[Math] Parameter Curves are Geodesics

Question

So let's suppose we have a surface $M$ that is embedded in $\mathbb{R}^3$ with an orthogonal parametrization. Further, assume that the parameter curves (i.e., $X(u$₀$, v)$ and $X(u, v$₀$)$ ) are geodesics that are unparametrized (i.e., not necessarily unit speed).

What can we say about the Gauss curvature of $M$?

Answer 1

The Gaussian curvature must be $0$.

Let us write $E = \langle X_u, X_u\rangle$ and $G = \langle X_v, X_v\rangle$ for the coefficients of the first fundamental form. We'll make use of two formulas:

Fact 1: For an orthogonal parametrization, the geodesic curvature of the $u$- and $v$-parameter curves are given by $$(\kappa_g)_{u = u_0} = \frac{-E_v}{2E\sqrt{G}}$$ $$(\kappa_g)_{v = v_0} = \frac{G_u}{2E\sqrt{G}}$$

Since the parameter curves are geodesics, it follows that $E_v = G_u = 0$.

Fact 2: For an orthogonal parametrization, the Gaussian curvature is given by $$K = \frac{-1}{2\sqrt{EG}}\left[ \frac{\partial}{\partial u}\left( \frac{G_u}{\sqrt{EG}} \right) + \frac{\partial}{\partial v}\left( \frac{E_v}{\sqrt{EG}} \right) \right].$$

From this, it follows that $K = 0$.