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How upper sheet of hyperboloid can be covered by a single coordinate system?

First let consider $f(x,y,z)=z^2-x^2-y^2-a^2$. Now i visualize from here that if we consider any open ball ($x^2+y^2<a^2$) then visually it can be covered whole upper sheet of hyperboloid by magnifying it .But i can't find any map or any type co-ordinate chart ... enter image description here

Edited: Now how can we prove that upper sheet of hyperboloid is a manifold in the sense that every point on it has a neighborhood diffeomorphic to an open subset of $\mathbb{R}^n$???(map is one-one but how prove it onto ????)

RAM_3R
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    For your added question: My explanation of the relation of $z$ to $x$ and $y$ can be reversed, so the map is also clearly "onto". In fact, for $x,y,z$ on the surface, the map $(x, y, z) \mapsto (x, y)$ is an inverse to the parameterization I wrote down, so the parameterization is 1-1 and onto its image. – John Hughes Aug 24 '19 at 22:37

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The problem is that $$ f(x,y,z)=z^2-x^2-y^2-a^2 $$ is not the equation of the hyperboloid. Rather $$ z^2-x^2-y^2-a^2 = 0 $$ is the equation. And for any point $(x, y, z)$ on the upper sheet, we'll have \begin{align} z^2 &= a^2 + x^2 + y^2 \\ z &> 0. \end{align} Taking the square root of the first equation gives \begin{align} z &= \pm \sqrt{a^2 + x^2 + y^2} \\ \end{align} but applying the second yields \begin{align} z &= \sqrt{a^2 + x^2 + y^2}. \end{align}

And now, because the set of $x,y,z$-triples satisfying that equation constitute a function-graph, it's easy. The paramaterization is $$ (x, y) \mapsto (x, y, \sqrt{a^2 + x^2 + y^2}). $$

John Hughes
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