21
$\begingroup$

So one of the major problems with the categories of schemes and algebraic spaces is that the "correct quotients" are oftentimes not schemes or algebraic spaces. The way I've seen this sort of thing rectified is either by moving a step up the categorical ladder or by defining some nonstandard quotient to "fix" things. That is, the (sheafy) quotient of a scheme by an étale equivalence relation is an algebraic space, and the (stacky) quotient of an algebraic space by a smooth groupoid action is an algebraic stack (and it is my understanding that these descriptions characterize alg. spaces and stacks up to equivalence).

Derived algebraic geometry gives us a number of powerful tools and has some very nice features: We have a whole array of new, higher dimensional, affine objects (coming from simplicial commutative rings), and a good supply of higher-categorical objects, which we get "all at once", as it were, rather than piecemeal one-level-at-a-time descriptions.

Does the theory of derived algebraic geometry give us enough "n-categorical headroom" (to quote a recent comment of Jim Borger) to take quotients of geometric objects "with reckless abandon" (not a quote of Jim Borger)?

$\endgroup$
5
  • 3
    $\begingroup$ Your guess about alg. spaces and stacks is pretty much on the money. But you are not just taking quotients when doing DAG, but homotopy colimits. The requirement that the groupoid action is smooth when describing an algebraic stack is necessary to make sure the resulting stack has a cover by a map in an appropriate pretopology. It may be that when considering quotients by arbitrary actions you are really 'doing' in homotopy quotients. I'm not sure though if one needs to worry about local representability by simplicial objects for derived stacks or something. $\endgroup$
    – David Roberts
    Feb 7, 2011 at 3:10
  • 1
    $\begingroup$ Yes, I realize that they should be homotopy quotients. That was implied. Taking strict quotients of derived stacks wouldn't make sense, since they're objects of the homotopy category. $\endgroup$ Feb 7, 2011 at 3:14
  • $\begingroup$ @Harry: You can't take a homotopy quotient of objects in the homotopy category. You need a diagram in the "honest" category. $\endgroup$
    – S. Carnahan
    Feb 7, 2011 at 3:39
  • $\begingroup$ @Scott: Yes, I'm familiar with the details. I meant what I said in a very nuanced way that I didn't feel was worth elaborating on. $\endgroup$ Feb 7, 2011 at 3:43
  • 1
    $\begingroup$ I know you are familiar with homotopy quotients, but I was more wondering whether the rationale behind the condition for a stack quotient implies there are any conditions necessary for homotopy quotients. Quite possibly my concerns are misplaced, and one doesn't care about local representability by a simplicial scheme, or what-have-you, but it would be interesting to see what happens in practice. $\endgroup$
    – David Roberts
    Feb 7, 2011 at 3:57

1 Answer 1

13
$\begingroup$

There is more than one way that derived algebraic geometry generalizes ordinary algebraic geometry. The new affines don't help you much with quotients, which are (homotopy) colimits, but they give you well-behaved intersections, which are (homotopy) limits. On the other hand, you can consider functors from affines (new or old) to a category like simplicial sets that has better quotient behavior than plain sets. This gives you a notion of derived stacks, and I believe they behave well under many colimits.

I'm not sure what you mean by "reckless abandon". I tend to make mistakes when I'm not careful with my mathematics, even if I'm looking at derived algebraic geometry.

$\endgroup$
6
  • 2
    $\begingroup$ By reckless abandon, I mean to ask if we can take homotopy quotients without requiring conditions like smoothness on the relation. $\endgroup$ Feb 7, 2011 at 3:39
  • 2
    $\begingroup$ In both old and new settings, you can take such quotients, but the resulting objects tend to be non-algebraic. $\endgroup$
    – S. Carnahan
    Feb 7, 2011 at 3:41
  • $\begingroup$ I guess the question is, do we have the ability (by homotopical black magic) to replace bad relations with good ones? $\endgroup$ Feb 7, 2011 at 3:44
  • 10
    $\begingroup$ I think the classifying stack of a randomly chosen infinite type group scheme is going to be about as badly behaved in derived land as it is in ordinary land, and no amount of black magic will resolve that. Of course such objects might not appear in people's everyday mathematical work. On the other hand, there appear to be moduli problems (e.g., stable maps stacks, J-holomorphic curves) that occur naturally where a good derived intersection theory could allow you to construct fundamental classes without using the word "virtual". $\endgroup$
    – S. Carnahan
    Feb 7, 2011 at 3:57
  • 2
    $\begingroup$ Yes, that is the kind of answer I was looking for. Thanks! $\endgroup$ Feb 7, 2011 at 4:01

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.