Significance of the vanishing of $K_{-1}(A)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T22:40:48Z http://mathoverflow.net/feeds/question/92633 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92633/significance-of-the-vanishing-of-k-1a Significance of the vanishing of $K_{-1}(A)$ Eric A. Bunch 2012-03-30T00:40:32Z 2012-03-30T22:31:22Z <p>In M. Schlichting's <a href="http://homepages.warwick.ac.uk/~masiap/research/frob5.pdf" rel="nofollow">paper</a>, he defines the negative $K$-theory for derived categories. In this he states that for $\mathcal{A}$ an idempotent complete (see below) triangulated category, $K_{-1}(\mathcal{A}) = 0$ is equivalent to the fact that for any full triangle embedding $\mathcal{A} \hookrightarrow \mathcal{B}$ of $\mathcal{A}$ into an idempotent complete triangulated category $\mathcal{B}$, we have that $\mathcal{B}/\mathcal{A}$ is also idempotent complete.</p> <p>An additive category is said to be idempotent complete (sometimes called Karoubian) if for every idempotent $e: A \rightarrow A, \hspace{2pt} e^2 = e$ defines a splitting $A = Im(e)\oplus Ker(e)$. </p> <p>My question is, what is the importance of this quotient category $\mathcal{B}/\mathcal{A}$ being idempotent complete?</p> <p>I would even be happy if the following, more specific question were answered. For a ring R, Bass defined the negative $K$-groups to be the cokernel of the map </p> <p>$K_{-(n-1)}(R[t]) \oplus K_{-(n-1)}(R[t^{-1}]) \rightarrow K_{-(n-1)}(R[t, t^{-1}])$, for $n \geq 0$. </p> <p>If $R$ is regular, by the homotopy invariance property of the $K$-groups, $K_{-1}(R) = 0$. It is proven in Schlichting's paper that the negative $K$-groups for a ring defined this way coincide with the negative $K$-groups of the category $Ch^b(Proj(R))$ of bounded chain complexes of the exact category of projective $R$-modules. The more specific question is what is the significance of the above quotient condition where $\mathcal{A}$ is the category $Ch^b(Proj(R))$?</p> http://mathoverflow.net/questions/92633/significance-of-the-vanishing-of-k-1a/92720#92720 Answer by Ben Wieland for Significance of the vanishing of $K_{-1}(A)$ Ben Wieland 2012-03-30T22:31:22Z 2012-03-30T22:31:22Z <p>There are several questions here. I'm not sure I understand them all. I'm going to try to address the role of idempotent completion. I suspect that if you reread Schlichting, you'll see everything I say was already there, but it may be helpful to have someone else say it.</p> <p>First, what is the point of $K$-theory? As I see it, the main point is $K_0$ and the other groups are mainly there to help compute it. Quillen or Waldhausen tells us that if we have a short exact sequence of nice categories $A \to B\to C$, then we get a fiber sequence of spaces (or connective spectra) and a long exact sequence of higher $K$-groups. In particular, $K_0(B)$ surjects onto $K_0(B/A)$ for the trivial reason the objects of the two categories are the same and so the equivalence relation can only get coarser; $B/A$ is defined by taking the objects of $B$ and modifying the morphisms. But $B/A$ is rarely what you really want. Most natural categories are Karoubian closed and $B/A$ might not be. In the most basic example, projective modules are the Karoubian envelope of free modules.</p> <p>In particular, if you have a localization of rings, you might expect such a short exact sequence of categories. With $B=R[t]$-modules and $C=R[t,t^{-1}]$-modules, the map $B\to C$ is a localization of categories, killing off the Serre subcategory of modules supported on $(t)$. So we get an LES of $K$-groups. But this is only true if we allowed infinitely generated modules, in which case all the groups are zero. If we impose a finiteness condition (bounded chain complexes of finitely generated projectives), it is not true that the localization of $R[t]$-modules is $R[t,t^{-1}]$-modules. If it were, then LES of higher $K$-groups would end with a surjection $K_0(R[t])\to K_0(R[t,t^{-1}])$, but as we see from Bass's definition, that is not surjective. But Thomason showed (with Trobaugh, but see also his other papers) that this is the only obstruction and the Karoubian envelope of the quotient $B/A$ is the category of $R[t,t^{-1}]$-modules.</p> <p>But when we combine positive and negative $K$-theory, we do get a full LES, including $K_0(B)\to K_0(C)\to K_{-1}(A)$. The purpose of $K_{-1}(A)$ is to fit in this sequence, to fix the failure of surjectivity, which stems from the fact that $C$ is not quite the same as $B/A$, but instead its Karoubian envelope. As to the characterization, if $K_{-1}(A)$ vanishes, then the LES shows that $K_0(B)$ surjects on $K_0(C)$, which is to say that $B/A$ is already closed and thus equivalent to $C$. Conversely, if $A$ had the property that $B/A$ were always closed, then no nonzero element of $K_{-1}(A)$ would be hit by the boundary map from an element of $K_0$ of some other category. Given our motivation, these elements would seem pointless, though that doesn't prove that they don't exist. To do that, embed $A$ in a category $B$ subject to a swindle; one is given, roughly, by infinite objects of $A$. This category has contractible $K$-theory, so $K_{n}(B/A)=K_{n-1}(A)$, in particular, $K_0$ (of the Karoubian closure) of $B/A$ is isomorphic to $K_{-1}(A)$. So that gives the characterization of vanishing, modulo the existence of a nice $K_{-1}$. </p> <p>What's so great about categories being Karoubian closed? As I said above, natural categories like modules are. But also, in the last paragraph, I secretly assumed that $A$ was. Swindle categories automatically are, so the swindle category containing $A$ also contained its closure. And it is only possible to take the quotient by thick subcategories, so $A$ had better be thick, hence closed. In particular, this definition of negative $K$-theory only really makes sense for closed categories. Even if we start with a closed category, if we want to iterate this procedure to define negative $K$-theory, we have to take the Karoubian envelope at each stage to continue. This is typical of how they are easier to work with.</p>