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I'm curious to find out where the viewpoint of higher categories may be useful so here is a somewhat vague question (which may or may not have a reasonable answer).

Given a triangulated category, one can consider the set of all possible t-structures on it. Simple examples where one can compute things by hand indicate that this is something complicated but not hopelessly so. See for example the paper http://arxiv.org/abs/0909.0552 by Jon Woolf which describes a three parametric family of t-structures on the constructible bounded derived category of $\mathbf{P}^1(\mathbf{C})$ stratified by a point and its complement. Some of these t-structures are more interesting than others and there is one that is the most interesting of them all since by taking the bounded derived category of its heart one gets back the triangulated category one started with. (For that t-structure the heart is the category of perverse sheaves on $\mathbf{P}^1(\mathbf{C})$.)

On the other hand, the set of t-structures on a triangulated category is interesting since there lurks somewhere the conjectural motivic t-structure whose existence implies Grothendieck's standard conjectures. See the recent paper http://arxiv.org/abs/1006.1116 by Beilinson.

On the triangulated categories page of the n-category lab website it says "Therefore, all the structure and properties of a triangulated category is best understood as a 1-categorical shadow of the corresponding properties of stable (infinity,1)-categories". See http://ncatlab.org/nlab/show/triangulated+category. Note that this is quite a strong statement, since it is referring to all, and not just some, properties and structure of a triangulated category.

So I'd like to ask: is there a higher categorical analog of a t-structure? More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable $(\infty,1)$ category?

upd: as Mike points out in the comments, the answer to the first question is yes and it is given by proposition 6.15 of Lurie's Stable Infinity Categories. The second, more "philosophical" question remains.

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I'm far from an expert, but I thought that stable infinity categories were the higher categorical analog of t-structures. My understanding is that it's actually a strength of the higher categorical theory that being stable is a property rather than an extra structure (as in the case of a t-structure on a triangulated category). You're probably already aware of arxiv.org/abs/math/0608228. –  Mike Skirvin Jul 4 '10 at 18:51
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Mike -- thanks! The first question seems to be answered by proposition 6.15 there. But even for stable $(\infty,1)$ categories a t-structure is still an extra structure, not property. –  algori Jul 4 '10 at 19:21
    
I guess you are implicitly assuming your triangulated categories are small? It is certainly not the case otherwise that there is necessarily a set of t-structures and things do seem a little hopeless in that case. –  Greg Stevenson Jul 4 '10 at 21:14
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Greg -- yes, to be completely precise one should fix a Grothendieck universe and require that set of objects belong to it; then so will the set of t-structures. But somehow it seems to me that these set theoretical subtleties are not very important here, which is why I'm willing to sweep them under the carpet. –  algori Jul 4 '10 at 21:52
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algori: I think I misread the point of your original question: actually you want to formulate things in a purely infinity-categorical way? If so, of course it's reasonable, as you wrote, to define a t-structure as a certain kind of localization as in DAG I Prop. 6.15. I think it's worth keeping in mind, though, that this structure is completely encoded at the homotopy level. Admittedly, I have no idea whether all the structure of the collection of all t-structures still lives there. –  Thomas Nevins Jul 6 '10 at 20:56

3 Answers 3

I wanted to contribute something because nobody's really explained why a t-structure on a stable $\infty$-category is the same thing as a t-structure on its homotopy category. It's not just because someone defined it as such in Definition 1.2.1.4 of Higher Algebra; it's because the most natural generalization is no generalization at all.

Roughly speaking, a t-structure on a triangulated category should have the following three properties:

  • h1. It determines a full subcategory $\mathcal D_{\geq 0}$ closed under the shift functor $\Sigma$.

  • h2. $hom(X,Y) = 0$ if $X \in \mathcal D_{\geq 0}$ and $Y \in \mathcal D_{\leq -1}$.

  • h3. Any object fits into an exact triangle, sandwiched between an object in $\mathcal D_{\geq 0}$ and $\mathcal D_{\leq -1}$.

If you think about the algebraic manipulations for which one utilizes t-structures, the natural generalization of a t-structure to a stable $\infty$-category ought to be as follows:

  • 1$.$ It determines a full subcategory $\mathcal C_{\geq 0}$ closed under the shift functor $\Sigma$.

  • 2$.$ $hom(X,Y)$ is contractible if $X \in \mathcal C_{\geq 0}$ and $Y \in \mathcal C_{\leq -1}$.

  • 3$.$ Any object fits into an fiber sequence, sandwiched between an object in $\mathcal C_{\geq 0}$ and $\mathcal C_{\leq -1}$.

If $\mathcal C$ is stable, one defines a triangulated structure on its homotopy category $\mathcal D = ho \mathcal C$ by demanding that its exact triangles are precisely those arising from fiber sequences. Then we see immediately that (1)-(3) imply (h1)-(h3). That is, we have a map from "oo-categorical" t-structures on $\mathcal C$ to t-structures on $ho \mathcal C$.

What might be surprising is that the other direction holds, but it's actually an easy exercise to show that any collection of data satisfying (h1)-(h3) uniquely determines data satisfying the properties (1)-(3). In a nutshell: (h1) implies (1) obviously. (h1) and (h2) together imply (2) because you can write a fiber sequence for $X[i]$ and $X[i+1]$, then apply the hom functor to get a fiber sequence of mapping spaces. The long exact sequence of homotopy groups will prove that every homotopy group of hom(X,Y) vanishes. Finally, by definition of exact triangles for $ho\mathcal C$, (h3) also implies (3).

So we have an interesting phenomenon where the set of t-structures can't distinguish between two oo-categories with equivalent homotopy categories. This is one of the main reasons that Bridgeland stability conditions don't have an obvious generalization to see higher-homotopical structures beyond the homotopy category.

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If I think about the algebraic manipulations for which one utilizes t-structures, what comes to my mind are the torsion and torsionfree classes of a torsion theory. And when I think to them, what comes to my mind are reflective factorization systems. I appreciate your explanation, but I have to disagree that as you say "the natural generalization [here] is no generalization at all": the ambient where axioms (1,2,3) live says $\Sigma$ is a genuine (homotopy) colimit, in opposition to the classical setting; this makes (has made, working out our paper) a incredible difference [continue] –  tetrapharmakon Aug 25 at 7:12
    
Since for example you don't have to ask that $\mathcal C_\ge$ / $\mathcal C_\le$ is closed under $\Sigma$ / $\Omega$ any more; it's automatic from its being coreflective / reflective. Another point, maybe a minor point: what is a t-structure? This could be a personal opinion (undoubtedly it is) but definitions should be self-explanatory; when they are not, something is missing. And, this is absolutely not an arrogant statement, but in our setting the definition of t-structure is self-explanatory [continue] –  tetrapharmakon Aug 25 at 7:18
    
t-structures are built up from those data (a pair of subcategories with properties blah blah) since they come from factorization systems: orthogonality of the objects come from orthogonality between arrows, closure under loop/suspension comes from (co)reflectivity, as well as the presence of suitable fiber sequences; t-structures abound since in a stable $\infty$-category it is extremely simple to be a simple factorizations system (Remark 3.12 of our work). Not to mention all the other results: the heart of a t-structure is abelian since you do the only possible thing [continue] –  tetrapharmakon Aug 25 at 7:22
    
like in the most self-explanatory Algebra proofs: you have some data, you use it in the only meaningful way. And "Harder-Narashiman filtrations"... These are blatantly Postnikov towers, which are a well known byproduct of factorization systems in $\omega$ stages. From our point of view it's rather obvious that these decompositions exist, induced from the given factorization system / t-structure: every object admits a "tower" $X\to A_1\to \dots A_n\to \dots \to Y$ where each $A_i$ has fiber $K_i[n_i]$ and $K_i$ lives in the heart. –  tetrapharmakon Aug 25 at 7:27
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here is Domenico's opinion (he has to post from his phone but didn't succeed): I don't think Hiro is meaning that the $\infty$-point of view is saying nothing here. On the contrary, I think he is pointing out that by speaking of the $\infty$-categorical version one is really speaking of the very same thing as the classical version on the homotopy category. Not something more general nor more special. It is precisely this that tells that one can use $\infty$-eyes to look at the old thing. And that with $\infty$-eyes one can see better than with 1-eyes, no doubt. [continue] –  tetrapharmakon Aug 25 at 10:22

So I'd like to ask: is there a higher categorical analog of a t-structure?

As Mike Skirvin pointed out in a comment, higher categorical analog of t-structures have been introduced by Lurie. A more up-to-date reference might be Higher Algebra ($\S$ 1.2.1).

More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable (∞,1) category?

I guess the answer can be found at the same place. There, Lurie says that "there is a bijective correspondence between $t$-localizations of $\mathcal C$ (a stable $\infty$-category) and $t$-structures on the triangulated category $h\mathcal C$.

The higher categorical point-of-view also seems to be useful to understand the yoga of derived functors in a more conceptual way. In Section 1.3 of the same reference (Higher Algebra) it is explained that if $\mathcal A$ is an abelian category with enough injectives, then its derived $\infty$-category $\mathcal D^-(\mathcal A)$ is stable, admits a $t$-structure, has homotopy category the standard derived category, and satisfies the following universal property: there is a canonical equivalence of abelian categories $\mathcal A\to \mathcal D^-(\mathcal A)^\heartsuit$, and if $\mathcal C$ is a stable $\infty$-category with a left-complete $t$-structure then any right exact functor $\mathcal A\to\mathcal C^\heartsuit$ extends (in an essentially unique way) to an exact functor $\mathcal D^-(\mathcal A)\to \mathcal C$.

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I feel the urge to answer you even if I'm not an expert (!), since now I can say that yes, one can do extremely better than referring a t-structure to its homotopy category.

In this work D. Fiorenza and I prove that t-structures on a stable $\infty$-category $\cal C$ are uniquely determined by a reflective, normal factorization system on $\cal C$.

In a few words, a factorization system $(\cal E,M)$ is reflective when both classes satisfy the 3-for-2 property (the same of weak equivalences in a model category), and normal when the class $\cal E$/$\cal M$ satisfy a rather peculiar pullback/pushout stability property, which can be packaged in a single self-dual request: denoting $R,S$ respectively the reflection and coreflection functors associated to the reflective factorization $(\cal E,M)$ (see nLab page for an explanation: the correspondence whose $\infty$-categorical analogue we exploit was first presented in the Cassidy-Hebert-Kelly paper "Reflective subcategories, localizations and factorization systems"), this factorization is normal in a stable category $\cal C$ iff the square $$ \begin{array}{ccc} SA & \to &A \\ \downarrow && \downarrow \\ 0 &\to & RA \end{array} $$ is a pullback/pushout. Read between the lines and notice that $R,S$ are nothing more than the truncation functors $\tau_{\ge 0}, \tau_{<0}$ of the t-structure, and the universality request for a normal, reflective factorization system is telling you that "every object lies in a distinguished triangle blah blah". And now, fortified by $\infty$-category theory, re-read the definition of t-structure with a new eye :)

Theorem 3.13 in the paper is the gist of the discussion: notice that the bijection of the correspondence established there holds for simple, well-motivated categorical reasons. Proofs are incredibly easy-to-follow, so easy that an entire section contains exercises for the reader!

And finally, there's no need to refer to the homotopy category even in the definition of the heart of the t-structure, which can be characterized in terms of the factorization system alone: abelianity of the heart is now automatic (see Exercises 4.1, 4.2) as a consequence of "adjacency" (Remark 4.2) of the torsion and torsion-free classes of the t-structure.

A more precise discussion of the points we leave as exercises will appear in a subsequent paper (denoted as [FLa] in the bibliography), together with examples from algebraic topology (Bousfield "localization" is a t-structure on spectra), algebraic geometry (I guess I have to thank the OP for giving the references above) and other realms: hope you'll be one of our readers!

PS: See also the discussion we opened on the nLab.

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I noticed that in one of the discussions it is mentioned that semi-orthogonal decompositions form an example of a t-structure. I don't understand, what is the heart of the t-structure in this case? –  bananastack Aug 24 at 16:22
    
Are you saying that a semiorthogonal decomposition is not a t-structure where the reflection and coreflection $\tau_\ge, \tau_\le$ are exact? –  tetrapharmakon Aug 24 at 17:08
    
I do not know what those words mean. In the nlab discussion I found the following: "(semi?)orthogonal decompositions are “well-behaved” t-structures". So my question is: if <A,B> is a semi-orthogonal decomposition, what is the heart of the relevant t-structure? I am just curious. –  bananastack Aug 24 at 17:16
    
Oh, I am sorry for the misunderstanding. I don't know the answer yet, but to the best of my knowledge a semiorthogonal deocmposition is simply a t-structure where the reflection and coreflection (the truncation in the two directions) are exact functors; this entails that the two "adjacent" subcategories of the t-structure are triangulated (or $\infty$-stable) subcategories. I warmly invite anybody to contradict me if I'm wrong, I'm still learning! –  tetrapharmakon Aug 24 at 17:26
    
If this is true, I don't know what this should entail for the heart. But thank you for the interesting question, I'll surely think about it. –  tetrapharmakon Aug 24 at 17:30

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