Let $C$ be (a full) exact subcategory of the category of $R$-modules. We suppose that $C$ is essentially small. If the Grothendieck group $K_{0}(C)=0$, what can be said about the higher groups $K_{n}(C)=0$ ? Is there a non-trivial example of such exact subcategory $C$ ?
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1$\begingroup$ could you say what is $R$? probably a ring, but hard to guess if you assume it to be commutative $\endgroup$– YCorCommented Mar 5, 2020 at 10:33
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$\begingroup$ @YCor $R$ is a ring with 1, that is all what we assume $\endgroup$– GSMCommented Mar 5, 2020 at 10:40
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2$\begingroup$ Doesn't it follow from the Freyd-Mitchell embedding theorem that every essentially small exact category is equivalent to a full exact subcategory of a module category? $\endgroup$– Jeremy RickardCommented Mar 5, 2020 at 10:58
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1$\begingroup$ Maybe we use two different definitions of an exact category. But where does that ambient abelian category come from? $\endgroup$– Martin BrandenburgCommented Mar 5, 2020 at 13:23
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4$\begingroup$ @MartinBrandenburg I'm using Quillen's definition en.wikipedia.org/wiki/Exact_category, which I think is what the OP meant. The ambient abelian category is the category of left exact functors from the exact category to abelian groups. $\endgroup$– Jeremy RickardCommented Mar 5, 2020 at 13:33
2 Answers
This is a long comment more than an answer.
If you think of $K_0$ as a universal domain for all kinds of functions that associate a "dimension" to a module, then $K_0(\mathscr{C})=0$ means that at least some of the modules in $\mathscr{C}$ are "the worst kind of infinite dimensional". For example they are not of finite length (otherwise the length would be a nontrivial dimension function) not even over subrings of $R$, they are not finitely generated over any commutative subring (otherwise dimension at some prime would be nontrivial) etc.
Because of the Eilenberg-Swindle the category of all $R$-modules $\mathscr{C}:=R\mathsf{-Mod}$ is always an example with $K_0(\mathscr{C})=0$. The offending modules are infinite direct sums $X\oplus X\oplus X \oplus \cdots$ because those fit into exact sequences $0\to A\to A \to X\to 0$ and $0\to X\to A\to A\to 0$ respectively so that $[X] =0$ in $K_0(R\mathsf{-mod})$. Is this "non-trivial" in the sense of your question? (The "essentially small" requirement can obviously be satisfied by taking any sufficiently large exact subcategory of R-mod instead, say the exact category generated by all finite length modules as well as a swindle module for each isomorphism class)
Conversely every category with $K_0(\mathscr{C})=0$ is similar in the sense that $[X]=0$ implies that $[X]$ is a finite sum of relations $\sum_{i=1}^n \pm([B_i]-[A_i]-[C_i])$ (viewed as an element in the free abelian group generated by isomorphism classes) with exact sequences $0\to A_i\to B_i\to C_i\to 0$ so that by taking direct sums over all positive and all negative signs, we have two sequences $0\to A\to B\to C\to 0$ and $0\to A'\to B'\to C'\to 0$ and $X$ is a direct summand of one of the modules. This is somewhat similar to a "swindle sequence". And in this sense the existence of "swindle sequences" is the common feature of all examples.
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1$\begingroup$ This doesn't answer the actual question about $K_n(C)$ (with $n>0$), also $C$ (which has two meanings in your post; I refer to the exact category) is supposed to be essentially small. $\endgroup$ Commented Mar 5, 2020 at 13:01
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4$\begingroup$ In the last paragraph, when you take direct sums, you're no longer working in the free abelian group generated by isomorphism classes, but in the quotient by relations coming from split short exact sequences. So $X$ may not be isomorphic to one of the six modules $A,B,C,A',B',C'$, but will be a direct summand of one of them. $\endgroup$ Commented Mar 5, 2020 at 13:02
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1$\begingroup$ @MartinBrandenburg Yes, I should have prefaced that with "too long for a comment". Sorry about that. And yes, I was indeed focused on the "is there a non-trivial example" part of the question because I wanted some clarification on what "non-trivial" means here. $\endgroup$ Commented Mar 5, 2020 at 13:03
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2$\begingroup$ @JeremyRickard Yes, you're right. I will correct that. $\endgroup$ Commented Mar 5, 2020 at 13:07
There is almost nothing one can deduce about the higher K-groups from just knowing $K_{0}(C)=0$.
As was pointed out in the comments, the restriction to modules over a general ring does not really restrict anything. Any exact category can (as explained) via the Quillen embedding be realized as an extension-closed subcategory of an abelian category and for the latter use Freyd-Mitchell (exactly as explained in the comments).
For producing nearly arbitrary examples: Let D be any idempotent complete exact category, e.g. any abelian category. Take (for example) the Tate category $\underleftrightarrow{\lim}D$ (an alternative notation is $\operatorname{Tate}(C)$) of Sho Saito's paper "On Previdi's delooping conjecture" (https://arxiv.org/abs/1203.0831). This is an exact category. By the previous remarks, it can be realized as a fully exact subcategory of a category of modules over a ring.
As proved in that paper, the nonconnective K-theory just shifts by one degree upwards (Theorem 1.2 of that paper). Since D was idempotent complete, the nonconnective K-theory will agree with usual Algebraic K-theory.
A direct computation will show that $K_0$ of this Tate category vanishes. So whatever K-groups you can find occurring in any idempotent complete exact category, you can make them appear (with a shift by one) in higher K-groups while simultaneously making $K_0$ vanish.
Such ideas were necessary to define nonconnective K-theory for exact categories in the first place, so you can alternatively find out about such constructs in Schlichting's paper "Delooping the K-theory of exact categories".
One does not have to use these Tate category constructions. There are others, e.g. the so-called Calkin category Ind(D)/D, which was used by Drinfeld in his paper on infinite-dimensional vector bundles (https://arxiv.org/pdf/math/0309155.pdf), see Section 3.3.1. The proofs for all these constructions rely on variants of Eilenberg swindles by the way, i.e. a machine to really make all K-groups vanish. One then re-assembles these contractible K-theory spectra in a way such that they become the loop space of an arbitrary input K-theory spectrum. That's the rough idea. In order to not having to do the "re-assembling" by hand (it's tough to exhibit bicartesian squares from scratch) one tries to make localization sequences "become" the squares one needs.
Finally, if your category $D$ happens to have a projective generator, you can also find such for the Tate category above (https://arxiv.org/pdf/1508.07880.pdf, Theorem 1.(2) for n=1). Thus, you can even realize all these examples simply as the projective modules over a certain ring $R$. This produces counter-examples even avoiding the Frey-Mitchell and Quillen embedding steps and one can reasonably explicitly describe these rings. By work of Wagoner, also various infinite matrix rings have these properties.