# What is the decategorification of a triangulated category?

The decategorification of an essentially small category $\mathcal C$ is the set $\lvert\mathcal C\rvert$ of isomorphism classes of $\mathcal C$.

If $\mathcal C$ carries additional structure, then so does $\lvert\mathcal C\rvert$. For example, the decategorification of a braided monoidal category is a commutative monoid.

Side Question 1: Is there more we can say about the commutative monoid $\lvert\mathcal C\rvert$ in the above example? Are there any general properties, or does every commutative monoid arise as $\lvert\mathcal C\rvert$ for some braided monoidal category $\mathcal C$?

Let $\mathcal T$ be a triangulated category. Then, by the above example, $\lvert\mathcal T\rvert$ is certainly a commutative monoid with respect to direct sum. This only uses the additive structure on $\mathcal T$, of course.

Main Question: How can we describe the additional structure on the commutative monoid $\lvert\mathcal T\rvert$ induced by the triangulated structure on $\mathcal T$?

Side Question 2: If $\mathcal A$ is an Abelian category, which additional properties does the commutative monoid $\lvert\mathcal A\rvert$ have?

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Translation should be some suitable of automorphism of monoid, isn't it? – Fosco Loregian Nov 14 '11 at 13:35
@tetrapharmakon: I suppose so, yes. – Rasmus Nov 14 '11 at 23:34
In fact I'm not answering your question but maybe you'll find it a good startpoit: sometimes you are interested in the Grothendieck group of the commutative monoid $|\cal T|$. – Fosco Loregian Nov 15 '11 at 15:37
@tetrapharmakon: Thanks for the reference. As is also described here, before taking the Grothendieck group, one divides out a set of relations coming from the distinguished triangles. These relations or probably the inherited structure on the monoid $\lvert\mathcal T\rvert$ I was looking for. – Rasmus Nov 15 '11 at 16:24
Q1: Consider discrete monoidal categories. – Martin Brandenburg Feb 3 '13 at 0:52

The additional structure is that there is a natural further quotient of the isomorphism classes you can take where you impose the additional relation $[X] - [Y] + [Z] = 0$ for every distinguished triangle $X \to Y \to Z \to \Sigma X$.
Indeed, that seems to be the correct formulation of the additional structure. In other words, there is a natural transformation $\lvert\mathcal T\rvert\to\mathrm{K}_0(\mathcal T)$. – Rasmus Jan 3 '15 at 11:53