Consider a projective plane with an absolute quadric, so that it is a hyperbolic plane.
Given a curve I wonder how the tangent to a curve is defined in a plane with constant positive curvature.

I am unsure whether the tangent in a space of constant positive curvature should be defined as a straight line or somehow as a geodesic and how would I describe such a geodesic geometrically?

When possible, I understand synthetic explanations much better.

  • $\begingroup$ Perhaps this will be something you can use: en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model $\endgroup$ – Alan Apr 29 '14 at 19:34
  • $\begingroup$ @Alan In the Beltrami-Klein model, do you think the line connection two points has also the smallest possible (hyperbolic) length, i.e. is it a geodesic line? $\endgroup$ – Gerard May 1 '14 at 20:56
  • $\begingroup$ Yes, I think the straight lines in this model of the hyperbolic plane are shortest lines in terms of the hyperbolic metric. The model that I've been working with recently gives a construction of geodesics on the hyperboloid which correspond to orthogonal circles in the disk model , $\endgroup$ – Alan May 1 '14 at 21:45
  • $\begingroup$ I'll leave a drawing of the model, (it's well known to many I'm sure). I was interested in it because it enabled me to produce geodesics on the hyperboloid through two given points. $\endgroup$ – Alan May 1 '14 at 21:56

One usually defines tangent line $L=T_p(C)$ to a curve $C$ in a surface $S$ at a point $p$ as a 1-dimensional linear subspace of $T_p(S)$. This definition is intrinsic to the topology of the surface $S$. You can also (but this is nonstandard) define tangent to $C$ at $p$ as the unique maximal geodesic in $S$ tangent to the line $L$. This of course requires a Riemannian metric $g$. Constant curvature is irrelevant here as well as the specific realization of $(S,g)$.

  • $\begingroup$ A projective plane P is not a surface is it? What would be the definition of $T_p(P)$? Is the "Riemannian" of a projective plane with absolute quadric known? $\endgroup$ – Gerard May 2 '14 at 7:46
  • $\begingroup$ Of course projective plane is a surface. Consider reading a book on differential geometry or topology. This should answer most if not all your questions. $\endgroup$ – Moishe Kohan May 2 '14 at 13:39

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Yet another model of the Hyperbolic plane that allows one to construct geodesics on the surface of the hyperboloid given two points on it. From an analytic point of view , it is not evident that the geodesic equations on the hyperboloid are satisfied, but I believe it can be shown that they are.

Again, this is not the Beltrami-Klein model suggested above, since the geodesics in this disk model are circles orthogonal to the unit circle. Yes, I know, a lot of models of the hyperbolic plane!

  • $\begingroup$ Nice. I would be interesting to see whether a projective plane with absolute quadric could also be used to construct the geodesic on the hyperboloid. In the plane you would then - per your suggestion - just have any straight line. The hyperboloid is projective as well, so a straightforward projection. Then compare whether the results are equal. $\endgroup$ – Gerard May 2 '14 at 7:43
  • $\begingroup$ Just as a note: It isn't a planar slice through the projection point and the two points on the cyan colored circle (geodesic) , It's an elliptic cone through the cyan colored circle intersection with the hyperboloid that fixes the geodesic on the hyperboloid. $\endgroup$ – Alan May 2 '14 at 15:53

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