2
$\begingroup$

Suppose $G$ is a classical matrix group over a finite field of order $q$.

If $C$ is a conjugacy class in $G$ , is $|C|$ a polynomial in $q$?

This question is supported by the fact that whenever I have calculated all conjugacy classes and their sizes for very small groups (for example, $GL_{2}(\mathbb{F}_{q})$, $GL_{3}(\mathbb{F}_{q})$, $SL_{2}(\mathbb{F}_{q})$, $SL_{3}(\mathbb{F}_{q})$) the sizes of conjugacy classes turn out to be polynomial in $q$.

So, does this property hold for all finite classical group, or at least the case of $GL_n$ or $SL_n$?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ I'm not sure how woud you define a conjugacy class for different $q$'s. Even the number of conjugacy classes of $PSL(2,q)$ dependes on the parity of q. $\endgroup$
    – LeechLattice
    Commented Apr 18, 2019 at 1:34
  • 3
    $\begingroup$ Your question doesn't quite make sense, because you aren't talking about one conjugacy class - you are talking about a family of conjugacy classes over a family of groups. You have to be careful how you choose these families for this to make sense. $\endgroup$
    – user44191
    Commented Apr 18, 2019 at 1:35

2 Answers 2

Reset to default
6
$\begingroup$

As already noted in the comments, as worded, this question does not quite make sense. Here is an attempt to say something about it anyway, since the context the OP is asking about is interesting.

In C. Chevalley. Classification des groupes algébriques semi-simples. Springer-Verlag, Berlin, 2005, it is shown that a split reductive algebraic group $G$ is polynomial count. Moreover, using the Bruhat Decomposition the explicit counting-polynomial can be written.

If $G$ is simply connected of rank $r$, then the number of semisimple conjuation classes is $q^r$ (this is a Theorem of Steinberg).

By Matsushima's Theorem if the conjugation orbit is of a semisimple element (closed orbit), then the stabilizer is a reductive subgroup $H$.

So now fix such a closed "stabilizer type" $H$. Then the conjugation orbit is a homogeneous space $G/H$. The counting polynomial for such homogeneous spaces is described in The virtual Poincaré polynomials of homogeneous spaces by M. Brion and E. Peyre.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ By the way, Chevalley's classification goes back to his Paris seminar in 1956-58 and was typeset (after editing) in 2005. $\endgroup$
    – Jim Humphreys
    Commented Apr 21, 2019 at 1:21
  • 1
    $\begingroup$ Also, the published version of the paper by Brion and Peyre is here,along with a review, if you have library access. :mathscinet.ams.org/mathscinet-getitem?mr=1943906 $\endgroup$
    – Jim Humphreys
    Commented Apr 21, 2019 at 1:30
2
$\begingroup$

I agree that the question needs a better formulation. In any case, an approach by Demetris Deriziotis might be useful because it's based on a different kind of analysis: see here (freely available online) and subsequent papers in journals like Communications in Algebra (less freely accessible). But he too aims for a count of all classes, including the recovery of Steinberg's count of semisimple classes. His method may make it easier to see that polynomials in $q$ count various types of classes.

Improve this answer
$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .