# K theory long exact sequence

(1) Suppose that $Z\subset X$ is a closed embedding, $U = X\setminus Z$ is the complement. If relevant, suppose that both $X, Z$ are smooth and even (if relevant) that the normal bundle of $Z\subset X$ is trivial. Then is it true that there is an exact triangle of complexes $K^* (Z)\to K^* (X)\to K^* (U) \to$? Here by $K$ theory I mean (connective) K theory of the category of perfect complexes. For what I need (this came up in the context of trace formulae), rational coefficients are enough. I found many places where very similar results are written down, but not this one explicitly.

(2) Suppose that, further, in the example above we have a (suitably flat) sheaf of smooth, compact (DG) algebras, $\mathcal{A}$, over $X$. Then we can study $K$ theory of categories of bundles of $\mathcal{A}$-modules over the three spaces. Again, is it true that these fall into an exact triangle as above?

It seems like both of these statements should follow from some colimit-compatibility of the $K$ theory functor, but I can't find a reference.

• Clark Barwick's answer to this question: mathoverflow.net/questions/5580/… might be of help, at least for the first question (your hypothesis seem to satisfy the conditions for TT 2.1.1, and hence the identification of the fiber sequence of spectra/exact triangle of complexes). But I don't have my copy of the ghost paper with me right now. Nov 25, 2015 at 6:23

Regarding the first question:

If $X$ is quasi-compact quasi-separated and $U \to X$ is a quasi-compact open immersion, then Thomason-Trobaugh showed that there is a "proto-localization sequence", i.e. a fibre sequence of spaces $$K_{Z}(X) \to K(X) \to K(U)$$ where $K_{Z}(X)$ is the K-theory of perfect complexes on $X$ that vanish on $U$. This follows from the fact that you have a corresponding exact sequence at the level of stable $(\infty,1)$-categories of perfect complexes, and that these categories are compactly generated, so that the sufficient conditions provided by Thomason's localization theorem, for the induced sequence on K-theory to be a fibre sequence, are verified.

On the other hand, for G-theory (= K-theory of coherent sheaves), one has an "honest" localization sequence. That is, for $X$ is noetherian, there is a fibre sequence $$G(Z) \to G(X) \to G(U).$$ This follows from Quillen's devissage theorem, which implies that $G(Z) \simeq G_{Z}(X)$, where $G_{Z}(X)$ is the K-theory of coherent sheaves on $X$ which vanish on $U$.

When $X$ is regular, K-theory is identified with G-theory (because any complex of quasi-coherent sheaves with coherent cohomology is quasi-isomorphic to a perfect complex).

If by "smooth" you meant regular, e.g. smooth over a field, then putting these together you get the fibre sequence $$K(Z) \to K(X) \to K(U)$$ desired.

For the second question, I am not sure. By the same type of argument as above you should get a proto-localization sequence, but I don't know if you can say more.

• Thanks! Is there a standard reference for colimit-compatibility of K theory? Nov 26, 2015 at 21:16
• @DmitryVaintrob, statements to this effect are in Blumberg-Gepner-Tabuada. They show that K-theory preserves filtered colimits and split exact sequences.
– AAK
Nov 26, 2015 at 23:13

You can combine Adeel Khan's answer with Proposition 6.9 of my paper with David Gepner to prove that there are these kinds of localization sequences in a great deal of generality. (Note that our proposition is simply an analogue for stable $\infty$-categories of a dg-categorical result of Toën.) So, if $A$ is a sheaf of (quasi-coherent) dg algebras on $X$, then this proposition shows that there is a fiber sequence $$K_Z(X,A)\rightarrow K(X,A)\rightarrow K(U,A).$$ Identifying the fiber term as $K(Z,i^*A)$, where $i:Z\rightarrow X$ is the inclusion and $i^*A$ is the derived pullback, is not something I've thought about. This kind of dévissage statement is much more difficult for dg algebras than it is for ordinary algebras, and it fails in some cases. There's a discussion of this in my paper with David and Tobias Barthel.