The Gauss map is angle preserving for minimal surface

Question

Let x:U $\rightarrow \mathbb{R}^3$ a be angle-preserving parametrization of a minimal surface with gauss curvature K(p) $\neq$ 0. Then, the unit normal vector field $N_{\textbf{x}}$:U$\rightarrow$ S$^2$ is a angle-preserving parametrization of S$^2$

My approach: My Idea is to show that one can write the first fundamental form as a multiple of the identity matrix and this then implies that the map is angle preserving. One can see $N_{\textbf{x}}$ as a parametrization of the unit Sphere $S^2$, so the first fundamental form $I_{p,N_{\textbf{x}}}$ can be interpreted as the thrid fundamental form $III_{p,\textbf{x}}$ of the initial parametrization.

This leads to $I_{p,N_{\textbf{x}}}$=$III_{p,\textbf{x}}$ Then I am using a Proposition that states:

For $q\in S$ $$III_q-trace(dN_q)II_q+det(dN_q)I_q=0$$ where dN is the differential of the gauss map.

Now I also know that the the mean curvature is defined as $H(q)=\frac{\kappa_1(q)}{2}+\frac{\kappa_2(q)}{2}$ and gauss curvature as $K(q)=\kappa_1(q)\kappa_2(q)$ so I get $I_{q,\textbf{x}}$=$2H(q)II_q-K(q)I_q$

because K(q) can also be seen as the determinant of $dN_q$

Thats how far I got. Now I got a hint, that this calculation so far would imply $\sqrt{ g_{N_x}}(p)=-K(x(p)) \sqrt{g_x(p)}$ but I don't know why

Thanks in advance

Answer 1

The results I got this far are, considering $S^2$, $\kappa_1=-\kappa_2$,so H(q)=0, which leads to ​$I_{q,\textbf{x}}=-K(x(q))I_q$

Now "taking the determinant on both sides"

$det(I_{q,\textbf{x}})=det(-K(q)I_q)$

$g_{N_x}(q)=-K(x(q))g_x(q)$

Taking now the square root gives

$\sqrt{g_{N_x}(q)}$=$\sqrt{-K(x(q))} \sqrt{g_x(q)}$

But this is slightly different from the hint I got $\sqrt{ g_{N_x}}(p)=-K(x(p)) \sqrt{g_x(p)}$

Answer 2

In fact the derivtaive of the Gauss map is an endomorphism of the tangent plane called the Weingarten map. (Note that $dN (p)$ send the tangent space at $p$ to the orthogonal of $N(p)$ which is the tangent space itself.

This endomorphism is symmetric and its eigenvalues are the principal curvatures.

Here these curvatures are opposite, so the Weingarten endomorphism is just $H$ times the map $(x,y)\to (x, -y)$. So it is a reflexion composed by a homothety, and conserves angles.