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Given a Riemannian surface with nonnegative Gaussian curvature, the area of a ball of radius $r$ around any point has area at most $\pi r^2$. I have a simple proof of this in the Euclidean cone case (a surface which is flat except at a discrete set of cone points), which shows that all points in the ball can be reached with geodesics that do not pass through a cone point.

My question: who would have proved this first, and does anyone know a good reference for such? All of my searches find much more complicated results involving sectional curvature in higher dimensional manifolds.

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For completeness, here's the link to our paper with a proof of the Euclidean cone case: naml.us/~irving/papers/irving_segerman2012_fractal.pdf –  Geoffrey Irving Dec 5 '12 at 18:46

2 Answers 2

Spivak credits Diquet (1848) with the result that

$$K(p) = \lim_{r\to 0} 12 \frac{\pi r^2-A(r)}{\pi r^4}.$$

This comes from an elementary argument with the Taylor expansion for the metric in geodesic normal coordinates. (See the Appendix for Chapter 3 of Volume II of Spivak's magnum opus Differential Geometry. Also look at a brief entry on Wikipedia: "Bertrand-Diquet-Puiseux Theorem.")

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Out of curiosity, do you know if authors around this time ever considered the relationship of $A(r)$ to $K$ when $r$ is large? (P.S. welcome to math.SE!) –  treble Dec 2 '12 at 23:00

Surely Gauss knew about this. Consider $S \subset \mathbb R^3$, a two dimensional oriented embedded surface. Recall that the Gauss map $G: S \to S^2$ is defined by $$G(p) = \mathbb n(p)$$ where $\mathbb n(p)$ is the unit normal of $S$ at $p$, translated to the origin (so that $\mathbb n \in S^2$, the unit sphere). Then the Gaussian curvature at $p$ effectively measures the signed area distortion of a small rectangle near $p$ under the map $G$. Thus the Gauss curvature $K$ is defined to be $$\textrm{det}(DG)$$ where $DG$ is the derivative of $G$, say, written with respect to a local parametrization of $S$ in some neighborhood of $p$. The hypothesis that $K := \textrm{det}(DG) \geq 0$ is thus saying something about area deformation. So it is not a long walk from here to the proof that you want (most proofs also involve one other concept: normal (or "geodesic") coordinates, also invented by Gauss). As far as a reference is concerned, consider looking at Ted Shifrin's great book Differential Geometry. I don't know if he proved the volume comparison result you wanted per se, but the "pieces" are certainly there.

Edit: After a little bit more thought, I guess it is more difficult than I thought to outright prove this from first principles based on what I said above. I don't know whether or not I've ever seen it worked out completely just for, say, surfaces in $\mathbb R^3$ (the more general results imply this case of course, but that is not what your question is about).

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