If $M^n$ is a submanifold of $\mathbb{R}^{n+k}$ such there are L.I. vector fields $v_1,\ldots,v_k$ such that $v_i \perp T_pM$ for every $p \in M$ then $M$ is orientable.

My attempt is:

Once $M^n$ is embedded on $\mathbb{R}^{n+k}$ take $dx_1\wedge \ldots dx_n \wedge dx_{n+1} \wedge \ldots \wedge dx_{n+k}$ the standard volume form on $\mathbb{R}^{n+k}.$

Lets define

$$\mu_M = \iota_{v_1,\ldots,v_k}dx_1 \wedge dx_2 \wedge \ldots \wedge dx_n\wedge \ldots\wedge dx_{n+k}.$$

This is a volume form to $M$!

We say that $M$ is positively oriented if $\mu_M (X_1,\ldots,X_n) > 0$ and negatively otherwise.

Is this right?


1 Answer 1


Yes, it's correct. (You should probably comment that $\mu_M$ is nowhere-zero because of the linear independence of the $v_i$. Can you check that?) You also have a trivialization of the normal bundle, so it's orientable, and that means the tangent bundle is orientable as well. (Any time two of three vector bundles in a short exact sequence are orientable, so is the remaining one.)

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    $\begingroup$ In fact, the tangent bundle is stably trivial, so $w_i(M) = 0$ for all $i > 0$. For $i = 1$, this is just the statement that $M$ is orientable. $\endgroup$ Commented Jul 25, 2016 at 21:33
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    $\begingroup$ Sure, @Michael, but that's using a cannon—and some unnecessarily advanced mathematics—to swat a fly. :) $\endgroup$ Commented Jul 25, 2016 at 21:35
  • $\begingroup$ thank you! Yes, I can verify that it is never zero. In fact it comes from the normality of $v_i$. Since they are L.I. and ortogonal to $T_pM$ the claim follows. $\endgroup$ Commented Jul 25, 2016 at 21:39
  • $\begingroup$ Of course Ted :) I just thought it was nice to point out that the hypotheses have greater implications than just orientability. $\endgroup$ Commented Jul 25, 2016 at 21:47

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