Identity with outer automorphism for finite group irrep.

Question

Let $ G $ be a finite group. Let $ \pi: G \to GL(V) $ be a degree $ d $ irreducible representation of $ G $. Let $ \sigma $ be an automorphism of $ G $ such that $ \sigma(\pi) $ is a distinct irreducible representation from $ \pi $. Is it the case that $$ 0= \sum_{g \in G} \pi(g) \pi(\sigma(g)) $$ Or perhaps $$ 0= \sum_{g \in G} \pi(g) \pi(\sigma(g))^* $$ where $ * $ denotes complex conjugate and we are now assuming that $ \pi: G \to GL_d(\mathbb{C}) $ is a complex representation. If so, is there any nice direct proof of this fact?

Answer 1

Let $\pi, \pi'$ be two distinct complex irreps of a finite group. The Great Orthogonality Theorem states that $$\sum_{g \in G} \pi(g)_{ij} \pi'(g)_{i'j'}^* = 0$$ where $i, j, i', j'$ are any matrix coordinates in the appropriate ranges, and $*$ denotes complex conjugation. (We need not even assume $\pi, \pi'$ are of the same dimension, only that they are distinct irreps.)

Now assume $\dim \pi = \dim \pi' = d$. We have $$\sum_{g \in G} \pi(g) \pi'(g)^* = \sum_{g \in G} \left(\sum_{k = 1}^d \pi(g)_{ik} \pi'(g)_{kj}^*\right)_{ij} = \left(\sum_{k = 1}^d \sum_{g \in G} \pi(g)_{ik} \pi'(g)_{kj}^*\right)_{ij} =0$$ where $(a_{ij})_{ij}$ denotes the matrix with $(i, j)$-entry $a_{ij}$, and we apply the Great Orthogonality Theorem to deduce that every coordinate is $0$. So the second version of your statement is the correct one.