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A category $\mathcal{A}$ is abelian if it has the following properties:

  • There is a zero object (both initial and terminal).
  • Every pair of objects has a product and a coproduct (which will be canonically isomorphic as a consequence of the rest of axioms).
  • Every morphism has a kernel and a cokernel.
  • Every monomorphism (resp. epimorphism) is normal, i.e. it is a kernel (resp. cokernel) of some morphism.

Every abelian category is naturally a (pre)additive category via either of the following two compositions (which are the same): $$ A \xrightarrow{(1,1)} A\oplus A \xrightarrow{\binom{f}{g}} B \quad \text{or}\quad A\xrightarrow{(f,g)} B\oplus B \xrightarrow{\binom{1}{1}} B $$

For an arbitrary category $\mathcal{C}$, being a (pre)additive category is not a property but rather some extra structure that we have to specify. But if $\mathcal{C}$ is abelian (which is a property), then the underlying additive structure is natural and doesn't have to be specified in an artificial way.

Now I am concerned with $k$-linear abelian cateogries, which means that the hom-sets are not only abelian groups but also $k$-vector spaces (and composition is $k$-bilinear). This made me wonder:

Is there a property that, whenever possessed by a category, gives raise to a natural $k$-linear structure?

I guess that the answer is no, because the integers seem much more natural than an arbitrary field $k$ (so I would expect such a propery to depend on $k$). So maybe the following question makes sense:

Can we reformulate the (absolute) property of being an abelian category as a relative property in terms of a category built up from $\mathbb{Z}$ in such a way that this generalizes to any other field (or even ring) to get the analogous conclusions?

I hope these questions make sense. Thanks.

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    $\begingroup$ This paper gives a characterization of categories that are $k$-linear for some $k$, but it requires some structure on the category. $\endgroup$ – Arnaud D. Apr 18 '18 at 10:47
  • $\begingroup$ Thank you for the reference! $\endgroup$ – Pedro Apr 20 '18 at 12:01

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