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It is well known to geometric analyst that the scalar curvature of a Riemannian manifold can be decomposed to two parts: one part has a divergence structure and the other part consists of lower order terms. My question is for the mean curvature of a hypersurface in a Euclidean space, dose it have a similar structure? Has any author considered a decomposition of this kind? Any reference will be useful to me. Thanks in advance.

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I just comment that for a graph in a Euclidean space the mean curvature has a divergence structure. –  Mathboy Nov 30 '12 at 16:16
    
The formula for graphical mean curvature is naturally in a divergence form. A hypersurfsce is locally graphical, so that gives what you want. Any book on minimal surfaces will give you the formula, the one I have in front of me is colding minicozzi eqn 1.6 on p 2. The 0 could replaced by mean curvature you're interested in nonminimal surfaces. –  Otis Chodosh Nov 30 '12 at 16:20
    
Thanks, Otis. Actually if we assume f is the immersion, I want to get a formula of the mean curvature which has two parts: one part consists of second derivatives of f and has a divergence structure and the other part consists of at most first derivatives of f –  Mathboy Nov 30 '12 at 16:34
    
Yong, you might want to take a look at the formula Otis has referred you to and then let us know whether it meets your needs or not. If not, you should perhaps present the formulas you do have and explain why they aren't what you want. –  Deane Yang Nov 30 '12 at 19:18
    
Thanks Deane. I have looked at the formula Otis told me. But I still do not know how it can satisfy my need. Actually I want to define some Mass for hypersurfaces, which is similar to the famous ADM mass. To do this I need a proper decomposition for the mean curvature. –  Mathboy Nov 30 '12 at 19:28
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I was unable to post an image without a big song and dance. If you go to MY STUFF and open the Michigan Math J. 1991, page 256 has the formula when the hypersurface is given as the level set of a smooth function. It is just a rotated version of the graph formula, as it must be. Meanwhile, the mean curvature is just a dimension constant times the divergence of the (oriented) unit normal, where it does not matter whther you take the divergence on the manifold or extend the unit normal field off the manifold and use the ambient divergence. All the same. I do not see how you are going to separate out first and second order derivatives.

Wait, I can try to typeset:

$$ n H = \frac{1}{|\nabla F|} \; \sum_{i=1}^{n+1} \sum_{j=1}^{n+1} \left( \delta_{ij} - \frac{F_i F_j}{|\nabla F|^2} \right) F_{ij} $$ That actually looks correct. Good for me. Level set of the function $F$ and $H$ is the mean curvature with one of the choices of unit normal field.

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Thanks a lot, Will. I will try to study your paper. If it works for me, I will tell you. Thanks again. –  Mathboy Nov 30 '12 at 22:34
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