Why is the inner product of a character with an irreducible character a non-negative integer?

I can see that by properties of the inner product it will be non-negative but I cannot see why it would be an integer?

  • 2
    $\begingroup$ I assume that you are talking about characters of finite groups, and that the former character is a character of a representation. Remember that the inner product of the characters of two representations of a group is the dimension of the $\operatorname{Hom}_G$ between these two representations. This is a nonnegative integer. $\endgroup$ – darij grinberg May 12 '15 at 14:49

I assume we're talking about finite groups and complex scalars (so, the semisimple situation).

Note that if $U$ and $V$ are reps of $G$, then $\hom(U,V)$ is also a rep: $g$ acts as $T\mapsto gTg^{-1}$ for all linear maps $T:U\to V$. The invariant subspace $\hom(U,V)^G$ is precisely the space of equivariant maps (aka intertwining operators) $\hom_G(U,V)$.

For any representation $V$, the linear map $\sigma=\sum_{g\in G}g$ satisfies $\sigma^2=|G|\sigma$, and we can deduce the linear map $P=\frac{1}{|G|}\sum_{g\in G}g$ is a projection $P=P^2$ whose image is $V^G$ (the image is clearly a subspace of the invariant space $V^G$, and $P$ acts as the identity on $V^G$). Taking traces, we conclude that ${\rm tr}(P)=\frac{1}{|G|}\sum_{g\in G}\chi_V(g)=\dim V^G$.

There is an obvious map $U^*\otimes V\to\hom(U,V)$ which is an isomorphism of representations. The character of $U^*\otimes V$ equals $\chi_{U^*\otimes V}(g)=\chi_{U^*}(g)\chi_V(g)=\overline{\chi_U(g)}\chi_V(g)$. Therefore,

$$\dim\hom_G(U,V)=\dim(U^*\otimes V)^G=\frac{1}{|G|}\sum_{g\in G}\overline{\chi_U(g)}\chi_V(g) $$

which is $\langle \chi_U,\chi_V\rangle$ (at least with the physicists' inner product).


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