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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...

Bernard
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  • 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 $X$? – Jesse Madnick Dec 10 '12 at 02:19
  • Yes, I meant $X$, sorry for that. – Bernard Dec 10 '12 at 02:22
  • 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. – Bernard Dec 10 '12 at 02:23
  • The concept of "normal" only makes sense when a Riemannian metric is present; the one from euclidean space should work. The concept of "transverse" always makes sense. Anyway, where are you getting this problem from? – Jesse Madnick Dec 10 '12 at 02:25
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    Guillemin and Pollack problem 18 p. 106 – Bernard Dec 10 '12 at 02:27
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    What's your definition of "oriented"? – Jason DeVito - on hiatus Dec 10 '12 at 02:39
  • That you can smoothly orient the tangent space at each point; i.e. for each point there is a local parametrization around it such that its differential at each point preserves orientation. – Bernard Dec 10 '12 at 02:42

2 Answers2

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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.

Jason DeVito - on hiatus
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  • 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? – Bernard Dec 10 '12 at 04:45
  • 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. – Jason DeVito - on hiatus Dec 10 '12 at 05:08
  • I was asking about the converse. – Bernard Dec 10 '12 at 05:44
  • Oh, I see! You start with slice charts whose differential (on $TX$) preserves orientation. Then the question is, do you pick $V = \partial_t$ or $V = -\partial_t$? Pick it in such a way that ${\partial_{x_i}, \pm\partial_t}$ is positively oriented (in $TY$). (I definitely could have given a better hint for that part.) – Jason DeVito - on hiatus Dec 10 '12 at 13:18
  • @JasonDeVito How do we then know that V is smooth? – gofvonx Jun 02 '13 at 10:21
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    @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.) – Jason DeVito - on hiatus Jun 03 '13 at 14:33
  • O guess you could also prove that X is orientable by doing the contraction of the volume form of Y by the normal vector field? – Guillermo Mosse Jun 28 '17 at 22:08
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    @SeñorBilly: Yes, that should work. – Jason DeVito - on hiatus Jun 29 '17 at 01:39
  • Sorry for a stupid question. Why $\partial_t$ is normal? – Ning Aug 30 '22 at 17:13
  • @Ning: Strictly speaking, the manifolds in question aren't Riemannian so there isn't a true notion of "normal" in the sense of "perpendicular to the surface". Rather, it's normal in the sense that it's non-zero in the vector bundle $TY|_X/TX$. As to why it's normal in this sense, this has to do with how slice coordinates work. The coordinates $(x_1,...,x_n,t)$ are coordinates for an open subset in $Y$ with the special property that $(x_1,....,x_n,0)$ form coordinates on an relatively open subset of $X$. Then, by definition of coordinates, $\partial_t$ is not in the span of all the... – Jason DeVito - on hiatus Aug 30 '22 at 17:37
  • $\partial_{x_i}$, so $\partial_t$ is not in $TX$, so it's non-zero in $TY|_X/TX$. – Jason DeVito - on hiatus Aug 30 '22 at 17:37
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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).

Michael Albanese
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