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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...

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  • $\begingroup$ 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, 2012 at 2:19
  • $\begingroup$ Yes, I meant $X$, sorry for that. $\endgroup$
    – Bernard
    Dec 10, 2012 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, 2012 at 2:23
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    $\begingroup$ Guillemin and Pollack problem 18 p. 106 $\endgroup$
    – Bernard
    Dec 10, 2012 at 2:27
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    $\begingroup$ What's your definition of "oriented"? $\endgroup$
    – Jason DeVito - on hiatus
    Dec 10, 2012 at 2:39

2 Answers 2

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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.

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  • $\begingroup$ 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, 2012 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 - on hiatus
    Dec 10, 2012 at 5:08
  • $\begingroup$ I was asking about the converse. $\endgroup$
    – Bernard
    Dec 10, 2012 at 5:44
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    $\begingroup$ @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 - on hiatus
    Jun 3, 2013 at 14:33
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    $\begingroup$ @SeñorBilly: Yes, that should work. $\endgroup$
    – Jason DeVito - on hiatus
    Jun 29, 2017 at 1:39
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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).

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