Global isothermal coordinates for particular surfaces of revolution 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 $\Phi:\mathbb{R}^2\to S$?
 A: In fact, more general statement is true: if $\phi$ is any smooth function (not necessary with compact support), the surface is conformally equivalent to the plane.
This follows from the result of Milnor,
J. Milnor, On deciding whether a surface is parabolic or hyperbolic, Amer. Math. Monthly, 84, 1, 43-46,
If the metric is $ds^2=r^2+g(r)^2d\theta^2$, then the surface is parabolic (=conformally equivalent to the plane) if and only if 
$$
\int^\infty \frac{dr}{g(r)}=\infty.$$
In our case we can take $g(r)=r/\sqrt{1+(\partial\phi/\partial r)^2}$
and the integral in Milnor's criterion is always divergent.
A: This is a consequence of the Uniformization Theorem (UT) and Riemann's theorem on removable singularities:

*

*By the $UT$, $S$ is conformal either to the open unit disk $\Delta$ or to the complex plane.


*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 for more references, in particular, to the example of Osserman of the graph of a function which is conformal to the unit disk.
