Indicator theorem

Statement

The following information about an irreducible linear representation over complex numbers can be garnered from the Frobenius-Schur indicator of its character $\chi$ (denoted $v(\chi )$) (viz, its inner product with the indicator character):

• $v(\chi )=0$ if and only if $\chi$ is not real-valued
• $v(\chi )=1$ if and only if $\chi$ is the character of a real representation
• $v(\chi )=-1$ if and only if $\chi$ is real-valued but does not arise as the character of a real representation

Proof

Let $\rho :G\to GL(V)$ be the linear representation giving the character $\chi$. Denote by $Sym^{2}(\rho )$ the corresponding representation on $Sym^{2}(V)$ and by $Alt^{2}(\rho )$ the corresponding representation on $Alt^{2}(V)$. Let $\chi _{A}$ be the character of $Alt^{2}(\rho )$.

Then, by some elementary computations:

$2\chi _{A}(g)=\chi (g)^{2}-\chi (g^{2})$

From this it follows that:

$v(\chi )=(1_{G},\chi ^{2}-2\chi _{A})=(1_{G},\chi ^{2})-2(1_{G},\chi _{A})$

Now since $Alt^{2}(\rho )$ is a direct summand of $\rho \otimes \rho$, we must have $1_{G},\chi _{A})\leq (1_{G},\chi ^{2})$. But if $\chi$ is not real-valued, then $(1_{G},\chi ^{2})=0$ so $v(\chi )=0$ and if $\chi$ is real-valued, then $(1_{G},\chi ^{2})=1$. Now two cases arise:

• $(1_{G},\chi _{A})=1$ and hence $v(\chi )=-1$. This happens only if $\chi$ is not the character of a real representation.
• $(1_{G},\chi _{A})=0$ and hence $v(\chi )=1$. This happens only if $\chi$ is the character of a real representation.

The idea behind proving the latter distinction is to relate this with the existence of a group-invariant symmetric bilinear form.