3
$\begingroup$

Consider a covering family $\{Y_i \to X\}$ and the induced sieve $R \subseteq X$, the subpresheaf of all maps to $X$ that factor through some $Y_i$. The family gives me an induced Cech nerve $C_\bullet$ with $C_n = \bigsqcup Y_{i_1} \times_X \cdots \times_X Y_{i_n}$. One could instead refine this by choosing any cover of the intersection $D_1 \to \bigsqcup Y_i \times_X Y_j$ and then iteratively picking a cover of the coskeleton of the piece of $D_\bullet$ constructed so far to get a hypercover.

I want to do the same thing with sieves. I have an actual application in mind -- a functor that is compatible with injections but not surjections. The naive guess doesn't seem to work: $R \times_X R = R$, so a cover is just another sieve $D_1 \subseteq R$ that also covers $X$. The goal is to compute cohomology by taking the limit of cohomology of the hypercovers as in Verdier's hypercovering theorem.

Since hypercovers modify the fiber products $Y_i \times_X Y_j$, maybe the right thing to do is to let $R$ no longer be a presheaf. Rather than demand $Y_i \times_X Y_j$ is in $R$, the cover $D_1$ would be in $R$. The condition ``if $S \to T$ and $T \in R$, then $S$ is covered by objects of $R$'' would be a replacement in degree 1. I'm not sure what the higher degree conditions should be, but maybe a compatibility of the choices of covering objects from $R$ for finite diagrams.

Q: Can one formulate the notion of hypercovers and descent with respect to them using sieves instead of covering families? Is there a version of the same hypercovering theorem?

I apologize if this is well known. I'm a little lost in the abstract nonsense and really just trying to get a workable version of Verdier's hypercovering theorem.

$\endgroup$
9
  • $\begingroup$ Coincidentally, I have been thinking about the same thing recently. I don't think it can be done with sieves, for the reason you worked out for yourself. But if you think of sieves not as subpresheaves of representable functors but rather as (special) full subcategories of slice categories then a generalisation presents itself: namely, arbitrary full subcategories. I don't know if that's enough to get a Verdier-like formula – if it isn't, generalising to arbitrary functors into slice categories should work, but I guess that's not practical for you. $\endgroup$
    – Zhen Lin
    Commented Mar 12, 2022 at 5:12
  • $\begingroup$ Anyway, I don't quite see a well-defined question in your post. What, precisely, would you accept as an answer? $\endgroup$
    – Zhen Lin
    Commented Mar 12, 2022 at 5:13
  • $\begingroup$ Good point. I edited to make an explicit question. What arbitrary functors into slice categories are you talking about exactly? Thank you. $\endgroup$
    – Leo Herr
    Commented Mar 12, 2022 at 8:23
  • $\begingroup$ I would be glad to explain, but since ultimately I believe the answer to your question as written is no I do not want to waste my time writing something won't satisfy you. $\endgroup$
    – Zhen Lin
    Commented Mar 12, 2022 at 8:57
  • $\begingroup$ If $(U_n)$ is a hypercover of $X$, then each $U_n$ can be seen as a cover of $X$, so this defines a sequence of sieves $(S_n)$ on $X$ with $S_n$ consisting of maps which factors through $U_n$, right? Is “what is a condition for a sequence $(S_n)$ of sieves on $X$ to come from a hypercover? or what extra data must be added to recover it?” a valid restatement of your question? (N.B.: maybe what I'm saying is completely stupid — in fact that's quite likely because each time I try to learn what a hypercover is, I realize I didn't understand it correctly the last time.) $\endgroup$
    – Gro-Tsen
    Commented Mar 12, 2022 at 11:35

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.