Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
So, I'm a little nervous that I may have actually lied here. At some point I had this nagging worry that the argument fell through, but I was busy with other things and hadn't gotten back to it. The idea is supposed to be that the quaternions tensor squared is M_2(R). –  Noah Snyder Jan 24 '11 at 23:56
    
Nope, it's fine. You're overthinking things. –  Ben Webster Jan 25 '11 at 2:07
add comment

4 Answers

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.

share|improve this answer
    
I don't understand what you mean by "which must preserve isotypic components". Aren't you restricting the tensor product form to irreducible subrepresentations? And the only thing you have to worry about is the form becoming degenerate (i.e. zero in this case). –  ndkrempel Jan 26 '11 at 2:32
    
If I just randomly restrict to irreducible representations, I believe the form could be become degenerate. At least it's not obvious to me that it's impossible. What is impossible is for it to become degenerate on an isotypic component. –  Ben Webster Jan 26 '11 at 16:55
    
Ok, but then I don't see how it rules out a direct sum of an indicator -1 irrep with itself, since that supports a non-degenerate symmetric form. –  ndkrempel Jan 26 '11 at 19:33
    
For the case of the tensor product of two indicator +1 irreps, it seems you could use positive definiteness + restriction to irreducible subreps to get the result. –  ndkrempel Jan 26 '11 at 19:45
    
I unaccepted this answer, since non-degenerateness on isotypic components is not enough, as ndkrempel pointed out. @ndkrempel: I think positive definiteness doesn't work either here since you do not have such a thing for complex reps, and every representation over the reals, no matter of which type, has a positive definite invariant form. –  Frieder Ladisch Feb 5 '11 at 18:59
add comment

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

share|improve this answer
    
@Zoltan: do you actually have a counterexample for the nonexistence of the FS-indicator grading? I suspected this but don't know of good examples. –  Qiaochu Yuan Feb 12 '11 at 1:25
    
Thanks for your answer, Zoltan. I figured out the extension to groups with complex irreps myself when trying to prove the FS-indicator-grading (I did this only in the context of finite groups, however). –  Frieder Ladisch Feb 12 '11 at 18:40
    
And do you know of any nice example, where there appears a symplectic irrep in the tensor product of two real irreps? Because I haven't found any yet (to answer also to Qiaochu's question), although I haven't thought much about this. –  Zoltan Zimboras Feb 13 '11 at 9:59
    
As you stated it, the extension to groups with complex irreps is actually wrong: The group $SL(2,3)$ has a unique symplectic irrep. Multiplying its character with the complex linear characters yields irreps with FS-indicator 0. True is the following: in a tensorproduct of a complex irrep with its dual, symplectic irreps occur with even multiplicity. As far as I can follow the paper you linked, their main result is the extension of this to more than two factors, and for more general categories. –  Frieder Ladisch Feb 13 '11 at 15:28
    
The counterexample Qiaochu asks for would actually settle the question, so I'm interested in that, too. However, your answer shows that the weaker statement we already have has interesting applications, too. –  Frieder Ladisch Feb 13 '11 at 15:34
show 1 more comment
up vote 7 down vote accepted

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.

share|improve this answer
    
Nice example! Sorry for the error, but at least people seemed to have learned something from it. –  Noah Snyder Feb 14 '11 at 19:03
1  
@Noah: Indeed, I learned a lot from your answer to the other question! –  Frieder Ladisch Feb 14 '11 at 19:33
add comment

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.

share|improve this answer
    
Thank you for your answer. As to your "It sometimes happens...", it seems that this happens more often than not (for finite groups). Of course, this may be wrong, since I only checked small groups. One condition, where it must happen, is contained in Zoltan's answer: If we know for some reason that the "fusion rules" are zero or odd numbers. –  Frieder Ladisch Feb 14 '11 at 18:09
    
The variation you mention is, I think, the argument Fein used to prove that the Schur index of a character $\chi$ divides $n[\chi^n,1_G]$ for any natural number $n$. One looks at $\chi\otimes \dotsb \otimes \chi$ on $G\times \dotsb \times G$ and then restricts to the diagonal subgroup. –  Frieder Ladisch Feb 14 '11 at 18:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.