Chain Complexes and Linear Infinity-Categories

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A statement I heard recently is that "chain complexes are the same thing as strict linear $\infty$-categories". Can someone explain how to see this?

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The cochain complex of a space $X$ is an $E_\infty$-algebra, i.e., product of cochains (cup product) are defined although it is only commutative up to homotopy, thereby turning it into an $E_\infty$-algebra. And I presume that an $E_\infty$-category (which is linear in your case because the hom sets are chain complexes over $\mathbf{Z}$) with one object is an $E_\infty$-algebra. I'm sure other users here would be able to elaborate more.  – Somnath Basu Sep 27 2010 at 5:36

3 Answers

The result you want is contained in

(BROWN, R. and HIGGINS, P.J.), `Cubical abelian groups with connections are equivalent to chain complexes', Homology, Homotopy and Applications, 5(1) (2003) 49-52.

which also deals with strict globular $\omega$-categories internal to abelian groups.

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You say 'how to see this', and for that I suggest that you first look at low dimensional cases. You also have to be a bit careful in the definition of chain complex as the ones in question have to be trivial in negative degree. You also, as hinted at by others, have to say what you mean by 'linear'.

(I will not assume that you know much on this, and excuse me if you do and I am saying obvious things.)

The first step is to see what $\infty$-cat structure, a `very short' chain complex gives, that is to take a chain complex with stuff only in dimension 0 and 1 and to stare at $C_1\oplus C_0$ and $C_0$ and to interpret them as the arrows and objects of a category with source given by the projection, target by the source plus the boundary $t(c_1,c_0) = c_0 + \partial c_1$. Composition is given by addition in $C_1$. Now just check the details to get a category. (In fact it is a 2-category with the additive structure on the groups giving the other composition.)

The next step is to try to see what a chain complex of length 2 (i.e. allowing $C_2$ to be non-zero) gives. Clearly you need to add an additional factor and to get some 2-cells. The formula is quite geometric so should be findable by fiddling around. Now you can hope to 'see' how to go further. (The actual construction is outlined in several different places, so once you think you know what is happening in low dimensions look for it in the literature. I have usually found it worth while working this sort of thing out myself first, knowing that the intuition is the really hard thing to grasp, whilst the technique of proof may be clear once the intuition is in place.)

If you want a linearised enriched version, now start working with the category of chain complexes. The chain maps etc, between chain complexes form a chain complex, which I presume you know well as it is 'classical', so now you can start applying the steps that are suggested above to that cochain complex. The objects will be the maps the 1-cells homotopies between them, the 2-cells homotopies between homotopies and so on.

You then have 'linear' $\infty$-categories as objects and the homs between them are also such objects.

Once all that seems routine then looking at the simplicial side and the Dold-Kan equivalence in the light of that is very well worth doing. Proofs of these things are reasonably easy to manufacture, once the intuitions are grasped, and that is what I am assuming you are asking for.

(Edit: if you want a full answer, look at Ronnie Brown and Phil Higgins' paper that he mentions in his reply. But I still think that trying to see what has to be done is very important.)

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(Edit: I was being a bit free and easy below. Dold-Kan is rigorous for $\infty$-groupoids = Kan complexes, but I'm only guessing when I say I can sweep details aka weak complicial sets under the carpet. I'll leave the answer as is, in case anyone else can fill in details)

If we take 'strict linear $\infty$-categories to mean '$\infty$-categories internal to $Ab$', and we take '$\infty$-category' to mean simplicial set (I'm sweeping some details under the carpet, but really you'd want to consider (weak) complicial sets to make this rigorous), then this is nothing but the Dold-Kan correspondence: simplicial abelian groups are the same thing as non-negatively graded chain complexes. Note that the simplicial set underlying a simplicial abelian group is automatically a Kan complex, so really this reduces to talking about $\infty$-groupoids.

If you want $\infty$-categories enriched in something linear, rather than internal categories, then you'd want some sort of weak enrichment in simplicial abelian groups. It sound like you might want some sort of $E_\infty$-category, but this is certainly not equivalent to chain complexes.

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