I am confused by this picture:


What is wrong with projecting from the origin, and using the disc at $t = 1$? After doing some computation to verify my gut, it doesn't seem like the projection in the drawing with give the correct asymptote cone of the hyperbloid - in this case it doesn't seem like the map is surjective onto the hyperbloid model.

So what is going on? I'm confused.

  • $\begingroup$ Projecting from the origin and using the disk at $t=1$ gives the Klein model instead of the Poincare model. The cone doesn't need to be asymptotic to the hyperboloid, it just needs to be parallel to the asymptotic cone, since then every line steeper than the sides of the cone will hit the hyperboloid. $\endgroup$
    – Jim Belk
    Commented Jun 16, 2015 at 3:50
  • $\begingroup$ See also en.wikipedia.org/wiki/… $\endgroup$
    – Willemien
    Commented Jun 16, 2015 at 7:58

1 Answer 1


alright, made a jpeg.

First method: start with point hy. Central projection to the Beltrami-Klein model at point k. Vertical projection to the upper hemisphere at point he. Stereographic projection from the South Pole $-1$ to point p in the Poincare disk.

Second method: project hy towards the South Pole directly to the Poincare Disk at point p.

The calculations needed to confirm that these agree can all be carried out in the illustration, let us call it the $xz$ plane, the plane defined by hy, $0,$ and $-1$ of the illustration. enter image description here

I took $x > 0$ and hy at $$ hy: \; \; \left( x, 0, \sqrt {1 + x^2} \right) $$ $$ k: \; \; \left( \frac{x}{\sqrt {1 + x^2}}, 0, 1 \right) $$ $$ he: \; \; \left( \frac{x}{\sqrt {1 + x^2}}, 0, \frac{1}{\sqrt {1 + x^2}} \right) $$ $$ p: \; \; \left( \frac{x}{1 +\sqrt {1 + x^2}}, 0, 0 \right) $$ You may want the identity $$ \frac{x}{1 +\sqrt {1 + x^2}} = \frac{\sqrt {1 + x^2} - 1}{x} $$


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