Isothermal parameterization, Inverse of the Gauss Map

Question

This problem is from Do Carmo's Differential Geometry of Curves and Surfaces. It is question 13 from chapter 3.5, to be specific.

Suppose that S is a minimal surface without any umbilical points (that is to say that $k_1 = -k_2$ everywhere). Let $\bar{x}$ be a parameterization of the unit sphere by stereograhic projection, and consider a neighborhood $V$ of a point $p$ on the surface so that the Gauss map $N: S \rightarrow S^2$ restricted to the neighborhood $V$ is a diffeomorphism.

The question is to show that the parameterization $q= N^{-1}$ o $\bar x$ is isothermal, meaning that $<q_u, q_u > = <q_v, q_v>$ and $<q_u,q_v> = 0$ everywhere on $S$.

Now when I saw this question, my first instinct was to try to verify that $q$ was isothermal directly by computing the parameterization in some kind of coordinates directly. I know that stereographic projection is given by the formulas:

$$ x= \frac {4u}{u^2+v^2+4}$$

$$ y = \frac {4v}{u^2+v^2+4}$$

$$ z = \frac {2(u^2+v^2)}{u^2+v^2+4} $$

I figured with this information, maybe I could find a way to write down what $q$ is in terms of $u$ and $v$ and take partials. The problem is that I don't know how to write down anything about $N^{-1}$. Like when working with $N$, I can say that a point on a surface $S$ goes to it's normal, which is something that I can compute in terms of cross products of partial derivatives of a parameterization of the surface.

Is this method workable? I don't know if I can proceed any further this way. If not, what can I do to approach this problem?

Answer 1

Stereographic projection is conformal and you can also show the Gauss map is conformal for a minimal surface.

This is not very hard; simply assume $\langle dN_p(t_1),dN_p(t_2)\rangle=\lambda(p)\langle t_1,t_2\rangle \forall t_1,t_2 \in T_pS$ and then take the basis of $T_pS$ consisting of the principal directions of the gauss map. A quick bit of algebra will show that the principal curvatures satisfy $k_1=-k_2$.

Hence the parameterisation you mention is a conformal mapping from the plane, hence isothermal.

Answer 2

The formulas of the stereographic projection give $\langle \bar{x}_u,\bar{x}_u\rangle = \langle \bar{x}_v,\bar{x}_v\rangle$ and $\langle \bar{x}_u,\bar{x}_v\rangle = 0$, and the first part of this exercise says $\langle \mathrm{d}N(q_u),\mathrm{d}N(q_u)\rangle = \lambda_p \langle q_u,q_u\rangle$, hence $$\begin{align*} \langle q_u,q_u\rangle &= \lambda_p^{-1}\langle \mathrm{d}N(q_u),\mathrm{d}N(q_u)\rangle \\ &= \lambda_p^{-1} \langle (N\circ q)_u,(N\circ q)_u\rangle\\ &= \lambda_p^{-1} \langle \bar{x}_u,\bar{x}_u\rangle \\ &= \lambda_p^{-1} \langle \bar{x}_v,\bar{x}_v\rangle \\ &= \langle q_v,q_v\rangle \end{align*}$$ Similarly $$\begin{align*} \langle q_u,q_v\rangle &= \lambda_p^{-1}\langle \mathrm{d}N(q_u),\mathrm{d}N(q_v)\rangle \\ &= \lambda_p^{-1} \langle (N\circ q)_u,(N\circ q)_v\rangle\\ &= \lambda_p^{-1} \langle \bar{x}_u,\bar{x}_v\rangle \\ &= 0 \end{align*}$$