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.



1 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.

  • $\begingroup$ Thank you very much. I'll look into it when I find time (probably tomorrow) and if the glueing now works, then I'll mark your answer as accepted. $\endgroup$
    – Florian R
    Mar 23, 2018 at 16:40
  • $\begingroup$ Did I understand correctly that a Beltrami form as defined is not the same as a Beltrami coefficient of a given map $f \colon X \rightarrow Y$? Because $\overline{\partial} f$ as well as $\partial f$ are maps from $TX$ to $TY$, so if $\mu$ is a map from $TX$ to $TX$, then the equation $\overline{\partial} f = \mu \partial f$ doesn't make sense. I ask because I was hoping to calrify the bijection between coefficients and solutions. $\endgroup$
    – Florian R
    Mar 24, 2018 at 13:15
  • $\begingroup$ As I mentioned in my answer, the concept of Beltrami coefficient does not really make sense on a Riemann surface. It makes sense on a given chart, but will change when you look at it in a different as you pointed out in your question. Take for example the sphere with a constant Beltrami differential on the lower half plane, when you look in the chart around infinity, the Beltrami coefficients will be different and won't be constant. Therefore, when you go on a surface, you need to adapt your definition as I mentioned above, the Beltrami forms represent that notion. $\endgroup$ Mar 25, 2018 at 2:18

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