Gaussian curvature radius - MathOverflow

Gaussian curvature radius

2011-02-08T09:47:07Z

In the paper Surface sampling and the intrinsic Voronoi diagram (2008), Ramsay Dyer defines the Gaussian curvature radius at a point $x$ of a surface $S$ to be $\rho_K(x) = 1/\sqrt{K(x)}$ where $K(x)=\kappa_1(x) \kappa_2(x)$ is the Gaussian curvature at $x$.

Trying to track back the notion in Berger's A panoramic view of Riemannian geometry, and in Lee's Riemannian manifolds and in Chavel's Riemannian Geometry yielded nothing.

My question is two-folded:

Where can I find more information about this notion?
Is there a reason not to define it as $\rho_K(x) = 1/|K(x)|$? Otherwise, this definition is only valid for non-negatively curved surfaces.

EDIT As pointed out by Deane Yang, there is no sense in the definition I suggested. Nevertheless, if one wants to relate the Gaussian curvature to a radius (for either negatively or positively curved surfaces) how about this alternative: $\rho_{K}(x)=1/\sqrt{|K(x)|}$?

Answer by Joseph O'Rourke for Gaussian curvature radius

2011-02-08T12:56:37Z

It so happens that $1/\sqrt{K}$, where $K$ is the Gaussian curvature, is, in a sense, the average of the arithmetic mean radius of curvature and the radius of harmonic mean curvature. The calculation is explained in the "Merged radius of curvature" subsection of the Wikipedia article on radius of curvature. It is called the arithmetic-harmonic mean radius of curvature.

Answer by Deane Yang for Gaussian curvature radius

2011-02-08T14:21:59Z

As for question #2, why does your definition make sense for a negatively curved surface? For a positively curved surface it does not give the right answer for spheres, since presumably you would want a sphere of radius $r$ to have a Gauss curvature radius of $r$. In particular, the word "radius" reflects a linear measurement and therefore should scale linearly if you rescale the surface.

The "radius of curvature" at a point on a curve is the radius of an osculating circle and turns out to be the reciprocal of the geodesic curvature.

On a point of a surface in $R^3$, you get a radius of curvature for each tangent direction, corresponding to the osculating circle in that direction. In particular, there are the two principal radii of curvature corresponding to the principal directions. The Gauss curvature radius, as defined above, is the geometric average. Since it can be defined in terms of Gauss curvature only, it has the advantage of being intrinsic. You could also define the "mean radius" by taking the arithmetic average. I don't recall seeing this before, but it also seems reasonable to study.

I recommend working out the example of $z = f(x,y)$ at the origin, where $f(0, 0) = \partial_xf(0,0) = \partial_yf(0,0) = 0$.

Answer by Ramsay for Gaussian curvature radius

2011-02-17T14:00:01Z

Deane's answer is similar to what I would have tried to say if I'd got here on time. I don't recall seeing the "Gaussian curvature radius" defined before, so I can't point you to other references. The definition is natural. On the one hand the bound on the distance to a conjugate point (Morse-Schönberg lemma) is given in terms of a bound on the Gaussian curvature radius, and on the other hand the Gaussian curvature radius provides an upper bound to the "maximal curvature radius"(reciprocal of the maximum of the absolute values of the principal curvatures). As Deane pointed out, these two curvature radii coincide on the sphere.

Since we are only using it as an upper bound, we just define it to be infinite if the curvature is non-positive. In flat or negatively curved spaces, conjugate points are not an issue; geodesics diverge.

As to your alternative, $\rho_K(x) = 1/\sqrt{|K(x)|}$, I guess it depends on what you want to do. You are making a smaller sizing function, but why? The spirit of the Morse-Schönberg lemma is better captured without the absolute value signs. If the infinite values disturb you, you have not avoided them when the Gaussian curvature vanishes.