I encounter the following claim in my general relativity research:
Given a regular surface $x(u,v)$ such that in a neighborhood $V$ os some point $p$ in $S$, the coordinate curves $x(u,v=v_{0})$ and $x(u=u_{0},v)$ are unit-speed geodesics. Does the Gaussian curvature always vanish on $V$? (In one of the papers I read, they took orthogonality between these two families of geodesics as another assumption, but I think this is actually unnecessary?)
Yes.
Consider segments of $u$-coordinate curves between two $v$-coordinate curves $\{u=u_1\}$ and $\{u=u_2\}$. These segments are a family of geodesics of the same length $|u_1-u_2|$. By the first variation formula, this implies that each of these segments meets the bounding lines $\{u=u_1\}$ and $\{u=u_2\}$ at equal angles. Since $u_1$ and $u_2$ are in fact arbitrary, one concludes that each $u$-coordinate curve intersects all $v$-coordinate curves at the same angle.
Then exchanging the roles of $u$ and $v$ shows that the angle between coordinate lines is constant everywhere on $V$. Thus the metric coefficients $(g_{ij})$ are constant. (In Gauss' notation, one has $E=G=1$ and $F$ is the cosine of that constant angle). It is well known and easy to prove that constant metric coefficients imply zero Gaussian curvature.
