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

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. :)

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.