Models of Hyperbolic Plane

Resurrecting this blog. Recently I have been giving lectures on hyperbolic geometry to my students including student trainees. The lectures allow me to brush up on basics of hyperbolic geometry for upcoming conference and also to help the trainees do their mini projects. I begin with the models of hyperbolic plane.

Hyperboloid Model

Everyone knows the unit (two-) sphere whose equation is x_1^2 + x_2^2 + x_3^2 = 1:


If one changes the sign in one variable of the sphere-equation, x_1^2 + x_2^2 - x_3^2 = 1, one obtains the one-sheeted hyperboloid:


If one changes the sign of the constant on the right hand side, x_1^2 + x_2^2 - x_3^2 = -1, one gets the two-sheeted hyperboloid:


These disconnected pieces remind us the hyperbola curves we plotted in schools x^2 - y^2 =\pm 1:


One should also mention the other possibility from the two-forms of the hyperboloid earlier, the one with zero on the right hand side, giving a double cone reminiscent of the light cones in relativity.


These null surfaces are the asymptotes of the hyperboloids (very much like the asymptotes of the hyperbola).

Returning to the double-sheeted hyperboloid; let us restrict to the positive sheet namely

L=\{(x_1,x_2,x_3)\in\mathbb{R}^3\vert x_1^2+x_2^2-x_3^2=-1,\ x_3>0\} ,

it gives the hyperboloid model of the hyperbolic plane. The surface can be parametrised in ‘angular’ coordinates much in the same way as the sphere via

x_1=\sinh\theta\cos\phi;\ x_2=\sinh\theta\sin\phi;\ x_3=\cosh\theta,

which leads to x_1^2+x_2^2-x_3^2=\sinh^2\theta-\cosh^2\theta= -1. Note the use of hyperbolic functions. Note also the form of bilinear form one can associate to \mathbb{R}^3 to form the equation of the hypwerboloid i.e.

(x,y)_{2,1} = x_1y_1 + x_2y_2 - x_3y_3\

Using this bilinear form, one could actually form the metric on L by forming the product (dx,dx)_{2,1} to give

ds^2 = dx_1^2+dx_2^2-dx_3^2\ .

Beltrami-Klein Model

In the hyperboloid model L above, it is tangent to the plane K of x_3=1. Next we project points on L to this plane by drawing a line to the origin (see Fig below).


The point l=(x_1,x_2,x_3)\in L is projected to the point k=(\eta_1,\eta_2)\in K. By using similar triangles, \eta_i=\eta_i/1=x_i/x_3 for i=1,2. Note that


Thus under this project the hyperboloid L is mapped to a unit disk K, which is the Beltrami-Klein disk model of the hyperbolic plane. By using the inverse coordinate transformation

x_i=\cfrac{\eta_i}{\sqrt{1-\eta_1^2-\eta_2^2}}\ ,\ (i=1,2);\quad x_3=\cfrac{1}{\sqrt{1-\eta_1^2-\eta_2^2}},

one can get the hyperboloid differentials in terms of these new coordinates as

dx_i=\cfrac{(1-\eta_1^2-\eta_2^2)d\eta_i+\eta_i(\eta_1 d\eta_1+\eta_2 d\eta_2)}{(1-\eta_1^2-\eta_2^2)^{3/2}}

for i=1,2\ and

dx_3=\cfrac{(\eta_1 d\eta_1+\eta_2 d\eta_2)}{(1-\eta_1^2-\eta_2^2)}.

Hence the metric for the B-K disk to be

ds^2=\cfrac{d\eta_1^2+d\eta_2^2}{1-\eta_1^2-\eta_2^2} + \cfrac{(\eta_1d\eta_1+\eta_2d\eta_2)^2}{(1-\eta_1^2-\eta_2^2)^2}.

Hemispherical Model

Another interesting model is to project the points of hyperboloid further to a hemisphere J (would aid us later to get other more convenient models).


The point l is now projected down to j=(\eta_1,\eta_2,\eta_3) from k of B-K model where the additional coordinate is given by \eta_3=1/x_3,\ x_3>0. It is easy to show that these coordinates now form the equation of a (hemi-)sphere


Hence, the hemisphere model of the hyperbolic plane. Note that this (hemi-)sphere has a different metric inherited from the hyperboloid i.e.


Poincare Disk Model

Using the hemisphere model, we can now make a stereographic projection from J to the plane \eta_3=0. The point j=(\eta_1,\eta_2,\eta_3) gets mapped to i=(\xi_1,\xi_2) on the Poincare disk I.


The relation between these coordinates are given by the similar triangle relations


One can work backwards (easier) to show that the Poincare disk metric

ds^2= \cfrac{4(d\xi_1^2+d\xi_2^2)}{(1-\xi_1^2-\xi_2^2)}

is equivalent to the metric of the hemisphere model earlier.

Upper Half-Plane Model

Now in the Poincare disk, we use the stereographic projection from (0,0,-1) to \eta_3=0 plane. We could use a different projection namely from (0,-1,0) on the hemisphere J to the plane \eta_2=0. The point j= (\eta_1,\eta_2,\eta_3) gets mapped to h=(x,y) on the plane mentioned.


The relation again between these coordinates are given by the similar triangles:


Unlike the earlier stereographic projection that squashes the hemisphere into a bounded (Poincare) disk, the present projection gives an unbounded plane due to the point (0,-1,0) gets mapped to infinity. Note also that \eta_3>0 implies that y>0. The metric can also be worked (as in the previous case) to

ds^2= \cfrac{dx^2+dy^2}{y^2}.

This upper half plane can be realised as


This is probably one of the most convenient model to work with. It can be realised as a homogeneous space. Consider the action of \textrm{SL}(2,\mathbb{R}) on H given by the fractional linear transformation

\begin{pmatrix} a&b\\c&d \end{pmatrix} \cdot z \mapsto \cfrac{az+b}{cz+d}\equiv z'

where a,b,c,d\in\mathbb{R};\ ad-bc=1. Note the action of \begin{pmatrix} -a&-b\\-c&-d \end{pmatrix} gives the same effect. Thus the effective group is really \textrm{PSL}(2,\mathbb{R}) = \textrm{SL}(2,\mathbb{R})/\pm I. The group action could also be conveniently put into matrix multiplication form in the following way:

\begin{pmatrix} a&b\\c&d \end{pmatrix}\cdot \begin{pmatrix} z\\1 \end{pmatrix} = \begin{pmatrix} az+b\\cz+d \end{pmatrix} \equiv \begin{pmatrix} \frac{az+b}{cz+d}\\1 \end{pmatrix}.

It would seem like that each point on H can be made in correspondence with each element of \textrm{PSL}(2,\mathbb{R}) but one ought to check if any of the group elements keep the point fixed. Consider for example the subgroup \textrm{SO}(2)\subset\textrm{SL}(2,\mathbb{R}). Elements \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}\in\textrm{SO}(2) keeps z=i fixed:

\begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}\ \begin{pmatrix} i\\1 \end{pmatrix}= \begin{pmatrix} i(i\sin\theta+\cos\theta)\\ i\sin\theta+\cos\theta \end{pmatrix}\equiv \begin{pmatrix} i\\1 \end{pmatrix}.

Thus in general for each point there will be effectively an (conjugated) \textrm{SO}(2) that keeps the point fixed. Thus the upper half plane can be realised as the quotient space \textrm{PSL}(2,\mathbb{R})/\textrm{PSO}(2).

In the next few forthcoming posts, we will investigate this model of the hyperbolic plane further


  1. Pages from
  2. S. Katok, “Fuchsian Groups, Geodesic Flows on Surfaces of Constant Negative Curvature and Symbolic Coding of Geodesics”, (
  3. J.W. Cannon, W.J. Floyd, R. Kenyon & W.R. Parry, “Hyperbolic Geometry” (
  4. R. Hayter, “The Hyperbolic Plane – A Strange New Universe” (
  5. J. Hilgert, “Maass Cusp Forms on \textrm{SL}(2,\mathbb{R})“, (

3 responses to this post.

  1. […] « Models of Hyperbolic Plane […]


  2. Posted by falaha on August 14, 2013 at 4:13 pm

    where do you get the elements of SO(2)? the matrix contains cos and sin…..


    • Posted by hishamuddinz on August 14, 2013 at 5:33 pm

      As stated in the post, SO(2) is a subgroup of SL(2,R). The group SL(2,R) is the group of invertible 2×2 matrices with determinant 1, while SO(2) adds a further condition of having the property M^tM=1 (orthogonal matrices with determinant one). The latter is none other than our rotation matrices that keeps the length of two-dimensional vectors fixed (and hence the cos and sin as their matrix elements).


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