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Let $k$ be a field and let $X$ be a smooth irreducible variety over $k$. Suppose that I know that the image of $X$ in the Grothendieck group of varieties over $k$ is equal to that of

a) ${\mathbb A}^n$ for some $n$

b) ${\mathbb A}^{n_1}\times {\mathbb G}_m^{n_2}$ for some $n_1, n_2$.

Does it follow that $X$ is isomorphic to the varieties appearing in a) and b)? If the answer is "yes", then can one drop the smoothness assumption on $X$?

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up vote 19 down vote accepted

The answer is no: Let $X$ the projective plane $\mathbb P^2$ minus a smooth quadric $Q$. Then $[X]=[\mathbb P^2]-[Q]=\mathbb L^2+\mathbb L+1-(\mathbb L+1)=[\mathbb A^2]$ but $X$ is not isomorphic to $\mathbb A^2$ as its Picard group is $\mathbb Z/2$. For b) just cross $X$ with $\mathbb G_m^n$, the Picard group is still $\mathbb Z/2$.

For the singular case the following is an interesting example: Let the symmetric group $\Sigma_n$ act on $(\mathbb A^m)^n$ by permuting the factors. Then it is a fact (proved by Totaro, see Lemma 4.4 of L. Göttsche, On the motive of the Hilbert scheme of points on a surface, Math. Res. Lett. 8 (2001), no.~5-6, 613--627.) that the class of $(\mathbb A^m)^n/\Sigma_n$ in the Grothendieck group of varieties is equal to that of $\mathbb A^{mn}$. However, when $m>1$ then the quotient is always singular and hence not isomorphic to $\mathbb A^{mn}$.

Addendum: In response to the further question by Alexander, I can't think of any strengthening that would make it true. For a moment I thought that if one instead asked that $X$ be smooth and proper and have the same class as $\mathbb P^n$ would imply that $X$ is isomorphic to $\mathbb P^n$. That however is counterexampled by a smooth quadric of odd dimension greater than $1$. We can project from a point to make the blowing up of one point isomorphic to the blowing up of a smooth quadric of $\mathbb P^n$ which gives that the class of $X$ is equal to $\mathbb P^n$. On the other hand $X$ is not isomorphic to $\mathbb P^n$ as its cohomology ring is not generated by its degree $2$ part.

Addendum 1: Some comments related to Alexander's latest question. It is known that the stable birational class of $X$ can be recovered from the class $[X]$. Let us assume that we actually get that it determines the birational class (this may very well always be true and is true in small dimensions).

  • For curves (we shall always assume that $X$ is smooth and irreducible) we get that we can recover the projective model of $X$ but as $[X]$ is equal to the class of the projective model minus a number of points we see that if $[X]$ is not proper we can not recover which points we have removed from the projective model to get $X$. Hence the reconstruction problem is reasonable only when $X$ is also proper which we shall assume.
  • Let us consider a ruled surface $X\to C$ with fibres $\mathbb P^1$'s. If we blow up a point on a fibre and then blow down the strict transform of the fibre containing the point, we get a new ruled surface which in general (ever?) is not isomorphic to $X$ but has the same class.
  • On the other hand in the case of non-ruled surfaces we have a minimal model such that all other surfaces birational to it is constructed by a succession of blowing ups. Hence we can distinguish the minimal model from all other birational surfaces. However, we can not tell which points we have blown up and by blowing up different points we get different surfaces.
  • In higher dimensions things seem worse as there almost never is a unique minimal model. For instance the two small resolutions of an ordinary double point in three dimensions have the same class but are in general non-isomorphic.
  • There are a few cases where a minimal model is unique. The one that comes to mind is when $X$ is an abelian variety in which case $X$ can actually be recovered from its class.

Hence I think that there will only be a very limited number of situations when one can hope to recover $X$ from its class.

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Thank you! Do you know if there is a way to put stronger conditions on $X$, so that the answer will be "yes"? – Alexander Braverman Oct 17 '10 at 14:29
By the definition of $K_0$, the class of $X$ only sees the "piece-wise geometry" of $X$: if $X$ and $Y$ are two varieties such that they can be stratified into locally closed subsets $X_1,...,X_n$ and $Y_1,..., Y_n$ with $X_i$ isomorphic to $Y_i$, then $[X]=[Y]$. The converse is a question of Larsen and Luntz. It is true over $\mathbb C$ if $X$ contains only finitely many rational curves (see Q.L & J. Sebag: {\it The Grothendieck ring of varieties and piecewise isomorphisms}, Math. Z., {\bf 265} (2010), 321-342, Theorem 5). But this is orthogonal to your situation. – Qing Liu Oct 18 '10 at 20:50
Thank you. My question is now kind of informal: what kind of conditions can one put on a variety so that the answer will be positive? The point is that the varieties I have in mind are very specific and I can probably prove many things about them if I only knew what to prove... – Alexander Braverman Oct 29 '10 at 11:51

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