So, I'm a little confused about one statement made in class today :

If M is a smooth manifold without boundary such that the tangent bundle of M is trivial, then M is orientable.

Is this always true ?


If you think of an orientation of a manifold as an orientation of the tangent space at each point which varies continuously, then if your tangent bundle is of the form $M\times \mathbb R^n$, you can use a fixed orientation on $\mathbb R^n$ to orient each $T_pM$. So yes, this is true in general. :)

  • $\begingroup$ Always ?, is it possible for me to find a counterexample to this statement ? $\endgroup$
    – user8169
    Mar 12 '11 at 23:14
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    $\begingroup$ @Danny: The proof implies it is always true, so there will not be a counterexample. Maybe if you explain why you think this statement is dubious, I (or someone else) could better address your concerns. Also, you probably realize that this is not an "if and only if." There are lots of orientable manifolds with twisted tangent bundles. $\endgroup$ Mar 13 '11 at 0:39
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    $\begingroup$ Like $S^2$ .... $\endgroup$ Mar 13 '11 at 3:35
  • $\begingroup$ @Derek: Did I call you Danny by accident? I'm so sorry about that! $\endgroup$ Mar 22 '11 at 17:50

Assume you have a smooth manifold $M$ whose tangent bundle is trivial and assume that $M$ is not orientable. Then since $M$ is not orientable there exists a loop $\gamma \in M$ such that when you go around the loop and come back where you started you obtain the reversed orientation. More explicitly, start at some point $x \in \gamma$ with some local orientation($\mu_x)$ and go around the loop by making continuous choice of orientations. When you come back to $x$ you obtain $- \mu_x$ which shows that neighborhood of $\gamma$ is Mobius band.

Now consider tangent bundle $TM$(which is trivial) and restrict it to neighborhood of $\gamma$, then you get a nontrivial(twisted) bundle which gives a contradiction. We must have obtained a cylinder($\gamma \times \mathbb{R}^n$) not a nontrivial bundle(open Mobius band in low dimension). This argument shows that we cannot have such loop $\gamma \in M \leadsto M$ must be orientable.

Another interesting theorem says that any simply connected manifold is orientable. You can try to prove this by similar argument.


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