Suppose $M^m$ is a manifold with boundary. If we are given an orientation for $M$, we can then derive an orientation for $\partial M$ by considering the orientation of $TM$ at $\partial M$ and then using an outward-pointing vector to get an orientation of $T(\partial M)$.

This made me wonder: is it possible that $M^\circ = M \setminus \partial M$ is orientable but $M$ is not? Is it possible that $M^\circ$ is orientable but $\partial M$ is not?


No, if $M^\circ$ is orientable then $M$ is orientable, and hence $\partial M$ is orientable. Here's two proofs. They're both secretly the same.

1) The obstruction to orientability is the first Stiefel-Whitney class $w_1(M) := w_1(TM) \in H^1(M;\mathbb Z/2)$. Fact: if $f$ is a smooth map, then $f^*w_1(\xi) = w_1(f^*\xi)$. Then $TM^\circ = \iota^* TM$, where $\iota$ is the inclusion map; and note that $\iota$ induces an isomorphism on cohomology, as it's a homotopy equivalence. So under the obvious identifications of cohomology rings, $w_1(TM^\circ) = w_1(TM)$, so $TM^\circ$ is orientable iff $TM$ is.

2) An orientation is a nonvanishing section of the top exterior power of the tangent bundle $\Lambda^n TM$. Given a nonvanishing section of $\Lambda^n TM^\circ$, put a fiberwise metric on $TM$ and suppose this section is of unit norm. Then in a small neighborhood of any point $p \in \partial M$, there is one and only one way of extending this to a section of $TM$. (Remember that locally in a chart $U$, the unit sphere bundle of $\Lambda^n TM$ looks like $\mathbb Z/2 \times U$.) So extend your section! There is no ambiguity, and this is again a smooth section by definition of the fiberwise metric varying smoothly. So one can always extend an orientation of $\Lambda^n TM^\circ$ to $\Lambda^n TM$ if you desire.

You can rewrite 2) in terms of nonvanishing differential forms of top degree, if you like. This may be more intuitive than thinking about it in terms of bundles. The replacement for the metric on $\Lambda^n TM$ is to scale your choice of form so that $\omega_p(X_1, \dots, X_n) = \pm 1$ if the $X_i$ are an orthonormal frame at $p$.

  • 1
    $\begingroup$ All of these proofs require a Riemannian structure. But you can localize the use of it to be near the boundary - so all you really need is that the boundary be paracompact. I have no idea what you can say when the boundary is not paracompact. If I had to hazard a guess, I would say it's false in this case. $\endgroup$ – user98602 Aug 18 '15 at 3:11
  • $\begingroup$ (The Riemannian structure shows up in the first proof because [$w_1(\xi) = 0$ iff $\xi$ is a trivial line bundle] is only true in the case that $\xi$ can be given a fiberwise Riemannian metric, and in the third because you need it to prove the existence of collar neighborhoods.) $\endgroup$ – user98602 Aug 18 '15 at 3:28
  • $\begingroup$ Collaring is really begging the question: It relies on the triviality of the normal bundle of $\partial M$ in $M$, which in turn is equivalent to orientability of $\partial M$ (as a line bundle is trivial if and only if it is orientable). $\endgroup$ – Ted Shifrin Aug 18 '15 at 3:57
  • $\begingroup$ @TedS: Good point, I somehow missed that. Deleted. $\endgroup$ – user98602 Aug 18 '15 at 3:59
  • $\begingroup$ @TedShifrin I do not understand your point. $\Bbb RP^2$ is not oriented but its normal bundle in $\Bbb RP^2\times I$ is. The fact that $\partial M$ is co-oriented by an outward (or inward) pointing vector field (after constructing one with a suitable metric) shows that $TM|_{\partial M}=T\partial M \oplus \Bbb R$. It is easy to see from that then $TM|_{\partial M}$ is orientable iff $T\partial M$ is. $\endgroup$ – PVAL-inactive Aug 18 '15 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.