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Let $F$ map points in $\mathbb{R}^3$ to points on the unit disk, $\Delta$, in the $xz$-plane (identified with $\mathbb{C}$) by projecting through $\Delta$ along lines that intersect at $(0,-1,0)$. Let $H$ be the forward sheet of the hyperboloid as in the hyperboloid model.

  1. Given an element $p:z\to(az+b)/(cz+d)$ of $\text{PSL}(2,\mathbb{R})$ is there a transformation $g$ of $H$ such that $F \circ g \circ F^{-1} = p$ on $\Delta$?

  2. If the answer to (1) is "yes," then is there a known explicit formula for $g$ in terms of $a$, $b$, $c$, and $d$?

To clarify the situation, in the attached figure $g$ is a transformation of $H$ that nearly satisfies the condition from (1) on the green test curve. I found $g$ in the figure by manually tweaking the elements of a 4x4 matrix until the red (projected) and green (correct) curves became visually close in the disk on the bottom right.

Figure showing transformations between the transformed hyperboloid and the unit disk

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That is a rather weird way to identify $\mathbb{H}^2$ with the disk, which has pretty bad properties, I think. The standard way is to take the disk at height $1$ (in your ordering, which is also weird, $y=1$) and use the central projection. –  Igor Rivin Aug 26 '12 at 1:12
    
Many of the various models of hyperbolic geometry, and the transformations between them, are discussed in the article "Hyperbolic geometry" in the book "Flavors of geometry". Here is a link: library.msri.org/books/Book31/index.html –  Sam Nead Aug 26 '12 at 7:17
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Yes, this is the standard identification of hyperboloid with the Poincare model of hyperbolic plane, see Wikipedia article. You can show first that geodesics on hyperboloid go to hyperbolic geodesics by direct computation. Or you can show that Lorentzian metric restricted to hyperboloid pulls back to the hyperbolic metric on the disk, also a direct computation. The rest is undergrad complex analysis identifying conformal group of the unit disk with PSL(2,R). Afterr you have done that you can express isomorphism explicitly in terms of matrix coefficients. –  Misha Aug 26 '12 at 10:07

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