Universal covers of punctured hyperbolic surfaces - MathOverflow

Universal covers of punctured hyperbolic surfaces

2011-04-12T18:17:11Z

Suppose S is a genus g surface with n punctures satisfying the hyperbolicity condition 2g + n - 2 > 0. If n > 0 the fundamental group of the surface is a free group on 2g + n - 1 := m generators.

If we look the universal covers of different punctured surfaces with the same m (e.g., thrice-punctured sphere and once-punctured torus for m = 2) in, say the hyperbolic plane or the Poincare disc model, how do they differ? The "only" apparent difference is in the number of punctures which should give rise to a difference in the lifts of the punctures to the boundary of the disc. The fundamental groups are isomorphic, but they must act differently to produce quotient surfaces of different genera. How?

How does the set of lifts of punctures on the boundary relate to the standard Farey set?

Thanks a lot in advance!

Answer by Sam Nead for Universal covers of punctured hyperbolic surfaces

2011-04-12T20:22:42Z

For the thrice-punctured sphere, there is a generating set where both generators are parabolic. For the once-punctured only the commutator (and its conjugates) is parabolic. Hence any element that can be extended to give a generating is hyperbolic.

Thinking in the upper half-plane model of $\mathbb H$, the Farey set is the rational points of the real line (plus the point at infinity). If $X$ is a punctured hyperbolic surface then the lifts of the ideal points of $X$ likewise form a dense set in the real line. If $X$ is a thrice-punctured sphere then, after possibly conjugating the deck group by an isometry of $\mathbb H$ the lifts of the punctures are the Farey set. However, for any other surface $X$ this only happens if the modulus of $X$ is carefully chosen.

Answer by Igor Rivin for Universal covers of punctured hyperbolic surfaces

2011-04-13T03:12:26Z

I strongly recommend you look at:

MR0795536 (86j:11069) Series, Caroline(F-IHES) The geometry of Markoff numbers. Math. Intelligencer 7 (1985), no. 3, 20–29. 11J06 (11J13 11J70)

Answer by S. Carnahan for Universal covers of punctured hyperbolic surfaces

2011-04-27T08:32:55Z

The simplest case, where $g=0, n=3$ and $g = n = 1$ yield isomorphic groups (free of rank 2), can be written explicitly using a little $2 \times 2$ matrix calculation. We choose a fundamental domain in the upper half-plane made out of two vertical lines with real parts $-1$ and $1$, and two semicircles whose diameters are the real intervals $[-1,0]$ and $[0,1]$. Since the boundaries are geodesics, it suffices to find Möbius transformations that transform the endpoints appropriately.

For $g=0, n=3$, we choose generators $\begin{pmatrix}1& 2 \\ 0 & 1 \end{pmatrix}$ to glue the vertical lines together, and $\begin{pmatrix}1& 0 \\ 2 & 1 \end{pmatrix}$ to glue the semi-circles. If you like modular curves, this quotient is called $Y(2)$, and classifies isomorphism classes of elliptic curves equipped with an ordered list of all 2-torsion points.

For $g=1, n=1$, we choose $\begin{pmatrix}1& 1 \\ 1 & 2 \end{pmatrix}$ to glue the left vertical line to the right semicircle, and $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ to glue the left semicircle to the right vertical line. This quotient is a leaky torus.

The above results form a special case of a general phenomenon (mentioned by Sam Nead), where the loops around punctures give unipotent (aka parabolic) generators of $\pi_1$, and handles give a pair of hyperbolic generators.

I don't have a good answer concerning the set of lifts of the boundary points - it will always be a disjoint union of $n$ orbits under the transformation group, but it can vary widely, since the transformation group has a continuous family of representations in $PSL_2(\mathbb{R}$.