necessary and sufficient condition for the lines of curvature

Question

I'm reading the book "Differential Geometry of Curves and Surfaces" of Manfredo Carmo, and this part confuses me:

We have the differential equation of the lines of the curvature: $$(fE-eF)(u')^2 +(gE-eG)u'v' + (gF - fG)(v')^2 = 0$$ Using the fact that the principal directions are orthogonal to each other, we conclude that a necessary and sufficient condition for the coordinate curves of a parametrization to be lines of curvature in a neighborhood of a nonumbilical point is that $F = f = 0$

I know that if the local parametrization of the surface is $x(u,v)$, then $F = <x_u, x_v>$, so to say that $F = 0$ meaning saying that $x_u$ is orthogonal to $x_v$, which I don't understand why we can state that. Can anyone help me clarify this? Thanks a lot for your time

Answer 1

Let $X\colon U\subset \mathbb{R}^2\longrightarrow S\subset \mathbb{R}^3$ be a parametrization of a regular surface $S$ that contains no umbilical points of $S$. Then we will show that:

the parametrized curves of $X$ are lines of curvature $\Longleftrightarrow$ $F=f=0$.

Proof:

• $(\Rightarrow)$ If the parametrized curves of $X$ are lines of curvature then we have that $X_u$ and $X_v$ are parallel to the principal directions at every point $X(u,v)$. Therefore $F=\langle X_u,X_v\rangle=0.$ Also $$f=-\langle dN(X_u),X_v\rangle=-\lambda\langle X_u,X_v\rangle=0,$$ where $\lambda$ is either $k_1$ and $k_2$. [By $k_1$ and $k_2$ we denote the principal curvatures]
• $(\Leftarrow)$ If $F=f=0$ then from the differential equation for the lines of curvature $$(fE-eF)(u')^2 +(gE-eG)u'v' + (gF - fG)(v')^2 = 0$$ we obtain $$(gE-eG)u'v'=0.$$ Since $gE-eG\neq 0$ ( this is due to the fact that $X(U)$ contains no umbilical points) it follows that $u'=0$ or $v'=0$. Therefore the parametrized curves are lines of curvature.