4
$\begingroup$

I'm asking this question with a particular regard to the geometric aspect(s) of the difference(s).

Let there be a sheaf $\mathcal{F}: Top(X)^{op} \rightarrow \mathfrak{Ab}$, with $Top(X)$ being the category whose objects are open subsets of X, and whose morphisms are inclusions, and $\mathfrak{Ab}$ is the usual category of abelian groups. What then, is the difference between the section $\mathcal{F}(P)$ and the stalk $\mathcal{F}_P$, where $P$ is some point in the topological space $X$. I suspect that they might be the same, perhaps up to isomorphism, but I don't have a good way of confirming this. Would I be correct in suspecting this, and even if not, why ?

Thank you for your time.

$\endgroup$
9
  • 7
    $\begingroup$ F(P) does not make any sense, your functor is defined on open sets not points in these. $\endgroup$
    – Riquelme
    Commented Jul 21, 2019 at 17:19
  • $\begingroup$ I'm rather new to sheaves but aren't points also open sets ? $\endgroup$ Commented Jul 21, 2019 at 17:20
  • 3
    $\begingroup$ @DatMinhHa In most topological spaces one come across, one-point subsets aren't open. $\endgroup$ Commented Jul 21, 2019 at 17:22
  • $\begingroup$ you even wrote that the objects in your category are open sets of some topological space. Are you sure you understand the definition of a functor? For every open subset $U $ of $X$, $F(U)$ is an abelian group. The stalk at a point $p $ in $X$ is obtained by taking the dircet limit over all open sets containing $p$.en.wikipedia.org/wiki/Stalk_(sheaf) $\endgroup$
    – Riquelme
    Commented Jul 21, 2019 at 17:25
  • $\begingroup$ @Riquelme Ah right I see where I was wrong now. Thank you. $\endgroup$ Commented Jul 21, 2019 at 17:27

2 Answers 2

3
$\begingroup$

Singletons are usually not open and therefore we cannot plug these into our sheaf. This means that we sort of have to approximate $\mathcal{F}$ at a singleton $\lbrace x \rbrace$. This is exatly what the stalk is. The stalk is given by $$\mathcal{F}_x = \varinjlim \mathcal{F}(U),$$ where the limit runs over $x \in U$. That means we approximate the sheaf $\mathcal{F}$ at the point $x$ by looking at smaller and smaller open neighborhoods of $x$ under $\mathcal{F}$.

$\endgroup$
1
$\begingroup$

As is already explained in the comments: It usually doesn't make sense to speak about $\mathcal F(P)$ for a point $P$ - that is: whenever $\{P\}$ is $\textit{not}$ open in $X$.

Note however that there always is an inclusion $\mathcal F(U) \longrightarrow \prod\limits_{P \in U}\mathcal F_P, s \longmapsto (s_P)_{P \in U}$ for any open set $U \subseteq X$. So arbitrary sections of a sheaf are still determined by their stalks and one uses these properties all the time when working with sheaves.

Related things you could want to take a look into are $\textit{sheafification}$ of presheaves and the $\textit{espace étalé}$ way of viewing sheaves.

Also note that $\mathcal F(P) \overset{\mathrm{def}}{=} \kappa_{X}(P) \otimes_{\mathcal O_{P}} \mathcal F_{P}$ is a well established notation for the $\textit{geometric fibre}$ of a sheaf of modules over a locally ringed space.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .