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Stream: learning: questions

Topic: basic sheaf question

John Baez (Nov 05 2021 at 14:54):

I want to say something in a paper in a way that doesn't sound stupid to algebraic geometers. Here's what I'd say naturally:

There are various interesting Grothendieck topologies on the category $\mathsf{AffSch}$ of affine schemes, but for any 'reasonable' topology, any sheaf $F: \mathsf{AffSch} \to \mathsf{Set}$ will have the property that if

$X \cong \coprod_\alpha X_\alpha$

then

$F(X) \cong \prod_\alpha F(X_\alpha)$

John Baez (Nov 05 2021 at 15:00):

So, my question is about how people, especially algebraic geometers, tend to think about Grothendieck topologies on the category of affine schemes. Do they automatically assume that if $X$ is a coproduct of $X_\alpha$'s then these $X_\alpha$ form a cover of $X$, or not? I.e., I'm asking if I can just scratch out the word 'reasonable', or if there's some official term I should put there.

Fawzi Hreiki (Nov 05 2021 at 15:58):

You’re just saying that any reasonable topology should be finer than the extensive topology

Fawzi Hreiki (Nov 05 2021 at 15:58):

Which is generally true since the Zariski topology is finer than the extensive topology

Morgan Rogers (he/him) (Nov 05 2021 at 16:01):

Umm depends on how big $\alpha$ is allowed to be. Infinite coproducts aren't respected as often by default.

Fawzi Hreiki (Nov 05 2021 at 16:02):

Ah. I had just assumed that the coproduct was finite.

Reid Barton (Nov 05 2021 at 16:02):

Specifically, infinite coproducts in schemes will be respected (because the summand inclusions are open embeddings) but affine schemes has its own infinite coproducts which are "wrong".

Reid Barton (Nov 05 2021 at 16:04):

(In the same sense that infinite coproducts of compact Hausdorff spaces exist but are "wrong", e.g., the coproduct of $S$ copies of a point is the space of ultrafilters on $S$)

John Baez (Nov 05 2021 at 18:52):

Okay, thanks! I believe I just need the finite case: so, I want a good description of Grothendieck topologies on $\mathsf{AffSch}$ such that any sheaf $F: \mathsf{AffSch} \to \mathsf{Set}$ has the property that if

$X \cong \coprod_{i=1}^n X_i$

then

$F(X) \cong \prod_{i=1}^n F(X_i)$

John Baez (Nov 05 2021 at 18:53):

It sounds like you're saying a good answer is: "Grothendieck topologies finer than the extensive topology".

John Baez (Nov 05 2021 at 18:55):

It sounds like this is something a topos theorist would say, not an algebraic geometer. That's okay, but is there are more algebraic-geometry friendly way to explain the class of topologies I'm talking about? I guess it'd probably help to list some topologies with this property.

John Baez (Nov 05 2021 at 18:58):

Is it true that the Zariski topology, the etale topology, the fpqc topology and the Nisnevich topology all have this property? (I have no idea what some of these topologies actually are - I've just heard people mention them.)

Fawzi Hreiki (Nov 05 2021 at 20:45):

Yes. They are all finer than the extensive topology.

Fawzi Hreiki (Nov 05 2021 at 20:47):

Another cool fact: you can prove descent for any of those topologies by first proving descent for the extensive topology and then proving descent for single map coverings. The idea is that once you have the extensive topology, you only really need to think about covers by single maps because you've already 'fixed coproducts'.

Fawzi Hreiki (Nov 05 2021 at 20:49):

For example, fpqc descent for affine schemes follows from extensive descent (which is obvious since they are representable by rings) and Beck's comonadicity theorem.

John Baez (Nov 05 2021 at 20:51):

Thanks!!!