Clarification on Beltrami Differentials

Question

I have troubles with the theory of existence of quasi-conformal homeomorphisms realizing Beltrami coefficients. Let $X$ be a (compact) Riemann surface and $f \colon X \rightarrow \mathbb{C}$ be smooth. Then on a coordinate chart $(U,z)$ of $X$, $x \in X$, the quotient $\left( \frac{\overline{\partial} f}{\partial f} \right)_x$ is of the form $\mu_{f,U}(x)\frac{d\overline{z}_x}{dz_x}$, where $\mu_{f,U}$ is the Beltrami coefficient of $f$ w.r.t. the chart $(U,z)$. This Beltrami coefficient is indeed local and depends on the chart as follows: if $(V,y)$ is another chart with $x \in V$, then $$\mu_{f,U}(x) = \mu_{f,V}(x) \frac{\left( \frac{\partial}{\partial \overline{z}} \overline{y} \right)(x)}{\left( \frac{\partial}{\partial z} y \right)(x)}.$$ In particular, $\mu_{f,U}(x) \ne \mu_{f,V}(x)$ for two different charts. However, the absolute value $|\mu_f|$ is independent of a chart and defines a global function $X \rightarrow \mathbb{R}$.

Now suppose we are given a Beltrami differential $\mu \in L^{\infty}$. Several sources (for instance, "An Introduction to Teichm{\"u}ller Spaces" by Imayoshi and Taniguchi, page 147) claim that there exists a unique quasi-conformal homeomorphism $f \colon X \rightarrow \mathbb{C}$ with Beltrami coefficient $\mu$ (which is supposed to follow from the usual measurable Riemann mapping theorem (mRmt) for $X=\mathbb{C}$). How do I need to understand this statement, because in that textbook they don't actually specify what they mean by a Beltrami differential? Do they mean (a) that $\mu \in L^{\infty}(X,[0,1))$ and that $|\mu_f|\equiv \mu$ or do they mean (b) that $\mu_{f,U}(x)=\mu(x)$ for $x \in U$?

The latter seems wrong to me because we established $\mu_{f,U}(x) \ne \mu_{f,V}(x)$. If it's the first, then is uniqueness really included? (Since we only consider the absolute value...) In this case, I also have difficulties constructing the map $f$ with coefficient $\mu$ with the mRmt. For me, it fails when trying to glue local solutions, which come from the usual mRmt on charts, together.

Cheers

Answer 1

I am not really familiar with Imayoshi and Taniguchi's book on Teichmüller theory, but here is my understanding of Beltrami differential, which I learned from Hubbard's book Theichmüller Theory and Applications to Geometry, Topology, and Dynamics Volume 1. He defines Beltrami differentials (forms) in Chapter $4$.

In the case of a map $f: X \to Y$ of Riemann surfaces, the equation $$\dfrac{\partial f}{\partial \overline{z}} = \mu \dfrac{\partial f}{\partial z}$$ does not really make sense, because of the chart conditions. An equation that would make sense is $$\dfrac{\partial f}{\partial \overline{z}}d\overline{z} = \mu \dfrac{\partial f}{\partial z}dz$$ which is equivalent to $\overline{\partial}f = \mu \partial f,$ or if you prefer $\mu = (\partial f)^{-1} \circ \overline{\partial} f.$ But then $\mu$ must be an antilinear form in order for these equations to make sense. This was a brief justification for the antilinear condition for the following definition from Hubbard's book.

Definition: An $L^{\infty}$ Beltrami form on $X$ is an element of the unit ball of $L^{\infty}_{\ast}(TX,TX),$ i.e. a measurable antilinear bundle map $\nu : TX \to TX$ with $\| \nu \|_{\infty} := \operatorname{esssup} |\nu| < 1.$

Then you can extend the measurable Riemann mapping theorem to the context of a Riemann surface using this proposition (Proposition $4.8.12$ from Hubbard's book).

Proposition: Let $X$ be a Riemann surface, and let $\mu$ be a Beltrami form on $X$. Let $\{U_i, i \in I\}$ be an open cover of $X$, and let analytic isomorphisms $(\varphi_i:U_i \to V_i)_{i \in I}$ be an atlas for $X$, where $V_i$ are open subsets of $\mathbb{C}$. We can then consider the function $\mu_i$ on $V_i$ such that $$\mu|_{U_i}=\varphi_i^{\ast}\left(\mu_i \dfrac{d\overline{z}}{dz}\right).$$ Then there exist mappings $\psi_i(\mu):V_i \to \mathbb{C}$ that are solutions of $$\dfrac{\partial \psi_i (\mu)}{\partial \overline{z}} = \mu_i \dfrac{\partial \psi_i (\mu)}{\partial z}$$ and that are homeomorphisms onto their images $W_i \subset \mathbb{C}.$ Moreover, the mappings $(\psi_i \circ \varphi_i : U_i \to \mathbb{C})_{i \in I}$ form an atlas defining a Riemann structure $X_{\mu}$ on $X$ independant of the choice of the atlas $(\varphi_i: U_i \to V_i)_{i \in I}$ for $X$ and of the choices of homeomorphisms $\psi_i.$

This proposition is proved in Hubbard and follows from the measurable Riemann mapping theorem (The mapping theorem in Hubbard). For more details, I suggest you read sections $4.6$ to $4.8$ in Hubbard. I hope this clarifies the meaning of a Beltrami differential on a Riemann surface and the existence of a unique solution.

Note that even though this is the definition I think about when I want to formalize the concept, the way I usually think about a Beltrami differential is by viewing it as a field of infinitesimal ellipses on the tangent space of $X$ which I want to straightened.