Giving parametrization of Hyperboloid

Question

I'd like to give a parametrization of an two sheet hyperboloid, such that this parametrization covers both sheets (Without using $±$ symbols).

Does that parametrization exist?

Answer 1

The answer is basically no for continuous functions: the key lies in the phrase "of two sheets". On the other hand, through some mild discontinuous devilry one can construct such a parametrisation: it suffices to look for a parametrisation $(x(t),y(t))$ of the hyperbola $$ x^2-y^2=a^2, $$ and then one can rotate this about the axis of $x$ to obtain the hyperboloid $$ x^2-(y^2+z^2)=a^2 $$ via the "cylindrical" coordinates $$ (x(t), y(t)\cos{\theta},y(t)\sin{\theta}). $$

So, how to parametrise $ x^2-y^2=a^2 $? The first thought is that $ t \mapsto (a\cosh{t},a\sinh{t}) $ maps $\mathbb{R}$ to the positive branch of the hyperbola. What if there were a way to only map $(0,\infty)$ to the positive branch? Then one could join it to a parametrisation of $(-\infty,0)$ that handled the negative branch.

There is an obvious map $(0,\infty) \to \mathbb{R}$: $x \mapsto \log{x}$. Then the parametrisation becomes $$ t \mapsto (a\cosh{\log{t}},a\sinh{\log{t}}) = \left( \frac{a}{2}\left( t + \frac{1}{t} \right),\frac{a}{2} \left( t - \frac{1}{t} \right) \right). $$ Now, it is easy to check that $$ \frac{a^2}{4}\left( \left( t + \frac{1}{t} \right)^2- \left( t - \frac{1}{t} \right)^2 \right) = a^2, $$ as it still should be, and hence every real number apart from $0$ lies on the hyperbola, and the positive reals are mapped to the positive branch. So what happens to the negative reals? $t=-1$ gives the point $(-a,0)$, which is on the negative branch! Continuity of the parametrisation for $t \neq 0$ then shows that $t<0$ cover the negative branch. Hence the closest you can get to a single parametrisation is $$ (t,\theta) \mapsto \left( \frac{a}{2}\left( t + \frac{1}{t} \right),\frac{a}{2} \left( t - \frac{1}{t} \right)\cos{\theta} , \frac{a}{2} \left( t - \frac{1}{t} \right)\sin{\theta} \right), $$ which is discontinuous at $t=0$, but does cover both sheets.