Vector Laplace Beltrami operator of the Gauss map

Consider an abstract surface $(M,g)$ embedded into $\mathbb{R}^3$ via $f:M \to \mathbb{R}^3$. Denote by $N:M \to \mathbb{R}^3$ the Gauss map (normal field) of the surface. Write the Laplace Beltrami operator as $\Delta^{(1)}$. Define the vector Laplace Beltrami operator acting on the normal field as

$$\Delta^{(3)} N := (\Delta^{(1)}N_1,\Delta^{(1)}N_2,\Delta^{(1)}N_3).$$

For Euclidean space, this concept is described e.g. in .

Is the extension to surfaces meaningful? What can we say about this operation on the Gauss map? Can we for example characterize normal and tangent part of $\Delta^{(3)} N$?

This is a standard formula in curves and surfaces, and you should be able to find it in most elementary differential geometry books or derive it yourself from the structure equations. The answer is $$\Delta N = (4H^2{-}2K)\ N + 2\ \nabla H$$ where $H$ and $K$ are the mean and Gaussian curvatures (respectively), and $\nabla H$ is the gradient of $H$, regarded as a vector field tangent to the surface. (Note that I am assuming that your Laplacian is the geometer's Laplacian, not the analyst's Laplacian.)
• @BuyangLI: Since the derivation of the formula is a straightforward calculation and since you know how to prove it, you could just say 'By a straightforward calculation...". I don't know why you need to cite a reference. If a referee objects, then you can just put in the calculation, which only takes a paragraph or so. Maybe it's true that the formula for only a surface in flat $3$-space is not written explicitly in many places, but a book that gives the calculation of $\Delta N$ for general hypersurfaces in general Riemannian manifolds will contain this formula as a special case. – Robert Bryant Jan 11 '18 at 16:20