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Let H be the hyperboloid model for the Poincare' disk.

Geodesics of H are given by intersections with H of planes { w | $<w , v>$ = 0}, v being a space like vector

Is it possible to represent hyperbolic circles as intersection of planes with the hyperboloid? Moreover is it possible to do so in such a way that is possible to see circles converging to an horocycle?

My guess would be the following: 1) The intersection of H with the plane {w | $<w , v> = c$}, v being a time-like vector this time and c is a constant, is an hyperbolic circle with center $\frac{v}{<v,v>}$ and radius $ArcCosh(c)$.

2) If we take for $c = <v,v>$ and we let v tend to a light-like vector, we get the horocycle condition and so the circles of point 1) tend to an horocycle tangent to the boundary point corresponding to the light-ray given by v.

Is my picture correct?

giulio bullsaver
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1 Answers1

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I'm going to ignore the $c$

Given $x^2 + y^2 - z^2 = -1$ with $z > 0,$ any plane through the origin that intersects the surface gives a geodesic. Any plane which intersects the surface in a closed curve gives a geodesic circle. Furthermore, if two such planes are parallel, the circles are concentric; the center is where a parallel plane is tangent to the surface at a single point.

Equidistant curves are made by planes parallel to a plane through the origin that gives a geodesic.

A plane through the origin that contains a line of the cone $z^2 = x^2 + y^2$ does not intersect the surface. However, a plane that is parallel to one of those, and does intersect the surface, gives a horocycle. Which means that rotating from a plane that defines a circle, through planes defining circles of larger radius, until we reach a plane that gives a horocycle, gives the limiting process you asked about. For example, if you wanted all the circles and the horocycle to have a common tangent, fix a line such as $x = \sqrt 3, z = 2, $ and take a family of planes containing that line with normals $(\sin \beta, 0, \cos \beta)$ and $0 \leq \beta \leq \pi/4$

Will Jagy
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