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.

  • $\begingroup$ the definition of the support function was deleted for some reason. here it is: h(u,v) = <X,N> where N is the unit normal. $\endgroup$ – DrDooglesworth Sep 29 '11 at 3:14
  • $\begingroup$ LaTeX can be included by placing the formulas in between dollar signs. I would recommend using langle and rangle instead of less than and greater than signs for angle brackets though. $\endgroup$ – j.c. Sep 29 '11 at 3:18
  • $\begingroup$ I fixed the LaTeX $\endgroup$ – David Roberts Sep 29 '11 at 3:53
  • $\begingroup$ I'm pretty sure there is such a formula. If the 2d case is as you say, it should follow that the Weingarten operator is $B^{-1}=\pm Hess(h)+hI$, and it should be possible to prove this by considering what happens in 2-planes containing the eigenvalues of the shape operator. $\endgroup$ – Jean-Marc Schlenker Sep 29 '11 at 7:25

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.


Oh, well. What you want is in my first article, HERE in the beginning of the section Notation and Methods.

  • $\begingroup$ I didn't see it there; I didn't even see the support function at all. Maybe you can explain how to use the support function to produce a description of the surface along the lines in your article, as a level set or in terms of a parameterization. $\endgroup$ – Ben McKay Sep 29 '11 at 7:52
  • $\begingroup$ You rotate and translate so that the point of interest is at the origin and the surface is horizontal at the origin. Then a neighborhood of the origin is the graph of a smooth function. Anyway, there is an answer by Deane Yang. $\endgroup$ – Will Jagy Sep 29 '11 at 18:14

See this:



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