8
$\begingroup$

Let $f: X \to Y$ a morphism of ringed spaces and $E$ a vector bundle over $Y$ for finite rank $r$.

My questions are:

  1. Why $E$ can be interpreted as locally free sheaf of rank $r$, therefore for each $y \in Y$ there exist an open set $U$ that contains $y$ such that $E \vert _U \cong \mathcal{O}_Y ^{\otimes r}\vert _U$

  2. Conversly, how can a locally free sheaf of rank $r$ turned to a vector bundle formally correct?

Background of my question: How to interpret $f^*E$? Problem:$f^*$ accepts as arguments only sheafs on $Y$...

$\endgroup$
4
  • 1
    $\begingroup$ What definition of vector bundle are you using? $\endgroup$
    – Mohan
    Commented Apr 6, 2018 at 22:43
  • $\begingroup$ That for each $y \in Y$ there exist an open enviroment such that $E \vert _U \cong U \times V$ where $V$ is a $r$ dimensional vector space. Futhermore the given "projection" map $p: E \to Y$ respects this local trivialisation $\endgroup$
    – user267839
    Commented Apr 6, 2018 at 22:53
  • $\begingroup$ Woah I've never heard an open neighborhood called an open environment before! $\endgroup$
    – Exit path
    Commented Apr 6, 2018 at 22:59
  • 1
    $\begingroup$ Well, till today:) $\endgroup$
    – user267839
    Commented Apr 6, 2018 at 23:01

2 Answers 2

10
$\begingroup$

The idea is to interpret the sheaf of section of the bundle $E$ as the locally free sheaf you mean. Since $E$ is a vector bundle, over small enough $U \subset Y$, $E|_U \simeq U \times k^n$, and the sections of the bundle $U \times k^n \to U$ correspond exactly to maps $U \to k^n$, whence we get isomorphism with $\mathcal{O}_U^n$.

Turning locally free sheaf $\mathcal{E}$ into a vector bundle $E \to Y$ is similarly simple: if our sheaf is free over each of $U_i$, and $\bigcup U_i = Y$, then we just need to glue $U_i \times k^n \to U_i$ over various $U_i$. Therefore, we need a family of gluing isomorphisms $u_{ij}: (U_i \cap U_j) \times k^n \to (U_j \cap U_i) \times k^n$.

Here's how we get them: we have local isomorphism $f_i: \mathcal{E}_{U_i} \to \mathcal{O}_{U_i}^n$ from the definition of "locally free". Restricting to intersection $U_i \cap U_j$, we get $f_{ij} = f_j|_{U_i \cap U_j} \circ f_i|_{U_i \cap U_j}^{-1}: \mathcal{O}_{U_i}^n|_{U_i \cap U_j} \to \mathcal{O}_{U_j}^n|_{U_i \cap U_j}$. Every such map is induced by a map $u_{ij}: (U_i \cap U_j) \times k^n \to (U_j \cap U_i) \times k^n$ (why?), and this is our gluing isomorphism.

$\endgroup$
8
  • $\begingroup$ Hi. Thank you for the answer. Two questions: Firstly: Do you mean the sections $U \to U \times k^n $? Secondly: Why the ring of maps $U \to k^n$ (= restricted sections) is isomorphical to $\mathcal{O}_U^n$? Does $\mathcal{O}(U) = k$ hold? Why? $\endgroup$
    – user267839
    Commented Apr 6, 2018 at 23:10
  • $\begingroup$ I mean sections $U \to U \times k^n$ of the bundle $U \times k^n \to U$. For second question, this is basically the definition of what the structural sheaf $\mathcal{O}_Y$ is: elements of $\mathcal{O}_Y(U)$ are in one-to-one correspondence with regular maps $U \to k$. $\endgroup$
    – xyzzyz
    Commented Apr 6, 2018 at 23:28
  • $\begingroup$ Are you sure? The strucure sheaf $\mathcal{O}_Y$ just garatees that $\mathcal{O}_Y(U)$ is a ring (if I consider wiki's definition of ringed space ). What implies that the local section must be regular maps mapping to $k$? Ore do you assump an extra structure of $Y$? $\endgroup$
    – user267839
    Commented Apr 6, 2018 at 23:40
  • $\begingroup$ There must be some extra structure -- otherwise, what $k$ even is? $\endgroup$
    – xyzzyz
    Commented Apr 6, 2018 at 23:55
  • $\begingroup$ @xyzzyz in fact you first used $k$ in your answer! In general judging by your previous comment with the one-to-one correspondence with regular maps, you are not working with ringed spaces but with varieties. The context may be clearer but the question only considered ringed spaces in which case, one should substitute for instance $U\times k^n$ with $U\times \mathcal{O}(U)^n$, where $\mathcal{O}(U)$ is a ring, not a field. $\endgroup$
    – user128787
    Commented Feb 13, 2023 at 9:58
7
$\begingroup$

Consider an arbitrary (topological) $k$-vector bundle $\pi: E \to X$. For each $U \subset X$, we'll assume that $\mathcal{F}_X(U)$ be the set of continuous maps $U \to k$. It isn't hard to see that $\mathcal{F}_X(U)$ is a sheaf of rings, for we can take sums, products, and restrictions of functions in the usual way.

We'll assume that $\mathcal{O}_X$ is a subsheaf of $\mathcal{F}_X$; otherwise it is not always possible to construct a locally free sheaf correponding to the vector bundle $\pi$.

Now if we let $\mathcal{E}(U)$ be the set of continuous sections $f: U \to \pi^{-1}(U)$ (where "section" means that $f(u)$ lies "above" $u$ in the sense that $\pi f(u) = u$) then I'll show $\mathcal E$ is an $\mathcal{O}_X$-module that's locally free of rank $\dim E$.

To see that it's an $\mathcal{O}_X$-module note that for $\psi \in \mathcal E(U)$, we have $$\psi(u) = (u,v) \in \{u\} \times k^n.$$ For $f \in \mathcal{O}_X(U)$, since $f(u) \in k$ we can use scalar multiplication in $k^n$ to define $$f\psi(u) = (u, f(u)\cdot v) \in \{u\} \times k^n.$$ Since $E$ is locally a cartesian product, $\psi$ is locally a cartesian product of continuous coordinate functions. Since multiplication of continuous functions is continuous, it follows that $f \psi \in \mathcal E(U)$, making $\mathcal E$ an $\mathcal O_X$-module.

Similarly, to check that the sheaf we have constructed is locally free, we'll use the fact that $E$ is locally homeomorphic to a product space. A continuous section $U \to U \times k^n$ is the same thing as $n$ continuous maps $U \to k$, hence $\mathcal{E}|_U \simeq \mathcal{O}_U^n$ as an $\mathcal{O}_U$-module.

The (possibly) surprising fact is that the converse is true: a locally-free $\mathcal{O}_X$-module of rank $r$ corresponds uniquely to a vector bundle $E$. To reiterate, all of this only makes sense if we can interpret $\mathcal{O}_X$ as a sheaf of continuous $k$-valued functions.

$\endgroup$
5
  • $\begingroup$ Hi. Thank you for the answer. One step stays unclear: Why can you "define" $\mathcal{O}_X(U)$? Is the structure sheaf $\mathcal{O}_X$ (here you mean $X =B$?, don't you) of $X$ already determined since that's a ringed space? $\endgroup$
    – user267839
    Commented Apr 6, 2018 at 23:26
  • $\begingroup$ Yes. The construction only makes sense when $\mathcal O_X$ can be regarded a subsheaf of the sheaf of continuous functions $X \to k$. This would happen for example if $X$ is a variety over $k$. I don't think it's possible to construct such a vector bundle given an arbitrary ringed space, like if we took $\mathcal O_X$ to be the constant sheaf of $\mathbb Z/4\mathbb Z$. Does this help? $\endgroup$ Commented Apr 6, 2018 at 23:33
  • $\begingroup$ Oh, also I corrected the typo and changed $B$ to $X$. Thanks! $\endgroup$ Commented Apr 6, 2018 at 23:35
  • $\begingroup$ So that's an extra claim that the local sections of the structure sheaf $\mathcal{O}_X$ must be of the shape "function from $U$ to $k$? Because if I consider $X$ just as an arbitrary regular space then $\mathcal{O}_Y(U)$ are just rings (compatible with restructions) $\endgroup$
    – user267839
    Commented Apr 6, 2018 at 23:49
  • $\begingroup$ Yes. There are all kinds of weird ringed spaces for which the above is not true. Since much of the motivation for ringed spaces comes from algebraic geometry, the writer probably has the structure sheaf of an algebraic variety in mind. Also the assumption that $\mathcal{O}_X$ is a subsheaf of $\mathcal F_X$ is stronger than any statement about local sections; all of the restriction maps must coincide with restriction of functions. $\endgroup$ Commented Apr 6, 2018 at 23:52

You must log in to answer this question.

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