2
$\begingroup$

I have a simple question about the conservativity of stalks of ind-constructible sheaves. Let $X$ be a topologically noetherian scheme, $S$ a set of geometric points of $X$ corresponding bijectively to the scheme-theoretic points of $X$, $\Lambda$ an algebraic extension of $\mathbf{Q}_\ell$ or the ring of integers of such, $I$ a filtered category (which I will assume is a directed poset), $\operatorname{Cons}_X(\Lambda)$ the abelian category of constructible $\Lambda$-sheaves on $X_{\text{pro-étale}}$ in the sense of Bhatt-Scholze §6.8, and let $(\mathcal F_i)_{i\in I}$ correspond to a functor $I\to \operatorname{Cons}_X(\Lambda)$ with colimit $\mathcal F$ computed in $\operatorname{Mod}(X_{\text{pro-étale}},\Lambda)$, the category of sheaves of $\Lambda$-modules, where $\Lambda$ here and often below also denotes the sheaf $\Lambda_X$ associated to the topological ring $\Lambda$.

My question is, if $\mathcal F_x=0$ for every $x\in S$, is $\mathcal F=0$? (Here $\mathcal F_x:=\Gamma(x_{\text{pro-étale}},x^*\mathcal F$.)

I believe I can prove this is true (argument below), but what is giving me some doubts is Warning 2.3.4.10 of the paper Weil’s conjecture for function fields [PDF], which says that there is a non-zero ind-constructible sheaf which vanishes at all closed points. This is relevant since Corollary 3.51 of the recent paper Constructible sheaves on schemes and a categorical Künneth formula connects $\operatorname{Ind}(D_{\mathrm{cons}}(X,\Lambda))$ to filtered colimits of pro-étale sheaves. Admittedly, this warning only deals with closed points, but so far I haven’t been able to construct a non-zero ind-constructible sheaf that vanishes at all closed points. Here is my purported (simple) ‘proof’ that the answer to my question is ‘yes’:

Proof It would suffice to show that for every $i\in I$, there is a $j\geq i$ so that $\varphi_{ij}:\mathcal F_i\to\mathcal F_j$ is zero. Suppose we could find a nonempty open $U\subset X$ over which this is true. Then we could replace $i$ by $j$ and $X$ by $\operatorname{supp}\operatorname{coker}(\ker\varphi_{ij}\to\mathcal F_i)$, and we'd be done by noetherian induction.

To find $U$: let $\eta$ be a geometric generic point of $X$. We can find a $j\geq i$ so that $\varphi_{ij}:\mathcal F_i\to\mathcal F_j$ induces zero on stalks at $\eta.^\dagger$ There is some connected neighborhood $U$ of $\eta$ so that both $\mathcal F_i$ and $\mathcal F_j$ are locally constant of finite presentation over $U$. Then the same is true of $\ker\varphi_{ij}$, and as $(\ker\varphi_{ij})_\eta=(\mathcal F_i)_\eta$, $\ker\varphi_{ij}|_U=\mathcal F_i|_U.^\ddagger$ $\square$

$\dagger$: Both $\eta^*$ and $\Gamma(\eta_\text{pro-étale},-)$ commute with colimits, the latter since $\eta$ is connected and w-contractible. For each $i$, there is some neighborhood $U_i$ of $\eta$ restricted to which $\mathcal F_i$ is (pro-étale) locally of the form $M_i\otimes_\Lambda\Lambda_X$, where $M$ is a $\Lambda$-module of finite presentation and I've dropped the underlines for constant sheaves ($\Lambda_X$ is not literally a constant sheaf). (This is what I mean by ‘locally constant of finite presentation.’) Note $(M_i\otimes_\Lambda\Lambda_X)_\eta=M_i$ as $\Lambda_X(\eta)=\Lambda$ and $\eta$ is w-contractible.

$\ddagger$: When $\Lambda$ is a field, this follows from Corollary 6.8.5 of [BS] and that $(-)_x$ is an exact functor (of abelian categories). Otherwise, modulo possibly shrinking $U$, replace Corollary 6.8.5 with Proposition 6.8.11.

It remains unclear to me how this doesn’t contradict Warning 2.3.4.10. Thanks for any clarification of the matter.

$\endgroup$
5
  • $\begingroup$ Is this direct for ind-lisse sheaves? My understanding is poor, but it seems that stratified versions are not too different from non-stratified versions. $\endgroup$
    – Z. M
    Commented Dec 8, 2022 at 6:56
  • $\begingroup$ There is no contradiction yet, right? Gaitsgory–Lurie only talk about stalks at closed points, and indicate (without any details) that this is not enough, and that the problem is related to $\ell$-completeness. Of course it would have been much better if they gave an actual example instead of a claim without any details ― you could write to them to ask for clarification. $\endgroup$ Commented Dec 8, 2022 at 9:21
  • $\begingroup$ @R.vanDobbendeBruyn And seemingly, G–L does not say that there are counterexamples for ind-constructible sheaves. $\endgroup$
    – Z. M
    Commented Dec 8, 2022 at 10:38
  • $\begingroup$ @Z.M I think their $\ell$-adic sheaves are by definition ind-constructible; see Definition 2.3.4.1. (To some extent, this is the most intuitive notion, as any usual étale sheaf is a filtered colimit of constructible ones. I don't know if $\operatorname{Ind}(X_{\text{ét}}^{\text{cons}}) \to X_{\text{ét}}$ is an equivalence, but I suspect not.) $\endgroup$ Commented Dec 8, 2022 at 14:52
  • $\begingroup$ @R.vanDobbendeBruyn Thanks. The last equivalence seems to hold, which is mentioned by OP: [ Remo–Richarz–Scholbach, Rem 3.52], also in Bhatt–Scholze. $\endgroup$
    – Z. M
    Commented Dec 8, 2022 at 15:01

1 Answer 1

0
$\begingroup$

I'm not sure how related this is with the ind-sheaf framework of Kashiwara-Schapira, but this question is emphasized several times in their 2001 book "Ind-Sheaves" ( e.g.: examples 4.2.17 & 4.2.18, 4.3.20, and exercise 4.8).

They define a notion of the "ind-stalk" of an ind-sheaf $K \in I(k_X) := Ind(Mod^c(k_X))$ at a point $x \in X$, denoted ${}_x K$ (in Section 4.3), and prove in Proposition 4.3.21 exactly the type of result you seem to be looking for: If $f: L \to K$ is a morphism of ind-sheaves of $k_X$-modules for which the induced morphism on ind-stalks ${}_x f: {}_x L \to {}_x K$ is zero for all $x \in X$, then $f$ is the zero morphism.

$\endgroup$
1
  • $\begingroup$ I don't think that it is the same. Kashiwara–Schapira does not seem to impose constructibility. $\endgroup$
    – Z. M
    Commented Dec 8, 2022 at 10:40

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.