Talk:Vector bundle

New To Advanced Math

Hi; I'm trying desperately to understand many of these advanced principals of mathematics, such as vector bundles, but no matter how many times I review the material, it doesn't sink in. Could someone please provide examples, problems to solve (with their solutions) and/or ways to visualize this? beno 26 Jan 2006

This definition seems a bit weird. The definition should be of either smooth vector bundle or the general notion of vector bundle without restricting cases where the base space is a smooth manifold. The definition of smooth vector bundle includes restrictions on the transition functions too.

Why don't you try and rework it? Be bold! :) Dysprosia 12:05, 23 Oct 2003 (UTC)

The phrasing is too dense: hard to edit from here. In what sense is the fibre 'isomorphic' to V? What category is that in?

Charles Matthews 09:15, 29 Nov 2003 (UTC)

category of vector spaces.--MarSch 11:15, 29 January 2007 (UTC)[reply]

Also: what is a "smooth projection from a vector space to a manifold"? The term "smooth" only applies to maps between two smooth manifolds, and the term "projection" only applies to a map from a direct product to one of the factors. AxelBoldt 15:05, 4 Dec 2003 (UTC)

The bundle projection is a smooth map from the total space to the base space. For every point of the base space there exists a neighbourhood such that the bundle projection (restricted to the inverse image of this neighbourhood) is a projection. --MarSch 11:21, 29 January 2007 (UTC)[reply]

I'm likewise finding this rather vague. In this context, what's to 'attach' to a topological space, and what's 'compatible' as opposed to a vector space that is not? Sojourner001 21:35, 5 August 2006 (UTC)[reply]

pretty late but here is a reply. "attach" means that to each point x of X, we assign to it a vector space V_x. think of a smooth manifold, where each point has a tangent space. "compatibility" doesn't refer to a single V_x but all of the total space E. more precisely, it means that, given x, you can always take some small neighborhood of x such that, in their totality, the collection of vector spaces {V_x| x in U} looks just like a disjoint collection (topologically) in E. put another way, locally the way you attach V_x to x is the trivial one. ("trivial" here means there is no glueing involved and the topology is simply the product topology.) this may not be true if U is replaced by all of X. take again the example of smooth manifolds. depending on the manifold, the collection of tangent space may not have the disjoint topology when topologized in the natural way. Mct mht 06:43, 18 January 2007 (UTC)[reply]

I've tried to rewrite the introduction to clarify some of these points. Let me know what you think. Geometry guy 14:40, 11 February 2007 (UTC)[reply]

What else needs to be done?

I've made quite a few edits to this article today, partly to link up with the (tidied-up) pullback bundle and (new) bundle map articles, but also to clarify a few points, such as bundle isomorphism and trivialization. There is clearly still work to be done here though. For instance the section on "Operations on vector bundles" needs to be expanded, and this is probably next on my agenda, unless someone else does it before me! Anything else on your wishlist for this article? Geometry guy 14:48, 11 February 2007 (UTC)[reply]

Dual bundle should be mentioned somewhere. Tensor product bundle still needs to be written, but it wouldn't hurt to provide a redlink. At some point it would be nice to mention the connection between vector bundles and their associated principal (frame) bundles, but this isn't the most pressing issue. -- Fropuff 16:56, 11 February 2007 (UTC)[reply]

Dual bundles and tensor product bundles are mentioned very briefly in vector bundle. This is the section I think should be expanded. The relation with frame bundles is another issue, and I agree it should be covered somewhere. Geometry guy 00:02, 13 February 2007 (UTC)[reply]

The section on operations on vector bundles has now been expanded. Please leave your comments and suggestions here. Geometry guy 00:11, 1 March 2007 (UTC)[reply]

I am not sure the definition of vector bundle morphism is correct here. The one given in Milnor's characteristic classes is such that there exists a morphism if and only if the domain is the pullback of the range. The amazing power of vector bundles as a tool flows from their ability to be classified using homotopy theory, there is not a hint of that here. Finally, characteristic classes are inseparable from the theory of vector bundles and there needs to be some hint of that.

Topoman

You raise a valid point. The definition of vector bundle morphism given here is correct (from my point of view). But it's also in some sense the weakest definition availiable, and generally not a very well-behaved one. The image of a vector bundle may fail to be locally trivial, for instance. I've seen the "pullback of the image" version elsewhere in topology as well (e.g., in Steenrod's Topology of Fibre Bundles). In this case, the image of a vector bundle is a vector bundle. There appears to be a distinction between the category of differentiable vector bundles (including the smooth ones), and the topological vector bundles. In the former, one can at least impose some rank conditions on the morphisms under consideration, but in the latter that is no longer possible. Silly rabbit 11:29, 23 April 2007 (UTC)[reply]

Bundles vs. Sections

The article says "Tangent bundles are not, in general, trivial bundles: for example, the tangent bundle of the (two dimensional) sphere is not trivial by the Hairy ball theorem." But doesn't the hairy ball theorem talk about tangent-vector fields (sections of the tangent vector bundle), not tangent vector bundles themselves? I see how the hairy ball theorem applies to vector fields, but how does it imply that the tangent bundle of a 2-D sphere is nontrivial? —Ben FrantzDale 03:07, 29 April 2007 (UTC)[reply]

On a related note, when would you talk about a vector bundle without immediately starting to talk about a vector field? —Ben FrantzDale 03:07, 29 April 2007 (UTC)[reply]

The hairy ball theorem says every vector field on the 2-sphere vanishes somewhere. Even though it talks about vector fields, this does actually tell you something about the tangent bundle. You see, a trivial bundle always has nowhere vanishing sections (even constant sections), therefore the tangent bundle of the 2-sphere can't possibly be trivial (that would contradict the hairy ball theorem). -- Fropuff 06:20, 29 April 2007 (UTC)[reply]

OK. At Fiber_bundle#Trivial_bundle, I don't see mention of this fact and hairy ball theorem also does not mention trivial bundles. Is that fact mentioned anywhere on Wikipedia? Thanks. —Ben FrantzDale 06:29, 29 April 2007 (UTC)[reply]

Well it's mentioned in passing at tangent bundle, but only briefly. In fairness the article on sections only got started very recently. It is still in a very stubby state with much left unsaid. The article on the hairy ball theorem should really mention this fact somewhere too. If I find the time I'll try to incorporate it. -- Fropuff 16:28, 29 April 2007 (UTC)[reply]

Fairness is good. I am just using my learning process as an opportunity to find and fix holes in Wikipedia's coverage of the topic. Thanks for your help. —Ben FrantzDale 16:47, 29 April 2007 (UTC)[reply]

Metrics in vector bundles

Anyone know where I can link for a metric in a vector bundle? Metric tensor seems to be populated exclusively with metrics in the tangent (and other tensor) bundles. Silly rabbit 03:30, 22 May 2007 (UTC)[reply]

You, sir, are an idiot. Please look in the most obvious place: metric (vector bundle), an article I myself wrote over a month ago! Silly rabbit 16:48, 31 May 2007 (UTC)[reply]

Is this a joke?

For a start, the definition given in the article is wrong! An additional condition, namely that the local trivialization p^(-1) (U) -> U X V is required to induce a k-linear transformation between vector spaces (for a field k) when restricted to each fibre, is required. The other parts of the definition do not necessarily imply this! Something has to be done...

Topology Expert (talk) 11:59, 2 October 2008 (UTC)[reply]

Can't you do something about it? Also, an example of a real life application of this would be greatly appreciated. NPOV-V-NOR 15:41, 10 October 2008 (UTC)[reply]

Attention point

In the first picture, the transversal must be an open interval to be locally homeomorphic to the real line. With a closed interval, homeomorphic to the extended real line. --kmath (talk) 14:06, 25 May 2009 (UTC)[reply]

Cardinality

Maybe I'm getting confused, but a finite dimensional real vector space must have continuum cardinality. What if some fibres have more guys? Then a finite dimensional structure won't exist... Money is tight (talk) 10:22, 24 May 2010 (UTC)[reply]

Sorry, I don't know what you mean by "guys" in the context, so I don't quite understand your concern. Does it resolve the matter to point out that the fibers of a vector bundle are all mutually isomorphic linear spaces? Sławomir Biały (talk) 12:27, 24 May 2010 (UTC)[reply]

Sorry for the childish remark I meant "guys" as elements, so what I'm saying is if some fibre has cardinality greater than continuum, it's meanless to speak of fibre bundles right? If that is so, it seems to me this is rather restrictive. Money is tight (talk) 13:15, 24 May 2010 (UTC)[reply]

Of course one can generalize the concept of vector bundle as presented in this article by slightly modifying part 3 of the formal definition:

3. For every x∈X there exists a structure of a topological vector space on the fiber π⁻¹({x}) so that the following compatibility condition is satisfied: around any point in X there is an open neighborhood U, a topological vector space V and a homeomorphism $\varphi :U\times V\to \pi ^{-1}(U)$ such that for every x∈U

$\pi \circ \varphi (x,v)=x$ for any v∈V.
$v\mapsto \varphi (x,v)$ is a homeomorphic vector space isomorphism between the topological vector spaces V and π⁻¹({x}).

This way you can have quite a few guys in each fibre. Frankly, cardinality is not much of an issue in this business. If one wants to take a "foundational approach", category theory is much more fruitful than bare set theory.Lapasotka (talk) 19:23, 25 May 2010 (UTC)[reply]

rank

Why the term "rank" rather than "dimension"? The latter seems more familiar when talking about vector spaces.. is there a reason why the different term is used here? Cesiumfrog (talk) 11:00, 17 October 2010 (UTC)[reply]

Dimension might refer to dimension of the total space, rather than the fibre. Sławomir Biały (talk) 11:59, 17 October 2010 (UTC)[reply]

Operations problem

There are problems with the operations section. Firstly, the notation $E_{x}$ is never defined (it clearly means the fiber space at x, but that's only clear if you already know about vector bundles, in which case you don't need this article...). Secondly we are in a context where both x and X mean something. When you see E_x it is not clear (because of the sans serif font) whether the subscript is big or little X; this makes the section potentially very confusing. They should be in math tags as $E_{x}$ and $E_{X}$ look totally different. I will fix this myself later when I have some time. Tinfoilcat (talk) 18:24, 2 February 2012 (UTC)[reply]

Problem with Real Vector Bundle definition

I myself was using this page as learning material on vector bundles so I'm not sure where to go in order to fix what's in the article, but:

In the definition of a real vector bundle, the function pi is defined from E to X, so that pi^-1 is from X to E. But later, phi is defined from U x R^k to pi^-1(U), which means that U is an element of X. But U is a neighborhood of points from X, which leads to an infinite number of sets of sets of sets ... being members of X.

Like I said I don't know what is supposed to be written, I just see a problem. — Preceding unsigned comment added by 74.243.202.152 (talk) 20:58, 31 August 2013 (UTC)[reply]

I don't see the problem. (Is it relevant that

\pi ^{-1}(U)

means the preimage of U under the function π?) Sławomir Biały (talk) 21:09, 31 August 2013 (UTC)[reply]

Algebraic Vector Bundles Needed

There should be a section discussing algebraic vector bundles, including the construction of vector bundles from locally free ${\mathcal {O}}_{X}$ -modules. This should include the (co)tangent bundles and (co)normal bundles with explicit examples.

That's in coherent sheaf#Basic constructions of coherent sheaves. -- Taku (talk) 02:35, 20 July 2017 (UTC)[reply]

Non-trivial example

A non-trivial example could be useful - perhaps the Möbius bundle (https://math.uchicago.edu/~chonoles/expository-notes/promys/promys2011-vectorbundles.pdf). Someone familiar with that should add it, or maybe I'll get around to it (if I can understand it...) Tomtheebomb (talk) 01:23, 11 October 2017 (UTC)[reply]

The continuous map $\pi$ need not to be surjective

Some documents require that $\pi$ must be a continuous surjection but in this definition, it's fine for the first condition to require that $\pi$ is a continuous map rather than a continuous surjection. If $\pi$ is not surjective, then $\pi ^{-1}(x)=\emptyset$ for some $x\in X$ but the second condition requires that $E_{x}=\pi ^{-1}(X)$ is a vector space, i.e. it must have at least an element. Hence the second condition implies that $\pi$ is a surjection.

The definition in Atiyah's (1994) lecture note about vector bundle and K-theory doesn't require that $\pi$ is a surjection. Duybao20 (talk) 08:57, 24 September 2022 (UTC)[reply]