Kac-Moody Lie Algebras
Chapter I

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

Last update: 16 August 2012

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

Construction of the algebras

The construction of the Kac-Moody algebras works for any matrix over K:

A= (aij) 1i,jn

– the aij needn't even be integers.

We begin with a lemma of linear algebra. Given a matrix A as above, a realization of A is a triple ( 𝔥, B, B ) where:

𝔥is a finite-dimensional vector space overk; B= ( α1, ,αn ) is a linearly independent set of vectors in 𝔥*=dual of𝔥; B= ( h1, ,hn ) is a linearly independent set of vectors in 𝔥;such that αj (hi) =aij ( 1i,jn ) .

A realization of A will be called minimal if dim 𝔥 is as small as possible (evidently it must be n).

Let l=rank(A)


  1. If ( 𝔥,B,B ) is a realization of A, then dim 𝔥2n-l.
  2. A has a minimal realization, of dimension 2n-l, which is unique up to isomorphism (but the isomorphism is not unique if l<n).

  1. Extend B to a basis h1,,hN of 𝔥 (so that dim 𝔥=N). The N×n matrix M= ( αj (hi) ) 1iN, 1jn is of the form M= ( A B ) and has rank n (because its columns are linearly independent). Let VA,Vb,VM denote the spaces spanned by the rows of A,B,M respectively. Then we have VM= VA+VB , and dimVA = rank(A) =l, dimVM = rank(M) =l . Hence

    N-n dimVB n-l

    i.e., N2n-l.

  2. By reordering the rows and the columns of A we may assume that the l×l minor of A in the top left-hand corner is nonsingular, say

    A= ( A1 A2 A3 A4 )

    with A1 a nonsingular l×l matrix. Let

    C= ( A1 A2 0 A3 A4 1n-l 0 1n-l 0 ) ,

    then detC=± detA1 0 ; hence the rows of C are linearly independent. Take 𝔥=k2n-l (row vectors); αj the jth coordinate function on 𝔥,hi the ith row of WHAT GOES HERE?. Then αj(hi) =aij and we have a realization of A.

    Conversely, let ( 𝔥,B,B ) be a minimal realization of A ( dim𝔥 =2n-l ) . Extend B to a basis h1 ,, h2n-l of 𝔥, and define αn+1 ,, α2n-l 𝔥* so that the matrix D= ( αj (hi) ) 1i, j2n-l has the form

    D= ( A1 A2 0 A3 A4 1n-l B1 B2 0 )

    B1,B2 at present unspecified. Then

    detD=±det ( A1 A2 B1 B2 ) ,

    and I claim that this matrix is nonsingular. For the submatrix

    M= ( A1 A2 A3 A4 B1 B2 ) = ( A B )

    of D has rank n, as before, and now we have VM=VA VB ; but VA has the first l rows of A as a basis, and the rows of B form a basis of VB; hence the rows of ( A1 A2 B1 B2 ) are linearly independent, as claimed.

    It follows that D is nonsingular, hence that α1 ,, α2n-l are a basis of 𝔥*. By adding to hn+1 ,, h2n-l suitable linear combinations of h1,,hl, we can make B1=0. But then detB20, and we can choose another basis of the subspace of 𝔥 spanned by hn+1 ,, h2n-l so as to make B2=1n-l, i.e. D=C. This completes the proof.

A matrix A as above will be said to be decomposable if we can partition the index set {1,,n} into two non-empty disjoint subsets I,J such that aij= aji=0 whenever iI and jJ. In other words, if after simultaneous permutation of rows and columns A becomes a nontrivial direct sum A1A2.

Clearly, if ( 𝔥i, Bi, Bi ) is a minimal realization of Ai (i=1,2) , then ( 𝔥,B,B ) is a minimal realization of A, where

𝔥 = 𝔥1 ×𝔥2 𝔥*= 𝔥1*× 𝔥2* (direct sum) B = (B1×0) (0×B2) B = (B1×0) (0×B2)

Again, if ( 𝔥,B,B ) is a minimal realization of A, then ( 𝔥*,B,B ) is a minimal realization of At.

Let A be any n×n matrix over k, as before; let ( 𝔥,B,B ) be a minimal realization of A ( B= ( α1 ,, αn ) ; B= ( h1 ,, hn ) ) . Let 𝔤 (A) denote the Lie algebra generated by 𝔥 and 2n elements ei,fi (1in) subject to the relations

(1.2) { [h,h] =0 ( all h,h𝔥 ) [ ei, fj ] = δij hi ( 1i, jn ) [ h,ei ] = αi (h) ei [ h,fi ] = -αi (h) fi } ( 1in; h𝔥 )

By (1.1), 𝔤(A) depends (up to isomorphism) only on the matrix A.

Objects defined by generators and relations are often not easy to handle directly. To get a group on 𝔤(A) we shall construct a family of representations ρλ of 𝔤(A), one for each λ𝔥*check the wedge?. These representations will act on the same vector space X:X is the free associative algebra over k on n generators x1,,xn, and we define an action of the generators of 𝔤(A) on X as follows: Let λ𝔥* and define

(a) h(1)= λ(h)·1 ; h(xjx) =xjh(x) -αj(h) xjx ( xX, h𝔥, 1jn ) (b) ei(1)=0 ; ei(xjx) =xjei(x) +δijhi (x) ( xX, 1i, jn ) (c) fi(x) =xix ( xX, 1in ) .

I claim that these formulas define a representation ρλ of 𝔤(A) on X. To verify this, we have to check the defining relations (1.2).

First, it follows from (a) that

h ( xj1 xjr ) = ( λ-αj1 -- αjr ) (h) xj1 xjr

by induction on r; hence each h𝔥 acts diagonally on X (relative to the basis of X formed by the monomials) and therefore [h,h]=0 for all h,h𝔥.

Next, we have

[ ei, fj ] = δij hj from (b) and (c) [ h, fj ] = -αj (h) fj from (a) and (c)

and therefore it remains to show that [ h,ei =αi (h)ei ] (as linear transformations of the vector space X). So let u=[h,ei] -αi(h)ei , then

[u,fj] = [ [h,ei] ,fj ] -αi(h) [ei,fj] = [ h, [ei,fj] ] - [ ei, [h,fj] ] -αi(h) [ei,fj] = [ h, δijhi ] + αj(h) δij hi - αi(h) δijhi = 0

Hence u(xjx) = xju(x) for all xX and 1jn; hence (induction on r) u ( xj1 xjr ) = xj1 xjr u(1) ; but

u(1) = h(ei(1)) =- ei(h(1)) -αi(h) ei(1) = -λ(h) ei(1) =0

and therefore u=0 as required.

Thus for each λ𝔥* the formulas (a) - (c) define a representation ρλ of 𝔤(A) on X.

This may look like a rabbit pulled out of a hat: in fact it is a standard construction (Verma module).

We may remark straightaway that the canonical mapping 𝔥𝔤(A) is injective. For if h𝔥 because zero in 𝔤(A), then from (a) we have λ(h)·1=0 for all λ𝔥*, and hence h=0.

Let 𝔫+ (resp. 𝔫-) denote the subalgebra of 𝔤(A) generated by e1,,en (resp. f1,,fn).


  1. 𝔤(A) =𝔫- 𝔥 𝔫+ (direct sum of vector spaces)
  2. 𝔫+ (resp. 𝔫-) is the free Lie algebra generated by e1,,en (resp. f1,,fn)
  3. unique involutory automorphsim ω of 𝔤(A) such that

    ω(ei) =-fi, ω(fi) =-ei, ω(h) =-h, (h𝔥) .

We shall take these in reverse order.
  1. is clear, since the relations (1.2) are stable under ω.
  2. Since X is the free associative algebra on x1,,xn, L(X) is the free Lie algebra on the same generators. Now the mapping φ: 𝔫- L(X) defined by φ(f)=f(1) takes fi to xi and is a Lie algebra homomorphism. Since L(X) is free, φ must be an isomorphism, and 𝔫- is the free Lie algebra on f1,,fn, and U(𝔫-) X . By applying ω, we see that 𝔫+ is the free Lie algebra on e1,,en.
  3. Let 𝔞=𝔫-+ 𝔥+𝔫+ . It follows easily from the defining relations (1.2) that 𝔞 is stable under ad ei, ad fi and ad h h𝔥. Hence it is an ideal in 𝔤(A), and since it contains the generators ei,fi,h𝔥 it is the whole of 𝔤(A). Remains to prove that the sum is direct. Suppose then that we have 𝔫-𝔫-, h𝔥, 𝔫+𝔫+ such that

    𝔫-+h+ 𝔫+=0

    Apply ρλ and evaluate at 1X. We have 𝔫+(1)=0 (because ei(1)=0), hence

    𝔫-(1)+ λ(h)1=0 inX

    whence λ(h)=0 and 𝔫-(1)=0. Since this is true for all λ𝔥*, it follows first that h=0; next, as we have seen, 𝔫-𝔫-(1) :𝔫-X is the embedding of 𝔫- in its universal enveloping algebra XU(𝔫-); hence 𝔫-=0, whence finally 𝔫+=0 and the proof is complete.


In general, if A is a k–algebra and G an abelian group, a G–grading of A is a decomposition

A= αG Aα (1)

of A into a direct sum of k–subspaces Aα, indexed by G, such that

AαAβ Aα+β ( α,βG ) .

The elements of Aα are said to be homogeneous of degree α; the decomposition (1) says that any xA can be written uniquely as the sum x=αxα of its homogeneous components (only finitely many of which can be 0).

An ideal 𝔞 in A is a graded ideal if

𝔞= α 𝔞α

where 𝔞α=𝔞Aα, that is to say if whenever x𝔞 all the homogeneous components xα of x lie in 𝔞. Any sum of graded ideals is graded; any ideal generated by homogeneous elements is graded.

If 𝔞 is a graded (two-sided) ideal, then A/𝔞 is a G–graded algebra:

A/𝔞= αG Aα/𝔞α .

In the present context, let

Q= i=1 n αi (n)

denote the Lattice generated by B in 𝔥* (the root lattice). Also let

Q+= i=1 n αi

For αQ we write α0 to mean αQ+, i.e. α= 1 n mi αi with all mi=0; also α>0 to mean αG+ and α0. Likewise α0,α<0. If α = 1 n mi αi Q , we define the height of α to be

ht(α)= 1 n mi .

Now the free Lie algebra generated by 𝔥 and e1,,en, f1,,fn is G–graded by assigning degree 0 to each h𝔥, degree αi to ei and degree -αi to fi (1in).

The relations (1.2) are homogeneous, hence 𝔤(A) is a Q–graded Lie algebra:

𝔤(A)= αQ 𝔤α, [ 𝔤α, 𝔤β ] 𝔤α+β

where 𝔤α consists of the homogeneous elements of degree α in 𝔤(A). By (1.3) we have 𝔤0=𝔥 and (for α0) 𝔤α=0 unless either α>0 or α<0, because

𝔫+= α>0 𝔤α, 𝔫-= α<0 𝔤α .

We can introduce other gradings on 𝔤(A). Let s:Q be any homomorphism of abelian groups, and for each m define

𝔤m(s)= s(α)=m 𝔤α

Then 𝔤(A)= m 𝔤m(s) , and [ gm(s), gm(s) ] 𝔤 m+m (s) , giving a –grading of 𝔤(A). The most important case of this is the principal grading, defined by s(αi)=1 (1in) . For this choice of s we have s(α)=ht(α) , and we therefore define

𝔤m= ht(α)=m 𝔤α (m)

We have 𝔤0=𝔥; 𝔤1 is spanned by e1,,en; 𝔤-1 by f1,,fn ; and 𝔫+= m1 𝔤m, 𝔫-= m1 𝔤-m .

Let α>0. Then 𝔤α is the α–component of the free Lie algebra 𝔫+ generated by e1,,en, hence it is spanned by all commutators

xα= [ ei1 eir ]

such that αi1 ++ αir =α (notation: [u1ur] means [ u1, [ u2,ur ] ] ). There are only finitely many of these (at most 𝔫h+α), hence 𝔤α is finite-dimensional. Likewise when α<0. In particular, 𝔤±αi are 1-dimensional.

For xα as above we have, for each h𝔥,

[h,xα] = p=1 r [ ei1 [ h,eip ] eir ] (because adhis a derivation) = p=1 r αip (h) [ ei1 eip eir ] by(1.2) = α(h) xd

Hence if temporarily we write

Mα= { x𝔤(A): [h,x]= α(h)xfor all h𝔥 }

for each αQ, then we have

𝔤 α Mα, allαQ .

(For the calculation above shows this is true when α>0; similarly when α<0 or α=0; and in other cases 𝔤α=0).

Moreover a standard argument shows that the sum αQMα is direct. For if there exist non-trivial relations

α xα =0 (1)

with xαMα (and only finitely many xα0), choose such a relation with as few non-zero terms as possible; by applying adh we conclude that

α α(h) xα =0 (2)

for each h𝔥. We can then subtract a multiple of (1) from (2) to obtain a shorter relation: contradiction. Hence we have

𝔤(A)= αQ 𝔤α αQ Mα 𝔤(A)

from which it follows that Mα=𝔤α for all αQ. To summarize:

(1.4) For each αQ, let 𝔤α denote the component of degree α in 𝔤(A). Then

  1. 𝔤α= { xg(A): [h,x]=α (h)xfor all h𝔥 }
  2. 𝔤α=0 unless α>0,α<0 or α=0; moreover 𝔤0=𝔥.
  3. each 𝔤α is finite-dimensional over k, and dim 𝔤 ±αi =1 .

Ideals in 𝔤(A)

We shall next prove that all ideals in 𝔤(A) are Q–graded ideals. This will be a consequence of the following lemma:

(1.5) Let 𝔥 be an abelian Lie Algebra, M an 𝔥–module. For each λ𝔥* let

Mλ= { xM: h·x=λ (h)x for allh𝔥 } .

Suppose that M= λ𝔥* Mλ , and let M be a submodule of M. Then

M= λ𝔥* Mλ, where Mλ= MMλ .


Each xM can be written in the form

x= i=1m xλi

where λ1,, λm are district elements of 𝔥*, and xλi Mλi . We have to show that each xλi M . The polynomial function i<j ( λi-λj ) on 𝔥 is not zero, hence h𝔥 such that λ1(h) ,, λm(h) are all distinct.

We have

hj·x= i=1m λi (h) j xλi ( 0jm-1 )

and we can solve these equations for xλ1 ,, xλm by Cramer's rule, since det ( λi (h) j ) 1jm 0jm-1 = i<j ( λi(h) - λj(h) ) 0 . Hence each xλi is a linear combination of the hj·x, hence lies in M.

We apply this lemma with M=𝔤(A) and M an ideal 𝔞 in 𝔤(A). Then 𝔞 is a 𝔥–submodule of 𝔤(A) under the adjoint action, hence by (1.4) and (1.5) we have

𝔞= αQ 𝔞α

where 𝔞α=𝔞 𝔤α i.e. 𝔞 is a Q–graded ideal. Hence also

𝔞= m 𝔞m

where 𝔞m= 𝔞 gm (principal grading).

Consider now ideals 𝔞 in 𝔤(h) such that 𝔞0=0, i.e. 𝔞𝔥=0. Any sum of such ideals has the same property, hence there is a unique largest ideal 𝔯 in 𝔤(A) such that 𝔯𝔥=0. We have



𝔯+= m>0 𝔯m= 𝔯 𝔫+
𝔯-= m<0 𝔯m= 𝔯 𝔫-

We have [fi,𝔯+] = m>0 [ fi, 𝔯m ] m>0 𝔯m-1 =𝔯+ ; and since clearly [ 𝔥,𝔯+ ] 𝔯+, [ ei, 𝔯+ ] 𝔯+ it follows that 𝔯+ is an ideal in 𝔤(A). Similarly, of course, for 𝔯-.

Next I claim that 𝔯1=𝔯-1 =0 . For 𝔯1= i=1 n 𝔯αi ; if 𝔯αi0 then 𝔯αi= 𝔤αi (because 𝔤αi is 1-dimensional, spanned by ei), hence ei𝔯; but then hi= [ ei, fi ] 𝔯𝔥 , contradiction. Hence 𝔯1=0, and similarly 𝔯-1=0.

Finally, we must have ω(𝔯) =𝔯 (for ω(𝔯) has the same properties as 𝔯).

To summarize:


  1. All ideals in 𝔤(A) are Q–graded.
  2. The set of ideals 𝔞 in 𝔤(A) such that 𝔞𝔥=0 has a unique maximal element of 𝔯.
  3. 𝔯+=𝔯 𝔫+ and 𝔯-=𝔯 𝔫- are ideals in 𝔤(A), and 𝔯=𝔯+𝔯- (direct sum)
  4. 𝔯1= 𝔯-1 =0
  5. ω(𝔯) =𝔯

Now define

𝔤(A)= 𝔤(A) /𝔯 .

It is this algebra which is the object of our investigations. If A is a Cartan matrix, 𝔤(A) is the Kac-Moody algebra defined by the matrix A.

(1.7) Remarks
Since the ideal 𝔯 is Q–graded (1.6), 𝔤(A) is a Q–graded Lie algebra:

𝔤(A)= αQ 𝔤α

where 𝔤α= 𝔤α /𝔯α ; thus

  1. 𝔤0= 𝔤0 =𝔥
  2. Since 𝔯1=𝔯-1 =0 (1.6), 𝔤1=𝔤1= i=1 n kei; 𝔤-1= 𝔤-1= i=1 n kfi; 𝔤αi= kei, 𝔤-αi= kfi, (1in)
    (Since the images of e1,,en, f1,,fn in 𝔤(A) remain linearly independent, we continue to denote them by the same symbols).
  3. 𝔤α=0 unless α-0 or α>0 or α<0, by (1.4)
  4. 𝔤α= { x𝔤(A) : [h,x]= α(h)x for allh𝔥 } (same proof as in (1.4)) and each 𝔤α is finite-dimensional. If α0 and 𝔤α=0 we say α is a root of 𝔤(A) with multiplicity mα= dim𝔤α . (In the classical case, all mα are 1).
  5. Let 𝔫+= 𝔫+/ 𝔯+= α>0 𝔤α, 𝔫-= 𝔫-/ 𝔯-= α<0 𝔤α

    These are subalgebras of 𝔤(A), generated by e1,,en and by f1,,fn respectively, and

    𝔤(A)=𝔫- 𝔥𝔫+ (vector space direct sum)

  6. All ideals 𝔞 in 𝔤(A) are Q–graded (1.6): 𝔞= αQ 𝔞α ; and 𝔤(A) has no ideal 𝔞0 such that 𝔞𝔥=0 (by construction)
  7. Since 𝔯 is stable under the involution ω (1.6) we have an involution ω:𝔤(A) 𝔤(A) under which ei-fi, fi-ei, h-h(h𝔥) we have ω(𝔤α)= 𝔤-α for all αQ, hence ω interchanges 𝔫+ and 𝔫-.

The following lemma is frequently useful:


  1. Let x𝔫+ be such that [x,fi]=0 (1in) . Then x=0.
  2. Let x𝔫- be such that [x,ei]=0 (1in). Then x=0.


We shall prove (i), and then (ii) will follow by use of the involution ω. Write x=α>0 xα , then we have α>0 [ xα, fi ] =0 for 1in. Hence [ xα,fi ] =0 for each α and each i; in other words we may assume x homogeneous. Consider the ideal U(𝔤)·x=𝔞 say, generated by x. (U(𝔤) acting via ad). Since 𝔤=𝔫- 𝔥𝔫+ (1.7) we have (corollary of P-B-W)

U(𝔤)= U(𝔫+) U(𝔥) U(𝔫-)

By assumption, U(𝔫-)·x =kx; U(𝔥)·x=kx , hence

𝔞= U(𝔤)·x= U(𝔫+)·x

has only positive components, hence 𝔞𝔥=0. Hence ((1.7)(vi)) 𝔞=0, i.e. x=0?

Example Suppose A=0 (the n×n zero matrix). What does 𝔤(0) look like?

We have αj(hi)= aij=0 for all i,j, hence the relations (1.2) give

[ hi,ej ] = αj(hi) ej = 0 (alli,j)

and likewise

[ hi,fj ] = 0 (alli,j)

Consider now [ei,ej]. We have

[ [ ei,ej ] , fk ] = [ ei, [ ej,fk ] ] - [ ej, [ ei,fk ] ] = [ ei, δjk hk ] - [ ej, δij hk ] =0

whence by (1.8) [ ei,ej ] =0 . Hence 𝔫+ is abelian, i.e. 𝔫+=𝔤1= 1 n kei . Similarly 𝔫-=𝔤-1= 1 n kfi , and

𝔤(0)=𝔤-1 𝔤0𝔤1

Note that 𝔤0=𝔥 has dimension 2n (because here l=0).

The algebra 𝔤(A)

Let 𝔤=D𝔤(A) be the derived algebra of 𝔤(A).

(1.9) 𝔤(A) is the subalgebra of 𝔤(A) generated by e1,,en, f1,,fn, and

𝔤(A)= 𝔫-𝔥 𝔫+

where 𝔥 is the subspace of 𝔥 generated by h1,,hn.

Thus 𝔤(A)= 𝔤(A) iff det (A)0.


Let 𝔞 denote the subalgebra of 𝔤(A) generated by e1,,fn , and let

𝔟=𝔫- 𝔥𝔫+ .

Since [h,ei]= αi(h)ei for all h𝔥, and since h𝔥 such that αi(h)0 , it follows that ei𝔤(A) ; similarly fi𝔤(A) , and therefore

𝔞𝔤(A) (1)

Next, by (1.7)(v), 𝔫+ and 𝔫- are subalgebras of 𝔞; and since hi= [ ei,fi ] 𝔞 , it follows that 𝔥𝔞, whence

𝔟𝔞 (2)

Finally, I claim that 𝔟 is an ideal in 𝔤(A). We have to check that

[h,b]𝔟 (h𝔥); [ei,𝔟]𝔟; [fi,b]𝔟

The first of these is obvious. As to the second, we have

[ ei,𝔫- ] 𝔫-+𝔥

(because [ ei,fj ] = δijhi (1.2)); [ ei,𝔥 ] 𝔫+; [ ei,𝔫+ ] 𝔫+; and [ fi,𝔟 ] 𝔟 is proved similarly. Since 𝔤(A)/𝔟 𝔥/𝔥 is abelian, it follows that

𝔤(A) 𝔟. (3)

(1), (2), (3) complete the proof.

Remark The algebra 𝔤(A) is also sometimes called the Kac-Moody algebra associated to the matrix A (if A is a Cartan matrix). We can give a more direct construction of 𝔤(A), as follows: Let 𝔤 (A) denote the Lie algebra with 3n generators ei,fi,hi (1in) subject to the relations

(1.2) { [ hi,hj ] =0 [ ei,fj ] =δij hi [ hi,ej ] =aij ej [ hi,fj ] =- aijfj (1i,jn)

Let Q(n) be a free abelian group on generators α1,,αn ; then 𝔤 (A) is Q–graded by assigning degrees 0, αi, -αi to hi,ei,fi respectively (1in), and there exists a unique maximal Q–graded ideal 𝔯 subject to 𝔯𝔥=0 (where 𝔥= 1n khi as above). Then 𝔤(A)= 𝔤 (A)/𝔯 .

We can then construct 𝔤(A) as a semidirect product of 𝔤(A) by a suitable algebra of derivations.

Semidirect products

In general, let 𝔤 be a Lie algebra, 𝔫 an ideal in 𝔤, 𝔞 a subalgebra of 𝔤, such that 𝔤=𝔫𝔞 (vector space direct sum). If x1,x2𝔤 , say xi=ni+ai ( ni𝔫, ai𝔞 ) then

[ x1,x2 ] = [ n1,n2 ] + [ a1.n2 ] - [ a2,n2 ] + [ a1,a2 ]

in which the first 3 terms on the right lie in 𝔫 (because 𝔫 is an ideal in 𝔤).

The Lie algebra 𝔞 acts (via ad) on 𝔫 as an algebra of derivations:

ad:𝔞 Der(𝔫)

Conversely, if we are given Lie algebras 𝔫,𝔞 and a Lie algebra homomorphism φ:𝔞Der (𝔫) , we construct the semidirect product 𝔤=𝔫𝔞 as follows: 𝔤=𝔫𝔞 as a vector space, and the Lie bracket in 𝔤 is defined by

[ n1+a1, n2+a2 ] = [ n1,n2 ] + φ(a1)n2 - φ(a2)n1 + [ a1,a2 ] .

One has of course to check the Jacobi identity, which is tedious but straightforward.

In the present case, let 𝔞 be a vector space complement of 𝔥 in 𝔥: then

𝔤(A)= 𝔤(A)𝔞

with 𝔤(A) an ideal (1.9) and 𝔞 a subalgebra. Hence 𝔤(A) may be constructed as the semidirect product 𝔤(A)𝔞, with 𝔞 acting as an (abelian) algebra of derivations.

The centre of 𝔤(A)

(1.10) The algebras 𝔤(A), 𝔤(A) have the same centre 𝔠:

𝔥0=𝔠= i=1 n Ker(αi) 𝔥 .

We have dim 𝔠=n-l, hence 𝔠=0 iff A is nonsingular.


Suppose x𝔤(A) commutes with e1,,fn . Say x= r xr (principal grading); then 0= [ x,fi ] = r [ xr,fi ] , so that [ xr,fi ] =0 for 1in and all r. By (1.8) it follows that xr=0 if r1, and similarly xr=0 for rwhat goes here?. Hence x=x0𝔥. But then (1.2)

0=[x,ei]= xi(x)ei

so that xi(x)=0 (1in), whence x 1nKer αi . Conversely, if αi(x)=0 for 1in, then by (1.2) we have [x,ei]= [x,fi]=0 , and of course [x,𝔥]=0. This shows that the centre of 𝔤(A) is

𝔠= i=1 n Ker(αi) ;

since the αi are independent linear forms on 𝔥, we have

dim 𝔠=dim 𝔥-𝔫=n-l.

Finally, I claim that 𝔠𝔥. For

1n μihi 𝔠𝔥 1n μiαi (hi)=0 (1jn) 1n μiaij =0 (1jn)

Since A=(aij) has rank l, it follows that 𝔠𝔥 has dimension n-l=dim𝔠. Hence 𝔠𝔥 as claimed and therefore 𝔠 is also the centre of 𝔤(A).


Let us say that two n×n matrices A=(aij) and A=aij are equivalent:


if wsn such that

a ij = a w(i), w(j) (1i,jn)

i.e. if A is obtained from A by applying the same permutation to rows and columns. Clearly AA 𝔤(A)𝔤 (A) : we have merely reindexed the generators. Now suppose that A satisfies the condition

aij=0 aji=0 ()

We associate with A a graph 𝔯(A), as follows: the vertices of A are the indices 1,2,,n, and distinct vertices i and j are joined by an edge iff aij0 or ajisomething

(1.11) Assume that A satisfies (). Then the following conditions on A are equivalent:

  1. A is equivalent to a nontrivial diagonal sum ( A1 0 0 A2 ) ;
  2. There exist non-empty complimentary subsets I,J of {1,2,,n} such that aij=0 for iI and jJ;
  3. 𝔯(A) is not connected.


If these equivalent conditions are satisfied, we say that A is decomposable.

(1.12) If A is decomposable, say A ( A1 0 0 A2 ) , then

𝔤(A)𝔤 (A1)× 𝔤(A2)

(direct product).


Consider 𝔤(A)=𝔤 (A1)× 𝔤(A2) , which is a Lie algebra generated by e1,,en, f1,,fn and 𝔥=𝔥1𝔥2 . Check that these generators satisfy relations (1.2) (for the Q–graded matrix A); hence 𝔤 is a homomorphic image of 𝔤(A) , i.e. we have a surjective homomorphism φ:𝔤 (A)𝔤 . Then φ(r)=𝔞 say is an ideal of 𝔤 such that 𝔞𝔥=0. But 𝔤 is a direct product, hence 𝔞=𝔞1×𝔞2 , where 𝔞i is an ideal in 𝔤(Ai) which intersects 𝔥i trivially (i=1,2). Hence (1.7) 𝔞1=𝔞2=0 and therefore 𝔞=0, consequently φ induces a surjective homomorphism φ:𝔤(A) 𝔤 . The kernel of φ is an ideal 𝔟 such that 𝔟𝔥=0 (because φ/𝔥 is injective), hence 𝔟=0 (1.7). Hence φ is an isomorphism.

αR , then Supp(α) connected.

Ideals in 𝔤(A)

Assume that A satisfies the condition

aij=0 aji=0 .


  1. Suppose A is indecomposable. Then every ideal in 𝔤(A) either contains 𝔤(A) or is contained in the centre 𝔠.
  2. 𝔤(A) is simple iff A is indecomposable and nonsingular.

  1. Let 𝔞 be an ideal in 𝔤(A). By (1.7)(vi), 𝔞 is Q–graded, hence we may write 𝔞= r 𝔞r (principal grading).

    Suppose first that 𝔞0𝔠. If 𝔞10, then ei𝔞1 for some i. But then hi= [ ei,fi ] 𝔞0 , hence hi𝔠 and therefore (1.10) aij=αj (hi)=0 for 1jsomething. This contradicts the assumption that A is indecomposable. Hence 𝔞1=0. It now follows by induction on r that 𝔞r=0 for all r1. For if x𝔞r where r2, then [x,fi] 𝔞r-1=0 by ind. hyp., hence x=0 by (1.8). Likewise 𝔞r=0 for r-1 and therefore 𝔞=𝔞0c.

    Now suppose that 𝔞0𝔠. Let h𝔞0, h𝔠. By (1.10) we have αi(h)0 for some i, hence ei=αi (h)-1 [ hi,ei ] 𝔞 ; similarly fi𝔞 (for this value of i), and hi= [ ei,fi ] 𝔞 . Since 𝔯(A) is connected (1.11) j[1,n] such that aij0; since [ hi,ej ] =aijej , it follows that ej𝔞, and likewise fj𝔞. It now follows that ej,fj𝔞 for every index j connected to i by a path in the graph 𝔯(A) – i.e. e1,,fn 𝔞 , hence (1.9) 𝔞𝔤(A).

  2. If A is decomposable, 𝔤(A) is not simple, by (1.12). Again, if A is singular, i.e. l<n, then 𝔠0 (1.10) and again 𝔤(A) is not simple.

    If A is nonsingular (l=n) then 𝔤(A)=𝔤(A) and 𝔠=0. Now use (i).

Note for later use the following corollary of (1.13):

(1.1312). Assume A indecomposable. Then the following conditions are equivalent:

  1. 𝔤(A) is infinite-dimensional
  2. R is infinite
  3. For each αR+ there exists i such that α+αiR+.


(i) (ii) is clear from the root space decomposition

𝔤(A)=𝔥 αR 𝔤α

(iii) (ii) is clear

(ii) (iii) If (iii) is false, there exists a positive root α such that α+αiR (1in). Let x𝔤α,x0. Then we have [x,ei]=0 (1in) , from which it follows that U(𝔫+)·x =kx and therefore the ideal 𝔞=U(𝔤)·x generated by x in 𝔤(A) is

𝔞=U(𝔫-) U(𝔥)U (𝔫+)·x= U(𝔫)·x

Hence 𝔞β=0 unless βα. But by (1.13) we have 𝔞𝔤(A) (because clearly 𝔞𝔠), hence in particular 𝔞𝔫+. It follows that all roots β are α, whence R+ and therefore R is finite.

Thus if R is finite there is a unique highest root q such that αφ for all αR.

The algebra 𝔤= 𝔤(A)/𝔠

We have

𝔤(A)= 𝔫-𝔥 𝔫+

by (1.9), and 𝔠𝔥 (1.10), hence

𝔤 (A)=𝔫- 𝔥 𝔫+

where 𝔥 =𝔥/𝔠 (so that dim 𝔥=n- (n-l)=l ).

Assume that A is indecomposable (and that aij=0 aji=0 ). The proof of (1.13)(i) shows that any Q–graded ideal 𝔞 in 𝔤(A) such that 𝔞0 (=𝔞𝔥) 𝔠 is contained in 𝔠. Hence 𝔤 (A) has no nontrivial Q–graded ideal 𝔞 such that 𝔞 0 =0 . From this it follows that (1.8) is valid for the algebra 𝔤 (A) .

We shall make use of this remark in the proof of the following proposition:

(1.14) Assume that A is indecomposable and that each root α has a nonzero restriction to 𝔥. Then the algebra 𝔤(A) is simple.


Let 𝔞0 be an ideal in 𝔤(A). Each x𝔞 is of the form

x= αS xα

where S is some finite subset of Q, each xα0 and xα𝔤α for α0,x0 𝔥 . Call |S| the length of x, and the number max αS ht(α) the height of x. Choose x0 in 𝔞 of minimal length.

Suppose that the chosen x has height r1, so that x=xα+ where ht(α)=r, xα0. Since xα0 we have [ xα,fi ] 0 for some i, by the remark above; hence [x,fi] is a nonzero element of 𝔞, of minimal length and height r-1. By proceeding in this way we shall obtain an element y0 of 𝔞 of minimal length and height 0, say

y=h0+ αS yα (1)

where h0 𝔥 (and 0), and |S|= |S|-1 . Similarly, if x has height <0, we use the ei rather than the fi to achieve the same result.

From (1) we have, for all h 𝔥 ,

[h,y]= αS α(h) yα

which is an element of 𝔞 of length <|S|, hence is 0. Hence α(h)=0 for all αS and all hh, i.e. αS α|𝔥=0 . By hypothesis, therefore S is empty and therefore y=h0 𝔥 . Since h00 we have αi(h0) 0 (by (1.something)) for some i, and therefore ei,fi𝔞 for this value of i (because αi (h0) ei= [ h0 ,ei ] 𝔞 ). But now it follows as in the proof of (1.13) that e1,,fn all lie in 𝔞, whence 𝔞=𝔤(A).

Remark: The converse of (1.14) is true if A is a Cartan matrix (proof later) I do not know whether it is so in general.

I shall conclude this chapter with some properties of 𝔤(A) that are valid only when A is a Cartan matrix. So assume now that the matrix A satisfies the condition (C):

(C) aij; aii=2; aij0 ifij; aij=0 aji=0 .

For each i=1,,n let si denote the subspace of 𝔤(A) spanned by ei,fi,hi . From (1.2) we have

[ ei,fi ] =hi, [ hi,ei ] =2ei, [ hi,fi ] =-2fi

(1.15) si is a subalgebra of 𝔤(A), isomorphic to 𝔰𝔩2(k).


The relations just written show that si is a 3-dimensional subalgebra of 𝔤(A). The mapping

ei ( 0 1 0 0 ) , fi ( 0 0 1 0 ) , hi ( 1 0 0 -1 )

is an isomorphism of si onto 𝔰𝔩2(k)

Next we require the following lemma:

(1.16) Let x,y be elements of an associative ring R. Then for each positive integer N we have

  1. xNy= r=0N (Nr) (adx) r yxN-r .
  2. xyN= r=0N (-1)r (Nr) yN-r (ady) r x .


We shall prove (ii); the proof of (i) is analogous. Let λy; py: R something denote respectively left and right multiplication by y in R. Since R is associative they commute with each other and hence also with ady=λy -py . Hence

xyN= pyN(x) = ( λy-ady ) N x = r=0N (-1)r (Nr) λy N-r (ady) r x.

Let us apply this formula with x=ei, y=fi, R=U(si) : we have

(adfi)ei= [ fi,ei ] =-hi (adfi) 2 ei=- [ fi,hi ] =-2fi (adfi) 3 ei =-2 ( fi,fi ) =0

and therefore

(1.17) eifiN = fiNei+ NfiN-1 hi+ (N2) fiN-2· -2fi = fiNei+ NfiN-1 ( hi-N+1 ) .

(1.18) In 𝔤(A) we have

(adei) 1-aij ej=0
(adfi) 1-aij fj=0

whenever ij.


It is enough to prove one of these relations, because the other then follows by applying the involution ω. Let fij= (adfi) 1-aij fj . By (1.8), in order to show that fij=0 it is enough to show that

[ ek,fij ] =0 (1kn) .

There are 3 cases to consider:

  1. ki,kj. Then ek commutes with fi and fj (1.2), hence with fij.
  2. k=j,ki. Then ej commutes with fi, hence

    [ ej,fij ] = (adfi) 1-aij [ ej,fj ] = (adfi) -aij [ fi,hj ] = aji (adfi) -aij fi.

    If aij0 this is zero, whilst if zij=0 then aji=0 (by (c)), so again it is 0.

  3. k=i,kj. We have, using the formula (1.17)

    [ ei,fij ] = (adei) (adfi) 1-aij fj = ( ad fi1-aji ) (adei) fj+ (1-aij) (adfi) -aji ( (hi)fj + aijfj ) = 0by (1.2).

Remark For an arbitrary Cartan matrix A, it is still an open question whether the relations (1.2) together with (1.17) are a complete set of defining relations for the algebra 𝔤(A): or, equivalently, whether the left sides of the relations (1.18) generate the ideal 𝔯 in 𝔤(A). At any rate this is known to be true (proof later, perhaps) if A is symmetrizable.

In general, a derivation d of a Lie algebra 𝔤 is said to be locally nilpotent if for each x𝔤 there exists a positive integer N(x) such that dN(x)x=0 : i.e. if for each x𝔤 is killed by some power of d. In that case ed:𝔤𝔤 is well defined, because the series

ed(x)x n0 dnx n!

terminates for each x𝔤. The Leibniz formula shows that ed[x,y]= [ edx,edy ] and hence that ed is an automorphism of the Lie algebra 𝔤 (with inverse what does this say??).

(1.19) adei and adfi (1in) are locally nilpotent derivations of 𝔤(A) (and of 𝔤(A)). (Consequently eadei, eadfi are automorphisms of 𝔤(A) and of 𝔤(A).)


It is enough to consider adei=φ, say. Let α be the subspace of 𝔤(A) consisting of all x𝔤(A) killed by some power of φ. Since φ is a derivation, 𝔞 is a subalgebra of 𝔤, by virtue of the Leibnitz formula: if φrx=0 and φsy=0, then φr+s-1 [x,y]=0 . Hence to show that 𝔞=𝔤(A) it is enough to show that the generators ej,fj,h𝔥 belong to 𝔞.

For ej this follows from (1.18). for h𝔥 we have

φ(h)= [ei,h]= -αi(h)ei (1.2)

whence φ2(h)=0. Finally, for fj we have φ(fj)= [ei,fj]=0 if ji, and φ(fi)=hi, φ2(fi) [ei,hi]=-2 ei, φ3(fi)=-2 [ei,ei]=0 .

The Lie algebra defined by a principal submatrix

Let A= (aij) 1i,jn be any n×n matrix with entries in k. For any non-empty subset J of {1,2,,n} let

AJ= (aij) i,jJ

be the principal submatrix defined by the subset J. Write

nJ=Card(J), lJ=rank(AJ) .

We wish to see how 𝔤(AJ) is related to 𝔤(A).

Let (𝔥,B,B) be a minimal realization of A, so that dim𝔥=2n-l=N say. Let

𝔥J= jJkhj
𝔠J= jJ Ker(αj)

which are subspaces of 𝔥.

(a) Let V be a vector subspace of 𝔥. Then the restrictions αj|V(jJ) are linearly independent (as linear forms on V) iff V+𝔠J=𝔥.


Take annihilators: V+𝔠J=𝔥 V0𝔠J0=0 (in 𝔥*).

But 𝔠J0 is the subspace of 𝔥* spanned by the αj(jJ), whence the result.

Note that 𝔠J= hαj;jJ 0 .

(b) Let 𝔥J be minimal among subspaces V of 𝔥 satisfying (i) V𝔥J; (ii) V+𝔠J=𝔥. Let check the following

BJ= { αj|𝔥J: jJ } , BJ= {hj:jJ}

Then ( 𝔥J,BJ, BJ ) is a minimal realization of AJ.


By (a), the elements of BJ are linearly independent in 𝔥J*. It remains to show that dim𝔥J= 2nJ-lJ .

Let π:𝔥𝔥/𝔥J be the projection. We have

π(𝔠J)= ( 𝔠J+𝔥J ) /𝔥J𝔠J/ ( 𝔠J𝔥J )


dim ( 𝔠J 𝔥J ) = nK-lJ

just as in the proof of (1.10); also dim𝔠J= N-nJ , so that

dimπ(𝔠J)= N-2nJ+lJ=N -NJ

say where NJ=2nJ-lJ.

Clearly 𝔥J must be such that 𝔥J/𝔥J=π (𝔥J) is a vector space complement of π(𝔠J) in 𝔥/𝔥J , and therefore

dimπ(𝔥J) = dim ( 𝔥/𝔥J ) -dimπ(𝔠J) = (N-nJ)- (N-NJ)= NJ-nJ

and finally dim𝔥J= nJ+ ( NJ-nJ ) =NJ .

(c) From (1.3) we have

𝔤(AJ)= 𝔫 J,- 𝔥J 𝔫 J,+

where 𝔫 J,+ (resp. 𝔫 J,- ) is the free Lie algebra generated by the ej,jJ (resp. by the fj,jJ). Hence 𝔤(AJ) is a subalgebra of 𝔤(A), and if we put QJ= jJ αj we have

𝔤(AJ)= 𝔥J+ βQJ β0 𝔤 β

with components 𝔤β ( βQJ,β0 ) the same as those in 𝔤(A) .

(d) Let 𝔯J be the unique largest ideal in 𝔤(AJ) satisfiying 𝔯J𝔥J=0 (so that 𝔤(AJ)=𝔤 (AJ)/𝔯J ). Then

𝔯J=𝔤(AJ) 𝔯

and hence

𝔯J= βQJ 𝔯β


Let 𝔯J=𝔤 (AJ)𝔯= 𝔯β . this is an ideal in 𝔤(AJ) which intersects 𝔥J trivially, hence certainly 𝔯J𝔯J .

Conversely, let φ:𝔤(AJ) 𝔤(A) 𝔤(A) , so that Ker(φ)=𝔯J . Let x𝔯J, where βQJ; I claim that φ(x)=0 in 𝔤(A). Suppose for example β>0, and proceed by induction on m=ht ( β ) . If m=1 then x=0 (1.6), so certainly φ(x)=0. If m>1, consider [ φ(x), fi ] . There are two cases:

  1. if iJ, then [ φ(x), fi ] =φ ( [x,fi] ) =0 by the inductive hypothesis, because [x,fi] 𝔯 J,β-αi and ht(β-αi) =m-1 ;
  2. if iJ, then since φ(x)𝔤β we have [ φ(x),fi ] 𝔤 β-αi ; but β-αi is not a root, because β= jJ mjαj (say) and iJ. Hence [ φ(x),fi ] =0 in both cases, and therefore by (1.8) φ(x)=0.
  3. Likewise if β<0. It follows that 𝔯J Ker(φ)= 𝔯J .

(e) From (d) it follows that the embedding of 𝔤(AJ) in 𝔤(A) induces an embedding of 𝔤(AJ) in 𝔤(A). We have

𝔤(AJ)=𝔥J+ βQJ β0 𝔤β /𝔯β

from (c) and (d); but 𝔤β /𝔯β=𝔤β (summand of 𝔤(A)). Hence

(1.20) 𝔤(AJ)=𝔥J+ βQJ β0 𝔤β

Hence is R (resp. RJ) is the set of roots of 𝔤(A) (resp. 𝔤(AJ)) we have RJ=RQJ, and the multiplicity of βRJ a root of 𝔤(AJ) is the same as its multiplicity as a root of R.


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

Version: October 30, 2001

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