-linear involution σ : g → g,σ(ei ) = fi , and σ : h→ h. Let

F : U (g)⊗U (g) →U (H) , F (x, y) =π(σ(x)y

).

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

Definition 21.2. The Shapovalov form F (λ) : M(λ)⊗M(λ) →C is defined by

F (λ)(xvλ, y vλ) = F (x, y)(λ) .

Exercise 21.3. Check that the Shapovalov form is well defined.

Lemma 21.4. The Shapovalov form has the following properties:

49

50 Iain Gordon

• F (λ)(xv, w

)= F (λ)(v,σ(v)w

)for x ∈U (g), v, w ∈ M(λ).

• F (λ)(M(λ)µ1 , M(λ)µ2

)= 0 if µ1 6=µ2.

• radF (λ) is the maximal submodule of M(λ).

We would like to find out about F (λ), or at least about detF (λ)η, the restriction of F (λ) toM(λ)λ−η, to help us understand simplicity or composition factors of the M(λ)’s.

Definition 21.5. Let Fη be the restriction of F to U (n−)−η.

Theorem 21.6 (Theorem A).

detFη =∏α∈∆+

∞∏n=1

(hα+

⟨ρ− nα

2 ,α⟩)mult(α)P (η−nα)

,

where P is the Kostant partition function, and detFη ∈U (H).

Theorem 21.7 (Theorem B). L(µ) is an irreducible subquotient of M(λ) if and only if there existsβ1, . . . ,βn ∈∆+ and n1, . . . ,nk ∈N such that⟨

λ+ρ−n1β1 −n2β2 −·· ·−ni−1βi−1,βi⟩= ni

2

⟨βi ,βi

⟩for all i = 1, . . . ,k (21.1)

and λ−µ=∑ki=1 niβi .

Observation. Pick λ ∈ H∗ such that ⟨λ,δ⟩ =−h∨, the dual Coxeter number, and ⟨λ+ρ,α⟩ 6∈Nfor each positive real root α. (Such weights exist, e.g. λ=−ρ.) This weight is called a KK weight(for “Kac-Kazhdan”). Then

⟨λ+ρ,δ,=⟩ n

2 ⟨δ,δ⟩ = 0 for all n ∈N, and so L(λ−nδ) is a factor ofM(λ) for all n ∈N. But for α ∈∆re+ ,⟨

λ+ρ−nδ,α⟩= ⟨

λ+ρ,α⟩ 6∈N

andn

2

⟨α,α

⟩= ndi , where α= w(αi ) ,

and so there are no subquotients of the form L(λ−nδ−mα). In particular, the Weyl group playsno rôle.

Conjecture 21.8 (Kac-Kazhdan, Theorem: Hayashi; Goodman, Wallach). Let λ be a KK weighton an affine Lie algebra. Then

chL(λ) = eλ∏α∈∆re+

(1−e−α

)−1 .

What are te ingredients of the proof?

• detFη is a polynomial in U (H ), so it can be factored. First calculate the “leading term” andget ∏

α∈∆re+

∞∏n=1

hmult(α)P (η−nα)α . (21.2)

• Use Kac’s Casimir to observe that if v ∈ M(λ)λ−β is a highest-weight vector, then∥∥λ+ρ∥∥2 −∥∥ρ∥∥2 = ∥∥λ−β+ρ∥∥2 −∥∥ρ∥∥2 ,

i.e.⟨λ+ρ,β

⟩= 12 ⟨β,β⟩.

So if this cannot be satisfied, then M(λ) is irreducible and so F (λ) is non-degenerate; hencedetFη is a product of linear factors

(hβ+⟨ρ− 1

2β,β⟩), and the formula for leading terms showsthat β= nα for some α ∈∆+, which we call a quasi-root. So we get products of

(hα+⟨ρ− 1

2α,α⟩).

50


The trick. We had already shown that the leading term of detFη (given by Equation (21.6)) is theexpression (21.2), and Fη is a product of finitely many linear terms of the form hα+

⟨ρ− nα

2 ,α⟩

.(Note that the Kostant partition function P (η−nα) vanishes for large n or large α.) Now extendeverything to C[t ]. For example, if we have the universal enveloping algebra U (g) extended overC[t ], then

M(λ) = U (g)⊗U (n+⊕H)C[t ]λ ,

where λ ∈ H∗⊗C[t ] is given by

λ(n+) = 0 and λ(h) =λ(h)+ t z(h) ,

where z ∈ H∗ is some generic element in the sense that that z(hα) 6= 0 for all α ∈Q+ \ 0. Thisproduces a form F t

η(λ) taking values in C[t ], and when we specialise t → 0, we recover Fη(λ),

and M(λ) → M(λ). FilterM(λ) = M 0 ⊇ M 1 ⊇ M 2 ⊇ ·· · ,

whereM i =

v ∈ M(λ) : F t (λ)(v, w) ∈ (t i ) for all v, w ∈ M(λ)

,

and specialise t → 0:M(λ) = M 0 ⊃ M 1 ⊇ M 2 ⊇ ·· ·

This is called Jantzen’s filtration; here M 1 is a maximal submodule.Pick λ such that λ(hβ)+⟨ρ− 1

2β,β⟩ vanishes for some quasi-root but does not vanish for allother quasi-roots. Then we see that

1. M(λ−β) is irreducible, and

2. M(λ) has a submodule U which is a direct sum of modules of the form M(λ−β).

Now by construction of λ we see that the highest power of t dividing detF tη(λ) equals the oerder

of the linear factor hα+⟨ρ− nα

2 ,α⟩

in detFη(λ). But M 1 =⊕M(λ−β), so if M(λ−β) appears in

M i with multiplicity si (β), we see that the order of t dividing detF tη(λ) is

∑i si (β)P (η−β).

Example. Suppose M(λ−β) contains three highest-weight modules of highest weight β, withmultiplicities respectively 1, 2 and 5. Then M 1 = M(λ−β)⊕3, M 2 = M(λ−β)⊕2, and M 3 = M 4 =M 5 = M(λ−β).

The functions φβ : Q+ → R, η 7→P (η−β) are all linearly independent. But comparing theleading term of Fη(λ) (cf. (21.2)) with the multiplicities in Jantzen’s filtration for the quasi-rootβ= nα, we get ∏

α∈∆re+

∞∏n=1

hmult(α)P (η−nα)α =∏

∆+

∞∏n=1

hP (η−nα)∑

i si (nα)α .

This proves Theorem A.The proof of Theorem B is an inductive variation on the proof of Theorem A, producing the

Jantzen Sum Formula ∑i≥1

ch M(λ)i = ∑(α,n)∈Dλ

ch M(λ−nα) , (21.3)

where the right-hand-side sum is over the set

Dλ :=

(α,n) ∈ ∆+×N :⟨λ+ρ− 1

2α,α⟩= 0

.

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

51

52 Iain Gordon

21.2 Bringing in the Weyl group

Definition 21.9. We define an equivalence relation ∼ on H∗ by λ∼µ if there exists a sequenceλ=λ0,λ1, . . . ,λn =µ such that for each 0 ≤ i < k, either λi ,λi +1 or λi+1,λi satisfy (21.1) forsome β1, . . . ,βn ∈∆+ and n1, . . . ,nk ∈N.

Definition 21.10. Given equivalence class [λ] ≡C (λ) =:Λ in H∗/∼, we say that a module M ∈O

has typeΛ if all its “composition factors” belong toΛ. (Cf. Section 18.1 for the category O .) Wewrite OΛ for the subcategory of modules of typeΛ.

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

Theorem 21.11 (Deodhar, Gabher, Kac). Let M ∈O . Then there exists a unique set MΛΛ suchthat M =⊕

ΛMΛ. Furthermore, Ext1O

(F,E) = 0 if E ∈OΛ, F ∈OΛ′ andΛ 6=Λ′.

Theorem 21.11 says that to understand category O , we only need to understand OΛ. Themoral will be that some choices ofΛ are better than others.

For w ∈W , we define an action w∗ on H∗ by w∗λ= w(λ+ρ)−ρ. We set

C := λ ∈ H∗ :

⟨λ+ρ,αi

⟩≥ 0 and⟨λ+ρ,α

⟩ 6= 0 if α ∈∆isotropic+

,

and define Tits’ cone of “good” elements

K g := ⋃w∈W

w∗C .

Define

Sα :=λ ∈ H∗ : ⟨λ−ρ,α⟩ = 1

2

⟨α,α

⟩,

S := ⋃α∈∆im+

Sα , and

K w.g. := H∗ \ S .

Elements of S are “critical hyperplanes”, and W acts on the set K w.g. of “weakly good” elements.

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

1. finite generalised Cartan matrices,

2. affine generalised Cartan matrices, and

3.

(2 −a−a 2

)for a ≥ 3.

(It is clear that K g ⊆ K w.g..)

Definition 21.13. Define an equivalence relation ≈ as for ∼, using K g instead of H∗, and anequivalence relation ≈ as for ∼, using K w.g. instead of H∗.

Proposition 21.14. If λ,µ ∈ K w.g., then λ ≈ µ if and only if there exists w ∈ W (λ), the integralWeyl group generated by

sβ :β ∈∆re+ such that

2⟨λ+ρ,β⟩⟨β,β⟩ ∈Z

,

such that w∗λ=µ.

52


Proof. We prove a better result. Let λ ∈ K w.g.. Then L(µ) is a subquotient of M(λ) if and only ifthere exist β1, . . . ,βt ∈∆re+ with λ> (sβ1 )∗λ> ·· · > (sβt · · · sβ1 )∗λ=µ.

We pick λ−µ ∈Q+ and do induction on ht(λ−µ): If ht(λ−µ) = 0, then λ= µ and there isnothing to do. Now assume ht(λ−µ) > 0. Then L(λ) = M(λ)

/M 1 from Jantzen’s filtration. Since

λ 6=µ, L(µ) is a subquitient of M 1, and hence by Jantzen’s Sum Formula (21.3) also a subquotientof M(λ−nα) for some (α,n) ∈ Dλ.

Now if we assume that α ∈ ∆im+ , we have⟨λ+ρ,α

⟩ = n2

⟨α,α

⟩. Set β := nα ∈ ∆im+ and⟨

λ+ρ,β⟩ = 1

2

⟨β,β

⟩; and so λ ∈ Sβ, contradiction. So α ∈ ∆re+ , and then

⟨λ+ρ,α

⟩ = n2

⟨α,α

⟩with

⟨α,α

⟩ 6= 0, and so n ≡ n(α) is uniquely determined.Finally we note that λ−n(α)α= (sα)∗λ ∈ K w.g., and by induction we conclude the “if” direc-

tion. The converse is just Jantzen’s Sum Formula (21.3).

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

Definition 21.16. Let Og be the full subcategory of O whose objects have composition factorsbelonging to K g. We say that M ∈Og has typeΛ if each composition factor of M belongs toΛ.

Example. If λ ∈ K g, then M(λ) ∈Og, and M(λ) has typeΛ.

There is a decompositionOg =⊕

Λ

OgΛ .

Moreover, the equivalence classes are labelled by⋃λ∈C W

/W (λ) thanks to Proposition 21.14.

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

21.3 The Kazhdan-Lusztig Conjecture

53

54 Iain Gordon

A References

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

• Kac: Infinite-dimensional Lie algebras.

• Carter: Lie algebras of finite and affine type.

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

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

• E. Frenkel: Langlands correspondence for loop groups.

• Pressley, Segal: Loop Groups.

• Humphreys: Lie algebras.

• Segal, Wilson: ? (probably IHES)

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

• Kumar: J. Algebra 108 (1987)

B Notational reference

Witt the Witt algebra over C

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

Lie(G) the Lie algebra of the Lie group G , g= TeG

g either a general Lie algebra or a finite-dimensional, simple one

g a central extension of a Lie algebra g; 0 → z→ g→ g→ 0

L g the loop algebra g⊗C[t , t−1]

L g =L goCd , where d = t ddt ∈ Der

(C[t , t−1]

)Note: H 2

(L g;C

)∼=C∼= H 2(L g;C

)L g =L g⊕Cd ⊕Cc, the central extension of L g, affine Kac-Moody Lie algebra

L (A) the Kac-Moody Lie algebra corresponding to the generalised Cartan matrix A

U (g) the universal enveloping algebra of the Lie algebra g;U (g) = T (g)

/(x ⊗ y − y ⊗x − [x, y])

⟨−,−⟩ a positive-definite sesquilinear form on a representation V ,non-degenerate if V is unitary

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

A the affine, twisted generalised Cartan matrix derived from A

54


C Generation of Lie algebras

Lie algebra Dynkin diagram GeneralisedCartan Matrix

Name Notation

simple g ∆, classical A, finite type

loop g[t±1

]n/a n/a

affine g[t±1

]⊕Cc n/a n/a

? g[t±1

]oC

( ddt ) n/a n/a

affine Kac-Moody g[t±1

]⊕Cd ⊕Cc ∆, untwisted affine A, untwisted affine

twisted affine Kac-Moody(g[t±1

]⊕Cd ⊕Cc)T

∆, twisted affine A, twisted affine

55

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