My question is somewhat related to this one. However I think it adds something new to the table so I decided to post it sperately.

There is a construction of K-theory for symmetric monoidal categories (smc) and one for exact categories. From what I understand they are (a priori) two different pair of shoes.

For a symmetric monoidal category $S$ there is the classifying space $BS$. We define the K-theory $K(S)$ to be the group completion of this space. If we consider $BS$ to be a one-object topological category it is up to an equivalence of categories a strict monoidal category and a group completion is given by $BS\to\Omega BBS$.

If we now take our favorite category $P(R)$ of projective modules over a ring then the above construction isn't very interesting ($BS$ is contractible since $0\in S$ is an initial object). But we can take $iP(R)$ the isomorphism category and then $K(iP(R))$ magically coincides with the K-theory of $P(R)$ as an exact category (I write magically since I can read the proof but still don't really know why taking the isomorphism category is the sensible thing to do).

Something I like about the monoidal approach is Thomasons mapping cone construction. Given a functor of smc $A\to B$, such that $A=iA$ one can construct a smc $C$ such that there is a long exact sequence:

$$\cdots \to K_{i+1}(C)\to K_{i}(A)\to K_{i}(B)\to K_{i}(C)\to\cdots$$

Now I am in the situation that $A$ and $B$ are in fact exact categories. $A=P(R)$ and $B=P(R,\mathbb G_m)$ the category of projective modules equipped with an automorphism. So to apply Thomason's construction I may just take $A=iP(R)$, fair enough. May I also take $B=iP(R,\mathbb G_m)$ and get the K-theory of $P(R,\mathbb G_m)$ as an exact category (I have the feeling that we pass to direct sum K-theory here, or something like that)? If I now construct $C$ then I notice that $iC=iP(R,\mathbb G_m)$, so I definitely may not just take the isomorphism category here. Somehow I got the feeling that I am just jumping between the definitions and never really now whether it's justified.

So the question: For an exact category $D$, how are $K(D)$ (K-theory of $D$ as an exact category) $K(D)$ (as a symmetric monoidal category) and $K(iD)$ (again monoidal) related?


1 Answer 1


As you note, the classifying space of any exact category is contractible because of the presence of an initial object, so your second K(D) (which I take to be the group completion of the classifying space of D under direct sum) is trivial. But the other two (namely K(D), meaning the Q-construction K-theory of D, and (iD)^{gp}, the group completion of iD under direct sum) are closely related. Indeed, there is always a map (iD)^{gp} --> K(D), and this map is an equivalence provided that every short exact sequence in D splits.

This is a non-trivial result. As far as I'm aware, it's due to Quillen, but the first proof of it was given in Grayson's paper Higher Algebraic K-theory: II [after Daniel Quillen]. Actually the result is not explicitly stated in that paper, but it follows immediately from the first two theorems on its page 11 (well, I suppose one also needs to compare the S^{-1}S construction of that paper with the usual group completion -- e.g. using the group completion theorem -- or else redo the arguments with the usual group completion). The proof is not that long, but it's sort of tricky, the trick being the theorem at the top of page 10.

Hope this helps!

  • $\begingroup$ Maybe I should add -- the motivation for taking the group completion of iD is not, like, "well, the group completion of D is trivial, so lets try something else". Actually that's sort of backwards. The operation of group completing iD is exactly the homotopical analog of the usual Grothendieck approach to direct-sum K_0, so it doesn't need any further justification as a model for higher K-theory. On the other hand, the moral reason why the Q-construction models higher K-theory is more delicate -- I'd recommend reading the nice answers at [cont'd] $\endgroup$ Commented Jul 18, 2012 at 20:47
  • $\begingroup$ mathoverflow.net/questions/1006/… $\endgroup$ Commented Jul 18, 2012 at 20:47
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    $\begingroup$ To clarify: the comparison theorem that Grayson states (bottom of p. 11) says that the $Q$-construction agrees with the plus construction, so at first glance it appears that his comparison between $S^{-1} S$ and $QP$ proceeds indirectly through the plus construction. But it's actually the other way around: Grayson shows that $S^{-1} S = \Omega QP$ directly via the homotopy pullback square at the top of p.11, and separately shows that $S^{-1} S$ agrees with the plus construction via a homology computation. The theorem at the bottom of p. 11 is the concatenation of these results. $\endgroup$
    – Tim Campion
    Commented Mar 14, 2018 at 19:44

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