## Kac-Moody Lie AlgebrasChapter III: Representation theory

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 $n×n$ 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

$𝔤=𝔤(A)=𝔥+ ∑α∈R𝔤α (direct sum)$

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 $\pi :\phantom{\rule{0.2em}{0ex}}𝔤\to 𝔤𝔩\left(M\right),$ which extends to $\pi :\phantom{\rule{0.2em}{0ex}}U\left(𝔤\right)\to \text{End}\phantom{\rule{0.2em}{0ex}}\left(M\right),$ i.e $M$ is a $U\left(𝔤\right)$–module. Notation $\left(M,\pi \right)$ when I want to be pedantic. More often then not I shall suppress $\pi$ and write $x.v$ or $xv$ for $\pi \left(x\right)v$ $\left(x\in 𝔤,\phantom{\rule{0.2em}{0ex}}v\in M\right)$.

## Weights

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

$Mλ≔ { v∈M:h.v= λ(h)vfor all h∈𝔥 } .$

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

$Mλ≅ HomU(𝔥) (Eλ,M) (✶)$

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 $\lambda \in {𝔥}^{*}$) $M↦{M}_{\lambda }$ is a left exact functor (from $𝔤$–modules to $𝔥$–modules).

Example:$\phantom{\rule{1em}{0ex}}$ $\left(M,\pi \right)=\left(𝔤,\text{ad}\right).$ 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

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

3. If $\phi :\phantom{\rule{0.2em}{0ex}}M\to N$ 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 $h.(x.v) = x.h.v+[h,x].v = λ(h)x.v+α (h)x.v = (λ+α)(h)x.v$ 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 $∑i=1m vλi=0 (1)$ where ${v}_{{\lambda }_{i}}\in {M}_{{\lambda }_{i}},\phantom{\rule{0.2em}{0ex}}{v}_{{\lambda }_{i}}\ne 0$ and ${\lambda }_{1},\dots ,{\lambda }_{m}\in {𝔥}^{*}$ are all distinct. Choose such a relation with $m$ $\left(\ge 2\right)$ as small as possible. By operating on (1) with an element $h\in 𝔥,$ we obtain $∑i=1mλi (h)vλi=0 (2)$ 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 $ 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

$D(λ)=λ-Q+$

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

$D(F)=⋃λ∈F D(λ).$

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.

$M=∑μ∈P(M) Mμ$

(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 $\phantom{\rule{1em}{0ex}}\left(E\right)\phantom{\rule{1em}{0ex}}0⟶{M}^{\prime }⟶M⟶{M}^{\prime \prime }⟶0\phantom{\rule{1em}{0ex}}$ 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 $\phantom{\rule{1em}{0ex}}\left({E}_{\lambda }\right)\phantom{\rule{1em}{0ex}}0⟶{M}_{\lambda }^{\prime }\stackrel{f}{⟶}{M}_{\lambda }\stackrel{g}{⟶}{M}_{\lambda }^{\prime \prime }⟶0\phantom{\rule{1em}{0ex}}$ 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 ${M}^{\prime }=\underset{\lambda }{\oplus }{M}_{\lambda }^{\prime }$ 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 $M′′=g(M)= ∑g(Mλ)⊂∑ Mλ′′⊂ M′′;$ 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 $\lambda -\mu -\sum _{1}^{n}{u}_{i}{\alpha }_{i}$ 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 $\text{ht}\phantom{\rule{0.2em}{0ex}}\left(\lambda -{v}_{i}\right)$ is a nonnegative integer. It follows that the sequence ${\left(\text{ht}\phantom{\rule{0.2em}{0ex}}\left(\lambda -{v}_{i}\right)\right)}_{i\ge 1}$ 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. $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{\lambda }=1;$

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) ${e}_{i}.{v}_{\lambda }=0$ $\left(1\le i\le n\right).$ 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 $M′⊂M+= ∑μ≠λMμ$ 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

$V(λ)= ind𝔟𝔤(Eλ) =U(𝔤)⊗U(𝔟) Eλ.$

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

$V(λ)≅U(𝔤)/ J(λ).$

For if $\pi$ is the representation of $U\left(𝔟\right)$ on ${E}_{\lambda },$ then $\pi :\phantom{\rule{0.2em}{0ex}}U\left(𝔟\right)\to k$ is such that $\pi \left({e}_{i}\right)=0\phantom{\rule{0.2em}{0ex}}\left(1\le i\le n\right)$ and $\pi \left(h\right)=\lambda \left(h\right),$ all $h\in 𝔥;$ hence $K=\text{Ker}\phantom{\rule{0.2em}{0ex}}\left(\pi \right)$ 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)

$0⟶K⟶U(𝔟)⟶ Eλ⟶0$

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

$U(𝔤)⊗U(𝔟) K⟶U(𝔤)⟶V(λ) ⟶0$

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 $\lambda +{\alpha }_{i}\notin P\left(M\right),\phantom{\rule{0.2em}{0ex}}1\le i\le n\right),$ 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.$\phantom{\rule{1em}{0ex}}$ When $\lambda =0,$ ${E}_{\lambda }=k$ with trivial $𝔥$–action $\left(h.1=0\right),$ 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 ${\text{End}}_{𝔤}\phantom{\rule{0.2em}{0ex}}\left(M\right)=k.$ Proof. Let ${v}_{\lambda }={M}_{\lambda }$ be a h.w. vector which generates $M.$ If $\phi :\phantom{\rule{0.2em}{0ex}}M\to M$ 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 $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{\lambda }=1$ (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 $f:\phantom{\rule{0.2em}{0ex}}{𝔥}^{*}\to ℤ$ such that $\text{Supp}\phantom{\rule{0.2em}{0ex}}\left(f\right)\subset D\left(F\right)$ 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:

$(fg)(v)= ∑λ+μ=v f(λ)g(μ) (finite sum) (1)$

If $\text{Supp}\phantom{\rule{0.2em}{0ex}}\left(f\right)\subset D\left(F\right),$ $\text{Supp}\phantom{\rule{0.2em}{0ex}}\left(g\right)\subset D\left(G\right),$ then $\text{Supp}\phantom{\rule{0.2em}{0ex}}\left(fg\right)\subset D\left(F+G\right).$ 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. $\text{Supp}\phantom{\rule{0.2em}{0ex}}\left(fj\right)\subset D\left(F\right)$ 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

$f(λ)=∑j∈J fj(λ)$

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

$f=∑λf(λ) eλ∈∑μ∈F eμℤ [ [ e-α1,…, e-αn ] ] (2)$

Now let $M\in 𝒪$ and define the formal character of $M$ to be the function $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)$ defined by

$ch(M)(λ)= dimMλ;$

thus $\text{Supp}\phantom{\rule{0.2em}{0ex}}\left(\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)\right)=P\left(M\right)\subset D\left(F\right)$ for some finite $F\subset {𝔥}^{*},$ so that $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)\in \epsilon$ and by (2) we have

$ch(M)= ∑λ∈P(M) dimMλ.eλ$

Thus $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)$ 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

$ch(M)= ch(M′)+ ch(M′′).$ 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

$∑i=0r (-1)i ch(Mi)=0$

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

$ch(V(λ))= eλ/ ∏α∈R+ (1-e-α) mα =eλ/πsay$

where as usual ${m}_{\alpha }=\text{dim}\phantom{\rule{0.2em}{0ex}}{𝔤}_{\alpha }=\text{dim}\phantom{\rule{0.2em}{0ex}}{𝔤}_{-\alpha }.$ (The product on the right is a unit in $\epsilon .\right)$ 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 $\sum {r}_{i}<\infty \right),$ 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 $n×n$ 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. $⟨\left[x,y\right],z⟩=⟨x,\left[y,z\right]⟩$ for all $x,y,z\in 𝔤$ (invariance)

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

Then for any $h\in 𝔥$ we have

$⟨h,hj⟩ = ⟨h,[ej,fj]⟩ = ⟨[h,ej],fj⟩ = αj(h)⟨ej,fj⟩$

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

$⟨hi,hj⟩= aijεj$

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 ${𝔤}_{\alpha }×{𝔤}_{\beta }$ (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

$[x,y]= ⟨x,y⟩hα∨$

where ${h}_{\alpha }^{\vee }\in 𝔥$ is defined by $⟨h,{h}_{\alpha }^{\vee }⟩=\alpha \left(h\right)$ for all $h\in 𝔥.$ Proof. Let $𝔞=\left\{x\in 𝔤\phantom{\rule{0.2em}{0ex}}:\phantom{\rule{0.2em}{0ex}}⟨x,𝔤⟩=0\right\}.$ 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 $\left[x,y\right]\in 𝔞.$ Now all ideals in $𝔞$ are graded (1.7), hence $𝔞=\sum {𝔞}_{\alpha },$ where ${𝔞}_{\alpha }=𝔞\cap {𝔤}_{\alpha }$ (and ${𝔞}_{0}=𝔞\cap 𝔥\right).$ 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 $⟨\left[x,h\right],y⟩=⟨x,\left[h,y\right]⟩$ and thus $-α(h) ⟨x,y⟩=β(h) ⟨x,y⟩$ 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 $⟨h,[x,y]⟩= ⟨[h,x],y⟩= α(h)⟨x,y⟩= ⟨h,hα∨⟩ ⟨x,y⟩$ 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 ${𝔥}^{\prime }=\sum _{1}^{n}k{h}_{i}$) and define $⟨x,y⟩$ on $𝔥×𝔥$ by $⟨x,hi⟩ = ⟨hi,x⟩ = εiαi(x) (x∈𝔥) ⟨y,z⟩ = 0 (y,z∈𝔥*)$ 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 $h\in \underset{1}{\overset{n}{\cap }}\text{Ker}\phantom{\rule{0.2em}{0ex}}{\alpha }_{i}=𝔠\subset {𝔥}^{\prime };$ thus $\sum {\lambda }_{i}{h}_{i}$ say, and then $∑1nλiεi αi(x)= ⟨h,x⟩=0$ 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 $ℤ$–grading of $𝔤:$ $𝔤r= ∑htα=r 𝔤α;𝔤= ∑r∈ℤ𝔤r; 𝔤0=𝔥$ Let ${G}_{n}=\sum _{\mid r\mid \le n}{𝔤}_{r}$ 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 $⟨[x,y],z⟩= ⟨x,[y,z]⟩ (1)$ whenever all 5 terms lie in ${G}_{1}.$ We shall now extend $⟨,⟩$ to a symmetric bilinear form on ${G}_{n}$ $\left(n\ge 2\right)$ by induction on $n,$ such that (1) holds whenever all 5 terms are in ${G}_{n},$ and such that $⟨𝔤i,𝔤j⟩=0 ifi+j≠0 (2)$ 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 ${𝔤}_{n}×{𝔤}_{-n}.$ Write $x = ∑i[si,ti] ∈ 𝔤n (3) y = ∑j[uj,vj] ∈ 𝔤-n (4)$ where ${s}_{i},{t}_{i}$ (resp. ${u}_{j},{v}_{j}\right)$ are homogeneous of positive (resp. negative) degree, hence lie in ${G}_{n-1}.$ Define now $⟨y,x⟩= ⟨x,y⟩= ∑j ⟨[x,uj],vj⟩. (5)$ 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, $⟨ [[s,t],u],v ⟩ = ⟨ [s,[t,u]],v ⟩ - ⟨ [t,[s,u]],v ⟩ (Jacobi) = - ⟨ [t,u],[s,v] ⟩ + ⟨ [s,u],[t,v] ⟩ (invariance) = - ⟨ [s,v],[t,u] ⟩ - ⟨ [s,u],[v,t] ⟩ (symmetry) = - ⟨ s,[v,[t,u]] ⟩ - ⟨ s,[u[v,t]] ⟩ (invariance) = ⟨ s,[t,[u,v]] ⟩ (Jacobi)$ i.e. we have $⟨ [[s,t],u] ,v ⟩ = ⟨ s,[t,[u,v]] ⟩ (6)$ From (3), (4) and (6) it follows that $∑j ⟨[x,uj],vj⟩ = ∑i,j ⟨ [[si,ti],uj] ,vj ⟩ = ∑i,j ⟨ si, [ti,[uj,vj]] ⟩ = ∑i ⟨si,[ti,y]⟩$ 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

$⟨hi,hj⟩= aijεj= ajiεi$

and an isomorphism $\theta :\phantom{\rule{0.2em}{0ex}}𝔥\to {𝔥}^{*}$ definted by

$θ(x)(y)= ⟨x,y⟩$

so that

$θ(hi)(x)= ⟨hi,x⟩=εi αi(x)$

for all $x\in 𝔥,$ whence

$θ(hi) = εiαi = αi∨ θ-1(αi) = εi-1hi = hi∨$

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

$⟨αi,αj⟩= ⟨ εi-1hi, εj-1hj ⟩ =εi-1 aij=εj-1 aji$

## 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 $B\in {\left(𝔤\otimes 𝔤\right)}^{*}={𝔤}^{*}\otimes {𝔤}^{*};$ 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 $T\left(𝔤\right)\right)$ 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 $⟨{x}_{i},{y}_{j}⟩={\delta }_{ij}\right);$ then $\omega =\sum _{1}^{n}{y}_{i}{x}_{i}.$

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 ${𝔤}_{\alpha }×{𝔤}_{-\alpha };$ choose a basis ${x}_{1},\dots ,{x}_{m}$ of ${𝔤}_{\alpha }$ $\left(m={m}_{\alpha }\right);$ let ${y}_{1},\dots ,{y}_{m}$ be the dual basis of ${𝔤}_{-\alpha },$ and define

$uα=∑i=1m yixi∈U(𝔤).$

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

$yi = ∑j ⟨ xj′,yi ⟩ yj′ xj′ = ∑i ⟨ xj′,yi ⟩ xi$

and therefore

$∑iyixi= ∑i,j ⟨ xj′,yi ⟩ yj′xi=∑j yj′xj′.$

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

Example:$\phantom{\rule{1em}{0ex}}$ We have $⟨{e}_{i},{f}_{i}⟩={\epsilon }_{i},$ hence ${u}_{{\alpha }_{i}}={\epsilon }_{i}^{-1}{f}_{i}{e}_{i}.\phantom{\rule{3em}{0ex}}\text{(1)}$

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

$[uα,x] = ∑i=1m ( yixix- xyixi ) = ∑i[yi,x] xi-∑iyi [x,xi] = vα,x- vα,x′say$

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 $\left[x,{x}_{i}\right]\in {𝔤}_{\alpha +\beta }.$ Let $\left({x}_{j}^{\prime }\right)$ be a basis of ${𝔤}_{\alpha +\beta },$ $\left({y}_{j}^{\prime }\right)$ the dual basis of ${𝔤}_{-\left(\alpha +\beta \right)}.$ Then

$[x,xi] = ∑j ⟨ yj′, [x,xi] ⟩ xj′ = ∑j ⟨ [yj′,x], xi ⟩ xj′ (invariance)$

and therefore

$vα,x′ = ∑iyi [x,xi] = ∑i,j ⟨ [yj′,x], xi ⟩ yixj′ = ∑j[yj′,x] xj′= vα+β,x$

i.e. we have the formula

$vα,x′= vα+β,x (x∈𝔤β) (2)$

Likewise

$vα,x= vα-β,x′ (x∈𝔤β) (3)$

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

$u=∑α∈R+ uα$

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

(3.13) We have

$[u,ei] = -hi∨ei [u,fi] = fihi∨ [u,h] = 0(h∈𝔥)$

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 ${h}_{i}^{\vee }={\epsilon }_{i}^{-1}{h}_{i}\right)$ Proof. We compute: $[u,ei] = ∑α∈R+ [uα,ei] = ∑α∈R+ vα,ei- ∑α∈R+ vα,ei′ = ∑α∈R+ vα,ei- ∑α∈R+ vα+αi,ei by (2)$ But ${v}_{\alpha ,{e}_{i}}=0$ unless $\alpha -{\alpha }_{i}\in {R}^{+}\cup \left\{0\right\},$ and therefore $[u,ei]= vαi,ei= εi-1 [fi,ei]ei =-εi-1hi ei=-hi∨ei.$ Similarly we have $[u,fi] = ∑αvα,fi -∑α∈R+ vα,fi′ = ∑α vα+αi,fi′ -∑α vα,fi′ = -vαi,fi′ =-εi-1fi [fi,ei]= εi-1fihi =fihsomething?$ Finally $\left[u,h\right]=\sum _{\alpha }\left({v}_{\alpha ,h}-{v}_{\alpha ,h}^{\prime }\right)=0$ by (2). $\square$

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

$ρ(hi)=12 aii (1≤i≤n)$

(thus $\rho \left({h}_{i}\right)=1$ if $A$ is a Cartan matrix). Then we have

$⟨ρ,αi⟩= ρ(hi∨)= εi-1ρ(hi) =12εi-1 aii=12 ⟨αi,αi⟩$

i.e.,

$⟨2ρ,αi⟩= ⟨αi,αi⟩.$

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

$Ω=ΩM:M⟶M$

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

$Ω(vλ)= ∣λ+ρ∣2 vλ+2u.vλ$

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

$u.vλ= ∑α∈R+ uα.vλ$

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: $Ω(ei.vλ) -eiΩ(vλ) = ( ∣λ+αi+ρ∣2 - ∣λ+ρ∣2 ) ei.vλ+2 [u,ei].vλ = ⟨ αi,2λ+2ρ+αi ⟩ ei.vλ-2 hi∨.ei.vλ by (3.1.something = ( ⟨ αi,2λ+2ρ +αi ⟩ -2 ⟨ αi,λ +αi ⟩ ) ei.vλ = ⟨ αi,2ρ-αi ⟩ ei.vλ=0.$ Likewise, $Ω(fi,vλ)- fi.Ω(vλ) = ( ∣λ-αi+ρ∣2 -∣λ+ρ∣2 ) fi.vλ+2 [u,fi].vλ = - ⟨ αi,2λ+ 2ρ-αi ⟩ fi.vλ+2fi hi∨.vλby (3.13) = ( - ⟨ αi,2λ+2ρ -αi ⟩ +2⟨αi,λ⟩ ) fi.vλ = - ⟨ αi,2ρ-αi ⟩ fi.vλ=0.$ Finally, $Ω(h.vλ)- h.Ω(vλ) = ( ∣λ+ρ∣2- ∣λ+ρ∣2 ) h.vλ+2 [u,h].vλ = 0by (3.13) again.$ $\square$

Remark: $\Omega$ is functorial, i.e. if $f:\phantom{\rule{0.2em}{0ex}}M\to N$ is a $𝔤$–module homomorphism (with $M,N\in 𝒪\right)$ then the diagram

$M ⟶f N ΩM ↓ ↓ ΩN M ⟶f N$

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 $M\right)$

$ΩM= ∣λ+ρ∣2. 1M.$

(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

$S= { λ∈D(F): ∣λ+ρ∣2=a } .$

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

$ch(M)= ∏-1∑λ∈S dλeλ$

where $\prod =\prod _{\alpha \in {R}^{+}}{\left(1-{e}^{-\alpha }\right)}^{{m}_{\alpha }}.$ Proof. If $\mu \in D\left(F\right)$ we have $\lambda -\mu \in {Q}^{+}$ for some $\lambda \in F,$ hence $\text{ht}\phantom{\rule{0.2em}{0ex}}\left(\lambda -\mu \right)$ is an integer $\ge 0.$ Define the depth of $\mu$ (relative to $F\right)$ to be $δ(μ)= max { ht(λ-μ): λ∈F,μ∈D(λ) }$ so that $\lambda \left(\mu \right)\in ℕ;$ also define $δ(M)=min { δ(μ):μ∈ P(M) } .$ 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 ${M}_{{\mu }_{i}}$ $\left(1\le i\le r\right).$ Let ${d}_{i}=\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{\mu }_{i}},$ and let $V=⊕i=1r V(μi)di.$ Choose a $k$–basis of each ${M}_{{\mu }_{i}}$ and let $\phi :\phantom{\rule{0.2em}{0ex}}V\to M$ 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 $0⟶M′⟶V ⟶φM⟶M′′ ⟶0.$ 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. ${\mu }_{i}\in S$ $\left(1\le i\le r\right).$ 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) $ch(M) = ch(V)+ ch(M′′) -ch(M′) = ∑i=1rdi chV(μi)+ ch(M′′) -ch(M′).$ Now repeat the same procedure on ${M}^{\prime }$ and ${M}^{\prime \prime }.$ After we have done it $m$ times we shall have say $ch(M)= ∑μ∈Smdμ chV(μ)+fm$ where ${S}_{m}$ is some finite subset of $S,$ and $\delta \left(v\right)>m$ for all $v\in \text{Supp}\phantom{\rule{0.2em}{0ex}}\left({f}_{m}\right).$ Now let $m\to \infty$ and we have $ch(M) = ∑μ∈Sdμ chV(μ) = ∏-1 ∑μ∈Sdμ eμby (3.10).$ $\square$

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

$ch(M) = ∏-1 ∑ μ∈D(λ) ∣μ+ρ∣2= ∣λ+ρ∣2 dμeμ$

with ${d}_{\mu }\in ℤ,$ and in particular ${d}_{\lambda }=1.$

## The Weyl-Kac character formula

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

Let $\left(M,\pi \right)$ 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 $\mu =\lambda -\sum _{i=1}^{n}{m}_{i}{\alpha }_{i},$ 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 $⇒$ simple).

(3.17) Let $\left(M,\pi \right)$ 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) $xNy=∑r=0N (Nr) (adx)ry xN-r$ $\left(x,y\in$ associative ring $R\right).$ 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 $x=\pi \left({f}_{i}\right),$ $y=\pi \left(u\right):$ $π(fi)Nv = π(fi)Nπ (u)x = ∑r=0N (Nr)π (adfiru) π(fi)N-rx.$ Now $\text{ad}\phantom{\rule{0.2em}{0ex}}{f}_{i}$ is locally nilpotent on $𝔤$ (1.19), hence also on $U\left(𝔤\right),$ so that ${\left(\text{ad}\phantom{\rule{0.2em}{0ex}}{f}_{i}\right)}^{m}u=0$ for some $m\ge 1.$ Hence if $N$ is large enough $\left(N=k+m-1$ would do) either $r\ge m$ or $N-r\ge k$ for each $r\in \left[0,N\right],$ and so $\pi {\left({f}_{i}\right)}^{N}$something $\square$

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

1. $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)$ is $W$–invariant (as a function on ${𝔥}^{*}\right)$

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 $\left[-p,q\right]$ in $ℤ,$ where $p,q\ge 0$ and $p-q=\mu \left({h}_{i}\right).$

3. If $\mu \in P\left(M\right),$ then $\mu \left({h}_{i}\right)\in ℤ$ $\left(1\le i\le n\right).$ Proof. We shall make use of the following formula: $eadxy= exye-x$ for elements $x,y$ of an associative $ℚ$–algebra, with $x$ nilpotent (so that ${e}^{x}$ is defined). The proof is very simple: we have $\text{ad}\phantom{\rule{0.2em}{0ex}}x={\lambda }_{x}-{\rho }_{x},$ and ${\lambda }_{x},{\rho }_{x}$ commute, hence $eadxy = eλx-ρxy = eλx e-ρxy = λex ρe-xy = exye-x.$ Let $x\in 𝔤.$ Since $\text{ad}\phantom{\rule{0.2em}{0ex}}{e}_{i}$ and $\pi \left({e}_{i}\right)$ are locally nilpotent, we have $π(eadeix) = eadπ(ei) .π(x) = eπ(ei) π(x) e-π(ei).$ by the formula above. Similarly with ${e}_{i}$ replaced by ${f}_{i}.$ Hence if (as in Ch. II) we write $w∼i= eadei e-adfi eadei$ then we have $π(w∼ix)= θiπ(x) θi-1 (1)$ where $θi= eπ(ei) e-π(fi) eπ(ei) ∈GL(M)$ Now recall (2.3) that ${\stackrel{\sim }{w}}_{i}h={w}_{i}h$ for $h\in 𝔥.$ It follows from (1) that $π(wih)= θiπ(h) θi-1. (2)$ Now let $\mu \in P\left(M\right),$ $v\in {M}_{\mu },$ $v\ne 0.$ Then $\pi \left(h\right)v=\mu \left(h\right)v$ $\left(h\in 𝔥\right)$ and therefore $π(h) (θi-1v) = θi-1π (wih)vby (2) = θi-1 (μ(wih)v) = (wiμ)(h) θi-1v$ (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 $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{\mu }\le \text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{{w}_{i}\mu };$ replacing $\mu$ by ${w}_{i}\mu$ we get the opposite inequality, hence $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)$ is $W$–invariant. Same proof as (2.31) (root strings). Follows from (b). $\square$

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

Let $P$ (resp. ${P}^{+}\right)$ denote the set of all integral (resp. dominant integral) $\lambda \in {𝔥}^{*}.$ Notice that each ${\alpha }_{j}\in P,$ because ${\alpha }_{j}\left({h}_{i}\right)={a}_{ij}\in ℤ:$ thus $Q\subset P.$ (Warning: ${Q}^{+}\not\subset {P}^{+}\right).$ Clearly ${P}^{+}=P\cap {C}^{\vee }$ $\left({C}^{\vee }$ 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) $eifiN+1= fiN+1ei+ (N+1)fiN (hi-N).$ 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 $0=eifiN+1 .vλ=(N+1) fiN (λ(hi)-N) vλ (1)$ 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 $\mu \in \lambda -{Q}^{+}\subset P.\right)$ 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 $\rho \left({h}_{i}\right)=\frac{1}{2}{a}_{ii}\phantom{\rule{0.2em}{0ex}}\left(1\le i\le n\right).$ Since $A$ is now a Cartan matrix, this condition now becomes

$ρ(hi)=1 (1≤i≤n).$

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

For $w\in W,$ let $\epsilon \left(w\right)=\text{det}\phantom{\rule{0.2em}{0ex}}\left(w\right)={\left(-1\right)}^{l\left(w\right)}.$ (sign character of $W\right).$

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

$w(eρ∏)= ε(w).eρ∏$

for all $w\in W.$ Proof. It is enough to verify this when $w={w}_{i}$ is a generator of $W.$ We have $wiρ=ρ-ρ(hi) αi=ρ-αi.$ On the other hand (2.6), ${w}_{i}$ sends ${\alpha }_{i}$ to $-{\alpha }_{i}$ and permutes the positive roots $\ne {\alpha }_{i}$. Thus $wi ( eρ∏α∈R+ (1-e-α)mα ) = eρ-αi (1-eαi) ∏ α∈R+ α≠αi (1-e-α)mα (sinceαihas multiplicity 1) = -eρ∏.$ $\square$

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

$⟨ρ,αi⟩= εi-1ρ (hi)= εi-1>0 (1≤i≤n) (1)$

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

$⟨λ,αi⟩= εi-1λ (hi)≥0. (2)$

(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

$ch(M)= ( ∑w∈Wε(w) ew(λ+ρ) ) /eρ∏α∈R+ (1-e-α)mα.$ Proof. From (3.16) we have, writing ${d}_{\mu }={c}_{\mu +\rho },$ $eρ∏.ch(M)= ∑μcμ+ρ eμ+ρ$ summed over $\mu \in D\left(\lambda \right)$ such that ${\mid \mu +\rho \mid }^{2}={\mid \lambda +\rho \mid }^{2},$ with coefficients ${c}_{\mu +\rho }\in ℤ$ and, in particular, ${c}_{\lambda +\rho }=1.$ Now $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)$ 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 $∑μcμ+ρ eμ+ρ=∑μ ε(w)cμ+ρ ew(μ+ρ)$ so that ${c}_{w\left(\mu +\rho \right)}=\epsilon \left(w\right){c}_{\mu +\rho }.$ Hence if ${c}_{\mu +\rho }\ne 0$ we have $w\left(\mu +\rho \right)\le \lambda +\rho$ for all $w\in W;$ choose $w$ so that $\text{ht}\phantom{\rule{0.2em}{0ex}}\left(\lambda +\rho -w\left(\mu +\rho \right)\right)$ is minimal and put $\nu =w\left(\mu +\rho \right).$ Then $\text{ht}\phantom{\rule{0.2em}{0ex}}\left(\lambda +\rho -{w}_{i}\nu \right)\ge \text{ht}\phantom{\rule{0.2em}{0ex}}\left(\lambda +\rho -\nu \right),$ i.e. $\text{ht}\phantom{\rule{0.2em}{0ex}}\left(\nu -{w}_{i}\nu \right)\ge 0$ and therefore $\nu \left({h}_{i}\right)\ge 0,$ or equivalently $⟨\nu ,{\alpha }_{i}⟩\ge 0.$ Thus $\nu$ satisfies $⟨\nu ,{\alpha }_{i}⟩\ge 0\phantom{\rule{0.2em}{0ex}}\left(1\le i\le n\right);$ $\nu \le \lambda +\rho ,$ i.e. $\lambda +\rho =\nu +\sum _{1}^{n}{m}_{i}{\alpha }_{i}$ with coefficients ${m}_{i}\ge 0;$ ${\mid \nu \mid }^{2}={\mid w\left(\mu +\rho \right)\mid }^{2}={\mid \mu +\rho \mid }^{2}={\mid \lambda +\rho \mid }^{2}.$ These three conditions force $\nu =\lambda +\rho ;$ for we have $0 = ∣λ+ρ∣2- ∣ν∣2 = ⟨ λ+ρ+ν,λ+ρ-ν ⟩ = ⟨ λ+ρ+ν,∑ miαi ⟩ = ∑mi ⟨ λ+ρ+ν,αi ⟩$ 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 $⟨\lambda +\rho +\nu ,{\alpha }_{i}>0⟩\phantom{\rule{0.2em}{0ex}}\left(1\le i\le n\right),$ and therefore all coefficients ${m}_{i}$ are 0, hence $\nu =\lambda +\rho$ and therefore $\left(\text{since}\phantom{\rule{0.2em}{0ex}}{c}_{\lambda +\rho }=1\right)$ $∑μcμ+ρ eμ+ρ=∑w∈W ε(w) ew(λ+ρ).$ $\square$

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 $\lambda$ $\left(\in {P}^{+},\phantom{\rule{0.2em}{0ex}}\text{by (3.19)}\right)$ then $\text{ch}\phantom{\rule{0.2em}{0ex}}\left(M\right)$ depends only on $\lambda .$ It follows that

$ch(M)=ch L(λ)$

i.e. $\text{dim}\phantom{\rule{0.2em}{0ex}}{M}_{\mu }=\text{dim}\phantom{\rule{0.2em}{0ex}}L{\left(\lambda \right)}_{\mu }$ 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

$∑w∈Wε(w) ewρ-ρ= ∏α∈R+ (1-e-α)mα$ 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 $\text{ch}\phantom{\rule{0.2em}{0ex}}L\left(0\right)=1.$ $\square$

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

$S(w) = { α∈R+: w-1α∈R- } = R+∩wR-$

is a finite set (2.10). Define

$s(w)= ∑α∈S(w)α$

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

(3.24) $s\left(w\right)=\rho -w\rho .$ Proof. If $w={w}_{{i}_{1}}\dots {w}_{{i}_{r}}$ is a reduced word for $w\left(r=l\left(w\right)\right)$ then (2.9) $S(w)= { αi1,wi1 αi2,…, wi1…wir-1 αir }$ from which it follows that if ${w}^{\prime }={w}_{{i}_{2}}\dots {w}_{{i}_{r}}$ $S(w)=αi1∪ wi1S(w′)$ i.e. if $w={w}_{i}{w}^{\prime }$ with $l\left({w}^{\prime }\right)=l\left(w\right)-1,$ then $S(w)=αi∪wi S(w1)$ and therefore $s(w)=αi+ wis(w′). (1)$ 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 $s(w) = αi+wi (ρ-w′ρ) by (1)←ind. hyp. = αi+ ( ρ-ρ(hi)αi ) -wρ = ρ-wρ$ since $\rho \left({h}_{i}\right)=1.$ $\square$

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

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

$∑w∈Wε(w) e-s(w)= ∏α∈R+ (1-e-α)mα .$

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

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

$chL(λ)= ∑w∈W ε(w) ew(λ+ρ) ∑w∈W ε(w)ewρ .$

(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 ${m}_{\alpha }>1\right).$

Examples

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

$ρ=12 ∑α∈R+α (1)$

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

$wiδ = 12∑α∈R+ wiα = 12 ( -αi+ ∑ α∈R+ α≠αi α ) =δ-αi;$

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

$∑w∈Wε(w) ewρ = eρ∏α∈R+ (1-e-α) = ∏α∈R+ ( eα/2- e-α/2 )$

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

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 ${u}_{i}-{u}_{j}$ $\left(i\ne j\right)$ and the positive roots ${u}_{i}-{u}_{j}$ $\left(i Thus

$ρ=12∑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 ${x}_{i}\right).$ We have

$eρ = x1n-12 xxn-32 … xn1-n2 = (x1…xn) 1-n2 x1n-1 x2n-2…$

and therefore

$∑w∈Wε(w) ewρ= (x1…xn) 1-n2 Δ(x1,…,xn)$

where $\Delta \left({x}_{1},\dots ,{x}_{n}\right)=\text{det}\phantom{\rule{0.2em}{0ex}}\left({x}_{i}^{n-j}\right)$ is the Vandermonde determinant. On the other side,

$eρ∏α∈R+ (1-e-α) = eρ∏i

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

$Δ(x1,…,xn) =∏i

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

$w1w2w1… torterms (all r≥1) w2w1w2… torterms (all r≥1)$

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

$s(w) = α1+w1α2+ w1w2α1+… torterms, by (2.8) = α1+ (2α1+α2) +(3α1+2α2) +… = 12r(r+1)α1 +12r(r-1)α2$

so that if we put ${x}_{1}={e}^{-{\alpha }_{1}},{x}_{2}={e}^{-{\alpha }_{2}}$ we have

$∑w∈Wε(w) e-s(w) = 1+∑r=1∞ (-1)r ( x112r(r+1) x212r(r-1) + x112r(r-1) x212r(r+1) ) = ∑r∈ℤ (-1)r x112r(r+1) x212r(r-1)$

On the other hand, the positive roots are

$rα1+(r-1)α2, rα1+rα2, (r-1)α1+rα2 (r≥1)$

$r{\alpha }_{1}+r{\alpha }_{2}=r\delta$ is an imaginary root; in fact (as we shall see later) it has multiplicity 1. So we obtain the identity

$∑r∈ℤ(-1)r x112r(r+1) x212r(r-1) = ∏r=1∞ ( 1-x1r x2r-1 ) ( 1-x1r-1 x2r ) ( 1-x1rx2r )$

in $ℤ\left[\left[{x}_{1},{x}_{2}\right]\right].$ If we put ${x}_{1}{x}_{2}=t,$ ${x}_{1}=x$ it takes the form

$∑r∈ℤ(-1)r xrt12r(r-1) =∏r=1∞ (1-xtr-1) (1-x-1tr) (1-tr)$

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

$∏r=1∞ (1-xn)= ∑r∈ℤ (-1)r x12r(3r-1)$

Euler's pentagonal number theorem.

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

$∑r≥1 (-1)r t12r(r-1) xr-x1-r 1-x → ∑r≥1 (-1)r+1 (2r-1) t12r(r-1) = ∑r≥0 (-1)r (2r+1) t12r(r+1)$

Thus we obtain another famous identity due to Jacobi:

$∏r=1∞ (1-tr)3= 1-3t+5t3-7 t6+9t10-11 t15+…$
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:

$w1(α1) =-α1 w2(α1) =α1+3α2 w1(α2) =α2+3α1 w2(α2) =-α2$

So the real roots are ${\alpha }_{1},\phantom{\rule{0.2em}{0ex}}{\alpha }_{2},\phantom{\rule{0.2em}{0ex}}{\alpha }_{1}+3{\alpha }_{2},\phantom{\rule{0.2em}{0ex}}{\alpha }_{2}+3{\alpha }_{1},\dots$

$w1(α1+3α2) = -α1+3 (α2+3α1) = 8α1+3α2 w1 (3α1+8α2) = -3α1+8 (α2+3α1) = 21α1+8α2$

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

$f2kα1+ f2k-2α2; f2k-2α1+ f2kα2 (k≥1)$

$\left({f}_{0}=0,{f}_{1}=1,\phantom{\rule{0.2em}{0ex}}{f}_{r}={f}_{r-1}+{f}_{r-2}\right)$

From this we calculate easily

$s ( w1w2… ⏟kterms ) = (f2k+1-1)α1 +(f2k-1-1) α2$

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

$1-x1-x2+x14 x2+x2x24- x112x24-x14 x212+x133x212 +x112x233-…$

$\left({x}_{i}={e}^{-{\alpha }_{i}}\right),$ i.e. it is a 'sparse' power series in ${x}_{1},{x}_{2}.$ By factorizing it as a product of factors ${\left(1-{x}_{1}^{{k}_{1}}{x}_{2}^{{k}_{2}}\right)}^{{m}_{{k}_{1},{k}_{2}}}$ we compute the multiplicities of the roots:

$=(1-x1) (1-x2) (1-x1x2)…$