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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.

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  • $\begingroup$ 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
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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 $(S,g)$.

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  • $\begingroup$ 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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enter image description here

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!

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  • $\begingroup$ 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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