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Let $\mathcal{D}$ be a triangulated category and a $t$-structure $(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$. The heart of the $t$-structure, $\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$, is an abelian category.

I know that in general there is not a natural functor from the derived category of the heart and the triangulated category .

But it's seem to be very linked ...

But if you consider stable infinite category or Grothendieck derivator is there such functor ?

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    $\begingroup$ You might be interested in Higher Algebra, theorem 1.3.3.2. and remark 1.3.3.3.. Depending on what you call "derived category", and with suitable hypotheses on $\mathcal D$ there will be a canonical functor $\mathcal D^{-}(\mathcal A)\to \mathcal D$. I think under more suitable hypotheses, this should extend to $\mathcal{D(A)\to D}$ $\endgroup$ May 18 '20 at 11:43
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    $\begingroup$ Also : see Dustin Clausen's second comment under his answer here : mathoverflow.net/questions/112412/… $\endgroup$ May 18 '20 at 12:34
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    $\begingroup$ I have studied some related questions in a general (strong and stable) Grothendieck derivator in a recent preprint: arxiv.org/abs/1807.01505 $\endgroup$ May 19 '20 at 10:21
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    $\begingroup$ For the case of the bounded derived category, this follows right away from Corollary 7.59 in this paper (for E an abelian category); see Remark 7.60 in loc. cit. For bounded above complexes or unbounded complexes, similar universal properties may be found in Lurie's Higher Algebra, aber you need some nice properties for the heart (being cocomplete and having enough projectives or being Grothendieck). $\endgroup$ May 19 '20 at 14:07
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    $\begingroup$ @MaximeRamzi I think this is this paper : mathematik.uni-regensburg.de/cisinski/unik.pdf (also arXiv:1911.02338) $\endgroup$ Nov 16 '20 at 18:25
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Assume $\mathcal D$ is a presentable stable $\infty$-category with a $\mathrm t$-structure (which is accessible and compatible with filtered colimits), and let $\mathcal A$ be its heart, $\mathcal{D(A)}$ its derived $\infty$-category.

Note that under those hypotheses, $\mathcal A$ is Grothendieck abelian (Higher Algebra, 1.3.5.23.).

You get a natural inclusion functor $\mathcal A\to \mathcal D$, which you can extend to $Fun(\Delta^{op},\mathcal A)\to \mathcal D$ (by geometric realization).

This functor preserves weak equivalences : indeed, (see HA, 1.2.4.4. and 1.2.4.5.), if $X$ is a simplicial object of $\mathcal A$ (and therefore $\mathcal D_{\geq 0}$), there is a spectral sequence with $E^2_{p,q}=\pi_p\pi_q(X)$ converging to $\pi_{p+q}(|X|)$ in $\mathcal A$.

Since $\pi_q(X) = 0$ for $q\neq 0$ in our situation (as $X$ takes values in $\mathcal A$), this spectral sequence degenerates, and $\pi_p(|X|) = \pi_p(X_\bullet)$ (where the latter are homotopy groups as computed in $\mathcal A$ via the classical Dold-Kan correspondance)

It follows that this functor yields a (unique up to a contractible space of choices) functor $\mathcal D_{\geq 0}(\mathcal A)\to \mathcal D$ (via, again the Dold-Kan correspondance).

Now $\mathcal D$ is presentable and stable, and this functor $\mathcal D_{\geq 0}(\mathcal A)\to \mathcal D$ preserves colimits, so it extends (again, uniquely) to a functor $\mathcal{D(A) \to D}$ which preserves colimits and preserves $\mathcal A$ (HA 1.3.5.21. : $\mathcal{D(A)}$ is right complete with its classical $\mathrm t$-structure, which implies in particular that $\mathcal{D(A)} = \lim(\dots \overset{\Omega}\to \mathcal D_{\geq 0}(\mathcal A)\overset{\Omega}\to \mathcal D_{\geq 0}(\mathcal A))$, and then we use 1.4.4.5. which says that this precisely has the universal property of "presentable stabilization")

Now these hypotheses on $\mathcal D$ may look pretty strong but they're reasonable ; and in fact you can't really hope for much better : if $\mathcal D$ isn't presentable, then it could be something like $\mathcal D^{-}(\mathcal A)$ and then there's no hope to get a sensible functor $\mathcal{D(A)}\to \mathcal D^{-}(\mathcal A)$ (this is something that won't change whether you're in an $\infty$-categorical setting or not).

This is probably already somewhere in HA but I couldn't find it written down completely.

As I pointed out in the comments, under different hypotheses (which may look weaker), you can get away with a natural functor $\mathcal D^{-}(\mathcal A)\to \mathcal D$ (morally, this is because $\mathcal D^{-}(\mathcal A) \subset \mathcal{D(A)}$ is $\bigcup_n \mathcal D_{\geq n}(\mathcal A)$, and so it is determined by $\mathcal D_{\geq 0}(\mathcal A)$ via finite limits, so you don't need a presentability hypothesis- you do, however need some hypothesis on $\mathcal D$ to be able to replace the first step, since you can't take colimits as easily).

I guess there may also be a more general statement about $\mathcal D^b(\mathcal A)$, the bounded derived $\infty$-category, which should require less hypotheses, since "everything is finite" but I couldn't tell you on the top of my head.

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  • $\begingroup$ This does not seem to answer the question. $\endgroup$ May 19 '20 at 13:06
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    $\begingroup$ @DavidWhite : Why not ? The question asks whether there is a functor from the derived category to $\mathcal D$ in the $\infty$-categorical setting. I give some reasonable conditions under which there is, and explain why we can't expect there to always be one - of course it does not settle the more general question "when is there one ?"; but it certainly adresses the question "is there such a functor ?" $\endgroup$ May 19 '20 at 13:28
  • $\begingroup$ Ok, I re-read the question and answer, and I apologize for my earlier comment. This is pretty close to an answer. I didn't get much sleep last night and interpreted the question incorrectly when I first read it. Nice job! $\endgroup$ May 19 '20 at 14:24
  • $\begingroup$ @DavidWhite : ah good, I was worried I had missed something - no need to apologize though, it's always good to make sure proposed answers are adequate ! $\endgroup$ May 19 '20 at 14:29
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I had reason to think about this a few years ago. When $\mathcal D$ arises as the derived category of an abelian category (with a possibly exotic $t$-structure), a construction of a realization functor $D^b(A) \to \mathcal D$ can be found already in Beilinson-Bernstein-Deligne-Gabber. (For them $\mathcal D$ is the derived category of constructible sheaves, equipped with the perverse $t$-structure, so $A$ is the category of perverse sheaves.) Their construction of the realization functor can in fact be imitated in the $\infty$-categorical setting, too, and in this case it works more generally for an arbitrary stable $\infty$-category $\mathcal D$. The way B-B-D-G construct the functor is they use the filtered derived category $\mathcal{D}F$. If $\mathcal D$ is the derived category of an abelian category $B$, then $\mathcal D F$ is the category of complexes in $B$ with a bounded filtration, localized at filtered quasi-isomorphisms. There is an induced $t$-structure on $\mathcal D F$ from a $t$-structure on $\mathcal D$, such that if the $t$-structure on $\mathcal D$ has heart $A$ then the heart of the $t$-structure on $\mathcal D F$ is isomorphic to the abelian category $\mathrm{Ch}^b(A)$ of bounded chain complexes in $A$. Thus we can consider the composition $\mathrm{Ch}^b(A) \to \mathcal D F \to \mathcal D$ where the first is the inclusion of the heart and the second forgets the filtration. A spectral sequence argument shows that this composition takes quasi-isomorphisms in $\mathrm{Ch}^b(A)$ to equivalences in $\mathcal D$, so there is an induced functor $D^b(A) \to \mathcal D$ which is the one we want.

The point of the above is that for a general triangulated category $\mathcal T$ there is no sensible triangulated category $\mathcal T F$ of filtered objects in $\mathcal T$, but if $\mathcal T$ happens to be the derived category of an abelian category we can write down the filtered derived category by hand. This is not a problem in the world of stable $\infty$-categories.

I believe that an analogous argument works also for complexes that are not necessarily bounded (using instead filtrations that are unbounded to the left or right or both). But if we want there to exist a functor $\mathcal DF \to \mathcal D$ that forgets the filtration then we need $\mathcal D$ to have sequential limits or colimits, and one should be careful with the spectral sequence argument.

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    $\begingroup$ I had tried some time ago to just consider unbounded filtrations and I could not make things work. There is another option, that I developed in a recent preprint: consider "tridimensional filtrations", that is, objects filtered by ZxNxN^{op}, where the Z-filtration is "point-wise bounded". The Beilinson-Bernstein-Deligne-Gabber realization then allows you to see such objects as "NxN^{op}"-shaped diagrams of bounded objects, and then you can take a sequential homotopy colimit in the N-direction, and a sequential homotopy limit in the N^{op}-direction. $\endgroup$ May 19 '20 at 14:13
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    $\begingroup$ Of course, you need sequential homotopy co/limits to exist, and that they are computed "pointwise" in a suitable sense. If the t-structure is taken on the base of a strong and stable derivator (e.g., the homotopy category of a bicomplete stable $\infty$-category), then all such requirements are satisfied. An alternative to derivators is probably to develop a theory of f-categories (like Beilinson did for bounded Z-filtrations) for these "tridimensional" filtrations, but I have not tried to do so. $\endgroup$ May 19 '20 at 14:18

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