# 5.1: The Poincaré Disk Model

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The **Poincaré disk model for hyperbolic geometry **is the pair \((\mathbb{D},{\cal H})\) where \(\mathbb{D}\) consists of all points \(z\) in \(\mathbb{C}\) such that \(|z| \lt 1\text{,}\) and \({\cal H}\) consists of all Möbius transformations \(T\) for which \(T(\mathbb{D}) = \mathbb{D}\text{.}\) The set \(\mathbb{D}\) is called the **hyperbolic plane**, and \({\cal H}\) is called the **transformation group in hyperbolic geometry.**

We note that \(\cal H\) does indeed form a group of transformations, a fact that is worked out in the exercises. Throughout this chapter the unit circle will be called **the circle at infinity**, denoted by \(\mathbb{S}^1_\infty\text{.}\) Of course, the circle at infinity is not included in the hyperbolic plane \(\mathbb{D}\) but bounds it. The circle at infinity will be used extensively in our investigations. We note here that any Möbius transformation that sends \(\mathbb{D}\) to itself also sends \(\mathbb{S}^1_\infty\) to itself.

Consider a cline \(C\) that is orthogonal to the circle at infinity \(\mathbb{S}^1_\infty\text{,}\) as in Figure \(5.1.1\). If we invert about \(C\text{,}\) \(\mathbb{S}^1_\infty\) is inverted to itself. Moreover, this inversion takes the interior of \(\mathbb{S}^1_\infty\text{,}\) namely the hyperbolic plane \(\mathbb{D}\text{,}\) to itself as well, (Exercise \(3.2.10\)). It follows that compositions of two such inversions is a Möbius transformation that sends \(\mathbb{D}\) to itself, and is thus in the group \({\cal H}\text{.}\) These inversions play an important role in hyperbolic geometry, and we give them a name.

An inversion in a cline \(C\) that is orthogonal to \(\mathbb{S}^1_\infty\) is called a** reflection of the hyperbolic plane**, or, a **hyperbolic reflection**.

It turns out that these reflections generate all the maps in \({\cal H}\text{.}\) For instance, rotation about the origin is a Möbius transformation that sends \(\mathbb{D}\) to itself, so it is in \(\cal H\text{.}\) But rotation about the origin is also the composition of two reflections about lines that intersect at the origin. Since any line through the origin meets the unit circle at right angles, reflection about such a line is a reflection of the hyperbolic plane, so rotation about the origin is the composition of two such reflections. We now prove the following general result.

Any Möbius transformation in \({\cal H}\) is the composition of two reflections of the hyperbolic plane.

**Proof**-
Suppose \(T\) is a Möbius transformation that sends \(\mathbb{D}\) to itself. This means some point in \(\mathbb{D}\text{,}\) say \(z_0\text{,}\) gets sent to the origin, \(0\). Let \(z_0^*\) be the point symmetric to \(z_0\) with respect to \(\mathbb{S}^1_\infty\text{.}\) Since \(T\) sends the unit circle to itself, and Möbius transformations preserve symmetry points, it follows that \(T\) sends \(z_0^*\) to \(\infty\text{.}\) Furthermore, some point \(z_1\) on \(\mathbb{S}^1_\infty\) gets sent to the point \(1\).

If \(z_0=0\text{,}\) then \(z_0^* = \infty\text{,}\) and \(T\) fixes \(0\) and \(\infty\text{.}\) Then by Example \(3.5.3\), \(T(z) = re^{i\theta}z\) is a dilation followed by a rotation. However, since \(T\) also sends \(\mathbb{D}\) onto \(\mathbb{D}\text{,}\) the dilation factor must be \(r = 1\text{.}\) So \(T\) is simply a rotation about the origin, which is the composition of two hyperbolic reflections about Euclidean lines through the origin.

Now assume \(z_0 \neq 0\text{.}\) In this case, by using \(z_0, z_0^*\text{,}\) and \(z_1\) as anchors, we may achieve \(T\) via two hyperbolic reflections, as follows:

First, invert about a circle \(C\) orthogonal to \(\mathbb{S}^1_\infty\) that sends \(z_0\) to the origin.

Such a circle does indeed exist, and we will construct it now. As in Figure \(5.1.2\), draw circle \(C_1\) with diameter \(0z_0^*\text{.}\) Let \(p\) be a point of intersection of \(C_1\) and the unit circle \(\mathbb{S}^1_\infty\text{.}\) Construct the circle \(C\) through \(p\) centered at \(z_0^*\text{.}\) Since \(\angle 0pz_0^*\) is right, \(\mathbb{S}^1_\infty\) is orthogonal to \(C\text{,}\) so inversion about \(C\) sends \(\mathbb{S}^1_\infty\) to itself. Furthermore, since \(z_0^*\) gets sent to \(\infty\) and symmetry points must be preserved, inversion in \(C\) sends \(z_0\) to \(0\).

Thus, the first inversion takes \(z_0\) to \(0\) and \(z_0^*\) to \(\infty\text{.}\) To build \(T\) we must also send \(z_1\) to \(1\). Note that inversion in circle \(C\) will have sent \(z_1\) to some point \(z_1^\prime\) on the unit circle (since \(\mathbb{S}^1_\infty\) is sent to itself). Now reflect across the line through the origin that bisects angle \(\angle 10z_1^\prime\text{.}\) This sends \(z_1^\prime\) to \(1\), sends \(\mathbb{D}\) to \(\mathbb{D}\text{,}\) and leaves \(0\) and \(\infty\) fixed. Composing these two inversions yields a Möbius transformation that sends \(z_0\) to \(0\), \(z_0^*\) to \(\infty\) and \(z_1\) to \(1\). Since a Möbius transformation is uniquely determined by the image of three points, this Möbius transformation is \(T\text{.}\)

Notice that in proving Theorem \(5.1.1\) we proved the following useful fact.

Given \(z_0\) in \(\mathbb{D}\) and \(z_1\) on \(\mathbb{S}^1_\infty\) there exists a transformation in \({\cal H}\) that sends \(z_0\) to \(0\) and \(z_1\) to \(1\text{.}\)

Therefore, one may view any transformation in \({\cal H}\) as the composition of two inversions about clines orthogonal to \(\mathbb{S}^1_\infty\text{.}\) Moreover, these maps may be categorized according to whether the two clines of inversion intersect zero times, once, or twice. In Figure \(5.1.3\) we illustrate these three cases. In each case, we build a transformation \(T\) in \({\cal H}\) by inverting about the solid clines in the figure (first about \(L_1\text{,}\) then about \(L_2\)). The figure also tracks the journey of a point \(z\) under these inversions, first to \(z^\prime\) by inverting about \(L_1\text{,}\) then onto \(T(z)\) by inverting \(z^\prime\) about \(L_2\text{.}\) The dashed clines in the figure represent some of the clines of motion, the clines along which points are moved by the transformation. Notice these clines of motion are orthogonal to *both* clines of inversion. Let's work through the three cases in some detail.

If the two clines of inversion, \(L_1\) and \(L_2\text{,}\) intersect inside \(\mathbb{D}\text{,}\) say at the point \(p\text{,}\) then they also intersect outside \(\mathbb{D}\) at the point \(p^*\) symmetric to \(p\) with respect to the unit circle since both clines are orthogonal to \(\mathbb{S}^1_\infty\text{.}\) This scene is shown in Figure \(5.1.3(a)\). The resulting Möbius transformation will fix \(p\) (and \(p^*\)), causing a rotation of points in \(\mathbb{D}\) around \(p\) along type II clines of \(p\) and \(p^*\text{.}\) Not surprisingly, we will call this type of map in \({\cal H}\) a **rotation of the hyperbolic plane **about the point \(p\text{;}\) or, if the context is clear, we call such a map a rotation about \(p\text{.}\)

If the clines of inversion intersect just once, then it must be at a point \(p\) on the unit circle. Otherwise, the symmetric point \(p^*\text{,}\) which would be distinct from \(p\text{,}\) would also belong to both clines, giving us two points of intersection. The resulting map moves points along circles in \(\mathbb{D}\) that are tangent to the unit circle at this point \(p\text{.}\) Circles in \(\mathbb{D}\) that are tangent to the unit circle are called **horocycles**, and this type of map is called a **parallel displacement**. See Figure \(5.1.3(b)\).

If the clines of inversion do not intersect, then at least one of the clines must be a circle; and according to Theorem \(3.2.7\), there are two points \(p\) and \(q\) symmetric with respect to both clines. These two points will be fixed by the resulting Möbius transformation, since each inversion sends \(p\) to \(q\) and \(q\) to \(p\text{.}\) Furthermore, these two fixed points must live on the unit circle by a symmetric points argument (the details of which are left as an exercise). In other words, if the clines of inversion \(L_1\) and \(L_2\) do not intersect, then they are type II clines of the fixed points \(p\) and \(q\) of the resulting transformation in \({\cal H}\text{.}\) Moreover, this transformation will push points along type I clines of \(p\) and \(q\text{.}\) We call such a Möbius transformation in \({\cal H}\) a **translation of the hyperbolic plane **or, simply, a **translation**. See Figure \(5.1.3(c)\).

Figure \(5.1.4(a)\) depicts the image of the figure \(M\) (resembling the letter ‘M’) under two applications of a rotation \(T\) of the hyperbolic plane about point \(p\text{.}\) This rotation is generated by two inversions about clines \(L_1\) and \(L_2\) that intersect at \(p\) (and meet \(\mathbb{S}^1_\infty\) at right angles). The figure \(M\) is pictured, as is \(T(M)\) and \(T(T(M))\text{.}\) The figure also tracks successive images of a point \(z\) on \(M\text{:}\) \(z\) gets mapped to \(T(z)\) which gets mapped to \(T^2(z) = T(T(z))\text{.}\) By this map \(T\text{,}\) any point \(z\) in \(\mathbb{D}\) rotates around \(p\) along the unique type II cline of \(p\) and \(p^*\) that contains \(z\text{.}\)

Figure \(5.1.4(b)\) depicts the image of \(M\) under two applications of a translation of the hyperbolic plane. The translation is generated by two inversions about non-intersecting clines \(L_1\) and \(L_2\) (that meet \(\mathbb{S}^1_\infty\) at right angles). The fixed points of the map are the points \(p\) and \(q\) on the circle at infinity. Under this translation, any point \(z\) in \(\mathbb{D}\) heads away from \(p\) and toward \(q\) on the unique cline through the three points \(p,q\text{,}\) and \(z\text{.}\)

We now derive the following algebraic description of transformations in \(\cal H\text{:}\)

Any transformation \(T\) in the group \(\cal H\) has the form

\[ T(z) = e^{i\theta}\frac{z - z_0}{1-\overline{z}_0z} \]

where \(\theta\) is some angle, and \(z_0\) is the point inside \(\mathbb{D}\) that gets sent to \(0\).

Assume \(z_0\) is in \(\mathbb{D}\text{,}\) \(z_1\) is on the unit circle \(\mathbb{S}^1_\infty\text{,}\) and that the map \(T\) in \({\cal H}\) sends \(z_0 \mapsto 0\text{,}\) \(z_0^* \mapsto \infty\) and \(z_1 \mapsto 1\text{.}\) Using the cross ratio,

\[ T(z) = (z,z_1;z_0,z_0^*) = \frac{z_1-z_0^*}{z_1-z_0}\cdot\frac{z-z_0}{z-z_0^*}\text{.} \]

But \(z_0^* = \dfrac{1}{\overline{z_0}}\text{,}\) so

\[ T(z) = \frac{z_1-1/\overline{z_0}}{z_1-z_0}\cdot\frac{z-z_0}{z-1/\overline{z_0}} = \frac{\overline{z_0}z_1 - 1}{z_1-z_0}\cdot\frac{z-z_0}{\overline{z_0}z- 1}\text{.} \]

Now, the quantity

\[ \frac{\overline{z_0}z_1 - 1}{z_1-z_0} \]

is a complex constant, and it has modulus equal to \(1\). To see this, observe that since \(1 = |z_1|^2 = z_1\overline{z_1}\text{,}\)

\begin{align*} \frac{\overline{z_0}z_1 - 1}{z_1 - z_0}& =\frac{\overline{z_0}z_1 - z_1\overline{z_1}}{z_1 - z_0}\\ & = \frac{-z_1(\overline{z_1 - z_0})}{z_1-z_0}\text{.} \end{align*}

Since \(|z_1| = 1\) and, in general \(|\beta| = |\overline{\beta}|\) we see that this expression has modulus \(1\), and can be expressed as \(e^{i\theta}\) for some \(\theta\text{.}\) Thus, if \(T\) is a transformation in \(\cal H\) it may be expressed as

\[ T(z) = e^{i\theta}\frac{z - z_0}{1-\overline{z}_0z} \]

where \(\theta\) is some angle, and \(z_0\) is the point inside \(\mathbb{D}\) that gets sent to \(0\).

Is the converse true? Is every transformation in the form above actually a member of \({\cal H}\text{?}\) The answer is yes, and the reader is asked to work through the details in the exercises.

## Exercises

Prove that \((\mathbb{D}, {\cal H})\) is homogeneous.

Suppose \(T\) is a Möbius transformation of the form

\[ T(z) = e^{i\theta}\frac{z - z_0}{1-\overline{z}_0z}, \]

where \(z_0\) is in \(\mathbb{D}\text{.}\) Prove that \(T\) maps \(\mathbb{D}\) to \(\mathbb{D}\) by showing that if \(|z| \lt 1\) then \(|T(z)| \lt 1\text{.}\)

**Hint**-
It is easier to prove that \(|z|^2 \lt 1\) implies \(|T(z)|^2 \lt 1\text{.}\)

Construct the fixed points of the hyperbolic translation defined by the inversion of two nonintersecting clines that intersect \(\mathbb{S}^1_\infty\) at right angles, as shown in the following diagram.

Suppose \(p\) and \(q\) are two points in \(\mathbb{D}\text{.}\) Construct two clines orthogonal to \(\mathbb{S}^1_\infty\) that, when inverted about in composition, send \(p\) to \(0\) and \(q\) to the positive real axis.

Prove that any two horocycles in \((\mathbb{D},{\cal H})\) are congruent.

Prove that \(\cal H\) is a group of transformations.