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Suppose I have a submanifold $M \subset \mathbb{R}^{n}$, of dimension $n-1$. Where a unit normal vector field is a section $\nu$ of the normal bundle $ TM^{\bot} \to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $\| \nu(p) \| =1$, for all $p$ in $M$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. And I tried to prove this, but I just get stuck everytime. I tried looking at the local flatness definition of a submanifold, and I know I need to do something with the orientation of $\mathbb{R}^n$ and the fact that $\nu$ is smooth. But at this point I'm just stuck and do not know how to prove it.

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    $\begingroup$ When you pull the tangent bunde of $\mathbb{R}^n$ back to $M$, it splits into the direct sum of the tangent bundle of $M$ and the normal bundle to $M$. The tangent bundle of $R^n$ is orientable hence so is the pullback, and the existence of a unit normal vector field is equivalent to the normal bundle being trivial (which is equivalent to orientability for line bundles). Now you could argue that a sum of a bundle with an orientable bundle is orientable iff the bundle is orientable (a fast argument would be with the first Stiefel-Whitney class, which is additive and 0 if the bundle is trivial). $\endgroup$ – Aleksandar Milivojevic Dec 2 '18 at 20:25

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