Curvature dependence of the Laplacian operator acting on a  n-1 dimensional compact submanifold in the n-dimensional Euclidian space Possibly a simple question in differential geometry (maybe not accurate but understandable in mathematical terms): Given an compact surface $ \mathbf {R} $ in $n$ Euclidean space parameterized by $n-1$ variables $ (x_1,x_2,...,x_{n-1}) $ in the following:
$ \mathbf {R} $={ $ X_1,X_2,X_3,...,X_n$ }, ($ X_i=X_i(x_1,x_2,...,x_{n-1}$ ) is the $i$-th Cartesian coordinate)
Then, what is the result of Laplacian operator $∇^2=(1/(\sqrt{g})\partial_{μ}g^{μυ}\sqrt{g} \partial_{υ} $ acting on the $ \mathbf {R} $ as $∇^2 \mathbf {R}$ ? I think that it should be a result that purely depends on the extrinsic curvatures, and also a geometric invariant. Please offer me the result together with a reference which is accessible to a physicist. 
Thanks.
 A: Its a pretty elementary computation (it's done in the Appendix of Klaus Ecker's book "Lectures on Regularity for Mean Curvature Flow" for instance) to see that if $f$ is a smooth function defined in a neighborhood of $R$, then
$$
\Delta_R f=\Delta_{\mathbb{R}^n} f -\nabla^2_{\mathbb{R}^n} f (\mathbf{n}, \mathbf{n}) +\mathbf{H}_R \cdot \nabla _{\mathbb{R}^n} f
$$
where here $\Delta$ is the negative definite laplace beltrami operator, $\nabla^2$ is the Hessian, $\mathbf{n}$ is a choice of normal to $R$ and $\mathbf{H}_R$ is the mean curvature vector of $R$.
A: Dear Rbega, Thank you! For a two dimensional surface, I can prove a much
simpler relation by direct computations: 
\begin{equation*}
1/\sqrt{g}\partial _{\mu }g^{\mu \nu }\sqrt{g}\partial _  {\nu }\mathbf{R}=2
\mathbf{\mathbf{H}_{R}}\text{.}
\end{equation*}
Now, Let me calculate explicitly in general according to your formula, with
use of the Einstein summation convention. Since we deal with a $n-1$
dimensional surface 
\begin{equation*}
{\mathbf{R}} = ({ X_ {1},X_ {2},...,X_ {n} } )= X_ {j}\mathbf{i}_{Xj}
\end{equation*}
with $\mathbf{i}_ {X_{j}}$ denoting the unit normal along $j$-th Cartesian
coordinate, we would have 
$ \Delta _ {\mathbf{R}^{n}} $ $ \mathbf{R} $ $=\partial _ {X_{i}} $ 
$ \partial _ {X_{i}} \mathbf{R} =0, $
and 
\begin{equation*}
\nabla _ {\mathbf{R}^{n}} \mathbf{R} 
\end{equation*}
\begin{equation*}
\equiv (\mathbf{i} _ {X_{i}} \partial _ {X_{i}}) (X_{j}\mathbf{i} _ {X_{j}}) 
\end{equation*}
\begin{equation*}
= \mathbf{i} _ {X_{i}} \delta _ {ij} \mathbf{i} _ {X_{j}},
\end{equation*}
and so 
\begin{equation*}
\mathbf{H}_{R}\mathbf{\cdot }\nabla _{\mathbf{R}^{n}}\mathbf{R=\mathbf{H} _{R}.}
\end{equation*}
If I\ am correct, please tell me what is $f(\mathbf{n},\mathbf{n})$ in our
problem 
\begin{equation*}
\mathbf{R}=X_{j}\mathbf{i}_{Xj}\text{,}
\end{equation*}
and what is $\nabla _{\mathbf{R}^{n}}^{2}$? In physics, we usually use two
forms: one is the 
\begin{equation*}
1/\sqrt{g}\partial _{\mu }g^{\mu \nu }\sqrt{g}\partial _{\nu }
\end{equation*}
on the surface, and another is $ \partial _ {X_{i}} \partial _ {X_{i}}$ 
in the $n$ dimensional Euclidean space, what is the $\nabla _{\mathbf{R}
^{n}}^{2}$? 
