# Hypersurfaces of R^n are Orientable

Gold Member
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.

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lavinia
Gold Member
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.

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Gold Member
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?

lavinia
Gold Member
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?
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.

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Gold Member
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

lavinia