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 $χ$ (denoted $v (χ)$ ) (viz, its inner product with the indicator character):

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

Proof

Let $ρ : G \to G L (V)$ be the linear representation giving the character $χ$ . Denote by $S y m^{2} (ρ)$ the corresponding representation on $S y m^{2} (V)$ and by $A l t^{2} (ρ)$ the corresponding representation on $A l t^{2} (V)$ . Let $χ_{A}$ be the character of $A l t^{2} (ρ)$ .

Then, by some elementary computations:

$2 χ_{A} (g) = χ (g)^{2} - χ (g^{2})$

From this it follows that:

$v (χ) = (1_{G}, χ^{2} - 2 χ_{A}) = (1_{G}, χ^{2}) - 2 (1_{G}, χ_{A})$

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

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