I am working with surfaces in Euclidean 3-space. If we let $X = X(u,v)$ denote a parameterization of such a surface, then the mean curvature, $H = H(u,v)$, can be computed in terms of the coefficients for the first and second fundamental forms.
My question is this: Is it possible to express the mean curvature, $H(u,v)$, in terms of the support function for this surface? The support function is defined to be $h = h(u,v) = \langle X, N\rangle$ where $N$ is a unit normal. (This function measures the oriented distance from a tangent plane to the origin.)
For curves in the plane there is a nifty result along these lines. If the curve has non-vanishing curvature its unit normal can be used for a parameterization, and in this situation the curvature satisfies $1/k = \pm (h''+ h)$ where, again, $h$ is the support function for the curve.
I'm hoping there is a similar result for convex surfaces in space, but, sadly, have been unable to find such a relationship. Any help would be greatly appreciated.
Best Answer
The following holds in any dimension: If $h$ is the support function, then the quadratic form given by $\nabla^2h + hg$, where $g$ is the Riemannian metric on the unit sphere, is the inverse to the second fundamental form. Its eigenvalues (with respect to an orthonormal basis) are the principal radii (reciprocals of the principal curvatures).
In 2-d, this means that the mean curvature can be written in terms of the trace and determinant of $\nabla^2h + hg$. I leave the details to be worked out by you.