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I think that every hypersurface is orientable.

Suppose I have a hypersurface $X$ that sits inside the ambient Euclidean space $\mathbb{R}^n.$ Then for every $x \in X, T_x(X)$ is an $n - 1$ dimensional vector space. Hence, we can find a normal vector to the vector space. We can simply do a similar procedure to the boundary orientation where we define the orientation of a basis of $T_x(X), \{v_1, \ldots, v_{n - 1}\},$ as just the orientation of $\{n_x, v_1, \ldots, v_{n - 1}\},$ where $n_x$ is a normal vector to the vector space. I doubt this is correct but I fail to see where this falls apart.

green frog
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    Moebius strip sits inside the three spaces though – Arctic Char May 15 '19 at 20:28
  • As far as I know, Klein bottle is a 2-dimensional manifold that doesn't embed in $R^3$. If a surface embeds as a hypersurface, then the proof seems legit. – Lada Dudnikova May 15 '19 at 20:29
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    the pointwise orientation you choose in this way may not be smooth (e.g. moebius strip) – asd May 15 '19 at 20:29
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    If $X$ is a closed smooth hypersurface of $\mathbb{R}^n$, then $X$ is orientable. – Michael Albanese May 15 '19 at 20:46
  • Sorry for being silly but isn't the Mobius strip a compact hypersurface? So in particular it is closed? Also I thought compact hypersurfaces were orientable... – green frog May 15 '19 at 20:49
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    "Closed" in this context is being used in the manifold sense, not in the topology sense of "a closed subset is a subset whose complement is open". A manifold is defined to be "closed" if it is compact and its manifold boundary is empty. In this sense the Möbius strip is most certainly not closed, because its manifold boundary is a circle. – Lee Mosher May 16 '19 at 02:40

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As people have pointed out above in the comments, you crucially need $X \subset \Bbb R^n$ to be closed (i.e., a compact manifold without boundary) hypersurface.

In that case, consider the restriction $\underline{\Bbb R}^n = T\Bbb R^n|_X$ of the tangent bundle to the ambient Euclidean space to $X$. $TX \subset \underline{\Bbb R}^n$ is a subbundle of codimension $1$, so we take it's orthogonal complement $NX$, which is the normal bundle of the submanifold. For your procedure to be coherent, you want a smooth section of the $1$-dimensional line bundle $NX$, which would be the family of normals $\{n_x\}_{x \in X}$ you describe.

Let $N_1 X = \{(x, v) \in NX : \|v\| = 1\}$ be the unit normal bundle; as $NX$ has rank $1$, this is a $S^0$-bundle on $X$ which is the same thing as a double cover on $X$. If this double cover wasn't trivial, there would be an embedded loop $\gamma$ in $X$ based at some point $x$ which lifts to a nontrivial embedded path $\tilde{\gamma}$ by the covering projection $N_1 X \to X$, in the sense that $\tilde{\gamma}(0) = y_0$ and $\tilde{\gamma}(1) = y_1$ where $\pi^{-1}(x) = \{y_0, y_1\}$. Let $\sigma$ be the segment in the fiber of the normal projection $NX \to X$ over $x$ joining $y_0$ and $y_1$. Then $C = \tilde{\gamma} \cup \sigma$ is a closed embedded loop in $NX$ which intersects $X$ transversely at a single point. By the tubular neighborhood theorem, we can embed $NX$ as an open neighborhood of $X$ in $\Bbb R^n$.

This would give a $1$-dimensional submanifold $C$ and a codimension $1$ submanifold $X$ in $\Bbb R^n$ with mod $2$ intersection number $1$. But $\Bbb R^n$ is simply connected, so $C$ is nullhomotopic, which is a contradiction as that would force the mod $2$ intersection number of $C$ and $X$ to be $0$.

Therefore, $N_1 X \to X$ is the trivial cover. In particular it admits a global section $n : X \to N_1 X$. $n(x)$ is a unit normal to $X$ at $x$ by construction, so this is the desired smooth family of normals to $X$ that you wanted. Now simply orient $T_x X$ according to the orientation of $T_x \Bbb R^n$ and the normal vector $n(x)$ by the "right-hand rule", following your idea.

Balarka Sen
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  • I think the statement is true even if we drop the compactness assumption on $X$ but require that it be boundaryless and embedded properly in $\mathbb R^n$ (i.e. as a closed subset). Sketch: associated to such an $X$, we can define a real line bundle $V$ on $\mathbb R^n$ which restricts on $X$ to $NX$ - using the real analogue of the usual way to go from a (smooth) divisor to line bundle in complex/algebraic geometry. Now, $V$ is trivial since $\mathbb R^n$ is contractible and so, $NX$ is too. – Mohan Swaminathan Jun 13 '19 at 01:04
  • @MohanSwaminathan That rings true to my ears; I like your sketch. I think it's possible to carry out the same proof idea as above: $NX$ can be embedded in $\Bbb R^n$ (we'll have varying normal radii) so the same construction above would give a loop $C$ intersecting $X$ transversely at a point. Take a big closed ball $B$ in which a nullhomotopy of $C$ fits in so that $F : C \times I \to \Bbb R^n$ is a nullhomotopy transverse to the compact submanifold with boundary $B \cap X$ which doesn't intersect $\partial(B \cap X)$. Usual transversality theory pushes through for a contradiction. – Balarka Sen Jun 13 '19 at 05:23
  • Alternatively, one could be more abstract: Take a large ball $B$ and consider the image of $X$ under $\Bbb R^n \to \Bbb R^n/B^c = S^n$. That's an embedded pseudomanifold, let's call it $Y$, in $S^n$ which gives a fundamental class $[Y] \in H_{n-1}(S^n)$ and similarly $[C] \in H_1(S^n)$. $\text{PD}[Y] \smile \text{PD}[C] = 1$ since $C$ and $Y$ intersect transversely at a point, but $[C] = 0$, contradiction (We collapsed a large ball complement instead of taking a one-point compactification since set of ends of $X$ may be badly behaved, so the notion of fundamental class of $X^+$ may get murky) – Balarka Sen Jun 13 '19 at 06:04
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    To pull the biggest nuke of all, one can dispense with the normal bundle altogether. By Alexander duality, $\tilde{H}_0(S^n \setminus Y) \cong H^{n-1}(Y)$. The above arguments really say $S^n \setminus Y$ is disconnected (if not, you can find a loop $C$ intersecting the top stratum of $Y$ transversely at a point, contradiction) so $H^{n-1}(Y) = \Bbb Z$. But $Y$ is $X/(X - K)$ where $K$ is a large compact set in $X$, so $H^{n-1}(X, X - K) \cong \Bbb Z$. Taking a direct limit over $K$, $H_c^{n-1}(X) \cong \Bbb Z$, so $X$ is orientable. I guess the takeaway is we can even drop smoothness. – Balarka Sen Jun 13 '19 at 06:14
  • Very nice! The argument in my sketch seems to be limited to $C^\infty$ submanifolds (it doesn't seem to extend to the case of $C^k$ submanifolds for $k<\infty$ in any obvious way, let alone locally flat submanifolds as your last comment does). – Mohan Swaminathan Jun 13 '19 at 12:57