5
$\begingroup$

I know that if $\alpha$ is geodesic and its curvature is never zero, and it's plane, then it's a line of curvature (i.e. the tangent is a principal direction). I can prove this using Frenet.

I want to show first that all points are umbilical, because then I know how to prove that the curvature is constant, so the surface must be in a plane, a pshere, or the pseudosphere (but it can't be the pseudosphere because of reasons).

Given a point in the surface, and a direction, there exists one and only one geodesic in that direction. If the curvature is never zero, the direction is principal. If I can do this with all points and all directions, all points are umbilical.

But...it can happen that the curvature of the geodesic is zero and I don't know what to do in that case. Can you help me?

$\endgroup$
3
  • $\begingroup$ oh, never mind. If the curvature is zero, then as $\alpha\ \cdot N = 0$, deriving that I get that the direction is principal and that the eigenvalue is zero! $\endgroup$ Nov 28, 2016 at 19:08
  • $\begingroup$ can u please explain the steps u take? $\endgroup$ Jan 31, 2018 at 1:00
  • $\begingroup$ If you are still interested, I will think of it in a few days. Sorry that I can't now $\endgroup$ Feb 2, 2018 at 12:32

1 Answer 1

1
$\begingroup$

HINT: Suppose the geodesic has $\kappa(P) = 0$ for some $P$. Then $P$ must be a planar point, a parabolic point, or a hyperbolic point of the surface. In the latter two cases, you get $\kappa_n\ne 0$ for all but one or two directions, and you're done. Now what if $P$ is a planar point?

Now it's a set-theoretic argument. You need to assume the surface is connected, of course.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .