TITLE:  Frobenius - Schur indicator and irreducible representations over R
QUESTION [5 upvotes]: from now on let $\mathbb{C}$ denote the complex number field, $G$ a finite group , and 
$T$ be a irreducible representation of G over $\mathbb{C}$ whose character is $\chi$.
Frobenius - Schur indicator is given by 
$ \mu = \frac{1}{|G|} \sum_{g\in G} \chi(g^2) $  
if $\mu$ =1 ,then $\chi$ can be realized over the real number field $\mathbb{R}$.
if $\mu$ =0 , then $\chi$ is not real.
if $\mu$ =-1 , then $\chi$ is real, but $T$ cannot be realized over $\mathbb{R}$.
also we know there are three different division algebra (finite dimension) over $\mathbb{R}$ : $\mathbb{R}$ , $\mathbb{C}$ and $\mathbb{H}$ (Hamilton quternions) , so the ring of endomorphisms commuting with the group action can be isomorphic only to either the real numbers, or the complex numbers, or the quaternions.
So I believe there are some connections between the three possible values of $\mu$ 
and the three kind of division algebra over $\mathbb{R}$. But I do not know why. Can anyone explain the connections between them? 
additionally, these three cases corresponde to where there is a symmetrc/skew symmetric $G$
-invariant bilinear form in V . Though I can go through the proofs, but I don't konw why the indicator can be controled by the existence of a bilinear form of a particular kind.
Is there something essential lying behind?
So far we have 
indicator--bilinear form--division algebra.
Also, for a given field $F$, and a division algebra $D$ over $F$, can we assign $D$ a “number invariant” $\mu_D$ such that $\mu_D$ completely determines $D$ (under isomorphism) ?

REPLY [5 votes]: I've always thought of this formula as an amusing accident. You say "I can go through the proofs", so I don't know if what I'll say helps any more than that, but here goes.

If $M$ is any endomorphism, it's easy to show $Trace(Sym^2(M)) = (Trace(M)^2 + Trace(M^2))/2$. Proof: check it for diagonal $M$, obtain for diagonalizable, extend by continuity to all. The same technique gives $Trace(Alt^2(M)) = (Trace(M)^2-Trace(M^2))/2$.
Basic character theory fact: the average of $Trace(g|_V)$ is $\dim V^G$.

If $V$ is a real representation, and $G$ is compact (yours being finite, apparently), then we can average a bilinear positive definite symmetric inner product to get a $G$-invariant one, which gives an isomorphism $V \equiv V^*$. That passes to the complexification $V_{\mathbb C}$. Hence there is an invariant vector (the inner product) in $ S ym^2(V)$.
If $V$ is a quaternionic representation, and $G$ is compact, then I admit I forget how to get an antisymmetric form on the complex representation $Forget(V)$. Maybe one can average a quaternionic-Hermitian form to get a $G$-invariant one?
Moreover one wants the converses of these: having the symmetric or antisymmetric inner products lets one realize a complex irrep as a complexification or $Forget()$ of a quaternionic rep. I don't remember this being too hard.
Then come a couple of minor miracles. Since $V$ is irreducible, Schur's lemma says there is at most a $1$-d space of invariant maps $V \to V^*$, i.e. $G$-invariant vectors in $V \otimes V$.  Since $V\otimes V$ is a $G\times Z_2$-rep, any $G$-invariant is either in $Sym^2(V)$ or $Alt^2(V)$, not some weird mix. Hence there are only three cases: no invariant at all, $\dim  (Sym^2(V))^G = 1$ and $\dim (Alt^2(V))^G = 0$, or vice versa. We can figure out which occurs by looking at $\dim (Sym^2(V))^G - \dim (Alt^2(V))^G = 0,1,-1$. By facts 1 & 2 above, that's the F-S indicator.