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Consider an algebraic manifold whose number of points is $q^n ([n+1]_q)$. Is there a geometric relation to $A^n (P^n)$? In particular, is there an equivalence in the Grothendieck ring of varieties or could there be a birational equivalence?

If there is no such equivalence in general, might some additional reasonable requirements on a manifold force there to be such an equivalence?

Motivation: one can see that some examples of identities on the level of enumerating $F_q$ points can be lifted to geometric relations:

Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?

Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$

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    $\begingroup$ For the projective case, it seems that fake projective spaces might provide counterexamples (but I don't know enough about these to say for sure). I have no idea what to expect when $X$ has $q^n$ points. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 23, 2018 at 21:10
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    $\begingroup$ If you don't require irreducibility, there are easy counterexamples. For example, if $E \subseteq \mathbb P^2$ is an elliptic curve and $E'$ is isogenous to $E$, then $X = (\mathbb P^2 \setminus E) \amalg E'$ has $q^n + \ldots + 1$ points over $\mathbb F_q$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 23, 2018 at 21:54
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    $\begingroup$ Among nonsingular projective varieties, a fake projective plane or odd-dimensional quadrics have the same point count as projective planes. In the case of a fake projective plane its class in the Grothendieck ring is not L^2 + L + 1 (and in fact not congruent to 1 (mod. L) because it's not stably rational). The class of a quadric is same as [P^n] (using projection from the point). $\endgroup$
    – Evgeny Shinder
    Commented May 23, 2018 at 22:07
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    $\begingroup$ Related: mathoverflow.net/questions/92657/…. Especially in the non-projective case, point count itself does not carry enough information to draw strong conclusions. Do you know something about the Hodge theory of your variety? $\endgroup$
    – Balazs
    Commented May 24, 2018 at 0:21
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    $\begingroup$ @EvgenyShinder It seems your comment gives an answer (negative). May be you can be you can write it as an answer ? (Providing some details would be great ... ) $\endgroup$
    – Alexander Chervov
    Commented May 24, 2018 at 7:38

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The Russell Cubic $R:=V(x + x^2 y + z^2 + t^3)\subset \mathbb{A}^4$ is not isomorphic to $\mathbb{A}^3$ although over $\mathbb{C}$ they are both diffeomorphic to $\mathbb{R}^{6}$ (see this Wikipedia page).

I ran a Mathematica program I quickly wrote to compute the number of solutions of $x + x^2 y + z^2 + t^3$ over $\mathbb{F}_p$. For the first twelve primes $p$ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37) I got $p^3$.

Based on this evidence I guessed that although $R$ and $\mathbb{A}^3$ are not isomorphic that they had the same counting polynomial.

In the comments below, Vladimir Dotsenko, provides an elementary proof of my guess:

"Consider the zero set of $x+x^2y+z^2+t^3$. Note that for $x\not=0$ we have the unique value for $y$, so this gives $(p−1)p^2$ points ($p−1$ choice for $x$, $p$ choices for $z$, $p$ choices for $t$). For $x=0$, the polynomial becomes $z^2+t^3$, so there are $p$ choices for $y$ and a choice of a zero $(z,t)$ of that polynomial. However, it clearly has $p$ zeros by the usual parametrization of a singular cubic curve: for $t=0$ there is just $z=0$, and for $t\not=0$, we have $t=−(z/t)^2$ so denoting $z/t=u$, we have $(p−1)$ solutions $(u^3,−u^2)$. "

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    $\begingroup$ I don't think Mathematica is needed. Consider the zero set of $x+x^2y+z^2+t^3$. Note that for $x\ne 0$ we have the unique value for $y$, so this gives $(p-1)p^2$ points ($p-1$ choice for $x$, $p$ choices for $z$, $p$ choices for $t$). For $x=0$, the polynomial becomes $z^2+t^3$, so there are $p$ choices for $y$ and a choice of a zero $(z,t)$ of that polynomial. However, it clealy has $p$ zeros by the usual parametrisation of a singular cubic curve: for $t=0$ there is just $z=0$, and for $t\ne 0$, we have $t=-(z/t)^2$ so denoting $z/t=u$, we have $(p-1)$ solutions $(u^3,-u^2)$. $\endgroup$
    – Vladimir Dotsenko
    Commented May 24, 2018 at 5:46
  • $\begingroup$ @VladimirDotsenko do not you know it is birational to A^3 ? Same class in Grothedieck ring of varieties ? $\endgroup$
    – Alexander Chervov
    Commented May 24, 2018 at 7:44
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    $\begingroup$ Vladimir's comment also applies to show that the class of this affine cubic is L^3 in the Grothendieck ring. $\endgroup$
    – Evgeny Shinder
    Commented May 24, 2018 at 8:11
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    $\begingroup$ Furthermore, the corresponding singular projective cubic 3-fold $XU^2 + X^2Y + Z^2U+T^3 = 0$ has the class $1 + L + L^2 + L^3 = [P^3]$ as the divisor at infinity is the cone over a cusp $X^2 Y + T^3 = 0$. $\endgroup$
    – Evgeny Shinder
    Commented May 24, 2018 at 8:13
  • $\begingroup$ @VladimirDotsenko Thanks for the proof! I will edit my post accordingly. $\endgroup$
    – Sean Lawton
    Commented May 24, 2018 at 11:22

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