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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!

MrT
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  • 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

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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 $K(p)\le 0$.

Ted Shifrin
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  • Thanks @TedShifrin I am trying to figure this out. So if we had a curve α and at some point there was a point where the curvature was 0 then the Gaussian curvature would be 0? I am having some trouble seeing this? Do you have an example of such curve? – MrT Sep 26 '14 at 20:59
  • 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
  • so if there is a curve and at some point the curvature was 0 then the Gaussian curvature would be <= 0. Would a Ruled hyperboloid follow this? – MrT Sep 26 '14 at 21:19
  • Indeed, a hyperboloid of one sheet has $K<0$ everywhere. – Ted Shifrin Sep 26 '14 at 21:29
  • Cool! Thanks @TedShifrin!! – MrT Sep 26 '14 at 21:48
  • so if a one sheet hyperbolic has K < 0 everywhere I could not use this as a counterexample could I? I would have to find a ruled surface with Gaussian curvature > 0 and show that C <= 0 ? Could I do this with a hyperboloid of one sheet? – MrT Sep 26 '14 at 22:38
  • MrT, there is no counterexample to the statement. I indicated a proof in my answer :) Any ruled surface, by what I said above, necessarily has $K\le 0$. – Ted Shifrin Sep 26 '14 at 22:39
  • How does Kn relate to Gaussian curvature. I see that if k=0 then Kn=0, but why does this force Gaussian curvature K <= 0? @TedShifrin – MrT Sep 27 '14 at 00:03
  • Because $k_n$ is between the principal curvatures, and their product is $K$. – Ted Shifrin Sep 27 '14 at 00:06
  • I think I am just confused that at a point where k = 0, I don't see how K would be a negative value. I see how it would be 0. @TedShifrin – MrT Sep 27 '14 at 00:19