# Piecewise isomorphism versus equivalence in Grothendieck ring

$$\DeclareMathOperator\Var{Var}$$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.

• 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 $\mathbf C^\times \times \mathbf A^1$. – R. van Dobben de Bruyn Jan 5 at 20:12
• "constructibly isomorphic" sounds like a strengthening of isomorphic, rather than a weakening. Why not "piecewise isomorphic"? – YCor Jan 6 at 0:37
• 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. – R. van Dobben de Bruyn Jan 6 at 1:48
• 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... – R. van Dobben de Bruyn Jan 6 at 2:00
• 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.) – R. van Dobben de Bruyn Jan 6 at 2:33

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.

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.

• 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. – EBz Jan 6 at 0:25
• @EBz totally agreed. I also think the other answer is better in terms of proper attribution. – R. van Dobben de Bruyn Jan 6 at 1:33