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Q: Consider an one sheet hyperboloid $S$ sitting in $\mathbb{R}^3$ which defined by $x^2+y^2-z^2 =1$. Show that there is a straight line in $S$ through every point of $S$. Also, deduce without any calculation, that the Gaussian curvature of $S$ is less than or equal to $0$ at each point.

What I done so far was, I know that the surface $S$ has regular parametrization: $$r(u,v)=(\cos u-v\sin u, \sin u+v \cos u,v)$$ $$r(u,v)=(\cos u,\sin u,0)+v(-\sin u,\cos u,1)$$ by looking at the second equation, if we keep u fixed, it is the standard straight line equation, and by varying u from 0 to 2$\pi$, we obtain the whole surface $S$. Hence, for every point in $S$, there is a straight line that go through it and lie in $S$.

However I don't know how to deduce the curvature of $S$ is $\le 0$ from the fact that one sheet hyperboloid is a ruled surface.

SamC
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1 Answers1

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Assume that the Gaussian curvature is positive. It means that the second fundamental form is positive (or negative, depends on the choice of orientation). It then follows, that the curvature of every curve on the surface is also positive, as it is at least equal to the second fundamental form of the derivative at each point. This contradicts, obviously, the existence of a straight line at every point.

Amitai Yuval
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  • isn't Gaussian curvature is independent of the choice of the orientation of the normal vector ? – SamC Oct 18 '14 at 14:53
  • It is, but the second fundamental form is not. – Amitai Yuval Oct 18 '14 at 14:55
  • After going through some notes and thinking, I still don't understand where is the contradiction.

    You first assume the Gaussian curvature of $S$ is positive. So there are two cases for the principal curvatures $k_1, k_2$ (belong to two smooth curves $c_1, c_2$), they either be both positive or negative. It doesn't require "the curvature of every curve on the surface is also positive". If so, it won't contradict the existence of straight lines.

     – SamC Oct 18 '14 at 16:35
  • Are you familiar with the second fundamental form? If not, then my answer is currently irrelevant for you. If you are, there is a theorem which guarantees that this form, evaluated on a tangent vector, is at most the curvature of $c$, where $c$ can be any curve that is tangent to this vector. All the rest is written in the answer. – Amitai Yuval Oct 18 '14 at 17:10
  • Ok, I might understand what you trying to say. Let me express in my own word (correct me if I am wrong). Gaussian curvature is defined to be $K = k_{1} k_{2}$. Where $k_{1}, k_{2}$ are the maximum/minimum values of $II(v)$, v is the tangent vector to a curve on $S$. So, if there exist a straight line on $S$ for every points, which mean $k_1 = 0$ and $K=0$ or else $0$ isn't the maximum/minimum values of $II(v)$ hence $k_{1},k_{2}$ will be one positive and one negative, $K \gt 0$. – SamC Oct 18 '14 at 17:33
  • Exactly! (I think in the end you meant "$K<0$". – Amitai Yuval Oct 18 '14 at 17:47
  • oh, yup. Just a typo there. And thanks :D – SamC Oct 18 '14 at 17:47