# Hypersurfaces of R^n are Orientable

1. Jan 28, 2015

### WWGD

Anyone know how to prove the result that every closed hypersurface of $\mathbb R^n$ , i.e., any closed (n-1)-submanifold of $\mathbb R^n$ is orientable? Note that if we assume this is true, this shows $\mathbb RP^n$ cannot be embedded in $\mathbb R^{n+1}$

EDIT: there is a result that every hypersurface can be represented as $f^{-1}(0)$ , where I think $f$ is at least an immersion. I heard this has to see too with Alexander duality but I dont see clearly where this duality comes into place, though.

Last edited: Jan 28, 2015
2. Jan 28, 2015

### lavinia

Correct this if it s wrong but I think it works.

If the compact smooth manifold $M$ is embedded in $R^{n+1}$ then it has a tubular neighborhood, $T$ ,with boundary another compact manifold,N.
The Z cohomology of the pair $(T,N)$ is equal to the cohomology of $S^{n+1}$ since $S^{n+1}$ is homeomorphic to the Thom space of the normal bundle of $M$ in $R^{n+1}$. This is proved using Excision.

The exact homology of the pair, (T,N), is

$0 ← H^{n+1}(S^n) ← H^n(N) ← H^n(M) ← H^n(S^{n+1}) ← ...$

which is $0 ← Z ← H^n(N) ← H^n(M) ← 0 ← ...$

If $M$ is non orientable then $N$ is connected so that $H^n(N)$ is equal to $Z$ and $H^n(M)$ is equal to $Z_{2}$

So the exact sequence is

$0 ← Z ← Z ← Z_{2} ← 0 ← ...$

which is impossible.

If $M$ is orientable then $N$ has two diffeomorphic components so that $H^n(N$ is equal to $Z ⊕ Z$ and $H^n(M)$ is equal to $Z$

So the exact sequence is

$0 ← Z ← Z ⊕Z← Z← 0 ← ...$

which is possible.

Last edited: Jan 28, 2015
3. Jan 28, 2015

### WWGD

Can you see why/how Alexander duality plays a role here? Sorry, I dont understand enough about Thom spaces to understand the answer. I am aware of excision and the LES associated to it, and I understand the argument otherwise, but maybe if you can see where/if Alexander duality is applied?

4. Jan 28, 2015

### lavinia

I don't really know the Alexander Duality theorem but here is a simplified version that is given as an exercise in Milnor's Characteristic Classes that answers your question.

BTW: I would be happy to learn the proof with you.

If K is a compact subset of the sphere, $S^n$ that is a retract of some neighborhood in $S^n$ then using ordinary homology, there is an isomorphism between
$H^{i-1}(K,x)$ and $H_{n-i}(S^n - K,y)$ where $x$ is a point of $K$ and $y$ is a point of $S^n-K$

An embedded compact smooth submanifold is a retract of a tubular neighborhood so the conditions are satisfied.

If $i - 1 = n - 1$ as in the case of a hyper-surface the isomorphism says
$H^{n-1}(K,x)$ and $H_{0}(S^n - K,y)$ are isomorphic.

But $H_{0}(S^n - K,y)$ is a direct sum of one less than the number of connected components of $S^n-K$ copies of $Z$

Since $H^{n-1}(K,x)$ is not zero the number of connected components must be greater than one so $H_{0}(S^n - K,y)$ is not the zero group but is a direct sum of at least one copy of $Z$..

But if $K$ is non orientable its top cohomology is $Z_{2}$ and so can not be isomorphic to a torsion free group.

Last edited: Jan 29, 2015
5. Jan 29, 2015

### WWGD

Nice, thanks. Sure, we can agree to read the proof together. My situation is a bit uncertain at this point, if you can be flexible, I would be glad to work it out with you. BTW, I read that Alexander Duality is a generalization of the Jordan
Curve thm., in that it deals with the topology of complementary subspaces ( with complementary meaning their union is the whole space). A.De studies the homological properties of complementary subspaces. The homological properties of a set can be defined in terms of those in the complement
...http://www.encyclopediaofmath.org/index.php/Alexander_duality

6. Jan 29, 2015

### lavinia

I am completely flexible. The Alexander Duality theorem does seem to generalize the Jordan curve theorem since for hypersurfaces of the sphere it says that the sphere is divided into at least two components. I wonder if the sphere is not special and that the theorem works for any simply connected manifold. I wonder this because the first proof I gave using tubular neighborhoods needed only that the sphere be simply connected (and also orientable).

Pick a source for the proof that we can read together