# Commuting matrix variety $[A,B]=0$ - can one geometrically explain divisibility of $F_ q$ point count by high powers of $q$?

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

• 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 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 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 at 2:53
• 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. – EBz Jan 13 at 11:19
• 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 at 21:06