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I have read in some lecture notes in the internet (without reference) the following result:

Let $E$ be a differentiable sphere bundle whose base $B$ has dimension $n\geq 2$ and whose fibers $F$ have dimension $n-1$. If $B$ is non-compact, then it admits a global smooth cross-section.

The proof is somewhat technical (involves sections with isolate singularities and how to remove them by sending them to infinity) and seems to be correct, but I would like to know a reference in the published literature for this result. In fact, the case $n=2$ would be enough. Thanks!

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How do they say that 'hence the bundle is trivial' works? A section is not enough. –  Michael Murray Oct 16 '12 at 12:49
    
Ok, I was mistaken. What they prove is that it admits a global section. Note that the fibers are circles. –  Manzano Oct 16 '12 at 12:57
    
The title should be "circle bundle" instead of "sphere bundle" –  John B Oct 16 '12 at 18:54

4 Answers 4

up vote 4 down vote accepted

I am guessing that you misunderstand the claim.

Here are some facts that seem to be in the right ballpark: a bundle whose fibers are spheres of dimension $n$ over a base whose dimension is $n$ or less will certainly have a section. And if the base $B$ has dimension $n+1$ then it will have a section if and only if the Euler class is zero; this is a cohomology class in $H^{n+1}(B)$, or a twisted version of this if the bundle is nonorientable. If the $(n+1)$-manifold $B$ is connected and noncompact then this cohomology group is zero, so a section exists.

So if you assume $n$ is meant to be dimension of fiber rather than total space then the claim makes sense but with $n+1$ rather than $n-1$ for the base. If the sphere bundle is the unit sphere bundle of a vector bundle and $n$ is the fiber dimension of the vector bundle, then you can change the $n+1$ to $n$, but that's still off by $1$.

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Thanks for your explicit answer! I only wanted the existence of a section for $n=2$ in order to simplify a proof but I know almost nothing about this theory, so I decided to pose my question here. Now, I have corrected the original statement. –  Manzano Oct 17 '12 at 11:12

This is NOT an answer but since the OP's origional statement is not clear, here is the theorem in the book of Greub, W., Halperin, S. and Vanstone that OP mentioned in his own answer:

Theorem III: Every sphere bundle with fibre dimension $n - 1$ over a connected base manifold of dimension $n\ge 2$ admits a cross-section with a single singularity. If the base is not compact, then the bundle admits a cross-section without singularities.

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Thanks for pointing that out. I was confused since I only wanted the result for $n=2$ and the actual statement and what I wrote coincide for $n=2$. Now, I have rewritten the original statement. –  Manzano Oct 17 '12 at 11:04

Note that the bundles on a product $\mathbb{R}^m\times S^1 $ are isomorphic to bundles obtained by pullback from bundles on $S^1$.

Bundles over $S^1$ with fiber a smooth manifold $F$ are classified buy the connected components of ${\rm Diffeo}\;(F)$. (Think monodromy.) The group ${\rm Diffeo}\; (S^n)$ has at least two components because the orientation preserving diffeomorphism are not homotopic to orientation reversing diffeomorphisms. Hence there are at least two topological types of sphere bundles over $S^1$, and thus at least two topological types on noncompact manifolds of product type $\mathbb{R}^m\times S^1$.

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Having a section does not imply triviality. –  Liviu Nicolaescu Oct 17 '12 at 0:56

The question has been solved because I found the reference after about 10 hours of searching!

Greub, W., Halperin, S. and Vanstone, R. Conections, Curvature and Cohomology.

It is Theorem III in Chapter VIII, Section 5.

Thanks, anyway!

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Hi Manzano, I'm a little confused by your claim now. The theorem you quote says "Every sphere bundle with fibre dimension $n-1 \geq 1$ and connected base manifold of dimension $n$ admits a cross-section with a single singularity. Morever if the base manifold is not compact then the bundle admits a cross-section with no singularity." So if you want a circle bundle, i.e. a fibration with fibre diffeomorphic to the circle, then you must have $n=2$ and hence this theorem tells you that a circle bundle over a non-compact manifold of dimension $2$ admits a cross-section. –  Michael Murray Oct 16 '12 at 23:31
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Notice that you can't do better than this as the Hopf fibration $S^3 \to S^2$ is a circle bundle not admitting a cross-section and hence $S^3 \times {\mathbb R}^m \to S^2 \times {\mathbb R}^m$ gives a circle bundle not admitting a cross-section over a non-compact, connected base manifold of any dimension greater than two by taking $m > 0$. –  Michael Murray Oct 16 '12 at 23:31
    
I am definitely not an expert on fiber bundles, so I messed everything up when I stated the theorem I wanted. As M. Murray says, I want existence of section for total space of dimension $3$, base of dimension $2$ and fiber a circle. Thanks for pointing out the counterexample so all is clearer now!! –  Manzano Oct 17 '12 at 10:59

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