Minimal surface and local minimizer of area functional

Asked Dec 12 '19 at 17:04

Active Dec 12 '19 at 17:04

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Let $(M,g)$ be a 3-dimensional Riemannian manifold and $\Sigma\subset M$ a closed surface. Is it equivalent to say:

1) $\Sigma$ is minimal, i.e. its mean curvature identically vanishes;

2) $\Sigma$ is a local minimizer for the area functional?

Does a characterization for minimal surface exists?

asked Dec 12 '19 at 17:04

Jules Binet

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As far as I recall, the gradient of the area functional (up to a constant multiplicative scalar depending on conventions) is the mean curvature vector. So... yes? – Ivo Terek Dec 12 '19 at 17:10
I've heard something like that, but I'm not sure. – Jules Binet Dec 12 '19 at 17:12
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Ok, found one reference. See Theorem 4 in page 19 of Henri Anciaux' Minimal Submanifolds in Pseudo-Riemannian Geometry. – Ivo Terek Dec 12 '19 at 17:15
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Does this answer your question? Is there a minimal graph in $\mathbb{R}^3$ which is not area-minimizing? – Arctic Char Dec 12 '19 at 17:54
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Also see this related post on MO – Arctic Char Dec 12 '19 at 17:55
Then a surface is minimal iff has zero mean curvature. Thank you all. – Jules Binet Dec 13 '19 at 10:30
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"a surface is minimal iff has zero mean curvature" This is THE definition of a minimal surface. @GiulioBinosi – Arctic Char Dec 14 '19 at 05:39
A minimal surface is a critical point of the area functional, not necessarily a local minimum. – Michael Albanese Dec 16 '19 at 02:51

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