We let $ G$ be a finite group.

If $ \chi$ is a complex character of $ G$ , we define $ \overline{\chi}:G \to \mathbb{C}$ by $ \overline{\chi}(g)=\overline{\chi(g)}$ for all $ g \in G$ , and define $ \chi^{(2)}:G \to \mathbb{C}$ by $ \chi^{(2)}(g) = \chi(g^2)$ . We write $ \chi_{S}$ and $ \chi_{A}$ for the symmetric and alternating part of $ \chi$ . We note that $ \chi_{S}$ and $ \chi_{A}$ are characters of $ G$ with $ \chi^{(2)}=\chi_{S} + \chi_{A}$ and $ \chi^{(2)}=\chi_{S} + \chi_{A}$ . We write

$ \nu(\chi):= \frac{1}{|G|}\displaystyle\sum_{g \in G}\chi(g^2)$

for the Frobenius Schur Indicator.

First, let $ \chi_{1}$ be the trivial character of $ G$ .=, i.e. $ \chi_{1}(g)=1$ for all $ g \in G$ . We want to show that $ \langle \chi , \overline{\chi} \rangle= \langle \chi_{S},\chi_{1}\rangle + \langle \chi_{A}, \chi_{1} \rangle$ .

We have: \begin{split} \langle \chi , \overline{\chi} \rangle &= \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\overline{\overline{\chi(g)}}\ &= \frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g)\chi(g)1\ &= \frac{1}{|G|}\displaystyle\sum_{g \in G} (\chi_{S}+\chi_{A})(g)1 \ &=\langle \chi_{S}+\chi_{A}, 1 \rangle \ &= \langle \chi_{S},1 \rangle +\langle \chi_{A} , 1 \rangle \ &= \langle \chi_{S},\chi_{1} \rangle +\langle \chi_{A} , \chi_{1} \rangle \end{split}

Is this correct?

Next, we let $ \chi$ be irreducible. We want to show that $ \nu(\chi) \in \{-1,1\}$ if $ \chi$ is real-valued, and that $ \nu(\chi)=0$ otherwise. Let us start from the ‘otherwise’ case first. We have:

\begin{split} \nu(\chi) &:=\frac{1}{|G|} \displaystyle\sum_{g \in G} \chi(g^2)\ &= \frac{1}{|G|} \displaystyle\sum_{g \in G} (\chi_{S}-\chi_{A})(g) \ &= \langle \chi_{S},\chi_{1} \rangle – \langle \chi_{A},\chi_{1} \rangle \ &= \langle \chi , \overline{\chi} \rangle – 2\langle \chi_{A} , \chi_{1} \rangle \end{split}

and I get stuck here. I think, for the ‘otherwise’ case, $ \langle \chi, \overline{\chi} \rangle = 0$ , because we assumed that $ \chi$ is irreducible and so it follows that $ \overline{\chi}$ is also irreducible and we also know the irreducible characters form an orthonormal basis (but is it for an arbitrary field?) and so it follows(?). For the real case we’d have that $ \langle \chi , \overline{\chi} \rangle = \langle \chi , \chi \rangle =1 $ from irreducibility of $ \chi$ , but then again, I am still not sure how to deal with $ 1- 2\langle \chi_{A} , \chi_{1} \rangle$ …I’d very much appreciate some help.