Build a Fuchsian group starting from punctures on a disk

Consider the moduli space of hyperbolic metrics on the disk with $n>3$ marked points on its boundary, $\mathcal{M}_{D,n}$.

$\mathcal{M}_{D,n}$ can be parametrised in terms of cross ratios of the punctures.

Equivalently, I can consider the double $D^d$ of $D$ wich is an hyperbolic surface with punctures. The lengths of the geodesic of this surface fully characterise it, and therefore are another way to study $\mathcal{M}_{D,n}$.

Finally, the double is obtained by quotienting $D$ by a suitable torsion free Fuchsian group $\Gamma$, which is a third way to study $\mathcal{M}_{D,n}$.

How can I explicitly build $\Gamma$ out of the cross ratios or of the geodesics?

• As for the cross-ratios, there is no simple description of $\Gamma$ in their terms, this is the notorious "accessory parameters" problem going back to Poincare. mathoverflow.net/questions/282846/… Feb 28, 2018 at 18:23

This is a very special case of the Fock-Goncharov construction.

Divide your ideal n-gon into n-2 ideal triangles. Given one cross ratio associated to each edge (i.e., to the 4 ideal vertices of the two ideal triangles adjacent to that edge) you want to reconstruct a representation $\Gamma\to PSL(2,{\mathbb R})$ from those cross ratios.

First you need some auxiliary construction. Fix the subdivision of your ideal n-gon into n-2 ideal triangles. It gives you an ideal triangulation of the double $D^d$. Let $T^\prime$ be the dual graph to this triangulation, i.e., it has one vertex inside each ideal triangle, and one edge transverse to each edge of the ideal triangulation. Let $T$ be the expanded dual graph, i.e., you expand each vertex of $T^\prime$ to a triangle, each of whose vertices is adjacent to one of the three outgoing edges. So $T$ has some „interior edges“ (which are inside an ideal triangle) and some „crossing edges“ (which cross an edge of the ideal triangulation).

Now every path in $D^d$ is homotopic to a path in $T$. So, a representation $\Gamma\to PSL(2,{\mathbb R})$ is given as soon as you associate a matrix in $PSL(2,{\mathbb R})$ to each edge of $T$, provided that along a 0-homotopic path the matrices multiply up to the identity matrix.

It turns out that the way to do this is to associate to each „interior edge“ the matrix $\left(\begin{array}{cc}0&-1\\ 1&1\end{array}\right)$, and to each „crossing edge“ the matrix $\left(\begin{array}{cc}0&-\sqrt{z}\\ \frac{1}{\sqrt{z}}&0\end{array}\right)$, where $z$ is the cross ratio associated to the crossed edge (i.e., to the 4 vertices of the adjacent ideal triangles). You can check that 0-homotopic loops are sent to the identity. More importantly, this representation yields an equivariant map of ideal boundaries such that the cross ratio associated to any edge is indeed the given $z$. It suffices to check this for adjacent triangles with vertices $\infty,-1,0,z$ which (in the Fock-Goncharov normalization) have cross ratio $z$ and where the above matrix switches $\infty$ and $0$, and sends $-1$ to $z$.

• Can you give a reference to this "Fock-Goncharov construction"? Feb 28, 2018 at 18:43
• The analogous construction for representations to PGL(3,R) (instead of PGL(2,R)) is nicely explained in old.i2m.univ-amu.fr/~fpalesi/index_files/positiverep.pdf Feb 28, 2018 at 19:03
• Thank you very much it seems very interesting! Could you point me at some reference for further details? Feb 28, 2018 at 19:05
• And the original source for the general case is of course arxiv.org/abs/math/0311149 . (In particular, section 9.) Feb 28, 2018 at 19:06
• Thilo: While what you wrote is fine, one should mention that the cross-ratios that you are using are not the same as the ones on the original punctured sphere (and which appeared in OP's question). I suspect that OP does not understand the difference between the two sets of cross-ratios. These two sets of cross-ratios are related by some transcendental function which nobody since Poincare can "explicitly" compute (say, compute its Taylor coefficients using some recursive formula). Feb 28, 2018 at 19:28