8
$\begingroup$

In the definition of an adic space, usually there is a presheaf defined by first saying what it is on a particular basis of the topology of the underlying space, the so called rational subsets. One then extends this to arbitrary opens by taking the limit over all the rational subsets inside the given open. However, in all the references I looked into so far, it is then quickly pointed out that this is in general not a sheaf, followed by a list of particular cases in which it is one.

In Algebraic Geometry, there are many occasions where some construction does not give a sheaf and one simply forces it to be one by saying "Sheafify!" So the Question is:

-Why doesn't one sheafify the structure presheaf of an adic space? Is there no sheafification functor in this case? If not, what goes wrong?

$\endgroup$
2
  • 2
    $\begingroup$ Presumably you can sheafify but you'll get the wrong global sections. $\endgroup$ Nov 11, 2015 at 11:45
  • $\begingroup$ Thank you, that sounds like a good idea. Do you (or somebody else) have an example at hand where this happens? Is this automatic if the presheaf is not a sheaf? $\endgroup$
    – jorst
    Nov 11, 2015 at 18:02

1 Answer 1

2
$\begingroup$

There is actually a sheafification process used in classical construction of rigid geometry, which is used to pass from the weak G-topology to the strong one (see section 9.2.2 of Bosch, Güntzer, Remmert's Non-Archimedean Analysis, for instance).

As to why people don't use it in the case of non-sheafy spaces, I would say that if depends on what you want to do. If it's enough for you to know the stalks of the structure sheaf, then fine. But, as Dylan Wilson pointed it out, you are likely not to end up with the global sections you expect. For an explicit example, see Theorem 3.15 in Mihara's paper "On Tate Acyclicity and Uniformity of Berkovich Spectra and Adic Spectra" (http://arxiv.org/abs/1403.7856).

$\endgroup$
3
  • $\begingroup$ Thank you for the answer and the example. One maybe trivial question as I'm new to this: Does the sheafification of rigid geometry carry over directly to adic spaces or is this just a plausibility argument? (It seems to be the "+" construction wich I know to work for set-valued presheaves on an arbitrary site, so I guess the question is if there are some technical subtleties with Huber rings instead of sets as the target category) $\endgroup$
    – jorst
    Nov 17, 2015 at 13:01
  • $\begingroup$ Well, I merely wanted to give you some reference where such kind of things are done in a setting close to yours. In general, if you want to sheafify with values in f-adic rings or Tate rings or other things, there certainly are some issues. You certainly need at least finite limits and colimits as well as filtered colimits in your category and probably more... I have to admit that I do not know about the precise conditions. $\endgroup$ Nov 18, 2015 at 10:03
  • 1
    $\begingroup$ Note: since posting of this answer Mihata has posted a new version of the paper in which numeration has changed; Theorem 3.15 is now Theorem 4.6. $\endgroup$
    – Wojowu
    Nov 30, 2019 at 19:07

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.