Suppose you are transported to an 2 dimensional hyperbolic world, ( a plane (2 dinensional) manifold with a constant negative curvature ) the only geometrical tools you have are a ruler, a pencil, a measuring stick (but with inches , not the absolute hyperbolic scale) a piece of string and an right angle,

How to measure the gaussian / total curvature of where you have landed?

(or "How many inches go into an absolute hyperbolic distance of 1 ?")


  • you can use the piece of string to make circles ,
  • you cannot use the piece of string and the ruler to measure curves. (just not

The Absolute hyperbolic distance is a measurement in hyperbolic geometry, if trilateral ABC has angles (measured in radians) $\angle ABC = \frac{\pi}{2}$ (right angle) , $\angle BAC = 1$ and $\angle ACB = 0$ ( $C$ is an Ideal point) then $AB$ has the absolute distance 1

  • $\begingroup$ I'm not sure about what you mean by an "inch" compared to the "absolute" hyperbolic distance, or even by a "plane manifold". If you mean a two-dimensional manifold with constant negative curvature, the definition of curvature already uses your hyperbolic distance, so there are no such things as "inches" in your world. Or "absolute" distance. You just have one distance defined on it. $\endgroup$ – Chocosup Oct 16 '14 at 13:29
  • $\begingroup$ @Chocosup updated my question hope this makes it clearer $\endgroup$ – Willemien Oct 16 '14 at 15:04

I don't know what you mean by "absolute hyperbolic scale" (perhaps you're referring to some model of hyperbolic space embedded in $\mathbb{R}^n$?), but distance is distance---part of the point of abstract Riemannian manifolds is that your metric (ruler) is part of the data of your space, and there are no other metrics sitting around.

Anyway, as for your question, one easy option would be to use the following characterization of scalar curvature: The circumference of a circle of radius $r$ on a Riemannian surface at a point with scalar curvature $\kappa$ satisfies $$C = 2 \pi\left(1 - \frac{S}{12} r^2 + O(r^4)\right).$$

So, measure off some small length $r$ of the string using the ruler, tie the string around the pencil and use it as a makeshift compass to draw a small circle of radius $r$ around any point, and use the rearranged formula $$S = \frac{1}{r^2} \left(12 - \frac{6C}{\pi} \right) + O(r^2).$$ The error term will be small when $r \ll |\kappa|$.

  • $\begingroup$ thanks for that but what is $O(r^2) $ and $O(r^4)$? your formulas boil down to $O(r^2) = -\frac {12 O(r^4) }{r^2} $ if i am correct not sure what to make of that, also sorry new constrictions in my question you cannot measure curves anymore. (forgot that bit) $\endgroup$ – Willemien Oct 16 '14 at 15:02
  • $\begingroup$ That's Big O notation: en.wikipedia.org/wiki/Big_O_notation#Infinitesimal_asymptotics This means that for small enough $r$, the contribution of the error represented by $O(r^2)$ is negligible compared to the terms written out explicitly. $\endgroup$ – Travis Oct 16 '14 at 15:15
  • $\begingroup$ As for your new restriction, one can use the same formula, but one must be a bit more clever about approximating the circumference: Simply construct a narrow triangle with long side $r$ and known small angle (say, by bisecting a right angle several times) and build congruent copies of it with adjacent long sides and a common vertex. Eventually, they'll overlap, giving you an estimate of $C$, and you can make the error in $C$ as small as you'd like by working with sufficiently small angles. It's anyway odd that you have a ruler but can't measure a particular linear object with it... $\endgroup$ – Travis Oct 16 '14 at 15:20
  • $\begingroup$ I suppose you'd also want to disallow laying a string out along a curve, holding it at its endpoints, stretching it taught, laying it on the surface, marking where the endpoints sit with the pencil, and measuring the distance between the endpoints with the ruler. With this restriction, it's not clear what a string allows you to do that a compass does not. $\endgroup$ – Travis Oct 16 '14 at 15:24

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