Talk:Fiber bundle

Contents

1 spelling 2 Fibers 3 Global sections 4 Clutching construction 5 Čech cohomology 6 G-atlas and atlas

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)