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I would like to understand the curvature of a hyperbolic plane better in relation to the underlying Euclidean model and intrinsically without a model.

I only consider the Beltrami-Klein model and the Cayley-Klein distance in that wikipedia article.

We have an Euclidean unit cirlce with radius $1$. When one of the points is the origin and Euclidean distance between the points is $r$ (where $|r|<1$) then the hyperbolic distance is: $$ r'= \frac{1}{2}\ln{\frac{1+r}{1-r}} $$ The article claims:

The factor of one half is needed to give the model the standard curvature of $−1$.

By solving $r'=1$, we see that the circle with Euclidean radius $0.7616$ is the unit circle in the hyperbolic plane.

One could define the distance also as: $$ r'= \frac{1}{2} c \cdot \ln{\frac{1+r}{1-r}} $$ with an arbitrary constant $c$ instead of one half.

Question 1) What is the formula that relates this constant $c$ to the curvature?
Apparantly when $c=1$ the curvature is $-1$.
I suspect the answer will be: curvature = $-1/c^2$?

Question 2) When choosing another constant $c$ according to the wikipedia article, the curvature changes. But, because the equation $r'=1$ will have a different solution, also the unit circle in the hyperbolic plane changes. What is the relationship between the curvature and the location of the hyperbolic unit circle?
Probably this just follows directly from the formula: $r=\frac{e^{2/c}-1}{e^{2/c} + 1} =\tanh{(\frac{1}{c})} \longleftrightarrow r'=1$.

Question 3) Suppose we live in a hyperbolic plane geometry without Euclidean model to back it up. Is it possible to determine its curvature and how?

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  • $\begingroup$ To compute sectional curvature, you'll need the curvature tensor. This may be simpler in polar coordinates at zero, but then to make this global, you'll need that the isometry group acts transitively. $\endgroup$ – Neal Jul 26 '16 at 0:37
  • $\begingroup$ @Neal I suppose it is somewhat more simple, because the curvatur is just a constant value in this version ot the hyperbolic plane. $\endgroup$ – Gerard Jul 26 '16 at 8:03

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