14
$\begingroup$

I'm trying to read (the introduction of) a survey by Toen on Derived Algebraic Geometry, specifically the "Simplicial Presheaves and Derived Algebraic Geometry" one.

He motivates the introduction of DAG as a means to construct moduli spaces. His example is the moduli of linear representations of a group admitting a finite presentation.

Now, DAG (AFAIU) enlarges the theory of stacks in two directions: a derived bit and a stacky bit. The derived bit concerns replacing rings with more general ring-like objects. The stacky bit comes from using stacks of oo-categories.

His motivation for the `derived' direction comes from the fact that in constructing a moduli space he gets a tensor product, which isn't considering higher Tor's. (I guess I'm sort of OK with that, but for no concrete reason)

His motivation for the `stacky' direction comes from taking quotients of a 'rigidified' moduli problem, which admits a moduli space (the aforementioned tensor product). The problem here is the usual fact that you want to remember the isomorphism groups of objects.

Unfortunately this only motivates us to introduce stacks, not oo-stacks.

So my question is: Can we complicate this example a bit more (but hopefully not too much) so that we need to use higher stacks?

I understand that if we have already enhanced our geometry along the derived direction then we actually need oo-stacks to keep track of notions of equivalence which are weaker than isomorphisms. So a parallel question might be:

if we don't derive our geometry first, do we need to introduce higher stacks?

Comments on the derived are also very much appreciated, thanks!

$\endgroup$
1
  • 1
    $\begingroup$ > if we don't derive our geometry first, do we need to introduce higher stacks? To get to the moduli you need to single out the solutions, what is usually picking from a larger space of candidates by requiring some conditions/equalities. This is a (projective) limit type construction hence right exact; if unsatisfactory one may repair the procedure by deriving on the left. Among the solutions, some are isomorphic, so one may need to quotient by automorphisms. This is a colimit type construction, hence left exact, so may need to right derive to get cocycles, hence sheaves, stacks, higher s. $\endgroup$ Feb 10, 2011 at 19:05

3 Answers 3

20
$\begingroup$

Higher stacks arise very naturally before deriving, whenever we consider for example moduli problems of objects that have not just automorphisms but where the automorphisms have automorphisms. Obviously for this to make sense our objects need to have some categorical structure already! For example, if you want to make a moduli space of stacks of some kind, you naturally find a two-stack -- stacks are 2-categorical in nature, and your automorphisms may well have automorphisms. Or another way to think about this - suppose you have a group acting on a stack, and the action has nontrivial stabilizers when acting on automorphisms of points in your stack. Then the quotient is a 2-stack.. (Or you can take the quotient of a scheme by the action of a group-stack! eg for $G$ abelian take the quotient of a point by $BG$, aka $BBG$.)

[Edit: my favorite example of this (an algebraic 2-stack appearing as a quotient of a stack) is due to Ng\^o in his work on the fundamental lemma (Drinfeld coined the name "Ngo 2-stack"). You take a reductive group G, and consider the stack $G/G$. This stack carries a natural action of the group scheme of centralizers, but this is poorly behaved (centralizers jump in dimensions). Ngo realized there's a FLAT group scheme of dimension = rank of G that acts, the group scheme of regular centralizers (for each $g\in G$ take a regular element in $G$ with the same invariant polynomials, a totally elementary argument shows this actually acts on the centralizer of $g$..) Now take the quotient.. you get an algebraic 2-stack, which contains an open dense piece which is just the SCHEME $h/W$. This construction is behind Ngo's analysis of the Hitchin fibration.]

Thinking this way you might get the feeling that you'll probably never practically need more than 2-stacks. But once you start working in a homotopical context you run into higher stacks very quickly. There's a very natural example where you run into the full structure of $\infty$-stacks immediately, as explained by To\"en and Vaqui\'e: moduli of objects in a derived category. If you have a [nice] abelian category you can define a moduli stack of objects in it --- since objects have automorphisms, families of objects are naturally groupoids so you find a stack. Suppose now you have a derived category [technically you need to "enhance" it to a dg or $A_\infty$ category or something equivalent, but let's ignore this]. In this context the presence of negative self-exts of objects means that automorphisms may have automorphisms which may have automorphisms.. i.e. you find higher homotopy groups of the natural moduli functor (which lands now in simplicial sets, or topological spaces). In any case you can make precise sense of moduli of objects in a derived category, and it is an $\infty$-stack!

(Another example of the same nature is the moduli problem for derived categories themselves! here again you have higher Ext groups in the form of Hochschild cohomologies, which are explained homotopically e.g. in To\"en's Inventiones paper on derived Morita theory.)

In summary once you're in a derived context, your moduli functors naturally land not in sets, not in groupoids (which are the same as 1-truncated homotopy types) but in higher groupoids, aka homotopy types, aka simplicial sets. Such moduli problems may be representable by higher stacks.

$\endgroup$
2
  • $\begingroup$ David, could you please elaborate a bit your example of BBG? What is BBG as a stack if G is finite abelian group? What are coherent sheaves on BBG? Maybe, you know answers on some of these questions: mathoverflow.net/questions/160025/… ? $\endgroup$
    – Klim Puhov
    Mar 30, 2014 at 1:23
  • 1
    $\begingroup$ BBG is the classifying (2-)stack of G-gerbes -- i.e., the functor assigning to a ring the 2-groupoid of G-gerbes (= BG-torsors) over spec R.. ie quotient stack of pt by the abelian group stack BG. I don't think it has nontrivial coherent sheaves (BG doesn't have interesting representations on vector spaces) - there are however interesting sheaves of categories on it... $\endgroup$ Mar 31, 2014 at 17:07
8
$\begingroup$

I think one place higher stacks come up is if you want to make a moduli space of objects in some derived category. Like if instead of moduli of vector bundles you want to do moduli of perfect complexes. Then your groups of automorphisms can have higher homotopy groups coming from possible negative self-Ext's of your objects.

Or again higher stacks could come up in the same classical way that higher categories come up: n-categories form an (n+1)-category. So for instance if you buy that stacks locally modelled on BG are interesting geometric objects, then if you want to consider a moduli of those it will have one more level of stackiness.

I'll put this as a tentative answer, but I still hope someone who's actually worked with this stuff will come along and have a say.

$\endgroup$
1
  • 1
    $\begingroup$ I just thought of another example, which I think Toen did some work on: Azumaya algebras. Since Morita theory makes them naturally into a 2-category, their moduli space could be viewed as a 2-stack. $\endgroup$ Feb 10, 2011 at 18:49
2
$\begingroup$

The moduli of linear categories (or abelian categories) in naturally a $2$-stack. You can take a look at the PhD thesis (in french) of Mathieu Anel (a former student of Bertrand Toën).

When you compute the tangent complex of this $2$-stack then you get the $2$-truncation of the Hochschild complex. If you consider the corresponding derived stack then you get the full Hochschild complex.

By the way, higher stacks are introduced to allow quotients, while derived schemes are introduced to allow fiber products.

$\endgroup$

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.