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Consider the horoball $$B_h=\left\{(z,t)\in\mathbb{H}^3\mid t>h>0\right\}.$$ If $T\in\mathbb{P}SL(2,\mathbb{C})$ is an isometry of the hyperbolic space which does not fix the point at infinity, then $T$ maps $B_h$ to a horoball which is tangent to the point $T(\infty)$. I would like to know whether there is a formula which expresses the radius of the horoball $T(B_h)$ in terms of $h$ and the matrix $\left(\begin{array}{cc}a & b \\ c & d \end{array}\right)$ associated to the transformation $T$. I tried to prove this in dimension 2 (just to deal with simpler computations). My guess is that the radius depends only on the parameters $c$ and $d$ of the matrix. To see this, I assumed that $T(\infty)=0$ so that $a=0$, $c\neq 0$ and $d=-c^{-1}$, that is $$T(z)=\frac{1}{-z-\frac{d}{c}}.$$ Now I'd like to find somehow the radius of the horoball, but I don't know how. Could you please help me?

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The transformation $$T(z)=\frac{az+b}{cz+d}$$ satisfies $T(-d/c)=\infty$ and $T(\infty)=a/c$. The original horoball contains the point $P=(-d/c,h)$. This point is the intersection of the geodesic ray $\gamma $ from $-d/c$ to $\infty$ and of the Euclidean sphere $S$ with center $-d/c$ and Euclidean radius $h$.

The image of $\gamma$ under $T$ is the geodesic ray from $\infty$ to $a/c$, that is, the vertical line through $a/c$. To find the image of $S$, consider that $-d/c$ and $\infty$ are symmetric about $S$. Their images $\infty$ and $a/c$ must be symmetric about $T(S)$. It follows that $T(S)$ is a Euclidean sphere centered at $a/c$. To find its radius, we take a point $-d/c+h\in S$ and calculate $$T(-d/c+h)=\frac{ach-1}{c^2h}$$ Hence the radius of $T(S)$ is $$|T(-d/c+h)-a/c|=\frac{1}{|c|^2h}$$

Conclusion: the image of the horoball under $T$ contains the point $T(P)=(a/c,|c|^{-2}h^{-1})$. Since the image is a Euclidean ball tangent to the boundary at $a/c$, its Euclidean radius is $$\frac{1}{2|c|^2h}.$$

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Thank you very much! Clear and quick. I didn't think about using an additional Euclidean sphere to make the computation. Thanks! – fatoddsun Jul 27 '12 at 10:13

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