14
$\begingroup$

I've heard that under certain assumptions on an algebraic variety $X$ there exist a quiver $Q$ for which there is an equivalence $$D^b(\mathsf{Coh}(X))\simeq D^b(\mathsf{Rep}(Q))$$ between the corresponding bounded derived categories. My (naive) questions about this fact are the following:

1) Is there any intuitive reason for the existence on this relation between the geometry of $X$ and the combinatorics of $Q$? For instance, given some variety, is there a way to make a guess of what quiver could realize this equivalence?

2) Is there a generalization for this result when $X$ is a Deligne–Mumford stack? If this is the case, how is the structure of the stabilizers reflected in the quiver?

$\endgroup$
  • 2
    $\begingroup$ What are these assumptions? $\endgroup$ – Martin Brandenburg Apr 28 '17 at 21:45
  • 3
    $\begingroup$ You probably want $Q$ to be a quiver with relations (i.e. a finite linear category). 1) When there is an equivalence like this, the a vertex $v$ of the quiver corresponds to a bounded complex of sheaves on $X$, $F_v$ such that ${\rm Hom}(F_v, F_w)$ has cohomology concentrated in degree $0$. And the number of arrows between $v$ and $w$ is the dimension of this ${\rm Hom}$ space. 2) You can get these sorts of equivalences from the formalism of (derived) morita theory, which works just as well for any derived category. $\endgroup$ – Phil Tosteson Apr 28 '17 at 21:55
  • $\begingroup$ Honestly, I don't know almost nothing about this stuff. I've heard that for the projective line the Kronecker quiver gives rise an equivalence like this, and that something similar happens for other varietites. I just wondering what could be the reason behind such a thing. $\endgroup$ – user95283 Apr 28 '17 at 22:10
  • 1
    $\begingroup$ A geometric property of $X$ which does give rise to a quiver with relations is having a strong exceptional collection. Maybe relevant post: mathoverflow.net/questions/114458/… $\endgroup$ – Andrea Ricolfi Apr 28 '17 at 22:58
6
$\begingroup$

This is more or less a longer version of Phil's comment. Unfortunately I don't know definitive references, so there might be some technical errors below. A possible reference is Keller's paper.

Here's the most high level statement: for any (pretriangulated, which just means formally add all cones, shifts, and sums) dg-category $C$ (there is a natural way to enrich $\mathsf{Coh}(X)$ as a dg-category) with a chosen set of compact generators $E_\alpha$ (in the sense of a weak triangulated generator of the category of compact objects), letting $Q$ be the associated full dg-category with these generators as objects, we have that the dg derived category of $Q$-modules is dg-equivalent to $C$. The functor $C \rightarrow Q\text{-mod}$ is $\operatorname{Hom}(E_\alpha,-)$. Note all $\operatorname{Hom}$ I write are derived.

Fixing this choice of generators $E_\alpha$, we get something "like" a quiver (a category "is like" a quiver, and a module over a category "is like" a representation) but is not a quiver in the classical sense. In particular (warning -- the further down on this list you go, the further out of my depth I go):

  • How small do you want the set of nodes to be? Each node of the quiver corresponds to a generator (i.e. one of the $E_\alpha$), so you might want finitely many (compact) generators. For example, the category $\mathsf{Rep}(G)$ for $G$ semisimple does not have finitely many compact generators (something like the direct sum of all irreducibles is not compact).
  • Do you want finiteness on the arrows? Then you might ask that the $\operatorname{Hom}$ spaces between generators $\text{Hom}(E_\alpha, E_\beta)$ be finite-dimensional. For example, for $A$ an infinite-dimensional algebra, $A$ itself is a compact generator for the category of $A$-modules, but its self-$\operatorname{Homs}$ will be infinite-dimensional. I'm not sure about precise statements, but examples where this actually happens tend to be proper varieties.
  • If by representations of a quiver $Q$ you want a dg derived category of some abelian catagory of representations, you want the dg $\operatorname{Hom}$ between your generators to be cohomologically concentrated in degree zero. In general, the dg derived category of modules over a dg-algebra $A$ might not be equivalent to the dg derived category of modules over any classical algebra. You can achieve this by insisting that the generators are projective or flat objects, for example.
  • Do you want the category of coherent sheaves, or the category of perfect complexes? The "large" versions of these are respectively, ind-coherent sheaves and the more familiar quasicoherent sheaves. For example, if $R = k[x]/x^2$, $R$ is a generator for perfect complexes, but $R/x \simeq k$ is a generator for the coherent category. If $X$ is smooth, these two categories coincide. On the right hand side, you always get $Q\text{-perf}$ as the compact objects.
  • Note that a given category might be equivalent to different quivers. It depends on your choice of generators.
  • For $X = \mathbb{P}^1$, you can check that $\mathcal{O}$ and $\mathcal{O}(-1)$ generate the category (i.e. that derived $\operatorname{Hom}(\mathcal{O} \oplus \mathcal{O}(-1), \mathcal{F}) \simeq 0$ if and only if $\mathcal{F} \simeq 0$), and then compute the derived $\operatorname{Homs}$ between these.
  • If you can define your category on a DM stack, then the same formalism should go through.
  • Also note that this equivalence is generally not monoidal (with the usual monoidal structure on quiver representations) nor $t$-exact. Reconstruction theorems from the derived category (e.g. Bondal-Orlov) that I know tend to require the monoidal (edit: or some other extra) structure, so I'm not sure how much geometry you can recover without it. See, for example, this MO question.
  • For example, the category of coherent sheaves on $BG$ is equivalent to the category of $G$-representations. If $G$ is semisimple (and the field is characteristic zero?) the resulting quiver is a bunch of nodes corresponding to each irreducible representation with no interesting arrows. You can't reconstruct $BG$ or $G$ from this not-very-interesting dg category without some extra structure.
  • There are probably results which say that certain geometric properties of $X$ will imply that such a quiver exists (e.g. see Andrea's comment), but I don't know them.
| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ A minor comment: the Bondal-Orlov reconstruction theorem does not require the monoidal structure (this is what makes it so surprising!). $\endgroup$ – Piotr Achinger Apr 29 '17 at 9:49
3
$\begingroup$

Let me try to answer in a slightly different way, with a few corrections (unless I'm the one that's wrong). It is a very general statement that, given a compact generator of a stable model or $\infty$-category, the categories of modules (appropriately defined) over the endomorphisms of that generator are equivalent to the original category. This goes back to Rickard's derived Morita equivalence, and the most general formulation is probably that of Schwede and Shipley-- it's also almost surely in Lurie's book.

So, in that sense, $Coh(X)$ is a bit of a red herring -- you don't need to know that your derived category comes from a variety, a scheme or whatever. You just need a compact generator. Once you have that, the question becomes, when do the endomorphisms of the generator look like the path algebra of a quiver. Specializing the the situation of a dg-category, some of the conditions are:

1) The dg-algebra of endorphisms must be quasi-equivalent to an ordinary algebra. In other words, the generator can't have any self-exts outside of degree zero.

2) The identity of the path algebra quiver decomposes into a set of idempotents associated with the nodes of the quiver. Associated to these idempotents are a set of projective modules. The direct sum of these projective modules is the algebra itself. Thus, the generator must decompose into a direct sum. It's the direct sum of these objects that is the generator; each is not a generator by itself.

3) Given two projective modules, $Hom(P_i,P_j)$ is isomorphic to the space of paths between those two nodes (not arrows). This need not be finite-dimensional if you are ok with your quiver having loops. However, a quiver also has a distinguished set of simple representations associated with the nodes that satisfy $Hom(P_i,S_j) = \mathbb{C}^{\delta_{ij}}$. These are just the one-dimensional representations associated with the nodes the quiver.

The dimension of the space of arrows between two nodes of the quiver are given by $Ext^1(S_i,S_j)$. So, if you want a finite number of arrows, that better be finite dimensional. In fact, you can recover the entire quiver algebra by looking at the self-ext dg/$A_\infty$-algebra.

Relating this to the comments, a full strong exceptional collection exactly gives you (1) and (2), but it's not a necessary condition -- you can get quivers for noncompact varieties, for example, a situation that is very common in studying D-branes at singularities in string theory.

One subtlety I've neglected so far is the difference between the unbounded derived category and the bounded derived category. I'm getting more out of my long-atrophied depth here, but I believe the equivalence from the unbounded category descends to the bounded one if perfect complexes are the same as compact objects. This is true, I think, for quasi-compact separated schemes. For stacks, you can see results in Ben-Zvi, Frances and Nadler.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think our answers basically agree :) A comment on the BZFN paper: what it says is that for perfect stacks, the compact objects in QCoh(X) (a categorical property) agree with the perfect objects (which are defined in terms of "functor of points" on X -- i.e. we know what perfect objects should be on an affine scheme S, so we say a qcoh sheaf on X is perfect if it pulls back to a perfect object along any S-point). This formalism of categorical generators naturally requires us to think about compact objects, so it's nice when compact obejcts have such a geometric characterization $\endgroup$ – Harrison Chen Apr 29 '17 at 23:42
  • $\begingroup$ This is related to the Coh(X) vs. Perf(X) I mentioned -- Coh(X) (meaning the bounded derived category with coherent cohomology) are not the compact objects of QCoh(X) when X is not smooth, even for classical varieties. Instead, Coh(X) are the compact objects of IndCoh(X), which is not such a familiar category (but useful!). So, when X is not smooth, taking ind-completions gives IndCoh(X) = Q-mod instead if we took compact generators for Coh(X) to build Q. On the other hand, in my experience when X is not smooth, I wouldn't expect a nice set of projective generators for Coh(X) anyway $\endgroup$ – Harrison Chen Apr 29 '17 at 23:45
  • $\begingroup$ Was really only objecting to two things in the answer. You start with a single generator, not a collection, and that generator has a direct sum decomposition. Each component isn't a generator by itself (unless I've really forgotten a lot). For example, on $\mathbb{P}^1$, you have the generator $\mathcal{O}\osum\mathcal{O}(1)$, but neither component is a generator by itself. $\endgroup$ – Aaron Bergman Apr 30 '17 at 1:11
  • $\begingroup$ I think the reason I want to allow for a collection of generators (which "generate together" -- I don't mean that each object individually generates the category) is that when only an infinite collection of compact objects can generate the category (e.g. for $X = BG$ for $G$ nonfinite reductive, the infinite collection of irred. rep), the (infinite) sum of these is no longer a compact object. For example, the "free" object corresponding to one of these generators should be (by definition) in $Q-perf$, so we really want the generators to be compact. $\endgroup$ – Harrison Chen Apr 30 '17 at 1:21
  • $\begingroup$ I think we just had a terminology mixup -- I interpreted you as saying that you had a collection of generators, not a collection that generates. $\endgroup$ – Aaron Bergman Apr 30 '17 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy