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:

  1. Where can I find more information about this notion?
  2. 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: $\rho_{K}(x)=1/\sqrt{|K(x)|}$?


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 $f(0, 0) = \partial_xf(0,0) = \partial_yf(0,0) = 0$.

  • $\begingroup$ @Deane: By mean radius, do you refer to $1/H(p)$ where $H(p)=\kappa_1(p)+\kappa_2(p)/2$ is the mean curvature at the point $p$. $\endgroup$ Feb 8 '11 at 15:53
  • $\begingroup$ I defined "mean radius" to be the mean of the principal radii of curvature. That's not the same as the reciprocal of the mean of the principal curvatures. $\endgroup$
    – Deane Yang
    Feb 8 '11 at 18:47
  • $\begingroup$ @Deane: Considering eq. (3) mathworld.wolfram.com/MeanCurvature.html what you actually defined is nothing but the quotient $H/K$. Is it correct? Is there some standard geometrical meaning of this quantity? $\endgroup$ Feb 10 '11 at 9:13
  • $\begingroup$ Dror, it's what I said it is, namely the average radius of curvature. It is indeed half of $H/K$. Any function of the principal radii of curvature can also be written as a function of $H$ and $K$. $\endgroup$
    – Deane Yang
    Feb 15 '11 at 16:29
  • $\begingroup$ @Deane: I don't know what is the hidden message in your example. Here, the Gaussian curvature at $(0,0)$ is nothing but the determinant of the Hessian of $f$ and the mean curvature is the trace. Were you aiming at something more specific? $\endgroup$ Feb 21 '11 at 15:39

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.

  • 1
    $\begingroup$ In short, and hopefully as correct, $\sqrt{\frac{1}{K}}$ is the arithmetic-harmonic mean of the two principal curvature radii $\frac{1}{\kappa_i}$. $\endgroup$ Feb 8 '11 at 15:15

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.

  • $\begingroup$ @Ramsay: Why is this smaller? Smaller then what? I agree that it is not very helpful definition w.r.t. Morse-Schoenberg lemma. But it is a sizing function which is well defined for non-flat surfaces. At the moment I'm not sure what I want to do with this definition; I found it a natural generalization to a negatively curved surface case and I was wondering if it was investigated in the literature. $\endgroup$ Feb 18 '11 at 10:22
  • $\begingroup$ By smaller I mean that $\rho_K(x) \leq \rho_G(x)$ for all $x$, where $\rho_G$ is the original definition. They agree everywhere except when the curvature is negative and $\rho_G$ provides no bound. I would be surprised if you found this to be a useful way to capture the geometry of negatively curved surfaces. Since it is purely intrinsic, you will never be able to control triangle normals this way, for example. It is not clear to me what is represented by the bound you're proposing to introduce. I don't recall seeing it anywhere previously. $\endgroup$
    – Ramsay
    Feb 18 '11 at 17:04

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