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It's well known that chain complexes are an abelian category, and in particular we can consider chain complexes of chain complexes, i.e. double complexes. Given a double complex $A^{\bullet\bullet} \in \mathrm{Kom}(\mathrm{Kom}(\mathcal A))$ we can form the total complex $\newcommand{\tot}{\mathrm{Tot}}\tot(A^{\bullet\bullet})$ which now lies "one level lower", in $\mathrm{Kom}(\mathcal A)$.

I can also try to consider chain complexes in the derived category of $\mathcal A$, but it is no longer clear (at least to me) how to build a total complex in the best way, since $d \circ d=0$ only has to hold up to homotopy in $D(\mathcal A)$.

But let me now consider more generally a triangulated category $\newcommand{\T}{\mathcal T}\T$ and a sequence of objects $A^0, \ldots, A^n$ with $d_i \colon A^i \to A^{i+1}$ such that $d_{i+1} \circ d_i = 0$. It seems to me that one can define a total complex $\tot(A^\bullet) \in \T$ as an iterated mapping cone: for instance, if $n=2$, then one can first consider $B = \mathrm{Cone}(d_0)$. Then we consider the diagram $$ \begin{matrix} A^0 & \to & A^1 & \to & B \\ \downarrow & & \downarrow & & \\ 0 & \to & A^2 & \to & A^2 & \end{matrix}$$ where both rows are distinguished triangles; by one of the axioms of triangulated categories there is a map $f \colon B \to A^2$ completing this to a map of triangles, and we define $\tot(A^\bullet)=\mathrm{Cone}(f)$.

However, this has a the drawback of not being functorial. For instance, I think that one would like to say that a map $A^\bullet \to B^\bullet$ of chain complexes in $\T$ is a quasi-isomorphism if $\tot(A^\bullet) \to \tot(B^\bullet)$ is an isomorphism, but this makes no sense unless $\tot$ is a functor. And one would like to say that if $f \colon \T \to \T'$ is a triangulated functor, then there is a natural equivalence between $f \circ \tot$ and $\tot \circ f$ as functors $\mathrm{Kom}(\T) \to \T'$ (am I right?), and again $\tot$ needs to be functorial.

First of all, I would like to know if what I've said so far is correct. Maybe there is a better way to set things up than this? Secondly, I've heard the slogan that stable $\infty$-categories solve all problems arising from the fact that triangulated categories don't have functorial mapping cones. Is there a better behaved notion of a chain complex in a stable $\infty$-category?

I suspect that I could answer these questions myself if I started reading Lurie's work, but it's a slightly intimidating amount of text and I thought I'd ask here first.

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  • $\begingroup$ The correct notion of a chain complex in a stable $\infty$-category is a simplicial object. Lurie proves a Dold-Kan theorem describing these in Higher Algebra (Theorem 1.2.4.1). $\endgroup$ – Qiaochu Yuan Oct 24 '14 at 7:42
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    $\begingroup$ You might find the concept of "Postnikov systems", as described for example in Orlov's paper "Equivalences of derived categories and K3 surfaces" useful. $\endgroup$ – ulrich Oct 24 '14 at 8:35
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    $\begingroup$ Bondal & Kapranov's "Enhanced triangulated categories" might be the most concise way to deal with this. $\endgroup$ – მამუკა ჯიბლაძე Oct 24 '14 at 12:11
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    $\begingroup$ I am skeptical that you can iterate the construction in the third paragraph beyond $n=2$. $\endgroup$ – Ben Wieland Oct 24 '14 at 20:37
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    $\begingroup$ ...and to summarize: a chain complex in a triangulated category can in general have many different totalizations, or none at all, and a theorem in Orlov's paper gives a sufficient condition for a chain complex to have a unique totalization up to canonical isomorphism. My guess is that Orlov's theorem can also be interpreted as describing when a chain complex in $hC$, where $C$ is a stable $\infty$-category, gives rise to a simplicial object in $C$ up to homotopy. $\endgroup$ – Dan Petersen Oct 25 '14 at 7:41
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There are different constructions of total complexes, using direct sums, direct products, etc. Each of them needs that the bicomplex satisfies some boundedness conditions. Your third paragraph indicates that you might be interested in cohomological positive bicomplexes $A^\bullet$ $$0\rightarrow A^0\stackrel{d_0}\longrightarrow A^1\rightarrow\cdots$$ Here each $A^i$ is an unbounded complex in the abelian category $\mathcal A$. Let us define $\operatorname{Tot}(A^\bullet)$ as $\operatorname{Tot}(A^\bullet)^{n}=\prod_{p+q=n}A^{pq}$ with the usual differential. This is a homotopy limit construction, i.e. you can consider the abelian category $\operatorname{Kom}^+(\mathcal A)$ of cohomological positive complexes in $\mathcal A$ and its derived category $D(\operatorname{Kom}^+(\mathcal A))$. The obvious degree $0$ inclusion $\mathcal A\rightarrow\operatorname{Kom}^+(\mathcal A)$ induces a functor $D(\mathcal A)\rightarrow D(\operatorname{Kom}^+(\mathcal A))$, and $\operatorname{Tot}$ can be characterized as its right adjoint. This answer is maybe not very satisfactory for you, but I think it's the best you can get. There is an obvious forgetful functor $D(\operatorname{Kom}^+(\mathcal A))\rightarrow\operatorname{Kom}^+(D(\mathcal A))$, but it is not well behaved.

The moral of the previous paragraph is that you cannot fully stay in the triangulated world if you want to consider totalization, but now I'll show you that you can stay close. The abelian category $\operatorname{Kom}^+(\mathcal A)$ is equivalent to $\mathcal A^{\Delta}$, i.e. cosimplicial objects in $\mathcal A$. In particular $D(\operatorname{Kom}^+(\mathcal A))\simeq D(\mathcal A^\Delta)$. This category is part of the triangulated derivator $\mathbb{D}(\mathcal A)$ associated to $\mathcal A$. This gadget is a 2- (or pseudo-) functor $$\mathbb{D}(\mathcal A)\colon \operatorname{Dia}^{\operatorname{op}}\longrightarrow\operatorname{CAT}\colon I\mapsto D(\mathcal A^I).$$ Here $\operatorname{CAT}$ is the big category of all categories and $\operatorname{Dia}$ is an appropriate full sub-2-category containing $\Delta$. Evaluating $\mathbb{D}(\mathcal A)$ at the inclusion $\Delta[0]\rightarrow\Delta$ we obtain the functor $D(\operatorname{Kom}^+(\mathcal A))\rightarrow D(\mathcal A)\colon A^\bullet\mapsto A^0$, its right adjoint is the degree $0$ inclusion $D(\mathcal A)\rightarrow D(\operatorname{Kom}^+(\mathcal A))$, and the right adjoint of this last one is totalization. The existence of such adjoints is part of the properties of a derivator.

Derivators, discovered by Grothendieck, are a user fiendly model for higher homotopy theory interpolating between $\infty$-categories and their homotopy categories. Their flavor is closer to the homotopy (triangulated) categories, the techniques are 2-categorical, but they encode higher information such as mapping spaces.

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  • $\begingroup$ I like very much your description of totalizations in the language of derivators, is this worked out in detail anywhere? Let me just ask you a couple of things: you take the inclusion $0:[0]\to \Delta$, this gives the evaluation $0^*:\mathbb{D}(\Delta)\to \mathbb{D}([0])$ and its left and right adjoints $0_!$ and $0_*$. You call "totalization" the right adjoint $0^!$ to $0_*$, why does this functor exist in general? It would if $[0]\to \Delta$ were a sieve but this is not the case... so how do you prove this? $\endgroup$ – Simone Virili Feb 9 '17 at 21:29

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