How does one prove that the Klein bottle cannot be embedded in $R^3$? I'm talking about smooth embeddings.
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2Have tried to build one? – draks ... Dec 18 '13 at 15:54
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Any compact, smooth hypersurface in $\Bbb R^n$ is necessarily orientable. (Without fancy algebraic topology, this is a consequence of the Jordan-Brouwer Separation Theorem.) Indeed, more is true: A compact hypersurface in any simply connected manifold is orientable.
Ted Shifrin
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1Isn't the mobius strip a compact smooth hypersurface in $\mathbb R^3$ which is not orientable? Or does the hypersurface need to be closed? – Cubi73 Jan 27 '21 at 13:12
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1@Cubi73: Yes, I of course mean a manifold with no boundary, but of course I should have been explicit. (I'm in the collection of geometers who insist on saying "with boundary" if boundary is allowed. Of course, saying "closed" as you did is just as ambiguous, except — again — for custom. – Ted Shifrin Jan 27 '21 at 17:56
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MO answer. In the link, there is a reference to a proof in Hatcher using Alexander duality as well as a more elementary proof.
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For a homological proof for all dimensions see my answer in http://math.stackexchange.com/questions/553939/are-closed-properly-embedded-manifolds-of-co-dimension-1-in-mathbbrn-orie/605060#605060 The method works for the general case of orientable simply connected manifolds. – lavinia Dec 18 '13 at 20:52