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I was studying the strong maximum principle for minimal surfaces and came across the statement that surface A lies on one side of the surface B. Can you please tell me what does it mean mathematically?

The Theorem Statement is: "If $\Sigma_1 ,\Sigma_2 \subset \mathbb{R}^n $ are complete connected minimal hypersurfaces (without boundaries), $\Sigma_1 \cap \Sigma_2 \neq \emptyset, $ and $\Sigma_2$ lies on one side of $\Sigma_1$, then $\Sigma_1 = \Sigma_2$.

  • While I might guess that this refers to something like the Jordan-Brouwer separation theorem, your question contains so little context that I am quite unsure. Can you supply more context? – Lee Mosher Feb 26 '16 at 13:48
  • @LeeMosher please see the edited question. – inquisitive Feb 26 '16 at 14:04

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What you need is version of the Jordan-Brouwer separation theorem, in which one is given $\Sigma \subset \mathbb{R}^n$ a connected hypersurface. The theorem says that if $\Sigma$ is compact then $\mathbb{R}^n-\Sigma$ has exactly two components $U,V$ and their closures are $\overline U = U \cup \Sigma$ and $\overline V = V \cup \Sigma$.

The theorem you need is more general, having the same conclusion but weaker hypotheses. I'm not sure of the best hypotheses. You definitely want to start with a connected hypersurface $\Sigma \subset \mathbb{R}^n$, but you also want to weaken the hypothesis of compactness, since minimal hypersurfaces are never compact. The minimum requirement is that $\Sigma$ be closed in $\mathbb{R}^n$; I am pretty sure that is sufficient in a purely topological context, but am just a tad unsure, and there might be easier versions with hypotheses of intermediate strength. The "complete" hypothesis in your quoted theorem is probably what you'll use for this purpose.

So in your quoted theorem, once you have appropriate hypotheses leading to the conclusion that $\mathbb{R}^n - \Sigma_1$ has two components $U,V$, then the two "sides" of $\Sigma_1$ are the two sets $\overline U = U \cup \Sigma_1$ and $\overline V = V \cup \Sigma_1$. Your theorem says that if $\Sigma_2$ is a subset of one of these two sides then $\Sigma_2=\Sigma_1$.

Lee Mosher
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