Talk:Fiber bundle

Local triviality

Should we make a section on the notion of "locally trivial" bundles? SpiralSource (talk) 10:06, 14 August 2023 (UTC)[reply]

Definition

What is the purpose of allowing a generic x in E in the definition, and then talking about a neighbourhood of pi(x)? I'm not confident enough in my understanding to change this immediately, but unless anyone points out something I've been misunderstanding in the next few days, I'd like to change the definition to talk about x in B, and U being a neighbourhood of x, which I think will be clearer. Tatterdemalion1983 (talk) 12:29, 23 November 2019 (UTC)[reply]

Chern

The article should some give some historical context. For example Chern who contributed hugely to the subject, is not mentioned in the article at all, only the Chern class is mentioned once. The first useful construction was in fact the sphere bundle, in the early 40's. Chern was able to give a purely intrinsic proof of Gauss-Bonnet Formula in Integral Formulas for the Characteristic Classes of Sphere Bundles, Shiing-Shen Chern, Proceedings of the National Academy of Sciences of the United States of America, Vol. 30, No. 9 (Sep. 15, 1944), pp. 269-273, [1] —Preceding unsigned comment added by Kotika98 (talk • contribs) 03:09, 8 November 2009 (UTC)[reply]

spelling

I find the inconsistency between the article's title and the body (fibre vs. fiber) somewhat annoying. I don't really care whether British or American spelling is used, though considering that there's likely lots of Americanisms in the body, renaming the article to "Fiber bundle" seems better. However, I'm new here, so I'd rather have someone more experienced comment first.

Rvollmert 23:58, 13 Nov 2003 (UTC)

I've moved it (I'm British, by the way ...).

Plenty of work left to do here. The use of intertwiner isn't consistent with the page (which is about linear representations): 'equivariant' is the more accurate term.

Charles Matthews 06:10, 14 Nov 2003 (UTC)

Fibers

Does one typically require the fiber F to be globally the same for the whole space B? If so, (and http://planetmath.org/encyclopedia/FiberBundle.html seems to agree), then covering maps are not fiber bundles, and vector bundles (which are defined as special fiber bundles) would not satisfy Swan's theorem for disconnected manifolds. AxelBoldt 19:09, 6 Dec 2003 (UTC)

I think there will be a problem somewhere. It is a version of the fundamental group/groupoid question; for example the covering map case is like sets on which the fundamental groupoid acts, if we don't take a connected base.

The intuition starts with identifications on F moving along paths; so when B isn't path-connected I suppose one should start again with greater care.

Perhaps it is best to define F first only for B connected. Then there should be an explanation why locally trivial fibrations can have different fibers, as a function depending on the components of B (locally constant). That ought to cover the most important cases. It isn't really satisfactory, for example for B totally disconnected. So, maybe some more extended discussion will be required.

Charles Matthews 11:22, 7 Dec 2003 (UTC)

Vector bundles would satisfy Swan's theorem over disconnected compact manifolds even if the fibers depend on the component. Phys 18:51, 6 Aug 2004 (UTC)

Yes, but with our current definitions, the fibers of fiber bundles (and therefore also of vector bundles) must not depend on the component, and I believe in that case Swan's theorem is false: not every projective module comes from a vector bundle with globally constant fiber. Maybe for now we should simply assume B to be connected in our fiber bundle article, and be done with it? AxelBoldt 11:31, 9 Aug 2004 (UTC)

Covering maps are most assuredly fibre bundles! Specifically, they are precisely those fibre bundles whose fibre is a discrete space.

And Yes, part of the definition of a fibre bundle is the topological space of the fibre. So for any given fibre bundle with bundle projection p: E → B and for any x ∈ B — regardless of any condition on B, the inverse image p^-1(x) is homeomorphic to that fibre F. In fact, x always has some open neighborhood U such that p^-1(U) is homeomorphic to U ×F.Daqu (talk) 06:52, 17 January 2015 (UTC)[reply]

In the current form there is an unnecessary incompatibility issue with vector bundle article, since here we insist that the model fiber is part of the data defining a fiber bundle. I don't have statistics on which convention is most widely used, but I find the following one more elegant:

(E,p,B) is a fiber bundle if p:E→B is a continuous surjection and there exists topological space S such that for every x∈B there is some open neighborhood U⊂B of x and homeomorphism ψぷさい:p^-1(U)→U×S such that p = pr₁∘ψぷさい.

This way the model fiber becomes defined only upto a homeomorphism, which I think is closer to the point.Lapasotka (talk) 16:47, 11 January 2018 (UTC)[reply]

Global sections

You claim that "Since bundles do not in general have sections, one of the purposes of the theory is to account for their existence". I assume that it should talk about "everywhere nonzero sections", right? Or what kind of bundle doesn't even allow a zero section? \Mikez 12:53, 20 Oct 2004 (UTC)

How do you define a zero section in general? You can't — they make sense on vector bundles but not on general fiber bundles. Take principal bundles for example. There is a theorem which states that a principal bundle is trivial if and only if it admits a global section (of any kind). So any nontrivial principal bundle (such as the Hopf bundle) doesn't admit a global section. -- Fropuff 13:34, 2004 Oct 20 (UTC)

Yes - one can get very elementary examples of this where the fiber is just tow points. For example, the edge(s) of a Moebius band. Charles Matthews 15:00, 20 Oct 2004 (UTC)

All right. Thanks for the answer. \Mikez 08:55, 26 Oct 2004 (UTC)

There is a theorem which states that a principal bundle is trivial if and only if it admits a global section (of any kind).

Is this true only for principle budles or more broadly? And can we add this to the article? Its seems like a key statement to me. Together with some simple illustrative statement like a mobius band does not have a globally defined section.

Hmm .. seems like every example given is a principle bundle (a sphere bundle, a tangent bundle and a covering space are all principle bundles, right?) Do we an example of a fibre that is not a principle bundle, or at least doesn't have some kind of natural/canonical group action? linas 03:52, 11 May 2005 (UTC)[reply]

Vector bundles are not generally principal bundles. They may have a group structure, but this group action does not commute with the coordinate transformations. All vector bundles, whether they are trivial or not, have at least one global section: namely the zero section. The Mobius band also has a global "zero section", the central circle. -- Fropuff 05:06, 2005 May 11 (UTC)

I was thinking of Mobius band as a bundle on the set {-1,1}; there is no middle, there's only the edges. Then there's no zero section. Otherwise, to explain non-triviality of mobius one has to appeal to Cech cohomology or something (see below).

As to vector bundles, couldn't one argue that vector bundles have a "natural" structure group GL(V)? This group is not commutative ...

Since GL(V) doesn't contain a zero element, that implies that there should be lots of non-trivial ways to assemble a vector bundle with a GL(V) structure group ... isn't that the whole point of the field strength? The field strength on a G-bundle tells you how the local sections are to be glued together, or something like that, that's the point of the clutching construction, and the cocyle condition, (the cocycle condition says that the field strength must be a closed form, right? Something like that...)

The only trivial G-bundle is that with vanishing field strength. If G=GL(V) and the base space is a manifold, then vector bundles with connections with non-vanishing field strengths are the same thing as saying the underlying base space has a metric and that metric has curvature. (One can say that such a vector bundle induces a metric on the base space, right? I think one can...) The group of coordinate transforms on the base space is just the set of gauge transforms on the GL(V)-bundle.

So ... consider a GL(V)-principle bundle ... it has a global section, the zero section, and so the statement that principle bundles with global section are trivial bundles ... I'm confused, because if the base space does not have a metric, then the bundles are not trivial. I'm confused... do you see why I'm confused? (I might be spouting nonsense here, its been a long time since I was in school, and I never mastered the topic in the first place.) linas 15:06, 11 May 2005 (UTC)[reply]

Never mind, I just un-confused myself. The point is twofold:

1) the article should say something about global sections and triviality, without falling into the pitfall I fell into (the hairy ball theorem says that the tangent bundle of the 2-sphere does not admit a global section ...err, well, not that, but you know what I mean. The point is that GL(2,R) has no zero element.).

2) we need an article on the field strength that covers the above. It would be a really fine place to explain the relation between geometry and topology. The topology of a 2-sphere can't be changed, but one can put metrics (aka connections on the associated GL(2,R)-bundle) on it that deform it in arbitary ways. linas 15:56, 11 May 2005 (UTC)[reply]

Dohh. I just remembered the theorem: a vector bundle is trivial if and only if it admits a section which is nowhere zero. Ditto for line bundles. linas 16:23, 11 May 2005 (UTC)[reply]

Obviously that is only true for line bundles. Otherwise you could prove that any vector bundle is trivial, by taking the direct sum with a trivial line bundle. Charles Matthews 18:45, 11 May 2005 (UTC)[reply]

Dohhh. Which has a nowhere-vanishing volume-element? i.e. so that the coordinate frames don't go degenerate? linas 23:47, 11 May 2005 (UTC)[reply]

Clutching construction

I believe that something called the clutching construction implies that a set of transition functions + open cover, obeying the cocycle condition, etc. uniquely determine the fiber bundle. Is that true, or are there bundles that can't be built this way? I think this is true ... I don't see how else things could be made to work, but thought I'd ask :) linas 14:38, 10 May 2005 (UTC)[reply]

Yes, up to bundle isomorphism, that is correct. This is also sometimes called the fiber bundle construction theorem. I meant to add a page on it, but I haven't gotten around to it yet. -- Fropuff 15:03, 2005 May 10 (UTC)

OK, I was debating with myself the writing of such a page; instead of doing the real work I should be doing ... linas 03:52, 11 May 2005 (UTC)[reply]

Čech cohomology

Another article to-do: I think there's a statement that isomorphism classes of principle bundles are in 1-1 correspondance to elements of the Čech cohomology group H^1(B,G) where B is the base space and G is the structure group. Or something like that. This follows from the cocycle condition making the bundle a part of an exact sequence. I think the article on covering map tries to say this in the specific case of coverings, but I don't know that it quite hits the mark; its pretty dense as written. I tried to make covering map more accessible, but doubt I suceeded. linas 14:21, 11 May 2005 (UTC)[reply]

Another remark on the link with Čech cohomology: in the wikipedia article it is constructed for abelian groups (with operation +) whereas here the cocycle condition seems to be expressed for cocycle with value in any Lie group Noix07 (talk) 16:06, 14 August 2014 (UTC)[reply]

The correct term, by the way, is principal bundles (not "principle" bundles).Daqu (talk) 04:32, 22 January 2015 (UTC)[reply]

G-atlas and atlas

Does the article (mean to) say that a fiber bundle has an atlas that might be given by a group action on fibers, but need not necessarily? If not why is there no name for "a fiber bundle with structure group G" or is this perhaps called a G-fiberbundle? Maybe that means the fiber is G. So it might be called fiber G-bundle instead.--MarSch 16:11, 12 Jun 2005 (UTC)

I think most authors (e.g. Steenrod) include the G-atlas in the definition of a fiber bundle. Other authors define a fiber bundle without a structure group. A fiber bundle with structure group G is then called a G-bundle. Note that the fiber of a G-bundle is not necessarily diffeomorphic to G (this happens only for principal bundles). Rather, the fiber is a G-space (i.e. a space on which G acts continuously). -- Fropuff 16:51, 2005 Jun 12 (UTC)

so there exist fiber bundles which are not G-bundles for any G? --MarSch 17:01, 12 Jun 2005 (UTC)

Can you think of some thing, any thing, T, that is not acted on any group G? If so, then you can make that thing T into a fiber. So if your base space is U, the fiber bundle is U x T. Unfortunately, it is rather boring because its a plain old product. To get something interesting, you need to define a way of twisting T in some way so that you can hook it up in some non-trivial way across the patches. But if you can twist, you can untwist. So you have identity, and inverses. Before long, you'll need closure too (product of twists is a twist) so bingo -- now T has a group of twists acting on it. So naively, its impossible to have a fiber bundle without a structure group. At least for a finite-dimensional T. Being less naive, or infinite-dimensional, as Fropuff is below, you have to work hard to come up with a set of 'twists' that fail to have a group structure; it gets subtle. For example, the limit of a sequence of continuous functions need not be continuous. You'd have to cook up something where the transition functions were continuous, but as they got assembled into a topology, something went to a non-continuous limit. linas 03:16, 13 Jun 2005 (UTC)

Presumably. I don't know of any off the top of my head. Naively, one can take the homeomorphism group of the fiber as the structure group, but I believe it may not always be possible to give this group a topology compatible with the natural action on F. Moreover, in the smooth case, the diffeomorphism group of F is almost never a Lie group. -- Fropuff 18:14, 2005 Jun 12 (UTC)

There are various topologies one can assign to the homeomorphism group of F. The least interesting topology of all makes this a continuous transformation group (the discrete topology):

G × G → G is continuous (obvious). Likewise for the inverse.
m : G × Y → Y is continuous, since m^-1(U) = ∪_g ({g}, g^-1U).

In the smooth case (and most topological cases of interest), the diffeo (resp. homeo) morphism group is a topological group with the compact-open topology. So there really isn't any loss of generality at all in considering only fibre bundles with structure group. However, I must admit that I've always been uneasy about the logic of such bundles, so I'm happy to see the structure group as an additional piece of structure. Silly rabbit 11:15, 5 May 2007 (UTC)[reply]

fibered manifold

Could someone confirm that this is the same as a fiber bundle?--MarSch 18:57, 26 November 2005 (UTC)[reply]

actually it isn't -MarSch 14:36, 25 January 2006 (UTC)[reply]

I believe a fibered manifold is a manfold with a fibration, not necessarily a fiber bundle. But I could be wrong. -- Fropuff 15:21, 25 January 2006 (UTC)[reply]

Style guidelines, red links, redirects

JA: Linas, I get confused by ½-2-alogs on user talk pages and I like to keep issues attached to specific articles, so I will copy your comments here and respond to them after I get some coffee in my veins. Jon Awbrey 11:32, 6 February 2006 (UTC)[reply]

I noticed several of your edits concerned "style", however, they seem to run against style guidelines. In particular, the see-also section should come before the references, not after. The external links should be merged with references, and external links should be given an author name, a title, a date. (so that they look like references).
Finally, I notice that you are making a lot of pipe constructions, i.e. [[something|somewhereelse]]. May I suggest that the correct way to do this is to create a redirect page instead. Do this by simply using [[somewhereelse]]. Follow the red link, and create a new page. After creating that page, add the following text: #REDIRECT [[something]]. This may not seem like a big difference, but it makes WP much more maintainable. The problem with using pipes is that it hides the true link. In particular, someday, the topic somewhereelse may become a full-fledged article, instead of being a redirect. It would then be very difficult to chase down all instances of [[something|somewhereelse]] and then fix them to make them point at somewhereelse. Thanks linas 06:29, 6 February 2006 (UTC)

JA: Ok, ½-awake now. First off, I didn't mess with the content of the refs or exts, as I was just scanning the article, and normally wouldn't do that on the first pass, and then only if I had the texts in hand, knew what country Paris (not Texas) or Cambridge was in, or had time to follow the links myself. I sort of agree with you about re-writing the external links as refs, and I've already re-done a few of those myself, but that conflicts with preferences expressed in some other parts of the Wiki City, and so I am presently of two minds about that. At any rate, I'm not accountable for what was there when I came in. Get to other issues later. Jon Awbrey 12:14, 6 February 2006 (UTC)[reply]

JA: Next item. I can't make sense of what you're saying about piping. There's already a redirect page, and the purpose of bypassing it is simply to save compute time. Unless I mis-grokked what I read, I was "just follwing orders" on that one. So please let me know if I misunderstood, either what you're saying or what the directive says. Jon Awbrey 17:08, 6 February 2006 (UTC)[reply]

Bypassing redirects is not always the right thing to do. In some cases it is unharmful; for example covering map and covering space will probably always point to the same page, so bypassing the redirect does no harm. But in many instances a redirect was created as a substitute for a proper article. For example, section (fiber bundle) currently redirects to fiber bundle, but someone may one day write an actual article on sections of fiber bundles. If someone had previously bypassed the redirect then the link would now point to the wrong article. In general, I try to avoid bypassing redirects altogether. A redirect is harmless and almost unnoticable to the reader. -- Fropuff 17:31, 6 February 2006 (UTC)[reply]

JA: Yes, I think I intuitively understood that, and only did it when it seemed like a synonym. Maybe I can dig up where I read that directive so somebody can correct it. Jon Awbrey 17:52, 6 February 2006 (UTC)[reply]

Hmm, the compute-time saved, if any, by avoiding a redirect is certainly lost several thousand-fold in having to review/edit/watchlist/discuss these thigs ;-) I doubt there's actual CPU-time saving at all: after all, there's some significatnt calculation that has to happen just to parse the wiki markup into something meaningful to begin with, ie. convert it to html. linas 00:11, 7 February 2006 (UTC)[reply]

JA: Well, meta-cognitively speaking, I'm pretty sure it's not the sort of thing I'd've thought of on my own, but until I run across it again I probably won't remember where I read to do that, maybe on the Redirect instruction blurb, or maybe I misinterpreted something it said about double redirects? Jon Awbrey 04:55, 7 February 2006 (UTC)[reply]

Questions about sections

It is well-known that most fiber bundles do not have global sections. The obstructions to building a global section are discussed (or should be discussed, but are not?) in the articles on universal bundle and equivariant cohomology. In particular, consider the following questions. Suppose I take a local section, and try to extend it, bit by bit, over the whole base space. Eventually, I run into a situation where I can't do this, and a further extension of the section forces it to be "multivalued" over some set of points. Now, for the barage of questions. Lets conduct this by analogy to complex analysis. The square root in complex analysis is a function that covers the complex plane twice, with singularities at 0 and infty, and a cut connecting these two that can be pushed around arbitrarily. It is known that analogous situations occur for fiber bundles: by attempting to promote a local section to a global section, one bumps into singluar points (the so-called Dirac monopole), which are connected by a "cut" (the so-called Dirac string), which can be arbitrarily repositioned. Question: are Dirac monopoles and Dirac string the only kind of singularities that can be encountered? Or are there others? If there are others, do these correspond, in some "clean" fashion, to H_2, H_3, etc? Is it correct to presume that, if I only knew group cohomology better, then the answers to my questions would be self-obvious?

A bundle over a contractible space is always trivial. If you can find a subspace that differs from your base space by a set of measure zero (can you always do this? certainly for surfaces you can) then you can have global sections with "branch cuts" or "Dirac strings" in some sense on that leftover set of measure zero. Oh, by the way, the square root isn't just analogous, it's an example: z --> z² is a principle Z2 bundle. -lethe ^{talk +} 23:51, 20 July 2006 (UTC)[reply]

Next question: In analogy to square-root in complex analysis, there are presumably fiber bundles for which double-valued or n-valued global sections are possible (up to a collection of critical points). Let me call such "multi-valued global sections" n-sections. The value of n here corresponds to number of elts in H_1, right? How many critical points are there? Is it illuminating to try to visualize these critical points in terms of Morse theory, or have I gone off the deep end?

In analogy to the logarithm on the complex plane, there are presumably cases where n becomes the countable infinity. In this case, does the n-section (actually, now an omega-section) have an accumulation point? Or does the section become dense in the total space? Are there omega-sections that ergodically cover the total space, in analogy to the way that 2^n x mod 1 ergodically covers the unit interval?

Last question: are there aleph-one-sections?

Last+1 question: under what situations can the space of connections on a fiber bundle be given a natural measure? linas 16:41, 20 July 2006 (UTC)[reply]

Too technical

After consulting some books, I'm starting to understand this topic, but it could be introduced in a much more accessible way. The Mobius strip example was particularly helpful to me: it is easy to visualize a literal bundle of literal fibers like a bottle brush being shaped into a Mobius strip or into a cylindar. I would love to see the article go directly to answering “what” and “why”. —Ben FrantzDale 04:12, 25 January 2007 (UTC)[reply]

I removed the tag. This article is already pretty accessible. Perhaps there is a more accessible intro possible, but having seen many presentations of this topic, I would say you seem to just be experiencing a typical difficulty in learning about this concept. "Why" could be worked on more (although there are some answers suggested), but that's not really an issue of accessibility per se. --C S (Talk) 05:00, 19 February 2007 (UTC)[reply]

I do not find the article's introduction very accessible, and I think I know what the problem is. First, it says a fibre bundle is a topological space, then it says a fibre bundle consists of a map. For an ordinary reader, that works like a contradiction. At that point, it is not at all clear what roles the sets B and E play in relation to the fibre bundle space or the product space that the previous statement talks about.

I think intro sections should not be so stringently correct. They should provide the basic idea, and be correct only in the sense that the exact meaning of the concept being explained, is one of the possible interpretations of the intro text. The exact definition may follow later. This would be similar to "The United Kingdom is an island nation in western Europe". That description fits Ireland too, yet it gives the reader a starting point.

I suggest it be stated that a fibre bundle is a space E that locally looks like a product space B x F, but globally may (or may not) be quite different. This similarity is reflected through the existence of a map pi:E->B that (fill in your preferred wording, explaining that pi instantiates the local similarity by behaving like a projection from subsets of E to subsets of B). Again, The role, the parallel to the projection in the case of the product space, is important, not the technicalities of the exact definition (which of course are also needed, but later, after the reader has a mental image). Cacadril (talk) 23:27, 12 January 2009 (UTC)[reply]

I think the lead already does state this, more or less. I disagree that lead sections should be exclusively "for dummies". Rather leads should be a summary of the content of the article, more like an abstract than an introduction. An introductory (or "motivation") section, written in more expository terms, can then follow the lead. That said, in essence I agree that the article would definitely benefit from such a section. However, such sections are probably the most difficult kind to write (well) in a mathematics article, so any assistance you can lend would be very much appreciated. Thanks, siℓℓy rabbit (talk) 00:28, 13 January 2009 (UTC)[reply]

Thanks so much. I should now eat my own words regarding the lack of suitability of expository material in the lead. Your edits to the lead look quite reasonable to me. siℓℓy rabbit (talk) 01:05, 13 January 2009 (UTC)[reply]

I think (in my opinion) that the reason why most people don't understand fibre bundles is because they are unfamiliar with the notion of a covering map. I think that one must have at least a good understanding of covering maps to understand fibre bundles (that is probably how fibre bundles were invented: 'generalize covering maps'). PST

{{citation needed}}, if you please. siℓℓy rabbit (talk) 12:59, 13 January 2009 (UTC)[reply]

Now I'm starting to feel a bit worried on closer examination. The fiber bundle isn't just the space E, but also the map πぱい. This edit diminishes the importance of the mapping &pi, treating it more as a matter of convenience than an essential part of the structure, which is, I think, a common mistake for beginners who first study fiber bundles. This problem persists in the second paragraph, where the article now points out that small parts of the Moebius strip are homeomorphic to parts of the circular cylinder. While true, local triviality implies a much stronger statement than just homeomorphism: locally there is a bundle map between the two spaces. siℓℓy rabbit (talk) 14:04, 13 January 2009 (UTC)[reply]

Yes, the article is pretty much rubbish like most other articles in WP. With these type of articles, rewrites generally should come into play (I think I will have a go at the lede now). PST

"Rubbish" is a too harsh. The main shortcomings are a lack of a decent "Introduction" section and an "Application" section. The lead is crippled by the lack of article contents to refer to, but it should still make an effort to summarize the article (and I hope it now at least does that). At any rate, the lead is generally not the place to begin a rewrite. Rather it is better to add sections to an article, and then adjust the lead when the article contents have matured somewhat. siℓℓy rabbit (talk) 16:44, 13 January 2009 (UTC)[reply]

Thanks for lending a hand. As to the importance of πぱい, I too worried about this, but ended up not including in my edit what I was thinking. The Mathworld article http://mathworld.wolfram.com/FiberBundle.html says a fiber bundle is a map πぱい, etc. This may have arisen after some authors lazily have short-circuited, writing "let πぱい:E→B be a fibre bundle" rather than "let E be a fiber bundle with the fiber bundle structure given by the map πぱい..."

In narrow fields, when some authors short-circuit the language finding it convenient, this may easily become standard usage. This may work OK for the insiders, who know what is actually meant, but tends to mystify outsiders. Writers of undergraduate and graduate courses often make an effort to clean up the language. The word "fiber bundle" suggests something similar to a cross product with a fiber as one factor. The map πぱい maps the fibers of the bundle to the base. It may be thought of as binding or attaching the fibers, but then it should be called a bundling or binding, not a bundle. I guess Mathworld's usage is a result of poor sensibility to language. Even if there should be a tendency toward accepting "the fiber bundle πぱい:E→B", the general awkwardness will also push users in the opposite direction. If you and I feel this awkwardness, this is no less part of the language than anything else.

I think the set E becomes a fibre bundle by virtue of the existence of πぱい with its properties, just like homeomorphic spaces become homeomorphic by virtue of the existence of a homeomorphism, a map with certain properties. We may, however, say that the map πぱい defines a fiber bundle structure on E. By analogy, a tangent bundle is a set comprising the disjoint union of all tangent spaces. Consider what it means that the union be disjoint, it actually means that not just the tangent space itself, but the pair (P, T_p) is a term of the union. Here P is a member of M, and T_P is the tangent space at P. This makes the union similar to a cross product with base M and fiber T, but the fibers may not be identical as they would be in a cross product. If the fibers are homeomorphic to each other, the tangent bundle becomes a fiber bundle.

But, as with ordinary English, words change meaning in illogical ways, and at some point one has to accept it as a new reality. Since I am no expert, I have to ask, perhaps the usage of Mathworld is so firmly established that it must be accepted. If so, imagine a dictionary entry "Fiber bundle:

1) A topological space, referred to as the total space, that is locally homeomorphic to a product space, whose factors are called the base space and the fiber space;

2) a map that defines the locality of such homeomorphism, by mapping e in the total space E to b in the base space B whenever for some open neighbourhood U of b, the preimage of U is homeomorphic with U × F, where F is the fiber space."

If none of us know the literature enough to feel competent, let us do whatever makes sense to us until somebody who knows more comes along. If we do something horrible, they sure will show up. But most questions of this kind have no hard correct answer, and the articles will never improve unless somebody dares do it. 81.237.206.192 (talk) 23:14, 13 January 2009 (UTC)[reply]

No, a fiber bundle definitely is the quadruple (E, B, πぱい, F), not just the topological space E. siℓℓy rabbit (talk) 23:23, 13 January 2009 (UTC)[reply]

Notice: The article on tangent bundle is also in desperate need of help. It does not even mention that the tangent bundle is a fibre bundle. The above IP poster seems to think that tangent bundles are not necessarily fibre bundles which is false. He/she also thinks that if all fibres are homeomorphic, the bundle is automatically a fibre bundle which is again false. PST

Actually, the article tangent bundle does indeed mention that it is a vector bundle (and also a fiber bundle). But this information is not contained in the lead of the article, so that it is not given the prominence it deserves. siℓℓy rabbit (talk) 13:38, 14 January 2009 (UTC)[reply]

Having another look at the article, suggests to me that the article is really conveying the wrong information. The tangent bundle is indeed composed of the disjoint union of the tangent spaces but it is certainly not just that. The 'disjoint union' bit conveys that the total space is disconnected (not so in general). The lede is rubbish. It seems that people spend their times improving simple articles but don't bother about these more intermediate ones articles. PST

Diagram needed

It is requested that a mathematical diagram or diagrams be included in this article to improve its quality. Specific illustrations, plots or diagrams can be requested at the Graphic Lab.
For more information, refer to discussion on this page and/or the listing at Wikipedia:Requested images.

This article needs a diagram or two in the bundle maps section. Also, this section needs a discussion of mappings of fibre bundles with structure group. Silly rabbit 10:49, 5 May 2007 (UTC)[reply]

I agree; the real definition of a fibre bundle utilizes the structure group. The article only seems to emphasise upon locally trivial fibrations and does not seem to emphasise upon the real definition of a fibre bundle.

Topology Expert (talk) 04:40, 27 August 2008 (UTC)[reply]

Problems with the article

I am a bit concerned with this article since it fails to explain some of the many important aspects of fibre bundles. For a start the article fails to:

State when two distinct fibre bundles could be isomorphic

Give more examples of fibre bundles and state some of their properties. For instance, in a fibre bundle with a locally trivial fibration (mapping) p, the preimage of any one point set in the base space is homeomorphic to the fibre F. This is already mentioned but such properties should be given a separate section in the article.

Instead of naming fibre bundles that are obvious, why not name some bundles which are not covering spaces, but are also quite interesting examples. I can give a few examples but I am not sure how to put them on Wikipedia since I don't know how to put images and write properly in LaTeX. If someone could help with this, I could improve the article in this aspect.

Also, the article does not really mention the importance of fibre bundles in mathematics (i.e how they are importand and not just why they are important). All the article seems to say currently, is that locally trivial fibrations are just an ordered quadruple of objects that mathematicians play around with. Therefore, the lede of the article needs improving.

I am not concerned with the level of explanation in the article; this seems fine to me. I think one cannot learn such concepts from someone else; he must rather think of examples and properties of fibre bundles himself. However, if anyone thinks that the explanation in the article is insufficient please respond with a reason.

In fact, if you agree or disagree, could you please reply to this with your opinions. I think that it is time that this article is significantly improved.

Topology Expert (talk) 04:37, 27 August 2008 (UTC)[reply]

Yah, the article is pretty cheezy. I changed the examples to be a bit less trivial and I'll poke around at the article some more. IMO one of the main failures is the introduction to the article. A fibre bundle is not a space, it is a map of spaces. Rybu (talk) 03:59, 21 September 2008 (UTC)[reply]

Not every submersion is a fiber bundle

For one thing, the projection map needs to be surjective, and typically submersions are not assumed surjective. Even so, it is still possible to have a submersion in which the fibers are not homeomorphic to each other, let alone not being locally trivial. There is a theorem of Ehresmann that if πぱい is a surjective submersion whose fibers are connected and compact, then it is a fiber bundle. But this result is clearly not optimal, and I think it would be very misleading indeed to state it without indicating that it is, in fact a (not so easy) theorem. siℓℓy rabbit (talk) 11:24, 22 September 2008 (UTC)[reply]

IMO your comments only indicate that the statement of the theorem wasn't fully-qualified -- ie: they should be qualified, rather than deleted. Indeed it is an easy theorem -- one that tends to be proven in a 1st course on manifolds, frequently as a homework assignment, or proven as a simple lemma in Milnor's differential topology notes. One of the problems with this article is that it assumes a rather rigid and not-historical definition of fibre bundle. A locally-trivial fiber bundle in one of the common historical senses of the word, need not have a fixed homeomorphism type for the fibre. In particular, the fiber over some points can be empty, provided the base-space is not connected. Anyhow, I'll see if I can fix up the statement. Rybu (talk) 20:02, 22 September 2008 (UTC)[reply]

I still find it inadequate. When I said "not so easy theorem", I meant that it doesn't follow immediately from the definitions of the terms involved that the thing is a fibre bundle. This makes it particularly poorly suited as an example of a fibre bundle, since there is a not-altogether-trivial theorem needed to give it the structure which it is intended to illustrate. siℓℓy rabbit (talk) 20:19, 22 September 2008 (UTC)[reply]

Where does it say that the only examples on Wikipedia can be trivial examples? And your last edit is wrong. Surjectivity is implied when the base space is connected and total space compact (unless you're being so pedantic to allow for the empty set to be a manifold). IMO now you're just making the example overly complicated by quoting the theorem in maximal generality. Isn't there any effort made towards making the examples easy to state & read? Rybu (talk) 20:43, 22 September 2008 (UTC)[reply]

Yes, I suppose it's true that surjectivity is implied by those two conditions. To be honest, I had a different theorem in mind from the beginning. Also, it doesn't say anywhere that only trivial examples should be presented. However, enough detail should be presented along with the example in order to show that it is actually an example of what it is supposed to illustrate. All I am saying is that I think we should be upfront about this being a theorem. Otherwise it is really misleading to someone still trying to puzzle out the definition of the thing. Perhaps this recent addition of yours should be moved to a different section altogether, where such implications can be discussed more fully. siℓℓy rabbit (talk) 20:55, 22 September 2008 (UTC)[reply]

I suppose that'd make sense. These kinds of standards do not seem to be uniformly applied throught the mathematics wikipedia pages -- sometimes examples follow from big theorems, sometimes they're simple and follow from the definition. Is there a concerted effort to keep examples that follow immediately after definitions to be essentially easy to see? Rybu (talk) 21:00, 22 September 2008 (UTC)[reply]

Did ______ write this article?

I don't want to fill the blank in the section heading (I will give you a hint: it starts with 'i') but something should seriously be done about amateurs (with good intentions) writing such "advanced" (compared to other articles) maths articles. The person who wrote most of this article obviously has no idea what a fibre bundle is (but recent edits by more knowledgeable users have improved the article) and nor has any idea as to why it is important. The first sentence is itself rubbish: "a fibre bundle is a space which looks locally like a product space". Nonsense (I don't think I need to explain why). The most appropriate description would be to note that in a sense a fibre bundle is a collection of spaces, all homeomorphic to a particular space F, parametrized by some space B. So in a sense, you are associating to each point of B, a particular space F in a manner that satisfies certain axioms. When there is a free, transitive action by a (topological) group G on the fibre, the bundle becomes a principal G bundle (again there have to be certain conditions met). There is nothing about such bundles in this article. From what I can see, there is absolutely nothing about the long-exact sequence of homotopy group associated to a fibration (another worrying fact) and nothing about Serre fibrations (although I am totally sure whether there should be something on that in this article). I will start with some improvements now but it would be helpful if other people could help out. PST

Principal bundles are mentioned in the article, both in the current lead and in the section which discussed fiber bundles with a structure group. As for the first sentence, there is a certain tension between the need to make technical articles accessible and to include meaningful mathematical information in them. The first sentence is a commonly repeated "introduction" to fiber bundles. (Actually, I think I wrote that sentence, or an early version of it, borrowing from the manifold article, which is one of the more highly-regarded mathematics articles for its accessibility.) siℓℓy rabbit (talk) 17:07, 13 January 2009 (UTC)[reply]

are there...

better first image to this page? The hairbruse is horripilant :) --kmath (talk) 03:41, 19 March 2009 (UTC)[reply]

Of course there are, see further down the article for examples. But there's not many fibre bundles that embed in 3-dimensional Euclidean space. If you add the further constraints that you want the fibre bundle not to be a covering space, and if you want it to be non-trivial and a manifold, then pretty much all the examples have been given in the article. In some sense the "first" non-trivial bundle which is not directly constructed via a covering space construction (such as the mapping cone of a covering space) is the Hopf fibration. This almost embeds in R^3. Probably the most pleasant non-manifold examples that are sketchable are things like the bundles over the circle with fibre the cone on a bunch of points. There's a lot of semi-visualizable highly non-trivial fibre bundles though that we might want to consider. You can think of RP^3 as this space { (C,p) : C is an oriented great circle on the 2-sphere and p is a point on C }. So you could think of all the lens spaces L_{2k,1} as being the space { (C, S) : C is an oriented great circle on the 2-sphere and S is a regular k-gon inscribed in the circle }, etc. These all fibre over the 2-sphere with fibre a circle, but with different monodromy. Rybu (talk) 04:47, 19 March 2009 (UTC)[reply]

Thanx. I think we should replace the first imagen for some more like-mathematical object, maybe the mobius band or the torus or the klein bottle be less horripilant--kmath (talk) 22:14, 20 March 2009 (UTC)[reply]

I think the hairbrush is an excellent lead picture, and I'd like to thank whoever added it, and say I'm glad it was kept. I'm a graduate electrical engineer, so I'm not totally unexposed to higher math, but its hard to visualize from abstract definitions in exactly what ways a "fiber bundle" is analogous to a fiber in physical experience. The hairbrush picture and its caption really nailed it for me, and I think it would be even more important for nontechnical readers. Pictures of the more interesting examples you mention could be added, but I think the hairbrush should be kept. Thanks again. --Chetvorno^TALK 02:47, 31 March 2018 (UTC)[reply]

F->B->E just notation or real math?

The article contains the following sentence:

"A fiber bundle (E, B, πぱい, F) is often denoted File:FiberBundle-02.png to indicate a short exact sequence of spaces."

I am aware that $F\to E \stackrel{\pi }{\to }B$ is a common way to notate fibre bundles. I think, however, that it's just meant to indicate that this is somewhat like a short exact sequence. I don't think that it really is a short exact sequence, as I don't see a canonic morphism $F\to E$. I could imagine that there are cases where $B\times F \to E \to B$ is a short exact sequence. Shouldn't the article be clear about the point, whether this either _is_ a short exact sequence, or whether it's not? -- 131.220.192.246 (talk) 13:58, 16 June 2009 (UTC)[reply]

IMO that comment is misleading. A fibre bundle is a fibre bundle. There is no notion of short exact sequence of spaces. The analogy is the homotopy long exact sequence of a fibration. No point being vague and misleading. Rybu (talk) 08:46, 17 June 2009 (UTC)[reply]

I think so too. But I fear that I just don't get it and that the problem lies with me. I think so mainly, because I find this notion in the literature. Just now I saw it in Lawson, Michelsohn: Spin Geometry. -- 78.54.224.191 (talk) 12:08, 17 June 2009 (UTC)[reply]

When you say you just saw it, do you mean Lawson has an explicit defininition of a short exact sequence of spaces? Rybu (talk) 04:22, 19 June 2009 (UTC)[reply]

No, I guess he's using it notationwise, but maybe I'm wrong. In his context, the $P$s are principal bundles, X is the manifold, E a vector bundle above X. See Spin Geometry, 1st edition from 1989 [there is a second edition from 1994], p79: "...Then from the fibration $O_n \to P_O(E) \to X$, there is a exact sequence. 0\to H^0(..)\to H^0...." and he's also using it in commutative diagrams as on page 80: Z_2 \to Spin _n \to SO_n and below that Z_2(same as above) \to P_{Spin _n}(E) \to P_{SO}(E) and below that the manifold X. Namely: There are arrows in the commutative diagram from $SO_n$ to $P_{SO}(E)$ and from $Spin _n$ to $P_{Spin _n}(E)$ -- 78.48.213.58 (talk) 10:27, 19 June 2009 (UTC)[reply]

ah, okay. IMO it looks like he's just being lazy. It's a very common thing for mathematicians to do. When there's such a strong analogy between objects in two different subject areas, and when they're so closely linked together you start to blur the notions. For a wikipedia article I strongly encourage avoidance of this kind of laziness. Because if you start calling fibre bundles short exact sequences of spaces you might as well start calling covering spaces galois extensions of spaces and then you have overlapping analogies that aren't entirely coherent. Rybu (talk) 17:19, 19 June 2009 (UTC)[reply]

Is this a common issue in mathematics? I wouldn't call it lazy to use an analogy at all - it's often more work finding a more accessible concept with threads to connect where you need them - only lazy to omit that it is an analogy, where the threads may unravel, etc. I guess I'm not sure if the intended audience is someone brushing up or learning from scratch, but without some kind of analogy I'm usually clicking through some links more than once before both the individual terms and their relationship to one another really stick with me. I mean... analogies can't be that detrimental if the standing thumbnail is a hairbrush, right? (For what it's worth, my name for 'fiber bundle' is 'hedgehog particle' - you're welcome.) I was swayed somewhat by Dijkstra's "On the Cruelty of Really Teaching Computing", but personally I've found that analogies pave the way to understanding without them. Sobeita (talk) 20:21, 31 March 2018 (UTC)[reply]

I don't see the problem here; there is in fact a notion of short exact sequences of mappings of (pointed) topological spaces. We can say that the kernel of some map f:X->Y is the preimage of the base point of Y. So here, the kernel of our fibre map F is the image of the inclusion of F. Also note that the inclusion of the fibre is 1-1 and that the fibre map is onto which ties in nicely with the idea of a SES.

I don't know if this really needs to be said in the article though, and I'm sure that the article is getting at the point that the induced maps of homotopy groups is a SES, not what I've just said... thinking about it, maybe this should be mentioned, the LES of homotopy groups is quite important. 81.148.49.125 (talk) 15:17, 24 April 2010 (UTC)[reply]

Principal bundles are a special case?

From the introduction: Fiber bundles can be generalized in a number of ways, the most common of which is requiring that the transition between the local trivial patches should lie in a certain topological group, known as the structure group, acting on the fiber F — surely this is a special case, not a generalisation, of a fibre bundle? Jowa fan (talk) 06:19, 19 February 2011 (UTC)[reply]

No, this is not necessarily a principal bundle, since the fibre does not need to be a group.

But you are entirely correct in saying this is not a "generalization" of the fibre bundle. Rather, this is a fibre bundle, exactly as defined in Steenrod. (Of course, the transition maps must satisfy a consistency condition as well.)

Simplest fibre bundle? I don't think so

The section titled Möbius strip begins as follows:

"Perhaps the simplest example of a nontrivial bundle E is the Möbius strip."

It would be hard to come up with a nontrivIal bundle simpler than the fibre bundle given by the circle double-covering the circle. This may be defined by the map f: S¹ → S¹ via f(z) = z², where S¹ is the unit circle in the complex plane ℂ. The fibre here is the simplest nontrivial group, the group of two elements, and this bundle is identical with its associated principal bundle, which is also the associated principal bundle of the Möbius strip.

This bundle is also only 1-dimensional with total space a manifold, whereas the Möbius strip is 2-dimensional with total space a manifold-with-boundary. It should definitely be presented here as the simplest bundle.

It is also very misleading to characterize a fibre bundle as being defined just by its total space. There are many examples to show that the same space can be the total space of various inequivalent nontrivial fibre bundles. For example, for any positive integer n, the circle is the total space of the fibre bundle ξくしー_n defined by the bundle projection p_n: S¹ → S¹ via p_n(z) = zⁿ, and all of these are inequivalent. (Of course, there is also the group of the bundle to consider, but it is not necessary to be that technical at this point.)

Therefore, I propose that the statement quoted be changed to read:

"One of the simplest fibre bundles has total space the Möbius strip, with bundle projection equal to the projection of the strip onto its core circle, and with fibre equal to the closed unit interval."

Daqu (talk) 00:59, 17 January 2015 (UTC)[reply]

External links modified

Hello fellow Wikipedians,

I have just modified one external link on Fiber bundle. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:

Added archive https://web.archive.org/web/20040808115056/http://planetmath.org/encyclopedia/FiberBundle.html to http://planetmath.org/encyclopedia/FiberBundle.html

When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.

This message was posted before February 2018. After February 2018, "External links modified" talk page sections are no longer generated or monitored by InternetArchiveBot. No special action is required regarding these talk page notices, other than regular verification using the archive tool instructions below. Editors have permission to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the RfC before doing mass systematic removals. This message is updated dynamically through the template {{source check}} (last update: 5 June 2024).

If you have discovered URLs which were erroneously considered dead by the bot, you can report them with this tool.
If you found an error with any archives or the URLs themselves, you can fix them with this tool.

Cheers.—InternetArchiveBot (Report bug) 08:53, 30 September 2017 (UTC)[reply]

"cutting" the Mobius strip

In the section on the Möbius strip, it reads: "locally the Möbius strip and the cylinder are identical (making a single vertical cut in either gives the same space)". It seems to me that the word "cut" here is ambiguous, and what may be clearer would be "creating a boundary on the surface". If one were really to cut the Möbius strip, that would be creating two separate spaces -- each side of the strip. Making a single cut in the cylinder would *not* yield two equivalent spaces. But setting up a boundary on (one side of) the Möbius strip would be equivalent to setting up a boundary on the cylinder.

Zeroparallax (talk) 20:34, 26 September 2018 (UTC)[reply]

Möbius figures

I think the example of the Möbius strip would be easier to follow with a labeled diagram, rather than identifying the relevant features using textual descriptions. Can someone please check whether these figures, created in Mathematica based on the existing Möbius strip diagram, are correct?

Stevvers (talk) 20:28, 11 May 2024 (UTC)[reply]