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 $$1_G,\chi_A) \le (1_G,\chi^2)$$. 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.