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Suppose $\mathcal{D}$ is a triangulated category and that we are given 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. Is it true in general that $\mathcal{D}=D(\mathcal{A})$ is the derived category of the heart of the given $t$-structure on $\mathcal{D}$? If not, is there an easy example that shows why not?

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math.lsu.edu/~pramod/tc/07s-7280/ps9.pdf - this is a problem set that gives a strategy to produce a counterexample. –  Julian Kuelshammer Nov 14 '12 at 21:03
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arxiv.org/abs/0809.4782v2 - and here is an example in the context of dg algebras (Example 27) –  Julian Kuelshammer Nov 14 '12 at 21:10
    
Cf. Pierre Deligne, "À quoi servent les motifs", in Motives, Part 1 (edited by Jannsen, Kleiman, Serre), especially page 154. –  ACL Nov 14 '12 at 22:32
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I think the easiest example I can think of is something like: take the category of dg modules for the dg algebra $Q[x]$, where $x$ is in non-zero degree. The heart of the natural $t$-structure on this is just the category of $Q$-vector spaces. Of course, this is just a special case of Dustin's example 2. –  Sam Gunningham Nov 14 '12 at 22:51
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I see that there is a vote to close with reason "off-topic", but no one has posted a comment with further explanation, and the question seems to be very much within the scope of MO. While we don't have rules about this, it seems like rather strange behavior. –  S. Carnahan Nov 15 '12 at 4:28
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2 Answers 2

up vote 11 down vote accepted

Some examples from topology:

1) If D is the homotopy category Sp of spectra, then D has a canonical t-structure where the truncations correspond to Postnikov towers, so that the heart is the category Ab of abelian groups. The resulting functor D(Ab) --> Sp is the "generalized Eilenberg-Maclane" functor, usually denoted H. It is not fully faithful, since maps Z/p --> Z/p[n] are zero for n>1 in D(Ab) but there are plenty of maps HZ/p --> HZ/p[n] in Sp corresponding to Steenrod operations. It is also not essentially surjective: the cones of such nontrivial maps Z/p --> Z/p[n] cannot be in the image of H, since otherwise they'd have to be isomorphic to Z/p[1] \oplus Z/p[n].

2) For a Q-linear example, let X be a simply connected space, and let D be "local systems of complexes of Q-vector spaces on X up to quasi-isomoprhism" (you can realize this as a full subcategory of the derived category of sheaves of Q-vector spaces on X if you like). Then truncation on fibers defines a t-structure on D with heart the category of local systems of abelian groups on X, which, in view of our hypotheses, is just Q-vector spaces. But the functor F: D(Q-vect) --> D is again neither fully faithful nor essentially surjective -- e.g. maps F(Q) --> F(Q)[n] biject with the n^{th} rational cohomology of X.

There are a couple of ways to make 2) less topological:

a) Combinatorially: you can replace X by a small category (even a poset) which realizes X, e.g. make a category C out of the some triangulation of S^2, then take D to be the full subcategory of the derived category of C-diagrams of Q-vector spaces consisting of objects where each map in C gets sent to a quasi-isomoprhism of complexes of vector spaces.

b) Algebraically: you can realize local systems on $X$ as modules over chains on the based loop space $\Omega X$, and choose $X$ so that $C_\ast(\Omega X)$ has a very simple model as a DGA. For instance if $X = CP^\infty$ then $C_\ast(\Omega X) = C_\ast(S^1) = Q[e]/e^2, |e|=1$; so an algebraic example is given by the derived category of $Q[e]/e^2$ - modules, or more generally modules over (almost?) any nontrivial rational DGA in homologically non-negative degrees with just $Q$ in degree zero,

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Thanks to Sasha's answer for pointing out that my answer is misleading in the places where I talk about "the resulting functor" from the derived category of the heart to D. As Sasha says, this resulting functor is not determined by the triangulated structure on D. However, in practice (e.g. in all the examples I talk about), D does carry enough structure to yield such a canonical functor. –  Dustin Clausen Nov 15 '12 at 15:56
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Here is a precise result which implies the existence of such a functor (the hypothesis on the heart can probably be weakened): suppose that D is a stable $\infty$-category admitting all colimits, and that D carries a t-structure such that the heart is equivalent to the category of A-modules for some associative ring A. Then there is a unique extension of this equivalence on hearts to a colimit-preserving functor Mod_{HA} --> D. Here Mod_{HA} stands for module spectra over the E-M spectrum HA; it is a stable $\infty$-category whose homotopy category is canonically equivalent to $D(A)$. –  Dustin Clausen Nov 15 '12 at 16:05
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The most serious problem is that in general there is no natural functor from $D(A)$ to $D$. To construct one you need an additional structure on $D$. There are several approaches here. One was suggested by Beilinson and gives a notion of a filtered triangulated category. Another important approach uses derivators.

On the other hand, if you have enough structure to construct a functor, then the criterion for it to be an equivalence is rather simple. If I remember right, the necessary and sufficient condition is that each morphism $A \to A'[n]$ in $D$ (with both $A$ and $A'$ in the heart) should be decomposable into a sequence $A \to A_1[1] \to A_2[2] \to \dots \to A_{n-1}[n-1] \to A'[n]$ with all $A_i$ being objects in the heart (in other words, the graded algebra of $Ext$'s should be 1-generated).

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@Sasha Can you give a reference for an approach of the realization functor via deviators? TIA. –  Leo Alonso Nov 15 '12 at 9:45
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@Leo Alonso: to my knowledge, there is, unfortunately, no published reference for the derivator point of view. However, in some sense which can be made precise, this is just a reformulation of the results in Bernhard Keller's paper "Derived categories and universal problems", Communications in Algebra 19 (1991), 699-747. This is because one can prove that any triangulated derivator defines a tower of triangulated categories in the sense of loc. cit. (or one can directly translate Keller's proof in the language of derivators), which is a nice exercise. –  Denis-Charles Cisinski Nov 15 '12 at 10:34
    
Definitely, Denis-Charles is the person to ask! I just know that this is one of the ways to use them. –  Sasha Nov 15 '12 at 10:47
    
Thank you both. In fact, I was wondering about an unbounded version of the realization functor. I'll check Keller's approach. –  Leo Alonso Nov 15 '12 at 11:43
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I just want to point out that if the triangulated category comes from a stable $\infty$-category, then you can produce a natural functor, using essentially the universal property of the derived category (in the $\infty$-categorical context). This is explained in chapter 1 of Lurie's "Higher Algebra." For instance, you can use this property to define the generalized Eilenberg-MacLane spectrum functor (from the derived category of abelian groups to spectra) in Dustin's answer. –  Akhil Mathew Nov 15 '12 at 14:58
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