42
$\begingroup$

The complex irreps of a finite group come in three types: self-dual by a symmetric form, self-dual by a symplectic form, and not self-dual at all. In the first two cases, the character is real-valued, and in the third it is sometimes only complex-valued. The cases can be distinguished by the value of the Schur indicator $\frac{1}{|G|} \sum_g \chi(g^2)$, necessarily $1$, $-1$, or $0$. They correspond to the cases that the representation is the complexification of a real one, the forgetful version of a quaternionic representation, or neither.

A conjugacy class $[g]$ is called "real" if all characters take real values on it, or equivalently, if $g\sim g^{-1}$. I vaguely recall the number of real conjugacy classes being equal to the number of real irreps.

  1. Do I remember that correctly?

  2. Can one split the real conjugacy classes into two types, "symmetric" vs. "symplectic"?

With #1 now granted, a criterion for a "good answer" would be that the number of symmetric real conjugacy classes should equal the number of symmetrically self-dual irreps.

(I don't have any application in mind; it's just bothered me off and on for a long time.)

$\endgroup$
10
  • 2
    $\begingroup$ Two types? or three? $\endgroup$ Nov 22, 2010 at 3:57
  • $\begingroup$ Perhaps whether the permutation of the conjugacy class by inversion is even or odd? $\endgroup$
    – Ben Webster
    Nov 22, 2010 at 4:30
  • 1
    $\begingroup$ Theorem 23.1 in James and Liebeck "Representation of Finite Groups" states that the number of real conjugacy classes is equal to the number of real irreps. I don't know about the second question though $\endgroup$ Nov 22, 2010 at 4:31
  • 5
    $\begingroup$ +1, neat question! @Ben, I don't think the sign of the permutation by inversion is correct. Under your definition, the conjugacy class of $(1 2 3)$ in $S_3$ would be symplectic. But, as everyone knows, all the conjugacy classes of $S_n$ should be symmetric, just like all the characters! $\endgroup$ Nov 22, 2010 at 7:58
  • 5
    $\begingroup$ @Allen: Maybe you will give us more of a hint of what you expect: the group $Q_8$ of quaternions has $4$ symmetrically self dual (real) representations and $1$ skew-symmetrically self-dual (quaternionic) representation, so we should expect exactly $1$ "quaternionic" conj. class. Given that the labeling should be stable by automorphisms, this special class must be either $1$ or $-1$. Which do you suppose it is? $\endgroup$ Nov 22, 2010 at 14:47

3 Answers 3

22
$\begingroup$

It's a great question! Disappointingly, I think the answer to (2) is No :

The only restriction on a `good' division into "symmetric" vs. "symplectic" conjugacy classes that I can see is that it should be intrinsic, depending only on $G$ and the class up to isomorphism. (You don't just want to split the self-dual classes randomly, right?) This means that the division must be preserved by all outer automorphisms of $G$, and this is what I'll use to construct a counterexample. Let me know if I got this wrong.

The group

My $G$ is $C_{11}\rtimes (C_4\times C_2\times C_2)$, with $C_2\times C_2\times C_2$ acting trivially on $C_{11}=\langle x\rangle$, and the generator of $C_4$ acting by $x\mapsto x^{-1}$. In Magma, this is G:=SmallGroup(176,35), and it has a huge group of outer automorphisms $C_5\times((C_2\times C_2\times C_2)\rtimes S_4)$, Magma's OuterFPGroup(AutomorphismGroup(G)). The reason for $C_5$ is that $x$ is only conjugate to $x,x^{-1}$ in $C_{11}\triangleleft G$, but there there are 5 pairs of possible generators like that in $C_{11}$, indistinguishable from each other; the other factor of $Out\ G$ is $Aut(C_2\times C_2\times C_4)$, all of these guys commute with the action.

The representations

The group has 28 orthogonal, 20 symplectic and 8 non-self-dual representations, according to Magma.

The conjugacy classes

There are 1+7+8+5+35=56 conjugacy classes, of elements of order 1,2,4,11,22 respectively. The elements of order 4 are (clearly) not conjugate to their inverses, so these 8 classes account for the 8 non-self-dual representations. We are interested in splitting the other 48 classes into two groups, 28 'orthogonal' and 20 'symplectic'.

The catch

The problem is that the way $Out\ G$ acts on the 35 classes of elements of order 22, it has two orbits according to Magma - one with 30 classes and one with 5. (I think I can see that these numbers must be multiples of 5 without Magma's help, but I don't see the full splitting at the moment; I can insert the Magma code if you guys want it.) Anyway, if I am correct, these 30 classes are indistinguishable from one another, so they must all be either 'orthogonal' or 'symplectic'. So a canonical splitting into 28 and 20 cannot exist.


Edit: However, as Jack Schmidt points out (see comment below), it is possible to predict the number of symplectic representations for this group!

$\endgroup$
1
  • 5
    $\begingroup$ G has 40 weakly real classes, 8 strongly real classes. Since it is 2-nilpotent with Abelian Sylow 2-subgroup, it has exactly 8+40/2 = 28 characters of indicator +1. In other words, the weakly real classes only count in pairs. $\endgroup$ Nov 22, 2010 at 19:16
16
$\begingroup$

A standard strengthening of "real element" is "strongly real element". An element is strongly real if it is conjugate to its inverse by an involution, or equivalently, if it is a product of involutions (equivalently, it sits nicely inside a dihedral group).

The quaternion group of order 8 shows that this strengthening is distinct: the only strongly real element is the identity, but Q8 has 4 representations with Frobenius–Schur indicator +1.

However, in:

Gow, R. "Real-valued and 2-rational group characters." J. Algebra 61 (1979), no. 2, 388–413 MR2222410 DOI:10.1016/0021-8693(79)90288-6

some inequalities relating the two ideas are given as well as a few reasonably strong results. If the Sylow 2-subgroups are dihedral or large enough semi-dihedral, then a quaternionic representation exists iff a non-strongly real but real element of odd order exists (theorems are spread out in the paper). In the case of a 2-nilpotent group (the last section) more precise relationships are given, with the number of quaternionic reps being a mixture of the number of strongly and weakly real elements.

Beware of wanting to generalize the real case too broadly. An F-rational character is a C-irreducible character whose values are in F. An F-rational element is an element conjugate to the correct powers given the (cyclotomic) Galois group of C/F. In p-groups for odd p, the F-rational classes are 1–1 with F-rational characters, but not in general for 2-groups or groups of odd order.

$\endgroup$
3
  • $\begingroup$ Why isn't -1 also a strongly real element? $\endgroup$ Mar 11, 2011 at 23:07
  • $\begingroup$ @DavidLHarden: −1 in Q8 is also strongly real $\endgroup$ Mar 12, 2011 at 16:41
  • $\begingroup$ Flip side of my original question: is there a corresponding notion of "strongly real character"? $\endgroup$ Feb 21, 2023 at 19:44
2
$\begingroup$

The recent paper, "A solution to Brauer's Problem 14" by John Murray and Benjamin Sambale, – J Algebra 621 (2023) 87-91, is relevant to this question. It gives a way of counting the number of characters with Frobenius-Schur indicator +1 just by counting solutions of $g_1^2\dotsm g_n^2=1$ with $g_1,\dotsc,g_n \in G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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