Teichmuller Theory question : Beltrami forms on hyperbolic Riemann surfaces whose lifts are smooth upto the boundary of $\mathbb{D}$

Question

Hello, my question is related to Teichmuller Theory. Let $D$ be the open unit disk and $X=D/{\Gamma}$ be a hyperbolic Riemann surface of the Fuchsian group $\Gamma$. In Teichmuller theory, we have the concept of $\Gamma$-compatible Beltrami cofficients $\mu$ on $D$ which satisfy: $\mu(z)=\mu(\gamma(z)).\frac{\bar{\gamma'(z)}}{\gamma'(z)}$ (which corresponds to Beltrami differential forms on the Riemann surface $X=D/{\Gamma}$.As in the standard literature,denote: $M(\Gamma)$:={ $\Gamma$ -compatible Beltrami coefficients on $D$ } = {Beltrami differential forms on Riemann surface $X$}= {measurable complex-antilinear bundle automorphims/self-maps of $TX$ of sup. norm $< 1$}.

Here is my question:

(I) Can we have a non-zero $\mu \in M(\Gamma)$ such that $\mu \in C^0(\bar{D})$, or even $C^k(\bar{D})$ ? Put in other words, is there a non-zero element in $M(\Gamma)\cap C^0(\bar{D})$ ? Note that if such a $\mu$ is constant,then obviously $\mu=0$, so the question is same as asking whether there is a non-constant $\mu$.

I was suspecting that in MANY cases there might not be any such $\mu$, because I was thinking the following might be correct, although couldn't prove it, so this could be my (related) second question :

(II) $\text{True or False ?}$ Let $X=D/{\Gamma}$ be a hyperbolic Riemann surface. Let $z\ne w \in D$, fix them. Then there is $\zeta \in S^1 = \partial{D}$ such that there are sequences $\gamma_n\in \Gamma, \nu_n \in \Gamma$ so that $\gamma_n(z)\to \zeta, \nu_n(w)\to \zeta$ as $n\to \infty.$

If (II) is true, then it partially answers (I), but implies that $|\mu|$= constant. Because if there is a non-absolutely constant $\mu \in M(\Gamma)\cap C^0(\bar{D})$, then there are $z\ne w \in D$ such that $|\mu(z)|\ne|\mu(w)|$. Now, as $\gamma_n(z)\to \zeta, \nu_n(w)\to \zeta$ as $n\to \infty,$ and $|\mu(\gamma_n(z))|=|\mu(z)|$, therefore passing to $\zeta, n \to\infty $, we have boundary values of $\mu$ not matching up at $\zeta$.

Any hints/solutions/references will be highly appreciated !

I would be happy if the answer to (I) is yes though, because then we might look at the "smooth" subset of Teichmuller spaces : {restriction of $\mu$-quasiconformal maps fixing $1,-1,i$ to $S^1,\mu \in C^0(\bar{D})$}, and we might consider its contractibility, complex structure etc. :)

Answer 1

Such continuous $\mu$ does not exist whenever $\Gamma$ is nonelementary. I will work in the upper half-plane model of the hyperbolic plane, let $\bar{B}$ be the compactification of ${\mathbb H}^2$ by its circle at infinity.

Let $\gamma_1, \gamma_2\in \Gamma$ be noncommuting hyperbolic elements. A continous Beltrami differential $\mu$ defines a continuous family of ellipsoids $E_z\subset T_z \bar{B}, z\in \bar{B}$. Unless $\mu=0$, there is a nonempty subset $U\subset \bar{B}$ where the ellipsoids are not circles. I assume that $U$ is the maximal set with this property. Thus, the action of $\Gamma$ preserves the line field $L_z$ on $U$ where the line $L_z\subset T_z U$ passes through the major axis of $E_z$. Conjugate $\Gamma$ so that $\gamma_1$ becomes a dilation $\gamma_1: z\mapsto \lambda z, \lambda>0$. Dilation $\gamma_1$ preserves slopes of the lines $L_z$. Therefore, by continuity of the family of lines $L_z$ at $z=0$, the slopes of all $L_z$'s are equal to the slope of $L_0$. In particular, all lines $L_z$ are parallel. Now, take another conjugation of $\Gamma$ (via an element $\alpha\in PSL(2, {\mathbb R})$), so that $\gamma_2$ is a dilation after this conjugation. The element $\alpha$ will send the parallel line field $L_z, z\in U$, to a family of lines $M_z:=\alpha'(L_z)$, which are no longer parallel (since $\alpha$ does not fix $\infty$). However, the same argument as before, implies that $M_z$ consists of parallel lines. Contradiction.

This type of arguments is used by Dennis Sullivan in his paper "On the ergodic theory at infinity of an arbitrary discrete group of hyperbolic motions", which you should consider reading.