Mean curvature of a level set 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
 A: As I recall, what I did originally was just to rotate everything. Put another way, if your level surface is a graph over the xy plane, $z=f(x,y),$ in such a way that $f(0,0)=0$ and $\nabla f(0,0) = (0,0),$ then the mean curvature at the origin is just the Laplacian $\Delta f(0,0),$ although I always liked to divide by 2, or $n$ in $\mathbb R^{n+1}.$ As soon as you need to rotate in order to get the level surface into that position, instead of just the Laplacian you get a mixture of products of first and second partials because of the rotation step. I think i say this in the paper, take the formula for a level set and just plug in $g(x,y,z) = f(x,y) - z$ and see what you get.
What is your background in Riemannian geometry? 
A: For future reference I needed this formula also; here's a derivation that doesn't require passing through curvature of graphs:
Let $\phi$ be your function; away from its critical points you can locally parameterize the level sets of $\phi$ by $r:\mathbb{R}^3\to \mathbb{R}^3$,
$$\phi(r(x_1,x_2,x_3,x_4)) = x_4,$$
with the partial derivatives of $r$ orthonormal.
Differentiating twice gives you
$$\frac{\partial r}{\partial x_i}^T \nabla^2\phi \frac{\partial r}{\partial x_i} + \nabla \phi \cdot \frac{\partial^2 r}{\partial x_i^2} = 0$$
for $i\in \{1,2,3\}$, where $\nabla^2$ is the Hessian, whence
$$\nabla \phi \cdot \Delta_{x_1,x_2,x_3} r = \frac{\partial r}{\partial x_4}^T \nabla^2\phi \frac{\partial r}{\partial x_4} - \Delta \phi.$$
Using the fact that the Laplacian of the embedding of a three-manifold is three times the mean curvature normal, and that $\nabla \phi$ is parallel to $\frac{\partial r}{\partial x_4}$, we arrive at Will's formula
$$H = \frac{\nabla \phi^T \nabla^2 \phi \nabla \phi - \|\nabla \phi\|^2\Delta \phi}{3\|\nabla \phi\|^3}$$
