Why is Riemann-Roch for stacks so hard? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:14:48Z http://mathoverflow.net/feeds/question/25218 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25218/why-is-riemann-roch-for-stacks-so-hard Why is Riemann-Roch for stacks so hard? Timo Schürg 2010-05-19T10:12:35Z 2010-10-22T07:52:23Z <p>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. </p> <p>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. </p> <p>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:</p> <ol> <li>Its hard to find such a factorization. <li>There's a problem with identifying the Chow-Ring of a stack with the K-Group equipped with the gamma-filtration. <li>The factorization exists, but the hands-on part is too difficult. <li>Maybe the K-Group doesn't have a lamda-ring structure? </ol> http://mathoverflow.net/questions/25218/why-is-riemann-roch-for-stacks-so-hard/41161#41161 Answer by Jon Skowera for Why is Riemann-Roch for stacks so hard? Jon Skowera 2010-10-05T14:46:00Z 2010-10-22T07:52:23Z <p>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).</p> <p>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:</p> <ul> <li><p>The morphism from a DM-stack to its coarse moduli space $p : F \to M$ induces an isomorphism of rational Chow rings,</p> <p>$p_{*} : CH(F) \otimes \mathbf{Q} \to CH(M) \otimes \mathbf{Q}$</p> <p>That's Thm 0.5 from Gillet's intersection theory paper.</p></li> <li><p>The same morphism induces a weak equivalence (now we're working with simplicial objects)</p> <p>$p_{*} : G(M) \otimes \mathbf{Q} \to H(F_{et}, G \otimes \mathbf{Q})$</p> <p>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). </p> <p>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).</p></li> </ul> <p>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</p> <p>$G_{*}(F) \otimes \mathbf{Q}(\mu_\infty) \cong H^{-*}(I_F, G \otimes \mathbf{Q}(\mu_\infty))$.</p> <p>I read the left side as the K-theory and the right side as the Chow ring.</p> <p>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 $H$:</p> <p>$\mathbf{K}_{*}(X, H) \otimes \mathbf{C} \cong \bigoplus_{h \in c(H)} \mathbf{K}_{*}(X^h)^{Z(h)}$,</p> <p>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)</p>