Finite groups in which every character has real values: grading the representations - MathOverflow

Finite groups in which every character has real values: grading the representations

2011-01-24T21:13:34Z

Let $G$ be a finite group. Then the irreducible complex representations of $G$ come in three sorts: real, complex and symplectic=quaternionic. The type of an irreducible character $\chi$ can be read of from the Frobenius-Schur indicator $$ s_2(\chi) = \frac{1}{|G|}\sum_{g\in G} \chi( g^2 ) \in \{ 1,0,-1 \}. $$ Now the following seems to be true (and has been used by Noah Snyder in his interesting answer to another question) but I can't see why: Suppose all characters of $G$ have real values. (Equivalently, every element of $G$ is conjugate to its inverse.) Then it seems that the Frobenius-Schur indicator defines a grading of the irreducible representations, and thus of the character ring. This means that if $\chi$ and $\psi$ are irreducible characters with $s_2 (\chi) = s_2(\psi)$ , then all the irreducible constitutents of $\chi\psi$ have indicator $1$ , and if $s_2(\chi) = -s_2(\psi)$ , then all constituents of $\chi\psi$ have indicator $-1$ . Why is this actually true? Of course, for example in the first case, $\chi\psi$ is afforded by a real representation, so the symplectic representations must occur with even multiplicity. But why can they not occur at all?

Moreover, I would like to know if this generalizes to other fields than $\mathbb{R}$ , using elements of a Brauer group instead of the Frobenius-Schur indicator.

EDIT: The statement turned out to be wrong in general (see below), so the original question is in some sense obsolete. A more appropriate questions would have been why this Frobenius-Schur indicator grading is there in some (many?) cases.

Answer by Ben Webster for Finite groups in which every character has real values: grading the representations

2011-01-25T02:06:03Z

Another way of saying Noah's point about quaternions is this. Every rep has a symmetric or antisymmetric invariant bilinear form, by assumption (this why you can't have complex representations; those aren't self dual at all). The tensor products of two such reps is again self-dual, just using the tensor product of the forms (which must preserve isotypic components) with symmetry vs. anti-symmetry following the obvious rules.

Unfortunately, this description doesn't tell me anything about the Brauer group more generally, but maybe someone who knows that story better can say something.

Answer by Zoltan Zimboras for Finite groups in which every character has real values: grading the representations

2011-02-11T19:50:54Z

Dear F. Ladisch,

I have come across this problem earlier, so let me add what I know. I agree with you and ndkrempel that Ben Webster's argument implies only that a symplectic irrep can occur with an even multiplicity in the product of two real irreps or of two symplectic irreps.

[Here I should add a short historical remark. Wigner basically used a similar argument to prove that for simply reducible finite (or even compact) groups the product of two irreps with the same FS-indicator cannot contain a symplectic representation. The term "simply reducible" means that the group has (i) no complex irreps, (ii) and the tensor product of any two irreps decomposes into a sum of irreps with multiplicities not exceeding 1. See E. P. Wigner, "On the Matrices Which Reduce the Kronecker Products of Representations of Simply Reducible Groups", reprinted in: L. C. Biedenharn and H. van Dam, Quantum Theory of Angular Momentum, Academic Press, New York (1965).]

But I don't think that the original form of the statement, that for finite groups with no complex irreps it is generally true that the Frobenius-Schur indicator gives a $\pm 1$ grading of $Rep(G)$, i.e., if the irreps $D_A$ and $D_B$ have the FS indicators $s_2^{A}$ and $s_2^B$, then $D_C$ can only be contained in $D_A \times D_B$ if $s_2^C =s_2^A \cdot s_2^B $. We only know that if $s_2^C \ne s_2^A \cdot s_2^B $ then the multiplicity of $D_C$ in $D_A \times D_B$ has to be even.

However, and here comes maybe the only new thing that I can add to the discussion, this latter statement can also be extended to groups with complex irreps (i.e., having FS indicator $0$). Namely, for every irrep $D_K$ of a group let us introduce a "new" indicator $\tilde{s}^K$, such that $\tilde{s}^K=s_2^K$ if $s_2^K= \pm 1$, and $\tilde{s}^K=1$ if $s_2^K= 0$ (or, $\tilde{s}^K=-(s_2^K)^2+s_2^K+1$ if you want :) ). It turns out, that again $\tilde{s}^C \ne \tilde{s}^A \cdot \tilde{s}^B$ implies that the multiplicity of $D_C$ in $D_A \times D_B $ is even. (A proof of this in the context of braided, semi-simple sovereign categories can be found here.)

And finally, a small personal remark :). The whole discussion started from Noah Snyder's remark on a previous question. He stated that $Rep(G)$ is graded basically by its center (this is true), and then he mentioned the non-trivial grading your question refers to, i.e. when there exist symplectic irreps of the group. As I mentioned, I don't think this grading by the FS-indicator is always there for any group. What is true, on the other hand, is that if the the multiplicities of irreps in any product of two irreps are zeros or odd numbers (like in the case of simply reducible groups, where the multiplicity is 0 or 1), then the existence of a symplectic irrep indeed implies the existence of a non-trivial grading of $Rep(G)$ and hence implies a nontrivial center of $G$. I actually noted this down in a very short review of representation rings a few years ago, see here.

All the best, Zoltan

Answer by F. Ladisch for Finite groups in which every character has real values: grading the representations

2011-02-14T16:06:12Z

As Zoltan suspected in his answer, the statement in my question is not true. I have now found counterexamples: Let $q$ be an odd prime power. Then $SL(2,q)$ contains a group isomorphic to the quaternion group $Q_8$, which yields a (semiregular) action of $Q_8$ on $V=(\mathbb{F}_q)^2$. Let $G = V \rtimes Q_8$ be the semidirect product. All faithful irreps of $G$ have degree $8$ (they are induced from nontrivial linear characters of $V$), and are of real type. The other irreps have $V$ in its kernel and come from irreps of $Q_8$. In particular, all irreps of $G$ are self-dual, and exactly one is of symplectic type, of degree $2$. The tensor product of a faithful irrep with itself contains the symplectic irrep as summand. Note that $\mathbf{Z}(G)=1$, so there's no nontrivial grading of the irreps of $G$.
Taking $q=3$ in the above yields a counterexample of order $72$, and I have to admit that I used GAP to find it. For the record, lets note that there are also four groups of order $64$, where all characters are real valued, but the FS-indicator doesn't define a grading of the irreps. These are the groups with identifiers [64, 218], [ 64, 224 ], [ 64, 243 ], [ 64, 245 ] in the SmallGroups library of GAP.

Answer by Greg Kuperberg for Finite groups in which every character has real values: grading the representations

2011-02-14T17:15:55Z

Here is a variation of the question which is true and generalizes readily to the Brauer group.

If $G$ and $H$ are two groups with two complex irreducible representations $V$ and $W$, then $V \otimes_{\mathbb{C}} W$ is an irrep of $G \times H$, and the Schur-Frobenius indicators multiply.

For the proof, I'll switch to the generalization that works for any Brauer group. Let $F$ be a field of characteristic zero and let $V$ and $W$ be two irreps of $G$ and $H$ over the field $F$. Then $V$ and $W$ might each represent an element of the Brauer group of $F$, because by Schur's lemma each of $\mathrm{End}_G(V)$ and $\mathrm{End}_H(W)$ is a division ring. I say "might" because it is not necessarily true that the centers of these division rings are $F$. If so, then $V \otimes_F W$ does not quite have to be $(G \times H)$-irreducible, but its summands land in the expected place in the Brauer group, using the equation $$\mathrm{End}_{G \times H}(V \otimes_F W) \cong \mathrm{End}_G(V) \otimes_F \mathrm{End}_H(V).$$ Taking the case $F = \mathbb{R}$, you get a relation that is equivalent to the previous paragraph.

Here also is a remark about Noah Snyder's answer to the other question. He points out that the representation ring of $G$ is graded by the character group $Z(G)^*$ of the center of $G$. It sometimes so happens, when the characters of $G$ are all real, that the Schur-Frobenius indicator is given by a group homomorphism from $Z(G)^*$ to $\{\pm 1\}$ . If so, then of course the indicator gives you a reduced grading of the representation ring of $G$. So, it would be interesting to know which finite or compact groups have this property, or an analogous property for another Brauer group. I think (not quite sure) that all compact simple Lie groups with real characters are examples.