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Gauss' Theorema egregium says that

the Gaussian curvature of a surface can be determined entirely by measuring angles, distances and their rates on the surface itself.

A surface looks like $\mathbb{R}^2$ locally so the angle sum of arbitrarily small triangles tends to $\pi$, doesn't it? Only when one considers bigger triangles - as Gauss did - one will find angle sums deviating from $\pi$. So I wonder how - concretely - an inhabitant of an arbitrarily (but smoothly) curved surface would measure the Gaussian curvature at a given point.

Hans-Peter Stricker
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  • Perhaps by measuring small things with high enough precision? – Alexander Shamov Jan 11 '13 at 13:43
  • Yes, but any triangle you use is not arbitrarily small, but has a certain definite size. Therefore, you can measure the sum of its angles and this in turn gives you a scale on which the surface is flat, since the curvature is product of principal curvatures ($K \sim 1/r^2$ on the sphere). In case you really live in the flatland, no matter how big a triangle you use, you will always obtain $\pi$ up to the precision of the measurement and therefore will have to conclude that the surface is flat at all scales of measurement (rather than just infinitesimal scales as a generic surface). – Marek Jan 11 '13 at 13:51
  • To avoid misunderstandings I shouldn't have mentioned Flatland: I meant any 2-dimensional surface, not only flat ones. – Hans-Peter Stricker Jan 11 '13 at 17:08

2 Answers2

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Fix an opening angle and draw an isosceles geodesic triangle with the two legs of length $r$ from that opening angle. Measure the angle defect.

Do this many times for a sequence of $r\to 0$, plot the angle defect versus $r^2$. The limiting slope would be the curvature at the point. As long as the curvature does not change too wildly from point to point, you can get a good approximation for small but finite $r$.

Willie Wong
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  • My error in reasoning has been that the angle defect must go to 0 with $r$. (I have overseen the difference between diffeomorphism and isometry.) But it remains true, that on the sphere the angle defect grows with $r$, doesn't it? – Hans-Peter Stricker Jan 11 '13 at 16:57
  • The angle defect does go to 0 with $r$. Hence we need to take slope of of angle defect / r². Basically, the curvature is an infinitesimal quantity (like velocity or acceleration) so macroscopically you can only measure an average, and you need to take a limit to get the value at one point. – Willie Wong Jan 14 '13 at 06:33
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Let's use normal coordinates. The circumference of a circle of radius r around 0 in these coordinates is then given by $$C(r)=\int\limits_{0}^{2\pi}\sqrt{g_{ij}\frac{dx^{i}}{d\phi}\frac{dx^{j}}{d\phi}}d\phi$$ where $x^{1}(\phi)=r \cos(\phi)$, $x^{2}(\phi)=r \sin(\phi)$. In these coordinates we have the expansion $g_{ij}=\delta_{ij}-\frac{1}{3}R_{ikjl}x^{k}x^{l}+O(\vert x \vert^{3})$, so by doing a little Taylor expansion $$C(r) = \int\limits_{0}^{2\pi}\sqrt{r^2-\frac{r^4}{3}R_{1221}+O(r^{5})}d\phi=2\pi\left(r-R_{1221}\frac{r^3}{6}\right)+O(r^{4})$$ with derivative $$\frac{dC(r)}{dr}=2\pi-\pi R_{1221}r^2+O(r^{3})$$ Therefore one could try to measure the sectional curvature (agrees with the Gaussian curvature in an local orthonormal frame like here, $K=R_{1221}$) by measuring the circumferences of circles around some point and investigate how it depends on the radius.

Dominik
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