Weyl Character Formula to find $M_\lambda(\mu)$

Question

In Introductory Lie algebra book by Humphreys, he has used Weyl Character Formula to find the dimension of $V(\lambda)$ in the examples followed by the proof of this formula. But how to find the dimensions of the weight spaces $M_\lambda(\mu)$ using the Weyl character formula?

There are other formulas to find the dimension of the weight spaces, but I want to do it using the Weyl Character Formula.

thanks in advance.

Answer 1

I am a little bit confused with Humphreys notations, correct me if I am proving the wrong thing

Let $M(\lambda)$ be the Verma modulus, $L(\lambda)$ -- irreducible factor. I am not shifting by $\rho$.

Then $ch(M(\lambda))=e^\lambda\sum_{\nu \in Q_+}p(\nu)e^{-\nu}$. Weil formula can be rewritten as $ch(L(\lambda))=\sum_{w \in W}\varepsilon(w)ch(M(w(\lambda+\rho)-\rho))$. Thus if $p$ is Kostant partition function you get $\dim L(\lambda)_\mu=\sum_{w \in W}\varepsilon(w) p(w(\lambda+\rho)-\rho-\mu)$. I don't know how to derive more useful formulas a la Freudenthal formula directly.