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Let $\mathcal{P}$ be a (small) exact category. Without delving into any homotopy theory, we can provide characterisations of $K_0(\mathcal{P})$ and $K_1(\mathcal{P})$ as plain categorical constructions:

$K_0(\mathcal{P})$ is the free abelian group on the objects of $\mathcal{P}$ under the relations $[P] = [P'] + [P'']$ if $0 \to P' \to P \to P'' \to 0$

$K_1(\mathcal{P})$ is the free abelian group on the pairs $(P, \alpha)$, where $P$ is an object and $\alpha$ one of its automotphisms, under suitable relations.

Does $K_2(\mathcal{P})$ admit a similar description?

My question has been answered (and then some!). It has been pointed out however that I was slightly mistaken: the $K_1(\mathcal{P})$ given above only coincides with the standard one ($\pi_2(Q \mathcal{P})$) when short exact sequence of $\mathcal{P}$ splits. So I actually refer to the `Bass $K_1$'.

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    $\begingroup$ I'm not expecting the answer to be a pretty one. $\endgroup$
    – Joshua Seaton
    Mar 21, 2013 at 3:55
  • $\begingroup$ can I ask what the relations are for $K_1$? $\endgroup$
    – Jacob Bell
    Mar 21, 2013 at 4:00
  • $\begingroup$ Sure: $[(P,\alpha \beta)] = [(P,\alpha)] + [(P, \beta)]$, and $[(P,\alpha)] = [(P,\alpha_1)] + [(P,\alpha_2)]$, where $0 \to P' \to P \to P'' \to 0$, $\alpha_1$ and $\alpha_2$ are automorphisms of $P'$ and $P''$ such that the translation from one short exact sequence to the other gives a commutative diagram. (One can find this near the beginning of chapter 3 in Rosenberg's algebraic K-theory text.) $\endgroup$
    – Joshua Seaton
    Mar 21, 2013 at 4:08

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Algebraic generators and relations for Quillen's K-group $K_n(P)$ are given in this paper: "Algebraic K-theory via binary complexes". Those you mention don't give the Quillen K-group $K_1(P)$ in general, but just for Quillen-exact categories where every short exact sequence splits; the group they give is known as Bass' $K_1$.

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  • $\begingroup$ @Dan: Thank you for the answer - and the paper. I will start making my way through it very soon. $\endgroup$
    – Joshua Seaton
    Mar 21, 2013 at 12:50
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The $K_1$ group you describe is the automorphism $K_1$, which is in general not isomorphic to Quillen's $K_1$ of an exact category. It coincides with Quillen's when exact sequences in $\mathcal P$ split. For a description by generators and relations of $K_1$ of any exact category see:

Nenashev, A. K1 by generators and relations. (English summary) J. Pure Appl. Algebra 131 (1998), no. 2, 195–212.

generators are pairs of short exact sequences on the same objects, and relations are given by $3\times 3$ diagrams. A generalization of this result to all $K_n$ is given in

Grayson, Daniel R. Algebraic K-theory via binary complexes. J. Amer. Math. Soc. 25 (2012), no. 4, 1149–1167.

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  • $\begingroup$ So what about $K_2$? $\endgroup$
    – Martin Brandenburg
    Mar 22, 2013 at 1:24

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