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Let $M$ be a hyperboloid of one sheet satisfying $x^2+y^2-z^2=1$. Show that $x(u,v)=(\frac{uv+1}{uv-1},\frac{u-v}{uv-1},\frac{u+v}{uv-1})$ gives a parametrization of $M$ where both sets of parameter curves are rulings.

I have been working on this for quite a while but have not been able to write $x(u,v)$ with the above parametrization as a ruled surface in $u$. I tried letting $u=\tan(\varphi)$ and $v=\tan(\psi)$ but it did not help.

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HINTS: First you should check that the points given by the parametrization do in fact lie on the hyperboloid. Next, you want to see that the $u$- and $v$-curves are in fact lines. You need to do some algebraic manipulations with the rational functions (like "long division"). It will help to fix $v=v_0$, say.

Ted Shifrin
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  • That was what confused me the most. Say a u-curve is $x(u,v_0)$. How do I long divide $x(u,v)$ coordinate by coordinate? For instance, I cannot find a real-valued function $f(u)$ and a real number $A$ such that $$\frac{uv+1}{uv-1}=f(u)+A\frac{uv_0+1}{uv_0-1}.$$ –  Feb 10 '16 at 10:42
  • I'm not quite sure where that equation came from. You should be considering $\dfrac{uv_0+1}{uv_0-1} = 1 + \dfrac 2{uv_0-1}$. – Ted Shifrin Feb 10 '16 at 15:43
  • Isn't a ruled surface $x(u,v)=\alpha(u)+v\beta(u)$ where $v\in\mathbb{R}$ with $\beta(u)$ being a ruling by definition? So if we want to show a $u$-curve is a ruling, then we should find the corresponding $\alpha(u)$ and constant $v$. Did I misunderstand the question or the definition? –  Feb 10 '16 at 18:55
  • You're correct, as far as the standard parametrization goes. But there's nothing that says that you have to move along the rulings with constant speed. Any reasonable function will do in the place of $v$ in that formula. – Ted Shifrin Feb 10 '16 at 18:59
  • Ted, do you know a traditional source for this parametrization? I cannot seem to get it as stereographic projection around one of its points. This relates to a continuing interest in an 1897 book by Fricke and Klein, which may or may not include this. Oh, a later English book with selections discussed is Noneuclidean Tessellations and their Groups by Wilhelm Magnus – Will Jagy Feb 10 '16 at 20:18
  • To Ted Shifrin: Are you suggesting writing the u-curve as $$\Big(1,\frac{1}{v_0},\frac{1}{v_0}\Big)+\frac{1}{uv_0-1}\Big(2,v_0+\frac{1}{v_0} ,-v_0+\frac{1}{v_0}\Big),$$ which can be seen as a line when $u$ varies? But then why are we showing the parameter curves are lines? How does it help finding the the $\alpha(u)$ and the function $f(u,v)$ so that $x(u,v)=\alpha(u)+f(u,v) x(u,v_0)$. –  Feb 10 '16 at 21:32
  • Does checking that the parameter curves are lines part of the verification that the parametrization maps onto the hyperboloid? –  Feb 10 '16 at 21:39
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    @WillJagy: I worked it out a number of years ago, never having seen it anywhere. I put it as an exercise in my differential geometry notes (which the OP may well be doing). I have recently run across it elsewhere. It basically comes from the fact that the hyperboloid of one sheet is projectively equivalent to the standard saddle surface. – Ted Shifrin Feb 10 '16 at 21:49
  • @user269715: So this does show that the $u$-curves are (parts of) lines. You're writing $x(u,v) = \alpha(v) + f(u,v)\beta(v)$. – Ted Shifrin Feb 10 '16 at 21:50
  • @Ted Shifrin: Sorry that I am still a bit confused. If in the above formula, I write $v$ instead of $v_0$, then it will be the hyperboloid but the ruling there is $(2, v+\frac{1}{v},-v+\frac{1}{v})$, which is not a $u$-curve of the form $x(u,v_0)$. Is that what the problem intends? Am I done here, except writing a similar formula by switching the $u$'s and $v$'s and changing a few signs? –  Feb 10 '16 at 22:12
  • You have to have different representations, as you're suggesting. This one shows that the $u$-curves are lines. You have to redo the algebra and write $x(u,v) = \tilde\alpha(u) + \tilde f(u,v)\tilde\beta(u)$ for different functions $\tilde\alpha, \tilde\beta, \tilde f$ to see that the $v$-curves are lines. – Ted Shifrin Feb 10 '16 at 22:17
  • @Ted Shifrin: So the problem is asking to show that the hyperboloid is a ruled surface with either parameter curves being its ruling. How does showing that the parameter curves are lines help with that? –  Feb 11 '16 at 03:29
  • Rulings are lines, by definition. Am I missing something? – Ted Shifrin Feb 11 '16 at 06:25
  • Rulings can be any smooth non-vanishing curves by the definition assumed for the problem. I was thinking I had to find the actual formula to express the parametrization of the hyperboloid as $$\alpha(u)+f(u,v)\cdot \text{(one parameter curve)},$$ so that I could prove that a parameter curve is a ruling of the hyperboloid as a ruled surface. –  Feb 11 '16 at 07:42
  • No, you're confused about the definition. The rulings of a ruled surface are always lines. – Ted Shifrin Feb 11 '16 at 08:36
  • Ted, used your parametrization, a bit revised, for solving $x^2 + y^2 = 1 + z^2$ in integers. Very nice. http://math.stackexchange.com/questions/1842447/solution-of-diophantine-equation/1843184#1843184 – Will Jagy Jun 29 '16 at 02:11