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This question was previously asked over at math.SE.

Let $X$ be a scheme. Let $\mathcal{E}$ be a sheaf of sets on $X$. Then we can define $\mathcal{O}_X\langle\mathcal{E}\rangle$, the free module over $\mathcal{E}$, as the sheafification of the presheaf $$ U \longmapsto \mathcal{O}_X(U)\langle\mathcal{E}(U)\rangle. $$ The stalk of $\mathcal{O}_X\langle\mathcal{E}\rangle$ at a point $x \in X$ is the free module $\mathcal{O}_{X,x}\langle\mathcal{E}_x\rangle$. It can also be described as $f_! f^{-1} \mathcal{O}_X$, where $f : \operatorname{Ét}(\mathcal{E}) \to X$ is the projection of the étalé space associated to $\mathcal{E}$.

In the case that $\mathcal{E}$ is a constant sheaf $\underline{M}$ with stalks some set $M$, the sheaf of modules $\mathcal{O}_X\langle\mathcal{E}\rangle$ coincides with $\mathcal{O}_X^{\oplus M}$ and is therefore free and in particular quasicoherent. In the case that $\mathcal{E}$ is a locally constant sheaf, the sheaf of modules $\mathcal{O}_X\langle\mathcal{E}\rangle$ is locally free and therefore too quasicoherent.

I'm wondering whether the converse holds: Does quasicoherence of $\mathcal{O}_X\langle\mathcal{E}\rangle$ imply that $\mathcal{E}$ is locally constant?

This looks like a simple statement, but I have been unable to prove it or come up with a counterexample. I know some indications that it's true:

  • If $X$ happens to be local as a topological space (spectrum of a local ring), then the converse holds: Using $(\mathcal{O}_X\langle\mathcal{E}\rangle)(X) \cong \mathcal{O}_X(X)\langle\mathcal{E}(X)\rangle$ (this requires locality) one can show that the canonical morphism $\underline{\mathcal{E}(X)} \to \mathcal{E}$ is an isomorphism.

  • As a consequence, if $\mathcal{O}_X\langle\mathcal{E}\rangle$ is quasicoherent, the pullback of $\mathcal{E}$ to any of the $\operatorname{Spec}(\mathcal{O}_{X,x})$ is constant. Thus $\mathcal{E}$ is "constant on all infinitesimal neighbourhoods".

  • If $\mathcal{O}_X\langle\mathcal{E}\rangle$ is not only quasicoherent, but even of finite presentation, then $\mathcal{E}$ is locally constant (with finite stalks).

  • If we assume that $\mathcal{O}_X\langle\mathcal{E}\rangle$ is not only quasicoherent, but even locally free (locally isomorphic to a module of the form $\mathcal{O}_X^{\oplus M}$), then locally we have $\mathcal{O}_{X,x}\langle\mathcal{E}_x\rangle \cong \mathcal{O}_{X,x}\langle M\rangle$, so $\mathcal{E}_x \cong M$, so at least the stalks are locally constant.

  • Let $j : V \hookrightarrow X$ be the inclusion of an open subset. Let $\mathcal{E}$ be $j_!(1)$, the extension of the terminal sheaf on $V$ by the empty set, explicitly given by $U \mapsto \{ \heartsuit \,|\, U \subseteq V \}$. This sheaf is locally constant iff $V$ is a clopen subset. Now furthermore assume that $X$ is integral. In this case we know that $\mathcal{O}_X\langle\mathcal{E}\rangle = j_!(\mathcal{O}_V)$ (extension by zero this time) is quasicoherent iff $V$ is a clopen subset. Therefore the converse holds in this case.

I'm interested in the question for the following reason: We can characterize quasicoherence in the internal language of the little and the big Zariski toposes of a scheme. (The first statement is already written up in these rough notes, Section 9, the case of the big Zariski topos will be soon.) We can also, quite trivially, characterize sheaves of the form $\mathcal{O}_X\langle\mathcal{E}\rangle$ in the internal language. If those sheaves turned out to be quasicoherent only if the basis sheaf is locally free, then we could characterize locally constant sheaves in the internal language. This would be quite nice and somewhat surprising.

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    $\begingroup$ Concerning $f_! f^{-1} \mathcal{O}_X$, where $f : \operatorname{Ét}(\mathcal{E}) \to X$ is the projection of the étalé space associated to $\mathcal{E}$: do you mean the functors $f_!$ and $f^{-1}$ acting on sheaves of abelian groups? $\endgroup$ – მამუკა ჯიბლაძე Mar 4 '17 at 10:29
  • $\begingroup$ Yes, and afterwards we can equip $f_! f^{-1} \mathcal{O}_X$ with the structure of an $\mathcal{O}_X$-module. But we also directly use the versions of these functors for modules instead of abelian groups. $\endgroup$ – Ingo Blechschmidt Mar 4 '17 at 10:37
  • $\begingroup$ For the latter you view $\operatorname{Ét}(\mathcal{E})$ as a ringed space with $f^{-1}\mathcal O_X$? Is it a scheme? $\endgroup$ – მამუკა ჯიბლაძე Mar 4 '17 at 10:44
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    $\begingroup$ Exactly. Yes, it is a scheme since $f$ is a local homeomorphism: For any point of $\mathrm{Ét}(\mathcal{E})$ there is an open neighbourhood $U$ which is homeomorphic to some open subset $V$ of $X$, and $(f^{-1} \mathcal{O}_X)|_U$ corresponds to $(\mathcal{O}_X)|_V$ under this homeomorphism. $\endgroup$ – Ingo Blechschmidt Mar 4 '17 at 10:51
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Let $X$ be a scheme over a field $k$ and $x \in X$ be a point with residue field $k$. Consider the skyscraper $\mathcal{E} = \mathrm{Sky}_x \{*\}$. Then $\mathcal{F} := \mathcal{O}_X\langle \mathcal{E} \rangle$ is the skyscraper $\mathrm{Sky}_x M$, where $M = \mathcal{O}_{X,x}$.

I claim $\mathcal{F}$ is quasicoherent. Consider the inclusion $ i : \{x\} = \mathrm{Spec}~k \hookrightarrow X$. Then $i_*i^{-1}\mathcal{F} \cong \mathcal{F}$. Since $i^{-1}\mathcal{F}$ is quasicoherent (indeed: $\mathrm{Vect}_k \simeq \mathrm{QCoh}(\mathrm{Spec}~k))$ so is its pushforward.

This gives a quasicoherent sheaf $\mathcal{O}_X\langle \mathcal{E} \rangle$ whose `defining sheaf' $\mathcal{E}$ is not locally constant.

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    $\begingroup$ It seems you mean that $\mathcal E$ is the sheaf of sets for which $\mathcal E(S) = \ast$ if $x \in S$ and $\mathcal E(S) = \varnothing$ otherwise. Unfortunately this doesn't even define a presheaf in general since there is no map $\ast \to \varnothing$. So I don't think your example works. (Not also that if the construction worked, it would contradict Ingo's first claim: that the result is always true if $X$ is the spectrum of a local ring.) $\endgroup$ – Dan Petersen Apr 6 '17 at 20:33
  • $\begingroup$ I agree with Dan (even though I would have quite liked to have my question resolved with a simple example!). The failure of $\mathcal{E}$ to be a sheaf can be explained geometrically: Recall that there is the familiar saying "a sheaf is a continuous family of sets". This is one of the few times where this saying is strinkingly vivid. The family which is $\{\star\}$ at $x$ and $\emptyset$ at all other points is not continuous in an intuitive sense of the word. $\endgroup$ – Ingo Blechschmidt Apr 6 '17 at 21:22
  • $\begingroup$ @Dan, thank you for pointing that out. Of course we shouldn't be taking values in the empty set. Sorry! By the way, had I of used the correct notion of skyscraper sheaf (replace $\emptyset$ with {*}) then, as written, this would have been the constant sheaf sending everything to the terminal object in $Ens$. $\endgroup$ – Artur Jackson Apr 7 '17 at 8:31
  • $\begingroup$ Dear @Ingo, I'm not sure this reasoning is correct. Certainly there are skyscraper sheaves. The issue here was that there are no maps to the initial object $\emptyset$. $\endgroup$ – Artur Jackson Apr 7 '17 at 8:32
  • $\begingroup$ Are we close to something interesting? Consider then $\mathcal{E} = \mathrm{Sky}_x 1_+$, where I'm using the notation $n_+ = \{*=0,1,\ldots,n\}$. The cardinality of the value set jumps on neighborhoods of $x$ from 1 to 2. Perhaps I'll look at this after some sleep. ^_^ $\endgroup$ – Artur Jackson Apr 7 '17 at 8:35

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