# Gaussian Curvature K > 0

If M is a surface with Gaussian curvature K > 0, then the curvature of any curve C ⊂ M is everywhere positive.

I was reading this in a textbook and I was trying to decide if this was true or not. I am leaning more towards it being false, but am trying to come up with a counter example. I can not think of a solid proof of this being true. Any hints? Thanks guys!

• do you mean the ambient curvature of $C$? – yess Sep 25 '14 at 15:26
• @yess it did not specifically say the ambient curvature. It just said the curvature k. Here, I believe k is defined as |a''(s)| or another variation with regards to the frenet frame. – MrT Sep 25 '14 at 15:57
• Then you are right. Consider the sphere and any geodesic. Then $K>0$ but the curvature of $C$ is $0$. – yess Sep 25 '14 at 16:06
• @yess: No, a great circle on the unit sphere has curvature $1$. It has geodesic ("intrinsic") curvature $0$ in the surface. – Ted Shifrin Sep 26 '14 at 20:44

Although the OP didn't specify that the surface sits in $\Bbb R^3$, I believe I recognize the question.
The statement is true, interpreting $\kappa$ as the curvature of the curve in the ambient $\Bbb R^3$. The hint is to consider Meusnier's Formula, $k_n = \kappa\cos\theta$, where $k_n$ is the normal curvature in the direction of the curve and $\theta$ is the angle between the surface normal and the principal normal. If you had a point $p$ with $\kappa=0$, this would force the Gaussian curvature $K(p)\le 0$.
• No, the Gaussian curvature certainly need not be $0$, but it does need to be $\le 0$. Try a line on a ruled surface, for example. – Ted Shifrin Sep 26 '14 at 21:02
• Indeed, a hyperboloid of one sheet has $K<0$ everywhere. – Ted Shifrin Sep 26 '14 at 21:29