Gaussian curvature of minimal surfaces expressed by torsion and curvature of its geodesics

Question

Let $S$ be a minimal surface, and let $\gamma$ be a geodesic parametrized by arc length, with curvature $k$ and torsion $\tau$,show that in the points of $\gamma$

$$ -K=k^2+\tau^2, $$

where, $K$ is the Gaussian curvature of $S$.

In this one I don't really know how to start, I thought about using $k^2=k^2_g+k^2_n$, or try to use the parametrization by asymptotic curves, but this took me nowhere.

Answer 1

First recall the relation of the three fundamental forms of a surface $S$ in $\mathbb{R}^3$: $$III-2HII+KI=0,$$ where $K$ and $H$ are the Gaussian and the mean curvature respectively. Since $S$ is minimal we obtain $$III+KI=0.$$ By evaluating along the geodesic $c\colon I\subset\mathbb{R}\longrightarrow S$, (parametrized by arclength) we obtain \begin{equation} III_{c(s)}(c'(s))+K(c(s))I_{c(s)}(c'(s))=0,\ \ \forall s\in I. \ \ \ (\dagger) \end{equation} But,

• $I_{c(s)}(c'(c)):=||c'(s)||^2=1,\ \forall s\in I$, &
• $III_{c(s)}(c'(s)):=I_{(N\circ c)(s)}((N\circ c)'(s))=||(N\circ c)'(s)||^2,\ \forall s\in I$, where $N$ is the Gauss map of $S$.
  Substituting to equation $(\dagger)$ we obtain $$-K(c(s))=||(N\circ c)'(s)||^2,\ \forall s\in I.\ \ \ \ (\ddagger)$$ Now, recall that $c$ is a geodesic. Therefore along $c$ the unit normal $N\circ c$ is parallel to the first normal of the curve $c$. Thus, $$(N\circ c)(s)=\pm \vec{n}(s),\ \forall s\in I.$$ By differentiating we obtain $$(N\circ c)'(s)=\pm(-k(s)\vec{t}(s)+\tau(s)\vec{b}(s)).$$ (Here we used the Frenet equation: $$\vec{n}'(s)=-k(s)\vec{t}(s)+\tau(s)\vec{b}(s),$$ where $\vec{t}(s)=c'(s), \vec{b}(s)$ is the second normal of $c$ and $k(s),\tau(s)$ are the curvature and the torsion of $c$ respectively.) Now, the desired result follows directly from the last equation and $(\ddagger)$, i.e. $$-K(c(s))=k^2(s)+\tau^2(s).$$