Let $\mathcal{C}$ be a stable $\infty$-category. Let $Fun(\mathbb{Z},\mathcal{C})$ be the category of sequences of objects in $\mathcal{C}$. Where the category $\mathbb{Z}$ stands for the nerve of the poset $\mathbb{Z}$. There's a canonical functor:

$$Gr:Fun(\mathbb{Z},\mathcal{C}) \to \underset{n \in \mathbb{Z}}{\coprod}{\mathcal{C}} \subset Fun(\mathbb{Z},\mathcal{C})$$

$$X \mapsto \underset{n \in \mathbb{Z}}{\coprod}{Cofib(X(n-1) \to X(n))}$$

Define the category $Fil(\mathcal{C})$ of filtered objects of $\mathcal{C}$ as the localization of $Fun(\mathbb{Z},\mathcal{C})$ w.r.t. morphisms $f: X \to Y$ which induce equivalences on graded pieces $Gr(f):Gr(X) \to Gr(Y)$.

(I've taken this definition from Enhancing the filtered derived category by Gwilliam and Pavlov. I have no claims to originality for this or for anything else in this question for that matter).

There's an obvious functor on $Fil(\mathcal{C})$ which shifts sequences. For notational convenience we will denote it $T: Fil(\mathcal{C}) \to Fil(\mathcal{C})$ with $T(X)(n):=X(n-1)$ (this convention will be useful later).

There's also the suspension functor coming from $\mathcal{C}$ which acts objectswise. We'll denote it as usual by $\Sigma$. Notice that $T$ and $\Sigma$ commute with each other.

The functor $Gr$ above sits in the fiber sequence of functors:

$$T \to Id \to Gr$$

Or equivalently:

$$Id \to Gr \to \Sigma T$$

Precomposing this sequence of functors with the tail $\Sigma T$ we get

$$Id \to Gr \to \Sigma T \to Gr \circ \Sigma T \to \Sigma^2 T^2$$

By repeating this process we get an infinite sequence of functors:

$$Id \to Gr \to \Sigma T \to Gr \circ \Sigma T \to \Sigma^2 T^2 \to Gr \circ \Sigma^2 T^2 \to \Sigma^3 T^3 \to \dots$$

One can just as easily extend this sequence in the other direction. Notice that inside this sequence sits the sequence:

$$\dots \to Gr \to Gr \circ \Sigma T \to Gr \circ \Sigma^2 T^2 \to Gr \circ \Sigma^3 T^3 \to \dots$$

Which is a "complex" in the sense that the composition of any two consecutive natural transformations is null-homotopic.

If $\mathcal{C}$ is provided with a t-structure then one should be able to make the sequence above the $E_1$ page of a spectral sequence with values in $\mathcal{C}^{\heartsuit}$.

In summary we have managed to make the construction of the $E_1$ page of a filtered object completely functorially and independent of the heart. Here's the (not so well defined) question:

How far can we take this? Can one make a functorial construction of an intermediate category between $Fil(\mathcal{C})$ (can be a functor, a sequence of functors etc.) and any target category of spectral which is independent from any t-structure on $\mathcal{C}$ s.t. choosing a t-structure induces a functor to spectral sequence with values in $\mathcal{C}^{\heartsuit}$ (with all the pages and all the differentials)?

Or even less precisely:

What is the "higher" structure underlying spectral sequences?

Edited: As was pointed out to me nothing is wrong with $Fil(\mathcal{C})$ as an answer to this question from this perspective so here's a refinement:

The Question: Let $\mathcal{C}$ be a stable $\infty$-category. Can we define a stable $\infty$-category $\mathcal{S(\mathcal{C})}$ associated with $\mathcal{C}$ having the following properties:

  1. There's a "natural" functor $Fil(\mathcal{C}) \to \mathcal{S(\mathcal{C})}$
  2. A t-structure on $\mathcal{C}$ induces a t-structure on $\mathcal{S(\mathcal{C})}$ whose heart is the category of spectral sequences with values in $\mathcal{C}^\heartsuit$ (or exact couples).
  3. For any t-structure the natural functor from (1) composed with the $\pi_0$ functor on $\mathcal{S(\mathcal{C})}$ gives the canonical functor $Fil(\mathcal{C}) \to SpSeq(\mathcal{C}^{\heartsuit})$ sending a filtered object to its associated spectral sequence (or exact couple).
  4. The assignment $\mathcal{C} \mapsto \mathcal{S(\mathcal{C})}$ is functorial.

Edited: One potential weakness in the precise version is that one could argue that spectral sequence should morally correspond to $\pi_*$ and not $\pi_0$. I have no idea what to expect.

  • $\begingroup$ Honestly I would say that $Fil(C)$ is already such a thing (given a t-structure from every object of $Fil(C)$ you get an exact couple, which is even better than a spectral sequence). What other kind of object are you looking for? $\endgroup$ Aug 19, 2017 at 17:20
  • $\begingroup$ Where is the exact couple here? I tried finding it but I got confused by indexing. More specifically how do you encode the data possesed by an "exact couple" before taking "cohomology"? $\endgroup$ Aug 19, 2017 at 17:21
  • $\begingroup$ More accurately: In order for "exact couple" to be a valid answer here I'd like to have (ideally) a stable $\infty$-category receiving a functor from $Fil(\mathcal{C})$ s.t. a choice of t-structure on $\mathcal{C}$ gives a t-structure on it whose heart is the category of exact couples with values in $\mathcal{C}^{\heartsuit}$ $\endgroup$ Aug 19, 2017 at 17:32
  • $\begingroup$ The question on your last comment is different (and clearer) than the question in the body. Maybe you could edit that in? $\endgroup$ Aug 19, 2017 at 19:23
  • $\begingroup$ @DenisNardin Edited. Thanks for the constructive criticism! $\endgroup$ Aug 20, 2017 at 13:59

1 Answer 1


Let me propose an answer to the question which isn't quite what you ask for. In fact, for the most part I agree with Denis that the correct object really is $\mathsf{Fil}(\mathcal{C})$. Also, I should mention that a really nice place to look for functorial discussions of spectral sequences is Verdier's thesis- most of what he does there carries over in a straightforward way to whatever fancy setting you'd like.

Ok. So the first issue is that the category of exact couples is not abelian, so we can't ask for it to appear as the heart of a t-structure.

As a warm-up, let's try to do the analogous thing for long exact sequences, and we'll get an answer which may or may not be satisfying. We'd like some functor from $\mathcal{C}^{[1]}$ to $\mathsf{LES}(\mathcal{C})$ which records `long exact sequences' without choosing a heart. Again, the category of long exact sequences is not abelian, so we can't expect to put a t-structure on $\mathsf{LES}(\mathcal{C})$ and have this be the heart. But whatever, it's sorta clear what we want: we should choose a sequence of pushouts and pullbacks that witness a bunch of triangles tied together. So let $\mathsf{Stair} \subset \mathbb{Z} \times \mathbb{Z}$ be the sub-poset consisting of pairs $(i,j)$ with $j=-i-1, -i, -i+1, -i+2$. An object of $\mathsf{LES}(\mathcal{C})$ will be a functor $X: \mathsf{Stair} \to \mathcal{C}$ such that $X(i, -i-1)$ and $X(i, -i+2)$ are zero objects and every square made up of adjacent vertices is a pushout square.

There is a forgetful functor $\mathsf{LES}(\mathcal{C}) \to \mathcal{C}^{[1]}$ recording the value at the arrow $(0,0) \to (1,0)$, and this is a trivial Kan fibration (a section can be obtained by iterated right and left Kan extensions). A section then gives a functor $\mathcal{C}^{[1]} \to \mathsf{LES}(\mathcal{C})$. Any $t$-structure on $\mathcal{C}$ yields a functor $\pi_0: \mathcal{C} \to \mathcal{C}^{\heartsuit}$ and postcomposition gives a functor $\mathsf{LES}(\mathcal{C}) \to \mathrm{Les}(\mathcal{C}^{\heartsuit})$ where the target is the ordinary category of long exact sequences in the heart. Actually, we could have used any homological functor, not just the $\pi_0$ of some heart. Again: this is sort of a waste of fanciness... in the end we're getting a functor to an ordinary category, so this passes through the homotopy category, and we could have just defined things there... but ok.

Now for spectral sequences. The indexing category here is more complicated, it's a variation of what Lurie calls "$\mathsf{Gap}(\mathbb{Z}, \mathcal{C})$" in HA.1.2.2. But the intuition isn't so bad. Basically, given a filtered object, we need to record a choice for every 'subquotient' (not just the adjacent ones), and then for every triple we need to build a staircase as above. Maybe I won't try to write this down... we'll call this category $\mathsf{SpObj}(\mathcal{C})$ for 'spectral objects in $\mathcal{C}$'. So an object in here is a functor from some horrible category of quadruples $(i,j,k;s)$ to $\mathcal{C}$ where $i\le j\le k$ and $s$ is in $\mathsf{Stair}$ and a bunch of these values are required to be zero objects, various squares are meant to be pushouts, etc. Basically, if we start with a filtered gadget $X(i)$ then let $X(i,j)$ be a choice of cone for $X(i) \to X(j)$, and then $X(i,j,k;(0,0))$ will be $X(i,j)$, $X(i,j,k;(1,0))$ will be $X(i, k)$, and the rest of the staircase records an extension of the map between these into a 'long exact sequence'.

This more elaborate version of taking the associated graded can be turned into a functor $\mathsf{Fil}(\mathcal{C}) \to \mathsf{SpObj}(\mathcal{C})$ which takes a filtered gadget and makes a choice of display like this (it can be built via lots of left and right Kan extensions; though this is no longer an equivalence!)

The upshot is that, given any homological functor $H: h\mathcal{C} \to \mathcal{A}$, where $\mathcal{A}$ is an abelian category, we can apply $H$ to every value taken by $X \in \mathsf{SpObj}(\mathcal{C})$ and unwinding we see that for every $i\le j$ we have a sequence of abelian groups that yield $H_n(X(i,j))$ and for every triple we have a chosen and functorial boundary map connecting these. This is precisely the data of what Verdier calls a "spectral object" and is enough to form a spectral sequence (the boundary maps are given, and the pages of the spectral sequence are formed by taking certain images of maps between then $H_{p+q}$'s). I think it's essentially an unrolled exact couple.

My personal opinion: this is all more trouble than it's worth...

  • $\begingroup$ It's possible you could find something with a t-structure whose heart is honestly the category of spectral sequences (which is abelian)... I think a warm-up question is the following: if $\mathcal{C}$ has a t-structure, is there a t-structure on $\mathcal{C}^{[1]}$ whose heart is equivalent to $\mathcal{C}^{\heartsuit}$ via the functor $(A \to B) \mapsto im(\pi_0A \to \pi_0B)$? If you could do that (maybe it's easy- I haven't tried), then I think a more elaborate version of that applied to smaller version of "SpObj(C)" would work. $\endgroup$ Aug 24, 2017 at 19:17
  • $\begingroup$ Thanks! This answer really helped me in realizing that the story is basically no more then that spectral sequences are to sequential limits/colimits as long exact sequences are to kernels/cokernels. Take my money! $\endgroup$ Aug 28, 2017 at 21:08

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