Hyperbolic Crown

Question

A $\textit{hyperbolic crown}$ is a hyperbolic annulus bounded by a closed geodesic $C$ on one side, and a chain of bi-infinite geodesics on the other. Each adjacent pair of bi-infinite geodesics bounds a “boundary cusp”.

Show that a hyperbolic crown with $m\geq2$ cusps is determined by $m$ real parameters.

What I think :

One parameter would be the length of the closed geodesic $C$. By an appropriate isometry we can demand our closed geodesic $C$ to lie along the real diameter of the Poincare disk symmetrically around $0$. Then, by the very definition of a crown, two ideal vertices are determined symmetrically on either side of the imaginary axis in the disc. The other $m-1$ ideal vertices can be chosen anywhere. I think that is how we will get $m$ real parameters that determine the crown. But can those $m-1$ ideal vertices be chosen anywhere? How?

Thank you in advance.

Answer 1

A different, and more intrinsic way to do this is to first notice that for each boundary cusp $b$ there exists a unique ray $\rho_b = [x_b,b)$ in the crown which is asymptotic to $b$ on its infinite end and which intersects the boundary circle at right angles at its base point $x_b$. If you enumerate the boundary cusps in circular order $b_1,...,b_m$ then the points $x_{b_1},...,x_{b_m}$ occur in circular order on the boundary circle $C$. The circular sequence of $m$ segment lengths $$l_1 = \text{Length}[x_{b_1},x_{b_2}] \quad , \quad ... \quad , \quad l_{m-1}=\text{Length}[x_{b_{m-1}},x_{b_m}] \quad , \quad l_m = \text{Length}[x_{b_m},x_{b_1}] $$ form parameters which determine the hyperbolic crown. Furthermore, any circularly ordered sequence of positive real numbers $l_1,...,l_{m-1},l_m$ can be realized as the parameters of a hyperbolic crown.

Answer 2

Alternatively, we can specify parameters in terms of cross-ratios as well. Let us take a hyperbolic crown with $m$ cusps and boundary geodesic length $l$ and cut it open in the Poincare disk model. Let the vertices of the $(m+3)$-polygon thus obtained be $z_1,z_2,\dots, z_{m+3}$, with $z_1$ and $z_2$ be the interior vertices and the others being ideal vertices. Then by an appropriate isometry $f$, we can map $z_1$ to $x$, $z_2$ to $0$ and $z_3$ to $i$, where $x$ is such that $d_{\mathbb{D}}(0,x)=d_{\mathbb{D}}(z_1,z_2)$. We have that $\left[z_1,z_2,z_3,z\right]=[x,0,i,f(z)]$. Let $\lambda_z=\left[z_1,z_2,z_3,z\right]$. Then it is easy to see that $f(z_i)$ is determined by the cross-ratio $\lambda_{z_i}$ for $i=4,5,\dots, m+2$. Note that $f(z_{m+3})$ is determined by the length $l$ itself. Then the $m$ parameter set $\{l,\lambda_{z_4},\lambda_{z_5},\dots,\lambda_{z_{m+2}}\}$ determines our hyperbolic crown.