Let $M$ be a smooth orientable surface in $\mathbb{R}^{3}$, and $N$ a unit normal vector field along $M$. Assume that the surface has no planar point, i.e., for any $p$ in $M$, the second fundamental form is non-vanishing.

If $\gamma$ is a smooth regular curve in $M$, I believe the following is true:

If $N$ is constant along $\gamma$, then $\gamma$ is a straight line segment.

How would you prove my claim?

  • $\begingroup$ The question is, how would you prove your claim? $\endgroup$ – Edu Aug 4 '17 at 10:06
  • $\begingroup$ What makes you believe that your claim is correct? I presume you've played around with lots and lots of examples? $\endgroup$ – Ted Shifrin Aug 4 '17 at 23:11

This claim is false. Just consider a torus sitting on a plane. The point of intersection of the torus and the plane is an honest circle which is certainly not a straight line. However the normal to the surface is constant along this curve.


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