Geodesics on surfaces of revolution about z axis with negative curvature

Question

This is a question in differential geometry of surfaces that I could not do We are given S a surface of revolution about the z axis with everywhere negative Gaussian curvature. We are to show that the intersection of S with the two planes z=0 and z=1 cannot both be geodesics

I have no idea on this one and need the help thanks all

Answer 1

Let $R$ be the region of $S$ between the intersection curves $c_0$ and $c_1$ ($c_i$ denotes the intersection curve of the surface of revolution $S$ with the plane $z=i,\ i=0,1$) Thus, if both were geodesics then by the Gauss-Bonnet theorem we would have $$\int_R K=2\pi\chi(R) , \ \ \ (\dagger)$$ where $K$ is the Gaussian curvature and $\chi(R)$ is the Euler-Poincare characteristic of the region $R$ which in our case is zero since $R$ is homeomorphic to an annulus region. Therefore, from $(\dagger)$ we obtain $$\int_R K=0,$$ which clearly is a contradiction since $K$ is everywhere negative!

Answer 2

Though you've got an excellent answer, here's a lower-tech argument for posterity: If the equation of the surface in cylindrical coordinates is $r = f(z)$ for some positive, $C^{2}$ function $f$, it's well-known that:

• A latitude $z = c$ is a geodesic if and only if $f'(c) = 0$.

• The Gaussian curvature and $f''$ have opposite sign. Particularly, if $K < 0$, then $f'' > 0$.

The desired result follows by elementary calculus.