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$\DeclareMathOperator\Comm{Comm}\DeclareMathOperator\Id{Id}$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.

Alexander Chervov
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  • 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? – LSpice Jan 12 '19 at 21:15
  • 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 $\mathbb Z$? – LSpice Jan 12 '19 at 21:18
  • Such an action would have to not commute with the natural $GL_n$ action, so at the least it wouldn't be very natural. – user44191 Jan 13 '19 at 02:53
  • @LSpice Agree we can talk about Z, not K. Yes, "n-tuples" means $n$ matrices such each pair commute - more conceptually it arises as considering all representations of $Z^n$ in matrices (as in https://mathoverflow.net/questions/272045/ ). About divisibility by (q-1) - I mean it holds true for every "n", since the only point which spoils freeness of action is the point when all matrices are zero-matrices. It is the same for all "n" : pairs, triples, n-tuples – Alexander Chervov Jan 13 '19 at 08:13
  • @user44191 may be you can extend your comment ? I also do not see some simple action, so it comes to my mind that something not very natural might happen, still there is some chance for geometric picture, for example Comm can be fibered over "exotic affine spaces" in some sense... – Alexander Chervov Jan 13 '19 at 08:17
  • I'm not saying anything very insightful; the $GL_n$ action leaves the tuple of $0$s invariant, so the $\mathbb{A}^3$ action would have to take it to other invariants; however, the space of invariants is two-dimensional (only pairs of scalar matrices). For an alternative view, the $\mathbb{A}^3$ action gives a one-dimensional family of commuting pairs of not-both-zero traceless matrices. – user44191 Jan 13 '19 at 09:10
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    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. –  Jan 13 '19 at 11:19
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    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 $2q^{k^2+1}-q^2$. – Richard Stanley Jan 13 '19 at 21:06
  • @richardstanley that seems K^* action is added, but indeed it is an idea to split comm to some pieces and think of separate actions. For scalar matrices as you mention it is easy – Alexander Chervov Jan 14 '19 at 07:17
  • @AlexanderChervov You are right, I shouldn't have called it a $K^3$-action. I am just defining an equivalence relation whose classes are all of size $|K|^3$. – Richard Stanley Jan 17 '19 at 15:07

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