# Are all exotic affine spaces count equivalent to affine space?

This question is inspired by this MO question; indeed it is a special case on which to focus.

An exotic affine space is an affine variety $V$ whose $\mathbb{C}$-points are diffeomorphic to $\mathbb{R}^{2n}$ yet $V$ is not algebraically isomorphic to $\mathbb{A}^n$.

Say that two varieties are count equivalent if they are both polynomial count varieties with the same counting polynomial.

As shown in the comments here, the Russell Cubic is count equivalent to $\mathbb{A}^3$ although it is not isomorphic to $\mathbb{A}^3$.

Question: Are all exotic affine spaces count equivalent to affine space?

• Don't you have to choose a model for this question to make sense? I'm pretty sure there are models of $\mathbb A^n$ whose point counts over finite fields are not always $q^n$... – R. van Dobben de Bruyn May 24 '18 at 12:17
• @R.vanDobbendeBruyn I thought the context made it pretty clear what I meant by "affine space". But I guess if you are confused, others might be too, so let me spell it out: I mean the usual one. – Sean Lawton May 24 '18 at 12:47
• Maybe R. van Dobben de Bruyn meant something in the sense of mathoverflow.net/questions/18747/… So the "model" of $\mathbb{A}^n_{\mathbb{C}}$ you're using to define what are its $\mathbb{F}_q$-points is the usual scheme $\mathbb{A}^n_{\mathbb{Z}}$, which is not a variety but okay... (Btw, I have no idea if there are other models of $\mathbb{A}^n_{\mathbb{C}}$ over $\mathbb{Z}$ that have a different number of $\mathbb{F}_q$-points) – Qfwfq May 24 '18 at 12:58
• @SeanLawton The issue is not with the model of affine space but the model of your affine variety. It all depends on the definition of "polynomial count", which you don't specify. The definition that is correct for most purposes is that the polynomial holds for all but finitely many primes. This is a perfectly well-defined concept for varieties over $\mathbb Q$. On the other hand, if you define "polynomial count" at every prime, then the answer is false, already in the case when $V_\mathbb C$ is algebraically isomorphic to $\mathbb A^n_{\mathbb C}$. – Will Sawin May 24 '18 at 13:11
• @Qfwfq It's easy to see that there are. For instance once can take $\operatorname{Spec} \mathbb Z[x_1,\dots,x_n, 1/2]$, which has no $\mathbb F_2$-points, or, more interestingly, $\mathbb Z[x_1,x_2, p x_1 - x_2 (x_2-1) ]$ which has $2p^2$ $\mathbb F_p$-points. – Will Sawin May 24 '18 at 13:13

For all but finitely many primes, yes. Any such $V$ has $V_{\mathbb C}$ smooth and has $H^i(V_{\mathbb C}, \mathbb Q_\ell)=0$ for $i\neq 0$ and $=\mathbb Q_\ell$ for $i=0$. Both these properties are known to be constructible, so they hold for $V_{\mathbb F_q}$ for all but finitely many $q$.
For any such $q$, the action of $\operatorname{Frob}_q$ on $H^0(V_{\overline{\mathbb F}_q}, \mathbb Q_\ell)$ must be trivial as it is one-dimensional and contains at least one invariant class. So its cohomology, with Frobenius action, matches affine space. Applying Poincare duality (using smoothness), the same is true for its cohomology with compact supports. Applying the Grothendieck-Lefschetz formula, its number of points matches affine space.