Now choose a dominant integral weight {\lambda}. By yesterday, we have:

\displaystyle \mathrm{ch} L(\lambda) = \sum_{\mu < \lambda} b(\lambda, \mu) \mathrm{ch} V(\mu).

Our first aim is to prove

Proposition 1 {b(\lambda, w \cdot \lambda) = (-1)^w} for {w \in W}, the Weyl group, and {\cdot} the dot action. For {\mu \notin W\lambda}, we have {b(\lambda, \mu)=0}.


After this, it will be relatively easy to obtain WCF using a few formal manipulations. To prove it, though, we use a few such formal manipulations already.

Manipulations in the group ring

I will now define something that is close to an “inverse” of the Verma module character {p \ast e(\lambda)} for {p(\lambda)} the Kostant partition function evaluated at {-\lambda} (inverse meaning in the group ring {\mathbb{Z}[L]}, where {L} is the weight lattice of {\beta} with {<\beta, \delta> \in \mathbb{Z} \ \forall \delta \in \Delta}). Define {q} by

\displaystyle q = \prod_{\alpha \in \Phi^+} \left( e(\alpha/2) - e(-\alpha/2) \right).

I claim that

\displaystyle q = e(\rho) \prod_{\alpha \in \Phi^+} (1 - e(-\alpha)), \ \ wp = (-1)^w p, \quad \forall w \in W.

(Note that since {w} acts on the weight lattice {L}, it clearly acts on the group ring. Here, as usual, {\rho = \frac{1}{2} \sum_{\gamma \in \Phi^+} \gamma}.)

The first claim is obvious. The second follows because the minimal expression of {w} as a product of reflections has precisely as many terms as the number of positive roots that get sent into negative roots by {w}, and a reflection has determinant {-1}.

Also, if {\alpha \in \Phi^+}, we have a “geometric series” result

\displaystyle (1-e(-\alpha))( 1 + e(-\alpha) + e(-2\alpha) + \dots ) = 1 = e(0).

Here the second term in parentheses must be considered in the same way that a character of an infinite-dimensional Verma module is considered: as a function {L \rightarrow \mathbb{C}} vanishing outside a finite union of sets of the form {\alpha - \sum_{\gamma \in \Phi^+} \mathbb{Z}_{\geq 0} \gamma }.

The reason this is important is that

\displaystyle p = \prod_{\alpha \in \Phi^+} ( 1 + e(-\alpha) + e(-2\alpha) + \dots )

as is easily checked. In particular, it follows that { p \ast q = e(\rho).} So, if {V(\mu)} is the Verma module,

\displaystyle \boxed{ \mathrm{ch} V(\mu) \ast q = e(\rho + \mu).}

If we apply this to the inverted formula, we find

\displaystyle \mathrm{ch} L(\lambda) \ast q = \sum b(\lambda, \mu) e(\mu + \rho).

Computation of {b(\lambda, \mu)}

Suppose now that {\lambda} is dominant integral.

First, I claim that {\rho} is a dominant integral weight. We need to check that {<\rho, \alpha> \in \mathbb{Z}_{\geq 0}} for all {\alpha \in \Phi^+}. There is a lemma:

Lemma 2 Suppose {\mu \in E} is such that {<\mu, \delta> \in \mathbb{Z}} for all {\delta \in \Delta}; then {\mu} is in the weight lattice, i.e. {<\mu, \alpha> \in \mathbb{Z}} for all {\alpha \in \Phi}.


This relies on some general facts about root systems, and I’m not particularly inclined to prove it here. I should probably have made this clearer earlier. It follows that if we choose fundamental weights {\omega_i} with {<\omega_i, \delta_j> = \delta_{ij}} where {\delta_j} ranges over {\Delta}, then the weight lattice is spanned by the {\omega_i}.

Anyway, we do know that for any {\delta \in \Delta},

\displaystyle s_{\delta}(\rho) = \rho - \delta \quad \mathrm{and} \quad s_{\delta}(\rho) = \rho - <\rho, \delta> \delta.

So {<\rho, \delta> = 1, \forall \delta}. By the lemma, it follows that {\rho} is dominant integral.

However, we actually don’t even need the lemma. All we need to know for the sequel is that {(\rho, \alpha) > 0} for all {\alpha \in \Phi^+}, which is now clear because {(\rho, \delta)>0} for all {\delta \in \Delta}.

Now, in the formula, {\mathrm{ch} L(\lambda) \ast q = \sum b(\lambda, \mu) e(\mu + \rho)}, we find that {L(\lambda)} is invariant under any {w \in W} (this is true for any finite-dimensional representation—the weights are invariant under the Weyl group, which may be seen using the representation theory of {\mathfrak{sl}_2}). Also, applying {w} to {q} changes the sign by {(-1)^w}. As a result, we find that

\displaystyle \sum b(\lambda, \mu) e(\mu+\rho) = (-1)^w \sum b(\lambda, \mu) e( w(\mu + \rho)).

It follows that if {\mu = w(\lambda+\rho)-\rho}, then {b(\lambda, \mu) = (-1)^w}. This, incidentally, is the same as saying that {\mu = w \cdot \lambda} in view of the definition of the dot action.

I claim now that this accounts for all the nonzero {b(\lambda, \mu)}. If not, since the set of weights {\mu+\rho} is invariant under {W}, we can find some {\mu + \rho} with {b(\lambda, \mu) \neq 0} and {\mu + \rho} in the first Weyl chamber, that satisfying {(\mu+\rho, \alpha)>0} if {\alpha \in \Phi^+}.

Now by the computation for the Casimir element, we know that:

\displaystyle (\lambda + \rho, \lambda + \rho) - (\mu+\rho, \mu+\rho) = 0.

We can write this as

\displaystyle 0 =(\lambda + \mu + 2\rho, \lambda - \mu) \geq (\lambda+2 \rho, \lambda - \mu).

since {\lambda-\mu} is a sum of positive roots ({\lambda} being the highest weight) and {\mu} is dominant.

Then since {\lambda} brackets nonnegatively with all positive roots and {\rho} brackets positively with all positive roots, we have that {(\lambda, \lambda - \mu) \geq 0} and {(2\rho, \lambda- \mu) \geq 0}, with equality holding in the latter iff {\lambda - \mu=0}. But this means equality must hold, and {\lambda = \mu}.

In particular, we have proved the formula

\displaystyle \boxed{ \mathrm{ch} L(\lambda) = \sum_w (-1)^w \mathrm{ch} V( w \cdot \lambda).}

Weyl character formula

We can rewrite the boxed formula as

\displaystyle \mathrm{ch} L(\lambda) \ast q = \sum_{w \in W} (-1)^w e( w( \lambda + \rho))

whence taking {\lambda=0} (since {q} is independent of {\lambda}!) and using {\mathrm{ch} L(0)=1 \in \mathbb{Z}[L]} (this is the trivial representation, everything acting by zero), we find:

\displaystyle q = \sum_{w \in W} (-1)^w e( w( \rho))

and thus:

Theorem 3 (Weyl) For {\lambda} dominant integral,\displaystyle \mathrm{ch} L(\lambda) = \frac{ \sum_{w \in W} (-1)^w e( w( \lambda + \rho)) }{ \sum_{w \in W} (-1)^w e( w( \rho)) }.