I'm looking for an explicit bijective function that maps the upper half of the hyperboloid $t^2 = x^2+y^2 +1 $ embedded in $R^3$ to the Poincaré disk with radius 1, explicitly by mapping a point passing through the xy-plane via a line element through $(t,x,y)=(-1,0,0)$. It makes sense that such a construction would be unique, since every point would have a unique line element, but this would constitute for a rather vague proof for bijectivity would it not. Any tips?

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    $\begingroup$ This function is a bijection, but more importantly, it is an isometry between Minkowski hyperboloid and the Poincaré disk. $\endgroup$
    – Zeno Rogue
    Mar 12, 2018 at 1:05

1 Answer 1


Geometrically it should be clear that the indicated map is a bijection, but one can write the map down explicitly and verify directly that you do indeed have a bijection to the interior of the unit disk in the $xy$-plane.

Namely, let $P(t_{0}, x_{0}, y_{0})$ be a point on the upper half of the hyperboloid $t^2 - x^2 - y^2 =1$ (so that we have $t_{0} \ge 1$ and $t_{0}^2 -1 =x_{0}^2 +y_{0}^2$ ).

One can write down explicitly the parametric equations of the line through the point $P$ and the point $Q(-1, 0, 0)$.

Using $s$ as the free parameter, the parametric equations of the indicated line are

\begin{align*} t = t(s) &= t_{0} + s(-1 - t_{0})\\ x = x(s) &= x_{0} -sx_{0} = x_{0}\left(1 - s \right)\\ y = y(s) &= y_{0} - sy_{0} = y_{0}\left(1 - s \right).\\ \end{align*}

This line intersects the $xy$-plane when $s = \frac{t_0}{1 + t_{0}}$. It follows that the point $R(t(s), x(s), y(s))$ where the line intersects the $xy$ plane satisfies \begin{align*} \left(x(s)\right)^2 + \left(y(s)\right)^2 &= x_{0}^2\left(1 - \frac{t_0}{1 + t_{0}}\right)^2 + y_{0}^2\left(1 - \frac{t_0}{1 + t_{0}}\right)^2\\ &= \frac{x_{0}^2}{\left(1 +t_{0}\right)^2} + \frac{y_{0}^2}{\left(1 +t_{0}\right)^2}\\ &= \frac{t_{0} - 1}{1 +t_{0}}\\ &<1. \end{align*}

This shows that the line through the points $P$ and $Q$ will intersect that $xy$-plane inside the circle $x^2 +y^2 = 1$ for all points $P(t_{0}, x_{0}, y_{0})$ on the upper half of the hyperboloid.

Denoting the upper half of the hyperboloid by $$\mathcal{U}\mathcal{H} = \left\{P(t, x, y) \in \mathbb{R}^{3} \Big \vert t^2 - 1 = x^2 +y^2 \textrm{ and } t \ge 1\right\}$$ and the open unit disk in the $xy$-plane by $$\mathcal{D} = \left\{Q(t, x, y) \in \mathbb{R}^{3} \Big \vert x^2 + y^2 < 1 \textrm{ and } t = 0\right\},$$ and putting it all together, you have nice function $f : \mathcal{U}\mathcal{H} \to \mathcal{D}$ defined by $$ f(t, x, y) = \left(0, \frac{x}{1 +t}, \frac{y}{1 + t}\right).$$

You can now fill in as much detail as you need to verify that the function $f$ is indeed a bijection.

  • $\begingroup$ I would like to again thank you for this beautiful answer. $\endgroup$
    – user459879
    May 24, 2019 at 22:03

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