0

In this question, it is asked how to prove that a doubly ruled surface by orthogonal lines is a plane.

I'm interested if this property is robust under perturbations of the angle the lines make; that is, suppose that we have a doubly ruled surface by lines which meet at angle $\theta$, where $\theta>0$ (so in the case of the question above, $\theta= \pi/2$). This $\theta$ is assumed to be fixed everywhere on the surface.

Question: with this assumption, can we conclude that the surface is a plane?

Any comment is appreciated!

ZenoCosini
  • 780
  • 1
    Compute the Gaussian curvature which has simple expression for ruled surfaces (https://math.stackexchange.com/q/2507529) – Jean Marie Oct 25 '21 at 09:12
  • This problem can be approached particularly effectively using differential forms and moving frames. – Ted Shifrin Oct 25 '21 at 18:30
  • @TedShifrin I'm not terribly familiar with these techniques. Could you point out a reference? Thanks! – ZenoCosini Oct 26 '21 at 08:56
  • 1
    There's a brief section on moving frames in my differential geometry text (linked in my profile). You might look at a wonderful new book by Jeanne Cleland, From Frenet to Cartan: The Method of Moving Frames. However, maybe you should just use Gauss-Bonnet in a straightforward manner to show that $K=0$ everywhere. Then you still have to deduce the second fundamental form is $0$. – Ted Shifrin Oct 26 '21 at 21:53

0 Answers0