19
$\begingroup$

It is well know that Karoubian categories (also called idempotent-complete categories) are living between additive and Abelian categories. While one of the most famous advantages to work with abelian cetegories is that they are closed under building kernels and cokernels of arbitrary morphism, the Karoubian categories have a slightly weaker property that only idempotent morphisms (i.e. $p$ with $p^2=p$) from there share this property.

Nevertheless it seems that in diverse constructions Karoubian categories or Karoubian envelopes of additive categories provide a more natural setting than Abelian categories.

I hope my question not becomes too broad: What is the philosophical meaning behind Karoubian categories or say in simpler words why in some constructions they are more prefered (e.g. category of pure motives and in K-theory) in contrast to say at first glance more 'flexible' abelian categories?

My natural guess is that if we think about the construction of of the category of pure motives we start with the category of smooth varieties over a base field and pass after application of this magic Karoubian completion functor to
idempotent-complete categories. Since it's not abelian it contains by definition only kernel and cokernels of idempotent morphisms but that's all we need there to proceed the constrution.

This lead me to conjecture that the main advantage of Karoubian categories in contrast to Abelian categories mights show when one have to perform a construction where one starts with a certain preadditive category but the construction requireres a category having at least some kernels and cokernels.

Now one can pass canonically to the Karoubian completion or extend the initial category to an Abelian category. But exactly here I see an obstacle with the secound and at first glance more 'natural' approach:

Does there always a way to embedd a preadditive category in an abelian category? If yes, seemingly the disadvantage of this approach seems to be that this Abelian category is much harder to control, while the Karoubian completion is constructed quite canonically and behaves more 'similar' to initial preadditive category.

Questions: Is what I tried to sketch above exactly the motivation why Karoubian categories are in some constructions more prefered then Abelian categories?

Are there more reasons making Karoubian categories 'interesting'? Is there any intuition or important example one should have in mind how to think about Karoubian categories?

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ May be the basic example is the category of vector bundles which is Karoubian, but not abelian? $\endgroup$
    – Mohan
    Commented Jan 5, 2021 at 17:38
  • 7
    $\begingroup$ Karoubian categories are more abundant than abelian ones, but I don't know if that makes them more interesting or in any sense preferred. By the way, although the category of Grothendieck motives is contructed as a Karoubian completion, one really wants it to be abelian... $\endgroup$
    – Donu Arapura
    Commented Jan 5, 2021 at 18:05
  • $\begingroup$ @DonuArapura: So the philosophy here with Karoubian completion is indeed: "Be satisfied with what you get", right? In case of pure motives we obtain in a relatively easy way our Karoubian category after completion which already have a lot of nice properties which an Abelian category would have, but there is just no canonical way known to extend it to an Abelian category, so that's a 'stay happy with what we have' philosophy? Then that's the whole moral? $\endgroup$
    – user267839
    Commented Jan 6, 2021 at 0:48
  • $\begingroup$ @GhostinGrothendieckuniverse It's not a question of moral/philosophical principles, it's about what can be proved. If we stick to pure motives modulo homological equivalence, then Grothendieck conjectured this is abelian. However, I don't think anyone knows how to prove it. If we switch to motives mod numerical equivalence, then Jannsen did prove it's abelian. $\endgroup$
    – Donu Arapura
    Commented Jan 6, 2021 at 13:05
  • 1
    $\begingroup$ Splitting idempotents has nothing at all to do with additivity or Abelianness. It's an easy construction that can be applied to any category at all. It's used in plenty of other subjects. For example it gives continuous lattices from algebraic ones and Scott-continuous maps. $\endgroup$
    – Paul Taylor
    Commented Jan 18, 2021 at 13:59

2 Answers 2

Reset to default
22
$\begingroup$

I don't know if this is going to answer your question but here's some relevant background. Splitting idempotents has a very special property from a categorical point of view: it is an absolute colimit (and also an absolute limit), meaning that it is preserved by any functor whatsoever. Two somewhat more familiar examples of absolute colimits, in enriched settings:

  • In categories enriched over pointed sets (equivalently, categories with zero morphisms), zero objects are an absolute colimit. This also applies to $\text{Ab}$-enriched categories.
  • In linear ($\text{Ab}$-enriched / preadditive) categories, (finite) biproducts are an absolute colimit.

In general we can ask for the completion of a (possibly enriched) category under absolute colimits; this is called its Cauchy completion because it turns out to specialize to Cauchy completion when thinking of metric spaces as enriched categories. We have:

  • For ordinary ($\text{Set}$-enriched) categories, the Cauchy completion of $C$ is given by splitting all idempotents, or equivalently by taking the full subcategory of the presheaf category $\widehat{C} = [C^{op}, \text{Set}]$ on the tiny objects (objects such that $\text{Hom}(F, -)$ preserves all colimits; these turn out to be exactly the retracts of representables).
  • For linear ($\text{Ab}$-enriched) categories, the Cauchy completion of $C$ is given by taking all formal direct sums and then splitting all idempotents, or equivalently by taking the full subcategory of the presheaf category $\widehat{C} = [C^{op}, \text{Ab}]$ on the tiny objects (these turn out to be exactly the retracts of finite direct sums of representables).

In particular, if we consider a ring $R$ as a one-object linear category $BR$, its Cauchy completion is exactly the category of finitely generated projective $R$-modules, which we famously use to define K-theory. This construction is a complete Morita invariant in the following sense: two rings $R, S$ are Morita equivalent (meaning $\text{Mod}(R) \cong \text{Mod}(S)$) iff their categories of finitely generated projective modules are equivalent (and this generalizes to enriched categories and Cauchy completions).

Cauchy completion is in some sense the "most harmless" and "most inevitable" completion: if you are ever going to apply a functor from your category $C$ to a category $D$ with colimits then every absolute colimit of objects in $C$ will appear in $D$ anyway (all idempotents will be split, etc.) so you might as well add them in first. Unlike the Yoneda embedding, which is the free cocompletion, Cauchy completion does not destroy colimits that may already exist in $C$. And because absolute colimits are preserved by all Hom functors $\text{Hom}(c, -)$, unlike adjoining colimits in general, the morphisms both into and out of an absolute colimit are already uniquely determined, so you have no choice how to do it anyway.

On the other hand attempting to write down some sort of completion of a linear category producing an abelian category seems quite tricky and potentially unwieldy. We can take the free cocompletion ($\text{Ab}$-valued presheaves) but again this destroys most existing colimits. Maybe the Isbell envelope has better properties but I don't know anything about it. Meanwhile the Cauchy completion is relatively easy to work with.

Improve this answer
$\endgroup$
11
  • 4
    $\begingroup$ Qiaochu, you're on fire today. The first paragraph is exactly what needs to be said. $\endgroup$
    – Todd Trimble
    Commented Jan 6, 2021 at 1:08
  • $\begingroup$ Sorry that I only now respond to your answer, I was quite bissy last days. One question: reading your answer I tried to understand the deeper meening what it would mean if we pass a (pre)additive category to it's idempotent completion. As you said then all idempotents would become absolute colimits and the point is at first glace how does it help. Abstractly, when passing to idempotent completion we therefore obtain a new category containing a distinguished class of morphisms (the idempotents) which are all absolute colimits. $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 1:30
  • $\begingroup$ If we thinking about concrete constructions of pure motives or $K$-theory then question might be arise why exactly this idempotent completion gives us a category which we need as 'intermediate' step (because it has now enough structure) to construct the final object/category we are looking for. I don't know the $K$-theoretic construction good enough but recently an analogy came into my mind which might draw a nearly similar picture. $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 1:30
  • $\begingroup$ Thinking about model categories these can be regarded as the 'most rigid category' where we can do homotopy theory. The main feature of model category is that it contains enough classes of distinguished morphisms ((co)fibirations, etc ...) to pass to homotopy theory. Can our case with idempotent completion regarded fro simmilar point of view: that this idempotent completion is some kind of 'most rigid category structure' (ie as in case of model category, a category with certain distinguished class of morphisms) which suffice to 'build' a category/ an object we are looking for. $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 1:31
  • $\begingroup$ Is this the right way to think about this issue? $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 1:44
2
$\begingroup$

I'd like to comment that idempotent completion is sometimes the right thing to do 'by design'. (I don't have enough rep to just leave a comment though.) I'm reminded of the category of motives, in which you would like to have some kind of 'cellular decomposition' of your varieties, the archetypical example being that one would like the motive $[\mathbb{P}^1]$ to split as $[\mathbb{A}^1] + [\text{pt}]$ or something along those lines. Idempotent completion does exactly that: it forces the idempotent map $\mathbb{P}^1 \to \operatorname{Spec} k \to \mathbb{P}^1$ to have a kernel, which leads to the desired decomposition.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes; of course, passing to these idempotent completions sometimes we indeed obtain the category we 'need' to develop 'good' theory like in the case of motives as you said. But my concern aimed primarily not at 'that' we obtain a category with which we can better work, but 'why' exactly this 'idempotent completion' gives the desired catogory where it's better to work. Up to now by Qiaochu Yuan's and Todd Trimble's answer / comment in all theories where the construction involves passing to idempotent completion the category which meets allmost all our requirements has to have $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 1:04
  • $\begingroup$ enough distinguished classes of morphisms, which are absolute colimits. The initial additive category might also have certain absolute colimits, but seemingly not enough. It's like in case of model categories, which have distinguished class of morphisms which suffice to also to do there homotopy theory. That's at least how I understood Qiaochu Yuan. (the analogy to model category came recently to my mind) $\endgroup$
    – user267839
    Commented Jan 20, 2021 at 1:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .