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Consider four $n\times n$ matrices $a,b,c,d$ over finite field $F_q$ (or $F_p$ for simplicity), such that they satisfy three equations: $ac=ca,bd=db, ad-da=cb-bc $. Thus an affine algebraic manifold is defined ("Manin matrix variety").

Question 1: Is it true that number of such matrices over $F_q$ is polynomial in $q$ ? (I.e. the variety is "polynomial count" - which is quite rare phenomena for algebraic manifolds).

Question 1b: If yes - can nice generating function be produced similar to "commuting variety" ?

Question 2: Additionally we may ask if variety is: irreducible, normal and Cohen-Macaulay ? Similar to results/expectations about its "younger brother" - the "commuting variety" - just: $ac=ca$.

Computational experiments:

  • For $2 \times 2$ matrices computations suggest the following polynomial: $ p^7 ( p^4+2p^3-p-1) $, checked for p=2,3,5,7,11,13,17,19.
  • For $3 \times 3$ we only computed number for p=2: 6078976

Variations - similar to known results for commuting variety we can restrict blocks $a,b,c,d$ to be invertible, nilpotent, (anti)-symmetric, triangular, etc. and still expect the polynomial count property:

  • Invertibility. Additional constraint $a,b,c,d$ are invertible - polynom fit for 2x2 and $p=2,3,5,7$ - is: $p(p+1)(p−1)^4(p^5+p^4−3p^3+4p^2−6p+4)$ (verified by fast code of Peter Taylor based on Grobner basis. PS I had some doubts about it is polynomial - but resolved now - thanks to Peter).

  • Nilpotency. With additional constraint 2-niltpotency: $0=a^2=b^2=c^2=d^2$ : 2x2 case checked for (p=2,3,5,7,11,13,17,19) : $2p^5+p^4-2p^3$, 3x3 case checked for p=2,3,5: $4p^{10}+5p^9+p^8-6p^7-6p^6-p^5+2p^4+2p^3$ . For 3x3 case with just nilpotency: $2p^{13}+p^{12}+2p^{10}+3p^9+3p^8+p^6+p^5$

  • (ANTIsymmetric): 2x2 $p^4$ (obviously), 3x3 : $p^7+2p^6-p^4-p^3$ checked for p=2,3,5,7,11 .

  • Symmetric: 2x2 $2p^9+p^8-2p^7$ (checked p=2,3,5,7,11,13,17), 3x3: $4p^{15} + 2p^{14}− 3p^{13}−p^{12}− 2p^{11}+p^{9}$ checked for p=2,3,5.

  • Upper triangular with zero diagonal: 2x2 $p^4$ obviously, $2p^9+p^8−2p^7$ (same as 2x2 block upper-triangular matrices - probably due to standard isomorphism ) .

  • Uppertriangular: 2x2 $2p^9+p^8−2p^7$ (same as above for antisymmetric 3x3). Suspicious: $10p^{15}-4p^{14}+2p^{13}-2p^{12}-19p^{11}-p^{10}-p^9$ - coefficients are big, fit changes from p=2,3 to p=2,3,5, some other bad signs. So there are doubts that it is polynomial count variety.

  • Uppertriangular and squared to zero (a la A.A.Kirillov - see references below) 3x3 case - $2p^8+2p^7−2p^6−2p^5+p^4$ - checked p=2,3,5,7,11,13,17,19. (2x2 case: $p^4$ obviously).

Motivation: Manin matrices are beautiful object which satisfies many interesting properties and have various applications in representation theory, knot theory, etc, so general (exaggerated) expectation - whatever reasonable property we propose it will be true for Manin matrices (see example of Koszulity here (thanks to Steven Sam). In particular - it is a natural generalization of the "commuting variety" - so motivation for the questions - whether the properties known/expected for the "commuting variety" generalize to "Manin matrix variety" ?

Thanks/Related MO/MSE questions/papers:

Generalizations: The question above appeals to $2\times 2$ Manin matrices, it is natural to expect similar polynomial count phenomena for $n\times m$ Manin matrices (but not normality - because normality break for n-tuples of commuting matrices). One can also consider super-and-q-analogues of Manin matrices and cases $M(u)$-polynomial Manin matrices (which produce relations like here). As well as some generalizations to some other quadratics algebras (and their B,C,D versions due to A.Molev).

PS

The questions above arise quite years ago in our paper, but I am not aware of any progress.

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    $\begingroup$ Very interesting question. I tried to think in terms of my techniques and organization in yifeng-huang-math.github.io/files/paper_comm_endo.pdf but I don't think I have an answer yet. It would be somewhat surprising if the invertible count does turn out to be not polynomial while the usual count and the nilpotent count are. There might be a hope for 2x2 and 3x3 matrices since commuting pairs should be in principle classifiable (suggested by the fact that commuting triple count is doable for up to 3x3). $\endgroup$ Commented Apr 2 at 6:02
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    $\begingroup$ I confirm your values for $p=2,3,5$, working on $p=7$. The guessed polynomial doesn't look so bad when you factor it as $p (p + 1) (p - 1)^4 (p^5 + p^4 - 3p^3 + 4p^2 - 6p + 4)$. $\endgroup$ Commented Apr 2 at 11:15
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    $\begingroup$ $p=7$ literally just completed and checks out. $\endgroup$ Commented Apr 2 at 11:44
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    $\begingroup$ Groebner bases. Code $\endgroup$ Commented Apr 2 at 11:53
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    $\begingroup$ Good news: for $p=11$ I get $227199720000$, so the guessed polynomial holds up. $\endgroup$ Commented Apr 10 at 9:07

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