# 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

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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?

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ah the author himself! firstly thanks for your reply. And ahh i see now, i was doing it the other way round. great that all works nicely now. and im doing a masters atm, and ive taken course on curvers and surfaces, manifolds and riemannian geometry. the thing is my lecturer last year proved some quite important result so his teaching suffered about so i only really learnt the basic definitions for the mean curvature etc., and in RG we havent really touched on it again. But i need this for my project (im proving the penrose inequality for graphs over R^3) May i ask another question of you? – user78410 May 18 '13 at 21:44
I don't understand the new question. In case this is it, there should be a line about how it does not matter whether we take the unit normal as part of the normal bundle, or instead extend it to an open neighborhood of the submanifold in the simplest way and take the ambient divergence. – Will Jagy May 18 '13 at 22:01
It's okay i figured it out now. thanks for your help there, needed that result. – user78410 May 18 '13 at 22:50

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 $$H = \frac{\nabla \phi^T \nabla^2 \phi \nabla \phi - \|\nabla \phi\|^2\Delta \phi}{3\|\nabla \phi\|^3}$$

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