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The primer on mapping class groups, by Farb and Margalit.

NEW ANSWER: As there has been much confusion on this point (some of it mine...): Definition: A Riemannian 2-manifold $S$ is of hyperbolic type if the universal cover of $S$ is conformally equiva …

The answer is "yes" if $S$ is the Riemann sphere. This is because a map $f$ of degree $d$ from the sphere to itself is homotopic to $z \mapsto z^d$. The answer is "basically no" if $S$ has genus two …

I don't have a precise answer, but the genus of $S$ has to grow like $n!$. To see this, note that when $n$ is large enough, $S$ cannot be the sphere or torus. So $S$ admits hyperbolic metrics. By Ni …

Looking at the link you give, it seems that you want an explicit representation of $\pi_1 = \pi_1(S)$, the fundamental group of the genus two surface, into $\mathrm{PSL}(2, \mathbb{C})$ so that the up …

$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = \Sigma_2$ be the genus two surface. In this case, $\ZZ^4$ is the deck group of the desired covering. Consider $\ZZ^4$ inside of $\R …

Fred Gardiner's book Teichmüller theory and quadratic differentials is a good reference. He (a) deals with the punctured case (called finite analytical type: see the first page of Chapter 2), (b) cov …

The answer to (1) is yes. Take $P$ a hyperbolic surface with one geodesic boundary, called $\delta$, and two punctures. Form $S$, a sphere with four punctures, by doubling $P$ across $\delta$. No …

In your description of moduli space you say: I can use a Moebius transformation to send three of the punctures at those location whilst the fourth puncture is free to move. That assumes that …

Collecting some of the answers above, and editing a bit: What can one say about its torsion elements? $\newcommand{\Mod}{\mathrm{Mod}}$As with lattices in Lie groups, and in word-hyperbolic groups, …

No, certainly not. Biholomorphic maps are very rare and very rigid. Generically, given Riemann surfaces $X$ and $Y$, there will be no biholomorphic maps between them. Yes. You will want to read u …

Let $S$ be the underlying surface and call the image of the dotted lines $\gamma$. If we cut $S$ along $\gamma$ then $S$ falls apart into two components $X$ and $Y$ (both tori with a single boundary …

For the thrice-punctured sphere, there is a generating set where both generators are parabolic. For the once-punctured torus only the commutator (and its conjugates) is parabolic. Hence any element …

You write "it appears that the action of the MCG on the curve complex is incomputable for closed surfaces." This is not correct. Geva Yashfe points out one approach in the comments. Here is another, …

In the closed or bounded case, there is a uniformization of a Riemann surface by a fuchsian group. In the closed case this is the content of the uniformization theorem. The bounded case follows from …