Smooth Manifold with Trivial Tangent Bundle

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 ?

2 Answers

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. :)

• Always ?, is it possible for me to find a counterexample to this statement ?
– user8169
Mar 12 '11 at 23:14
• @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. Mar 13 '11 at 0:39
• Like $S^2$ .... Mar 13 '11 at 3:35
• @Derek: Did I call you Danny by accident? I'm so sorry about that! 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.