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I have two involved questions, firstly, I know that the gauss map sends a surface to the unit sphere, so for a surface $\Sigma\subset\Bbb R^3$, parametrised by $u:U\subset\Bbb R^2\to \Bbb R^3$.

Would the gauss map be $N:\Sigma\to S^2$, $N(p)=\frac{u_x\times u_y}{\|u_x\times u_y\|}(p)$

I.e. the unit outward normal evaluated at $p\in\Sigma$.

Then I have read that if $N$ is harmonic, i.e. $\Delta N=0$, then $\Sigma$ is a surface of constant mean curvature, is this all true?

Thanks very much.

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up vote 1 down vote accepted

Minimal hypersusfaces do have the property that their Gauss map is harmonic. A more general result is

$\mathbf{Theorem (Ruh, Vilms)}$. Let $M$ a $m$-dimensional submanifold of $\mathbb{R}^n$. The Gauss map of $M$ is a harmonic map if and only if $M$ has parallel mean curvature vector.


Parallel mean curvature vector is equivalent to costant mean curvature, and define the class of hypersurfaces whose Gauss map is harmonic.

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