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$\newcommand{\GFq}{\mathbf F_q}$ Let $r_n(q)$ denote the number of isomorphism classes of $n$-dimensional modules of the $\GFq$-algebra $\GFq[x,y]$. Is it known whether there exists a polynomial $p_n(t)$ (with say, integer coefficients) such that $r_n(q) = p_n(q)$ for all prime powers $q$?

Here are the values that I know (see arXiv:1212.6157v2):

  • $r_1(q) = q^2$
  • $r_2(q) = q^4 + q^3 + q^2$
  • $r_3(q) = q^6 + q^5 + 2q^4 + q^3 + 2q^2$
  • $r_4(q) = q^8 + q^7 + 3q^6 + 3q^5 + 5q^4 + 3q^3 + 3q^2$

Note: for the free algebra $\GFq\langle x, y\rangle$ this is well known, since it is the number of $n$-dimensional representations of the quiver with one vertex and two loops. In fact, it is also known now that this polynomial has non-negative integer coefficients (this follows from the work of Mozgovoy or Hausel, Letellier and Rodriguez-Villegas).

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The number you seek is closely related to the number of commuting pairs of $n\times n$ matrices over $\mathbf{F}_q$. This is a polynomial and the formula was first given by Feit and Fine in a 1960 Duke paper. The stack of $\mathbf{F}_q[x,y]$ modules is the stack quotient of the commuting variety

$$C(n) = \{(A,B)\in End(\mathbf{F}_q^n)^2 : [A,B]=0\}$$

by $Gl(n)$ acting by conjugation. The number you have asked for is the number of points in the coarse space of this stack. I don't know much about that number, but the "stacky" number of points is much better behaved --- in other words, instead of counting modules, you should be counting modules by 1 over the number of automorphisms of the module. That number is equal to the number of commuting pairs of matrices over $\mathbf{F}_q$ divided by the number of elements in $Gl(n,\mathbf{F}_q)$. The generating function of that number has a nice product expansion:

$$\sum_{n=0}^{\infty} \frac{|C(n)|}{|Gl(n)|} t^n = \prod_{k=1}^\infty \prod_{m=1}^\infty (1-q^{2-m}t^k)^{-1}$$

This is in Feit and Fine's paper or you can find a more modern motivic derivation of this in my paper with Morrison. http://arxiv.org/abs/1206.5864

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  • $\begingroup$ If the stack is Deligne-Mumford, the number of $\mathbf F_q$-points is the same for the stack and the coarse space. Is this true here? $\endgroup$
    – Dan Petersen
    Commented Sep 12, 2013 at 6:38
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    $\begingroup$ @Amritanshu : Just to clarify, the number you are asking for is $|C(n)/Gl(n)|$, which is the number of isomorphism classes of modules, which is the cardinality of the coarse moduli space of the stack $[C(n)/Gl(n)]$. The number that is given by the above formula is $|C(n)|/|Gl(n)|$, which is the number of isomorphism classes of modules counted by the reciprocal number of automorphisms, which is (in some sense) the cardinality of the stack $[C(n)/Gl(n)]$. The stack is often better behaved than its coarse space so maybe your original question doesn't have a good answer, but I don't know. $\endgroup$
    – Jim Bryan
    Commented Sep 12, 2013 at 15:02
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    $\begingroup$ You might want to edit your question so that the tags "commuting variety" and/or "algebraic geometry" appear. The question is really about the class of the coarse space $C(n)/Gl(n)$ in the Grothendieck group of varieties --- there is a lemma of Katz which says that a class in the Grothendieck group will be a polynomial in $[\mathbb{A}^1]$ iff the point count over $\mathbf{F}_q$ is polynomial in $q$. So you should look for an affine paving of $C(n)/Gl(n)$, i.e. a stratification where the strata are all affine spaces. You could find this over $\mathbb{C}$ and that would suffice. $\endgroup$
    – Jim Bryan
    Commented Sep 13, 2013 at 14:48
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    $\begingroup$ It might well be easier (and good enough) to pave by tori instead of affine spaces. But anyway, this begins to sound too much like the wild problem of classifying representations of $k[x,y]$. I doubt you're going to get results for general $n$. $\endgroup$
    – Allen Knutson
    Commented Sep 13, 2013 at 20:11
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    $\begingroup$ Is it just a coincidence that $r_n(1)$ is the middle coefficient of $(1+x+x^2)^n$ for $1\leq n\leq 4$? $\endgroup$
    – Richard Stanley
    Commented Jul 14, 2014 at 23:57

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