# Generalization of the Great Orthogonality Theorem?

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).

• 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 $\pi$. Jun 18, 2023 at 17:53
• @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? Jun 18, 2023 at 23:36