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)?

even less precisely:Or

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:

- There's a "natural" functor $Fil(\mathcal{C}) \to \mathcal{S(\mathcal{C})}$
- 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).
- 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).
- 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.