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,$\displaystyle \mathrm{ch} L(\lambda) = \frac{ \sum_{w \in W} (-1)^w e( w( \lambda + \rho)) }{ \sum_{w \in W} (-1)^w e( w( \rho)) }.$