Orientation of hypersurface

Question

Some books on mean curvature flow (e.g. Mantegazza, Ecker) state that an embedded hypersurface in $\mathbb{R}^{k+1}$ is orientable (Mantegazza page 3, Ecker page 110). In other words, they assume the existence of a globally defined (unit) normal field along the embedded hypersurface.

Does anyone know how to prove this? Isn't the Mobius strip able to be embedded in Euclidean space, yet it isn't orientable?

For some definitions, let $M$ be a smooth $k$-dimensional manifold. A smooth map $\varphi:M\rightarrow\mathbb{R}^{k+1}$ is a hypersurface (an immersion) if its differential is injective. It is an embedding if it is also a homeomorphism onto its image $\varphi(M)$. An immersed hypersurface is a subset $S\subset\mathbb{R}^{k+1}$ which is a $k$-dimensional manifold such that the inclusion map $\iota:S\hookrightarrow\mathbb{R}^{k+1}$ is an immersion. An embedded hypersurface is a subset $S\subset\mathbb{R}^{k+1}$ which is a $k$-dimensional manifold whose topology conicides with the subspace topology inherited from $\mathbb{R}^{k+1}$ and the inclusion is an embedding.

We can also characterise an embedded hypersurface using single-slice adapted charts: Each point $p\in S$ is contained in a coordinate neighborhood of a single-slice chart $(U,\mathsf{u})$ on $\mathbb{R}^{k+1}$ adapted to $S$ such that $U\cap S=\{(u^1,...,u^{k+1})\;|\;u^{k+1}=0\}$. The pair $(U\cap S,\mathsf{proj}_{\mathbb{R}^k}\circ\mathsf{u}|_{U\cap S})$ is then a chart for $S$ and as we range over $p\in S$ we get an atlas for $S$.

Answer 1

Theorem
Every closed embedded hypersurface of $S\subset \mathbb R^{k+1}$ is orientable.
Proof
The hypersurface $S$ is defined by family $(U_i,f_i)$ where the $U_i$'s are an open covering of $\mathbb R^{k+1}$ and the $f_i:U_i\to \mathbb R$ are $C^\infty$ functions subject the condition that there exist $C^\infty$ functions $g_{ij}:U_i\cap U_j\to \mathbb R^\star$ satisfying $f_i=g_{ij}f_j$ on $U_i\cap U_j$.
We then have $S\cap U_i=f_i^{-1}(0)$ (here we used $S$ is closed).

Those $g_{ij}$ define a line bundle $L$ on $\mathbb R^{k+1}$, which is necessarily trivial like all line bundles on $\mathbb R^{k+1}$, since $\mathbb R^{k+1}$ is contractible.
But then the restriction $L|S$ of $L$ to $S$ is trivial too and since that restriction is the normal bundle of the embedding $S\hookrightarrow \mathbb R^{k+1}$, that normal bundle is trivial, which implies that $S$ is orientable.

[Note carefully that connectedness or compactness of $S$ is not assumed]

Edit
In this more recent answer I prove independently that the smooth hypersurface $S$ has a global equation, which implies that it is orientable.

Answer 2

Using differential topology, it's easy to see that a non-orientable hypersurface, embbed in an Euclidian space$\mathbb{R}^n$, would lead to a contradiction.

Briefly, an oriented hypersurface admits a global normal vector field, so non-orientability would lead to a closed curve on the hypersurface, say $\gamma:[0,1] \to M, \gamma (0)=\gamma (1)=p\in M$, and a normal vector field on it will “come back with the opposite direction”.

Take an $\epsilon$ positive and define a smooth curve as following: for $t \in(0+\epsilon,1-\epsilon), \theta(t)=\gamma(t)+\epsilon N(\gamma(t))\in \mathbb{R}^n$, where N is the normal vector mentioned above. And connecting $\theta(1-\epsilon)\ and\ \theta(0+\epsilon)$ smoothly, we got a smooth closed curve in $\mathbb{R}^n:\theta:S^1=[0,1]/\sim \to \mathbb{R^n}$, that intersect your hypersurface M only once, i.e. $I_2(\theta,M)=1$. And by transversally approximation, we can assume that the curve transversal to M.

Now $\pi_1(\mathbb{R}^n)$ is trival, so the homotopy gives a map $f:D^2 \to\mathbb{R}^n$, with $f(\partial D^2)=\theta(S^1)$, which could be seen as an extension. Again we can assume that f is transversal to M. But the fact that $\theta$ could extend to the hole $D^2$ iff the mod 2 intersection number $I_2(\theta,M)=0$, which lead to a contradiction.

All above is from H.Samelson's paper :https://www.ams.org/journals/proc/1969-022-01/S0002-9939-1969-0245026-9/S0002-9939-1969-0245026-9.pdf, you could follow his paper as well.

Answer 3

Another solution that uses some algebraic topology: without loss of generality assume $M$ is connected of dimension $k-1$, and therefore is a compact connected hypersurface without boundary in the compactification $S^{k}$. Using $\mathbb{Z}_2$ coefficients everything is oriented, so we can use Poincare duality to get that $H^{k-1}(M) \cong \mathbb{Z}_2$. The top degree part of the long exact sequence of relative cohomology of the pair $(S^k, M)$ looks like $$ 0 \to H^{k-1}(M) \to H^k(S^k, M) \to H^k(S^k) \to H^k(M) \to 0$$Replacing the groups that we know gives $$0 \to \mathbb{Z}_2 \to H^k(S^k, M) \to \mathbb{Z}_2 \to 0$$ which shows that $H^k(S^k, M) \cong \mathbb{Z}_2^{2}$. A straightforward argument using excision and tubular neighbourhoods shows that $H^*(S^k, M) \cong H^*_c(S^k \setminus M)$ where $H^*_c$ denotes the compactly supported cohomology, so we have that $$H^k_c(S^k\setminus M) \cong \mathbb{Z}_2^2$$ By Poincare duality (noting that $S^k \setminus M$ is an open subset of $S^k$ and hence oriented) we have $H_0(S^k \setminus M) \cong \mathbb{Z}_2^2$, so $S^k \setminus M$ has two components, one of which must contain the point $\infty$. The normal bundle of $M$ in $S^k$ can therefore be trivialised by choosing a section that always points towards infinity. Therefore $M$ itself is orientable because it has a trivial normal bundle in $S^k$. Looking at the situation in $\mathbb{R}^k$, the section that we defined is essentially the Gauss map for $M$.