# Orientation of hypersurface

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 $S$.

• 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. Jul 11, 2014 at 5:28
• Yes I'm familiar with the basics of algebraic topology, an explanation would be welcomed.
– mdg
Jul 11, 2014 at 7:31
• +1 for studiosus. However compactness and connectedness are irrelevant for the theorem. Jul 11, 2014 at 8:07

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.

• 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.
– mdg
Jul 11, 2014 at 20:25
• 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 $S\cap U\cap V$ Jul 11, 2014 at 20:51
• 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. Jul 13, 2014 at 7:36
• @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 :-) Feb 2, 2020 at 20:05
• @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. Feb 3, 2020 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.

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 $$M$$.

• You need some assumptions on $M$ for it to be compact, without a boundary in $\mathbb{S}^k$. Either it should be compact and without boundary in $\mathbb{R}^k$ or it must be "nice" at infinity so that its compactification is a manifold. Mar 10, 2023 at 22:08
• Thanks for pointing that out! Mar 12, 2023 at 0:40