3
$\begingroup$

Let $k$ be a field, and let $K_0(V_k)$ be the Grothendieck ring of $k$-varieties. One type of relation which defines $K_0(V_k)$ is the following: if $A$ is a $k$-variety and $C$ a closed subset of $A$, then $$ [A] = [A \setminus C] + [C]. $$ (Here, $[B]$ denotes the class of $B$ in $K_0(V_k)$.)

It is well known that $K_0(V_k)$ is a toy model for understanding certain (nonabelian) aspects of motives.

My question: why relations of type $ [A] = [A \setminus C] + [C]$ ? What is the connection with the theory of motives ?

Improve this question
$\endgroup$
5
  • 8
    $\begingroup$ Anything that behaves like an Euler characteristic should statisfy the scissor relation, so it factors through $K_0$. In particular, assuming a good category of mixed motives, the motivic Euler characteristic would be an example. $\endgroup$
    – Donu Arapura
    Commented May 18, 2020 at 15:55
  • 3
    $\begingroup$ Quick google search gives (it's stated here) that in characteristic 0 there is a homomorphism from the Grothendieck ring to the $K_0$ of the category of pure motives. This means that these classes in the $K_0$ of motives satisfy the corresponding scissor relations. $\endgroup$
    – Wojowu
    Commented May 18, 2020 at 15:59
  • 2
    $\begingroup$ On a basic level, this relation obviously holds when you consider varieties over finite fields, and count the number of points over the field. This already hints at a connection to motives via the Weil conjectures. For more on this point of view, you might want to look at Thomas Hales' wonderful Bull. AMS article "What is motivic measure" ams.org/journals/bull/2005-42-02/S0273-0979-05-01053-0/… $\endgroup$
    – Balazs
    Commented May 18, 2020 at 19:26
  • 1
    $\begingroup$ @DonuArapura Is that true? I would expect a Tate shift to appear in the closed piece $\endgroup$
    – Denis Nardin
    Commented May 18, 2020 at 20:59
  • 1
    $\begingroup$ @DenisNardin I probably wasn't very precise. I meant to use compactly supported cohomology and the localization sequence $\ldots h_c^i(U)\to h_c^i(X)\to h_c^i(C)\ldots$ But I agree there would be twist for the other one. $\endgroup$
    – Donu Arapura
    Commented May 18, 2020 at 21:13

0

Reset to default

You must log in to answer this question.