Let $K_0(V_k)$ be the Grothendieck ring of $k$-varieties, and consider the scheme $X = \mathrm{Spec}(k[x]/(x^2))$.

I understand that this scheme has one point, but I am missing the fact that in $K_0(V_k)$, the class of $X$ is $[X] = 1$ (which is something I read at this site in the context of Grothendieck classes of quadrics).

Why is that ? Why is $X$ in the same class as $\mathrm{Spec}(k[x]/(x))$ ?

  • 4
    $\begingroup$ $[X]=[X_{\mathrm{red}}]$ for any $k$-scheme $X$, just because the Grothendieck ring is defined by saying that for any closed subscheme $Z$ of $X$ we have $[Z]+[X-Z]=[X]$, including for $Z=X_{\mathrm{red}}$. $\endgroup$ – Devlin Mallory Mar 10 at 18:09
  • 2
    $\begingroup$ Sure: $\mathrm{Spec}\, k[x]/x$ is a closed subscheme of $\mathrm{Spec}\, k[x]/x^2$ (corresponding to the surjection $k[x]/x^2\to k[x]/x$, and thus definitionally $[\mathrm{Spec}\, k[x]/x]=[\mathrm{Spec}\, k[x]/x^2]$ in the Grothendieck ring. Then, since for any $k$-scheme $X$ we have $X\times_k k \cong X$ we have that $[\mathrm{Spec}\,k]$ is the identity in the Grothendieck ring. $\endgroup$ – Devlin Mallory Mar 10 at 18:43
  • 1
    $\begingroup$ @DevlinMallory: And by "definitionally," you mean that the complement of $\mathrm{Spec}(k[x]/(x^2))$ in $\mathrm{Spec}(k[x]/(x))$ is empty ? $\endgroup$ – THC Mar 10 at 19:01
  • 3
    $\begingroup$ Yes, that's exactly it (and more generally the same is true for the reduced subscheme $X_{\mathrm{red}}$ of any scheme $X$: the complement is empty, so the classes of $X$ and $X_{\mathrm{red}}$ agree in the Grothendieck ring). $\endgroup$ – Devlin Mallory Mar 10 at 19:11
  • 1
    $\begingroup$ Maybe you want to ask whether there is some "enriched" Grothendieck ring mapping onto $K_0(V_k)$, which takes into account non-reducedness (so that the class of "$X-X_{\mathrm{red}}$", in some sense, would be meaningful and possibly nonzero...) I don't know how to formalize this properly. $\endgroup$ – YCor Mar 11 at 11:44

An answer has already been given in the comments: the definition of $𝐾_0(𝑉_π‘˜)$ includes the relations $[𝑋]=[𝑋_{\rm red}]$ so fat points are equal to points by definition. The intuition is that $𝐾_0(𝑉_π‘˜)$ was primarily conceived to study phenomena involving classical topological invariants (Euler characteristic of the underlying topological space, Hodge structure, etc), or point counts over finite fields, which are all insensitive to the scheme structure anyway.

There is however a theory which encodes richer data, though it (at least a priori) depends on more input than just the scheme structure of $X$, and is only defined for certain kinds of schemes. This is the notion of a motivic vanishing cycle attached to a pair $(M,f)$ where $f\colon M\to f$ is a regular function on a smooth variety $M$. The intuition is that this should be thought of as a motivic invariant attached to the degeneracy locus $X=\{df=0\}$, thought of as a subscheme of $M$, which is sensitive to the scheme structure of $X$. In your example, we would consider the pair $(M,f)=({\mathbb A}^1, x^3)$ which indeed has the double point $X$ as degeneracy locus. This motivic vanishing cycle takes values in a ring which is larger than $𝐾_0(𝑉_π‘˜)$, remembering also monodromy data about the function; in the example at hand, you would need to worry about order-three monodromy. But there is still an Euler characteristic map to the integers, which returns the answer $2$ for this example, indicating that indeed we are talking about a double point.

The motivic vanishing cycle was defined by Denef-Loeser, and the interpretation I gave above arose in works of Joyce and Kontsevich-Soibelman in connection with motivic Donaldson-Thomas theory. A starting reference might be On motivic vanishing cycles of critical loci, arXiv:1305.6428, by Bussi-Joyce-Meinhardt.

| cite | improve this answer | |
  • $\begingroup$ This is a neat idea. Is there any sense in which this is independent of the way in which you describe $X$ as a critical locus? Here is an example that is analogous to what I'm looking for. If $X$ is a critical locus, it carries a symmetric perfect obstruction theory, which is enough to define $[X]_{Virt} \in A_0(X)$ independent of the choice of action functional. Does something similar hold for the class in $K(Var)$? $\endgroup$ – Phil Tosteson Mar 11 at 17:12
  • $\begingroup$ Actually, I see that the paper you mention contains at least one result along these lines (Theorem 3.2, if anyone else is interested) $\endgroup$ – Phil Tosteson Mar 11 at 18:54

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.