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    $\begingroup$ Do any of the 300-odd remaining groups have complex irreps? I think I have an idea for showing that no complex irreps implies that the maximum is obtained at some specific central element. $\endgroup$ – Noah Snyder Oct 18 '10 at 20:05
  • $\begingroup$ Representations come in three sorts: real=orthogonal, complex, pseudoreal=symplectic=quaternionic. Complex representations are indeed the ones whose characters are non-real. I was asking about what happens in the case where there are no complex reps, so all representations are either real or quaternionic. (I couldn't quite get the argument I had in mind to work, but I'm still curious about the question.) $\endgroup$ – Noah Snyder Oct 19 '10 at 1:49
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    $\begingroup$ In your other answer, you had 1912 groups instead of 1911. Did something happen? $\endgroup$ – S. Carnahan Oct 19 '10 at 4:00
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As pointed out in comments and at this question, this answer is not entirely correct because real/quaternionic does not always give a $\mathbb{Z}/2$ grading. Nonetheless in practice this answer seems to be "mostly right." I'm going to leave it up unmodified because changing it would confuse the comments and other answers too much. Sorry for the mistake!

If $G$ has a quaternionic representation and $G$ has no complex representations (i.e. reps whose character is complex) then for certain nontrivial central elements the square root function is $\sum_\chi \chi(1)$ which is clearly the maximum possible value and clearly larger than the value on the identity.

The key idea in the proof is the fact that $Z(G)^*$ (the group of characters of the center of $G$) is the universal grading group for the category of $G$ representations.

Let me unpack that a little bit. To every irrep of $G$ you can assign it's "central character", that is an element of $Z(G)^*$. This assignment is multiplicative in the sense if $U$ is a subrep of $V \otimes W$ then the central character of $U$ is the product of the central characters of $V$ and $W$ (this is just because central elements act by scalars). In other words, $\operatorname{Rep}(G)$ is graded by $Z(G)^*$. The claim is that an $H$-grading of $\operatorname{Rep}(G)$ is the same thing as a map $Z(G)^* \to H$.

Why is this true? I'm just going to sketch the proof, in particular I'm just going to talk about the case where $Z(G)$ is trivial, but I'll indicate how the general case works. Since the center is trivial you can find a faithful representation $V$. But then let's look at a really high tensor power of $V$ and ask how it breaks up. We can compute that using character theory. Since the rep is faithful and there's no center, the character of a high tensor power of $V$ is dominated by the value on $1$. Hence any high tensor power of $V$ contains all other irreps. In particular, the $n$-th and $(n+1)$-th powers contain the same irreps and this tells you that the grading group is trivial. In general you want to argue that the contribution of the center dominates the values of inner products (but you need to be a bit careful about non-faithful representations).


Ok, how is this relevant to anything? Well, if you have a group with a quaternionic representation but no complex representations, then the Frobenius-Schur indicator gives a $\pm 1$ grading of $Rep(G)$, namely put the real reps in grade $1$, and the quaternionic ones in grade $-1$. Hence the Frobenius-Schur indicator $s(\chi)$ must be given by the central character $\chi(z)/\chi(1)$ for some central element $z$. Clearly these central elements maximize the square root function.

  • $\begingroup$ This argument about universal grading groups also works for compact groups, in particular for simply connected simple compact Lie groups you can see that the universal grading group is the weight lattice mod the root lattice and thus see that the weight lattice mod the root lattice is isomorphic to characters of the center. $\endgroup$ – Noah Snyder Oct 19 '10 at 20:52
  • $\begingroup$ That's really pretty, and it takes care of a good proportion of small groups whose square root counting function is not maximised at the identity. Thanks! Actually, your idea prompts me to test, whether the maximum is always attained at a central element, whenever the centre is non-trivial. I will report back on this. $\endgroup$ – Alex B. Oct 20 '10 at 1:39
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    $\begingroup$ Dear Noah, your idea seems rather more successful than meets the eye. Out of the first 1675 groups that do not maximise the square root counting function at the identity, all but 23 maximise it at a central element! Two of these 23 have trivial centre, so no surprise there, but 20 of them are actually 2-groups. Your argument explains 202 of the 1652 remaining ones. $\endgroup$ – Alex B. Oct 21 '10 at 3:00
  • $\begingroup$ By the way, a minor correction: in your last sentence, not the $z$ that gives rise to the Frobenius-Schur indicator maximises $r_2$, but its inverse. $\endgroup$ – Alex B. Oct 21 '10 at 3:01
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    $\begingroup$ The FS-indicator does not alway give a grading. For example, in the group $(C_3)^2 \rtimes Q_8$, all irreps are self-dual, there is a symplectic irrep, and the center is trivial. Also, the number of square roots is maximal for the identity element. On the positive side, the Frobenius-Schur indicator grading seems to be there for most of the groups that have no complex irreps. $\endgroup$ – Frieder Ladisch Feb 14 '11 at 16:18

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