3
$\begingroup$

Let $X$ a codimension 1 smooth submanifold of the n-dimensional smooth manifold $Y$. Assume $Y$ is oriented. We want to show that $X$ is orientable if and only if it admits a global smooth normal vector field (in Y).

How can we prove this? I have no idea how to even begin...

$\endgroup$
  • $\begingroup$ Also, what do you mean by "normal vector field"? Are you assuming that $Y$ has a Riemannian metric? Or do you just mean that the vector field is transverse to $X$? $\endgroup$ – Jesse Madnick Dec 10 '12 at 2:19
  • $\begingroup$ Yes, I meant $X$, sorry for that. $\endgroup$ – Bernard Dec 10 '12 at 2:22
  • $\begingroup$ Um, I guess so, although I don't know too much about Riemannian metrics. Basically I'm thinking of $Y$ being embedded in some Euclidean space. $\endgroup$ – Bernard Dec 10 '12 at 2:23
  • 1
    $\begingroup$ Guillemin and Pollack problem 18 p. 106 $\endgroup$ – Bernard Dec 10 '12 at 2:27
  • 1
    $\begingroup$ What's your definition of "oriented"? $\endgroup$ – Jason DeVito Dec 10 '12 at 2:39
5
$\begingroup$

Hint: For $p\in X$, let $U_p$, with coordinates $(x_1,...,x_n,t)$, be a slice chart around $p$ (meaning around $p$, $X$ corresponds to points where $t=0$).

Now, given your normal vector field $V$, orient $X\cap U_p$ by declaring the ordered basis $\{\partial_{x_i}\}$ to be positively oriented iff the ordered basis $\{\partial_{x_i}, v\}$ is positively oriented in $Y$.

Conversely, if $X$ is oriented, define $V = \partial_t$.

I'll leave it to you to prove that all this works.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see... but don't you need to pick the slice chart so that its differential (on TY) preserves orientation? How can we do that? $\endgroup$ – Bernard Dec 10 '12 at 4:45
  • $\begingroup$ On each chart, you pick an individual orientation. If you can pick them so that on overlaps they agree, you've oriented $X$. I'm telling you how to pick them on slice charts - you still have to verify that on overlaps, the choice agrees. $\endgroup$ – Jason DeVito Dec 10 '12 at 5:08
  • $\begingroup$ I was asking about the converse. $\endgroup$ – Bernard Dec 10 '12 at 5:44
  • 1
    $\begingroup$ @gofvonx: $V$ is a coordinate vector field and coordinate vector fields are always smooth: If $f$ is smooth then, $\partial_i f = \frac{d}{dx^1} f\circ x^{-1}$ is the derivative of a compositioin of smooth functions. (You pick $V = \partial_t$ or $V = -\partial_t$ once for the whole chart: Using an argument relying on the disconnectedness of $GL_n$, one shows the choice at one point uniquely determines the choice at every point.) $\endgroup$ – Jason DeVito Jun 3 '13 at 14:33
  • 1
    $\begingroup$ @SeñorBilly: Yes, that should work. $\endgroup$ – Jason DeVito Jun 29 '17 at 1:39
4
$\begingroup$

Here's an alternative proof using some facts about the first Stiefel-Whitney class $w_1$.

We have a short exact sequence of vector bundles on $X$:

$$0 \to TX \to i^*TY \to \nu \to 0$$

where $i : X \to Y$ is the inclusion, and $\nu$ denotes the normal bundle of $X$ in $Y$. Therefore $w_1(i^*TY) = w_1(TX) + w_1(\nu)$. As $Y$ is orientable, $w_1(TY) = 0$ so $w_1(i^*TY) = i^*w_1(TY) = 0$ and hence $w_1(TX) = w_1(\nu)$. So $X$ is orientable if and only if $w_1(\nu) = 0$, but as $\nu$ is a line bundle ($X$ has codimension one), this is equivalent to $\nu$ being trivial. Therefore $X$ is orientable if and only if $\nu$ has a nowhere-zero section (i.e. $X$ admits a nowhere-zero normal vector field).

| cite | improve this answer | |
$\endgroup$

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.