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I have been trying to figure this out as it would seem that it should be so. I have been search though, and the only solution seems to treat the compact case with homology beyond what I know. I believe that it is true, but I could imagine a counter-example may exist.

I frankly have no idea how to go about proving this. I am able to show that orientablility is equivalent to the existence of smooth function on $\mathbb{R}^n$ with the manifold as the preimage of a regular value, but then I become stuck.

Does any have any proofs or counterexamples?

Pax
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2 Answers2

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There is a fairly elementary treatment in Guillemin and Pollack, Differential Topology. Pages 85-91 take you through the Jordan-Brouwer Separation Theorem. Your hypersurface separates $\mathbb R^n$ into an inside and an outside. There is a consistent choice of normal vector pointing inside. Depending how you are learning orientation, there are various ways a normal field in codimension one gives an orientation

Will Jagy
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  • Thank you. I had not thought of that. Is there an argument for this without using the separation theorem? – Pax Nov 06 '13 at 06:22
  • Because it appears pollock only treats the compact case. – Pax Nov 06 '13 at 06:26
  • This is about as simple as it gets. G+P assume $C^\infty$ throughout. take a look at these reviews: http://www.amazon.com/Differential-Topology-Graduate-Texts-Mathematics/dp/0387901485 – Will Jagy Nov 06 '13 at 06:29
  • Noncompact should be alright if the immersion is a proper mapping and a finite number of ends. – Will Jagy Nov 06 '13 at 06:37
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Here is a homological argument.

Suppose an unorientable n dimensional smooth compact manifold, M, is embedded as a hypersurface of Euclidean space.

Since its tangent bundle Whitney sum its normal bundle is trivial, its normal bundle must be non-trivial since its first Stiefel-Whitney class must be non-zero. This means that the boundary of a tubuluar neighborhood of M is a connected orientable manifold. Call it N.

Further, the tubular neighborhood, T, has the homotopy type of M since M is a strong deformation retract of T.

Think of M as embedded in the n+1-sphere by adding the point at infinity.

The Mayer-Vietoris sequence in dimensions $n$ and $n+1$ is

$H^{n+1}(T) \oplus H^{n+1}(S^{n+1}-T) \leftarrow H^{n+1}(S^{n+1}) \leftarrow H^n(N) \leftarrow H^n(T) \oplus H^n(S^{n+1}-T) \leftarrow H^n(S^{n+1})$

The first two terms on the left are zero because T and $S^{n+1}-T$ are (n+1)-dimensional manifolds with boundary. $H^{n+1}(S^{n+1})$ and $H^n(N)$ are both isomorphic to $\mathbb{Z}$ since these are the top dimensional $\mathbb{Z}$-cohomologies of compact orientable manifolds without boundary. Similarly, $H^n(T)= \mathbb{Z}_2$ because it is homotopically equivalent to a non-orientable $n$-manifold. Finally, $H^n(S^{n+1})=0$.

So the sequence is

$0 \leftarrow \mathbb{Z} \leftarrow \mathbb{Z} \leftarrow \mathbb{Z}_2 \oplus H^n(S^{n+1} - T) \leftarrow 0$

This sequence can not be exact.

evgeny
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lavinia
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