# Frobenius - Schur indicator and irreducible representations over R

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) ?

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## 2 Answers

You asked a lot about the F–S indicator. I'll answer part of it for now, so that there is a smaller piece to answer later.

Your first division algebra question makes more sense in a slightly different context: if T is an irreducible real representation of G (an irreducible ℝG-module), then three cases occur: EndℝG(T) = ℝ is real, = ℂ is complex, or = ℍ is quaternionic. In the first case T is also irreducible over the complexes and its Schur indicator is +1. In the second case T is not irreducible over the complexes, but is instead the sum of two complex conjugate representations, both of whose Schur indicators are 0. In the third case, the representation "ramifies" when extended to the complexes, and you get the sum of two isomorphic representations (both of) whose Schur indicator is -1.

I like to think of this in terms of complex endomorphism rings, but then it no longer specifically mentions the quaternions: EndℂG(T) = ℂ is complex when the Schur indicator is +1, = ℂ×ℂ when the Schur indicator (of the summands) is +0, = M2(ℂ) when the Schur indicator (of the summand) is -1. These correspond to applying –⊗ℂ to the reals, complexes, and quaternions.

Your numerical invariant of the division algebra is probably best understood as a quest for the "Brauer group" of F. There is a very well-defined and reasonably well-understood theory for this. That is usually phrased in terms of "central simple algebras", but it turns out to mostly be a matter of notation. For F a p-adic field, then yes, you mostly just get a number, but for a general field it is a little bit of a stretch (but not insane) to say you get a number.

On the off chance someone is interested, this is one of the places where the Schur index is easily understood. If the Schur indicator is +1, then the field of values and the splitting field are equal to the reals. If the Schur indicator is 0, then the field of values and the splitting field are equal to the complexes. If the Schur indicator is -1, then the field of values is the reals, but the splitting field is degree 2 (the complexes), so the Schur index is 2. This is also the smallest multiple of the character that has a representation realizable over the field of values, and so this is is why in the -1 case, T splits into two (self-conjugate) copies of itself. In terms of blocks of the group ring, this block is a matrix ring over the quaternion algebra, and it contains many copies of the irreducible. Each copy of the irreducible has a centralizer isomorphic to the complexes, a splitting field of the block and the irreducible.

Edit: A gentle book for character theory is G. James and M. Liebeck's Representations and Characters of Groups. Its chapter 23 is entirely to devoted to all of these different ideas (as well as Frobenius's original motivation) without using anything fancy. It has many examples, many exercises, and solutions to many exercises. I think it is useful for both group theorists and people who just want to use representations.

For fancier things: I. Reiner's Maximal Orders has nice coverage of the division algebras associated to finite group algebras (and their maximal orders). My favorite textbook treatment of the numerical invariant of the division algebras are Albert's Modern Higher Algebra and Structure of Algebras, but some people think they are old fashioned. Recent treatments will often get lost in technicalities that are irrelevant to group algebras. You might be OK with one of large textbooks on algebra; many have a chapter on Central Simple Algebras and the Brauer Group. MathOverflow readers will probably like Rowen's Algebra: A Non-commutative View.

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thank you, Schmidt. Are there any text books cantain this material? –  zhaoliang Apr 9 '10 at 5:26
I recommend James&Liebeck. Frobenius–Schur is not until chapter 23, but the book is very gentle. I listed a couple of other books, in case J&L is not to your taste. –  Jack Schmidt Apr 9 '10 at 14:59

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.

1. 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 $Trace(Alt^2(M)) = (Trace(M)^2-Trace(M^2))/2$.

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.

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