# Proof of the Frobenius Schur indicator

I am trying to prove the Frobenius-Schur indicator for $\chi$ irreducible character.

\begin{equation} i_{\chi} = \begin{cases} 0, & \text{if $\chi$ is not real valued} \\ \pm1, & \text{if $\chi$ is real valued} \end{cases} \end{equation}

Now starting from definition

\begin{align}i_{\chi} &= \frac{1}{|G|}\sum_{g \in G}\chi(g^2) \\ &= \frac{1}{|G|}\sum_{g \in G}\chi_{S^2V}(g)-\chi_{A^2V}(g) \\ &= \frac{1}{|G|}\sum_{g \in G}\chi_{S^2V}(g)-\frac{1}{|G|}\sum_{g \in G}\chi_{A^2V}(g) \\ &= \dim (S^2V)^G-\dim (A^2V)^G \end{align}

which is the number of trivial representations in $S^2V$ - number of trivial representations in $A^2V$.

Somehow this is related to the fact that $\chi$ is an irreducible character.

I'm pretty sure I am correct so far but I cannot see how this relates to the final answer.

I have looked at this online but I am not familiar with invariant bilinear form so could these please be avoided.

• By $\text{dim}(S^2V)^G$, do you mean $\text{dim}(\text{Hom}_{\mathbb{C}G}(S^2V,\mathbb{C}G)$? Or do you mean $\text{dim}_{\mathbb{C}G}(S^2V)$? – Tanner Strunk Oct 26 '17 at 14:33

I think this is causing trouble because people are confused as to what level of linear algebra you have such that representation theory, $S^2 V$ and $A^2 V$ make sense to you, but invariant bilinear forms don't. But I'll take a try.
We have isomorphisms of $G$-representations: $$S^2 (V) \oplus A^2(V) \cong V \otimes V \cong \mathrm{Hom}(V,V^{\ast}).$$ So $$\dim S^2 (V)^G + \dim A^2(V)^{G} = \dim \mathrm{Hom}(V^{\ast},V)^G.$$ Schur's Lemma states that, if $X$ and $Y$ are irreducible complex representations of $G$, then $\mathrm{Hom}(X,Y)^G$ has dimension either $0$ or $1$. So the right hand side is $\leq 1$. So $(\dim S^2 (V)^G, \dim A^2(V)^{G})$ is $(1,0)$, $(0,0)$ or $(0,1)$, and thus the difference $\dim S^2 (V)^G - \dim A^2(V)^{G}=1$, $0$ or $-1$.
• I was missing $V \otimes V \cong \mathrm{Hom}(V,V^{\ast})$ – Permian May 3 '15 at 14:00