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The length of a module is well-known to be an additive invariant of finite-length modules. That is, if $R$ is a ring and $Art(R)$ its category of finite-length modules, then $length : Ob (Art(R)) \to \mathbb N$ is a function which is additive in the sense that for any short exact sequence $0 \to A \to B \to C \to 0$, we have

$$length(B) = length(A) + length(C).$$

Correspondingly, there is a map $length : K_0(Art(R)) \to \mathbb Z$. But this statement forgets the positivity property that the length of a module is nonnegative.

Question 1: Let $R$ be a ring and $Art(R)$ is category of finite-length modules. Let $l : K_0(Art(R)) \to \mathbb Z$ be a map which carries every (non-virtual) $M \in Art(R)$ to some (nonnegative) $l(M) \in \mathbb N$. Then is $l$ a multiple of the usual length function?

Question 2: Are there other interesting examples of categories $\mathcal C$ with maps $K_0(\mathcal C) \to \mathbb Z$ which carry every $C \in \mathcal C$ into $\mathbb N$? How about for other ordered abelian groups as the target?

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    $\begingroup$ For question 1, try $k \times k$ for a field k. For question 2 in a slightly different setting, there's Elliott's theorem. $\endgroup$
    – John Wiltshire-Gordon
    Commented Nov 17, 2023 at 1:32
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    $\begingroup$ Question 1: the objection raised by @JohnWiltshire-Gordon applies whenever $R$ is not local (and nonzero). On the other hand, if $R$ is local with residue field $k$, $Art(R)$ is generated by $k$, so the answer is yes. $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 17, 2023 at 7:56
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    $\begingroup$ For question 2, you can have a look at Lemma 5.3 of this preprint, where the group is the value group of a valuation ring. $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 17, 2023 at 8:07
  • $\begingroup$ @JohnWiltshire-Gordon Thanks, good points! What is Elliott's theorem? I'm searching and seeing some stuff on $C^\ast$-algebras, but I'm not sure that's what you're referring to. $\endgroup$
    – Tim Campion
    Commented Nov 17, 2023 at 23:32
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    $\begingroup$ @TimCampion Yes, I'm referring to the C^* algebra stuff. There's some characterization of all possible ordered abelian groups $K_0 A$ where $A$ is an AF-algebra. Since all these questions are trivial for semisimple categories, the AF case is a nice next example. $\endgroup$
    – John Wiltshire-Gordon
    Commented Nov 19, 2023 at 0:56

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