Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,846 answers , 20,597 comments
1,470 users with positive rep
501 active unimported users
More ...

Fibre bundles vs ringed spaces

+ 1 like - 0 dislike
95 views

An fibre bundle consists of

  1. a topological space $X$ which is the base space of the vector bundle $B$
  2. a suitable bundle projection $\pi: B \to X$
  3. for every $x \in X$ the fibre $F = \pi^{-1}(x)$ is some topological space too

The following compatibility condition has to be satisfied:

For every point $x\in X$ there is a suitable  environment $U \subset X$ and a mapping $\phi: U \times F \to \pi^{-1}(U)$
such that for all $x\in U$,

  1. $(\pi \circ \phi)(x,v) = x$ for all vectors $v \in F$
  2. The map $v \mapsto \phi(x,v)$ from $F$ to $\pi^{-1}(\{x\})$ is linear and bijective.

The pair $(U,\phi)$ is called a local trivialization.

A ringed space is a pair $(X,\mathcal{O}_X)$ where $X$ is a topological space
and $\mathcal{O}_X$ is a sheaf of commutative rings on X.

A locally ringed space is a ringed space $(X,\mathcal{O}_X)$  such that all stalks of $\mathcal{O}_X$ are local rings (i.e. they have unique maximal ideals). Note that it is not required that $\mathcal{O}_X(U)$ be a local ring for every open set $U$. In fact, that is almost never going to be the case.

For example, $X$ can be a Hausdorff second countable space and $C^{\infty}(U\to C^{\infty}(U))$ the sheaf of rings where, for each open set $U \subset X$, $ C^{\infty}(U)$ is the $R$- algebra of smooth functions.

At a first glance, the concept of ringed spaces looks somehow similar two the concept of fibre bundles to me. Can these to notions be related to each other, such that a ringed space can be seen as a fibre bundle where the space $X$ of the ringed space takes the role of the base space and the sheaf of commutative rings $\mathcal{O}_X$  takes the role of the fibre?

I am interested in this because to me it seems relevant to understand the relationship between different approaches to describe supermanifolds for example ...

asked Aug 16 in Mathematics by Dilaton (4,295 points) [ revision history ]

1 Answer

+ 2 like - 0 dislike

It is natural, I think, to initially get the feeling that sheaves and fiber bundles are very related.  However, the two are actually quite different, despite having some important overlap.  Put another way, there are tons of fiber bundles on $X$ which are not sheaves on $X$ and visa versa, but there is an important class of fiber bundles which coincides with an important class of sheaves, as I hope to explain. 

First of all, the category of sheaves on $X$ is way too big; we find connections to bundles by fixing a structure sheaf $\mathcal{O}_{X}$, defining a locally-ringed space $(X, \mathcal{O}_{X})$, as you describe.  We now don't want to consider all sheaves on the topological space $X$, but rather sheaves of $\mathcal{O}_{X}$-modules.  We say a sheaf $\mathscr{F}$ is a sheaf of $\mathcal{O}_{X}$-modules if for all open sets $U \subseteq X$, then $\mathscr{F}(U)$ is an $\mathcal{O}_{X}(U)$-module (plus a compatibility with restriction morphisms).  

There are certain very special $\mathcal{O}_{X}$-modules called locally-free sheaves.  These are sheaves $\mathscr{F}$ such that there exists an open cover $\{U_{i}\}$ of $X$ satisfying $\mathscr{F}(U_{i}) \cong \mathcal{O}_{X}(U_{i})^{\oplus N}$, for some $N$.  Notice that a general sheaf need not have any local trivialization property...this is one of the reasons they cannot possibly be related to a fiber bundle.  To find a relation, you need some sort of local trivialization property.  This comes naturally in the form of locally-free sheaves.  

However, note that the structure sheaf $\mathcal{O}_{X}$ must be fixed throughout!  In other words, it doesn't make sense to talk in full generality about "locally-free sheaves."  Rather, one must talk about "locally-free sheaves inside the category of $\mathcal{O}_{X}$-modules."  

Anyway, the punchline here is that for a fixed locally-ringed space $(X, \mathcal{O}_{X})$, deep within the category of $\mathcal{O}_{X}$-modules, you find a collection of locally-free sheaves which are indeed fiber bundles.  One perhaps intuitive way to think about the category of all sheaves on $X$ is by starting with those that define fiber bundles and allowing more and more degenerative behaviour.  For example, allow them to not be locally trivial, then maybe allow them to jump in rank, then allow for them to only be supported on part of $X$, etc.  

I hope this was somewhat helpful!  

answered Aug 17 by scpietromonaco (150 points) [ revision history ]
edited Aug 17 by scpietromonaco

Thanks for these nice and clear explanations, it is helpful indeed!

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...