It's a well-known but difficult theorem that any $C^2$ surface in $R^3$ with parameter domain the unit disk can be put into isothermal parameters. The Wikipedia article on isothermal coordinates references several proofs (none of which I can access immediately). Suppose given a family of surfaces $X^t$ such that the map $t \mapsto X^t$ is continuous from some interval of $t$ values to $C^n(D,R^3)$. Then can we find surfaces $Y^t$ such that $Y^t$ gives isothermal parameters for $X^t$ and the map $t \mapsto Y^t$ is continuous into $C^n(D,R^3)$?

It looks like this might follow if we know (a) the isothermal parameters are solutions of some differential equation and (b) solutions of that differential equation depend continuously in the $C^n$ metric on parameters. That proof promises to involve checking a lot of details in two complicated proofs and will look like hand-waving. So perhaps this is stated somewhere in the literature?

  • $\begingroup$ Do you need some boundedness conditions on the "$C^2$ surface in $\mathbb{R}^3$ with parameter domain the unit disk" you permit? Otherwise, you might end up with surfaces like the plane, which cannot be put into global isothermal co-ordinates with domain the disc (although of course it admits local isothermal co-ordinates). $\endgroup$ – macbeth Mar 22 '11 at 3:17

Choose $o^t\in X^t$ so that $t\mapsto o^t$ is smooth and yet choose a famity of unit vectors $u^t\in T_{o^t}X^t$, so that $t\mapsto u^t$ is smooth.

Parametrize $X^t$ isothermaly by unit disc $f^t:D\to X^t$ in such a way that $f^t(0)=o^t$ and $(d_0f^t)(1)$ is proportional to $u^t$. Such a parametrization is unique. [The later follows since conformal diffeomorphism $h:D\to D$ such that $h(0)=0$ and $h'(0)\in\mathbb R_+$ has to be identity.]

It follows that $t\mapsto f^t$ is continuous; otherwise different partial limits would give different isothermal coordinate with chousen origin and real direction.

With a bit more work, one can show that $t\mapsto f^t$ is smooth.

  • $\begingroup$ This seems like an excellent idea, and a big improvement over trying to check the detailed proofs. As soon as I can work out the details I will accept your answer. $\endgroup$ – Michael Beeson Mar 15 '11 at 19:57
  • $\begingroup$ I tried to work out the details. There is a gap, though. Suppose we have a sequence of isothermal parametrizations $Y^{t_n}$ that are bounded in $C^n$ norm away from $X^0$. Now we need to argue by compactness that some subsequence converges. For that we need a uniform bound on the isothermal parametrizations $Y^t$. Where will we get a bound on the $C^n$ norms of the $Y^t$? $\endgroup$ – Michael Beeson Mar 16 '11 at 20:26
  • $\begingroup$ The isothermal coordinate is a solution of some elliptic PDE with coefficients taken from the first fundamental form. Such solutions are known to be uniformly smooth in a compact subdomain. From this you should get $C^\infty$ case. For $C^n$ you have to be more careful, the first fundamental form is in class $C^{n-1}$, but if I remember right solution of elliptic equation with such coefficient should be $C^n$... (It was a while since I study this stuff.) $\endgroup$ – Anton Petrunin Mar 17 '11 at 16:36
  • $\begingroup$ Now I have another difficulty with your answer. Let's check the uniqueness. Suppose $W$ and $V$ are two different isothermal parametrizations of $X$ (taking 0 to the same point and having the same x-derivative at 0). Then if I understand you, you want to define $\varphi:= W^{-1} V$ and say that $\varphi$ is a conformal mapping of the disk, fixing 0 and with $\varphi_x(0) = 1$, and therefore $\varphi$ is the identity. Nice, but, the domain of $W^{-1}$ is a geometric surface, so how do I define it if $W$ is not one-one? It seems this will only work to get LOCAL isothermal coordinates. $\endgroup$ – Michael Beeson Mar 21 '11 at 23:38
  • 1
    $\begingroup$ @Michael Beeson: Since $W$ and $V$ are two different parametrizations of $X$, we can write $V=W\circ\varphi$, for some self-diffeomorphism $\varphi$ of the disc. Since $W$ and $V$ are both conformal, we know that $\varphi$ must be locally conformal, hence conformal. $\endgroup$ – macbeth Mar 22 '11 at 3:02

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