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The Jordan-Brouwer Separation Theorem says that any connected compact hypersurface in Euclidean space divides the space into two connected components with the hypersurface their common boundary, cf. the lecture notes: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Schmaltz.pdf

However, in the notes https://www.jstor.org/stable/2323445 the author proved the theorem with two restrictions for oriented and smooth hypersurface. Although they are equivalent, say, any smooth hypersurface in Euclidean space is orientable, I don't know if this restriction can be dropped anyway. And the author said $C^1$- or $C^2$-smooth also satisfies the proof instead of $C^\infty$, but is $C^1$- or $C^2$-smoothness equivalent to orientability?

In summary, what are the necessary conditions for a hypersurface separates the Euclidean space?

A. connected

B. compact

C. $C^k$-smooth (and what $k$?)

D. oriented

Analysis Newbie
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    C and D are not required but A and B are needed. Consider reading an algebraic topology textbook covering the subject rather than online notes that deal with special cases. See also here. – Moishe Kohan Dec 07 '21 at 16:56
  • @MoisheKohan Thank you very much. I am just a beginner studying differential geometry. I am wondering if orientability is not required, how about the Möbius strip and the Klein bottle which are compact but obviously non-orientable. They do not separate the Euclidean space. – Analysis Newbie Dec 07 '21 at 18:12
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    You need a compact manifold without boundary, sometimes these are called closed manifolds. The key thing is that you need is a $Z_2$-fundamental class. You have it for closed nonorientable manifolds. As for Klein bottle, it simply does not embed in $E^3$. – Moishe Kohan Dec 07 '21 at 18:22
  • Yes, I know the Klein bottle is embedded in $\mathbb{R}^4$, but does it separate $\mathbb{R}^4$ into two components? @MoisheKohan – Analysis Newbie Dec 07 '21 at 19:25
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    No, for the separation theorem you have to have a codimension 1 submanifold. Otherwise, it never separates. – Moishe Kohan Dec 07 '21 at 19:33
  • @MoisheKohan Why do we need compactness? – Anonymous Oct 18 '22 at 15:31
  • @Anonymous: Think of an open interval $J$ in $R^2\subset S^2$. Is $S^2\setminus J$ connected? – Moishe Kohan Oct 18 '22 at 17:12
  • @MoisheKohan Thanks! This example seems to work in Euclidean space too, like in the question. i.e. if $J$ is an open interval embedded in $\mathbb{R}^2$, then $J$ is a connected hypersurface whose complement is also connected. I realized though that the “compact” requirement can be weakened to “closed” since then the submanifold is properly embedded which is sufficient for the Jordan-Brouwer separation theorem. – Anonymous Oct 20 '22 at 18:19

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Suppose that $X\subset S^n$ is a compact subset which is homeomorphic to a topological manifold (possibly with boundary). Then Alexander Duality says: $$ \tilde{H}_0(S^n -X)\cong \check{H}^{n-1}(X)\cong H^{n-1}(X). $$ Therefore, if $X$ separates $S^n$, then $\tilde{H}_0(S^n -X)\ne 0$, hence, $dim(X)=n-1$ and at least one component of $X$ is orientable and has empty boundary. Conversely, if $X$ has codimension 1 in $S^n$ and has at least one component which is orientable and with empty boundary, then the same isomorphism says that the complement to $X$ is disconnected. If you want to get exactly two connected components, as in JST, then you have to assume that $X$ itself is connected, has empty boundary and is orientable.

Lastly, here is a proof that a closed (i.e. compact and with empty boundary) connected non-orientable $n-1$-dimensional manifold $X$ cannot be embedded in $S^n$:

Otherwise, $\tilde{H}_0(S^n-X, {\mathbb Z}_2)\cong H^{n-1}(X, {\mathbb Z}_2)\ne 0$, implying that $X$ separates $S^n$. But then, the argument above with integer (co)homology leads to a contradiction.

Moishe Kohan
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