Is the following true?
Let $\Sigma$ be a compact orientable hypersurface without boundary in $R^n$. Then $R^n\setminus\Sigma$ has at least two connected components.
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$\begingroup$ No, for example, if $\Sigma$ is the empty set then the complement has only one component. Have you tried looking up the Jordan-Brouwer separation theorem? $\endgroup$– Ryan BudneyCommented Mar 17, 2013 at 20:20
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1$\begingroup$ Ryan: Normally the word "hypersurface" implies codimension 1, no? $\endgroup$– Steven LandsburgCommented Mar 18, 2013 at 2:45
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$\begingroup$ @Steven: The empty set is a manifold of any dimension (this convention being forced onto those of us who use bordism groups). $\endgroup$– Mark GrantCommented Mar 18, 2013 at 13:44
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$\begingroup$ The definition of orientability is another matter--in cobordism theory for orientable manifolds, the two possible conventions are both strange: Either the empty manifold has a unique orientation, (any orientable manifold should have two!) or the empty manifold can be formally given two different orientations whilst the empty cobordism between them realizes an isomorphism. (When is an orientable manifold invertibly cobordant to its opposite?) $\endgroup$– Hiro Lee TanakaCommented Mar 20, 2013 at 1:13
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3$\begingroup$ @Mark: The old-fashioned ("pre-cobordism-theory") convention is that empty set has topological dimension $-1$: This convention is most suitable for the purposes of separation theorems, like the one in the current question. $\endgroup$– MishaCommented Mar 20, 2013 at 3:33
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Yes: whenever $U \subset M$ is an open subspace with complement $Z$, then there is a long exact sequence $$ \ldots\to H^\bullet_c(U) \to H^\bullet_c(M) \to H^\bullet_c(Z) \to H^{\bullet+1}_c(U)\to \ldots $$ which in this case gives $$ H^{n-1}_c(\mathbf R^n) = 0 \to H^{n-1}_c(\Sigma) \to H^n_c(\mathbf R^n \setminus \Sigma) \to H^n_c(\mathbf R^n) \to 0.$$ By Poincaré duality we have $H^n_c(\mathbf R^n \setminus \Sigma) \cong H^0(\mathbf R^n \setminus \Sigma)^\vee$ and $H^{n-1}_c(\Sigma) = H^0(\Sigma)^\vee$, so the ranks of these cohomology groups are just the numbers of connected components of $\mathbf R^n \setminus \Sigma$ resp. $\Sigma$. Since $H^n_c(\mathbf R^n)$ is one-dimensional this shows that $\mathbf R^n \setminus \Sigma$ has exactly one more connected component than $\Sigma$. Compactness of $\Sigma$ was an unnecessary hypothesis.
answered Mar 17, 2013 at 20:28
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1$\begingroup$ It's not true if the surface isn't compact. You need the extra assumption that it's a closed subspace. $\endgroup$ Commented Mar 17, 2013 at 21:10
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2$\begingroup$ In my proof it is a closed, not necessarily compact, oriented submanifold (e.g. a line in the plane). Is it incorrect? $\endgroup$ Commented Mar 17, 2013 at 23:47
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$\begingroup$ Thank you very much Dan. What are the minimal regularity assumptions on $\Sigma$? Can $\Sigma$ have self intersections? For instance if $\Sigma$ is the boundary of two spheres touching each other at one point. Then $\Sigma$ is connected and $R^n \setminus \Sigma$ has three connected components. $\endgroup$– user32291Commented Mar 18, 2013 at 2:31
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$\begingroup$ John: Dan's LES shows quite generally that the number of connected components of the complement is one more than the dimension of $H^{n-1}(\Sigma)$. (Note that this gives the right answer in your two-spheres-touching case.) When $\Sigma$ is a closed oriented manifold (which your $\Sigma$ is not), $H^{n-1}$ counts the connected components of $\Sigma$. $\endgroup$ Commented Mar 18, 2013 at 2:51
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1$\begingroup$ @Taras Some point-set hypotheses are necessary for what I say below to be true. Let me denote the one-point compactification by an overline. Then $H^\bullet_c(U)=H^\bullet(\overline M,\overline Z)$, $H^\bullet_c(M)=H^\bullet(\overline M,\{\ast\})$ and $H^\bullet_c(Z) = H^\bullet(\overline Z, \{\ast\})$ and it is a special case of the long exact sequence of a triple. $\endgroup$ Commented Sep 5, 2018 at 14:15