Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $F$ be a Deligne-Mumford stack that is of finite type, smooth and proper over $\mathrm{Spec~}k$ for a perfect field $k$. Consider $K_m$, the presheaf of $m$-th $K$-groups on $F_{et}$, the etale site of $F$:

$K_m : F_{et} \to Ab$

$(U \to F) \mapsto K_m(U)$

$(f : U \to V) \mapsto (f^{*} : K_m(V) \to K_m(U))$

My question is, what are some simple cases when this is already a sheaf? For example, is it a sheaf when $F = BG$ for a finite group $G$?

Background

My question is aimed at a computation of motives of DM-stacks. The sheaffification $\mathcal{K}_m = K_m^{++}$ is one way to define the Chow groups of $F$:

$A^m(f) := H^m(F_{et}, \mathcal{K}_m \otimes {\bf Q})$

A twist on this definition leads to a well-behaved theory of motives for DM-stacks described by Toen

Etale site

Someone might be able to confirm that the cohomology can be computed using the etale site whose objects are etale morphisms from affine schemes, since Laumon and Moret-Bailly show it's equivalent (by the inclusion) to the larger site which contains all etale morphisms from algebraic spaces (Champs algebriques, p.102). This might simplify working with the $K$-groups.

share|improve this question
2  
In some sense, $K$-theory is a global invariant exactly because it's not a sheaf. Consider $K_0$ as a warm-up. If it were a sheaf, it would be zero much too often. –  Minhyong Kim Dec 12 '10 at 2:56
1  
Right, but there are some descent properties for presheaves of K-theory spectra aren't there ? –  Zoran Skoda Dec 25 '10 at 20:34
1  
Zoran: Yes, Zariski (or better, Nisnevich) but not etale in general. –  Dustin Clausen Dec 25 '10 at 21:17

1 Answer 1

In general, these presheaves are not sheaves, even on the etale sites of fields. As an easy example, $K_2(\mathbb{C})$ is non-torsion divisible, but $K_2(\mathbb{R})$ has a $2$-torsion element given in symbols by $(-1,-1)$ in Milnor K-theory. But, $K_2(\mathbb{R})$, if $K_2$ were a sheaf, would be the $\mathbb{Z}/2$-fixed points of $K_2(\mathbb{C})$. This cannot happen in this example.

Using the fact that $K_{2i}$ of an algebraically closed field is a non-torsion uniquely divisible group, I imagine one can construct counter-examples for any even K-group.

I would imagine that odd K-groups are also not sheaves.

However, for finite fields, the situation might be different, by Quillen's computation. There, it looks as if the K-groups might be sheaves.

For details on $K_2$ and Milnor $K$-theory, look up Matsumoto's Theorem. For other K-groups of algebraically closed fields, see Suslin's paper On the K-theory of algebraically closed fields.

In general, the place to start thinking about the etale site and algebraic K-theory would be Thomason's paper Algebraic K-theory and etale cohomology.

share|improve this answer
    
Thanks for resolving the question in the case of Milnor's K-theory. Unfortunately, my question was about the K-theory defined for schemes by the $Q$ construction. I think it agrees for $K_0$ and $K_1$, but not for the higher K-groups. –  Jon Skowera Jan 25 '11 at 16:31
1  
For a field, it agrees for K_0, K_1, and K_2. –  Benjamin Antieau Jan 31 '11 at 5:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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