Kac-Moody Lie Algebras
Chapter III: Representation theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 10 September 2012

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

To begin with, let $A=\left(a_{ij}\right)$ be any $n\times n$ matrix over the field $k$. Eventually $A$ will have to be a symmetrizable Cartan matrix, but we shall bring in that assumption only when it becomes necessary.

Recall that

$$𝔤=𝔤\left(A\right)=𝔥+\sum _{\alpha \in R}{𝔤}_{\alpha }\text{(direct sum)}$$

and that each root space ${𝔤}_{\alpha }$ is a finite-dimensional (1.7).

Let $M$ be a $𝔤$–module, i.e. a $k$–vector space on which $𝔤$ acts, so that we are given a Lie algebra homomorphism $\pi :𝔤\to 𝔤𝔩\left(M\right),$ which extends to $\pi :U\left(𝔤\right)\to \text{End}\left(M\right),$ i.e $M$ is a $U\left(𝔤\right)$–module. Notation $(M,\pi )$ when I want to be pedantic. More often then not I shall suppress $\pi$ and write $x.v$ or $xv$ for $\pi \left(x\right)v$ $(x\in 𝔤,v\in M)$.

Weights

For any $𝔤$–module $M$ and any $\lambda \in {𝔥}^{*}$ we define

$$M_{\lambda }≔\{v\in M:h.v=\lambda \left(h\right)v\text{for all}h\in 𝔥\}\text{.}$$

If $M_{\lambda }\ne 0$ we say that $\lambda$ is a weight of $M$, that $M_{\lambda }$ is the weight space and that the elements of $M_{\lambda }$ are the weight vectors for the weight $\lambda$. We have

$$\begin{array}{cc}{M}_{\lambda }\cong {\text{Hom}}_{U\left(𝔥\right)}(E_{\lambda },M)& \left(✶\right)\end{array}$$

where $E_{\lambda }$ is the $1$–dimensional $𝔥$–module defined by $\lambda ,$ that is to say $E_{\lambda }=k{e}_{\lambda }$ where $h.e_{\lambda }=\lambda \left(h\right)e_{\lambda }$ for all $h\in 𝔥.$ The isomorphism $\left(✶\right)$ associates to each $v\in M_{\lambda }$ the homomorphism $E_{\lambda }\to M$ which takes $e_{\lambda }$ to $v.$ From $\left(✶\right)$ it follows that (for a fixed $\lambda \in {𝔥}^{*}$) $M\mapsto M_{\lambda }$ is a left exact functor (from $𝔤$–modules to $𝔥$–modules).

Example:$$ $(M,\pi )=(𝔤,\text{ad}).$ The weight spaces are $𝔥$ and the ${𝔤}_{\alpha },$ and the set of weights is $R\cup \left\{0\right\}.$

(3.1)

1. For $\alpha \in R\cup \left\{0\right\}$ and $\lambda \in {𝔥}^{*}$ we have

   $${𝔤}_{\alpha }.M_{\lambda }\subset M_{\lambda +\alpha }$$

2. The sum ${\displaystyle M^{\prime }=\sum _{\lambda \in {𝔥}^{*}}{M}_{\lambda }}$ is direct, and $M^{\prime }$ is a $𝔤$–submodule of $M.$

3. If $\phi :M\to N$ is a $𝔤$–module homomorphism, then $\phi \left(M_{\lambda }\right)\subset N_{\lambda }$ for all $\lambda .$

		Proof.
	Let $x\in {𝔤}_{\alpha },$ $v\in M_{\lambda },$ $h\in 𝔥.$ Then we calculate $$\begin{array}{ccc}h.(x.v)& =& x.h.v+[h,x].v\\ & =& \lambda \left(h\right)x.v+\alpha \left(h\right)x.v\\ & =& (\lambda +\alpha )\left(h\right)x.v\end{array}$$ so that $x.v\in M_{\lambda +\alpha }.$ If the sum $\sum M_{\lambda }$ is not direct, there will be nontrivial relations of the form $$\begin{array}{cc}\sum _{i=1}^m{v}_{{\lambda }_i}=0& \text{(1)}\end{array}$$ where $v_{{\lambda }_i}\in M_{{\lambda }_i},v_{{\lambda }_i}\ne 0$ and ${\lambda }_1,\dots ,{\lambda }_m\in {𝔥}^{*}$ are all distinct. Choose such a relation with $m$ $(\ge 2)$ as small as possible. By operating on (1) with an element $h\in 𝔥,$ we obtain $$\begin{array}{cc}\sum _{i=1}^m{\lambda }_i\left(h\right)v_{{\lambda }_i}=0& \text{(2)}\end{array}$$ Choose $h\in 𝔥$ such that ${\lambda }_1\left(h\right)\ne {\lambda }_2\left(h\right),$ multiply (1) by ${\lambda }_1\left(h\right)$ and subtract from (2). This produces a nontrivial relation of length $<m:$ contradiction. Also it is clear from (i) that ${𝔤}_{\alpha }.M^{\prime }\subset M^{\prime }$ for each $\alpha \in R\cup \left\{0\right\},$ whence $𝔤.M^{\prime }\subset$M with a prime? Obvious. $\square$

If $M$ is any $𝔤$–module, let $P\left(M\right)\subset {𝔥}^{*}$ denote the set of weights of $M.$ (It might be empty.) Also, for each $\lambda \in {𝔥}^{*},$ let

$$D\left(\lambda \right)=\lambda -Q^{+}$$

and for any subset $F$ of ${𝔥}^{*}$ let

$$D\left(F\right)=\bigcup _{\lambda \in F}D\left(\lambda \right)\text{.}$$

We shall use this notation only for finite subsets $F$ of ${𝔥}^{*}\text{.}$

Let $𝒪$ denote the category of $𝔤$–modules $M$ which satisfy the following two conditions:

1. $M$ is $𝔥$–diagonalizable with finite dimensional weight spaces, i.e.

   $$M=\sum _{\mu \in P\left(M\right)}{M}_{\mu }$$

   (direct sum, by (3.1)), with each $M_{\mu }$ finite-dimensional;

2. $P\left(M\right)\subset D\left(F\right)$ for some finite $F\subset {𝔥}^{*}.$

   The morphisms in $𝒪$ are $𝔤$–module homomorphisms.

(3.2) Let $\left(E\right)0⟶M^{\prime }⟶M⟶M^{\prime \prime }⟶0$ be a short exact sequences of $𝔤$–modules, with $M\in 𝒪$. Then

1. $M^{\prime },M^{\prime \prime }\in 𝒪;$

2. For each $\lambda \in {𝔥}^{*}$ the sequence $\left(E_{\lambda }\right)0⟶M_{\lambda }^{\prime }\stackrel{f}{⟶}{M}_{\lambda }\stackrel{g}{⟶}{M}_{\lambda }^{\prime \prime }⟶0$ is exact;

3. $P\left(M\right)=P\left(M^{\prime }\right)\cup P\left(M^{\prime \prime }\right).$

		Proof.
	Since $M$ is $𝔥$–diagonalizable we have $M^{\prime }={\displaystyle \underset{\lambda }{\oplus }}{M}_{\lambda }^{\prime }$ by (1.5), and $f\left(M_{\lambda }^{\prime }\right)\subset M_{\lambda }$ by (3.1), so that $M_{\lambda }^{\prime }$ is finite-dimensional and $P\left(M^{\prime }\right)\subset P\left(M\right)\subset D\left(F\right).$ Hence $M^{\prime }\in 𝒪.$ Next, we have $g\left(M_{\lambda }\right)\subset M_{\lambda }^{\prime \prime }$ by (3.1), for all $\lambda \in {𝔥}^{\prime },$ hence $$M^{\prime \prime }=g\left(M\right)=\sum g\left(M_{\lambda }\right)\subset \sum M_{\lambda }^{\prime \prime }\subset M^{\prime \prime };$$ consequently we have equality throughout, whence $g\left(M_{\lambda }\right)=M_{\lambda }^{\prime \prime }$ for each $\lambda \in {𝔥}^{*}$ and the sequence $\left(E_{\lambda }\right)$ is therefore exact. Finally, $M_{\lambda }^{\prime \prime }$ is finite-dimensional and $P\left(M^{\prime \prime }\right)\subset P\left(M\right),$ so that $M^{\prime \prime }\in 𝒪,$ and (iii) is now obvious. $\square$

Recall the partial order $\lambda \ge \mu$ on ${𝔥}^{*}:$ $\lambda \ge \mu$ iff $\lambda -\mu -{\displaystyle \sum _1^n}{u}_i{\alpha }_i$ with each $u_i\ge 0.$

(3.3) Each module $M\in 𝒪$ has at least one maximal weight.

		Proof.
	Suppose $M$ has no maximal weight. Then $P\left(M\right)$ contains an infinite strictly increasing sequence ${\mu }_1<{\mu }_2<\dots .$ For each $\lambda \in F,$ the ${\mu }_i\in D\left(\lambda \right)$ form a subsequence. Since $F$ is finite, at least one of these subsequences is infinite, say $v_1<v_2<\dots$ in $D\left(\lambda \right).$ Each $v_i\in D\left(\lambda \right)=\lambda -Q^{+},$ hence $\text{ht}(\lambda -v_i)$ is a nonnegative integer. It follows that the sequence ${\left(\text{ht}(\lambda -v_i)\right)}_{i\ge 1}$ is an infinite strictly decreasing sequence of integers $\ge 0,$ which is absurd. $\square$

If $M$ has a unique maximal weight $\lambda ,$ then $\lambda$ is called the highest weight of $M.$

Highest weight $𝔤$–modules

We shall say that a $𝔤$–module $M$ is a highest weight (h.w.) module if

1. $M$ has a highest weight, say $\lambda ;$

2. $M$ is generated (as $U\left(𝔤\right)$–module) by some $v_{\lambda }\in M_{\lambda }.$

(3.4) Let $M$ be a h.w. $𝔤$–module, with highest weight $\lambda .$ Then

1. $M\in 𝒪;$

2. $\text{dim}{M}_{\lambda }=1;$

3. $P\left(M\right)\subset D\left(\lambda \right);$

4. $M$ has a unique maximal submodule, hence a unique simple quotient;

5. If $M^{\prime }$ is a nonzero homomorphic image of $M,$ then $M^{\prime }$ is h.w. with h.w.

		Proof.
	We have $\lambda +{\alpha }_i\notin P\left(M\right),$ hence by (3.1) $e_i.v_{\lambda }=0$ $(1\le i\le n).$ It follows that ${𝔫}_{+}.v_{\lambda }=0,$ i.e. $U\left({𝔫}_{+}\right).v_{\lambda }=k{v}_{\lambda }.$ Since $U\left(𝔤\right)=U\left({𝔫}_{-}\right)U\left(𝔥\right)U\left({𝔫}_{+}\right),$ we have $M=U\left(𝔤\right)v_{\lambda }=U\left({𝔫}_{-}\right)v_{\lambda }.$ Let $y_1,y_2,\dots$ be a $k$–basis of ${𝔫}_{-}$ consisting of root vectors. By Poincaré-Birkhoff-Witt, the monomials $y_1^{r_1}{y}_2^{r_2}\dots$ form a $k$–basis of $U\left({𝔫}_{-}\right),$ hence the vectors $y_1^{r_1}{y}_2^{r_2}\dots v_{\lambda }$ span $M$ (as a $k$–vector space). But each such vector is a weight vector, for if $y_i\in {𝔤}_{-{\beta }_i}$ then $y_1^{r_1}{y}_2^{r_2}\dots v_{\lambda }\in M_{\lambda -r_1{\beta }_1-r_2{\beta }_2-\dots }.$ It follows that $M$ is the sum of its weight spaces and that each weight space $M_{\mu }$ is finite-dimensional, for there are only finitely many solutions of the equation $\mu =\lambda -\sum r_i{\beta }_i$ in non-negative integers $r_i.$ Moreover each such $\mu \in D\left(\lambda \right),$ and in particular $M_{\lambda }$ is 1–dimensional, generated by $v_{\lambda }.$ So we have proved (i) – (iii). Let $M^{\prime }$ be a proper submodule of $M.$ Then $M^{\prime }\in 𝒪$ (3.2), hence $M^{\prime }=\sum M_{\mu }^{\prime }$ where $M_{\mu }^{\prime }=M^{\prime }\cap M_{\mu }.$ But $M_{\lambda }^{\prime }=0,$ otherwise by (ii) $M_{\lambda }^{\prime }$ would contain and hence $M^{\prime }=M.$ It follows that $$M^{\prime }\subset M_{+}=\sum _{\mu \ne \lambda }{M}_{\mu }$$ and hence the sum of all proper submodules of $M$ is contained in $M_{+},$ hence is a proper submodule. This proves (iv), and (v) is clear. $\square$

We shall now show how to construct all h.w. $𝔤$–modules.

Verma modules

Let $\lambda \in {𝔥}^{*}$ and let $E_{\lambda }$ as before denote the 1–dimensional $𝔥$–module corresponding to $\lambda :$ $E_{\lambda }=k{u}_{\lambda }$ where $h.u_{\lambda }=\lambda \left(h\right)u_{\lambda }$ for all $h\in 𝔥.$

Let $𝔟=𝔥+{𝔫}_{+}$ be the subalgebra of $𝔤$ generated by $𝔥$ and $e_1,\dots ,e_n.$ The subalgebra $𝔟$ is a semidirect product ${𝔫}_{+}⋊𝔥,$ because ${𝔫}_{+}$ is an ideal in $𝔟$ and $𝔟/{𝔫}_{+}=𝔥.$ We may regard $E_{\lambda }$ as a $𝔟$–module by making ${𝔫}_{+}$ act trivially, i.e. ${𝔫}_{+}.u_{\lambda }=0.$ The Verma module $V\left(\lambda \right)$ is defined to be the induced $𝔤$–module

$$V\left(\lambda \right)={\text{ind}}_{𝔟}^{𝔤}\left(E_{\lambda }\right)=U\left(𝔤\right){\otimes }_{U\left(𝔟\right)}{E}_{\lambda }.$$

Let $v_{\lambda }=1\otimes u_{\lambda }\in V\left(\lambda \right).$ Clearly $v_{\lambda }$ generates $V\left(\lambda \right),$ and since $U\left(𝔤\right)=U\left({𝔫}_{-}\right)\otimes U\left(\text{something}\right)$ we have $V\left(\lambda \right)=U\left({𝔫}_{-}\right).v_{\lambda },$ showing that $V\left(\lambda \right)$ is a h.w. $𝔤$–module with highest weight $\lambda ,$ and that it is free of rank 1 as a $U\left({𝔫}_{-}\right)$–module.

Alternative description of $V\left(\lambda \right):$ let $J\left(\lambda \right)$ denote the left ideal in $U\left(𝔤\right)$ generated by something and all $h-\lambda \left(h\right),$ $h\in 𝔥.$ Then

$$V\left(\lambda \right)\cong U\left(𝔤\right)/J\left(\lambda \right).$$

For if $\pi$ is the representation of $U\left(𝔟\right)$ on $E_{\lambda },$ then $\pi :U\left(𝔟\right)\to k$ is such that $\pi \left(e_i\right)=0(1\le i\le n)$ and $\pi \left(h\right)=\lambda \left(h\right),$ all $h\in 𝔥;$ hence $K=\text{Ker}\left(\pi \right)$ is the left ideal of codimension 1 in $U\left(𝔟\right)$ generated by ${𝔫}_{+}$ and all $h-\lambda \left(h\right);$ tensoring the exact sequence (of left $U\left(𝔟\right)$–modules)

$$0⟶K⟶U\left(𝔟\right)⟶E_{\lambda }⟶0$$

with $U\left(𝔤\right)$ (over $U\left(𝔟\right)$) gives

$$U\left(𝔤\right){\otimes }_{U\left(𝔟\right)}K⟶U\left(𝔤\right)⟶V\left(\lambda \right)⟶0$$

and the image of $U\left(𝔤\right){\otimes }_{U\left(𝔟\right)}K$ in $U\left(𝔤\right)$ is $J\left(\lambda \right).$

The Verma modules are the "universal" h.w. $𝔤$–modules:

(3.5)

1. $V\left(\lambda \right)$ is a h.w. $𝔤$–module with highest weight $\lambda .$

2. Every h.w. $𝔤$–module with highest weight $\lambda$ is a homomorphic image of $V\left(\lambda \right).$

		Proof.
	Already observed above. Let $M$ be a h.w. with generator $x\in M_{\lambda }.$ Then the ideal $J\left(\lambda \right)$ kills $x,$ hence $M$ is a homomorphic image of $U\left(𝔤\right)/J\left(\lambda \right)=V\left(\lambda \right).$ $\square$

By (3.4)(iv) it follows that $V\left(\lambda \right)$ has a unique simple quotient $L\left(\lambda \right):$ by (3.2) and (3.4)(i), we have $L\left(\lambda \right)\in 𝒪.$ Moreover, the $L\left(\lambda \right)$ are precisely the simple objects in the category $𝒪:$

(3.6) If $M\in 𝒪$ is simple, then $M\cong L\left(\lambda \right)$ for a unique $\lambda \in {𝔥}^{*}.$

		Proof.
	By (3.3), $M$ has at least one maximal weight, say $\lambda .$ Let $x\in M_{\lambda },$ $x\ne 0.$ Then ${𝔫}_{+}.x=0.$ (because $\lambda +{\alpha }_i\notin P\left(M\right),1\le i\le n),$ hence $x$ is killed by $J\left(\lambda \right),$ and therefore the submodule $U\left(𝔤\right).x=M^{\prime }$ generated by $x$ is a quotient of $V\left(\lambda \right).$ Since $M^{\prime }\ne 0$ and $M$ is simple, we have $M^{\prime }=M;$ hence $M$ is a simple quotient of $V\left(\lambda \right),$ hence $M\cong L\left(\lambda \right).$ Suppose also that $M\cong L\left(\mu \right).$ Then we have a $𝔤$–isomorphism $\text{something}\left(\lambda \right)\stackrel{\sim }{\to }L\left(\mu \right),$ under which weight spaces correspond (3.1). Hence $\lambda$ is a weight of $L\left(\mu \right),$ whence $\lambda \le \mu ;$ similarly $\mu \le \lambda$ and therefore $\lambda =\mu .$ $\square$

(3.7) Example.$$ When $\lambda =0,$ $E_{\lambda }=k$ with trivial $𝔥$–action $(h.1=0),$ and $V\left(0\right)=U\left({𝔫}_{-}\right).$ The maximal submodule of $V\left(0\right)$ is the augmentation ideal of $U\left({𝔫}_{-}\right),$ hence $L\left(0\right)$ is the trivial 1–dimensional $𝔤$–module.

(3.8) Ket $M$ be a h.w. module. Then ${\text{End}}_{𝔤}\left(M\right)=k.$

		Proof.
	Let $v_{\lambda }=M_{\lambda }$ be a h.w. vector which generates $M.$ If $\phi :M\to M$ is a $𝔤$–module homomorphism something have $\phi \left(v_{\lambda }\right)\in M_{\lambda },$ hence $=a{v}_{\lambda }$ for some $a\in k$ (because $\text{dim}{M}_{\lambda }=1$ (3.4)). The Kernel of $\phi -a.1$ is a submodule of $M$ which contains $v_{\lambda },$ hence is the whole of $M,$ i.e. $\phi -a.1=0.$ $\square$

Characters

Let $\epsilon$ be the set of all functions $f:{𝔥}^{*}\to \mathbb{Z}$ such that $\text{Supp}\left(f\right)\subset D\left(F\right)$ for some finite subset $F$ of ${𝔥}^{*}:$ i.e. $f\left(\mu \right)=0$ unless $\lambda -\mu \in Q^{+}$ for some $\lambda \in F.$ Clearly $\epsilon$ is closed under addition and subtraction of functions; define multiplication by convolution:

$$\begin{array}{cc}\left(fg\right)\left(v\right)=\sum _{\lambda +\mu =v}f\left(\lambda \right)g\left(\mu \right)\text{(finite sum)}& \text{(1)}\end{array}$$

If $\text{Supp}\left(f\right)\subset D\left(F\right),$ $\text{Supp}\left(g\right)\subset D\left(G\right),$ then $\text{Supp}\left(fg\right)\subset D(F+G).$ Thus $\epsilon$ is a commutative ring.

A family ${\left(fj\right)}_{j\in J}$ of functions in $\epsilon$ is summable if $\exists$ finite $F\subset {𝔥}^{*}$ such that

1. $\text{Supp}\left(fj\right)\subset D\left(F\right)$ for all $j\in J;$

2. for each $\lambda \in {𝔥}^{*}$ we have $fj\left(\lambda \right)=0$ for almost all $j\in J.$

In that case the function $f$ defined by

$$f\left(\lambda \right)=\sum _{j\in J}{f}_j\left(\lambda \right)$$

is well defined and belongs to $\epsilon ,$ and we write $f=\sum f_j.$

For each $\lambda \in {𝔥}^{*}$ let $e^{\lambda }$ denote the characteristic function of $\lambda .$ Then $e^{\lambda }{e}^{\mu }=e^{\lambda +\mu }$ from the rule (1) defining multiplication. For any $f\in \epsilon ,$ the family ${\left(f\left(\lambda \right)e^{\lambda }\right)}_{\lambda \in {𝔥}^{*}}$ is summable, and we have

$$\begin{array}{cc}f=\sum _{\lambda }f\left(\lambda \right)e^{\lambda }\in \sum _{\mu \in F}{e}^{\mu }\mathbb{Z}\left[[e^{-{\alpha }_1},\dots ,e^{-{\alpha }_n}]\right]& \text{(2)}\end{array}$$

Now let $M\in 𝒪$ and define the formal character of $M$ to be the function $\text{ch}\left(M\right)$ defined by

$$\text{ch}\left(M\right)\left(\lambda \right)=\text{dim}{M}_{\lambda };$$

thus $\text{Supp}\left(\text{ch}\left(M\right)\right)=P\left(M\right)\subset D\left(F\right)$ for some finite $F\subset {𝔥}^{*},$ so that $\text{ch}\left(M\right)\in \epsilon$ and by (2) we have

$$\text{ch}\left(M\right)=\sum _{\lambda \in P\left(M\right)}\text{dim}{M}_{\lambda }.e^{\lambda }$$

Thus $\text{ch}\left(M\right)$ is nothing but the generating function for the multiplicities of the weights $\lambda \in P\left(M\right).$

(3.9) ch is an additive function on the category $𝒪,$ i.e. if $0⟶M^{\prime }⟶M⟶M^{\prime \prime }⟶0$ is an exact sequence in $ℳ,$ then

$$\text{ch}\left(M\right)=\text{ch}\left(M^{\prime }\right)+\text{ch}\left(M^{\prime \prime }\right).$$

		Proof.
	This follows from (3.2)(ii) by counting dimensions $\square$

More generally, if $0⟶M_0⟶M_1⟶\dots ⟶M_r⟶0$ is an exact sequence in $𝒪,$ we have

$$\sum _{i=0}^r{(-1)}^i\text{ch}\left(M_i\right)=0$$

by breaking up the exact sequence into short exact sequences and applying (3.9).

We shall first compute the character of a Verma module $V\left(\lambda \right):$

(3.10) Let $\lambda \in {𝔥}^{*}.$ Then

$$\text{ch}\left(V\left(\lambda \right)\right)=e^{\lambda }/\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{m_{\alpha }}=e^{\lambda }/\pi \text{say}$$

where as usual $m_{\alpha }=\text{dim}{𝔤}_{\alpha }=\text{dim}{𝔤}_{-\alpha }.$ (The product on the right is a unit in $\epsilon .)$

		Proof.
	We saw earlier that $V\left(\lambda \right)$ is a free $U\left({𝔫}_{-}\right)$–module of rank 1. As before, let $y_1,y_2,\dots$ be a $k$–basis of ${𝔫}_{-}$ consisting of root vectors, say $y_i\in {𝔤}_{-{\beta }_i}.$ Then $V\left(\lambda \right)$ has a $k$–basis consisting of weight vectors $y_1^{r_1}{y}_2^{r_2}\dots v_{\lambda },$ where each $r_i\ge 0$ (and $\sum r_i<\infty ),$ the vector just written being of weight $\lambda -r_1{\beta }_1-r_2{\beta }_2-\dots$ $\square$

Invariant bilinear form

As before, $A=\left(a_{ij}\right)$ is any $n\times n$ matrix $/k,$ and $𝔤=𝔤\left(A\right).$ Suppose that there exists a symmetric ($k$–valued) bilinear form $⟨x,y⟩$ on $𝔤$ such that

1. $⟨[x,y],z⟩=⟨x,[y,z]⟩$ for all $x,y,z\in 𝔤$ (invariance)

2. the restriction of $⟨,⟩$ to $𝔥$ is nondegenerate.

Then for any $h\in 𝔥$ we have

$$\begin{array}{ccccc}⟨h,h_j⟩& =& ⟨h,[e_j,f_j]⟩& =& ⟨[h,e_j],f_j⟩\\ & & & =& {\alpha }_j\left(h\right)⟨e_j,f_j⟩\end{array}$$

Let ${\epsilon }_j=⟨e_j,f_j⟩\in k.$ Condition (ii) ensures that ${\epsilon }_j\ne 0;$ taking $h=h_i$ in the calculation above we have

$$⟨h_i,h_j⟩=a_{ij}{\epsilon }_j$$

and therefore the matrix $AE=\left(a_{ij}{\epsilon }_j\right)$ is symmetric, i.e. $A$ is symmetrizable ($E$ is a nonsingular diagonal matrix).

This proves the first part of

(3.11) Let $⟨x,y⟩$ be an invariant symmetric bilinear form on $𝔤,$ whose restriction to $𝔥$ is nondegenerate. Then

1. the matrix $A$ is symmetrizable

2. $⟨x,y⟩$ is nondegenerate on $𝔤$

3. $⟨x,y⟩$ restricted to ${𝔤}_{\alpha }\times {𝔤}_{\beta }$ (where $\alpha ,\beta \in R\cup \left\{0\right\}$) is

   1. zero if $\alpha +\beta \ne 0$

   2. nondegernerate if $\alpha +\beta =0$

4. If $\alpha \in R$ and $x\in {𝔤}_x,$ $y\in {𝔤}_{-\alpha },$ then

   $$[x,y]=⟨x,y⟩h_{\alpha }^{\vee }$$

   where $h_{\alpha }^{\vee }\in 𝔥$ is defined by $⟨h,h_{\alpha }^{\vee }⟩=\alpha \left(h\right)$ for all $h\in 𝔥.$

		Proof.
	Let $𝔞=\{x\in 𝔤:⟨x,𝔤⟩=0\}.$ Since the form is invariant, $𝔞$ is an ideal in $𝔤:$ for if $x\in 𝔞$ and $y,z\in 𝔤$ we have $$⟨[x,y],z⟩=⟨x,[y,z]⟩=0$$ whence $[x,y]\in 𝔞.$ Now all ideals in $𝔞$ are graded (1.7), hence $𝔞=\sum {𝔞}_{\alpha },$ where ${𝔞}_{\alpha }=𝔞\cap {𝔤}_{\alpha }$ (and ${𝔞}_0=𝔞\cap 𝔥).$ But ${𝔞}_0=0,$ because if $h\in {𝔞}_0$ then certainly $⟨h,𝔥⟩=\text{something}$ whence $h=0.$ But $𝔤$ has no nontrivial ideals with trivial $𝔥$–component, hence $𝔞=0.$ Let $x\in {𝔤}_{\alpha },$ $y\in {𝔤}_{\beta },$ $h\in 𝔥.$ Then $⟨[x,h],y⟩=⟨x,[h,y]⟩$ and thus $$-\alpha \left(h\right)⟨x,y⟩=\beta \left(h\right)⟨x,y⟩$$ If $\alpha +\beta \ne 0,$ choose $h$ such that $\alpha \left(h\right)+\beta \left(h\right)\ne 0.$ It follows that $⟨x,y⟩=0,$ which proves (a). Next, suppose $x\in {𝔤}_{\alpha }$ is such that $⟨x,{𝔤}_{-\alpha }⟩=0.$ Then $⟨x,𝔤⟩=0$ by (a), whence $x=0$ by (ii). We have $$⟨h,[x,y]⟩=⟨[h,x],y⟩=\alpha \left(h\right)⟨x,y⟩=⟨h,h_{\alpha }^{\vee }⟩⟨x,y⟩$$ whence the result, by nondegeneracy. $\square$

Proposition (3.11) has a converse:

(3.12) Suppose that the matrix $A$ is symmetrizable. Then there exists a nondegenerate symmetric invariant bilinear form $⟨x,y⟩$ on $𝔤$ (which therefore has the properties listed in (3.11)).

		Proof.
	By assumption, there exist non-zero scalars ${\epsilon }_j$ such that $a_{ij}{\epsilon }_j=a_{ji}{\epsilon }_i.$ We shall first construct the form on $𝔥$ (of (2.23)) and then extend it to $𝔤.$ Choose a vector space complement ${𝔥}^{\prime \prime }$ of ${𝔥}^{\prime }$ in $𝔥$ (where as usual ${𝔥}^{\prime }={\displaystyle \sum _1^n}k{h}_i$) and define $⟨x,y⟩$ on $𝔥\times 𝔥$ by $$\begin{array}{ccccccc}⟨x,h_i⟩& =& ⟨h_i,x⟩& =& {\epsilon }_i{\alpha }_i\left(x\right)& & (x\in 𝔥)\\ ⟨y,z⟩& =& 0& & & & (y,z\in {𝔥}^{*})\end{array}$$ To see that this form is nondegenerate, suppose that $h\in 𝔥$ is such that $⟨h,𝔥⟩=0.$ Then in particular we have ${\epsilon }_i{\alpha }_i\left(h\right)=⟨h,h_i⟩=0,$ whence $h\in {\displaystyle \underset{1}{\overset{n}{\cap }}}\text{Ker}{\alpha }_i=𝔠\subset {𝔥}^{\prime };$ thus $\sum {\lambda }_i{h}_i$ say, and then $$\sum _1^n{\lambda }_i{\epsilon }_i{\alpha }_i\left(x\right)=⟨h,x⟩=0$$ for all $x\in 𝔥,$ so that $\sum {\lambda }_i{\epsilon }_i{\alpha }_i=0$ in ${𝔥}^{*},$ hence ${\lambda }_1=\dots ={\lambda }_n=0$ and so $h=$something. Recall the principal $\mathbb{Z}$–grading of $𝔤:$ $${𝔤}_r=\sum _{\text{ht}\alpha =r}{𝔤}_{\alpha };𝔤=\sum _{r\in \mathbb{Z}}{𝔤}_r;{𝔤}_0=𝔥$$ Let ${\displaystyle G_n=\sum _{\mid r\mid \le n}{𝔤}_r}$ for $n\ge 0.$ The extension of $⟨x,y⟩$ to $G_1$ is unique, for by (3.11)(iii) we must have $⟨{𝔤}_{\alpha },{𝔤}_{\beta }⟩=0$ if $\alpha +\beta \ne 0,$ and also $⟨e_j,f_j⟩={\epsilon }_j.$ It is straightforward to verify that $$\begin{array}{cc}⟨[x,y],z⟩=⟨x,[y,z]⟩& \text{(1)}\end{array}$$ whenever all 5 terms lie in $G_1.$ We shall now extend $⟨,⟩$ to a symmetric bilinear form on $G_n$ $(n\ge 2)$ by induction on $n,$ such that (1) holds whenever all 5 terms are in $G_n,$ and such that $$\begin{array}{cc}⟨{𝔤}_i,{𝔤}_j⟩=0\text{if}i+j\ne 0& \text{(2)}\end{array}$$ whenever $\mid i\mid \le n$ and $\mid j\mid \le n.$ So assume $n\ge 2$ and $⟨,⟩$ defined on $G_{n-1},$ satisfying (1) and (2). To extend the form to $G_n$ we have, in view of (2), only to define $⟨x,y⟩$ on ${𝔤}_n\times {𝔤}_{-n}.$ Write $$\begin{array}{cccccccc}& x& =& \sum _i[s_i,t_i]& \in & {𝔤}_n& & \text{(3)}\\ & y& =& \sum _j[u_j,v_j]& \in & {𝔤}_{-n}& & \text{(4)}\end{array}$$ where $s_i,t_i$ (resp. $u_j,v_j)$ are homogeneous of positive (resp. negative) degree, hence lie in $G_{n-1}.$ Define now $$\begin{array}{cc}⟨y,x⟩=⟨x,y⟩=\sum _j⟨[x,u_j],v_j⟩.& \text{(5)}\end{array}$$ The whole point is to show that this is well defined, i.e. that it does not depend on the expression (4) for $y.$ For this purpose we make the following calculation: dropping the suffixes, $$\begin{array}{cccc}⟨[[s,t],u],v⟩& =& ⟨[s,[t,u]],v⟩-⟨[t,[s,u]],v⟩& \text{(Jacobi)}\\ & =& -⟨[t,u],[s,v]⟩+⟨[s,u],[t,v]⟩& \text{(invariance)}\\ & =& -⟨[s,v],[t,u]⟩-⟨[s,u],[v,t]⟩& \text{(symmetry)}\\ & =& -⟨s,[v,[t,u]]⟩-⟨s,\left[u[v,t]\right]⟩& \text{(invariance)}\\ & =& ⟨s,[t,[u,v]]⟩& \text{(Jacobi)}\end{array}$$ i.e. we have $$\begin{array}{cccccc}& ⟨[[s,t],u],v⟩& =& ⟨s,[t,[u,v]]⟩& & \text{(6)}\end{array}$$ From (3), (4) and (6) it follows that $$\begin{array}{ccc}\sum _j⟨[x,u_j],v_j⟩& =& \sum _{i,j}⟨[[s_i,t_i],u_j],v_j⟩\\ & =& \sum _{i,j}⟨s_i,[t_i,[u_j,v_j]]⟩\\ & =& \sum _i⟨s_i,[t_i,y]⟩\end{array}$$ and therefore $⟨x,y⟩$ (as defined by (5)) is well-defined, and satisfies the invariance condition (1) by our definition (5). $\square$

Notation. In $𝔥$ we have $⟨h_i,x⟩={\epsilon }_i{\alpha }_i\left(x\right),$ in particular

$$⟨h_i,h_j⟩=a_{ij}{\epsilon }_j=a_{ji}{\epsilon }_i$$

and an isomorphism $\theta :𝔥\to {𝔥}^{*}$ definted by

$$\theta \left(x\right)\left(y\right)=⟨x,y⟩$$

so that

$$\theta \left(h_i\right)\left(x\right)=⟨h_i,x⟩={\epsilon }_i{\alpha }_i\left(x\right)$$

for all $x\in 𝔥,$ whence

$$\begin{array}{ccccc}\theta \left(h_i\right)& =& {\epsilon }_i{\alpha }_i& =& {\alpha }_i^{\vee }\\ {\theta }^{-1}\left({\alpha }_i\right)& =& {\epsilon }_i^{-1}{h}_i& =& h_i^{\vee }\end{array}$$

We use $\theta$ to transport the scalar product from $𝔥$ to $𝔥*:$ thus

$$⟨{\alpha }_i,{\alpha }_j⟩=⟨{\epsilon }_i^{-1}{h}_i,{\epsilon }_j^{-1}{h}_j⟩={\epsilon }_i^{-1}{a}_{ij}={\epsilon }_j^{-1}{a}_{ji}$$

Casimir operator

In the classical situation, where $𝔤$ is finite-dimensional, the Casimir operator plays an important role in representation theory. The invariant bilinear form may be regarded as an element $B\in {(𝔤\otimes 𝔤)}^{*}={𝔤}^{*}\otimes {𝔤}^{*};$ since it is non-degenerate it induces an isomorphism of ${𝔤}^{*}$ with $𝔤,$ hence determines an element of $𝔤\otimes 𝔤.$ The image of this in $U\left(𝔤\right)$ (which is a quotient of the tensor algebra $T\left(𝔤\right))$ is the Casimir element $\omega$. Since $B$ is invariant it follows that $\omega$ is in the centre of $U\left(𝔤\right),$ hence acts as a scalar on any simple $𝔤$–module. Explicitly, if $x_1,\dots ,x_n$ is any $k$–basis of $𝔤,$ let $y_1,\dots ,y_n$ be the dual basis (so that $⟨x_i,y_j⟩={\delta }_{ij});$ then $\omega ={\displaystyle \sum _1^n}{y}_i{x}_i.$

In the present situation, where $A$ is any symmetrizable matrix, we proceed as follows. Let $\alpha \in R^{+}\cup \left\{0\right\};$ by (3.11), the bilinear form $⟨x,y⟩$ is nondegenerate on ${𝔤}_{\alpha }\times {𝔤}_{-\alpha };$ choose a basis $x_1,\dots ,x_m$ of ${𝔤}_{\alpha }$ $(m=m_{\alpha });$ let $y_1,\dots ,y_m$ be the dual basis of ${𝔤}_{-\alpha },$ and define

$$u_{\alpha }=\sum _{i=1}^m{y}_i{x}_i\in U\left(𝔤\right).$$

Then $u_{\alpha }$ is independent of the choice of dual bases, for if $x_1^{\prime },\dots ,x_m^{\prime };$ $y_1^{\prime },\dots ,y_m^{\prime }$ is another pair of dual bases, we have

$$\begin{array}{ccc}{y}_i& =& \sum _j⟨x_j^{\prime },y_i⟩y_j^{\prime }\\ x_j^{\prime }& =& \sum _i⟨x_j^{\prime },y_i⟩x_i\end{array}$$

and therefore

$$\sum _i{y}_i{x}_i=\sum _{i,j}⟨x_j^{\prime },y_i⟩y_j^{\prime }{x}_i=\sum _j{y}_j^{\prime }{x}_j^{\prime }.$$

If $\alpha \in Q^{+}$ is not a root (or zero) we define $u_{\alpha }=0$ (the sum is empty).

Example:$$ We have $⟨e_i,f_i⟩={\epsilon }_i,$ hence $u_{{\alpha }_i}={\epsilon }_i^{-1}{f}_i{e}_i.\text{(1)}$

Let $x\in {𝔤}_{\beta }$ $(\beta \in R\cup \left\{0\right\}),$ then we have

$$\begin{array}{ccc}[u_{\alpha },x]& =& \sum _{i=1}^m(y_i{x}_{i}x-x{y}_i{x}_i)\\ & =& \sum _i[y_i,x]x_i-\sum _i{y}_i[x,x_i]\\ & =& v_{\alpha ,x}-v_{\alpha ,x}^{\prime }\text{say}\end{array}$$

where, for the same reason as before, $v_{\alpha ,x}$ and $v_{\alpha ,x}^{\prime }$ are independent of the choice of dual bases. Since $x\in {𝔤}_{\beta }$ and $x_i\in {𝔤}_{\alpha }$ we have $[x,x_i]\in {𝔤}_{\alpha +\beta }.$ Let $\left(x_j^{\prime }\right)$ be a basis of ${𝔤}_{\alpha +\beta },$ $\left(y_j^{\prime }\right)$ the dual basis of ${𝔤}_{-(\alpha +\beta )}.$ Then

$$\begin{array}{ccc}[x,x_i]& =& \sum _j⟨y_j^{\prime },[x,x_i]⟩x_j^{\prime }\\ & =& \sum _j⟨[y_j^{\prime },x],x_i⟩x_j^{\prime }\text{(invariance)}\end{array}$$

and therefore

$$\begin{array}{ccc}{v}_{\alpha ,x}^{\prime }& =& \sum _i{y}_i[x,x_i]\\ & =& \sum _{i,j}⟨[y_j^{\prime },x],x_i⟩y_i{x}_j^{\prime }\\ & =& \sum _j[y_j^{\prime },x]x_j^{\prime }=v_{\alpha +\beta ,x}\end{array}$$

i.e. we have the formula

$$\begin{array}{cccc}& \begin{array}{|c|}\hline v_{\alpha ,x}^{\prime }=v_{\alpha +\beta ,x}\\ \hline\end{array}(x\in {𝔤}_{\beta })& & \text{(2)}\end{array}$$

Likewise

$$\begin{array}{cccc}& \begin{array}{|c|}\hline v_{\alpha ,x}=v_{\alpha -\beta ,x}^{\prime }\\ \hline\end{array}\phantom{(x\in {𝔤}_{\beta })}& & \text{(3)}\end{array}$$

In particular, $v_{\alpha ,x}^{\prime }=0$ unless both $\alpha$ and $\alpha +\beta$ are positive roots (or 0); and likewise $v_{\alpha ,x}=0$ unless both $\alpha ,\alpha -\beta \in R^{+}\cup \left\{0\right\}.$

Now let

$$u=\sum _{\alpha \in R^{+}}{u}_{\alpha }$$

(in some completion of $U\left(𝔤\right)$...)

(3.13) We have

$$\begin{array}{ccc}[u,e_i]& =& -h_i^{\vee }{e}_i\\ [u,f_i]& =& f_i{h}_i^{\vee }\\ [u,h]& =& 0(h\in 𝔥)\end{array}$$

where $h_i^{\vee }=$ image of ${\alpha }_i$ under the isomorphism ${𝔥}^{*}\to 𝔥$ induced by the bilinear form. (i.e. $⟨h_i^{\vee },h⟩={\alpha }_i\left(h\right),$ so that $h_i^{\vee }={\epsilon }_i^{-1}{h}_i)$

		Proof.
	We compute: $$\begin{array}{ccc}[u,e_i]& =& \sum _{\alpha \in R^{+}}[u_{\alpha },e_i]\\ & =& \sum _{\alpha \in R^{+}}{v}_{\alpha ,e_i}-\sum _{\alpha \in R^{+}}{v}_{\alpha ,e_i}^{\prime }\\ & =& \sum _{\alpha \in R^{+}}{v}_{\alpha ,e_i}-\sum _{\alpha \in R^{+}}{v}_{\alpha +{\alpha }_i,e_i}\text{by (2)}\end{array}$$ But $v_{\alpha ,e_i}=0$ unless $\alpha -{\alpha }_i\in R^{+}\cup \left\{0\right\},$ and therefore $$[u,e_i]=v_{{\alpha }_i,e_i}={\epsilon }_i^{-1}[f_i,e_i]e_i=-{\epsilon }_i^{-1}{h}_i{e}_i=-h_i^{\vee }{e}_i.$$ Similarly we have $$\begin{array}{ccc}[u,f_i]& =& \sum _{\alpha }{v}_{\alpha ,f_i}-\sum _{\alpha \in R^{+}}{v}_{\alpha ,f_i}^{\prime }\\ & =& \sum _{\alpha }{v}_{\alpha +{\alpha }_i,f_i}^{\prime }-\sum _{\alpha }{v}_{\alpha ,f_i}^{\prime }\\ & =& -v_{{\alpha }_i,f_i}^{\prime }=-{\epsilon }_i^{-1}{f}_i[f_i,e_i]={\epsilon }_i^{-1}{f}_i{h}_i=f_{i}h\text{something?}\end{array}$$ Finally $[u,h]=\sum _{\alpha }(v_{\alpha ,h}-v_{\alpha ,h}^{\prime })=0$ by (2). $\square$

Choose an element $\rho \in {𝔥}^{*}$ such that

$$\rho \left(h_i\right)=\frac{1}{2}{a}_{ii}(1\le i\le n)$$

(thus $\rho \left(h_i\right)=1$ if $A$ is a Cartan matrix). Then we have

$$⟨\rho ,{\alpha }_i⟩=\rho \left(h_i^{\vee }\right)={\epsilon }_i^{-1}\rho \left(h_i\right)=\frac{1}{2}{\epsilon }_i^{-1}{a}_{ii}=\frac{1}{2}⟨{\alpha }_i,{\alpha }_i⟩$$

i.e.,

$$\begin{array}{|c|}\hline ⟨2\rho ,{\alpha }_i⟩=⟨{\alpha }_i,{\alpha }_i⟩.\\ \hline\end{array}$$

Now let $M\in 𝒪$ and define a $k$–linear map

$$\Omega ={\Omega }_M:M⟶M$$

as follows: if $v_{\lambda }\in M_{\lambda }$ $(\lambda \in P\left(M\right))$ then

$$\Omega \left(v_{\lambda }\right)={\mid \lambda +\rho \mid }^2{v}_{\lambda }+2u.v_{\lambda }$$

where ${\mid \lambda +\rho \mid }^2=⟨\lambda +\rho ,\lambda +\rho ⟩$ and

$$u.v_{\lambda }=\sum _{\alpha \in R^{+}}{u}_{\alpha }.v_{\lambda }$$

is a finite sum, because ${𝔤}_{\alpha }.v_{\lambda }=0$ for almost all $\alpha \in R^{+}.$

(3.14) ${\Omega }_M$ is a $𝔤$–module homomorphism.

		Proof.
	Since $𝔤=𝔤\left(A\right)$ is generated by the $e_i,$ the $f_i$ and $𝔥,$ it is enough to verify that $\Omega$ commutes with the action of each of these elements. So we calculate: $$\begin{array}{ccc}\Omega (e_i.v_{\lambda })-e_i\Omega \left(v_{\lambda }\right)& =& ({\mid \lambda +{\alpha }_i+\rho \mid }^2-{\mid \lambda +\rho \mid }^2)e_i.v_{\lambda }+2[u,e_i].v_{\lambda }\\ & =& ⟨{\alpha }_i,2\lambda +2\rho +{\alpha }_i⟩e_i.v_{\lambda }-2h_i^{\vee }.e_i.v_{\lambda }\text{by (3.1.}\text{something}\\ & =& (⟨{\alpha }_i,2\lambda +2\rho +{\alpha }_i⟩-2⟨{\alpha }_i,\lambda +{\alpha }_i⟩)e_i.v_{\lambda }\\ & =& ⟨{\alpha }_i,2\rho -{\alpha }_i⟩e_i.v_{\lambda }=0.\end{array}$$ Likewise, $$\begin{array}{ccc}\Omega (f_i,v_{\lambda })-f_i.\Omega \left(v_{\lambda }\right)& =& ({\mid \lambda -{\alpha }_i+\rho \mid }^2-{\mid \lambda +\rho \mid }^2)f_i.v_{\lambda }+2[u,f_i].v_{\lambda }\\ & =& -⟨{\alpha }_i,2\lambda +2\rho -{\alpha }_i⟩f_i.v_{\lambda }+2f_i{h}_i^{\vee }.v_{\lambda }\text{by (3.13)}\\ & =& (-⟨{\alpha }_i,2\lambda +2\rho -{\alpha }_i⟩+2⟨{\alpha }_i,\lambda ⟩)f_i.v_{\lambda }\\ & =& -⟨{\alpha }_i,2\rho -{\alpha }_i⟩f_i.v_{\lambda }=0.\end{array}$$ Finally, $$\begin{array}{ccc}\Omega (h.v_{\lambda })-h.\Omega \left(v_{\lambda }\right)& =& ({\mid \lambda +\rho \mid }^2-{\mid \lambda +\rho \mid }^2)h.v_{\lambda }+2[u,h].v_{\lambda }\\ & =& 0\text{by (3.13) again.}\end{array}$$ $\square$

Remark: $\Omega$ is functorial, i.e. if $f:M\to N$ is a $𝔤$–module homomorphism (with $M,N\in 𝒪)$ then the diagram

$$\begin{array}{ccc}M& \stackrel{f}{⟶}& N\\ {\Omega }_M\downarrow & & \downarrow {\Omega }_N\\ M& \stackrel{f}{⟶}& N\end{array}$$

commutes. For $f$ commutes with the action of $u,$ and preserves weight spaces.

(3.15) Example. Let $M$ be a h.w. module with h. wt. $\lambda .$ If $v_{\lambda }\in M_{\lambda }$ is a generator of $M$ (3.4), we have ${𝔤}_{\alpha }.v_{\lambda }=0$ for all $\alpha \in R^{+},$ hence $u.v_{\lambda }=0$ and therefore ${\Omega }_M.v_{\lambda }={\mid \lambda +\rho \mid }^2{v}_{\lambda }.$ Hence by (3.14) (since $v_{\lambda }$ generates $M)$

$${\Omega }_M={\mid \lambda +\rho \mid }^2.1_M.$$

(3.16) Let $M\in 𝒪$ be such that ${\Omega }_M=a.1_M$ for some scalar $a.$ Let $F$ be a finite subset of ${𝔥}^{*}$ such that $P\left(M\right)\subset D\left(F\right),$ and let

$$S=\{\lambda \in D\left(F\right):{\mid \lambda +\rho \mid }^2=a\}.$$

Then there exist integers $d_{\lambda },\lambda \in S$ such that

$$\text{ch}\left(M\right)={\prod }^{-1}\sum _{\lambda \in S}{d}_{\lambda }{e}^{\lambda }$$

where $\prod ={\displaystyle \prod _{\alpha \in R^{+}}}{(1-e^{-\alpha })}^{m_{\alpha }}.$

		Proof.
	If $\mu \in D\left(F\right)$ we have $\lambda -\mu \in Q^{+}$ for some $\lambda \in F,$ hence $\text{ht}(\lambda -\mu )$ is an integer $\ge 0.$ Define the depth of $\mu$ (relative to $F)$ to be $$\delta \left(\mu \right)=\text{max}\{\text{ht}(\lambda -\mu ):\lambda \in F,\mu \in D\left(\lambda \right)\}$$ so that $\lambda \left(\mu \right)\in \mathbb{N};$ also define $$\delta \left(M\right)=\text{min}\{\delta \left(\mu \right):\mu \in P\left(M\right)\}.$$ Since $F$ is finite there are only finitely many $\mu \in D\left(F\right)$ of given depth; in particular, $M$ has only finitely many weights $\mu$ of least depth $\delta \left(M\right),$ and they are all maximal weights. Call them ${\mu }_1,\dots ,{\mu }_r.$ We shall kill the weight spaces $M_{{\mu }_i}$ $(1\le i\le r).$ Let $d_i=\text{dim}{M}_{{\mu }_i},$ and let $$V=\underset{i=1}{\overset{r}{\oplus }}V{\left({\mu }_i\right)}^{d_i}.$$ Choose a $k$–basis of each $M_{{\mu }_i}$ and let $\phi :V\to M$ be the $𝔤$–homomorphism which maps the generators of the summands of $V$ to the chosen basis elements of the $M_{{\mu }_i}.$ Let $M^{\prime },M^{\prime \prime }$ be the kernel and cokernel of $\phi ,$ so that we have an exact sequence $$0⟶M^{\prime }⟶V\stackrel{\phi }{⟶}M⟶M^{\prime \prime }⟶0.$$ Then $M^{\prime }\in 𝒪$ because it is a submodule of $V,$ and $M^{\prime \prime }\in 𝒪$ because it is a quotient of $M.$ Now $\Omega$ acts as scalar multiplication by ${\mid {\mu }_i+\rho \mid }^2$ on $V{\left({\mu }_i\right)}^{d_i}$ (3.15), and hence also on the image $\phi \left(V{\left({\mu }_i\right)}^{d_i}\right),$ which is a non zero submodule of $M.$ Since by hypothesis $\Omega$ acts as scalar multiplication by $a$ on $M,$ it follows that ${\mid {\mu }_i+\rho \mid }^2=a,$ i.e. ${\mu }_i\in S$ $(1\le i\le r).$ Hence $\Omega$ acts as $a.1$ on $V,$ and hence on $M^{\prime };$ also on $M^{\prime \prime }.$ By construction we have $\delta \left(M^{\prime }\right)>\delta \left(M\right)$ and $\delta \left(M^{\prime \prime }\right)>\delta \left(M\right),$ and by additivity of ch (3.9) $$\begin{array}{ccc}\text{ch}\left(M\right)& =& \text{ch}\left(V\right)+\text{ch}\left(M^{\prime \prime }\right)-\text{ch}\left(M^{\prime }\right)\\ & =& \sum _{i=1}^r{d}_i\text{ch}V\left({\mu }_i\right)+\text{ch}\left(M^{\prime \prime }\right)-\text{ch}\left(M^{\prime }\right).\end{array}$$ Now repeat the same procedure on $M^{\prime }$ and $M^{\prime \prime }.$ After we have done it $m$ times we shall have say $$\text{ch}\left(M\right)=\sum _{\mu \in S_m}{d}_{\mu }\text{ch}V\left(\mu \right)+f_m$$ where $S_m$ is some finite subset of $S,$ and $\delta \left(v\right)>m$ for all $v\in \text{Supp}\left(f_m\right).$ Now let $m\to \infty$ and we have $$\begin{array}{ccc}\text{ch}\left(M\right)& =& \sum _{\mu \in S}{d}_{\mu }\text{ch}V\left(\mu \right)\\ & =& {\prod }^{-1}\sum _{\mu \in S}{d}_{\mu }{e}^{\mu }\text{by (3.10).}\end{array}$$ $\square$

Remark. Suppose in particular that $M$ is a h.w. module, with highest weight $\lambda .$ Then

$$\begin{array}{ccc}\text{ch}\left(M\right)& =& {\prod }^{-1}\sum _{\genfrac{}{}{0ex}{}{\mu \in D\left(\lambda \right)}{{\mid \mu +\rho \mid }^2={\mid \lambda +\rho \mid }^2}}{d}_{\mu }{e}^{\mu }\end{array}$$

with $d_{\mu }\in \mathbb{Z},$ and in particular $d_{\lambda }=1.$

The Weyl-Kac character formula

From now on, $A$ is a Cartan matrix.

Let $(M,\pi )$ be a h.w. module, with highest weight $\lambda \in {𝔥}^{*}.$ Then each $\pi \left(e_i\right)$ is a locally nilpotent endomorphism of $M.$ For if $\mu \in P\left(M\right),$ say $\mu =\lambda -{\displaystyle \sum _{i=1}^n}{m}_i{\alpha }_i,$ and if $x\in M_{\mu },$ then $\pi {\left(e_i\right)}^{m}x\in M_{\mu +m{\alpha }_i}=0$ if $m>m_i.$

If also each $\pi \left(f_i\right)$ is locally nilpotent on $M,$ we shall say that $M$ is a quasi-simple $𝔤$–module. (Later we shall see that quasi-simple $\Rightarrow$ simple).

(3.17) Let $(M,\pi )$ be a h.w. module with highest weight $\lambda ,$ and generator $x\in M_{\lambda }.$ If $\exists k\ge 1$ such that $\pi {\left(f_i\right)}^{k}x=0$ for $1\le i\le n,$ then $M$ is quasi-simple.

		Proof.
	Recall the formula (1.16) $$x^{N}y=\sum _{r=0}^N\left(\genfrac{}{}{0ex}{}{N}{r}\right){\left(\text{ad}x\right)}^{r}y{x}^{N-r}$$ $(x,y\in$ associative ring $R).$ Let $v\in M,$ so that $v=\pi \left(u\right)x$ for some $u\in U\left(𝔤\right),$ and apply (1.16) with $x=\pi \left(f_i\right),$ $y=\pi \left(u\right):$ $$\begin{array}{ccc}\pi {\left(f_i\right)}^{N}v& =& \pi {\left(f_i\right)}^N\pi \left(u\right)x\\ & =& \sum _{r=0}^N\left(\genfrac{}{}{0ex}{}{N}{r}\right)\pi \left({\text{ad}{f}_i}^{r}u\right)\pi {\left(f_i\right)}^{N-r}x.\end{array}$$ Now $\text{ad}{f}_i$ is locally nilpotent on $𝔤$ (1.19), hence also on $U\left(𝔤\right),$ so that ${\left(\text{ad}{f}_i\right)}^{m}u=0$ for some $m\ge 1.$ Hence if $N$ is large enough $(N=k+m-1$ would do) either $r\ge m$ or $N-r\ge k$ for each $r\in [0,N],$ and so $\pi {\left(f_i\right)}^N$something $\square$

(3.18) Let $(M,\pi )$ be a quasi-simple $𝔤$–module with highest weight $\lambda .$ Then:

1. $\text{ch}\left(M\right)$ is $W$–invariant (as a function on ${𝔥}^{*})$

2. If $\mu \in P\left(M\right)$ and $1\le i\le n,$ then the set of integers $r$ such that $\mu +r{\alpha }_i\in P\left(M\right)$ is a finite interval $[-p,q]$ in $\mathbb{Z},$ where $p,q\ge 0$ and $p-q=\mu \left(h_i\right).$

3. If $\mu \in P\left(M\right),$ then $\mu \left(h_i\right)\in \mathbb{Z}$ $(1\le i\le n).$

		Proof.
	We shall make use of the following formula: $$e^{\text{ad}x}y=e^{x}y{e}^{-x}$$ for elements $x,y$ of an associative $\mathbb{Q}$–algebra, with $x$ nilpotent (so that $e^x$ is defined). The proof is very simple: we have $\text{ad}x={\lambda }_x-{\rho }_x,$ and ${\lambda }_x,{\rho }_x$ commute, hence $$\begin{array}{ccccc}{e}^{\text{ad}x}y& =& e^{{\lambda }_x-{\rho }_x}y& =& e^{{\lambda }_x}{e}^{-{\rho }_x}y\\ & =& {\lambda }_{e^x}{\rho }_{e^{-x}}y& =& e^{x}y{e}^{-x}.\end{array}$$ Let $x\in 𝔤.$ Since $\text{ad}{e}_i$ and $\pi \left(e_i\right)$ are locally nilpotent, we have $$\begin{array}{ccc}\pi \left(e^{\text{ad}{e}_i}x\right)& =& e^{\text{ad}\pi \left(e_i\right)}.\pi \left(x\right)\\ & =& e^{\pi \left(e_i\right)}\pi \left(x\right)e^{-\pi \left(e_i\right)}.\end{array}$$ by the formula above. Similarly with $e_i$ replaced by $f_i.$ Hence if (as in Ch. II) we write $${\stackrel{\sim }{w}}_i=e^{\text{ad}{e}_i}{e}^{-\text{ad}{f}_i}{e}^{\text{ad}{e}_i}$$ then we have $$\begin{array}{cc}\pi \left({\stackrel{\sim }{w}}_{i}x\right)={\theta }_i\pi \left(x\right){\theta }_i^{-1}& \text{(1)}\end{array}$$ where $${\theta }_i=e^{\pi \left(e_i\right)}{e}^{-\pi \left(f_i\right)}{e}^{\pi \left(e_i\right)}\in \text{GL}\left(M\right)$$ Now recall (2.3) that ${\stackrel{\sim }{w}}_{i}h=w_{i}h$ for $h\in 𝔥.$ It follows from (1) that $$\begin{array}{cc}\pi \left(w_{i}h\right)={\theta }_i\pi \left(h\right){\theta }_i^{-1}.& \text{(2)}\end{array}$$ Now let $\mu \in P\left(M\right),$ $v\in M_{\mu },$ $v\ne 0.$ Then $\pi \left(h\right)v=\mu \left(h\right)v$ $(h\in 𝔥)$ and therefore $$\begin{array}{ccc}\pi \left(h\right)\left({\theta }_i^{-1}v\right)& =& {\theta }_i^{-1}\pi \left(w_{i}h\right)v\text{by (2)}\\ & =& {\theta }_i^{-1}\left(\mu \left(w_{i}h\right)v\right)\\ & =& \left(w_i\mu \right)\left(h\right){\theta }_i^{-1}v\end{array}$$ (since ${\theta }_i^{-1}$ is $k$–linear). This calculation shows that ${\theta }_i^{-1}v\in M_{w_i\mu },$ and hence that $w_i\mu \in P\left(M\right).$ Consequently $P\left(M\right)$ is $W$–stable; also ${\theta }_i^{-1}$ takes $M_{\mu }$ into $M_{w_i\mu },$ so that $\text{dim}{M}_{\mu }\le \text{dim}{M}_{w_i\mu };$ replacing $\mu$ by $w_i\mu$ we get the opposite inequality, hence $\text{ch}\left(M\right)$ is $W$–invariant. Same proof as (2.31) (root strings). Follows from (b). $\square$

A linear form $\lambda \in {𝔥}^{*}$ is integral if $\lambda \left(h_i\right)\in \mathbb{Z}$ $(1\le i\le n);$ dominant integral if $\lambda \left(h_i\right)\in \mathbb{N}$ for $1\le i\le n.$

Let $P$ (resp. $P^{+})$ denote the set of all integral (resp. dominant integral) $\lambda \in {𝔥}^{*}.$ Notice that each ${\alpha }_j\in P,$ because ${\alpha }_j\left(h_i\right)=a_{ij}\in \mathbb{Z}:$ thus $Q\subset P.$ (Warning: $Q^{+}\not\subset P^{+}).$ Clearly $P^{+}=P\cap C^{\vee }$ $(C^{\vee }$ the dual fundamental chamber).

(3.19) Let $M$ be a quasi-simple $𝔤$–module with highest weight $\lambda .$ Then $\lambda \in P^{+}.$ Conversely, if $\lambda \in P^{+}$ then $L\left(\lambda \right)$ is quasi-simple.

		Proof.
	Recall (1.17) $$e_i{f}_i^{N+1}=f_i^{N+1}{e}_i+(N+1)f_i^N(h_i-N).$$ Let $v_{\lambda }\in M_{\lambda }$ be a generator of $M.$ Since $M$ is quasi-simple, $\exists N\ge 0$ such that $f_i^N.v_{\lambda }\ne 0,$ $f_i^{N+1}.v_{\lambda }=0;$ also $e_i.v_{\lambda }=0,$ whence $$\begin{array}{cc}0=e_i{f}_i^{N+1}.v_{\lambda }=(N+1)f_i^N(\lambda \left(h_i\right)-N)v_{\lambda }& \text{(1)}\end{array}$$ and therefore $\lambda \left(h_i\right)=N\ge 0.$ Thus $\lambda \in P^{+}.$ (Notice that this gives another proof of (3.18)(c), namely that $P\left(M\right)\subset P:$ for if $\mu \in P\left(M\right),$ then $\mu \in \lambda -Q^{+}\subset P.)$ For the second part, let $v_{\lambda }$ be the generator of $L\left(\lambda \right)$ and let $x_i=f_i^{\lambda \left(h_i\right)+1}{v}_{\lambda }.$ I claim that $x_i=0.$ For now we have from (1) that $e_i{f}_i^{N+1}{v}_{\lambda }=0$ if $N=\lambda \left(h_i\right),$ i.e. $e_i{x}_i=0;$ also $e_j.x_i=f_i^{\lambda \left(h_i\right)+1}{e}_j.x_i=0$ if $j\ne i$ (because $e_j,f_i$ then commute). Hence $x_i$ generates a proper submodule of $L\left(\lambda \right).$ Since $L\left(\lambda \right)$ is simple, we must have $x_i=0.$ By (3.17), it follows that $L\left(\lambda \right)$ is quasi-simple. $\square$

Recall that $\rho \in {𝔥}^{*}$ was chosen such that $\rho \left(h_i\right)=\frac{1}{2}{a}_{ii}(1\le i\le n).$ Since $A$ is now a Cartan matrix, this condition now becomes

$$\rho \left(h_i\right)=1(1\le i\le n).$$

Thus $\rho \in P^{+}.$

For $w\in W,$ let $\epsilon \left(w\right)=\text{det}\left(w\right)={(-1)}^{l\left(w\right)}.$ (sign character of $W).$

(3.20) $e^{\rho }\prod$ is $W$–skew, i.e.

$$w(e^{\rho }\prod )=\epsilon \left(w\right).e^{\rho }\prod$$

for all $w\in W.$

		Proof.
	It is enough to verify this when $w=w_i$ is a generator of $W.$ We have $$w_i\rho =\rho -\rho \left(h_i\right){\alpha }_i=\rho -{\alpha }_i.$$ On the other hand (2.6), $w_i$ sends ${\alpha }_i$ to $-{\alpha }_i$ and permutes the positive roots $\ne {\alpha }_i$. Thus $$\begin{array}{ccc}{w}_i\left(e^{\rho }\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{m_{\alpha }}\right)& =& e^{\rho -{\alpha }_i}(1-e^{{\alpha }_i})\prod _{\genfrac{}{}{0ex}{}{\alpha \in R^{+}}{\alpha \ne {\alpha }_i}}{(1-e^{-\alpha })}^{m_{\alpha }}\left(\text{since}{\alpha }_i\text{has multiplicity 1}\right)\\ & =& -e^{\rho }\prod .\end{array}$$ $\square$

Now assume that the Cartan matrix $A$ is symmetrizable. Then the scalar product on $𝔥$ and ${𝔥}^{*}$ is $W$–invariant (2.23), and we have

$$\begin{array}{cc}⟨\rho ,{\alpha }_i⟩={\epsilon }_i^{-1}\rho \left(h_i\right)={\epsilon }_i^{-1}>0(1\le i\le n)& \text{(1)}\end{array}$$

For the same reason, if $\lambda \in P^{+}$ we have

$$\begin{array}{cc}⟨\lambda ,{\alpha }_i⟩={\epsilon }_i^{-1}\lambda \left(h_i\right)\ge 0.& \text{(2)}\end{array}$$

(3.21) Theorem (V. Kac) Let $A$ be a symmetrizable Cartan matrix and let $M$ be a quasi-simple $𝔤\left(A\right)$–module with highest weight $\lambda .$ Then

$$\text{ch}\left(M\right)=\left(\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )}\right)/e^{\rho }\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{m_{\alpha }}.$$

		Proof.
	From (3.16) we have, writing $d_{\mu }=c_{\mu +\rho },$ $$e^{\rho }\prod .\text{ch}\left(M\right)=\sum _{\mu }{c}_{\mu +\rho }{e}^{\mu +\rho }$$ summed over $\mu \in D\left(\lambda \right)$ such that ${\mid \mu +\rho \mid }^2={\mid \lambda +\rho \mid }^2,$ with coefficients $c_{\mu +\rho }\in \mathbb{Z}$ and, in particular, $c_{\lambda +\rho }=1.$ Now $\text{ch}\left(M\right)$ is $W$–invariant (3.18) and $e^{\rho }\prod$ is $W$–skew (3.20). Hence there product is $W$–skew, and therefore for each $w\in W$ we have $$\sum _{\mu }{c}_{\mu +\rho }{e}^{\mu +\rho }=\sum _{\mu }\epsilon \left(w\right)c_{\mu +\rho }{e}^{w(\mu +\rho )}$$ so that $c_{w(\mu +\rho )}=\epsilon \left(w\right)c_{\mu +\rho }.$ Hence if $c_{\mu +\rho }\ne 0$ we have $w(\mu +\rho )\le \lambda +\rho$ for all $w\in W;$ choose $w$ so that $\text{ht}(\lambda +\rho -w(\mu +\rho ))$ is minimal and put $\nu =w(\mu +\rho ).$ Then $\text{ht}(\lambda +\rho -w_i\nu )\ge \text{ht}(\lambda +\rho -\nu ),$ i.e. $\text{ht}(\nu -w_i\nu )\ge 0$ and therefore $\nu \left(h_i\right)\ge 0,$ or equivalently $⟨\nu ,{\alpha }_i⟩\ge 0.$ Thus $\nu$ satisfies $⟨\nu ,{\alpha }_i⟩\ge 0(1\le i\le n);$ $\nu \le \lambda +\rho ,$ i.e. $\lambda +\rho =\nu +{\displaystyle \sum _1^n}{m}_i{\alpha }_i$ with coefficients $m_i\ge 0;$ ${\mid \nu \mid }^2={\mid w(\mu +\rho )\mid }^2={\mid \mu +\rho \mid }^2={\mid \lambda +\rho \mid }^2.$ These three conditions force $\nu =\lambda +\rho ;$ for we have $$\begin{array}{ccccc}0& =& {\mid \lambda +\rho \mid }^2-{\mid \nu \mid }^2& =& ⟨\lambda +\rho +\nu ,\lambda +\rho -\nu ⟩\\ & & & =& ⟨\lambda +\rho +\nu ,\sum m_i{\alpha }_i⟩\\ & & & =& \sum m_i⟨\lambda +\rho +\nu ,{\alpha }_i⟩\end{array}$$ But $⟨\lambda ,{\alpha }_i⟩\ge 0$ by (2) because $\lambda \in P^{+}$ (3.19); $⟨\rho ,{\alpha }_i⟩>0$ (1); and $⟨\nu ,{\alpha }_i⟩\ge 0$ ((i) above). Hence $⟨\lambda +\rho +\nu ,{\alpha }_i>0⟩(1\le i\le n),$ and therefore all coefficients $m_i$ are 0, hence $\nu =\lambda +\rho$ and therefore $(\text{since}{c}_{\lambda +\rho }=1)$ $$\sum _{\mu }{c}_{\mu +\rho }{e}^{\mu +\rho }=\sum _{w\in W}\epsilon \left(w\right)e^{w(\lambda +\rho )}.$$ $\square$

Recall (3.19) that $L\left(\lambda \right)$ is quasi-simple if $\lambda \in P^{+}.$ The character formula (3.21) shows that if $M$ is a quasi-simple $𝔤$–module with highest weight $\lambda$ $(\in P^{+},\text{by (3.19)})$ then $\text{ch}\left(M\right)$ depends only on $\lambda .$ It follows that

$$\text{ch}\left(M\right)=\text{ch}L\left(\lambda \right)$$

i.e. $\text{dim}{M}_{\mu }=\text{dim}L{\left(\lambda \right)}_{\mu }$ for all $\mu .$ But $L\left(\lambda \right)$ is in any case a homomorphic image of $M,$ and so we conclude that $M=L\left(\lambda \right):$

(3.22) Every quasi-simple $𝔤$–module is simple.

Another corollary of (3.21) is the "denominator formula":

(3.23) For any symmetrizable Cartan matrix we have

$$\sum _{w\in W}\epsilon \left(w\right)e^{w\rho -\rho }=\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{m_{\alpha }}$$

		Proof.
	Take $\lambda =0$ in (3.21) and observe that $L\left(0\right)$ is the trivial 1–dimensional $𝔤$–module (3.7), so that $\text{ch}L\left(0\right)=1.$ $\square$

We can write (3.23) in another form, as follows. Recall that for $w\in W,$

$$\begin{array}{ccc}S\left(w\right)& =& \{\alpha \in R^{+}:w^{-1}\alpha \in R^{-}\}\\ & =& R^{+}\cap w{R}^{-}\end{array}$$

is a finite set (2.10). Define

$$s\left(w\right)=\sum _{\alpha \in S\left(w\right)}\alpha$$

a finite sum of positive roots. We then have the formula (for any Cartan matrix A)

(3.24) $s(w)=\rho -w\rho .$

		Proof.
	If $w=w_{i_1}\dots w_{i_r}$ is a reduced word for $w(r=l\left(w\right))$ then (2.9) $$S\left(w\right)=\{{\alpha }_{i_1},w_{i_1}{\alpha }_{i_2},\dots ,w_{i_1}\dots w_{i_{r-1}}{\alpha }_{i_r}\}$$ from which it follows that if $w^{\prime }=w_{i_2}\dots w_{i_r}$ $$S\left(w\right)={\alpha }_{i_1}\cup w_{i_1}S\left(w^{\prime }\right)$$ i.e. if $w=w_i{w}^{\prime }$ with $l\left(w^{\prime }\right)=l\left(w\right)-1,$ then $$S\left(w\right)={\alpha }_i\cup w_{i}S\left(w^1\right)$$ and therefore $$\begin{array}{cc}s\left(w\right)={\alpha }_i+w_{i}s\left(w^{\prime }\right).& \text{(1)}\end{array}$$ To prove (3.24), we proceed by induction on $l\left(w\right).$ The result is clearly true when $l\left(w\right)=0,$ for then $w=1,$ $s\left(w\right)=0.$ Assume $l\left(w\right)>0$ and write $w=w_i{w}^{\prime }$ as above, then $$\begin{array}{ccc}s\left(w\right)& =& {\alpha }_i+w_i(\rho -w^{\prime }\rho )\text{by (1)}\leftarrow \text{ind. hyp.}\\ & =& {\alpha }_i+(\rho -\rho \left(h_i\right){\alpha }_i)-w\rho \\ & =& \rho -w\rho \end{array}$$ since $\rho \left(h_i\right)=1.$ $\square$

By virtue of (3.24) we can rewrite (3.23) in the form

(3.23${}^{\prime }$)

$$\sum _{w\in W}\epsilon \left(w\right)e^{-s\left(w\right)}=\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{m_{\alpha }}.$$

Also form (3.23) we can rewrite the character formula (3.21) in the form

(3.21${}^{\prime }$) Let $\lambda \in P^{+},$ then

$$\text{ch}L\left(\lambda \right)=\frac{{\displaystyle \sum _{w\in W}}\epsilon \left(w\right)e^{w(\lambda +\rho )}}{{\displaystyle \sum _{w\in W}}\epsilon \left(w\right)e^{w\rho }}.$$

(3.23${}^{\prime }$) is a statement about the root system $R$ and the Weyl group $W;$ it may be formally inverted to give a formula for the multiplicities $m_{\alpha }$ (recall that all real roots have multiplicity 1; the imaginary roots may have multiplicities $m_{\alpha }>1).$

Examples

1. Suppose $A$ is of finite type, so that $𝔤\left(A\right)$ is finite-dimensional and $R$ is finite. In that case

   $$\begin{array}{cc}\rho =\frac{1}{2}\sum _{\alpha \in R^{+}}\alpha & \text{(1)}\end{array}$$

   For if $\delta$ is $\frac{1}{2}$ the sum of the positive roots then by (2.6)

   $$\begin{array}{ccc}{w}_i\delta & =& \frac{1}{2}\sum _{\alpha \in R^{+}}{w}_i\alpha \\ & =& \frac{1}{2}(-{\alpha }_i+\sum _{\genfrac{}{}{0ex}{}{\alpha \in R^{+}}{\alpha \ne {\alpha }_i}}\alpha )=\delta -{\alpha }_i;\end{array}$$

   but on the other hand $w_i\delta =\delta -\delta \left(h_i\right){\alpha }_i,$ so that $\delta \left(h_i\right)=1$ for all $i.$ Since $A$ is nonsingular, $𝔥$ is spanned by $h_1,\dots ,h_n$ and therefore $\delta =\rho .$

   The Denominator formula (3.23) now reads

   $$\begin{array}{ccc}\sum _{w\in W}\epsilon \left(w\right)e^{w\rho }& =& e^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })\\ & =& \prod _{\alpha \in R^{+}}(e^{\alpha /2}-e^{-\alpha /2})\end{array}$$

   by virtue of $\left(✶\right).$where is this from? It is a polynomial identity in the group ring $\mathbb{Z}\left[\frac{1}{2}Q\right].$

   For a specific example, take $A$ of type $A_{n-1}.$ Let $u_1,\dots ,u_n$ be the standard basis of ${ℝ}^n,$ then the roots may be taken to be $u_i-u_j$ $(i\ne j)$ and the positive roots $u_i-u_j$ $(i<j).$ Thus

   $$\rho =\frac{1}{2}\sum _{i<j}(u_i-u_j)=\frac{1}{2}\sum _{i=1}^n(n+1-2i)u_i$$

   Put $x_i=e^{u_i};$ the Weyl group $W$ is here the symmetric group $S_n$ acting by permuting the $u_i$ (or, equivalently, the $x_i).$ We have

   $$\begin{array}{ccc}{e}^{\rho }& =& x_1^{\frac{n-1}{2}}{x}_{\mathrm{x}}^{\frac{n-3}{2}}\dots x_n^{\frac{1-n}{2}}\\ & =& {\left(x_1\dots x_n\right)}^{\frac{1-n}{2}}{x}_1^{n-1}{x}_2^{n-2}\dots \end{array}$$

   and therefore

   $$\sum _{w\in W}\epsilon \left(w\right)e^{w\rho }={\left(x_1\dots x_n\right)}^{\frac{1-n}{2}}\Delta (x_1,\dots ,x_n)$$

   where $\Delta (x_1,\dots ,x_n)=\text{det}\left(x_i^{n-j}\right)$ is the Vandermonde determinant. On the other side,

   $$\begin{array}{ccc}{e}^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })& =& e^{\rho }\prod _{i<j}(1-x_i^{-1}{x}_j)\\ & =& {\left(x_1\dots x_n\right)}^{\frac{1-n}{2}}\prod _{i<j}(x_i-x_j)\end{array}$$

   and therefore the "denominator formula" in this case reduces to the familiar factorization of the Vanermonde determinant:

   $$\Delta (x_1,\dots ,x_n)=\prod _{i<j}(x_i-x_j).$$

   So it is an essentially trivial polynomial identity in this case.

2. Let $A=\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right),$ of affine type. The Weyl group $W$ is infinite dihedral, generated by reflections $w_1,w_2.$ So its elements are 1 and

   $$\begin{array}{c}{w}_1{w}_2{w}_1\dots \text{to}r\text{terms (all}r\ge 1\text{)}\\ w_2{w}_1{w}_2\dots \text{to}r\text{terms (all}r\ge 1\text{)}\end{array}$$

   We have $w_1\left({\alpha }_1\right)=-{\alpha }_1,$ $w_2\left({\alpha }_1\right)={\alpha }_1-{\alpha }_1\left(h_2\right){\alpha }_2={\alpha }_1+2{\alpha }_2$ and likewise $w_1\left({\alpha }_2\right)=2{\alpha }_1+{\alpha }_2,$ $w_2\left({\alpha }_2\right)=-{\alpha }_2.$ So we get the following picture:

   Now if $w=w_1{w}_2{w}_1\dots$ to $r$ terms then

   $$\begin{array}{ccc}s\left(w\right)& =& {\alpha }_1+w_1{\alpha }_2+w_1{w}_2{\alpha }_1+\dots \text{to}r\text{terms, by (2.8)}\\ & =& {\alpha }_1+(2{\alpha }_1+{\alpha }_2)+(3{\alpha }_1+2{\alpha }_2)+\dots \\ & =& \frac{1}{2}r(r+1){\alpha }_1+\frac{1}{2}r(r-1){\alpha }_2\end{array}$$

   so that if we put $x_1=e^{-{\alpha }_1},x_2=e^{-{\alpha }_2}$ we have

   $$\begin{array}{ccc}\sum _{w\in W}\epsilon \left(w\right)e^{-s\left(w\right)}& =& 1+\sum _{r=1}^{\infty }{(-1)}^r(x_1^{\frac{1}{2}r(r+1)}{x}_2^{\frac{1}{2}r(r-1)}+x_1^{\frac{1}{2}r(r-1)}{x}_2^{\frac{1}{2}r(r+1)})\\ & =& \sum _{r\in \mathbb{Z}}{(-1)}^r{x}_1^{\frac{1}{2}r(r+1)}{x}_2^{\frac{1}{2}r(r-1)}\end{array}$$

   On the other hand, the positive roots are

   $$r{\alpha }_1+(r-1){\alpha }_2,r{\alpha }_1+r{\alpha }_2,(r-1){\alpha }_1+r{\alpha }_2(r\ge 1)$$

   $r{\alpha }_1+r{\alpha }_2=r\delta$ is an imaginary root; in fact (as we shall see later) it has multiplicity 1. So we obtain the identity

   $$\sum _{r\in \mathbb{Z}}{(-1)}^r{x}_1^{\frac{1}{2}r(r+1)}{x}_2^{\frac{1}{2}r(r-1)}=\prod _{r=1}^{\infty }(1-x_1^r{x}_2^{r-1})(1-x_1^{r-1}{x}_2^r)(1-x_1^r{x}_2^r)$$

   in $\mathbb{Z}\left[[x_1,x_2]\right].$ If we put $x_1{x}_2=t,$ $x_1=x$ it takes the form

   $$\sum _{r\in \mathbb{Z}}{(-1)}^r{x}^r{t}^{\frac{1}{2}r(r-1)}=\prod _{r=1}^{\infty }(1-x{t}^{r-1})(1-x^{-1}{t}^r)(1-t^r)$$

   and is due to Jacobi (and earlier, unpublished, to Gauss): it is called Jacobi's triple product identity and it can be specialized in various ways, at least two of which are worth notice:

   1. Put $t=x^3$ and we get

      $$\prod _{r=1}^{\infty }(1-x^n)=\sum _{r\in \mathbb{Z}}{(-1)}^r{x}^{\frac{1}{2}r(3r-1)}$$

      Euler's pentagonal number theorem.

   2. Divide both sides by $1-x$ and then set $x=1.$ On the product side we get ${\displaystyle \prod _{r=1}^{\infty }}{(1-t^r)}^3,$ and on the sum side

      $$\begin{array}{ccc}\sum _{r\ge 1}{(-1)}^r{t}^{\frac{1}{2}r(r-1)}\frac{x^r-x^{1-r}}{1-x}& \to & \sum _{r\ge 1}{(-1)}^{r+1}(2r-1)t^{\frac{1}{2}r(r-1)}\\ & =& \sum _{r\ge 0}{(-1)}^r(2r+1)t^{\frac{1}{2}r(r+1)}\end{array}$$

      Thus we obtain another famous identity due to Jacobi:

      $$\prod _{r=1}^{\infty }{(1-t^r)}^3=1-3t+5t^3-7t^6+9t^{10}-11t^{15}+\dots$$

3. Finally, if $A$ is symmetrizable and of indefinite type, the denominator formula can be used to compute the multiplicities of the imaginary roots. For a simple example take $A=\left(\begin{array}{cc}2& -3\\ -3& 2\end{array}\right).$ The Weyl group $W$ is infinite dihedral:

   $$\begin{array}{ccccccc}{w}_1\left({\alpha }_1\right)& =& -{\alpha }_1& & w_2\left({\alpha }_1\right)& =& {\alpha }_1+3{\alpha }_2\\ w_1\left({\alpha }_2\right)& =& {\alpha }_2+3{\alpha }_1& & w_2\left({\alpha }_2\right)& =& -{\alpha }_2\end{array}$$

   So the real roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+3{\alpha }_2,{\alpha }_2+3{\alpha }_1,\dots$

   $$\begin{array}{ccccc}{w}_1({\alpha }_1+3{\alpha }_2)& =& -{\alpha }_1+3({\alpha }_2+3{\alpha }_1)& =& 8{\alpha }_1+3{\alpha }_2\\ w_1(3{\alpha }_1+8{\alpha }_2)& =& -3{\alpha }_1+8({\alpha }_2+3{\alpha }_1)& =& 21{\alpha }_1+8{\alpha }_2\end{array}$$

   the coefficients of which we recognise as Fibonacci numbers: the real roots are

   $$f_{2k}{\alpha }_1+f_{2k-2}{\alpha }_2;f_{2k-2}{\alpha }_1+f_{2k}{\alpha }_2(k\ge 1)$$

   $(f_0=0,f_1=1,f_r=f_{r-1}+f_{r-2})$

   From this we calculate easily

   $$s(\underset{\underset{k\text{terms}}{⏟}}{w_1{w}_2\dots })=(f_{2k+1}-1){\alpha }_1+(f_{2k-1}-1){\alpha }_2$$

   so that the series ${\displaystyle \sum _{w\in W}}\epsilon \left(w\right)e^{-s\left(w\right)}$ is

   $$1-x_1-x_2+x_1^4{x}_2+x_2{x}_2^4-x_1^{12}{x}_2^4-x_1^4{x}_2^{12}+x_1^{33}{x}_2^{12}+x_1^{12}{x}_2^{33}-\dots$$

   $(x_i=e^{-{\alpha }_i}),$ i.e. it is a 'sparse' power series in $x_1,x_2.$ By factorizing it as a product of factors ${(1-x_1^{k_1}{x}_2^{k_2})}^{m_{k_1,k_2}}$ we compute the multiplicities of the roots:

   $$=(1-x_1)(1-x_2)(1-x_1{x}_2)\dots$$

References

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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