Why is Riemann-Roch for stacks so hard?

2010-05-19T10:12:35Z

First some indication that it really is a difficult problem: Both Vistoli and Gillet in their classics on intersection theory on stacks remark that their should be a Riemann-Roch theorem for proper representable morphisms, but that they are not able to prove it. I think thats more than enough evidence.

The two existing proofs by Toen and Joshua both involve using not the naive Chow ring, but a modified version. This makes both proofs quite heavy on K-Theory, and I don't really get them.

So what makes the proof using the naive Chow-Ring so difficult? If I remember correctly from reading Fulton-Langs "Riemann-Roch Algebra", the basic technique is to factor a morphism as a regular imedding followed by a projection. The cases of a regular imbeddings and projections are treated by a hands-on methods. Here are some reasons I can think of why this might not work for stacks:

Its hard to find such a factorization.
There's a problem with identifying the Chow-Ring of a stack with the K-Group equipped with the gamma-filtration.
The factorization exists, but the hands-on part is too difficult.
Maybe the K-Group doesn't have a lamda-ring structure?

Answer by Jon Skowera for Why is Riemann-Roch for stacks so hard?

2010-10-05T14:46:00Z

Ben gave much of the answer, but I'll try to make it precise. Toen says there is no Riemann-Roch for the naive (rational) chow ring (Remark 4.3 in Theoremes de R-R) (EDIT: unless you also take the naive K-theory).

He says the problem is the chow rings missing the stack structure (p.1), like Ben says. Like you point out, the morphism to a point won't be representable. If a stack has a representable morphism to a scheme, it must be an algebraic space. (Just pull-back by the identity. Representable means this will be an algebraic space.) There's also a problem with the G-theory. The problems are:

The morphism from a DM-stack to its coarse moduli space $p : F \to M$ induces an isomorphism of rational Chow rings,

$p_{*} : CH(F) \otimes \mathbf{Q} \to CH(M) \otimes \mathbf{Q}$

That's Thm 0.5 from Gillet's intersection theory paper.
The same morphism induces a weak equivalence (now we're working with simplicial objects)

$p_{*} : G(M) \otimes \mathbf{Q} \to H(F_{et}, G \otimes \mathbf{Q})$

of the cohomology of the G-theory sheaf with the G-theory of the coarse moduli space. (Corollary 3.8 of Toen's R-R paper).

Note that G-theory is formed from the K-groups of the category of coherent sheaves, K-theory from the category of vector bundles. The natural morphism $K \to G$ gives a "Poincare" duality. It is an isomorphism in the case of algebraic spaces, but not in general for stacks (Prop 2.2 of Toen's thesis).

The workaround considers the ramification stack (aka. classifying stack of cyclic subgroups) of F, denoted $I_F$, and has been known in the case of complex orbifolds (V-varieties) since Kawasaki wrote on it in 1979. Toen's proof seems to center on proving

$G_{*}(F) \otimes \mathbf{Q}(\mu_\infty) \cong H^{-*}(I_F, G \otimes \mathbf{Q}(\mu_\infty))$.

I read the left side as the K-theory and the right side as the Chow ring.

He eventually reduces to the known case $F = [X / H]$ of equivariant K-theory with a smooth projective variety $X$ quotiented by a finite group $H$:

$\mathbf{K}_{*}(X, H) \otimes \mathbf{C} \cong \bigoplus_{h \in c(H)} \mathbf{K}_{*}(X^h)^{Z(h)}$,

where the sum is over the set $c(H)$ of conjugacy classes of $H$, $X^h$ is the fixed point subscheme, and $Z(h)$ is the centralizer of $h$ in $H$. (That's Vistoli 1991, maybe also Angeniol, Lejeune-Jalabert 1985)

Why is Riemann-Roch for stacks so hard? - MathOverflow

Why is Riemann-Roch for stacks so hard?

Answer by Jon Skowera for Why is Riemann-Roch for stacks so hard?