6
$\begingroup$

I am reading a paper that proves the representability for certain functors whose domain is the category of superschemes. The paper claims that to prove representability of functors (or possibly just the functor of interest) in this category we must first place an etale topology on the category.

More broadly, why is this the case? What about certain categories (or functors) requires an etale topology for proving representability problems?

I don't understand the motivation behind this, possibly because I have never really worked with representability problems in much depth.

$\endgroup$
1
  • 2
    $\begingroup$ It would help to share the title of the secret paper you're reading, as then we wouldn't have to guess the context or the particular approach in which to coach an answer. $\endgroup$ Jan 19, 2019 at 21:59

1 Answer 1

7
$\begingroup$

I don't know much about superschemes, but would like to share some viewpoints for representability of functors on ordinary schemes. A theorem of Grothedieck states that a representable functor is a sheaf in fpqc topology, hence a priori it is also a sheaf in coarser topologies such as fppf topology and étale topology (of course in Zariski topology, too). This has two folds of implications:

  • By passing from a coarser topology to a finer topology, it is harder and harder for your functor to be a sheaf, because there are more and more open covers and it is harder and harder to satisfy the gluability that being a sheaf requires. You can view this as a series of harder and harder "trials" for your functor to be representable. If in étale, fppf, or fpqc topology you find out your functor stops being a sheaf, then it is definitely not representable. For example, let $k$ be a field, the functor that associates $\Gamma(X, \Omega_{X/k}^1)$ to a $k$-scheme $X$ is a sheaf even in étale topology, but it stops being a sheaf if you go as far as fppf topology because inseparable covers are allowed, hence it is not representable.
  • On the other hand, if you "sheafify" your functor in a finer and finer topology, then the sheaf gets bigger and bigger, and is closer and closer to being representable. Sheafification in some topology finer than Zariski topology sometimes is inevitable. For example, if $X$ is the conic $x^2+y^2+z^2 = 0$ in real projective plane, then the relative Picard functor $\mathrm{Pic}_{X/\mathbb{R}}$ is not an étale sheaf, but its étale sheafification $(\mathrm{Pic}_{X/\mathbb{R}})_{\mathrm{\acute{e}t}}$ is representable.
$\endgroup$

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.