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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 …

The answer to question two is also "no". The right way to minimize the dilatation is vary both the metric and the representative $f'$ of the mapping class $[f]$. There is a large family of pairs $(f …

This is done, for shearing coordinates (both for pants decompositions and triangulations), by Theorem 1.3 of the paper "Shearing coordinates and convexity of length functions on Teichmüller space" by …

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 …

$\newcommand{\Sym}{\operatorname{Sym}}$The answer is above, in the comment by Stefan Behrens -- I am just adding this to make some robot happy. The $k$-fold symmetric product $\Sym^k$ of a space $C$ …

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 …

Suppose that $S$ is your Riemannian surface and $X \subset S$ is a flat subsurface (that is, locally isometric to $\mathbb{R}$ with the usual metric). Let's suppose that $X$ has some nontrivial topol …

You have misunderstood the definition of Teichmuller space. You might want to look at "A primer on mapping class groups" by Farb and Margalit (in particular Chapters 10 through 14).

Edit: as Moishe points out, my answer (just below) is for a different, and easier, question. I will leave the answer up, as it does feel “related”. The answer is no. I find it conceptually easier to …

I interpret your question as follows: Suppose that a Riemann surface $X$ is given as a flat surface (say, by gluing together euclidean polygons by local isometries, satisfying a few nice conditions). …

Here is Moishe Kohan’s comment turned into an answer. Suppose that $S$ is a closed, connected, oriented surface of genus at least two. Let $\rho \colon \mathrm{Homeo}^+(S) \to \mathrm{MCG}^+(S)$ be …

I think that there are counterexamples, but I am not an expert.... Here is my attempt. Let $S$ be the unit square in the complex plane. We obtain $R$ by gluing opposite sides by translation. So $R …

The Riemann sphere isn't conformally equivalent to the others because it is not homeomorphic to them. :)

Here is a more "topological" counterexample. Let $C = \mathbb{C}^\times$ be the punctured plane. Let $D = \{ z \in \mathbb{C}^\times \mid |z| < 1 \}$ be the punctured disk. Define $\iota \colon D \ …

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 …