Differential Geometry and Its Application - Exercise 2.1.22

Problem:

Solution:

Disclaimer, I thought about the solution with the hint from wikipedia, in particular, this spinning model. But not the hint text in the problem.

There I started to think, maybe the lines are just joining points on the circle with a phase shift. So I tried:

$(0, 0, -1) \to (\cos \theta, \sin \theta, 1)$, the mid point is $\frac{1}{2}(\cos \theta, \sin\theta, 0)$, so that is indeed on a circle in the $z = 0$ plane.

With that in mind, now I generalize, for the general hyperboloid, when $z = \pm c$, it is a ellipse with major radius $\sqrt{2} a$ and minor radius $\sqrt{2} b$.

Consider the 'phase shift' lines:

$\sqrt{2}((a \cos u, b \sin u, -c) + v(a \cos (u + s), b \sin (u + s), c))$

Our goal is to find the unknown $s$, the phase shift required, to fit the formula. To do that, we just check if all these points is in fact on the hyperboloid.

$\begin{eqnarray*} & & \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} \\ &=& \frac{(\sqrt{2}(a\cos u + va\cos(u + s)))^2}{a^2} + \frac{\sqrt{2}(b\sin u + vb\sin(u + s)))^2}{b^2} - \frac{(\sqrt{2}((-c) + vc)^2}{c^2} \\ &=& 2(\cos u + v\cos(u + s))^2 + 2(\sin u + v\sin(u + s))^2 - 2(v - 1)^2 \\ &=& 2\cos^2 u + 4v \cos u \cos(u + s) + 2v^2\cos^2(u + s) + 2\sin^2 u + 4v \sin u \sin(u + s) + 2v^2\sin^2(u + s) - 2v^2 + 4v - 2 \\ &=& 2\cos^2 u + 2\sin^2 u + 2v^2\cos^2(u + s) + 2v^2\sin^2(u + s) + 4v \cos u \cos(u + s) + 4v \sin u \sin(u + s) - 2v^2 + 4v - 2 \\ &=& 2\ + 2v^2 + 4v \cos (s) - 2v^2 + 4v - 2 \\ &=& 4v \cos (s) + 4v \\ \end{eqnarray*}$

At this point it should be obvious that $s = \pm \pi$, that correspond to the two ruling patch for the surface, and the surface is doubly ruled.

Now looking at the hint in the text, I think I can simplify this by making the ellipse at $z = 0$ the directix instead, to do so, we just move $v$ by 1 as follow:

$\sqrt{2}((a \cos u, b \sin u, -c) + 1(a \cos (u + s), b \sin (u + s), c) + v(a \cos (u + s), b \sin (u + s), c))$

$\sqrt{2}((a (\cos u + \cos (u + s)), b (\sin u +\sin (u + s)),  0) + v(a \cos (u + s), b \sin (u + s), c))$

Now we can use the sum to product formula to simplify this to:

$\sqrt{2}((a (2\cos(u + \frac{s}{2})\cos \frac{s}{2}), b (2\sin(u + \frac{s}{2})\cos \frac{s}{2}),  0) + v(a \cos (u + s), b \sin (u + s), c))$

Remember $s = \pm \pi$, so all these simplifies to simply:

$(a \cos(u + \frac{s}{2}), b \sin(u + \frac{s}{2}),  0) + v(a \cos (u + s), b \sin (u + s), c)$

Just shift the definition of $u$ by $\frac{s}{2}$, we get

$(a \cos(u), b \sin(u),  0) + v(a \cos (u  + \frac{s}{2}), b \sin (u  + \frac{s}{2}), c)$

So finally we have these two ruling patches:

$(a \cos(u), b \sin(u),  0) + v(-a \sin u, b \cos u, c)$

$(a \cos(u), b \sin(u),  0) + v(a \sin u, -b \cos u, c)$

Now we get back to the full circle to the problem text hint!