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 $S$.

  • 6
    $\begingroup$ When they say that a hypersurface is orientable they mean a connected closed hypersurface (compact, no boundary). This is a natural setting for the MCF. to prove orientability you verify that it separates, which in turn follows from Alexander duality. I can write a detailed explanation if you are familiar with algebraic topology. $\endgroup$ – Moishe Kohan Jul 11 '14 at 5:28
  • 1
    $\begingroup$ Yes I'm familiar with the basics of algebraic topology, an explanation would be welcomed. $\endgroup$ – mdg Jul 11 '14 at 7:31
  • $\begingroup$ +1 for studiosus. However compactness and connectedness are irrelevant for the theorem. $\endgroup$ – Georges Elencwajg Jul 11 '14 at 8:07

Every closed subset embedded hypersurface of $S\subset \mathbb R^{k+1}$ is orientable.
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 $S\cap U_i=f_i^{-1}(0)$

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]

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

| cite | improve this answer | |
  • 1
    $\begingroup$ I don't see how your definition of a hypersurface relates to the definitions given above (inclusion is an embedding or in terms of single-slice adapted charts). Can you please show the definitions are equivalent. $\endgroup$ – mdg Jul 11 '14 at 20:25
  • 1
    $\begingroup$ Dear G.S., my $(U_i,f_i)$ corresponds to $(U, u^{k+1})$ in your single slice chart $(U,\mathsf u)$. My $g_{ij}$ corresponds to your $\frac{u^{k+1}}{v^{k+1}}$ where $(V,\mathsf v)$ is another single slice chart . There is a little technicality in showing that $\frac{u^{k+1}}{v^{k+1}}$ makes sense and is non-zero on $S\cap U\cap V$ $\endgroup$ – Georges Elencwajg Jul 11 '14 at 20:51
  • $\begingroup$ By the way, I'm curious to know if the proof above is to be found somewhere in the literature. It is quite natural for an algebraic geometer but involves a different point of view from that of differential geometers, as can be seen for example in the choice of the family $(U_i, f_i)$ and the $g_{ij}$'s as data for defining a submanifold. $\endgroup$ – Georges Elencwajg Jul 13 '14 at 7:36
  • 1
    $\begingroup$ @ziggurism: the Möbius band is not a closed embedded hypersurface of $\mathbb R^3$. I said that compactness is not assumed in the Theorem, but closedness is assumed. I really wonder why so many people keep believing that the Möbius strip is a counterexample to the Theorem I wrote: a theorem doesn't say anything about objects that don't satisfy its hypotheses :-) $\endgroup$ – Georges Elencwajg Feb 2 at 20:05
  • 1
    $\begingroup$ @ziggurism: closed = complement of open subset of $\mathbb R^{k+1}$. This is the last time I'm telling you that I'm not assuming compactness. $\endgroup$ – Georges Elencwajg Feb 3 at 8:30

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ You did not say that the non-orientable hypersurface must be closed (i.e., compact without boundary). The proof definitely uses that hypothesis, even though you never said where. Note that, as the OP said, the Möbius strip is a counterexample if the hypersurface has nontrivial boundary or if the hypersurface is non-compact. $\endgroup$ – Ted Shifrin Nov 6 '19 at 21:05

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.