Let $M$ be a surface in $\mathbb{R}^3$ given by a regular chart, say $X:M \longrightarrow \mathbb{R}^3$, with its first fundamental form $g$, Gauss map $N$, Gaussian curvature $K$ and mean curvature $H$. I want to show that $\Delta N = -2g^{ij} \partial_i H \partial_j X - 2(2H^2-K)N,$ where $\Delta$ denotes the Laplace-Beltrami operator on $M$.

Expanding $\Delta N$ in terms of the basis $(\partial_1 X,\partial_2 X, N)$, I have already shown that the $N$ component is in fact $-2(2H^2-K)$, but unfortunately I seem to unable to calculate the rest. I've been looking for the result in several books on differential geometry but have not been successful so far. Can someone give a hint or just a reference for a proof of this equation?

  • 2
    $\begingroup$ Have you tried using moving frames? Your formula (which is correct, of course) follows by an elementary calculation if you do it that way. This kind of question really belongs more in MathStackExchange than here, since it is routine using moving frames. If you aren't familiar with moving frames, you might consult Barrett O'Neill's "Elementary Differential Geometry". $\endgroup$ – Robert Bryant Feb 13 '16 at 22:24

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