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So I would like to show that any closed hypersurface (a smooth, compact, and boundaryless manifold of dimension $n$ embedded in $\mathbb{R}^{n+1}$) is orientable. This is an exercise in Guillemin and Pollack's book and the book suggests that the reader use the separation theorem of Jordan and Brouwer.

I understand a fair some of the argument: At each point $p$ in the manifold $M$ consider the tangent space $T_pM$ which can be identified as $\{p\} \times \mathbb{R}^{n} \subset \{p\}\times \mathbb{R}^{n+1}$. We now may assign to each point on the Manifold a normal vector, given by (normalized) vector $v_p$ that spaces the space $\left(\{p\}\times \mathbb{R}^{n}\right)^{\perp}$. Of course, this choice could be multivalued, we might pick $v_p$ or $-v_p$, depending on how we feel, but the point is that we can make a continuous choice of normal vector. To do this, most proofs say something like the following: the Jordan Brouwer theorem allows us to write: $$ \mathbb{R}^{n+1}\setminus M = U_1 \sqcup U_2 $$ where $U_1,U_2$ are both connected components of $\mathbb{R}^{n+1}$ and $U_1$ is bounded whereas $U_2$ is unbounded. It is then said that one should choose at each point the "inward" or "outward" pointing normal (intuitively, the normal whose arrow points towards $U_1$ or the normal whose arrow points towards $U_2$). However, this is seems difficult for me to state rigorously. Does anyone have any ideas as to how one formalizes a normal "pointing inwards" or "pointing outwards"?

rubikscube09
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  • Yes, if you take actual points $p+tv_p$ for small positive values of $t$, are you in $U_2$? If so, $v_p$ is outward-pointing. (The actual proof of J-B separation in Guillemin & Pollack is based on such notions.) – Ted Shifrin Jan 21 '20 at 00:04
  • This makes sense, (and unfortunately I did not go over the actual proof of the theorem in the text) I was thinking of the case of say, two spheres that "almost" intersect transversally/are tangent at a point. However I assume this will violate some assumptions of the theorem. – rubikscube09 Jan 21 '20 at 00:15
  • What you just described is not a manifold, so you'd better go back to basics. – Ted Shifrin Jan 21 '20 at 00:16
  • Right. How silly of me. – rubikscube09 Jan 21 '20 at 00:18

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