It seems that there is no g.r.r for stack yet according to dejong. Does anyone know anything about it? But as you might know, there are some complex manifold which is not scheme having atiyah singer index theorem. So I was wondering if there exists some analogue of g.r.r for special stack.

  • 1
    $\begingroup$ The Atiyah-Singer index theorem also holds for manifolds with no complex structure. Could you explain why it is relevant? $\endgroup$ – S. Carnahan Nov 17 '09 at 16:00

If you work with the naive Chow-groups and allow non-representable morphisms the GRR-Theorem does not hold! In the paper by Toen quoted above and in some of the papers by Joshua there are explicit counterexamples. They always involve non-representable morphisms.

There are two ways to get around this.

The first is to modify the definition of the Chow-groups. This is what Toen does. He takes Chow groups with coefficients in the characters of the stack, which is quite an involved definition. But it leads to a GRR-theorem for DM-stacks.

The second approach by Joshua is modify the topology to keep track of the stabilizer groups. He introduces a topology which he calls the isovariant etale site, which is motivated by ideas of Thomason. This gives a different kind of Chow groups. For this recall that you can define the Chow groups as cohomology of some sheaves using higher K-theory. For stacks this was done by Gillet. You can then get new kinds of Chow groups by calculating the cohomology of these sheaves in the isovariant etale topology. In a series of papers Joshua proves GRR-theorems in this context.


It seems that Bertrand Toen's thesis gives an answer to this question. It's long but appears to be quite thorough.

Toen also has an earlier paper where he does Grothendieck-Riemann-Roch for Deligne-Mumford stacks. This material seems to have been subsumed in the paper above, but perhaps it is easier to read here.


Roy Joshua also has a paper on this topic: Check out http://www.math.ohio-state.edu/~joshua/pub.html


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.