1. ## Mean Curvature

On page 256 http://zakuski.utsa.edu/~jagy/papers/Michigan_1991.pdf there is a formula for the mean curvature of a level set of a function. Im interested in the case n=3. How do you prove this formula? I can prove it for the case of a graph, so i thought i would use the implicit function theorem to make it a graph locally, but i dont see how I will get the original function in the final answer. Alternatively, I'm aware this equation is just the divergence of the normal vector, but how do you show that H can be defined this way? I seem to need to refer to a chart, but i can't seem to find one that gives a nice answer. Any help would be much appreciated
Tom

2. ## Re: Mean Curvature

Yes since grad(F) is not zero we can assume that( without loss of generality) $F_{n+1} \ne 0$, according to the implicit function theorem $x_{n+1}=f(x_1, x_2, \ldots , x_n)$ where f is a smooth function. Then the remaining is trivial.

To see the relationship between mean curvature and the normal vector field, still let $n$ be the unit normal vector of the hyper-surface M in $R^{m+1}$. $\mathrm{I\!I}(X, Y)=\langle D_X n, Y \rangle$ be the second fundamental form of M, where $D$ is the ordinary differential operator of $R^{m+1}$. The mean curvature $H$ is the mean value of the principal curvatures, that is, eigenvalues of $\mathrm{I\!I}$.
So $H = \frac{1}{m} \mathrm{tr}(\mathrm{I\!I})$.
Let $e_1, e_2, \ldots, e_m$ be an orthonormal basis of the tangent plane, we have
$H = \frac{1}{m} \mathrm{tr}(\mathrm{I\!I}) = \frac{1}{m} \sum_{i=1}^m \mathrm{I\!I}(e_i, e_i) = \frac{1}{m} \sum_{i=1}^m \langle D_{e_i} n, e_i\rangle$.
The last term is nothing but $\frac{1}{m}\mathrm{div}(n)$. You may find a proof in some standard text book. But if you couldn't find one let me know, and I'll post one here.

3. ## Re: Mean Curvature

Hi. Thanks for your reply, and sorry for my late reply, I only just realised there was a comment! Im still a bit confused. Say for example I have M as a level set, so that n = gradf \ |gradf|. How does the mean curvature expression give me the "divergence in the flat metric" ie the standad \nabla = (d\dx, d\dy,... etc.). Wouldnt my basis have to be the standaard basis for R^3? but then, this isnt neccessarily a basis for the tangent plane of M? indeed, there would only be two vectors in that case. This is where im getting quite confused. any help would be much appreciated

4. ## Re: Mean Curvature

together with n the normal vector of the surface you got 3, which is a basis of R^3.

5. ## Re: Mean Curvature

So then why is it that i can then use the normal divergence from multivariable calculus? is it because the divergence doesent depend on the orthonormal basis chosen, so that i can then take the standard orthonormal basis on R^3? Because using the levi civita connection with the 3 basis vectors (n, e_1, e_2) where e_1, e_2 are the orthonormal basis for the tangent plane of M gives me a different expression than the "standard" divergence of R^3

6. ## Re: Mean Curvature

let e_3=n, div(v)=\sum <D_{e_i}v, e_i>