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The following statement seems to be taken as given in papers I'm reading:

Let $\mathcal{M}^n$ be a compact, embedded hypersurface in $\mathbb{R}^{n+1}$. Then $\mathcal{M}$ is the boundary of some compact domain $\Omega \subset \mathbb{R}^ {n+1}$.

Is this an elementary result? I feel there must be some algebraic topology argument here. Any suggestions?

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What is your definition of domain? I thought it included 'connected', but that doesn't work here. –  ndkrempel Feb 25 '11 at 16:35

3 Answers 3

up vote 4 down vote accepted

There are several ways to see this fact, which is a simple instance of Alexander duality. Here is the simplest I know.

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.

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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.

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+1: nice and slick argument ! And it points moreover to the fact that the statement of OP is false if $n=0$ : one point is an hypersurface in $\mathbb{R}$, after all. And your answer also works in the locally flat $C^0$ case (the conclusion of the implicit function theorem becomes the assumption, essentialy). Your answer should definitely be the accepted one. –  BS. Feb 25 '11 at 22:30

This sounds like a higher dimensional version of the Jordan curve theorem, known as the Jordan-Brouwer separation theorem. See here.

Already the Jordan curve theorem is highly non-trivial, so I would say this is also.

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The key point as in the answer by BS, is to assume that the hypersurface is smooth---otherwise as you say it generalizes the JCT and is much harder. Another, perhaps more intuitive, way to tell if a point p not on a smooth hypersurface M is inside or outside M is to choose a ray starting at p that is nowhere tangent to M and count the number of intersections with M. Odd signifies inside and even outside. Of course you have to show that such rays exist (in fact almost any ray works) and that the parity is independent of the choice of ray, but neither is difficult. –  Dick Palais Feb 25 '11 at 18:47
    
I see. I did not know that hypersurface implies smooth. Does it? –  Sándor Kovács Feb 25 '11 at 19:30
    
Sorry, I should have specified the smoothness in the statement. –  T-' Feb 25 '11 at 20:03
    
@Dick : in fact, your argument about odd/even intersections with a ray was what I had in mind, I just reformulated it as a degree mod 2. @Sandor : I asssumed that hypersurface meant smooth (at least $C^1$). I agree that there is some implicit here, although I would be surprised to see Alexander's horned sphere called an (hyper)surface. Anyway, you're right that the $C^0$ situation is quite a bit more difficult. –  BS. Feb 25 '11 at 22:22

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