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This question is from an old publicly available qualifying exam that I’m using to learn this field (Stanford geometry and topology). All work is my own.

Let $M \subset \mathbb R^3$ be a closed, connected, 2-dimensional submanifold. Show that $M$ is orientable.

Answer: From the classification theorem for closed surfaces, we know that $M$ must be homeomorphic to the sphere, the connected sum of $g$ tori for $g \ge 1$, or the connected sum of $k$ real projective planes for $k \ge 1$.

But if $M$ were to be the connected sum of any real projective planes $\mathbb RP^2$, then by the Whitney embedding theorem we would not be able to embed $M$ in $ \mathbb R^3$. Then $M$ must be homeomorphic to a sphere or the connected sum of $g$ tori, and thus must be orientable.

(a) Was this application of the classification theorem for surfaces followed by the Whitney embedding theorem correct?

As a bonus, even if I'm correct (or not), if there's a stronger tool, method, or theorem for proving this, then I'd love to see it.

Nate
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    How do you apply the Whitney embedding theorem exactly? – psl2Z Sep 21 '24 at 10:29
  • The standard way is to prove that the complement of the surface consists of two disconnected open sets and one is bounded. – Deane Sep 21 '24 at 18:04
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    It does not seem you actually know what Whitney's embedding theorem says. – Moishe Kohan Sep 21 '24 at 18:17
  • You can embed an orientable surface in $\mathbb R^3$ but you can’t embed real projective space in $\mathbb R^3$. The Whitney embedding theorem gives the general case for embedding in $\mathbb R^{2n}$, but has corollaries that allow certain manifolds to be embeddable in lower dimensions. So the argument just uses the nature of it being an embedding to filter down the surface classification theorem to only sphere and tori connected sums. – Nate Sep 21 '24 at 20:19
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    The Whitney embedding theorem tells you what number of dimension suffices to have an embedding of a manifold in Euclidean space. It does not at all tell you what number of dimension is necessary to obtain such a thing for any given manifold. – Thorgott Sep 22 '24 at 17:04
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    Look at the Jordan-Brouwer Separation Theorem in G&P. – Ted Shifrin Sep 22 '24 at 19:16
  • As @Thorgott mentioned you don't need to use any Whitney embedding theorem. Also, I think one can drop the connectedness hypothesis. If $M$ is a $2$-dimensional submanifold of $\mathbb{R}^3$ then it has a trivial normal bundle. This tells you that $M$ is orientable. You can find the proof here. This shows that the above result is true for hypersurfaces in $\mathbb{R}^n$. – Trishan Mondal Sep 29 '24 at 18:22

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