Kac-Moody Lie Algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 13 August 2012

This is a typed version of I.G. Macdonald's lecture notes on Kac-Moody Lie algebras from 1983.


Before we get down to serious business, let me begin by telling you a little in general terms about this subject, and what I propose to do (and what I do not propose to do) is this course of lectures. Basically it is an outgrowth of the theory of finite-dimensional complex simple (or semisimple) Lie algebras developed by Killing and E. Cartan nearly 100 years ago. If 𝔤 is a complex finite-dimensional semisimple Lie algebra, one can associate canonically with 𝔤 (I will go into details later) a certain matrix C=(Cij) of integers, called the Cartan matrix of 𝔤, which determines 𝔤 up to isomorphism. This matrix satisfies the following conditions:

(C) cii=2; cij0 ifij; cij=0 cji=0
(P) All principal minors of C are positive.

Conversely, any matrix of integers satisfying these two conditions is the Cartan matrix of some semisimple Lie algebra 𝔤, and one can write down generators and relations for 𝔤 which involve only the integers cij. (SOMETHING GOES HERE) (I will write them down explicitly a little later.)

In the late 1960's, V. Kac and R. Moody more or less simultaneously and independently, had the idea of starting with a "generalized Cartan matrix" (satisfying (C) but not (P)), writing down the same set of generators and relations as in Serr's theorem, and looking at the resulting Lie algebra (which is now infinite-dimensional). These are the so-called Kac-Moody Lie algebras, or Lie algebras defined by generalized Cartain matrices.

Of course, nearly always when one takes an attractive and elegant piece of mathematics, such as the classical theory of finite-dimensional semisimple Lie algebras over , and starts tinkering with it by weakening the axioms, the result is usually of no interest at all. The surprising thing in this case is that one does continue to get a coherent theory (which of course includes the classical theory as a special case) in which all the main features of the classical theory have their counterparts: root system, Weyl group, representations and characters – culminating in a generalization (due to V. Kac) of Weyl's character formula.

Moreover it has become clear during the last 10 years or so that these Kac-Moody Lie algebras impinge on many other areas of mathematics:

Number Theory(Modular forms)
Combinatorics(Partitions, Rogers-Ramanujan identities)
Topology(Loop spaces and Loop groups)
Linear Algebra(Representations of Quivers)
Completely integrable systems
Mechanics and Particle Physics

My aim in this course of lectures is to cover the basic structure and representation theory of Kac-Moody Lie algebras, but not any of the applications listed above. Also I don't propose to assume any particular knowledge of Lie algebras, so I will begin by briefly reviewing some of the basic notions. We shall work always over a field k of characteristic 0 (and fairly soon k will be , the field of complex numbers). A Lie algebra 𝔤 is then a vector space over k endowed with a bilinear multiplication

(x,y) [x,y] (Lie bracket)


(1) [x,x]=0 for all x𝔤
(2) [x,[y,z]]+ [y,[z,x]]+ [z,[x,y]]=0

for all x,y,z𝔤 (Jacobi identity).

By applying (1) to x+y and using the bilinearity of the bracket we have

(1) [x,y] = -[y,x]

and conversely (1) (1) (take x=y).


  1. V any vector space over k, define [x,y]=0 for all x,yV. This is an abelian Lie algebra.
  2. A any associative k–algebra, define [x,y]=xy-yx. Check that (2) holds. We have a Lie algebra L(A).
  3. A any k–algebra – i.e. A is a k–vector space endowed with a bilinear multiplication (x,y)xy: A×AA .
    A derivation d:AA is a k–linear mapping which satisfies
    d(xy)= (dx)y+ x(dy) (x,yA)
    If d1,d2 are derivations, so is [d1,d2]= d1d2-d2d1 (check this). Again the Jacobi identity is satisfied (the verification is the same as in Ex. 2), so we have a Lie algebra Der(A).
    It follows from () that
    dn(xy)= p+q=n n! p!q! (dpx) (dqy) (Leibniz)
    by induction on n; hence if d is nilpotent ( dN=0 for some N>0 )
    ed = n0 dn n!
    is well-defined (because the sum is finite) and is an automorphism of A:
    ed(xy)= n0 dn(xy) n! = p,q0 dpx p! · dqy q! = ed(x) ed(y)
    so that ed:AA is a k–algebra homomorphism, hence an automorphism because ed·e-d=1.
  4. Let V be a finite-dimensional k–vector space, A=End(V). Then L(A) (Ex. 2) is a Lie algebra denoted by 𝔤𝔩(V). If V=kn, so that A=Mn(k) is the algebra of n×n matrices over k, we write 𝔤𝔩n(k) in place of 𝔤𝔩(kn). 𝔰𝔩n(k)= { X𝔤𝔩n(k) : traceX=0 } is a subalgebra of 𝔤𝔩n(k), because trace[X,Y]= traceXY- traceYX=0 .
  5. G a (real or complex) Lie group, 𝔤=tangent space Te(G) to G at the identity element e (k= or here). 𝔤 inherits from the group G a Lie algebra structure: roughly speaking, addition X+Y in 𝔤 corresponds to multiplication in G near the identity, and the bracket [X,Y] ERROR HERE? to formation of the commutator xyx-1y-1 (for x,y near e). Here of course 𝔤 is finite-dimensional: dim𝔤=dimG. This is the origin of the subject: Lie algebra 𝔤=linear approximation to G at e.

Basic concepts

Many notions for groups have counterparts for Lie algebras. This is hardly surprising, given the origin of the subject.

Let 𝔤 be a Lie algebra. If 𝔞,𝔟 are vector subspaces (or just subsets) of 𝔤, let [𝔞,𝔟] denote the subspace of 𝔤 spanned by all [x,y] with x𝔞,y𝔟. Observe that [𝔞,𝔟] = [𝔟,𝔞] (because [x,y] =- [y,x] ).

Subalgebra: a vector subspace 𝔞 of 𝔤 is a subalgebra if [𝔞,𝔞]𝔞 (so that 𝔞 is a Lie algebra in it's own right).

Ideal: a vector subspace 𝔫 of 𝔤 is an ideal of 𝔤 if [𝔤,𝔫]𝔫 (normal subgroup).

Quotient algebra: Let 𝔫 be an ideal in 𝔤, and form the vector space quotient 𝔤/𝔫, whose elements are the cosets x=x+𝔫. Define [x,y] = [x,y] , this does not depend on the choice of representations, and makes 𝔤/𝔫 into a Lie Algebra. (G/N)

Homomorphism: a homomorphism from 𝔤 to 𝔥 is a k–linear map f:𝔤𝔥 such that f ([x,y]) = [ f(x), f(y) ] for all x,y𝔤.
It's kernel 𝔫=f-1(0) is an ideal in 𝔤, it's image 𝔞=f(𝔤) is a subalgebra of 𝔥, and f induces an isomorphism 𝔤/𝔫𝔞

If x,y𝔤 are such that [x,y]=0, we say that x,y commute. In particular, if any two elements of 𝔤 commute, ie if [𝔤,𝔤]=0, we say that 𝔤 is abelian (Ex. 1 above).

Centre of 𝔤=z= { x𝔤 : [x,𝔤]=0 } . It is an ideal in 𝔤.

Derived algebra 𝔤=D𝔤= [𝔤,𝔤] consists of all linear combinations of brackets [x,y]. D𝔤 is an ideal in 𝔤 (by virtue of the Jacobi identity):

[[x,y],z]=- [[y,z],x]- [[z,x],y] D𝔤,

hence [D𝔤,𝔤]D𝔤. Moreover 𝔤/D𝔤 is abelian (and D𝔤 is the smallest ideal with abelian quotient).
Derived series, upper and lower central series; nilpotent, solvable Lie algebras.

Adjoint representation

For each x𝔤 we define ad(x):𝔤𝔤 by ad(x)y=[x,y]. Then


is a homomorphism of Lie algebras, because for all x,y,z𝔤 we have

ad[x,y]·z = [[x,y],z] = -[z,[x,y]] = [x,[y,z]] - [y,[x,z]] (Jacobi) = (adx) (ady) z - (ady) (adx) z = [ adx, ady ] ·z

Moreover each adx is a derivation of 𝔤:

(adx) [y,z] = [(adx)y,z] + [y,(adx)z]

This is another equivalent form of the Jacobi identity.

The kernel of ad is the centre z of 𝔤.

Inner automorphisms

If x𝔤 is such that adx is nilpotent ((x)N=0 for someN>0) then we can form eadx (Ex. 3 above) which is an automorphism of 𝔤. The subgroup Int(𝔤) of Aut(𝔤) generated by these eadx is the group of inner automorphisms of 𝔤. It is a normal subgroup of Aut(𝔤), because if φAut(𝔤) we have φ(adx)φ-1 =adφ(x) and therefore also

φ(eadx) φ-1=eadφx .


Let 𝔤 be a Lie algebra. A representation ρ of 𝔤 on a k–vector space V is by definition a Lie algebra homomorphism ρ:𝔤𝔤𝔩(V). In other words, for each x𝔤 we have a linear transformation ρ(x):VV depending on linearity on x:

ρ(αx+βy) = αρ(x)+ βρ(y) ( x,y𝔤; α,βk )

and satisfying

ρ([x,y])= ρ(x)ρ(y)- ρ(y)ρ(x) .

An equivalent notion is that of a 𝔤module, which is a vector space V on which 𝔤 acts linearly, i.e. we are given a bilinear mapping

(x,v)x·v: 𝔤×VV


[x,y]·v= x·y·v-y·x·v ( x,y𝔤; vV )

To connect the two notions, define x·v=ρ(x)v.

Usual notions of irreducibility, direct sums etc.

Universal enveloping algebra of a Lie algebra

If G is a group, a G–module (or representation of G) is the same thing as a kG–module, where kG is the group algebra of G over k. The analogue of this for Lie algebras is the universal enveloping algebra U(𝔤) of a Lie algebra 𝔤, which may be defined as follows: for the tensor algebra of the vector space 𝔤

T(𝔤)= n0 Tn(𝔤)

where T0(𝔤)=k, T1(𝔤)=𝔤, Tn(𝔤)= 𝔤𝔤 (nfactors) forn2 .
Let J𝔤 be the two-sided ideal of IS THIS A T? U(𝔤) generated by all

xy-yx-[x,y] (x,y𝔤)

and define

U(𝔤)= T(𝔤)/J𝔤 .

U(𝔤) is functional in 𝔤: if φ:𝔤𝔥 is a homomorphism of Lie algebras, it induces T(φ):T(𝔤) T(𝔥) , and

T(φ) ( xy -yx- [x,y] )
φxφy-φyφx -[φx,φy]J𝔥

so that T(φ) maps J𝔤 into J𝔥 and hence induces

U(φ):U(𝔤) (𝔥)

U is the left adjoint of the functor L (Ex. 2) from associative algebras to Lie algebras (over k): for each Lie algebra 𝔤 and each associative algebra A, there is a canonical bijection

Hom assoc. alg (U(𝔤),A) Hom Lie alg. (𝔤,L(A))

For if φ:𝔤L(A) is a Lie algebra homomorphism, it is a k–linear mapping 𝔤A such that

φ[x,y] =φ(x)φ(y) -φ(y)φ(x) . (1)

Extend φ to a homomorphism φ:T(𝔤)A by defining

φ (x1xn) = φ(x1) φ(xn),

then the kernel of φ contains the ideal J𝔤, by virtue of (1). Hence φ induces a homomorphism of associative algebras φ#: U(𝔤)A .

In the other direction, let θ:U(𝔤)A be a homomorphism of associative algebras, and form the linear mapping

θb:gT(𝔤) U(𝔤)θA

which is a Lie algebra homomorphism 𝔤L(A). Finally verify that the mappings φφ#,θθb are inverses of each other.

In particular, if ρ:𝔤𝔤𝔩(V) =L(End(V)) is a representation, we have ρ#:U(𝔤)End(V) , i.e. V is a U(𝔤)–module.

The Poincaré-Birkhoff-Witt Theorem

Recall that U(𝔤)= T(𝔤)/ J𝔤; T(𝔤)= n0 Tn(𝔤) is a graded algebra, but J𝔤 is not a graded ideal, because the generators xy-yx-[x,y] are not homogeneous: xy-yxT2 (𝔤), [x,y]T1(𝔤) . So U(𝔤) is not a graded algebra; but it does carry a filtration, defined as follows: Let

Tn= i=0 n Ti(𝔤)

Let π: T(𝔤) U(𝔤) be the canonical homomorphism, and let

Un =π(Tn)

The Un are vector subspaces of U:

k= U0 U1 ; U= n0 Un

and since TmTn Tm+n we have Um·Un Um+n , ie U(𝔤) is a strong filtered associative k–algebra. Now form the associated graded algebra: define

Gn= Un/ Un-1 ( n0; U-1=0 )

then the multiplication in U(𝔤) induces bilinear mappings

Gm×Gn Gm+n

namely (for xUm,yUn )

( x+ Um-1 ) ( y+ Un-1 ) = xy+ U m+n-1

which make G=Gr(U(𝔤)) = n0 Gn into a graded associated k–algebra. For each n1 we have

0/math> 0 Tn-1 Un-1 Tn Un Tn(𝔤) Gn 0/math> 0 π π φ

commutative with exact rows; φ: Tn(𝔤) Gn is surjective, hence we have a surjective algebra homomorphism


defined by

φ(x)= π(x)+ Un-1 (xTn)

In particular, if x,y𝔤 we have

π ( xy-yx ) = π([x,y]) U1

and therefore

φ ( xy-yx ) =0 in G2

Hence the kernel of φ contains the two-sides ideal I of T(𝔤) generated by all xy-yx ( x,y𝔤 ) ; now T(𝔤)/I is by definition the symmetric algebra S=S(𝔤) of the vector space 𝔤; hence φ induces a surjective homomorphism

ω:S(𝔤) Gr(U(𝔤))

(P – B – W) ω is an isomorphism.

For the proof, see the standard texts (Bourbaki (Ch. I), Humphreys, Jacoborn)

Let σ:T(𝔤) S(𝔤) be the canonical homomorphism.

Let V be a vector subspace of Tn (𝔤) which is mapped isomorphically by σ onto Sn (𝔤) . Then π(V)(V) is a complement of Un-1 in Un.


The diagram

Un Sn Gn VTn θ ω π σ

commutes, hence θ maps V isomorphically onto Gn=Un/ Un-1 (because ω is an isomorphism, by P-B-W). Hence the result.

The canonical map gT(𝔤) πU(𝔤) is injective.

We may therefore identify 𝔤 with it's image in U(𝔤).

Let (xλ) λL be a totally ordered k–basis of 𝔤. Then the elements

x λ 1 x λ n = π ( x λ 1 x λ n )

such that λ1 λn ( for all n0 ) form a k–basis of U(𝔤).


Let Vn be the subspace of Vn(𝔤) spanned by all xλ1 xλn with λ1λn . Clearly σ maps Vn isomorphically onto Sn(𝔤), hence by Corollary 7.2 π(Vn) is a complement of Un-1 in Un. By induction on n it follows that U(𝔤) is the direct sum of the π(Vn) for all n0.

(Corollary 7.4 is also known as the P-B-W theorem).


To recapitulate from last time:– to each Lie algebra 𝔤 we associate U(𝔤), its universal enveloping algebra: U(𝔤)= T(𝔤)/J𝔤 where T(𝔤) generated by all xy-yx- [x,y] (x,y𝔤) . 𝔤 embeds in U(𝔤) (by virtue of the P-B-W theorem) and we identify 𝔤 with its image in U(𝔤). In U(𝔤) we have [x,y]=xy-yx (x,y𝔤) . Moreover U(𝔤) is universal in the following sense: if φ:𝔤A is any k–linear mapping of 𝔤 into an associative k–algebra A such that φ[x,y]= φ(x)φ(y)- φ(y)φ(x) – i.e. if φ:𝔤L(A) is a Lie algebra homomorphism, then φ extends uniquely to a homomorphism φ#: U(𝔤)A , as follows :– first extend φ to φ: T(𝔤)A in the obvious way:

φ ( x1xn ) = φ(x1) φ(xn) ( x1, ,xn 𝔤 )

and then observe that J𝔤Kerφ, so that φ induces φ#: U(𝔤)A as desired. Thus φφ# is a mapping

Hom Lie alg ( 𝔤, L(A) ) Hom assoc. alg ( U(𝔤)A )

which one easily verifies to be bijective (i.e. U is a left adjoint of the functor L, as I said last time).

Recall also (Corollary 7.4 of P-B-W th.) that if (xλ) λL is an ordered k–basis of 𝔤, then the monomials xλ1 xλn with λ1 λn and n0 (if n=0, the product is empy and conventionally is said to be read as 1, the identity element of U(𝔤)) form a k–basis of U(𝔤). This has the following consequence: if


where 𝔞,𝔟 are subalgebras and 𝔤 is the direct sum of the vector spaces 𝔞,𝔟, then U(𝔞),U(𝔟) are subalgebras of U(𝔤) and

U(𝔤)= U(𝔞) U(𝔟)= U(𝔟) U(𝔞)

(Take ordered bases (yμ), (zν) of 𝔞,𝔟 respectively; the monomials yμ1 yμm with μ1μm form a k–basis of U(𝔞), the monomials zν1 zνn with ν1νn form a k–basis of U(𝔟), and the monomials yμ1 yμm zν1 zνn with μ1μm and ν1νn form a k–basis of U(𝔤).)

Free Lie algebras

Let X be a set, k a field. We want to define the free Lie algebra Lie(X) on the set X. There are two ways of proceeding: one involves P-B-W, the other doesn't.

(1) Form the free non-associative algebra F(X) on X. How does one do this?

Define inductively sets Xn,n1 by

X1 = X X2 = X1 × X1 X3 = ( X2×X1 ) ( X1 × X2 ) and in general Xn = p=1 n-1 Xp×Xn-p

and put M(X)= n1 Xn (disjoint union) (the "free magma" on X).

If a,bM(X), say aXp and bXq, then (a,b) Xp×Xq Xp+qM so we have a multiplication ab=(a,b) in M(X). Then F(X) is the k–algebra with M(X) as basis, i.e. it consists of all finite linear combinations λiai with λik and aiM(X), and multiplication defined in the obvious way:

( λi ai ) ( μj bj ) = i,j λi μj ai bj .

Now let J be the 2-sided ideal in F(X) generated by all

xx, x(yz)+ y(zx)+ z(xy) ( x,y,z F(X) )

and define

Lie(X)= F(X)/J

where F(X) is a graded algebra and J is a homogeneous ideal. It is clear that Lie(X) is a Lie algebra and that if

j:XF(X) Lie(X)

is the canonical embedding, then any mapping φ of the set X into a Lie algebra 𝔤 extends uniquely to a Lie algebra homomorphism

φ#:Lie(X) 𝔤

X Lie(X) 𝔤 φ# φ j

(extend φ in the obvious way to φ:F(X) 𝔤 and observe that the generators of J in the kernel of φ, by definition).

In other words, we have a bijection

Homsets (X,𝔤) HomLie alg. ( Lie(X),𝔤 )

i.e. the functor Lie (from sets to Lie algebras) is a left adjoint of the forgetful functor Φ (from Lie algebras to sets).

(2) Let A(X) be the free associative algebra on X (=F(X)/I, where I is the 2-sided ideal generated by all (xy)z -x(yz) ( x,y,zF(X) ) ). The embedding

XA(X) =L(A(X))

induces, as we have just seen, a Lie algebra homomorphism

α:Lie(X) L(A(X))

hence also

β:U (Lie(X)) A(X)

But also we have a mapping XLie(X) U(Lie(X)) , hence (by the universal property of A(X)) a homomorphism of associative algebras

γ:A(X) U(Lie(X))

Check that these two homomorphisms β,γ are inverses of each other, hence that

U(Lie(X)) A(X)

By P-B-W, Lie(X) embeds in U(Lie(X)), hence in A(X), so that the mapping α above is injective. In other words, the free Lie algebra Lie(X) may be described as the subalgebra of L(A(X)) generated by X.

We have βk=alpha and γα=k, also αj=i, hence γβk=γα=k, hence γβ=1 (because k injective); βγi= βγαj= βkj= αj=i , hence βγ=1 (because i is injective).

X U(Lie(X)) A(X) Lie(X) α γ β j i k

Finally, if R is any subject of Lie(X), the Lie algebra generated by X subject to the relations R is by definition Lie(X)/𝔞, where 𝔞 is the ideal of the Lie(X) generated by R (i.e. the intersection of all ideals of Lie(X) which contain R).

Finite-dimensional simple Lie algebras /

This is the classical theory we intend to generalise. A Lie algebra 𝔤 is said to be simple if its only ideals are 0 and 𝔤, and if also 𝔤 is non-abelian (thus deliberately excluding the 1-dimensional abelian Lie algebra). Take k=, and dim 𝔤<.

An element x𝔤 is semisimple if ad x:𝔤𝔤 is a semisimple (i.e., diagonalizable) linear transformation.

Let 𝔤 be simple, finite-dimensional. Then 𝔤 has nonzero subalgebras consisting of semisimple elements (toral subalgebras); they are necessarily abelian.

Let 𝔥 be a maximal toral subalgebra (or Cartan subalgebra) of 𝔤. Certainly such exist, for dimensional reasons. Moreover (a non-trivial fact) any two such are conjugate in 𝔤 (i.e. transforms of each other under the group Int(𝔤) of inner automorphisms). Fix 𝔥 once for all. l=dim𝔥 is called the rank of 𝔤.

Let 𝔥* be the vector space dual of 𝔥. Introduce the killing form

x,y= trace (adx) (ady) (x,y𝔤)

This is symmetric, nondegenerate and invariant, ie

[x,z] ,y = x, [z,y] ( x,y,z𝔤 )

Moreover its restriction to 𝔥 is nondegenerate, hence defines an isomorphism ω:𝔥 𝔥* ( ω(x)(y) = x,y ) and a symmetric bilinear form λ,μ on 𝔥* ( λ,μ = ω-1λ, ω-1μ ) .

Example 𝔤=𝔰𝔩n ()= Lie algebras of n×n matrices with trace 0. Here we may take 𝔥 to consist of the diagonal matrices

k= ( h1 hn ) with 1 n hi =0

(so that l=rank(g)=n-1).


Consider the adjoint representation ad𝔤 of 𝔤, restricted to 𝔥: this is a representation of 𝔥 on 𝔤. Since 𝔥 is abelian, all its irreducible representations are 1-dimensional, so that 𝔤 splits up into a direct sum of 1-dimensional 𝔥–modules. explicitly, for each α𝔥* define

𝔤α= { x𝔤: (adx)= α(h)x for allh𝔥 }

Then it turns out that 𝔤0=𝔥; the nonzero α𝔥* such that 𝔤α0 are called the roots of 𝔤 (relative to 𝔥). They form a finite subject R of 𝔥*, called the root system of (𝔤,𝔥), and we have

𝔤=𝔥+ αR 𝔤α (direct sum)

Moreover each 𝔤α(αR) is 1-dimensional, and [ 𝔤α, 𝔤β ] 𝔤α+β (hence is 0 if α+βR{0}). If α is a root, so is -α.

In the case of 𝔰𝔩n(), let eij ( 1i, jn ) be the matrix units, and for 1in let ui:𝔥 be the ith projection: ui(h)=hi. Then

(adh)eij = [h,eij] = heij- eijh = ( hi-hj ) eij = ( ui-uj ) (h)·eij (h𝔥)

which shows that the roots are α=ui-uj (ij) ; 𝔤α= eij , and the root space decomposition is clear. We compute the Killing form on 𝔥 as follows from above

(adh) (adh) eij = ( hi-hj ) ( hi- hj ) eij

so that

h,h = i,j ( hi-hj ) ( hi- hj ) = 2n 1 n hi hi

(remember that hi hi =0 ). So it is a multiple of the obvious scalar product.

It is possible to choose roots α1,,αl (l=dim𝔥) such that each root αR is of the form α= 1 l ni αi with coefficients ni and either ni0 (positive roots) or all ni0 (negative roots). The αi are called a set of simple roots or a basis B of R (they are also a basis of 𝔥*). Choose such a basis once for all. There is then a unique highest root, for which ni is a maximum; and a unique lowest root, for which ni is a minimum.

Weyl group

For each αR, let wα denote the reflection in the hyperplane orthogonal to α in 𝔥*, so that

wα(λ)= λ- λ,α α ( λ 𝔥* )

where α=2α/ || α || 2 is the coroot of α. The reflections wαi corresponding to the simple roots generate a finite group of isometries of 𝔥*, called the Weyl group W of (𝔤,𝔥). Each reflection wα lies in W; R is stable under W; and each root αR is of the form wαi for some wW and some simple root αi. Moreover, any other basis of R is of the form ( wα1 ,, wαl ) =wB for some (unique) wW.

Cartan matrix

The numbers

αij= αi, αj = 2 αi, αj αi, αj

are integers, and the matrix (aij) 1i, jl is called the Cartan matrix of. It is independent of the choices of 𝔥 and of basis of R. It satisfies the following conditions:

(C) aii=2 (1il); aij0 ifij; aij=0 aji=0
(P) All principal minors of A are positive.

In the case of 𝔰𝔩n() we may take αi=ui- ui+1 ( 1in-1 ) . The Weyl group W is the symmetric group Sn (for wαi interchanges ui and ui+1 and leaves the other uj fixed). Here the Cartan matrix is

A= 2 -1 -1 2 -1 -1 2 -1 -1 2

(with l=n-1 rows and columns).

Generators and relations

The Cartan matrix A determines 𝔤 up to isomorphism.

Choose generators ei𝔤αi, fi𝔤-αi ( 1il ) such that ei,fi =1, and elements h1,,hl𝔥 such that hi,h = αi (h) , so that

αj(hi) = αi, αj =aij

Then the 3l elements ei,fi,hi generate 𝔤 subject to the following relations (Serre):

[hi,hj] =0for alli,j [ei,fj] =δijhi [hi,ej] =aijej ; [hi,fj] =-aijfj ( adei ) 1-aij ej = ( adfi ) 1-aij fj =0 (ij).

The idea now is (roughly) the following: start with any matrix A of integers satisfying (C), and form the Lie algebra with the above generators and relations.

However there is one remark that should be made at this point. In the classical set up (which I have just been describing) the Cartan matrix A is nonsingular, the hi (1il) form a basis of the Cartan subalgebra 𝔥, and the simple roots αj𝔥*. Now a generalized Cartan matrix may well be singular (and it would be foolish to exclude this possibility, because for the affine Lie algebras the Cartan matrix is singular).


I.G. Macdonald
Issac Newton Institute for the Mathematical Sciences
20 Clarkson Road
Cambridge CB3 OEH U.K.

Version: October 30, 2001

page history