Weyl’s character formula (to be proved shortly) gives an expression for the character of  a finite-dimensional simple quotient of a Verma module. In here, we will express the character of the simple quotient using Verma module characters.  Next time, we will calculate the coefficients involved.

Filtration on highest weight modules

Let ${W(\lambda)}$ be any highest weight module with highest weight ${\lambda}$. Then ${W(\lambda)}$ is a quotient of ${V(\lambda)}$, so the Casimir ${C}$ acts on ${W(\lambda)}$ by scalar multiplication by ${(\lambda + \rho, \lambda+\rho) - (\rho, \rho)}$.

Suppose we have a composition series $\displaystyle 0 \subset W^0 \subset W^1 \subset \dots \subset W^t = W(\lambda)$

with successive quotients simple module ${L(\mu)}$. Then ${C}$ acts on the successive quotients by scalars that we compute in two different ways, whence by yesterday’s formula: $\displaystyle \boxed{ (\mu + \rho, \mu + \rho) = (\lambda + \rho, \lambda+ \rho).}$

In fact, such a filtration exists:

Proposition 1 ${W(\lambda)}$ has a finite filtration whose quotients are isomorphic to ${L(\mu)}$, where ${\mu \in \lambda - \sum_{\delta \in \Delta} \mathbb{Z}_{\geq 0} \delta}$ (which we write as ${\mu \leq \lambda}$) and ${\mu}$ satisfies the boxed formula.

In general, this follows simply because every element in ${\mathcal{O}}$ has finite length, and the ${L(\mu)}$ are the only candidates for simple modules!

Theorem 2 The category ${\mathcal{O}}$ is artinian.

The only proofs I can find of this use Harish-Chandra’s theorem on characters though, so I’ll follow Sternberg in proving the proposition directly. (I hope later I’ll come back to it.)

For the proposition, induct on the sum $\displaystyle \sum_{\mu: (\mu +\rho, \mu+\rho) = (\lambda+\rho, \lambda+\rho), \mu \leq \lambda} \dim W(\lambda)_{\mu} \quad (*).$

This sum is finite because there are only finitely many possibilities for ${\mu}$. If it’s ${1}$, then ${W(\lambda)}$ cannot have any nontrivial submodules. If it did, then any submodule is a weight module and belongs to ${\mathcal{O}}$, so has a highest weight vector ${w_+ \notin W(\lambda)_{\lambda}}$, say of weight ${\lambda'}$. Then, by yesterday’s remarks on Casimir elements, ${\lambda'}$ satisfies the condition to be in the above sum, with ${\dim W(\lambda)_{\lambda'} > 0}$, contradiction. So in this case we have simplicity and the result is evident.

Now suppose this sum is ${>1}$. Then if ${W(\lambda) \simeq L(\lambda)}$, then the result is obvious, so assume we have a highest weight proper submodule ${W' \subset W}$, of weight ${\mu}$ satisfying the boxed condition. There is an exact sequence $\displaystyle 0 \rightarrow W' \rightarrow W \rightarrow W/W' \rightarrow 0$

and all three objects here belong to ${\mathcal{O}}$ as one may easily check. Then ${W', W/W'}$ have smaller sums as in (*), and are both highest weight modules (with weights ${\mu, \lambda}$ respectively), so we can apply the inductive hypothesis.

Expression of the character of ${L(\lambda)}$ via Verma module characters

We have that $\displaystyle \mathrm{ch} {V(\lambda)} = \sum_{\mu < \lambda } c(\lambda, \mu) \mathrm{ch} {L(\mu)}$

Here ${\mu}$ ranges over the finite set of integral points ${<\lambda}$ satisfying ${(\mu+\rho, \mu+\rho) = (\lambda+\rho, \lambda+\rho)}$. Incidentally, recall that the inner product ${(.,.)}$ is positive-definite on ${\sum_{\alpha \in \Phi} \mathbb{R} \alpha}$.

Also ${c(\lambda, \lambda)=1}$ since the weight space of ${\lambda}$ must be one-dimensional, and ${L(\mu)}$ for ${\mu \neq \lambda}$ contributes nothing there.

So, we have a triangular expression for ${\mathrm{ch} {V(\lambda)}}$ in terms of ${\mathrm{ch} L(\mu)}$ with entries one on the diagonal. Inverting:

Proposition 3 There are ${b(\lambda, \mu) \in \mathbb{Z}}$ with $\displaystyle \mathrm{ch} L(\lambda) = \sum_{\mu < \lambda} b(\lambda, \mu) \mathrm{ch} V(\mu).$

Also, ${b(\lambda, \lambda)=1}$

Next, we’ll actually compute these ${b(\lambda, \mu)}$.

The value of the ${b(\lambda, \mu)}$

This turns out to be very simple.

Proposition 4 ${b(\lambda, w \lambda) = (-1)^w}$ for ${w \in W}$, the Weyl group. For ${\mu \notin W\lambda}$, we have ${b(\lambda, \mu)=0}$.

Here ${(-1)^w}$ is the determinant of ${w}$ as an orthogonal transformation.

In particular, we can obtain the character of ${L(\lambda)}$ $\displaystyle \boxed{ \mathrm{ch} L(\lambda) = \sum_{w \in W} (-1)^w \mathrm{ch} V(w \lambda). }$

Next time, we shall prove this proposition and associated character formulas.