# Are vector bundles principal bundles?

1. Apr 27, 2008

### mma

As far as I know, principal bundle is a fiber bundle with a fiber beeing a principal homogeneous space (or a topological group). According this definition vector bundle is a special principal bundle, because vector space with vector addition as group operation is a topological group.
But I feel that something is wrong with this, because there is a theorem that a principal bundle is trivial if and only if it has a global section (see C. Nash, S. Sen: Topology and Geometry for Physicists, p. 152). Möbius strip (regarded as vector bundle) has a global section (the identically 0 section) but it isn't trivial. Where is the error?

2. Apr 27, 2008

### Hurkyl

Staff Emeritus
Are you sure the Möbius strip is an R-bundle over the circle? How does R act on it?

3. Apr 27, 2008

### Doodle Bob

A key element of the definition of principal fibre bundles is that the transition mappings of a principal fibre bundle are left multiplications of elements from the group G. For a vector space viewed as an additive group, these are translations. But the transition mappings of a vector bundle are contained in the general linear group of the relevant dimension and in fact will *never* be nontrivial translations.

So, nontrivial vector bundles are not principal fibre bundles. Although, of course, every vector bundle induces a principal Gl(n,R)-fibre bundle consisting of its fibre-wise endomorphisms.

4. Apr 28, 2008

### mma

Yes, this is what I forgot. In other words, in the case of principal bundles, G (the symmetry group of the fiber) must be the structural group of the fiber bundle, and cannot be an arbitrary group. The Möbius vector bundle doesn't comply this. But for example the cylinder does. Thank you!

5. Apr 28, 2008

### mma

I think that my trouble caused that some sources make a difference between principal bundles and principal G-bundles. (e.g. http://en.wikipedia.org/wiki/Principal_bundle), while others call principal bundle only principal G-bundles.
The equality of the symmetry group of the fiber and the structure group of the fiber bundle is required only in the case of principal G-bundles, but for the more general general principal bundles isn't. My book talks about principal bundles but it means principal G-bundles. The theorem cited in the first post is valid only for principal G-bundles.

So, I think that the correct answer to my question is that vector bundles are all principal bundles, but not all of them are principal G-bundles.

6. Apr 28, 2008

### Doodle Bob

I would say that in the vast majority of the literature the following phrases mean the same thing: principal bundles, principal fibre bundles, principal G-bundles. So, it's best to assume that this is the case when reading a refereed paper or text.

Wikipedia is famous for not cohering very well with the current state of mathematical jargon. My favorite is "polychoron" which is supposedly the term for 4-dimensional polytopes. But a quick search of the literature (in MathSciNet) reveals that not a single refereed paper uses it.

7. Apr 28, 2008

### OrderOfThings

This reminds somewhat of the fact that parallelizability of a manifold is equivalent to the existence of a global section in the frame bundle over the manifold. The frame bundle is a principal bundle with group GL(n).

For a one-dimensional manifold the frame bundle and tangent bundle are the same.

8. Apr 30, 2008

### OrderOfThings

The cylinder and the Möbius strip are principal bundles with group $(\mathbf{R},+)$. They are also vector bundles. The cylinder and the Möbius strip with the zero section removed are principal bundles with group $(\mathbf{R}\backslash 0,\cdot)$. The cylinder without the zero section is the frame bundle of the circle.

9. Apr 30, 2008

### Hurkyl

Staff Emeritus
Are you sure the Möbius strip is an R-bundle over the circle? How does R act on it?

10. May 1, 2008

### OrderOfThings

Erh, well,...

11. May 1, 2008

### Doodle Bob

I'm not sure that I understand the concern here. Clearly, it's not the actual bounded Moebius strip that is being discussed but it's unbounded cousin, which can be defined piecewise over the circle by letting U and V be open sets on S^1 such that their intersection is two disconnected arcs, setting U'=UxR and V'=VxR, defining an equivalence relation on the disconnected union of U' and V' by: For u' in U' and v' in V', u'~v' iff u'=(x,t) and v'=(x,-t) (and otherwise an element is equivalent just to itself), then setting M=(U' union V')/~.

This both defines the unbounded Moebius band and gives it a vector bundle structure over S^1. We have a local R-actions on both trivializations with a certain amount of compatibility, which is all that we would want anyway. I don't think that there has to be a *global* G-action on a principal fibre bundle, only a very well-behaved local action. Similarly, there isn't necessarily a global Gl(V)-action on a vector bundle with fibres isomorphic to V.

12. May 1, 2008

### Hurkyl

Staff Emeritus
The notion of a global G action on a bundle over a manifold M is equivalent to the notion of a local action by the trivial bundle GxM on that same bundle.

A G-bundle is not "a bundle upon which one can locally define an action of G, and it's not even "a bundle upon which one can globally define an action of G"... a G-bundle is "a bundle for which we have chosen a specific action of G"... and the Möbius strip does not have a global G action. Nor can you define a consistent 'local' G-action -- as you piece together the local G-actions, you will either have to run into a discontinuity someplace, or a fiber where the action is trivial (which violates the condition that G acts freely and transitively)

Now, what we do have is that Möbius strip can act on itself by addition. (For comparison, it doesn't have a multiplicative action on itself! The natural multiplication operation for sections of the Möbius strip actually has its values in the cylinder.

Last edited: May 1, 2008
13. May 1, 2008

### Doodle Bob

Sorry, Hurkyl, I was getting confused over the terminology. When I say "R-bundle", I was thinking of R as a 1-diml. vector space and thus was trying to describe the Moebius strip as a *vector* bundle over S^1, which is what I did. This must be one of the reasons for the "principal" in the phrase "principal fibre bundle."

14. May 1, 2008

### Hurkyl

Staff Emeritus
The relevant part is being a G-bundle -- that terminology (like G-set or G-module) is used for a bundle together with an action by the group G. The "principal" part just adds a bit more about the behavior of the action.

Yes, I agree that the Möbius strip is, in fact, a vector bundle over R.

15. May 2, 2008

### Doodle Bob

It's that damn R, which can mean anything from a vector space to a group to a field to just a simple 1-diml. manifold -- I've read many a paper referring to "S^2-bundles," which of course do not have a global S^2-action on them.

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