Given a bundle whose base space is a "cylinder", i.e. $\mathbb{S}^1 \times \mathbb{B}^n$ where $\mathbb{B}^n$ is an $n$-ball, is orientability (of the total space) enough to ensure that the bundle is trivial, and how does one go about showing this.

I would basically like to understand the next simplest case after that of bundles with contractible base space, which are all trivial.

It feels like there should be a simple answer, that doesn't involve the diffoemorphism class of the fibres, possibly using some homotopy-type argument to reduce the question to a bundle with base space $\mathbb{S}^1$.

Also, are there any good references I should be looking at?


  • $\begingroup$ The classic reference is Steenrod's Topology of Fiber Bundles. $\endgroup$ – Ted Shifrin Sep 18 '14 at 14:55

Steenrod's Topology of fiber bundles is indeed the place to look - thanks Ted Shifrin for the suggestion.

Theorem 11.4: If $X$ is a $C_\sigma$-space [normal and locally compact], then any bundle $\mathcal{B}'$ over the base space $X \times \mathbb{B}^1$ is equivalent to a bundle of the form $\mathcal{B} \times \mathbb{B}^1$.

So my question reduces to the same question for a bundle over a circle, and then one can find examples of orientable non-trivial bundles.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.