If X is an algebraic scheme, K_0(X) has a filtration by taking the subgroups generated by coherent sheaves whose support as at most dimension k. The associated graded groups are the quotients, and there exists a natural map from the k-th Chow group A_k(X) to the k-th graded part Gr_k K_0(X), just by mapping [V] to [O_V]. This is example 15.1.5 in Fulton's book.

This even becomes an isomorphism after tensoring with Q! That's Corollary 18.3.2 in Fulton's book.

All of the definitions surely make sense for DM-stacks. We've got Chow-groups and K_0 can be graded in the same way. I'm not sure whether the natural map above actually passes to rational equivalence, but I'll just assume this to be true.

Here's the question: Is the map still an isomorphism?


Maybe you might have a look at Toën's paper: "On motives for Deligne-Mumford stacks", IMRN No. 17 (2000), 909-928. He defines Chow groups of a Deligne-Mumford stack X with coefficients in the characters of X, and proves that the corresponding graded ring of Chow groups is isomorphic to K_0 (see the Remark following def. 3.3 of loc. cit.).

  • $\begingroup$ Thanks for the reference, that clarified a lot! From what I understood it does not make much sense to compare K_0 and the naive Chow-groups of a DM-stack. In "Theoremes de Riemann-Roch pour les Champs des Deligne-Mumford" Toen gives an explicit example where the two are not isomorphic. The only possible way to compare the class of a sheaf under the RR-map to something in the naive Chow-ring seems to be to compare the underlying Chow-motives, since these two theories are equiv. for the two different definitions of A_*. $\endgroup$ – Timo Schürg Nov 9 '09 at 9:50
  • $\begingroup$ It might of course still be that the graded parts are isomorphic in special cases since there are the obvious maps taking a sheaf to it's support and an integral closed substack to it's structure sheaf, but even if it did work it probably were cheating and not the right thing to do. $\endgroup$ – Timo Schürg Nov 9 '09 at 9:53

Your Answer

By clicking "Post 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.