Compact Hypersurfaces Bounding Compact Domains - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:16:50Z http://mathoverflow.net/feeds/question/56642 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains Compact Hypersurfaces Bounding Compact Domains T-' 2011-02-25T15:54:05Z 2011-02-25T22:38:02Z <p>The following statement seems to be taken as given in papers I'm reading:</p> <blockquote> <p>Let $\mathcal{M}^n$ be a compact, embedded hypersurface in $\mathbb{R}^{n+1}$. Then $\mathcal{M}$ is the boundary of some compact domain $\Omega \subset \mathbb{R}^ {n+1}$.</p> </blockquote> <p>Is this an elementary result? I feel there must be some algebraic topology argument here. Any suggestions?</p> http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains/56654#56654 Answer by BS for Compact Hypersurfaces Bounding Compact Domains BS 2011-02-25T17:49:31Z 2011-02-25T22:38:02Z <p>There are several ways to see this fact, which is a simple instance of <a href="https://secure.wikimedia.org/wikipedia/en/wiki/Alexander_duality" rel="nofollow">Alexander duality</a>. Here is the simplest I know.</p> <p>Let $H$ be a compact smooth hypersurface in $\mathbb{R}^n$, whithout boundary, and $x\in \mathbb{R}^n\setminus H$. Then the radial projection $p_x : H \to \mathbb{S}^{n-1}$, $y\mapsto (y-x)/\|y-x\|$ has a degree mod $2$, say $d_x$, which may be defined as the number of elements mod $2$ of $p_x^{-1}(u)$ for almost all $u\in\mathbb{S}^{n-1}$ (this is well-defined by transversality theory). Then the subset of $x\in \mathbb{R}^n$ such that $d_x=1$ is your $\Omega$, a relatively compact open set with boundary $H$. EDIT : here one must assume $n>1$, as in Mohan's answer, otherwise the relative compactness might not hold.</p> http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains/56656#56656 Answer by Sándor Kovács for Compact Hypersurfaces Bounding Compact Domains Sándor Kovács 2011-02-25T17:51:49Z 2011-02-25T17:51:49Z <p>This sounds like a higher dimensional version of the Jordan curve theorem, known as the <em>Jordan-Brouwer separation theorem</em>. See <a href="http://en.wikipedia.org/wiki/Jordan_curve_theorem" rel="nofollow">here</a>.</p> <p>Already the Jordan curve theorem is highly non-trivial, so I would say this is also.</p> http://mathoverflow.net/questions/56642/compact-hypersurfaces-bounding-compact-domains/56669#56669 Answer by Mohan Ramachandran for Compact Hypersurfaces Bounding Compact Domains Mohan Ramachandran 2011-02-25T19:27:43Z 2011-02-25T20:14:42Z <p>An alternate argument to the one given by BS in the smooth case is to note that by the implicit funcion theorem the hypersurface is locally two sided.If it is not globally two sided then one can construct a connected two sheeted covering of euclidean space which is a contradiction.Then we note that euclidean space in dimension atleast two has one end.This implies that the complement of the hypersurface has one nonrelatively compact component and the result follows.This proof works for any closed embedded locally two sided hpersurface.</p>