Let $f:X\longrightarrow S$ be a morphism of schemes. What is the link between sheaf-sections of $O_X$ over an open set of $X$ and morphism-sections of $f$. Is there a kind of correspondence?
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$\begingroup$ Consider the case S = Spec k ... I suggest that you follow the advice of your username ;-) $\endgroup$– Kevin H. LinCommented Aug 31, 2010 at 22:43
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$\begingroup$ This is amusing, Kevin: in my answer (which I wrote before seeing your comment) I mention a case where $X$=Spec k... $\endgroup$– Georges ElencwajgCommented Aug 31, 2010 at 23:16
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8$\begingroup$ Since $O_X$ has nothing to do with $S$, you're looking for the wrong dictionary. The correspondence is that sections of $O_X$ over an open $U$ correspond to sections of $\mathbf{A}^1_X \rightarrow X$ over $U$. In that respect, the two general concepts are related. To see that, I second Kevin's wise advice. Rather than ask the obvious follow-up with more general sheaves, I again refer back to your name. $\endgroup$– BCnrdCommented Aug 31, 2010 at 23:27
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$\begingroup$ Let's take $f$ of relative dimension 1. In that case one sheaf-section correspond to one (positive) divisor, and, on the other hand, the schematic image of one morphism-section seems to come from one (positive) (relative? and usually more specific) divisor. (In other rel dim, a correspondence could involve one s-section and family of m-sections.) $\endgroup$– WorkitoutCommented Sep 1, 2010 at 10:48
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5 Answers
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Dear Workitout, of course I can't prove there is no link, but I'm rather pessimistic . Here is a fuzzy argument in support of my feeling.
The set of sections of $\mathcal O_X$ is never empty (after all it is a ring and so contains zero!) .But I would say that "in general" (in a non technical sense), $f:X\to S$ has an empty set of sections. For example if $X=Spec A$ is affine and $f:X\to S=Spec (\mathbb Z ) $ is the unique morphism, sections of $f$ correspond to ring morphisms $A\to \mathbb Z$, and my feeling is that there is no particular reason why they should exist: if $A$ is a field (say), certainly no ring morphism $A\to \mathbb Z$ exists .
answered Aug 31, 2010 at 23:05
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The name "sections" of a sheaf comes from the old viewpoint of a sheaf as its espace étalé. That is, they are sections of the canonical map $\acute Et(\mathcal{O}_X)\to X$ (this is a continuous map, not a map of schemes).
answered Sep 22, 2010 at 9:03
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In my answer here I indicated how sheaf sections are sections of maps. This doesn't involve the structure map, but clearly motivates the name, if that's what you were looking for.
A more detailed account is in "Sheaves in Geometry and Logic" (from p. 88) by MacLane/Moerdijk, and an even more detailed and intuition-emphasizing one in Goldblatt's "Topoi", online viewable under the link.
answered Sep 21, 2010 at 17:18
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Sections to the morphism $X \to S$ are more or less $S$-rational points of $X$: if for example $S = Spec(R)$ and $X = Spec(R[x_1, \dots, x_n]/(f_1, \dots, f_m))$ with polynomials $f_1, \dots, f_m \in R[x_1, \dots, x_n]$, then sections to $X \to S$ correspond to points $(a_1, \dots, a_n) \in R^n$ with $f_i(a_1, \dots, a_n) = 0$ for all $i$.
On the other hand, the elements of $\mathcal{O}_X(U)$ can be seen as holomorphic functions on $U$.
So the one kind of sections can be seen as "points" of the geometric object, the others can be seen as "functions" on the geometric object.
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At least in the case that $f:X\to S$ is a vector bundle the answer is given in Ex 5.18 Chapter 2 of "algebraic Geometry" by Hartshorne.
