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Let $\Sigma$ be a connected noncompact orientable surface. I'm not assuming that $\Sigma$ is of finite type or anything -- for instance, I'm allowing $\Sigma$ to be the $2$-sphere minus a Cantor set. I'm pretty sure that the tangent bundle of $\Sigma$ is trivial. Here are two pieces of evidence.

  1. This is true if $\Sigma$ can be embedded in a closed surface. Proof : it is easy to see that a closed surface minus a point has a trivial tangent bundle, so $\Sigma$ must have one too.

  2. If $U\Sigma$ is the unit tangent bundle, then we have the standard short exact sequence $$1 \longrightarrow \mathbb{Z} \longrightarrow \pi_1(U\Sigma) \longrightarrow \pi_1(\Sigma) \longrightarrow 1,$$ where the $\mathbb{Z}$ is the loop around the fiber. Since $\Sigma$ is noncompact, the group $\pi_1(\Sigma)$ is free, so this short exact sequence splits (just like you would have if the tangent bundle was trivial).

Does anyone know how to prove this in general? Thanks!

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1 Answer 1

up vote 11 down vote accepted

In general, the tangent bundle of a smooth $n$-manifold $M$ is classified by a (homotopy class of) map $\phi:M\rightarrow BO(n)$. The manifold $M$ is orientable iff there is a lift of this map to $BSO(n)$.

For $n=2$, and $M$ orientable, we see see that the tangent bundle to $M$ is classified by a map $\phi:M\rightarrow BSO(2) = BS^1 = \mathbb{C}P^\infty$. But homotopy classes of maps from $M$ into $\mathbb{C}P^\infty$ is canonically isomorphic to $H^2(M)$, and in this case this group is trivial since $M$ is not compact. It follows that there is a unique (up to isomorphism) rank 2 orientable vector bundle on $M$. Since both $TM$ and the trivial bundle are orientable, they must be isomorphic.

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