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Contemporary MathematicsVolume 00, 0000

An Exposition of Generalized Kac-Moody algebras

Elizabeth Jurisich

Abstract. We present a detailed exposition of the theory of generalized Kac-Moody algebras associated to symmetrizable matrices. A proof of the character

formula for a standard module is given, generalizing the argument of Garland

and Lepowsky for the Kac-Moody case. A short proof of the theorem that anygeneralized Kac-Moody algebra can be decomposed into direct (vector space)

sums of free subalgebras and a Kac-Moody subalgebra is given.

Introduction

This paper provides a detailed exposition of the theory of generalized Kac-Moody algebras, whose theory was originally developed by R. Borcherds. Thetheory of generalized Kac-Moody Lie algebras has an important application inBorcherds’ proof [B3] of the “Monstrous Moonshine” conjectures of Conway andNorton for the moonshine module of [FLM], where the Monster Lie algebra (ageneralized Kac-Moody algebra) plays an important role (see also [J] and [JLW]).

It is shown in [J], that generalized Kac-Moody algebras with no mutually or-thogonal simple imaginary roots have a decomposition in terms of a Kac-Moodysubalgebra and large free subalgebras over certain modules of the Kac-Moody sub-algebra. Using this decomposition some of the theory of this type of generalizedKac-Moody algebra becomes simplified (esp. in the case of the Monster Lie alge-bra, see [J] and [JLW]). As an application of the character formula for standardmodules a short proof is provided of a more general theorem of E. Jurisich and R.Wilson [JW] decomposing any generalized Kac-Moody algebra in a similar way.

Results such as character formulas and a denominator identity known for Kac-Moody algebras are stated in [B1] for generalized Kac-Moody algebras. We foundit necessary to do some additional work in order to understand fully the precisedefinitions and also the reasoning which are implicit in Borcherds’ work on thissubject. V. Kac in [K] gives an outline (without detail) of how to rigorously de-velop the theory of generalized Kac-Moody algebras by indicating that one shouldfollow the arguments presented there for Kac-Moody algebras (this is the approach

1991 Mathematics Subject Classification. Primary 17B65, 17B67; Secondary 17B55.

c©0000 American Mathematical Society0271-4132/00 $1.00 + $.25 per page

1

2 Elizabeth Jurisich

of [HMY]). Here we develop the theory of generalized Kac-Moody algebras, gen-eralizing arguments appearing in the exposition [L] (see also [K] and [M]), andby extending the homology results of [GL] (not covered in [K]) to these new Liealgebras. The treatment here closely resembles that in [L] and [GL] and is some-times essentially identical, but here additional arguments are necessary to handlethe presence of simple imaginary roots. Proofs of the character formula for stan-dard modules and the denominator formula for a generalized Kac-Moody algebraassociated to a symmetrizable matrix are thus presented. That the results of [GL]hold for generalized Kac-Moody algebras is mentioned and used in [B3]. The workdone here generalizing [GL] has also already been carried out in the work of S.Naito [N] (where the matrices are assumed finite).

I would like to thank J. Lepowsky for his extremely helpful commentary, andfor providing some improved arguments involved in the generalizations given hereof [GL] and [L].

1. Definition of a Generalized Kac-Moody Algebra

1.1 Construction of the algebra associated to a matrix.In [B1] Borcherds defines the generalized Kac-Moody algebra associated to a

matrix, which for convenience he takes to be symmetric. Two further definitions(different from the first and from one another) are given in [B2] and [B3]. Of thesedefinitions the one given below most resembles the one appearing in [B1], wherethere is a definition in terms of generators consisting of ei’s, fi’s, for i in some indexset I, and a vector space H, and relations determined by square matrices whichagree with the ones given here. Borcherds does not divide out by the radical (thatis, the largest graded ideal having trivial intersection with h) in his definition ofgeneralized Kac-Moody algebra.

The Lie algebra denoted g′(A) in [K], which is defined from an arbitrary matrixA, is equal to the generalized Kac-Moody algebra g(A), defined below, when thematrix A satisfies conditions C1–C3 given below.

Let E be a field of characteristic 0. Let F ⊂ E be a subfield which is equal tosome subfield of R, for example E = C or R, F = R or Q. Vector spaces will beover E unless otherwise indicated.

Let I a finite or countably infinite set, which we identify with 1, 2, . . . , k orZ+ (the positive integers) for notational purposes. Let A = (aij)i,j∈I be a matrixwith entries in E.

Let gF be the free Lie algebra on the generators hi, ei, fi, where i ∈ I. Let sbe the ideal generated by the elements:

[hi, hj ]

[hi, ek]− aikek[hi, fk] + aikfk

[ei, fj ]− δijhi

for all i, j, k ∈ I.

Definition. Let g0 = gF /s.

AN EXPOSITION OF GENERALIZED KAC-MOODY ALGEBRAS 3

The free associative algebra F over the hi, ei, fi has an associative algebragrading determined by giving hi degree (0, 0, . . . ), ei degree (0, . . . 0, 1, 0, . . . ), andfi degree (0, . . . 0,−1, 0, . . . ), where ±1 appears in the ith position. Restrictingthis grading to the natural copy of gF in F gives gF a ZI -Lie algebra grading.For a sequence of integers (n1, n2, . . . ) let gF (n1, n2, . . . ) denote the correspondingsubspace of gF . Note that gF (n1, n2, . . . ) = 0 unless only finitely many of the niare nonzero.

Let Di (i ∈ I) be the linear endomorphism that acts on gF (n1, n2, . . . ) asmultiplication by the scalar ni. Then Di is a derivation. Call it the ith degreederivation of gF . Since the ideal s is homogeneous under the above grading, g0 canalso be graded by ZI , and the Di can be transferred naturally to g0. Let η be theautomorphism of order 2 of gF taking hi to −hi and interchanging ei and fi for alli ∈ I. Since η(s) = s, η is well defined on g0.

We use the same symbols hi, ei,and fi for their images in g0. Let h be theabelian subalgebra spanned by the hi for all i ∈ I. Let g+

0 be the subalgebragenerated by the ei for i ∈ I and let g−0 be the subalgebra generated by the fi fori ∈ I. The space g−0 ⊕h⊕g

+0 is a subalgebra of g0, and since g−0 ⊕h⊕g

+0 contains the

generators ei, fi, hi, it follows that g0 = g−0 ⊕h⊕g+0 . If the ni are nonnegative (resp.

nonpositive) and only finitely many are nonzero, then g0(n1, n2, . . . ) is spannedby the elements [ei1 [ei2 [· · · [eir−1

, eir ] · · · ] (respectively, [fi1 [fi2 [· · · [fir−1, fir ] · · · ])

where ej (or fj) occurs |nj | times.

Proposition 1.1. The Lie algebra g0 has triangular decomposition

g0 = g−0 ⊕ h⊕ g+0 .

The abelian subalgebra h has basis consisting of the hi, i ∈ I. Furthermore, g±0 isthe free Lie algebra generated by the ei (respectively, the fi) for i ∈ I. In particular,ei, fi, hii∈I is a linearly independent set in g0.

proof. As in the classical case, we will construct a sufficiently large represen-tation of the Lie algebra. Let hF be the span of the hi, i ∈ I. Define αj ∈ (hF )∗ asfollows :

αj(hi) = aij .

Let X be the free associative algebra on the set xii∈I . Let λ ∈ (hF )∗. A rep-resentation of gF on X is given by the following action of the generators of gF onX:

h · 1 = λ(h) for all h ∈ hFh · xi1 · · ·xir = (λ− αi1 − · · · − αir )(h)xi1 · · ·xir for all h ∈ hFfi · 1 = xifi · xi1xi2 · · ·xir = xixi1xi2 · · ·xirei · 1 = 0ei · xi1 · · ·xir = xi1ei · xi2 · · ·xir + δii1(λ− αi2 − · · · − αir )(hi)xi2xi3 · · ·xir .

To show that X can be regarded as a g0-module we verify that the ideal s annihilatesthe module X. It is clear that [h, h′] is 0 on X for h, h′ ∈ h. The element [ei, fj ]−


δijhi also acts as 0 on X by the following computation :

[ei, fj ] · xi1 · · ·xir = eifj · xi1 · · ·xir − fjei · xi1 · · ·xir= ei · xjxi1 · · ·xir − xjei · xi1 · · ·xir= δij(λ− αi1 − · · · − αir )(hi)xi1 · · ·xir= δijhi · xi1 · · ·xir .

By a similar computation [hi, fk] + aikfk annihilates X. Now consider theaction of [hi, ek]− aikek on X:

[hi, ek] · 1 = hiek · 1− ekhi · 1= ek · λ(hi)1 = 0

and

aikek · 1 = 0.

Thus [hi, ek]− aikek annihilates 1. Furthermore, using the above results

[[hi, ek]− aikek, fl] = [[hi, ek], fl]− [aikek, fl]

= [ek[fl, hi]] + [hi[ek, fl]]− [aikek, fl]

= 0.

Thus [hi, ek]−aikek commutes with the action of fl for all l. Since any xi1 · · ·xir =fi1 · · · fir ·1, this means that [hi, ek]−aikek annihilates X. Thus X can be regardedas an g0-module. That the hi’s are linearly independent follows from the firstrelation, since λ ∈ hF is allowed to vary.

From the definition of the module we see that U(g−0 ) generates the regularrepresentation of X, so there is a homomorphism (of associative algebras):

U(g−0 )→ X

fi 7→ xi.

If gX is the free Lie algebra generated by xii∈I then there is an associative algebrahomomorphism:

U(gX)Φ→U(g−0 )

xi 7→ fi.

But U(gX) = X, so the map Φ is an isomorphism. Thus gX is isomorphic to g−0 ,the Lie subalgebra in U(g−0 ) = U(gX) generated by the fi, i ∈ I. Applying η showsthat g+

0 is the free Lie algebra generated by eii∈I . Thus the set of ei’s (and fi’s)are linearly independent, and since g0 = g−0 ⊕ h⊕ g+

0 any finite collection of the hi,ei and fi for i ∈ I together is linearly independent.

Definition. Let ui denote the subalgebra of g0 spanned by the elements hi, ei,and fi.


Remark. If aii ∈ E is nonzero, then the Lie algebra ui is isomorphic to sl2.This is easily seen by rescaling hi and ei, so that

hi =2hiaii

, ei =2eiaii

.

If aii = 0 then ui is a Heisenberg algebra.

Assume that we have a distinguished subset I0 ⊂ I such that for all i ∈ I0,1. aii 6= 02. 2aij/aii ∈ −N (i 6= j ∈ I)3. aij = 0 implies aji = 0 for all j ∈ I.For all i 6= j and i ∈ I0 define

d+ij = (ad ei)

−2aijaii

+1ej ∈ g+

0

d−ij = (ad fi)−2aijaii

+1fj ∈ g−0

Proposition 1.2. In g0, for all i, j, k ∈ I with i ∈ I0 and i 6= j, we have[ek, d

−ij ] = 0 and [fk, d

+ij ] = 0.

proof. By the above remark, spanhi, ei, fi = sl2. It is sufficient to show[ek, d

−ij ] = 0:

First assume k 6= i and k 6= j.

(1.1) (ad ek)(ad fi)x = [fi, [ek, x]] = (ad fi)(ad ek)x

so

[ek, (ad fi)−2aijaii

+1fj ] = (ad fi)

−2aijaii

+1[ek, fj ] = 0

Now assume k = i. Consider the sl2-module generated by the highest weightvector fj , with the adjoint action. Let n = 2aij/aii,

(ad ei)(ad fi)1−nfj

=

(ad fi)(ad ei)fj + (adhi)fj , if n = 0

(ad fi)1−n(ad ei)fj + (−n+ 1)(ad fi)

−n(adhi)fj+ 1

2 (−n+ 1)(−n)(ad fi)−n−1(ad (−aiifi))fj , if n < 0

=

[hi, fj ] if n = 0

(−n+ 1)(−aij)(ad fi)−nfj

+aii2 (−n+ 1)(n)(ad fi)

−nfj if n < 0.

By the definition n = 0 if and only if aij = 0, furthermore aii2 (n) = aij . Thus

both of the above terms are zero, and ad ei annihilates (ad fi)−2aijaii

+1fj .

Finally assume k = j. By equation (1.1),

(1.2) [ej , (ad fi)−2aijaii

+1fj ] = (ad fi)

−2aijaii

+1hj

Since [fi, hj ] = ajifi, it follows immediately that (1.2) is zero if 2aij/aii < 0. Ifaij = 0 then aji = 0 and (1.2) is also zero.


Let k±0 be the ideal of g±0 , respectively, generated by the elements:

d±ij for all i ∈ I0, i 6= j ∈ I[ei, ej ] if aij = aji = 0 for k+0

[fi, fj ] if aij = aji = 0 for k−0 .(1.3)

Note that the last elements are not redundant if i ∈ I\I0.Let n1 = g+

0 /k+0 and n−1 = g−0 /k

−0 . Define k0 = k+0 ⊕ k−0 .

Proposition 1.3. The subalgebras k±0 and k0 are ideals of g0.

proof. Let g0 act on itself by the adjoint representation. To see that k−0 is anideal, first consider the action on the generators d−ij . By Proposition 1.1:∑

i 6=j

U(g0) · d−ij =∑i 6=j

U(g−0 )U(h)U(g+0 ) · d−ij

=∑i 6=j

U(g−0 ) · d−ij ⊂ k−0 .

The last equality holds by Proposition 1.2 and the fact that for h ∈ h,

[h, (ad fi)Nfj ] = λ(ad fi)

Nfj

where λ is a scalar.We must also consider the generators of the form (1.3). By the Jacobi identity,

and the assumption that aij = aji = 0:

[ek[fi, fj ]] = [δkihi, fj ] + [fi, δkjhj ]

= −δkiaijfj + δkjajifi

= 0.

Thus we can use the same argument appearing above for d−ij to conclude∑i6=j

aij=aji=0

U(g0) · [fi, fj ] ⊂ k−0 .

By a symmetric argument, k+0 is an ideal of g0, therefore k0 = k+0 ⊕ k−0 is also anideal.

Define the Lie algebra g1 = g0/k0. The Lie algebra g1 is given by the cor-responding generators and relations. Since the ideal k0 is homogeneous, g1 has aZI -gradation induced by the given grading of g0, and η is well defined on g1. Weuse the same letters ei, fi, hi and h to denote their images in g1. The Lie algebra g1

has a unique graded ideal, r1, maximal among graded ideals not intersecting h. Theideal r1 does not intersect the span of the hi, ei, and fis. The fact that η(r1) ⊂ r1is easy to verify.


Definition. The Lie algebra g = g(A) is defined to be g1/r1.

We use the same letters hi, ei, fi, ui and h for their images in g. The map ηand the Di’s are well defined on g. It is still true that if the ni’s are all nonnegative(resp. nonpositive), and only finitely many of the ni are nonzero, then g(n1, n2 . . . )is spanned by the elements:

[ei1 [ei2 [· · · [eir−1 , eir ] · · · ]

(resp. [fi1 [fi2 [· · · [fir−1, fir ] · · · ])

where ej (resp. fj) occurs |nj | times. The space g(0 . . . , 0,±1, 0, . . . ) where ±1occurs in the ith position is spanned by ei (resp. fi).

Define the subalgebras

n+ = n = n1/(n1 ∩ r1) and n− = n−1 /(n−1 ∩ r1).

Then, since g0 = g−0 ⊕h⊕g+0 , we have the triangular decomposition g1 = n−1 ⊕h⊕n1,

and therefore g = n− ⊕ h ⊕ n, using the same notation for h in the Lie algebrasg0, g1 and g.

Proposition 1.4. For each i ∈ I0, g is a direct sum of finite-dimensionalirreducible ui-modules.

proof. Since ui is isomorphic to sl2 the usual proof for Kac-Moody algebrasworks in this case.

1.2 Definition of a generalized Kac-Moody algebra.First we construct the extended Lie algebra ge. Let d0 denote the (possibly

infinite-dimensional) space of commuting derivations of g spanned by the Di fori ∈ I. Let d be a subspace of d0. Form the semidirect product Lie algebra ge = dng.Then he = dn h is an abelian subalgebra of ge which acts via scalar multiplicationon each space g(n1, n2, . . . ).

Let αi ∈ (he)∗, i ∈ I, be defined by the conditions:

[h, ei] = αi(h)ei for all h ∈ he.

Note that αj(hi) = aij for all i, j ∈ I. Choose d so that the αi for i ∈ I are linearlyindependent. This is possible because this condition holds for d = d0. If t is ad-invariant subalgebra of g, denote by te the subalgebra dn t of ge.

For all ϕ ∈ (he)∗ define

gϕ = x ∈ g|[h, x] = ϕ(h)x for all h ∈ he.

If ϕ,ψ ∈ (he)∗ then [gϕ, gψ] ⊂ gϕ+ψ. By definition ei ∈ gαi , fi ∈ g−αi for all i ∈ I,so that if all ni ≤ 0 or all ni ≥ 0 (only finitely many nonzero) it is clear that

g(n1, n2, . . . ) ⊂ gn1α1+n2α2+···.

In fact, since the αi for i ∈ I are linearly independent

gn1α1+n2α2+··· = g(n1, n2, . . . )

and g0 = h. Therefore,


(1.4) g =∐

(n1,n2,... )∈ZIgn1α1+n2α2+···.

Remark. The center of g is contained in h. We observe that gnαi = 0 unlessn = ±1.

Definition. The roots of g are the nonzero elements ϕ of (he)∗ such thatgϕ 6= 0. The elements αi are the simple roots. Let ∆ be the set of roots, ∆+ theset of roots which are non-negative integral linear combinations of αi’s. The rootsin ∆+ are called the positive roots. Let ∆− = −∆+ be the set of negative roots.All of the roots are either positive or negative.

The elementary properties of ge and g are summarized in the following propo-sition.

Proposition 1.5. The Lie algebra g has the following properties:1. g = n− ⊕ h⊕ n+

2. h = g0 and h has a basis hi, i ∈ I3. n+ =

∐ϕ∈∆+

gϕ and n− =∐ϕ∈∆−

gϕ

4. ∆ = ∆+

⋃∆− (disjoint union)

5. [gϕ, gψ] ⊂ gϕ+ψ for all ϕ,ψ ∈ (he)∗

6. η(gϕ) = g−ϕ for all ϕ ∈ (he)∗, and dim gϕ = dim g−ϕ <∞7. g±αi has basis ei resp., fi for all i ∈ I8. For each i ∈ I0 every ad ui-stable subspace of g is a direct sum of finite-

dimensional ui-modules.9. αj(hi) = aij for all i, j ∈ I

10. Every nonzero ideal of ge meets he.11. If the (possibly infinite) matrix A′ is obtained from the matrix A by a permu-

tation of the rows and a corresponding permutation of columns, then thereis a natural isomorphism g(A) ∼= g(A′).

12. If A has the form

(B 00 C

), there is a natural isomorphism g(A) ∼= g(B)×

g(C).

For the definition of generalized Kac-Moody algebra we assume that A =

(aij)i,j∈I is a matrix with entries in F, satisfying the following conditions:(C1) A is symmetrizable.(C2) If aii > 0, then aii = 2. For all k 6= j, ajk ≤ 0, and for all i ∈ I0, aij ∈ −N

(j, k ∈ I).(C3) If aij = 0, then aji = 0 (i, j ∈ I).

(Recall a matrix A is called symmetrizable if there are positive nonzero numbersqi ∈ F, i ∈ I, such that DA is symmetric, where D is the diagonal matrix withentries qi. Note that this condition implies (C3).)

We also assume that I0 is chosen to be as large as possible, so that I0 = i ∈I | aii > 0.

Definition. The generalized Kac-Moody (Lie) algebra associated to the ma-trix A is g = g(A).


1.3 An invariant bilinear form on ge.

Definition. We define a symmetric bilinear form (·, ·) taking values in E onR by defining (αi, αj) = qiaij and extending linearly to any element of ∆ (or R).Notice that on ∆ the form (·, ·) actually takes values in F.

For all i ∈ I, let xαi = qihi ∈ h. For ϕ ∈ R with ϕ =∑aiαi, define

xϕ =∑aixαi ∈ h. We can extend the E-valued symmetric form (·, ·) defined on R

to a symmetric form on (he)∗ satisfying the condition:

(λ, αi) = λ(xαi) for all αi and λ ∈ (he)∗.

Fix such a form.The map given by ϕ 7→ xϕ defines a linear isomorphism from R to h. The

bilinear form on R can be transferred to a form (·, ·)h on h by letting (xαi , xαj )h =(αi, αj) = qiaij for all i, j ∈ I. Note that (xψ, xϕ)h = (ψ,ϕ) = ψ(xϕ) = ϕ(xψ),since this is true for the xαi , i ∈ I.

To extend the above form to a symmetric bilinear form on he = d ⊕ h, let(Di, xαj )he = αj(Di) = δij for all i, j ∈ I, and pick an arbitrary symmetric bilinearform on d. This determines a form (·, ·)he on he that satisfies

(1.5) (h, xϕ)he = ϕ(h)

for all h ∈ he and ϕ ∈ R.

Remark. Note that x ∈ h is determined by the values (h, x)he as h rangesthrough he.

Recall that a symmetric bilinear form (·, ·) on a Lie algebra g is called invariantif

([a, b], c) = −(b, [a, c])

for all a, b, c ∈ g.The bilinear form on he will be extended to a unique ge-invariant bilinear form

by using induction.

Theorem 1.6. There is a unique invariant symmetric bilinear form (·, ·) onge which extends (·, ·)he . For all ϕ,ψ ∈ ∆ such that ϕ+ ψ 6= 0,

(gϕ, gψ) = (he, gϕ) = 0

and[a, b] = (a, b)xϕ

for all a ∈ gϕ and b ∈ g−ϕ. In particular, [gϕ, g−ϕ] ⊂ Exϕ. Moreover, (ei, fj) =1qiδij for all i, j ∈ I.

Proposition 1.7. A form (·, ·) on ge that extends (·, ·)he is he-invariant if andonly if for all ϕ,ψ ∈ ∆ such that ϕ+ ψ 6= 0, we have

(gϕ, gψ) = (he, gϕ) = 0.

proof. Given α, β ∈ ∆∪0, h ∈ he, and a ∈ gα, b ∈ gβ . If (·, ·) is he-invariant,then ([h, a], b) = −(a, [h, b]). Therefore, (α + β)(h)(a, b) = 0. If α + β 6= 0, chooseh such that (α+ β)(h) 6= 0 so that (a, b) = 0.


Proposition 1.8. If ge has an invariant bilinear form (·, ·) which extends(·, ·)he then it is unique. For all ϕ ∈ ∆, a ∈ gϕ, and b ∈ g−ϕ,

[a, b] = (a, b)xϕ.

In particular, the existence of (·, ·) implies that [gϕ, g−ϕ] ⊂ Exϕ.

proof. By Proposition 1.7 it is sufficient to show that [a, b] = (a, b)xϕ. Leth ∈ he. Then

(h, [a, b])he = (h, [a, b]) = ([h, a], b)

= ϕ(h)(a, b)

= (h, xϕ)he(a, b)

= (h, (a, b)xϕ)he .

Thus (·, [a, b] − (a, b)xϕ) corresponds to an element of R∗ which vanishes on allh ∈ he, so we have [a, b] = (a, b)xϕ.

Consider the following grading of ge: For ϕ ∈ R such that ϕ =∑i∈I niαi

where ni ∈ Z, define the height ht (ϕ) to be the integer∑i∈I ni. If ϕ ∈ ∆+ then

ht (ϕ) > 0, and if ϕ ∈ ∆− then ht (ϕ) < 0. For all n ∈ Z+, define

∆(n) = ϕ ∈ ∆ | |ht (ϕ)| ≤ n.

The subspace ge(n) of ge is defined by:

ge(n) = he ⊕∐

ϕ∈∆(n)

gϕ.

A symmetric bilinear form (·, ·) on ge(n) is called invariant if for all ϕ1, ϕ2, ϕ3 ∈∆(n) ∪ 0 such that |ht (ϕ1 + ϕ2)| ≤ n and |ht (ϕ1 + ϕ3)| ≤ n, we have

([a1, a2], a3) = −(a2, [a1, a3])

for all ai ∈ gϕi or in he if ϕi = 0.The invariant form will now be defined inductively on ge(n). Define (·, ·)1 on

ge(1) by letting (·, ·)1 = (·, ·)he on he, and by the conditions

(ei, ej)1 = (fi, fj)1 = 0 for all i, j ∈ I,

(h, gϕ) = 0 for all h ∈ he, ϕ ∈ ∆(1),

and

(ei, fj)1 = δij1

qifor all i, j ∈ I.

Then (·, ·)1 is invariant, and clearly satisfies [a, b] = (a, b)1xϕ for all ϕ ∈ ∆(1),a ∈ gϕb ∈ g−ϕ.

Now assume that n > 0 and that (·, ·)n is an invariant form on ge(n) whichextends (·, ·)he . We will extend the form (·, ·)n to an invariant form on the spacege(n+ 1).


Lemma 1.9. For all ϕ,ψ ∈ ∆(n), (gϕ, gψ)n = 0 unless ϕ + ψ = 0 and(he, gϕ)n = 0. Furthermore,

[a, b] = (a, b)nxϕ for all a ∈ gϕ, b ∈ g−ϕ.

proof. See the proofs of Propositions 1.7 and Proposition 1.8.

Lemma 1.10. For all ϕ ∈ ∆(n+ 1), [gϕ, g−ϕ] ⊂ Exϕ.

proof. By Lemma 1.9, we may assume ϕ ∈ (∆(n+ 1)\∆(n))∩∆+. Let a ∈ gϕ

and b ∈ g−ϕ. Without loss of generality we may assume, for some i, j ∈ I,

a = [ei, a′]

b = [fj , b′]

where a′ ∈ gϕ−αi and b′ ∈ g−ϕ+αi (recall that gϕ is spanned by elements of theform [ei1 [ei2 [· · · [eir−1

, eir ] · · · ]]] with∑αim = ϕ).

Then

[a, b] = [[[ei, a′], fj ], b

′] + [fj , [[ei, a′], b′]]

= [[[ei, fj ], a′], b′] + [[ei, [a, fj ]], b

′] + [fj , [[ei, b′], a′]] + [fj , [ei, [a

′, b′]]]

= δij(ϕ− αi)(hi)[a′, b′] + ([ei, [a′, fj ]], b

′)nxϕ−αj

+ (fj , [[ei, b′], a′])nx−αj + δijαi([a

′, b′])hi

= δij(a′, b′)n (ϕ− αi)(hi)xϕ−αi + αi(xϕ−αi)hi

+ ([ei, [a′, fj ]], b

′)nxϕ + ([ei, [a′, fj ]], b′)n + (fj , [[ei, b′], a′])nx−αj .

Considering the first term, we see

(ϕ− αi)(hi)xϕ + (hi, xϕ−αi)x−αi + (xϕ−αi , xαi)hi

= (ϕ− αi)(hi)xϕ,

and by the invariance of (·, ·)n,

([ei, [a′, fj ]], b

′)n + (fj , [[ei, b′], a′])n = 0.

Proof of Theorem 1.6. By Lemmas 1.9 and 1.10, we can define (·, ·)n+1

to be the symmetric form on ge(n + 1) which extends (·, ·)n, such that [a, b] =(a, b)n+1xϕ for all a ∈ gϕ, b ∈ g−ϕ with ϕ ∈ ∆(n + 1), and such that for ϕ,ψ ∈∆(n+ 1),

(gϕ, gψ)n+1 = 0

unless ϕ+ ψ = 0, and (he, gϕ) = 0.All that remains to be shown in order to prove Theorem 1.6 is that this form

(·, ·)n+1 is invariant. Since no confusion will arise, we now write (·, ·)n+1 = (·, ·).Let ϕ1, ϕ2, ϕ3 ∈ ∆(n+ 1) ∪ 0 such that |ht (ϕ1 + ϕ2)| ≤ n+ 1 and |ht (ϕ1 +

ϕ3)| ≤ n+ 1.We claim that if ai ∈ gϕi or he (in the case ϕi = 0), for i = 1, 2, 3, then

(1.6) ([a1, a2], a3) = −(a2, [a1, a3]).


To prove this claim, we may assume that ϕ1 + ϕ2 + ϕ3 = 0, and by the proofsof Propositions 1.7 and 1.8 that each ϕi 6= 0.

First suppose ϕ1 and ϕ2 are linearly independent. Let a1 = [a2, a3], a−2 =[a1, a3] and a−3 = [a1, a2], so that ai ∈ g−ϕi (i = 1, 2, 3). Then

[a1, a1] = [a−3, a3] + [a2, a

−2],

so that,

(1.7) (a1, a1)xϕ1

= (a−3, a3)x−ϕ3+ (a2, a

−2)xϕ2.

But x−ϕ3= xϕ1

+ xϕ2by the assumption ϕ1 + ϕ2 + ϕ3 = 0 and so (1.7) gives(

(a1, a1)− (a−3, a3)

)xϕ1

=((a2, a

−2) + (a−3, a3))xϕ2

.

By the definition of xϕ we have((a1, a

1)− (a−3, a3))ϕ1 =

((a2, a

−2) + (a−3, a3))ϕ2.

Thus linear independence of ϕ1 and ϕ2 gives:

(a2, a−2) + (a−3, a3) = 0

which is equivalent to equation (1.6).Now suppose ϕ1 and ϕ2 are linearly dependent. Then for each i = 1, 2, 3 and

j ∈ I, ϕi /∈ Eαj . Assume that ϕi0 is positive for i0 one of 1, 2, 3 and is negativeotherwise; this is enough because the automorphism η can be applied. We may alsoassume that ht (ϕi0) = n+ 1, so that |ht (ϕi)| ≤ n for i 6= i0. Since the roles of ϕ2

and ϕ3 are symmetric it is enough to show the claim for i0 = 1 or 2. We may alsoassume that ai0 = [ek, a

′i0

] for some k ∈ I and some a′i0 ∈ gϕi0−αk .First let i0 = 1. Then

([a1, a2], a3) + (a2, [a1, a3]) = ([[ek, a′1], a2], a3) + (a2, [[ek, a

′1], a3])

= ([[ek, a2], a′1], a3) + ([ek, [a′1, a2]], a3)

+ (a2, [[ek, a3], a′1]) + (a2, [ek, [a′1, a3]]).

By the invariance of (·, ·)n this expression is zero.Now suppose that i0 = 2. Then

(1.8) ([a1, a2], a3) + (a2, [a1, a3]) = ([a1, [ek, a′2]], a3) + ([ek, a

′2], [a1, a3]).

Because αk and ϕ2 − αk are linearly independent, the previous argument appliesto the second term on the right. Using invariance of (·, ·)n

([ek, a′2], [a1, a3]) = −(a′2, [ek, [a1, a3]])

= −(a′2, [[ek, a1], a3])− (a′2, [a1, [ek, a3]])

= −([a′2, [ek, a1]], a3)− ([a′2, a1], [ek, a3])

= −([[a′2, ek], a1], a3)− ([ek, [a′2, a1]], a3)− ([a′2, a1], [ek, a3])

= −([[a′2, ek], a1], a3).

Thus equation (1.8) equals zero and we are done.


1.4 The radical of g1 is zero.The theorem below follows from Theorem 1 in [GK] or Proposition 9.11 in [K],

and a lemma in [B1].The only generalization of [K] being that g0 has a countablenumber of generators. This theorem is also proven in [HMY] in the same manner.

The Lie algebra g0 defined in §1.1, which is denoted g(A) in [K], can be gradedas in [K] by the lattice Q =

∑i∈I Zαi (agreeing with the ZI -grading in §1.1), and

g1 has an induced Q-grading, we will call this the root grading of g1. Denote byΠ the set of simple roots of g0. Let r0 be the radical of g0. Then note that if πis the natural map π : g0 → g1 then π(r0) = r1. Let r±0 = r0 ∩ n±0 , also denote(ri)

α = ri ∩ gαi for α ∈ Q, i = 0, 1. Let Q+ =∑i∈I Nαi.

Theorem 1.11. The ideal r1 ⊂ g1 is zero.

We restate the following result as it appears in [K], where ρ ∈ h∗ satisfies(ρ, αi) = 1

2 (αi, αi):

Lemma 1.12. The ideal r+0 (resp. r−0 ) is generated as an ideal in n+0 (resp.

n−0 ) by those (r0)α (resp. (r0)−α) for which α ∈ Q+\Π and 2(ρ, α) = (α, α).

Lemma 1.13. Let α =∑i∈I kiαi, where ki ∈ N, be such that (g1)α 6= 0. Then

supp(α) cannot be written as a disjoint union, S1 ∪ S2, of two proper subsets suchthat (αi, αj) = 0 for all i ∈ S1, j ∈ S2.

proof. Given α =∑i∈I kiαi as in the lemma, we know that gα1 is spanned by

elements of the form [ei1 , [ei2 [· · · [ein−1, ein ]] · · · ]. The lemma is obvious if ht(α) < 2.

If ht(α) ≥ 2 the lemma is proven by induction on ht(α). Suppose ht(α) = 2, andthat supp(α) = S1 ∪ S2 where S1 and S2 are as in the lemma. Any [ei1 , ei2 ] ∈ gα

is zero because we can assume i1 ∈ S1 and i2 ∈ S2 so that (αi1 , αi2) = qi1ai1i2 = 0thus α is not a root. If the lemma is true for ht(α) = n− 1, then ht(α) = n followsfrom the Jacobi identity.

The following lemma appears in [B1].

Lemma 1.14. Let α be an element of Q+ \ 0 such that (α, αi) ≤ 0 for alli ∈ I0. Then (α, α) − 2(α, ρ) ≤ 0 and equality holds if and only if α = αi where(αi, αi) = 0.

proof. Let α =∑i∈I kiαi, ki ≥ 0. Then

(1.9) (α, α)− 2(α, ρ) =∑i∈I

ki(αi, α− αi).

If αi is such that (αi, αi) ≤ 0, since aij ≤ 0 for i 6= j it follows that

ki(αi, α− αi) ≤ 0.

If αi is such that ki 6= 0 and (αi, αi) > 0, then −ki(αi, αi) < 0 and since (α, αi) ≤ 0we conclude ki(α, αi) ≤ 0. Thus equation (1.9) implies the desired inequality.

For equality to hold all the αi must satisfy (αi, αi) ≤ 0 and (αi, α − αi) = 0.Thus ∑

i 6=j

kj(αi, αj) + (ki − 1)(αi, αi) = 0.


Since all of the above summands are nonpositive, we conclude that (αi, αj) = 0 ifkj 6= 0, and if ki > 1 then (αi, αi) = 0. Thus by Lemma 1.13, equality holds if andonly if α = αi where (αi, αi) = 0.

Proof of Theorem 1.11. (See the proof of Theorem 2 in [GK] or Theorem9.11 in [K]). Let r1 =

∐α∈Q rα1 be the grading induced by the Q-grading on g1.

If i ∈ I0 there exists automorphisms of Q given by

ri(αj) = αj − aijαi

and for each ri there exists ri ∈ Aut(g1) such that

ri(gα1 ) = griα1 .

Assume that r+1 6= 0, then we can choose α =∑i∈I kiαi ∈ Q+ \ 0 such that

rα1 6= 0 and ht (α) is minimal. Thus ht (ri(α)) ≥ ht (α) for all i ∈ I0. This meansthat (α, αi) ≤ 0 for all i ∈ I0. Hence we can apply Lemma 1.14. Now consider(α, α)−2(α, ρ), by Lemma 1.12, we have (α, α)−2(α, ρ) = 0. By Lemma 1.14 sucha root must be equal to one of the αi. This is a contradiction because Lemma 1.12states that α 6= αi. That r−1 = 0 follows from applying η.

2. The Root System

2.1 The Weyl group.Because the reflections generating the Weyl group are defined only with respect

to each αi with i ∈ I0, in this section much of our attention is restricted to these“real” simple roots αi. As a result, many of the arguments are unchanged from thesemisimple case (except that we allow infinitely many reflections ri). We includethese proofs for the sake of completeness.

Let R be the E-linear subspace of (he)∗ spanned by ∆ , so that R has basisαi | i ∈ I. For each i ∈ I0 define the linear transformation ri : R→ R by:

riαj = αj − aijαi for all j ∈ I.

Equivalently:

riϕ = ϕ− ϕ(hi)αi for all ϕ ∈ R.It is clear that riαi = −αi, and ri acts as the identity on the subspace

ϕ ∈ R | ϕ(hi) = 0. The Weyl group, W , is defined to be the group of linearautomorphisms of R generated by the reflections ri (i ∈ I0). Note that, unlikeKac-Moody algebras, generalized Kac-Moody algebras may have W generated bya set of cardinality much smaller than the number of αi (i ∈ I).

Definition. A root, α, is defined to be real if (α, α) > 0 and imaginary if(α, α) ≤ 0. Let ∆R denote the set of real roots and ∆I the set of imaginary roots.If a root α =

∑i∈I niαi, ni ∈ Z, then define the height ht (α) =

∑i∈I ni.

Proposition 2.1. W preserves ∆ , and in fact dim gϕ = dim gwϕ for w ∈Wand ϕ ∈ ∆.


proof. It is sufficient to prove the Proposition for a reflection ri. Since i ∈ I0(so αi is real and ui is isomorphic to sl2), we can apply Proposition 1.5 (8) to theui-module

∐n∈Z g

ϕ+nαi .

Proposition 2.2. For all i ∈ I0, the reflection ri permutes the elements of∆+\αi.

proof. Let ϕ ∈ ∆+\αi, then ϕ =∑j∈I njαj where nj ∈ Z (only finitely

many nj are nonzero) and for some j0 6= i, nj0 > 0. (Such a j0 exists becausegnαi 6= 0 means n = ±1.) Then

riϕ = ϕ− ϕ(hi)αi.

Thus

riϕ =

∑j 6=i

njαj

+ (ni − ϕ(hi))αi,

and since riϕ ∈ ∆ and nj0 > 0, it follows that riϕ ∈ ∆+.

Proposition 2.3. For all ϕ ∈ ∆R, dim gϕ = 1 and in fact ϕ ∈ W · αi | αi ∈∆R. Furthermore:

W∆R = ∆R(2.1)

W∆I = ∆I

∆R = −∆R(2.2)

∆I = −∆I(2.3)

W (∆I ∩∆+) = ∆I ∩∆+

proof. First we show that for all ϕ ∈ ∆R, dim gϕ = 1. Let α be an element inW ·ϕ such that |htα| is minimal. If α = ±αi for some i then dim gα = dim gαi = 1.Let α =

∑njαj . We can assume without loss of generality that α is positive, i.e.,

each nj > 0.Since α is real,

(α, α) =∑

ni(α, αi) > 0.

Therefore,

(2.4) (α, αi) > 0

for some αi. This αi must be real; otherwise, (αj , αi) ≤ 0 for all j, so∑nj(αj , αi) =

(α, αi) ≤ 0 contradicting equation (2.4).If α 6= αi, then ri reduces the height of α, since

riα = α− α(hi)αi, and α(hi) =(α, αi)

qi> 0.

Thus α cannot be of minimal height unless α = αi for some simple real root αi.This shows wαi = ϕ for some w ∈W and by Proposition 2.1 dim gϕ = 1.

It is easy to check that W preserves the form (·, ·), so W∆R = ∆R and W∆I =∆I . The next two statements, (2.2) and (2.3), are obvious from the definitions ofreal and imaginary roots. To show W (∆I ∩∆+) = ∆I ∩∆+, let ϕ ∈ ∆I ∩∆+ and


let ri be a reflection. Then as above riϕ ∈ ∆I . Since αi is real, ϕ ∈ ∆+\αi. ByProposition 2.2, riϕ ∈ ∆+ so that riϕ ∈ ∆I ∩∆+.

For all w ∈W define

Φw = ∆+ ∩ w∆− = ψ ∈ ∆+ | w−1ψ ∈ ∆−.

Note that Φw ⊂ ∆R.The following statements are obvious:

Φw ⊂ ∆R ∩∆+

Φ1 = ∅Φri = αi

Definition. Given w ∈W , let n(w) be the number of elements in Φw. Definethe length of w, denoted l(w), to be the smallest integer k such that w can be writtenas the product of k of the reflections ri, i ∈ I0. An expression w = ri1ri2 · · · rik ,ij ∈ I such that αij is real, is called reduced if k = l(w). By convention l(1) = 0.

Proposition 2.4. Let w ∈W and i ∈ I0. Then1. ri(Φw\αi) = Φriw\αi2. If αi ∈ Φw, then αi /∈ Φriw and n(riw) = n(w) − 1. If αi /∈ Φw then

αi ∈ Φriw and n(riw) = n(w) + 1.

proof. First we will show (1). Let ϕ ∈ Φw\αi. By Proposition 2.2, riϕ ∈∆+\αi. Consider (riw)−1riϕ = w−1ϕ. This is in ∆− by assumption. Thusri(Φw\αi) ⊂ Φriw\αi, but this implies ri(Φriw\αi) ⊂ Φw\αi (using riwin place of w). Thus equality holds.

For (2), let αi ∈ Φw. Then αi /∈ Φriw because (riw)−1αi = −w−1αi ∈ ∆+. By(1), n(riw) = n(w) − 1. Suppose αi /∈ Φw, then αi ∈ Φriw because (riw)−1αi =−w−1αi ∈ ∆−. Using (1) again, we conclude n(riw) = n(w) + 1.

Proposition 2.5. For all w ∈W

n(w) ≤ l(w) <∞.

proof. By induction on l(w). Suppose l(w) = 1. Then w = ri, so n(w) =1 = l(w). Assume that n(w) ≤ l(w) for all w ∈ W such that l(w) = n. Letw ∈ W such that l(w) = n + 1. Then w = riw

′ where w′ ∈ W has l(w′) = n.Then by Proposition 2.4, n(riw

′) ≤ n(w′) + 1, and by the induction assumptionn(w′) + 1 ≤ l(w′) + 1 = l(w).

We will show that in fact l(w) = n(w).

Proposition 2.6. Let w ∈ W , i ∈ I0, and suppose that wαi = αj for somej ∈ I. Then wriw

−1 = rj.


proof. (Note that rj exists because αj is necessarily real.) Let x = rjwriw−1.

Then x preserves α1, α2, . . . by the following computation: If ϕ ∈ ∆+\αj, thenfor some n ∈ Z

xϕ = rjw(w−1ϕ+ nαi) = rjϕ− nαj ∈ ∆+,

as rjϕ ∈ ∆+\αj. Moreover, x fixes αj , so x preserves ∆+. Thus x preserves∆− as well, and so x∆+ = ∆+. Since x is a linear automorphism it permutesα1, α2, . . . , the set of simple roots.

For all k ∈ I x preserves the span of αj and αk. Since xαk is a simple root ,xαk = αk.

Proposition 2.7. Let w = ri1ri2 · · · rij ,where ik ∈ I0, be a reduced expressionof w ∈W . Then

ri1ri2 · · · rij−1αij ∈ ∆+.

proof. Assume ri1ri2 · · · rij−1αij /∈ ∆+, then choose m ∈ i1, . . . , ij−1 maxi-

mal such that rimrim+1 · · · rij−1αij ∈ ∆−. Then ϕ = rim+1 · · · rij−1αij ∈ ∆+, and soϕ = αim by Proposition 2.2. Thus setting y = rim+1 · · · rij−1 we have yαij = αim .By Proposition 2.6,

(2.5) rimyrij = y,

but this is a contradiction because the expression (2.5) equals rimrim+1· · · rij , which

is reduced.

Proposition 2.8. For all w ∈ W, l(w) = n(w). In fact, if w = ri1ri2 · · · rijis a reduced expression, then Φw consists of the distinct elements

αi1 , ri1αi2 , ri1ri2αi3 , . . . , ri1ri2 · · · rij−1αij .

proof. That all of the elements listed in the proposition are in ∆+ followsfrom applying Proposition 2.7 to w = ri1 , w = ri1ri2 , etc. It is easy to see that w−1

applied to each of the elements is in ∆−, by using rinαin = −αin , and applyingProposition 2.7 again. If they are not all distinct, then

rik · · · rinαin+1 = αik ∈ ∆+

where rik · · · rinrin+1 is some substring of ri1ri2 · · · rij . Then

rik+1· · · rinαin+1

= −αik ∈ ∆−,

which is a contradiction, because we can apply Proposition 2.7 to show

rik+1· · · rinαin+1

∈ ∆+.

Hence by Proposition 2.5, the elements listed must exhaust Φw.

Proposition 2.9. If w ∈ W preserves ∆+ (or equivalently, if Φw = ∅) thenw = 1.


proof. Apply Proposition 2.8 to the case n(w) = 0.

We extend the action of W from R to all of (he)∗. For i ∈ I0, define the linearautomorphism r′i of (he)∗ by the conditions

r′iϕ = ϕ− ϕ(hi)αi for all ϕ ∈ (he)∗

Then r′i|R = ri, and the restriction map to R provides a homomorphism fromthe group W ′ generated by the r′i (i ∈ I0) onto W . Note that both ri and r′i haveorder 2.

Proposition 2.10. For all i 6= j (with i, j ∈ I0) the order mij of rirj equalsthe order of r′ir

′j and is related to aij as follows:

aijaji 0 1 2 3 ≥ 4mij 2 3 4 6 ∞

proof. The matrix of rirj and also of r′ir′j on the 2-dimensional space S =

Eαi ⊕ Eαj is

T =

(aijaji − 1 aij−aji −1

)with characteristic polynomial:

(2.6) λ2 + (2− aijaji)λ+ 1.

If aij = aji = 0, then T = −I, and so has order 2. If aijaji = 1, then the polynomial(2.6) is λ2 + λ+ 1, hence T has order 3. If aijaji = 2, then (2.6) is λ2 + 1, and has

roots ±√−1 and hence T has order 4. If aijaji = 3, then (2.6) is λ2 − λ + 1 with

roots two primitive 6th roots of one, which means T has order 6. If aijaji > 4,then T has infinite order because the characteristic polynomial (writing a = aijaji)

has roots 12

(a− 2±

√a(a− 4)

), which are real and not ±1. If aijaji = 4 the

characteristic polynomial is (λ− 1)2, but T 6= 1, so T has infinite order.Now we must show that the order of T on S equals the order of T . If aijaji 6= 4

then the matrix (αi(hi) αi(hj)αj(hi) αj(hj)

)is nonsingular, so the annihilator of hi and hj in R (or in (he)∗) is a linear comple-ment S′ of S in R (or in (he)∗). But rirj (or r′ir

′j) is 1 on S′; thus the order of T

equals the order of rirj (or r′ir′j). In the case aijaji = 4 then rirj (or r′ir

′j) must

have infinite order because its restriction to S does.

Recall the definition of Coxeter group from [Bo]; note that it allows infinitelymany generators.

Proposition 2.11. The restriction map

W ′ →W

r′i 7→ ri

is an isomorphism. W is a Coxeter group generated by the set rii∈I0 with definingrelations r2

i = 1 and (rirj)mij = 1 (i 6= j), where mij is the order of rirj, given in

Proposition 2.10, and (rirj)∞ = 1 is interpreted to be the vacuous relation.


proof. It is sufficient to prove the “exchange condition” , even in this casewhere the group W is not finitely generated [Bo]. The “exchange condition” is:For every w ∈W , every reduced expression w = ri1 · · · rij (ik ∈ I0) and every i ∈ I0with l(riw) < l(w), there exists k ∈ 1, . . . , j such that

riri1 · · · rik−1= ri1 · · · rik .

We use induction on l(w). If l(w) = 1 then the condition is clearly true. Assumethe condition for all w such that l(w) < j. let w′ = ri1 · · · rij−1

. If l(riw′) < l(w′),

then we can apply the induction hypothesis to w′ since l(w′) = j − 1 and we aredone. Otherwise, l(riw

′) = j (see Propositions 2.4 and 2.8). But

n(riw) = l(riw) < l(w) = n(w),

which implies (Proposition 2.4) αi ∈ Φw and rij (w′)−1αi ∈ ∆−. Similarly,

n(riw′) = l(riw

′) > l(w′) = n(w′),

which implies αi /∈ Φ′w, so that (w′)−1αi ∈ ∆+. This means (w′)−1αi = αij(Proposition 2.3), so that w′rij (w

′)−1 = ri by Proposition 2.6. Therefore,

riri1 · · · rij−1 = ri1 · · · rij ,

which is the exchange condition with k = j.

Identify the groups W ⊂ Aut(R) and W ′ ⊂ Aut((he)∗), and write ri for r′i.

Definition. Let ρ be any fixed element of (he)∗ such that ρ(hi) = 12aii for all

i ∈ I. Note (ρ, αi) = 12 (αi, αi)

For every finite subset Φ of R define 〈Φ〉 =∑ψ∈Φ ψ, which is an element of R.

Note that 〈Φw〉 is defined for all w ∈W and that 〈Φri〉 = αi for all i ∈ I0.

Proposition 2.12. For all w ∈W

〈Φw〉 = ρ− wρ.

In particular ρ− wρ is a nonnegative integral linear combination of (finitely manyof) the (real) αi, i ∈ I0.

proof. By induction on l(w). The result is clear for l(w) = 0 or 1. Supposethat 〈Φ′w〉 = ρ − w′ρ for all w′ ∈ W with l(w′) < l(w). Let w = ri1 · · · rij be areduced expression for w and let w′ = ri2 · · · rij , then l(w′) = l(w)− 1. Now,

ρ− wρ = ρ− ri1w′ρ= ρ+ ri1(ρ− w′ρ)− ri1ρ= ri1〈Φw′〉+ 〈Φri1 〉

and by Proposition 2.4 this equals 〈Φw〉.

Proposition 2.13. For all w1, w2 ∈W

〈Φw1w2〉 = 〈Φw1

〉+ w1〈Φw2〉.


proof. This follows immediately from Proposition 2.12.

Proposition 2.14. The only Weyl group element that fixes ρ is the identity.Equivalently, if w1ρ = w2ρ where w1, w2 ∈W , then w1 = w2.

proof. If wρ = ρ for w ∈ W then 〈Φw〉 = 0 by Proposition 2.12. Then byProposition 2.9, w = 1.

Proposition 2.15. If w1, w2 ∈ W and Φw1= Φw2

or if 〈Φw1〉 = 〈Φw2

〉 thenw1 = w2.

proof. Proposition 2.12 implies that if 〈Φw1〉 = 〈Φw2〉 then

ρ− w1ρ = ρ− w2ρ.

So w1ρ = w2ρ. Thus by Proposition 2.14, w1 = w2.

2.2 Some inequalities involving the root system.

Definitions.1. Let λ ∈ (he)∗; then λ is called integral if λ(hi) ∈ Z when i ∈ I0 (i.e., when

aii > 0). The set of integral weights will be denoted by (he)∗Z.2. Let λ ∈ (he)∗ be such that λ takes values in F on the hi, for i ∈ I0. We

say λ is in the distinguished Weyl chamber if λ(hi) ≥ 0, or equivalently if(λ, αi) ≥ 0 for all such i. This distinguished Weyl chamber will denoted D.

3. An element λ ∈ (he)∗ taking values in F on the hi, i ∈ I, is dominant ifλ(hi) ≥ 0, or equivalently if (λ, αi) ≥ 0 for all i ∈ I. Denote by P+ the setof dominant integral elements.

Remark.1. All roots are integral (since all simple roots are integral).2. W preserves the set of integral elements.3. ρ is in D, and is integral, but ρ may not be dominant.4. All imaginary simple roots are in −D.

Note that the form (·, ·) on (he)∗ is W -invariant. Note that the condition that λtakes values in F on the hi means that assertions such as λ(hi) ≥ 0 and (λ, αi) ≥ 0are meaningful.

Definition. Given µ ∈ P+, for n > 0, let Ω(µ, n) denote the set of all sums ofn distinct pairwise orthogonal imaginary simple roots, each orthogonal to µ. Forn = 0 set Ω(µ, 0) = 0. Let Ω(µ) = ∪n≥0Ω(µ, n). For η ∈ Ω(µ), let η denote thesimple roots appearing in the sum η.

Lemma 2.16. Let ψ be a collection of not necessarily distinct pairwise orthogo-nal simple imaginary roots. Let ϕ be a collection of not necessarily distinct positiveroots. If 〈ψ〉 = 〈ϕ〉, then ψ = ϕ.

proof. Let ψ = αi1 , . . . , αik, and let ϕ = ϕ1, . . . , ϕr. We show there isa bijection between αi1 , . . . , αik and ϕ1, . . . , ϕr. Assume that 〈ψ〉 = 〈ϕ〉. If νis any subset of ψ containing at least two elements, then 〈ν〉 /∈ ∆+ by Lemma 1.13(and the fact that nαi ∈ ∆+ implies that n = ±1). In particular, 〈ν〉 6= ϕj for allj. Thus k ≤ r.


Furthermore, because the simple roots are linearly independent, no collectionof elements in ϕ ⊂ ∆+ can sum to an αim unless one of the elements equals αimand the rest are zero. Thus 〈ψ〉 = 〈ϕ〉 implies ψ = ϕ.

It will be useful to distinguish between imaginary roots that can be written aswαi for some w ∈ W and i ∈ I\I0, and those that cannot. In what follows, rootsof the form wαi for some w ∈W and i ∈ I often assume the role played by the realroots in arguments involving Kac-Moody algebras.

Proposition 2.17. Let w ∈ W , Φ a finite subset of ∆+ and η ∈ Ω(0). Letγ ∈ R be a finite sum of not necessarily distinct positive imaginary roots in ∆I \σαi | σ ∈W, i ∈ I \ I0. If 〈Φw〉+wη = 〈Φ〉+ γ then γ = 0 and Φ = Φw ∪wη.(Here wη has the obvious meaning.) In particular, Φ consists of Weyl grouptransforms of simple roots.

proof. Let β1, . . . , βn be the distinct elements of Φ. Let γ1, . . . , γk be thedistinct elements of Φw and let αi1 , . . . , αij be the distinct elements of η. Theproposition is clear if both k and j are zero, so assume k 6= 0 or j 6= 0.

Case 1. Assume that w−1βi ∈ ∆+ for all i = 1, . . . , n. Then

w−1γ1 + · · ·+ w−1γk + αi1 + · · ·+ αij = w−1β1 + · · ·+ w−1βn + w−1γ.

Equivalently,

αi1 + · · ·+ αij = −w−1γ1 − · · · − w−1γk + w−1β1 + · · ·+ w−1βn + w−1γ.

By the assumption on the βi, i = 1, · · · , n, and the definitions of Φw and γ theabove expression is of the form

(2.7) αi1 + · · ·+ αij = ϕ1 + · · ·+ ϕr,

where each ϕi ∈ ∆+. Note that of the ϕi might not be distinct, because of thedefinition of γ.

Thus by Lemma 2.16

η = ϕ1, . . . , ϕr=−w−1γ1, . . . ,−w−1γk, w

−1β1, . . . , w−1βn ∪ w−1γ.

By the definition of γ, w−1γ ∩ η = ∅. Thus γ = 0. Furthermore, since theelements γm (m = 1, . . . , k) of Φw are real, η ∩ −w−1γ1, . . . ,−w−1γk = ∅,,so αi1 , . . . , αij = w−1β1, . . . , w

−1βn and γ1, . . . , γk = ∅. Thus 〈Φw〉 = 0,w = 1 and αi1 , . . . , αij = β1, . . . , βn.

Case 2. Assume that some βi ∈ Φ satisfies w−1βi ∈ ∆−. Then βi ∈ Φw, soβi = γs for some s, 1 ≤ s ≤ k. Now consider the sum

〈Φw \ γs〉+ wη = 〈Φ \ βi〉+ γ.

If all of the remaining βi satisfy w−1βi ∈ ∆+, then we apply the argument of Case1 above (except that w 6= 1 here). This establishes γ = 0, wαi1 , . . . , wαij =β1, . . . , βn \ βi and γ1, . . . , γk \ γs = ∅. Otherwise, there is some βrsuch that w−1βr ∈ ∆−, i.e, βr = γt for some t, 1 ≤ t ≤ k and we consider


〈Φw \ γs, γt 〉+wη = 〈Φ \ βi, βr〉+ γ. We iterate this procedure until there areno more elements βi of Φ satisfying w−1βi ∈ ∆−, and Φw is exhausted.

Proposition 2.18. Let T be the subset of (he)∗ consisting of the elements ofthe form −〈Φ〉 − γ where Φ is a finite subset of ∆+ and γ is a finite sum of notnecessarily distinct positive imaginary roots in ∆I \ σαi | σ ∈W, i ∈ I \ I0. Thenρ+ T is W -invariant.

proof. It suffices to show that ri(ρ− 〈Φ〉 − γ)− ρ ∈ T for all reflections ri. Ifαi ∈ Φ then Φ = Φ′ ∪ αi where Φ′ = Φ\αi. We have ρ(hi) = 1 and

ri(ρ− 〈Φ〉 − γ)− ρ= ρ− ρ(hi)αi − ri〈Φ′〉+ αi − riγ − ρ= −ri〈Φ′〉 − riγ ∈ T

by Proposition 2.2 and 2.3. If αi /∈ Φ then

ri(ρ− 〈Φ〉 − γ)− ρ = −αi − ri〈Φ〉 − riγ ∈ T

as above.

Proposition 2.19. Let λ ∈ (he)∗ be integral, and let U be a W -invariantsubset of (he)∗, all of whose elements are of the form λ−

∑i∈I niαi, ni ∈ N. Then

every element of U is conjugate to an element in D.

proof. Let µ ∈ U . Choose w ∈ W so that in the expression wµ = λ −∑i∈I niαi the sum

∑i∈I ni is minimal. Then wµ is in D because if (wµ)(hi) < 0

for some real αi, then riwµ = wµ −mαi, where m = wµ(hi) < 0. Since U is W -invariant riwµ = λ−

∑j∈I njαj−mαi and ni+m < ni. So riwµ = λ−

∑j∈I mjαj

where∑i∈I mi <

∑i∈I ni, which is a contradiction.

Proposition 2.20. Let µ ∈ (he)∗ be dominant and let ν = µ −∑i∈I niαi,

where ni ∈ N. If ν + ρ is in D, then

(µ+ ρ, µ+ ρ)− (ν + ρ, ν + ρ) ≥ 0.

Furthermore, equality holds if and only if ν = µ − γ, where γ is a sum of notnecessarily distinct pairwise orthogonal simple imaginary roots, each of which isorthogonal to µ.

proof. First we show (µ+ ρ, µ+ ρ)− (ν + ρ, ν + ρ) ≥ 0. Consider

(µ+ ρ, µ+ ρ)− (ν + ρ, ν + ρ)

= (µ+ ρ, µ+ ρ)− (µ+ ρ−∑

niαi, µ+ ρ−∑

niαi)

= (µ,∑

niαi) + (ν + 2ρ,∑

niαi).(2.8)

By the assumption on µ, we have (µ,∑niαi) ≥ 0. So it is sufficient to show

(2.9) (ν + 2ρ,∑

niαi) ≥ 0.


Let αk be a root appearing in µ − ν, so nk > 0 in the expression ν = µ −∑i∈I niαi. If αk ∈ ∆R then nk(ν + ρ + ρ, αk) ≥ 0, because ν + ρ and ρ are in D.

Thus we have the desired inequality contributing to (2.9) for real αk.If αk ∈ ∆I then, by definition of ρ, (ν + 2ρ, αk) = (ν + αk, αk) and

ν + αk = µ−∑i∈I

niαi + αk

= µ−∑i∈I

miαi

where nk ≥ 1 and mi ∈ N. So

(ν + αk, αk) = (µ−∑i∈I

miαi, αk)

= (µ, αk)−∑i∈I

mi(αi, αk) ≥ 0(2.10)

by assumption on µ, and since αk ∈ ∆I . Thus (ν + 2ρ, αk) = (ν + αk, αk) ≥ 0for αk ∈ ∆I and (2.9) holds for imaginary αik . Summing over real and imaginaryroots gives the inequality (2.9). This in conjunction with equation (2.8) gives thedesired result.

Equality holds if and only if ni(µ+ ν + 2ρ, αi) = 0 for all i ∈ I, since all of the

terms in (2.9) have the same sign. If αi ∈ ∆R we have ni(ρ, αi) = ni(αi,αi)2 ≥ 0,

ni(ν + ρ, αi) ≥ 0 and

ni(µ+ ν + 2ρ, αi) = ni(µ, αi) + ni(ν + ρ, αi) + ni(ρ, αi)

= 0

which implies ni(αi,αi)2 = 0, so that ni = 0. Thus no real roots αi appear in the

expression

ν = µ−∑i∈I

niαi.

If αi ∈ ∆I then we have from the previous calculation (2.10):

(2.11) ni[(µ, αi) + (µ, αi)−∑j∈I

mj(αj , αi)] = 0,

where mj = nj if i 6= j and mi = ni − 1. If ni > 0 then (2.11) holds if and only ifthe conditions

1. (µ, αi) = 02. for all j 6= i, nj > 0 implies (αj , αi) = 03. if ni > 1, then (αi, αi) = 0

are satisfied.Thus ν = µ−

∑i∈I niαi where the αi’s appearing satisfy the statement in the

theorem. This condition is clearly necessary and sufficient.


Proposition 2.21. Let T ⊂ (he)∗ be as in Proposition 2.18, let µ ∈ (he)∗ bedominant integral and let T ′ be a W -invariant subset of (he)∗ all of whose elementsare of the form µ−

∑i∈I niαi where ni ∈ N, excluding the elements (other than µ)

of the form µ−∑niαi where (µ, αi) = 0 if ni > 0. Suppose λ = τ + τ ′ with τ ∈ T ,

τ ′ ∈ T ′. Then(µ+ ρ, µ+ ρ)− (λ+ ρ, λ+ ρ) ≥ 0.

Equality holds if and only if there exists w ∈W and an η ∈ Ω(µ) such that λ+ ρ =w(µ + ρ − η). Equivalently, equality holds if and only if τ = −〈Φw〉 − wη andτ ′ = wµ. In the case of equality, both η and w are uniquely determined by λ, andso are τ and τ ′.

proof. By Proposition 2.18, ρ + T + T ′ is W -invariant. Also, every elementof ρ + T + T ′ is of the form µ + ρ −

∑i∈I niαi. Applying Proposition 2.19 there

exists w ∈W such that

w−1(λ+ ρ) ∈ D ∩ (ρ+ T + T ′).

Then we may write

w−1(λ+ ρ)− ρ = µ−∑i∈I

niαi ni ∈ N

and apply Proposition 2.20 to ν = w−1(λ+ ρ)− ρ. So

(µ+ ρ, µ+ ρ)− (w−1(λ+ ρ), w−1(λ+ ρ)) ≥ 0

which is equivalent to

(µ+ ρ, µ+ ρ)− (λ+ ρ, λ+ ρ) ≥ 0

by W -invariance of (·, ·). Equality holds if and only if w−1(λ + ρ) = µ + ρ −∑k∈I nkαk where αk with nk > 0 are simple imaginary, orthogonal to µ, and

(αk, αj) = 0 for k 6= j, (αk, αk) = 0 if nk > 1. Thus equality holds only if thereexists such an element w, and conversely, if there is such a w, then as in (2.8)

(µ+ ρ, µ+ ρ)− (λ+ ρ, λ+ ρ) = (2µ+ 2ρ−∑k∈I

nkαk,∑k∈I

nkαk)

= (2ρ−∑k∈I

nkαk,∑k∈I

nkαk)

=∑k∈I

nk(∑i 6=k∈I

ni(αi, αk) + (nk − 1)(αk, αk))

= 0.

Let γ =∑k∈I nkαk where the αk are as above, and fix w. Then equality holds

if and only if w−1(λ+ρ) = µ+ρ−γ, or equivalently w−1(τ+ρ)−ρ+w−1τ ′ = µ−γ.This holds if and only if w−1(τ+ρ)−ρ = −

∑k∈I mkαk and w−1τ ′ = µ−

∑k∈I qkαk,

where mk + qk = nk and mk = nk or 0 and qk = nk or 0. By definition of T ′,and the given properties of the αk, nk 6= 0, it follows that qk = 0. Thus mk = nk.Furthermore,

∑k∈I nkαk ∈ T by W -invariance of T + ρ, thus

(2.12)∑k∈I

nkαk = 〈Φ〉+ ψ


where Φ ⊂ ∆+ and ψ is a finite sum of not necessarily distinct positive imaginaryroots in ∆I \ σαi | σ ∈ W, i ∈ I \ I0. Applying Lemma 2.16 to the expression(2.12), we conclude ψ = 0, and that the collection consisting of the αk with eachαk listed nk times is equal to the set Φ. Since Φ ⊂ ∆+ each nk = 0 or 1. Thisshows γ = η for some η ∈ Ω(µ).

We have shown equality holds if and only if

w−1(τ + ρ)− ρ = −η and w−1τ ′ = µ

if and only ifτ + ρ− wρ = −wη and τ ′ = wµ

by the W -invariance of T ′ and T + ρ, and by the definitions of T and T ′.It remains only to show the uniqueness of w and η. Suppose now that

λ+ ρ = w′(µ+ ρ+ η′)

= w(µ+ ρ+ η)

(with w′ and η′ having the same properties as w and γ). Let w0 = w−1w′. Consider

w0(µ+ ρ− η′) = µ+ ρ− η.

Thus

(2.13) w0µ− w0η′ + w0ρ− ρ = µ− η.

Since τ ′ = wµ and T ′ is W -invariant, µ ∈ T ′, so w0µ ∈ T ′. We decompose w0µ as

w0µ = µ−∑i∈I

miαi −∑j∈I

njαj −∑k∈I

pkαk

where mi, nj , pk ∈ N and the αi are real, the αj are imaginary with (αj , µ) 6= 0,and the αk are imaginary with (αk, µ) = 0.

Also,

−w0η′ = −η′ −

∑l∈I

qlαl

with αl real, ql ∈ Z. In fact, ql ∈ N because for an imaginary simple αm, w0αm =αm +

∑snαn (αn real) is a positive root, so sn ∈ N, and η′ is a sum of such α′ms.

By Proposition 2.12

w0ρ− ρ = −〈Φw0〉 = −

∑r∈I

trαr

with αr real and tr ∈ N. Furthermore, η =∑s∈I usαs with αs imaginary, us ∈ N

and (αs, µ) = 0. Equating coefficients in (2.13) gives:

−∑i∈I

miαi −∑l∈I

qlαl −∑r∈I

trαr = 0.

and−∑

njαj = 0,

so each mi = 0, ql = 0, tr = 0, as well as each nj = 0. Thus w0µ = µ−∑pkαk ∈ T ′,

so by the definition of T ′ each pk = 0. Thus w0µ = µ, w0η′ = η and w0ρ− ρ = 0.

The latter equality gives w0ρ = ρ, i.e., w0 = 1. Equation (2.13) gives µ−η′ = µ−η.We conclude that w = w′ and η′ = η.


3. Standard Modules and a Character Formula

We shall prove the character formula (due to Borcherds) for “standard mod-ules”, which are defined below. The method used here for obtaining the characterformula for a standard ge-module and denominator identity for g is to generalizethe results of [GL]. The arguments presented here are very close to those in [GL],additional arguments and lemmas are provided to deal with the presence of simpleimaginary roots. The character and denominator formulas appear in [B1] and [K];in both of these references the indicated proof is to follow the methods of [K]. Ex-amples of standard modules for generalized Kac-Moody algebras with one simpleimaginary root can be found in [GT], where they are called basic modules. Otherexamples of standard modules are found in [JLW].

3.1 Definitions and preliminary results.We define “standard modules” for a generalized Kac-Moody algebra ge. It is

for this class of modules that a character formula can be proven using methodsanalogous to those used in the Kac-Moody case. The definition of standard modulegiven here agrees with the definition of a standard module for the case when ge isa Kac-Moody algebra. Integrable modules for a generalized Kac-Moody algebra,as they are defined in [K], that are also highest weight modules do not have to bestandard modules by the definition given here.

Let X be an he-module, e.g., a ge-module considered an he-module by restric-tion, and let ν ∈ (he)∗. Define the weight space

Xν = x ∈ X|h · x = ν(h)x for all h ∈ he.

An element ν ∈ (he)∗ is a weight if Xν 6= 0. X is called a weight module if X is adirect sum of its weight spaces.

Some examples of weight modules that we encounter are: ge (under the adjointoperation), X ⊗ Y if X and Y are weight modules,

∧(n±) and U(ge).

Note that if X is a ge-module, ϕ ∈ ∆ and ν ∈ (he)∗ then

gϕ ·Xν ⊂ Xν+ϕ.

A ge-module is called a highest weight module if it is generated by a weightvector x which is annihilated by n+. The vector x is called a highest weight vectorand x is uniquely determined up to nonzero scalar. If x has weight Λ then Λ iscalled the highest weight of X. For any highest weight module X:

U(n−) · x = X

X =⊕λ≤Λ

Xλ, XΛ = Ex, dimXλ <∞

where Λ is the highest weight, x is the highest weight vector, and “λ ≤ Λ” meansλ is of the form Λ −

∑i∈I niαi, ni ∈ N. Lowest weight modules can be defined

analogously.Let S ⊂ I0, define ∆S = ∆ ∩

∐i∈S Zαi, ∆S

+ = ∆+ ∩∆S and ∆S− = ∆− ∩∆S .

Denote by hS the span of the hi, i ∈ S. Let gS be the subalgebra of g generatedby the ei and fi with i ∈ S. Note that gS is isomorphic to the derived subalgebra


of the Kac-Moody algebra associated to the matrix (aij)i,j∈S . We have the rootspace decomposition:

gS =∐ϕ∈∆S

+

gϕ ⊕ hS ⊕∐ϕ∈∆S

−

gϕ.

Define the following subalgebras of g = g(A):

n+ =∐ϕ∈∆+

gϕ; n− =∐ϕ∈∆−

gϕ; n+S =

∐ϕ∈∆S

+

gϕ;

n−S =∐ϕ∈∆S

−

gϕ; u+ =∐

ϕ∈∆+\∆S+

gϕ; u− =∐

ϕ∈∆−\∆S−

gϕ; r = gS + h.

Let p = r ⊕ u+, the “parabolic subalgebra” of g defined by S. Note thatg = u− ⊕ p. The subalgebras pe, re and geS are defined in the obvious ways. Notethat if S = ∅ one obtains the usual Borel subalgebra b = h⊕ n+.

Let ∆+(S) = ∆+ \∆S+. Define W (S) to be the set w ∈W | Φw ⊂ ∆+(S).

Given λ ∈ (he)∗ such that λ(hi) ∈ Z+ for i ∈ S, define a universal highestweight module as follows: Let L(λ) be the (unique up to isomorphism) irreduciblegeS-module with highest weight λ. Then the generalized Verma module V L(λ) withhighest weight λ is defined to be the induced ge-module

V L(λ) = U(ge)⊗U(pe) L(λ).

It is clear from the definition that V L(λ) is a highest weight module with highestweight λ, and the highest weight space can be identified with the highest weightspace of the gS-module L(λ). By the Poincare-Birkhoff-Witt theorem there is anisomorphism of vector spaces

U(u−)⊗ L(λ) ∼= V L(λ).

The ge-module induced from the one-dimensional he weight module of weight λis the Verma module V λ. Given any highest weight module X, with highest weightµ and highest weight vector x, then there is a ge-module surjection:

V λ → X.

Let X be a ge-module. Recall P+ denotes the set of dominant integral elements(see §2.2).

Definition. A module X is a standard module if X is a highest weight modulewith highest weight µ ∈ P+ and highest weight vector x satisfying:

1. for i ∈ I if µ(hi) = 0 then fi · x = 02. if αi ( i ∈ I) is real then fni+1

i · x = 0, where ni = µ(hi).

Proposition 3.1. Let µ ∈ P+. Then there exist standard ge-modules Xµmax

and Xµmin of highest weight µ, such that given any standard module X with highest

weight µ, there are ge-module surjections

Xµmax → X → Xµ

min.

The module Xµmin is the unique irreducible highest weight module with highest weight

µ.


proof. These standard modules are constructed from Verma modules as theyare for Kac-Moody algebras. Given µ ∈ P+ let

J = I0 ∪ i ∈ I | µ(hi) = 0 ⊂ I.

For i ∈ J , let ni = µ(hi). Let v0 = 1 ⊗ 1 ∈ V µ. If aii > 0 (i.e., if i ∈ I0), then bythe representation theory of sl2,

ei · (fni+1i · v0) = 0.

If i ∈ J is such that µ(hi) = 0, then

ei · fni+1i · v0 = ei · fi · v0 = µ(hi) · v0 = 0.

In fact, fni+1i · v0 is annihilated by n+, because it is also true that if j 6= i, then

[ej , fi] = 0. Let

Y = V µ/∑i∈JU(ge)fni+1

i · v0;

then Y is a standard module of highest weight µ.If W is the maximal submodule of Y not intersecting the highest weight space

of Y , then Z = Y/W is an irreducible standard highest weight module of highestweight µ.

Let X be a standard module with highest weight µ and let x be a highestweight vector. Then fni+1

i · x = 0 for all i ∈ J , and by the universal property ofV µ and the definition of Z we have surjections:

Y → X → Z.

Thus we can take Y = Xµmax and Z = Xµ

min.The module Z is the unique (up to isomorphism) irreducible highest weight

module of highest weight µ: If X is another such module, with highest weightvector x, then for each i ∈ J the elements fni+1

i · x are n+-invariant (using anargument similar to the above), so each generates a proper ge-submodule of X, andmust therefore be zero. Thus X is also standard. Hence X is isomorphic to Z.

Definition. Let PS = λ ∈ (he)∗|λ(hi) ∈ N for all i ∈ S.

Remark. Proposition 3.1 of [GL] holds in this setting without change, so thereis a natural bijection between PS and the set of (isomorphism classes) of irreduciblefinite-dimensional re-modules which are irreducible as gS-modules.

Proposition 3.2. The set of weights of a highest weight module X satisfyingcondition (2) above is stable under W acting on (he)∗. For every weight µ of Xand every w ∈ W , dimXµ = dimXwµ. In particular, if µ is the highest weight ofX, then dimXwµ = 1 for all w ∈W .

proof. Let ui be the Lie algebra isomorphic to sl2 spanned by ei, fi, hifor aii > 0. The ei-invariant highest weight vector x is contained in the finite-dimensional irreducible ui-module V =

∑m≥0 Efmi · x. Since X is generated by x,

any v ∈ X is in a finite dimensional ui-module, so X is completely reducible underui.


For any weight µ we can apply sl2 theory to the ui-stable space∐n∈ZXµ−nαi

to conclude that dimXµ = dimXµ−µ(hi)αi = dimXriµ. The rest of the propositionis clear.

We will need the following lemma and corollary.

Lemma 3.3. Let X be a standard module with highest weight µ. Then the setof weights of X does not include elements other than µ of the form µ−

∑i∈I niαi

(ni ≥ 0) where (µ, αi) = 0 if ni > 0. That is, for every weight ν, µ − ν is not anonzero linear combination of simple roots all orthogonal to µ.

proof. Suppose v ∈ X such that h · v = (µ −∑i∈I niαi)(h) · v where the

αi and ni are as above. Then v is a linear combination of elements of the formfi1fi2 · · · fin · x for some i1, i2, · · · , in ∈ I. We may assume that each of the fikand in particular, fin , must correspond to one of the above αi’s where ni 6= 0 (i.e.[h, fik ] = −αi(h)fik), because v has the given weight. Thus fin corresponds tosome αi such that (µ, αi) = 0, so by assumption fin · x = 0. Thus each summandin v is 0, so v itself must be 0.

Since the set of weights of X is stable under the action of W we have thefollowing corollary.

Corollary 3.4. The set of weights of X does not include W - transforms ofelements of the form given in the lemma.

3.2 The Casimir operator.Define for ν ∈ (he)∗

D(ν) =

ν −

∑i∈I

niαi|ni ∈ N+

⊂ (he)∗.

Let O be the category of ge-modules X such that X is a weight module with finite-dimensional weight spaces, and whose set of weights lies in a finite union of sets ofthe form D(ν).

Recall from §1.3 that ge has a symmetric invariant bilinear form (·, ·), and thatthis form satisfies:

[a, b] = (a, b)hϕ ϕ ∈ ∆, a ∈ gϕ, b ∈ g−ϕ

(h, hϕ) = ϕ(h) ∀h ∈ he, ϕ ∈ R.The form (·, ·) induces a nonsingular pairing between gϕ and g−ϕ, (see [K] and [L]where the arguments work in this generality). Furthermore, (gϕ, gψ) = 0 unlessϕ = −ψ, (gϕ, he) = 0, and (ei, fj) = δij

1qi

for i, j ∈ I.

Let s1, . . . , sm be a basis of gϕ, ϕ ∈ ∆. Then there is a unique dual basist1, . . . , tm of g−ϕ relative to (·, ·). Set wϕ =

∑mi=1 tisi, which is an element of

U(ge), the universal enveloping algebra of ge.The element wϕ is independent of the basis simi=1, because it can be written

as wϕ = f (κϕ ⊗ 1)(ιϕ), where ιϕ ∈ (gϕ)∗ ⊗ gϕ ' End gϕ is the unique elementcorresponding to 1 ∈ End gϕ, where κϕ : (gϕ)∗ → g−ϕ is the linear isomorphism


given by (·, ·)|gϕ×g−ϕ and f : g⊗ g→ U(ge) is the map induced by multiplication.It is clear that the element f (κϕ ⊗ 1)(ιϕ) of U(ge) is the same as wϕ, becausewith respect to the basis sjmj=1 , ιϕ =

∑mi=1 s

∗i ⊗ si where s∗i is the dual element

to si.On X ∈ O

Γ1 = 2∑ϕ∈∆+

wϕ

acts as a well-defined operator, because once it is applied to some x ∈ X the sumbecomes a finite one. Note that wαi = qifiei (these qi are the diagonal entries ofthe matrix D that symmetrizes A). Define Γ2 on X ∈ O by the condition that Γ2

acts on the weight space Xν as scalar multiplication by (ν + ρ, ν + ρ).

Definition. The Casimir operator ΓX is defined by ΓX = Γ1 + Γ2.

Proposition 3.5. Let X,Y ∈ O, f : X → Y be a ge-module map. Thenf ΓX = ΓY f .

proof. Since f is a module map it is easy to see that Γ1 f = f Γ1. It isobvious that Γ2 commutes with f .

Proposition 3.6. For X ∈ O the Casimir operator ΓX commutes with theaction of ge on X.

proof. It is easy to see that ΓXh = hΓX for h ∈ he. So it is sufficient to showthat eiΓX = ΓXei and fiΓX = ΓXfi, for all i ∈ I.

Let Φ = ∆+\αi be a union of αi-root strings, so that Φ =⋃ϕ+ niαi|ni ∈

Z∩∆+, where ϕ ∈ (he)∗. Recall ui is Lie subalgebra spanned by ei, fi, hi. ThenM =

∐ϕ∈Φ gϕ and N =

∐ϕ∈−Φ gϕ are ui (not necessarily sl2) modules that are

contragredient under the (ge-invariant) form (·, ·). The formal “canonical element”∑ϕ∈Φ wϕ is formally ui-invariant, and is a well-defined operator on any vector in

M , so acts as a ui-invariant operator.Let ν ∈ (he)∗ and let x ∈ Xν . Take Φ = ∆+\αi. Then

Γ1(x) = 2wαix+ 2∑ϕ∈Φ

wϕx = 2qifieix+ 2∑ϕ∈Φ

wϕx

Γ1(fix) = 2qifieifix+ 2∑ϕ∈Φ

wϕfix

Γ1(eix) = 2qifiei2x+ 2

∑ϕ∈Φ

wϕeix.

Thus

fiΓ1(x)− Γ1(fix) = 2qi(fi2ei − fieifi)x

= 2qifi(−hi)x = −2qifiν(hi)x

= −2qiν(hi)fix

and

eiΓ1(x)− Γ1(eix) = 2qi(eifiei − fiei2)x

= 2qihieix = 2qi(eihi + aiiei)x

= 2qi(ν(hi) + aii)eix.


Now consider Γ2:

fiΓ2(x)− Γ2(fix) = (ν + ρ, ν + ρ)fix− (ν − αi + ρ, ν − αi + ρ)fix

= (2(ν + ρ, αi)− (αi, αi)) fix

= (2(ν, αi) + (αi, αi)− (αi, αi)) fix

= 2ν(xαi)fix = 2qiν(hi)fix

and

eiΓ2(x)− Γ2(eix) = (ν + ρ, ν + ρ)eix− (ν + αi + ρ, ν + αi + ρ)eix

= (−2(ν + ρ, αi)− (αi, αi)) eix

= (−2ν(xαi)− (αi, αi)− (αi, αi)) eix

= −2 (ν(xαi) + (αi, αi)) eix = −2qi(ν(hi) + aii).

Thus fiΓX(x) = ΓX(fix) and eiΓX(x) = ΓX(eix) for all x ∈ Xν and hence for allx ∈ X.

Corollary 3.7. Let X be a highest weight module for ge with highest weightλ. Then X lies in the category O, and the Casimir operator of X acts on X asscalar multiplication by (λ+ρ, λ+ρ). In particular, if X contains an n+-annihilatedweight vector of weight η ∈ (he)∗, then (η + ρ, η + ρ) = (λ+ ρ, λ+ ρ).

proof. Let x be the highest weight vector of X, let y ∈ ge. Then ΓX · (y ·x) =y · ΓX · x = (λ+ ρ, λ+ ρ)y · x by Proposition 3.6.

3.3 Computation of the homology H∗(u−, Xt).

A general exposition of Lie algebra homology is contained in Section 1 of [GL].We will apply this to the Lie algebra ge = pe ⊕ u−, where pe = u+ ⊕ gS ⊕ he. LetX be a standard ge-module with highest weight µ ∈ P+. A U(u−)-free resolutionof X is given below.

Proposition 3.8. Let X be a standard ge-module. For each j ∈ N, let DXj be

the ge-module U(ge) ⊗U(pe) (∧j

(ge/pe) ⊗X), where X is regarded as a pe-moduleby restriction. Then there is an exact sequence of ge-modules and ge-module maps

· · · dX2−→DX

1

dX1−→DX0

εX0−→X −→ 0,

with this complex naturally isomorphic to V (ge, pe, X) (using the notation of [GL]).Each DX

j is isomorphic as an U(u−)-module and as an re-module to

U(u−)⊗E (∧j

(u−)⊗X),

with U(u−) acting by left multiplication on the first factor, and re acting via thetensor product action on the tensor product of the three re-modules.

proof. This follows directly from Proposition 1.9 [GL].

The re-module complex C∗(x) given by

· · · 1⊗dX2−→ E⊗U(u−) D

X1

1⊗dX1−→ E⊗U(u−) DX0 −→ 0


is naturally isomorphic to the standard re complex, see [GL], for computing the

homology H∗(u−, Xt) as an re-module, and for each j ∈ N, Cj(X) '

∧j(u−)⊗X.

Since X and all the DXj lie in the categoryO, we can form the Casimir operators

ΓX and Γ(j) = ΓDXj . For any vector space Y let IY denote the identity transfor-

mation on Y . Then ΓX = (µ + ρ, µ + ρ)IX by Corollary 3.7, and Proposition 3.5implies that

dXj+1 Γ(j + 1) = Γ(j) dXj+1 for all j ∈ N+

and thatεX0 Γ(0) = (µ+ ρ, µ+ ρ)εX0 .

By Proposition 3.6 the Γ(j) (j ∈ N) are ge-module maps, so they are U(u−)-module maps. Since the DX

j are U(u−)-free, and form an exact sequence, thereis a chain homotopy between the maps (µ + ρ, µ + ρ)IDXj and Γ(j) (for j ∈ N).

More explicitly, we can construct U(u−)-module homomorphisms hj : DXj → DX

j+1

(j ∈ N) such that

hj−1 dXj + dj+1 hj = Γ(j)− (µ+ ρ, µ+ ρ)IDXj ,

for all j ∈ N where if j = 0 the leftmost term is omitted.Then the operator

1⊗ Γ(j)− (µ+ ρ, µ+ ρ)ICj(X) ∈ EndCj(X)

induces the zero map on the jth homology Hj(u−, Xt) of C∗(X), for each j ∈ N.

As in [GL], we shall decompose the Casimir operator ΓY for Y ∈ O. Set

ΓY,1 = 2∑

ϕ∈∆+(S)

wϕ ∈ EndY.

Then ΓY,1 commutes with the action of re on Y . Thus ΓY,2 ∈ EndY defined byΓY,2 = ΓY − ΓY,1 commutes with the action of re by Proposition 3.6.

Set Γ2(j) = ΓDXj ,2 (j ∈ N). Then as re-module endomorphisms of Cj(X) =

E⊗U(u−) DXj ,

1⊗ Γ(j) = 1⊗ Γ2(j).

Thus1⊗ Γ2(j)− (µ+ ρ, µ+ ρ)ICj(X)

induces the zero map on the jth homology of C∗(X).For λ ∈ PS denote by Cj(X)(λ) the sum of all the re-submodules of Cj(X)

isomorphic to L(λ). Set

Bj(X) =∐λ∈PS

(λ+ρ,λ+ρ)=(µ+ρ,µ+ρ)

Cj(X)(λ),

and also letB′j(X) =

∐λ∈PS

(λ+ρ,λ+ρ)6=(µ+ρ,µ+ρ)

Cj(X)(λ),

for all j ∈ N. Then Cj(X) = Bj(X) ⊕ B′j(X), because Cj(X) is a completelyreducible gS-module, and C∗(X) is the direct sum of the re-module subcomplexesB∗(X) and B′∗(X). When restricted to Cj(X)(λ), the operator 1 ⊗ Γ2(j) acts as


the scalar (λ + ρ, λ + ρ). Thus 1 ⊗ Γ2(j) − (µ + ρ, µ + ρ)ICj(X) is a nonsingularoperator on B′j(X), this shows:

Proposition 3.9. The homology of B′∗(X) is zero. In particular, the homol-ogy H∗(u

−, Xt) of C∗(X) is naturally re-module isomorphic to the homology of itssubcomplex B∗(X).

To finish the computation of H∗(u

−, Xt) we must understand the weight spaces

of Cj(X)(λ), and the weight spaces of∧j

(u−).

Proposition 3.10. For each j ∈ N, the weights of∧j

(u−) lie in the subsetT of (he)∗ defined in Proposition 2.18. Let w ∈ W (S), and let η ∈ Ω(0). Suppose

that l(w) ≤ j. If η ∈ Ω(0, j − l(w)), then −〈Φw〉 − wη is a weight of∧j

(u−) andthe corresponding weight space is one-dimensional. If w ∈W and either w /∈W (S)

or l(w) + ht (η) 6= j then −〈Φw〉 − wη is not a weight of∧j

(u−).

proof. We have the decomposition n− =∐ϕ∈∆+

g−ϕ. Choose a basis bii∈Jof n− such that for each i ∈ J , bi ∈ g−ϕi . Let J ′ ⊂ J be such that bii∈J′ is abasis of u− ⊂ n−. Since we can assume the indexing set J is linearly ordered, a

basis for∧j

(n−) is given by bi1 ∧ · · · ∧ biji1<···<ij where im ∈ J , and a basis for∧j(u−) is bi1 ∧ · · · ∧ biji1<···<ij with im ∈ J ′. For each sequence i1 < · · · < ij ,

bi1 ∧ · · · ∧ bij is a weight vector with weight −∑jm=1 ϕim , here bik ∈ g−ϕik . Since

dim g−ϕ = 1 for any root ϕ in wαi|w ∈ W, i ∈ I the ϕik in wαi|w ∈ W, i ∈ Imust be distinct so the weights are in T .

Let w ∈ W with l(w) ≤ j, then Φw = ϕi1 , . . . , ϕil(w) for i1 < · · · < ij and

ϕik ∈ ∆R∩∆+ by Proposition 2.7. Let η ∈ Ω(0, j−l(w)) (so η = αil(w)+1+· · ·+αij ,

where the αik are j − l(w) distinct simple imaginary roots, which are mutuallyorthogonal). Then −〈Φw〉 − wη is a weight of bi1 ∧ · · · ∧ bij , where if 1 ≤ k ≤ l(w)

then bik is in g−ϕik and if l(w) < k ≤ j then bik is in g−wαik .Proposition 2.17 and the one-dimensionality of the root spaces gwαi for w ∈W

and i ∈ I imply that every other basis vector of∧

(n−) has a weight different from−〈Φw〉 − wη. Thus the weight space

∧(n−)−〈Φw〉−wη has dimension one and is in∧j

(n−).If w ∈ W (S) then Φw ⊂ ∆+(S) and so bi1 , . . . , bil(w)

∈ u−. Furthermore,

bil(w)+1, . . . , bij ∈ u− by definition, so

∧(n−)−〈Φw〉−wη ⊂

∧(u−).

If w /∈ W (S) then ϕim /∈ ∆+(S) so bim /∈ u− for some m = 1, . . . , l(w). So,

regardless of the choice of η, bi1 ∧· · ·∧bij /∈∧j

(u−) and −〈Φw〉−wη is not a weight

of∧j

(u−). If ht η 6= j − l(w) then we’ve seen above −〈Φw〉 −wη is not a weight of∧j(u−).

Proposition 3.11. Let X be a standard ge-module with highest weight µ ∈ P+.

For all j ∈ N, define Wj to be the set of weights λ of∧j

(u−) ⊗ X such that(λ + ρ, λ + ρ) = (µ + ρ, µ + ρ). Set W =

⋃jWj. Then there is a bijection

between W (S) × Ω(µ) and W, and for each j ∈ N, Wj corresponds bijectivelyto (w, η) ∈ W (S) × Ω(µ) | l(w) + ht (η) = j. Thus the sets Wj are disjoint.The correspondence is given as follows: If w ∈ W (S) and η ∈ Ω(µ) such that


l(w) + ht (η) = j then λ = −〈Φw〉 − wη + wµ = w(µ+ ρ− η)− ρ is the element ofWj associated to the pair (w, η).

The weight space (∧j

(u−)⊗X)λ is one-dimensional, and is the tensor product∧j(u−)−〈Φw〉−wη ⊗Xwµ of one-dimensional weight spaces.

proof. Let w ∈W (S) and let η ∈ Ω(µ, j − l(w)). Set

λ = −〈Φw〉 − wη + wµ

= w(µ+ ρ− η)− ρ.

By Proposition 3.10∧j

(u−)−〈Φw〉−wη is one-dimensional, and by Proposition 3.2

so is Xwµ. We have∧j

(u−)−〈Φw〉−wη⊗Xwµ ⊂ (∧j

(u−)⊗X)λ, and by construction

(λ+ρ, λ+ρ) = (µ+ρ, µ+ρ). Thus λ ∈ Wj and we have a map from W (S)×Ω(µ)to W such that elements satisfying l(w) + ht (η) = j map to Wj .

Let T and T ′ be as in Proposition 2.21. By Proposition 3.10 the set of weightsof∧

(u−) lie in the subset T ; by Proposition 3.2 and Lemma 3.3 the set of weightsof X is of type T ′. Let λ ∈ W. Then the weight space (

∧(u−) ⊗ X)λ is a finite

direct sum of tensor products∧

(u−)τ ⊗ Xτ ′ where τ ∈ T is a weight of∧

(u−),τ ′ ∈ T ′ is a weight of X, and λ = τ + τ ′. By Proposition 2.21 there is a uniquew ∈W and η ∈ Ω(µ) such that

λ = w(µ+ ρ− η)− ρ= −〈Φw〉 − wη + wµ

and τ = −〈Φw〉 − wη and τ ′ = wµ. Since τ is a weight of∧

(u−) we see byProposition 3.10 that in fact w ∈W (S). Thus we have a bijection between W (S)×Ω(µ) and W.

If we consider∧j

(u−) and λ ∈ Wj in the above argument, we conclude that

τ = −〈Φw〉 − wη is a weight of∧j

(u−), so that l(w) + ht (η) = j. Thus the abovebijection restricts to bijections (w, η) ∈ W (S)× Ω(µ) | l(w) + ht (η) = j ↔ Wj ,for j ∈ N

Since the decomposition λ = τ + τ ′ is unique∧j(u−)−〈Φw〉−wη ⊗Xwµ = (

∧j(u−)⊗X)λ,

where the two factors on the left are one-dimensional.

The gS-module∧j

(u−)⊗X is completely reducible, see [K]. Note that this istrue even though S is not necessarily of “finite type”, as in the case of [GL], oreven finite. Furthermore, by the same argument appearing in [GL], the subspace

(∧j

(u−)⊗X)λ is annihilated by ei for all i ∈ S. In fact, the submodule generated

by the one-dimensional space (∧j

(u−)⊗X)λ is isomorphic to the irreducible moduleL(λ), see also [K].

Thus we have shown the following generalization of Theorem 8.5 of [GL]:

Theorem 3.12. Let X be a standard ge-module with highest weight µ ∈ P+.Let S ⊂ I0. The correspondence

(w, η) 7→ w(µ+ ρ− η)− ρ


is a bijection from the set (w, η) ∈ W (S)× Ω(µ) | l(w) + ht (η) = j onto the set

of all weights λ of∧j

(u−) ⊗ X such that (λ + ρ, λ + ρ) = (µ + ρ, µ + ρ). Eachsuch weight w(µ+ ρ− η)− ρ occurs with multiplicity one in

∧(u−)⊗X with one-

dimensional weight space∧j

(u−)−〈Φw〉−wη ⊗Xwµ. Furthermore, w(µ+ ρ− η)− ρis in PS, and the weight space of w(µ+ρ−η)−ρ generates a copy of the irreduciblere-module L(w(µ + ρ − η) − ρ). In particular, in any direct sum decomposition ofthe re-module ∧j

(u−)⊗X =∐λi

L(λi)

where λi ∈ PS, the λi for which (λi+ρ, λi+ρ) = (µ+ρ, µ+ρ) must coincide with thew(µ+ρ−η)−ρ as (w, η) ranges through (w, η) ∈W (S)×Ω(µ) | l(w)+ht (η) = j.Each such λi occuring exactly once in the decomposition

∐λiL(λi). In fact, the

above correspondence (w, η) 7→ w(µ+ρ−η)−ρ is a bijection, so all of the irreduciblere-modules L(w(µ+ ρ− η)− ρ) are inequivalent as w ranges through W (S) and ηranges through Ω(µ).

Recall the complex C∗(X) defined after Proposition 3.8 is the standard re-

module complex for computing H∗(u−, Xt). Also C∗(X) = B∗(x) ⊕ B′∗(X) where

the homology of B′∗(X) is zero by Proposition 3.9. The complex B∗(x) has all zeromaps because Theorem 3.12 and Proposition 3.11 imply that each λ can appear asa weight of Cj(X) for only one j, so that the sequence

· · · −→ Cj−1(X)(λ) −→ Cj(X)(λ) −→ Cj+1(X)(λ) −→ · · ·

is equal to· · · −→ 0 −→ Cj(X)(λ) −→ 0 −→ · · · .

Therefore Hj(u−, Xt) can be naturally identified with the re-submodule Bj(X) of

Cj(X), and

Bj(X) =∐λ∈PS

(λ+ρ,λ+ρ)=(µ+ρ,µ+ρ)

Cj(X)(λ) =∐

w∈W (S),l(w)≤jη∈Ω(µ,j−l(w))

L(w(µ+ ρ− η)− ρ).

This gives:

Theorem 3.13. Let X,µ, j be as in Theorem 3.12 . The jth homology spaceHj(u

−, Xt) is isomorphic as an re-module to the direct sum∐w∈W (S),l(w)≤jη∈Ω(µ,j−l(w))

L(w(µ+ ρ− η)− ρ)

of inequivalent irreducible re-modules.

3.4 The character formula for standard modules.Let (he)∗Z denote the set of integral linear functionals. Let A be the abelian

group of all possibly infinite formal integral linear combinations of elements of (he)∗Z.Denote by eλ the element of A corresponding to λ ∈ (he)∗Z. Then Z[(he)∗Z], the inte-gral group algebra of (he)∗Z, is contained in A, we have eλeµ = eλ+µ, and elementsof A can be multiplied provided that the resulting expression is well defined.


Let E denote the category of all he-modules X such that X has a direct sumdecomposition

∐λ∈(he)∗Z

Xλ where each Xλ is finite dimensional.

For an object X in the category E , define its formal character X (X) as anelement of A as follows:

X (X) =∑

λ∈(he)∗Z

(dimXλ)eλ

Lemma 3.14. Let

0→ X1 → X2 → X3 → 0

be an exact sequence in E. Then

X (X2) = X (X1) + X (X3).

proof. Obvious because dim(X2)λ = dim(X1)λ + dim(X3)λ.

If C∗ is a chain complex in E , then the homology groups Hj of C∗ are in E .The chain complex C∗ is admissible if

∐j

Cj is in E .

Lemma 3.15. [Euler-Poincare] Let C∗ be an admissible chain complex in E.Then

∐j

Hj is in E and

∑j∈N

(−1)jX (Cj) =∑j∈N

(−1)jX (Hj).

Now we apply the Euler-Poincare principle to the chain complex C∗, so that

Cj = Cj(X) =∧j

(u−)⊗X and Hj = Hj(u−, Xt). For the trivial one-dimensional

ge-module X = E and the special case S = ∅ (so re = he), Theorem 3.13 yields:

Theorem 3.16. We have the denominator identity∏ϕ∈∆+

(1− e−ϕ)dim g−ϕ =∑w∈W

(detw)∑

η∈Ω(0)

(−1)ht(η)ew(ρ−η)−ρ

=∑w∈W

(detw)∑

η∈Ω(0)

(−1)ht(η)e−〈Φw〉−wη.(3.1)

proof. There are dim(∧j

(n−))λ = # of ways of writing λ as a sum of j

elements in ∆+, so that in the alternating sum,∑j∈N

(−1)j∑

λ∈(he)∗Z

dim(∧j

(n−))λeλ,

the coefficient of eλ = (# of ways of writing λ as the sum of an even number ofelements of ∆+)−(# of ways of writing λ as an odd number of elements of ∆+).

This is exactly the coefficient of eλ in∏ϕ∈∆+

(1− e−ϕ)dim g−ϕ .


On the right hand side of the Euler-Poincare formula:∑j∈N

(−1)jX (Hj) =∑j

(−1)j∑w∈Wl(w)≤j

∑η∈Ω(0,j−l(w))

ew(ρ−η)−ρ

=∑w∈W

∑η∈

⋃n≥0

Ω(0,n)

(−1)n+l(w)ew(ρ−η)−ρ

=∑w∈W

(detw)∑

η∈Ω(0)

(−1)ht(η)ew(ρ−η)−ρ.

For X an arbitrary standard ge-module with highest weight µ ∈ P+ we obtaina character formula:

Theorem 3.17. We have the following character formula:

(3.2) X (X) =

∑w∈W (detw)

∑η∈Ω(µ)(−1)ht(η)ew(µ+ρ−η)∑

w∈W (detw)∑η∈Ω(0)(−1)ht(η)ew(ρ−η)

.

proof. For any standard ge-module X Theorem 3.13 gives :∑j∈N

(−1)jX (Hj(X)) =∑w∈W

(detw)∑

η∈Ω(µ)

(−1)ht(η)ew(µ+ρ−η)−ρ.

We also have:∑j∈N

(−1)jX (Cj(X)) =∑j∈N

(−1)jX (∧j

(n−)⊗X) = X (X)∑j∈N

(−1)jX (Cj(E)),

which can be seen by writing (∧j

(n−) ⊗ X)λ as the direct sum of spaces of the

form (∧j

(n−))τ ⊗Xτ ′ where τ + τ ′ = λ. Applying the formula in Theorem 3.16,and using the Euler-Poincare lemma, we obtain the character formula.

Since the character formula given in Theorem 3.17 depends only on the highestweight of the standard module X, we conclude that all standard modules of thesame highest weight are isomorphic. We have established:

Corollary 3.18. Every standard ge-module is irreducible.

As an application of the character formula for standard modules, one can give

a short proof of Theorem 3.19, below. Theorem 3.19 is a generalization of themain theorem of [J], and the direct proof given there works in this case also, withappropriate changes (see also [JW]).

Let I = I1 ∪ I2 (disjoint union), satisfying1. If i, j ∈ I2 and i 6= j then (αi, αj) 6= 02. I0 ⊂ I1Let g1 be the subalgebra of g generated by the ei and fi with i ∈ I1. This

algebra is isomorphic to the generalized Kac-Moody algebra associated to thematrix (aij)i,j∈I1 , by the results of §1.4. There is a triangular decompositiong1 = n+

1 ⊕ h1 ⊕ n−1 .


Theorem 3.19 (Jurisich-Wilson). Let g(A) = g be a generalized Kac-Moody algebra, associated to a symmetrizable matrix A. Then g = u+⊕(g1+h)⊕u−,where u− is the free Lie algebra on the direct sum of the standard highest weightg1-modules U(n−1 ) · fj for j ∈ I2 and u+ is the free Lie algebra on the direct sum ofthe standard lowest weight g1-modules U(n+

1 ) · ej for j ∈ I2.

proof. The subalgebra g1 of g and hence its universal enveloping algebra U(g1)act via the adjoint action on g. We identify the root −αj of g with the weight ofthe highest weight vector fj of the standard g1-module U(g1) · fj ⊂ g, which wedenote by L1(−αi).

By our assumptions on I2, the set Ω(0) of all sums of mutually orthogonaldistinct imaginary simple roots of g can be written as the disjoint union of thethe subset consisting of elements of Ω(0) having no terms αj with j ∈ I2, andthe subset consisting of elements of Ω(0) where there is exactly one term αj withj ∈ I2. Denote by Ω1(µ, n), n ≥ 0, the set of sums of n distinct, pairwise orthogonal,imaginary simple roots of g1, each orthogonal to µ ∈ P+. Then as in the characterformula, let Ω1(µ) = ∪nΩ1(µ, n). Thus Ω(0) =

∐j∈I2(αj +Ω1(−αj))∪Ω1(0). Here

αj + Ω1(−αj) = αj + η | η ∈ Ω1(−αj).Since I0 ⊂ I1 we can decompose the right side of equation (3.1), the denomi-

nator identity, (call this D) of g as

D =∑w∈W

(detw)∑

η∈Ω1(0)

(−1)ht(η)ew(ρ−η)−ρ

−∑w∈W

(detw)∑

j∈I2,η∈Ω1(−αj)

(−1)ht(η)ew(ρ−η−αj)−ρ.(3.3)

The second term of equation (3.3) is precisely e−ρ times the sum of the “numera-tors” ∑

w∈W(detw)

∑η∈Ω(−αj)

(−1)ht(η)ew(−αj+ρ−η)

appearing in the character formula (3.2) of the standard highest weight g1-modulesL1(−αi) as j ranges through I2. Let

Dg1=∑w∈W

(detw)∑

η∈Ω(0)

(−1)ht(η)ew(ρ−η)−ρ,

the “denominator” (times eρ) of the character formula for a standard g1-module.Then equation (3.3) becomes

D = Dg1

1−∑j∈I2

X (L1(−αj))

.Let F (V ) denote the free Lie algebra over the vector space V =

∐j∈I2 L1(−αj).

Let ∆1+ be the set of positive roots of g1, which can be considered as a subset

of ∆+. By a generalization of Witt’s formula for computing the dimension of ahomogeneous subspace of a graded free Lie algebra (see [J] or [Ka])we have:

1−∑j∈I2

X (L1(−αj)) =∏

ϕ∈∆+\∆1+

(1− e−ϕ)dimF (V )−ϕ .


Thus dimF (V )−ϕ = dim g−ϕ. The inclusion V →∐ϕ∈∆+\∆1

+g−ϕ extends to

a grading preserving Lie algebra homomorphism from F (V ) to∐ϕ∈∆+\∆1

+g−ϕ.

The free Lie algebra F (V ) is isomorphic to the ideal of g−0 (see Proposition 1.1)generated by the fj with j ∈ I2. Furthermore this ideal maps onto

∐ϕ∈∆+\∆1

+g−ϕ

by construction of g. Thus the grading preserving homomorphism is a surjection,and by considering dimensions we conclude that there is an isomorphism betweenF (V )−ϕ and g−ϕ.

Thus there is an isomorphism

n− = n−1 n u−,

where u− = F (V ) (the semidirect product is the obvious one given by the actionof n−1 on u−).

By iterating this procedure one can decompose n− further, eventually obtaininga decomposition involving a series of free subalgebras and a Kac-Moody subalgebra(see also [JW]).

References

[B1] R. Borcherds, Generalized Kac-Moody Lie algebras, J. Algebra 115 (1988), 501–512.[B2] , Central extensions of generalized Kac-Moody Lie algebras, J. Algebra 140 (1991),

330–335.

[B3] , Monstrous moonshine and monstrous Lie superalgebras, Invent. Math. 109 (1992),405–444.

[Bo] N. Bourbaki, Groupes et Algebras de Lie ch.IV, Masson, Paris, 1981.

[CN] J. H. Conway and S. P. Norton, Monstrous Moonshine, Bull. London Math. Soc. 11(1979), 308-339.

[FLM] I. Frenkel, J. Lepowsky, and A. Meurman, Vertex Operator Algebras and the Monster,

Academic Press, Boston, 1988.[GK] O. Gabber and V. Kac, On defining relations of certain infinite-dimensional Lie algebras,

Bull. Amer. Math. Soc. 5; no.2 (1981).[GL] H. Garland and J. Lepowsky, Lie algebra homology and the Macdonald-Kac formulas,

Invent. Math. 34 (1976), 37-76.

[GT] R. Gerbert and J. Teschner, On the fundamental representation of Borcherds algebraswith one simple imaginary root, Lett. Math. Phys (to appear).

[HMY] K. Harada, M. Miyamoto, H. Yamada, A generalization of Kac-Moody Lie algebras (to

appear).[J] E. Jurisich, Generalized Kac-Moody Lie algebras, free Lie algebras and the structure of

the Monster Lie algebra, Journal of pure and applied algebra (to appear).[JLW] E. Jurisich, J. Lepowsky, R. Wilson, Realizations of the Monster Lie algebra, Selecta

Mathematica, New Series 1 (1995), 129 - 161.[JW] E. Jurisich, R. Wilson, On the structure of generalized Kac-Moody algebras (to appear).

[K] V. Kac, Infinite dimensional Lie algebras, third edition, Cambridge University Press,1990.

[Ka] Seok-Jin Kang, Kac-Moody Lie algebras, spectral sequences, and the Witt formula, Trans.Amer. Math. Soc. 339 (1993), 463-493.

[L] J. Lepowsky, Lectures on Kac-Moody Lie algebras, Universite Paris VI (1978), unpub-lished preprint.

[M] R. Moody, A new class of Lie algebras, Journal of algebra 10 (1968), 211-230.

[N] S. Naito, The strong Bernstein-Gelfand-Gelfand resolution for generalized Kac-Moody

algebras, I, The existence of the resolution, Publ. of the Research Institute for Mathe-matical Sciences 29 (1993), 709-730.


Department of Mathematics, The University of Chicago, Chicago IL 60637E-mail address: [email protected]

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