What is the trace of this map?

Question

Let $g$ be a Lie algebra and $Q^+$ the set of dominant weights. For every $\lambda \in Q^+$, there is an irreducible $g$-module $V_{\lambda}$ with highest weight $\lambda$. Let $\lambda, \mu \in Q^+$, $\varphi: Hom_g(V_{\mu}, V_{\lambda} \otimes V_{\lambda}) \to Hom_g(V_{\mu}, V_{\lambda} \otimes V_{\lambda})$ be a linear map given by $f \mapsto \tau f$, where $\tau$ is the flip map. We have $V_{\lambda} \otimes V_{\lambda} = S^2(V_{\lambda}) \oplus \Lambda^2(V_{\lambda})$.

Do we have $tr \varphi = m_{\mu}(S^2(V_{\lambda}))-m_{\mu}(\Lambda^2(V_{\lambda}))$? Here $m_{\mu}(S^2(V_{\lambda}))=\dim Hom_{g}(V_{\mu}, S^2(V_{\lambda}))$, $m_{\mu}(\Lambda^2(V_{\lambda}))=\dim Hom_{g}(V_{\mu}, \Lambda^2(V_{\lambda}))$.

Answer 1

Yes. We can choose basis of $\mathrm{Hom}_\mathfrak{g}(V_{\mu}, S^2V_{\lambda})$
and extend it by a basis of $\mathrm{Hom}_\mathfrak{g}(V_{\mu}, \Lambda^2V_{\lambda}).$ Since we have $f = f^S + f^A$ for all homomorphisms into $\bigotimes V_\lambda$, we obtain a basis of the whole $\mathrm{Hom}_\mathfrak{g}(V_{\mu}, \bigotimes^2V_{\lambda})$. Then the trace of $\varphi$ (which is just the flip $\tau$ restricted to $\mathrm{Hom}_\mathfrak{g}(V_{\mu}, \bigotimes^2V_{\lambda})$) is easily seen to be what you suggested.