I noticed that for faithful representations of some groups sum of corresponding matrices is degenerate. E.g. for a representation of $S_2$ which permutes basis vectors we have $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix} $$ Analogously for other permutation groups.

For the representation of cyclic groups by rotations we have sum of matrices to be zero. The same is for the quaternionic group,for the $M16$ group and standard representations of permutation groups.

Is it just a coincidence or not?

I know that any finite dimensional representation of a finite group decomposes uniquely to a direct sum of irreducible ones. So the problem can be reduced to irreducible representations only. Also I know that representations of finite groups are equivalent to unitary ones. But all that doesn't help.


1 Answer 1


Let $\rho : G \to GL_n(\mathbb{C})$ be a complex representation of a nontrivial finite group $G$, and put \begin{align*} \pi = \frac{1}{\#G}\sum_{g\in G} \rho(g)\in M_n(\mathbb{C}). \end{align*} Then \begin{align*} \gamma \pi(x) = \frac{1}{\#G} \sum_{g\in G} \rho(\gamma g) x = \frac{1}{\#G} \sum_{g\in G} \rho(g) x = \pi(x) \end{align*} for any $\gamma\in G$. Thus $\pi$ is the projection onto the subspace $(\mathbb{C}^n)^G\subset \mathbb{C}^n$ fixed pointwise by $G$ (acting by $\rho$). If $G$ does not act trivially on $\mathbb{C}^n$ (e.g., $\rho$ is faithful), that subspace is a proper one, and so $\pi$ is degenerate.


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