2
$\begingroup$

Let $X$ be a topological space, and let $\mathcal{F}$ be a (pre-)sheaf of sets (groups, rings, etc.) on $X$. It seems to me that several authors in algebraic geometry prefer to use the global section functor $\Gamma$ to express the set (resp. group, ring, etc.) of sections $\mathcal{F}(U)$ as $\Gamma(U,\mathcal{F})$, for an open subset $U$ in $X$. For instance, see the fifth line in this Tag on the Stacks Project.

Is there a good reason to prefer the expression $\Gamma(U,\mathcal{F})$ to $\mathcal{F}(U)$, apart from emphasising being contravariant in $U$ and covariant in $\mathcal{F}$?

$\endgroup$
3
$\begingroup$

One reason may be that the notation $\mathcal F(U)$ isn't always standard to mean sections of $\mathcal F$ over $U$. For instance, Serre's FAC is an example of an important paper where $\mathcal F(U)$ means something totally different; it actually refers to the restriction of the sheaf $\mathcal F$ to $U$, which a lot of people now would write $\mathcal F|_U$.

As for using the notation $\Gamma$ in general: You may or may not be aware that sheaves $\mathcal F$ on $X$ are in bijection with surjective maps $\pi:\mathscr F\to X$, where $\mathscr F$ is a topological space, such that for each $f\in\mathscr F$ there is a neighborhood $U$ of $f$ and $V$ of $\pi(f)$ such that $\pi|_U:U\to V$ is a homeomorphism. In this sense, "sections of a sheaf $\mathcal F$" are literally sections of the map $\pi:\mathscr F\to X$, i.e. functions $s:X\to\mathscr F$ such that $\pi\circ s=\text{id}$. Furthermore, this is analogous to sections of a vector bundle $\pi:E\to X$, and when talking about vector bundles we always denote global sections by $\Gamma(E)$. So I guess this notation just carried over to sheaves.

$\endgroup$
  • $\begingroup$ Would you know the answer to this question maybe? $\endgroup$ – Watson Jul 22 '17 at 19:17
  • 1
    $\begingroup$ @Watson I've always assumed the $\Gamma$ stood for "global" in "global sections" like you say in your comment, and have never really questioned it. Could be wrong though.. $\endgroup$ – Alex Mathers Jul 24 '17 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.