This is the question 17 of the section 3.2 in Do Carmo's book about Differential Geometry.
Let $S$ a regular surface, $K$ its Gaussian Curvature, $\kappa_{1},\kappa_{2}$ its maximal (and minimal) curvatures, and $H$ its mean curvature.
I need to show that, if $H\equiv0$ in $S$ and $S$ has no planar points, then the Gauss map $N:S\rightarrow S^{2}$ satisfies
$$\left<dN_{p}(w_{1}),dN_{p}(w_{2})\right>=-K(p)\left<w_{1},w_{2}\right>, $$
where $p\in S$ and $w_{1},w_{2}\in T_{p}(S)$
I know that $H\equiv 0\implies \kappa_{1}=-\kappa_{2}$, and $\kappa_{1},\kappa_{2}\neq 0$ because $S$ has no planar points. I know too that $K=\kappa_{1}\kappa_{2}.$
What should I do to solve this? I taught:
$$\text{II}{p}(dN_{p}(w_2))=-\left<dN_{p}(w_{1}),dN_{p}(w_{2}\right>, $$
but this leads me nowhere.
Best Answer
Let $e_1$ and $e_2 $ be the principal directions in $p\in S$. Since $(e_1 ,e_2)$ is an orthonormal basis of $T_pS$ we have that :
$w_1 = a_1e_1+a_2e_2$ and $w_2= b_1e_1+b_2e_2$
Then $dN_p(w_1) = dN_p (a_1e_1+a_2e_2)= (-a_1)(k_1)(e_1)-(a_2)(k_2)(e_2)$ and $dN_p(w_2) = dN_p (b_1e_1+b_2e_2)= (-b_1)(k_1)(e_1)-(b_2)(k_2)(e_2)$.
So we have $\langle dN_p(w_1),dN_p(w_2)\rangle = a_1b_1k_1^2+ a_2b_2k_2^2$.
Because $H=0 $ in S , $k_1=-k_2\implies k= -k_1^2=-k_2^2$ . Since S has no planar points, k is not $0$. Then:
$\langle dN_p(w_1),dN_p(w_2)\rangle =-k(a_1b_1+a_2b_2)= -k \langle w_1,w_2\rangle$.