Asymptotic parametrization for negatively curved surfaces

Question

Let $S$ be a complete simply connected negatively curved surface immersed in Euclidean space $\textbf{R}^3$. Does there exist a parametrization $f\colon\textbf{R}^2\to\textbf{R}^3$ for $S$ such that the curves $t\mapsto f(t,s)$ and $s\mapsto f(t,s)$ are asymptotic, i.e., their second derivatives are tangent to $S$?

Asymptotic parametrizations exist locally, because a pair of transverse line fields on a surface can always be integrated locally, e.g., see p. 167 of Spivak Vol I. It is also well known that when $S$ has constant curvature, an asymptotic parametrization may be constructed globally. This is the basis for the proof of Hilbert's theorem on the nonexistence of complete surfaces of constant negative curvature in $\textbf{R}^3$.

If it is not possible in general to construct a global asymptotic parametrization, then when does one exist?

Answer 1

As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^{z} \cos x = \cos y$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.

There are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$. Explicitly, set $$ u(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{-}y)}{\cos x {+} \cos y}\right)\ \text{and}\ \ v(x,y,z) = \frac2\pi\,\arcsin\left(\frac{\sin(x{+}y)}{\cos x {+} \cos y}\right) $$ when $|x|,|y| <\pi/2$ and $(x,y,z)$ lies in $S$. These functions then extend analytically to the entire surface $S$ and have the claimed properties.

It may help to observe that the surface $S$ is preserved by transformations $X$,$Y$,$D$ and $T$, where $X(x,y,z) = (-x,y,z)$, $Y(x,y,z)= (x,-y,z)$, $D(x,y,z)=(y,x,-z)$, and $T(x,y,z)=(x{+}\pi,y{+}\pi,z)$ and that $(u,v)\circ X = (-v,-u)$, $(u,v)\circ Y = (v,u)$, $(u,v)\circ D = (-u,v)$ and $(u,v)\circ T = (u,v{+}2)$ while $$ (u,v)(\pi/2,\pi/2,z) = \left(\frac2\pi\,\arcsin\bigl(\tanh(z/2)\bigr),1\right). $$ Application of these rules suffices to extend $u$ and $v$ to the entire surface $S$.

Now, $S$ is not simply-connected. The image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the points of the form $(2m{+}1,2n{+}1)$ where $m$ and $n$ are integers. Note, for example, that the lines $u=c$ and $v=c$ are asymptotic curves, but are disconnected if $c$ is an odd integer. (The case of $c$ an integer corresponds to vertical straight lines in $S$, i.e., where $\cos x = \cos y = 0$.)

It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points in $\mathbb{R}^2$. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.

In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.