Consider the sum $$ \Pi_\tau = \frac{1}{|G|} \sum_{g \in G} \chi_\tau^*(g) \, (\mu^\star \otimes \nu)(g), $$ where $G$ is a finite group, $\mu$, $\nu$, and $\tau$ are irreps of $G$, and $\chi_\tau$ is the character of $\tau$. Here $\mu^\star$ is the dual representation, i.e., $\mu^\star(g) = (\mu(g^{-1})^T$.

When $\tau$ is trivial then this is the "Great orthogonality theorem": $$ \Pi_1 = \frac{1}{|G|} \sum_{g \in G} \mu_{ij}(g^{-1}) \nu_{k \ell }(g) = \frac{1}{dim(\mu)}\delta_{\mu \nu} \delta_{ik}\delta_{j\ell} . $$

Thus $\Pi_\tau$ can be seen as a generalization of the great orthogonality theorem. Is there a similar result for $\Pi_\tau$ in general?

Edit: $\Pi_\tau$ can be seen to be a projection matrix so that $Tr(\Pi_\tau) = 0 \iff \Pi_\tau = 0$. But $Tr(\Pi_\tau) = \langle \mu^\star \otimes \nu , \tau \rangle $. This is the multiplicity of $\tau$ in the reducible rep $\mu^\star \otimes \nu$.

When $\tau$ is trivial, there are nice results, for example, $\mu^\star \otimes \nu$ contains a single copy of the trivial rep iff $\mu = \nu$ (Schur's lemma).

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    $\begingroup$ You seem to have answered the question in your edit. If you put $\dim(\mu)$ in the numerator of $\frac{1}{|G|}$ and replace $\mu^\ast\otimes\nu$ with $\pi$ (an arbitrary rep), then $\Pi$ is the projection onto the $\tau$-isotypic component of $\pi$. $\endgroup$
    – coiso
    Jun 18, 2023 at 17:53
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    $\begingroup$ @elemelons I agree with your comment, but I am not sure I am seeing how that says anything about the matrix elements and their orthogonality. Maybe you could say more? $\endgroup$ Jun 18, 2023 at 23:36


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