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$\DeclareMathOperator\Var{Var}$Let $K_{0}(\Var_{\mathbb{C}})$ be the Grothendieck ring of varieties over $\mathbb{C}$. The class of a variety, $X$, in $K_{0}$ is denoted $[\,X\,]$. If $X$ and $Y$ are varieties then we say that they are piecewise isomorphic if there are finite locally closed stratifications, $\{X_{i}\}$ of $X$, and $\{Y_{i}\}$ of $Y$, so that there exist isomorphisms $X_{i}\rightarrow Y_{i}$.

Note, if $X$ and $Y$ are piecewise isomorphic then we have $[\,X\,]=[\,Y\,]$ in $K_{0}(\Var_{\mathbb{C}})$. I expect the converse should be false, perhaps even generically so (in some sense).

Question. What is a simple example of a pair of varieties with equivalent classes in $K_{0}(\Var_{\mathbb{C}})$ which are not piecewise isomorphic?

Here is a candidate example; let $C$ be the affine cone inside $\mathbb{A}^{3}$ given by the equation $x^{2}+y^{2}+z^{2}=0$, and denote by $q\mathrel{:=}[\,\mathbb{A}^{1}\,]$ the Lefschetz motive. Then we have $[\,C\,]=q^{2}$, since removing the cone point we obtain a (Zariski locally trivial) $\mathbb{C}^{*}$ bundle over $\mathbb{P}^{1}$. I expect that $C$ is not piecewise isomorphic to $\mathbb{A}^{2}$. Another example is given by $\operatorname{SL}(2)$, which has class $q^{3}-q$, and which I presume is not piecewise isomorphic to $\mathbb{A}^{3}$ with a line removed.

Edit. As noted in the comments the example $C$ above is in fact piecewise isomorphic to the affine plane.

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    $\begingroup$ Isn't the $\mathbf C^\times$-bundle trivial over $\mathbf A^1$? Then you can break up both $C$ and $\mathbf A^2$ into a point, a $\mathbf C^\times$, and a $\mathbf C^\times \times \mathbf A^1$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 5, 2021 at 20:12
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    $\begingroup$ "constructibly isomorphic" sounds like a strengthening of isomorphic, rather than a weakening. Why not "piecewise isomorphic"? $\endgroup$
    – YCor
    Commented Jan 6, 2021 at 0:37
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    $\begingroup$ Without loss of generality, no $V_i$ appears as a $V_{ij}$ and no $W_i$ as a $W_{ij}$. Matching up the terms, possibly adding $[X] - [X]$ on the right hand side or $[Y] - [Y]$ on the left hand side, there is an isomorphism $\coprod_i V_i \cong \coprod_i W_i$ and exactly one of the $W_{ij}$ (resp. $V_{ij}$) is isomorphic to $X$ (resp. $Y$). The $W_{ij}$ and $V_{ij}$ give two stratifications of this space with matching pieces, except one piece is $X$ in the former and one piece is $Y$ in the latter. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 6, 2021 at 1:48
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    $\begingroup$ By 'matching pieces' I mean up to isomorphism, not as subschemes of $\coprod_i V_i \cong \coprod_i W_i$. A good first example to keep in mind is $\mathbf P^2$ minus a line and $\mathbf P^2$ minus a circle: the most obvious space to take is $\mathbf P^2$ (but in this case you can also find a piecewise isomorphism). With the K3 example, what has to happen is that $X \amalg Z$ and $Y \amalg Z$ have $Z$ as the big open piece, for otherwise the same birationality argument would apply. I find it kind of hard to picture... $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 6, 2021 at 2:00
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    $\begingroup$ A refinement that is conceptually much more satisfying and that solves a lot of these issues is the graded Grothendieck ring of this paper of Nicaise and Ottem. Basically, when looking at $n$-dimensional varieties, you do not want to allow scissors relations that involve higher-dimensional stuff. (The usefulness to birational questions of keeping track of the dimension was already clear from this paper of Kotschick and Schreieder, and Nicaise–Ottem lift Theorem 2 of Kotschick–Schreieder to the Grothendieck ring.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 6, 2021 at 2:33

2 Answers 2

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There are no simple examples as yet; it's been an open question going back to at least Larsen and Lunts - Motivic measures and stable birational geometry, which has been open for about 15 years, and some of us believed that it should be true.

The first counterexample for smooth non-projective varieties was constructed by Borisov as a consequence on his work on L-zero divisors: Borisov - The class of the affine line is a zero divisor in the Grothendieck ring.

There are currently no counterexamples known for smooth projective varieties. Specifically it is not known if $X$, $Y$ are smooth connected projective varieties over a field of characteristic zero such that $[X] = [Y]$, whether $X$ and $Y$ must be birational; they are stably birational by the work of Larsen and Lunts above.

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After a little digging in the literature, I found the following example:

Theorem. [KS18, Thm. 1.9] There exist non-isomorphic K3 surfaces $X$ and $Y$ over $\mathbf C$ such that $$[X \times \mathbf A^1] = [Y \times \mathbf A^1].$$

Now if $X \times \mathbf A^1$ and $Y \times \mathbf A^1$ were constructibly isomorphic, then in particular they would be birational. But stably birational surfaces are birational (in this case I believe this follows simply by considering minimal models), which in the case of K3 surfaces would imply $X \cong Y$.

References.

[KS18] A. Kuznetsov and E. Shinder, Grothendieck ring of varieties, D- and L-equivalence, and families of quadrics. Selecta Math. (N.S.) 24.4 (2018), p. 3475–3500. ZBL06941785.

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    $\begingroup$ Thank u very much, I accepted the answer above (by one of the authors of the paper u cite!) bc it was two minutes prior to this one, which I would have been happy to accept as well. $\endgroup$
    – user108998
    Commented Jan 6, 2021 at 0:25
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    $\begingroup$ @EBz totally agreed. I also think the other answer is better in terms of proper attribution. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 6, 2021 at 1:33

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