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So, I am trying to solve the following problem. Suppose you have a nowhere zero smooth vector field on a 2 dimensional oriented compact manifold. Prove that the unit tangent bundle $TU$ is diffeomorphic to $M\times S^1$.

So, I can see that at each point of $M$ the unit tangent space is isomorphic to $S^1$, but I don't see how to use the nowhere vanishing vector field.

First, I need to convince myself that over small enough neighborhoods we have a local trivialization. Then maybe by compactness I could cover the manifold with finitely many trivially covered neighborhoods. And maybe then use the nowhere vanishing vector field to connect them somehow, but I am not sure.

Or, maybe there is a clever way of writing the diffeomorphism using the vector field.

Any hints or thoughts would be appreciated, thanks in advance!

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  • $\begingroup$ Was $M$ assumed to be orientable? If not, the result doesn't hold (see aes' comment below). $\endgroup$ – Michael Albanese Jan 3 '15 at 23:00
  • $\begingroup$ Yes, I fixed it. $\endgroup$ – TheManWhoNeverSleeps Jan 3 '15 at 23:55
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First note that if $V$ is a nowhere vanishing vector field, then $\widehat{V} = \frac{V}{\|V\|}$ is a global section of the unit tangent bundle.

Let $p \in T_mU \cong S^1$, then there is $z \in S^1$ such that $p = z\widehat{V}(m)$. The diffeomorphism $TU \cong M\times S^1$ is given by $p \mapsto (\pi(p), z)$.

This is very similar to the proof that a line bundle which admits a nowhere zero section is trivial.

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    $\begingroup$ Maybe I missed a condition in OP, but this doesn't work if $M$ isn't orientable (then $TM$ doesn't reduce to a $\mathbb{C}$-bundle and your argument doesn't work). In fact it's false, as, letting $K$ be the Klein bundle, the total space of $UK$ is orientable but $K \times S^1$ is not. $\endgroup$ – aes Jan 3 '15 at 19:14
  • $\begingroup$ @aes: You're correct of course. I've asked the OP for clarification. $\endgroup$ – Michael Albanese Jan 3 '15 at 22:59
  • $\begingroup$ @aes Could you clarify why the above proof doesn't work without orientability? $\endgroup$ – TheManWhoNeverSleeps Jan 3 '15 at 23:56
  • $\begingroup$ @TheManWhoNeverSleeps The formula $p = z\hat V(m)$ makes no sense unless $TM_m$ has a complex structure. The condition that an $\mathbb{R}^2$ vector bundle has a complex structure is the same as the condition that it has an orientation. $\endgroup$ – aes Jan 4 '15 at 5:02

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