# Are exotic affine spaces motivic/whatever equivalent to affine space?

This question is inspired by this MO question; in turn by this MO; in turn by these MO, MO.

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 $\mathbb{A}^n$. Will Sawin showed in his answer MO that the number of $F_q$ points is the same for fake affine spaces as for non-fake (for generic $q$).

Question: Are exotic affine spaces equivalent to affine space in Grothendieck ring of varietes ? Or may be there are some other simple geometric equivalence which is stronger than just point counting and weaker than isomorphism ?

PS

Evgeny Shinder in comments to MO states that " 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). "

• I think the first question to figure out is whether such a variety $V$ is necessarily rational (that is has a Zariski open subset isomorphic to a Zariski open subset of $A^n$). This may be known, but I don't know the answer. If $V$ is rational, and has same cohomology as an affine space, one would think that $[V] = [A^n]$ in the Grothendieck ring, and if $V$ is not rational, most likely the classes won't be equal. In the case of Russel's cubic, $V$ is rational and $V = [A^3]$. – Evgeny Shinder May 26 '18 at 23:19
• Note that fake projective planes cannot give an example here because they are not simply-connected. Also note that my answer works equally well for mixed Hodge structures. – Will Sawin May 27 '18 at 7:08
• Since the map from $V$ to a point induces an isomorphism on Betti cohomology, the induced map $[V] \to [0]$ of Nori motives is an isomorphism. Assuming a long list of conjectures, that would also imply ($A^1$,etale)-equivalence (so equivalent Voevodsky motives). – Jon Pridham May 27 '18 at 8:40
• @JonPridham may be you can extent your comment to an answer ? – Alexander Chervov May 27 '18 at 9:14
• Alexander: just be aware that there are (at least) three meanings of a "motivic equivalence": (1) having same cohomology theories, e.g. point count over finite fields and Hodge structures over C, (2) isomorphism of motives in some category, e.g. that of Voevodsky's motives, (3) equality of classes in the Grothendieck ring. Typical implications between these are (2) => (1), (3) => (1). In the question you are asking (1) is true, (2) is more difficult, but probably still true and (3) is unclear, since it has to do with rationality as I explained above. – Evgeny Shinder May 27 '18 at 22:00

Here is an argument showing that if $V$ is a smooth complex surface with trivial integral homology groups (note that exotic $\mathbf{A}^2$ do not exist, as explained in the comments), then $[V] = \mathbb{L}^2$ in the Grothendieck ring of varieties. We use the notation $\mathbb{L} = [\mathbb{A}^1]$.

It follows from this paper: https://projecteuclid.org/euclid.jmsj/1230128845 (which I learnt about from this MO post Topologically contractible algebraic varieties) that $V$ is a rational surface.

Let $X$ be a non-singular compactification of $V$ such that the divisor $D = X \setminus V$ has simple normal crossings with smooth components $C_1, \dots, C_t$.

Since $X$ is a non-singular projective rational surface, we have $[X] = 1 + k\mathbb{L} + \mathbb{L}^2$. We have $[D] = [C_1] + \dots + [C_t] + r$, where $r \in \mathbb{Z}$ ($r$ depends on the intersection graph of these curves).

Altogether $[V] = 1 + k\mathbb{L} + \mathbb{L}^2 - [C_1] - \dots - [C_t] - r$. Applying Hodge realization to this equality, using the fact that $V$ has a trivial Hodge structure, we deduce that all curves $C_i$ are rational so that $[C_i] = [\mathbb{P}^1] = 1 + \mathbb{L}$. Thus $V$ is a polynomial in $\mathbb{L}$ and the only possibility is $[V] = \mathbb{L}^2$.

P.S. If $V$ an exotic affine space of dimension one, then $V$ is isomorphic to $\mathbb{A}^1$, because it has to be a genus zero curve. If $V$ is an exotic affine space of dimension $3$ or higher, rationality of $V$ is an open question.

• My understanding is that there are no exotic 2 dimensional affine spaces by a theorem of Ramanujam. See arxiv.org/pdf/alg-geom/9506005.pdf – Sean Lawton May 28 '18 at 10:50
• Sean: This would be a shame for my proof! But what is a Ramanujam surface then? – Evgeny Shinder May 28 '18 at 10:53
• It is a contractible affine variety not isomorphic to affine space (in dimension 3 and higher this is sufficient to prove it is exotic). However, to be exotic it needs to be diffeomorphic to R^4. This is not the case since its fundamental group at infinity is nontrivial (so it is not even homeomorphic to R^4). – Sean Lawton May 28 '18 at 11:04
• Okay, thank you, I will edit the proof. – Evgeny Shinder May 29 '18 at 8:01