$\DeclareMathOperator\Comm{Comm}\DeclareMathOperator\Id{Id}$Consider the variety $\Comm$ of commuting matrices $[A,B]=0$ over some field $K$. It is much studied, and interesting for various reasons.

One has an obvious free action of $K^2$: $A\to A+a\Id$, $B\to B+b\Id$ on it. Hence the number of $F_q$ points is divisible by $q^2$. However actually it is divisible by a higher power of $q$: for $2\times2$ matrices by $q^3$, for $3\times3$ by $q^5$, etc, so:

Question Is there some free action of $K^3$ on $\Comm$, or any other geometric explanation for the divisibility above?

Similar higher divisibility seems to hold true for triples, $n$-tuples of commutting matrices, so we have similar questions.

Remark 1 Number of $F_q$ points of $\Comm$ has been calculated W.Feit, N. Fine, Pairs of commuting matrices over a finite field, 1960. For each matrix size $n$ it is given by polynomial in $q$ with integer coefficicents.

E.g., for $2\times2$ matrices, it is $ q^3(q^3+q^2-q)$.

For $3\times3$, it is $q^5(q^7+q^2(q^2-1)(q^3-1)/(q-1) + (q^2-1)(q^3-1) ) $

For general $n$, one has a sum over partions of $n$, with the summand corresponding to the partition $1^{b_1}2^{b_2}\cdots$ being $[n]_q!/ \prod_i [b_i]_q! q^{some~power}$. The leading term $q^{n^2+n}$ comes from partion $1^n$, and the "least term" comes from partion $n^1$ (see Feit-Fine for details).

Remark 2 In general count equivalence even to $K^n$ does not imply algebraic equivalence as discussed here: MO300946, MO301249. Though in that particular case there might exist some geometric reason.

Remark 3 If my notes are correct, the commuting-triples count for $n=2$ is $q^4(q^4+q^3+q^2-q-1)$, and for quadruples is $q^5(q^5+q^4+q^3-q-1)$. There should be nice generating functions for commuting tuples: MO271752, MO272045.

One may also observe for any $n$, for $n$-tuples of commuting matrices multiplication by $K^*$ acts freely , except of one point - all matrices are zero, and hence number of $F_q$ points $N$ $(N-1)$ is divisible by $(q-1)$ - for all $n$-tuples. There are also some $Z/2Z$ actions comming from $(A,B)->(B,A)$ and similar, which are free on certain easy to describe part of a scheme.

  • $\begingroup$ Does "commuting $n$-tuples" mean each pair of $\binom n 2$ matrices commutes? When you say "$(N - 1)$ is divisible by $(q - 1)$", is $N$ the count of commuting $n$-tuples? $\endgroup$ – LSpice Jan 12 at 21:15
  • $\begingroup$ Also, purely as a language matter, it seems strange to talk of the variety over an unidentified field $K$, unrelated to $\mathbb F_q$, and then to take $\mathbb F_q$-points. Why not talk about $\operatorname{Comm}$ as a variety over $\mathbb Z$? $\endgroup$ – LSpice Jan 12 at 21:18
  • $\begingroup$ Such an action would have to not commute with the natural $GL_n$ action, so at the least it wouldn't be very natural. $\endgroup$ – user44191 Jan 13 at 2:53
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    $\begingroup$ According to the paper "degree 0 motivic Donaldson Thomas invariants" (behrend, bryan, szendröi) the feit-fine paper computes the cut-and-paste motive of the commuting variety as a polynomial in the lefschetz motive L. Perhaps this will help find some geom explanation, even if it's not in the "show it's a torsor for a suitable vector group" direction. $\endgroup$ – EBz Jan 13 at 11:19
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    $\begingroup$ One can eke out another power of $q$ (for $k\times k$ matrices with $k>1$) as follows. The number of commuting pairs $(A,B)$ where $A$ is a scalar matrix is $q^{k^2+1}$. When $A$ is not a scalar there is a free $K^3$ action $A\mapsto cA+bI$ and $B\mapsto B+bI$. Note that we cannot extend this to $B\mapsto dB+bI$ since the number of pairs $(A,B)$ where at least one of $A,B$ is a scalar is $2q^{k^2+1}-q^2$. $\endgroup$ – Richard Stanley Jan 13 at 21:06

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