20
$\begingroup$

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?

$\endgroup$
6
  • 1
    $\begingroup$ math.lsu.edu/~pramod/tc/07s-7280/ps9.pdf - this is a problem set that gives a strategy to produce a counterexample. $\endgroup$ Nov 14, 2012 at 21:03
  • 1
    $\begingroup$ arxiv.org/abs/0809.4782v2 - and here is an example in the context of dg algebras (Example 27) $\endgroup$ Nov 14, 2012 at 21:10
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 14, 2012 at 22:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – S. Carnahan
    Nov 15, 2012 at 4:28
  • 4
    $\begingroup$ A "stupid" counterexample: take $\mathcal D^{\ge 0} = \mathcal D$, $\mathcal D^{\le 0} = 0$, hence $\mathcal A = 0$. $\endgroup$
    – Jakob
    Nov 2, 2016 at 22:26

2 Answers 2

22
$\begingroup$

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,

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$ Nov 15, 2012 at 15:56
  • 5
    $\begingroup$ 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 $D(A)$. $\endgroup$ Nov 15, 2012 at 16:05
  • $\begingroup$ Are there references (introductory, if possible) for the description of (higher?) local systems in terms of modules over chains on the based loop space given in 2b)? Google didn't turn up much that was of use to me. $\endgroup$
    – bavajee
    Feb 7, 2016 at 8:36
16
$\begingroup$

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).

$\endgroup$
10
  • $\begingroup$ @Sasha Can you give a reference for an approach of the realization functor via deviators? TIA. $\endgroup$
    – Leo Alonso
    Nov 15, 2012 at 9:45
  • 3
    $\begingroup$ @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. $\endgroup$ Nov 15, 2012 at 10:34
  • 8
    $\begingroup$ 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. $\endgroup$ Nov 15, 2012 at 14:58
  • 1
    $\begingroup$ Neeman (1991) has a modification of the definition of a triangulated category, where there always exists a functor from the (bounded) derived category of the heart to the ambient triangulated category. $\endgroup$
    – ACL
    Dec 12, 2016 at 22:52
  • 1
    $\begingroup$ @AT0: No, sorry, I don't remember this (I have read this 20 years ago). It could be in one of the versions of Gelfand-Manin, or maybe in the Beilinson's paper. $\endgroup$
    – Sasha
    Jun 10, 2021 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.