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One may define a canonical round metric conformally equivalent to $m$ in the following way: the pullback $f^*(dvol_m)$ of the area measure of $m$ gives a finite measure on $S^2$. Regarding $S^2$ as th …

By the uniformization theorem, there is a unique conformally equivalent constant curvature metric up to scaling which is equivalent to a Riemann surface. A conformal Killing vector field would give a …

For a), when $\chi(F)<0$ $F$-bundles over $B$ are classified by maps $\pi_1(B)\to Mod(F)$, where $Mod(F)$ is the mapping class group of $F$. This boils down to the fact that $Diff_0(F)$ is contractibl …

I wondered this myself, I made some similar pictures to approximate the stable lamination of a pseudo-Anosov map (the blue curve is a geodesic, the other colors horocycles). Every geodesic on the mod …

One can answer your question positively, although I haven't tried to compute $g_R$ as a function of $R$. One can tessellate a certain right-angled pentagon by $238$ triangles. Moreover, a genus 2 s …

If the points are (setwise) invariant under reflection through a plane perpendicular to the sphere, then the the great circle of the reflection plane intersecting the sphere will be geodesic in the hy …

The braid groups are good (which are mapping class groups of punctured disks) by Proposition 3.5 of Grunewald, F.; Jaikin-Zapirain, A.; Zalesskii, P. A., Cohomological goodness and the profinite com …

I think that $A=2d$ will work, basically by applying Morse theory to the distance function from $x$. Morse theory for distance functions was originally considered by Gromov (and then Cheeger). For the …

I think Scott's argument is that the lengths of $6g-6$ curves can't form coordinates for Teichmuller space. If one has $6g-6$ geodesics which parameterize, then they must be filling (they meet every s …

There's a slight issue I believe with the other answers. If we consider moduli space as an orbifold (of complex dimension $3g-3$), and the hyperelliptic locus an immersed suborbifold (of complex dimen …

A result of Behrstock and Minsky (cf. Hamenstadt too) implies that the rank of mapping class groups is the maximal rank of abelian subgroups, which is $3g+p-3$ for a connected hyperbolic surface of ge …

There's a remarkable theorem of Margulis that pertains to your question. Let $G$ be a semisimple Lie group (in your case, $PSL_2(\mathbb{R})$), and let $\Gamma$ be an irreducible lattice in $G$. The c …

(Deleted incorrect suggestion). I think one can use uniformization and the construction of automorphic functions on the universal cover to produce meromorphic functions. A google search for these ter …

I think the map $F$ is injective, but not surjective. There is a unique conformal map of the interior of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$, up to the act …

For a hyperbolic element $A\in SL(2,\mathbb{Z})$, we have the length of the closed geodesic is given by $\ln[(tr^2(A)-2+\sqrt{tr^4(A)-4tr^2(A)})/4]$, and this is monotonic in $|tr(A)|$ for $|tr(A)|>2$ …