# 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.

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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 $R$.

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Thank you for your answer. But in my question, the Laplacian(-Beltrami) operator takes a definite form, corresponding to your $\Delta$. Then what does the difference between $\Delta$ and $\Delta _ {\mathbb {R}^n}$? It appears a compact form of the result for the definite $\mathbf{R}=\{X_{1},X_{2},...,X_{n}\}$. – QHLIU Apr 15 '13 at 15:39
I'm not sure I completely understand your question. In any case, $\Delta_{\mathbb{R}^n}=\sum_{i=1}^n \partial_i^2$ is the usual Euclidean Laplacian. $\Delta_R$ is the Laplacian of the metric $g$ induced on $R$ from the euclidean metric and so is the operator you are interested in. – Rbega Apr 15 '13 at 19:34
my clarification sees in the form of answer below. – QHLIU Apr 16 '13 at 3:07

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 $\nabla _{\mathbf{R} ^{n}}^{2}$?

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