My question is related to the uniformization theorem. But not being an expert on this, I am not sure if my question is covered by this result. Before going deeply into the question, I would like to know if one of you already has an answer to my question.


Here comes the question.


Let $\varphi \in C^\infty\_c(\mathbb{R})$ be an *even* function. Consider the graph of $\varphi$ in the $x z$-plane and the surface $S$ obtained by rotating the graph of $\varphi$ around the $z$ axis. A global orthogonal parameterization of $S$ is given by
$$
(u,v)\in\mathbb{R}^2 \mapsto (x(u) \mathrm{sech}(v),x(u) \mathrm{tanh} (v), z(u))
$$
where $(x(u),z(u))$ is a parameterization of the graph of $\varphi$. But this parameterization of $S$ is, in general, not isothermal.


Is it true that $S$ admits a global isothermal parameterization $\Phi:\mathbb{R}^2\to S$?



Global isothermal coordinates for particular surfaces of revolution
This is a consequence of the Uniformization Theorem (UT) and Riemann's theorem on removable singularities:


1. By the $UT$, $S$ is conformal either to the open unit disk $\Delta$ or to the complex plane.
2. Suppose that $S$ is conformal to $\Delta$. Observe that the complement to a large disk in $S$ is contained in the $xy$-plane (by the compact support assumption), hence, is conformal to the punctured disk $\Delta^\*=\Delta \setminus \{0\}$. Suppose that $f: S\to \Delta$ is a conformal mapping. Restricting $f$ to $C$, we obtain a conformal mapping of $\Delta^\*$ onto $\Delta \setminus K$, where $K\subset \Delta$ is a compact. Since this restriction is bounded, by Riemann's theorem it extends holomorphically to $0$. But this extension cannot exist since it would contradict, for instance, the Maximum Principle. qed


See my answer [here](https://math.stackexchange.com/questions/2561435/simply-connected-noncompact-complete-2-dimensional-submanifold-of-mathbbr3/2564854#2564854) for more references, in particular, to the example of Osserman of the graph of a function which is conformal to the unit disk.


