Let $G$ be a compact semisimple group and let $\Gamma$ be a finite subgroup of $G$. I am interested, for $(\pi,V)\in \widehat G$ (irred rep of $G$), in a formula for $\mathrm{dim} V^\Gamma$, the dimension of the invariant space of $V$ by $\Gamma$. When $\Gamma$ is the trivial group, Weyl dimension formula says that $$ \mathrm{dim}(V) = \prod_{\alpha\in\Phi^+} \frac{\langle\lambda+\rho,\alpha\rangle}{\langle\rho,\alpha\rangle}, $$ where $\lambda$ is the highest weight of $\pi$ (I think the notation is usual).

One idea: let $P:V\to V$ given by $$ P(v)=\frac1{|\Gamma|} \sum_{\gamma\in\Gamma} \pi(\gamma)(v), $$ then $P$ is surjective onto $V^\Gamma$, thus $\mathrm{dim}(V^\Gamma)= \mathrm{Tr}(P) = |\Gamma|^{-1} \sum_{\gamma\in\Gamma} \chi_\pi(\gamma)$, where $\chi_\pi$ is the character of $\pi$. Hence, by applying the Weyl character formula, we obtain a formula for $\mathrm{dim}(V^\Gamma)$. However, I expect that there is a more explicit formula (like Weyl dimension formula, without a sum over the Weyl group) in terms of the root system of $\mathfrak g$.

Question: Is there a formula for $\mathrm{dim}(V^\Gamma)$ similar to Weyl Dimension Formula?