# 4.2: Möbius Geometry

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- 23316

We spent a fair amount of time studying Möbius transformations in Chapter 3, and this will pay dividends now.

**Möbius Geometry **is the geometry \((\mathbb{C}^+,{\cal M}),\) where \(\cal M\) denotes the group of all Möbius transformations.

Without actually stating it, we essentially proved that \(\cal{M}\) is a group of transformations. Namely, we proved that the inverse of a Möbius transformation is again a Möbius transformation, and that the composition of two Möbius transformations is a Möbius transformation. We remark that the identity map on \(\mathbb{C}^+\text{,}\) \(T(z) = z\text{,}\) is a Möbius transformation (of the form \(T(z) = \dfrac{(az+b)}{(cz+d)}\text{,}\) where \(a = d = 1,\) and \(b = c = 0)\text{,}\) so \(\cal M\) is a group.

Below we recast the key results from Chapter 3 in geometric terms:

- Any two clines are congruent in Möbius Geometry (Theorem \(3.4.10\)).
- The set of all clines is a minimally invariant set of Möbius Geometry (Theorems \(3.4.5\) and Theorem \(3.4.10\)).
- The cross ratio is an invariant of Möbius Geometry (Theorem \(3.4.8\)).
- Angle measure is an invariant of Möbius Geometry (Theorem \(3.4.5\)).

While we're at it, let's restate three other facts about Möbius transformations:

- Any transformation in \(\cal{M}\) is uniquely determined by the image of three points.
- If \(T\) in \(\cal{M}\) is not the identity map, then \(T\) fixes exactly \(1\) or \(2\) points.
- Möbius transformations preserve symmetry points.

What else? Euclidean distance is not an invariant function of Möbius Geometry. To see this, one need look no further than the map \(T(z) = \dfrac{1}{z}\text{.}\) If \(p = 2\) and \(q = 3\) (two points on the real axis) then \(d(p,q)=|p - q| = 1\text{.}\) However, their image points \(T(p) = \dfrac{1}{2}\) and \(T(q) = \dfrac{1}{3}\) have a Euclidean distance between them of \(\dfrac{1}{6}\). So our old-fashioned notion of distance goes out the window in Möbius geometry.

We emphasize that angles *are preserved* in Möbius geometry, which is a good thing. Why is this a good thing? Remember that in the distant past, humanity set out looking for a geometry in which Euclid's first 4 postulates hold true, but the 5^{th} one fails. The 4^{th} postulate states that all right angles equal one another. This means that if Ralph is holding a right angle over in the corner, and Randy is holding one down the block somewhere, we ought to be able to transform one onto the other and see that the angles are the same. Transformations do not change angles.

Rather than pursue the very general Möbius geometry, we take the preceeding facts and apply them straight away to two of its special “subgeometries,” hyperbolic geometry and elliptic geometry.

## Exercises

Which figures in Figure \(4.2.1\) are congruent in \((\mathbb{C}^+,\cal{M})\text{.}\)

**Answer**-
\(A \cong F\text{;}\) \(B \cong G\text{;}\) \(C \cong E\text{.}\)

Describe a minimally invariant set in \((\mathbb{C}^+,\cal{M})\) containing the “triangle” comprised of the three vertices \(0\), \(1\), and \(i\) and the three Euclidean line segments connecting them. Be as specific as possible about the members of this set.

Suppose \(p\) and \(q\) are distinct, finite points in \(\mathbb{C}^+\text{.}\) Let \(G\) consist of all elliptic Möbius transformations that fix \(p\) and \(q\text{.}\) We consider the geometry \((\mathbb{C}^+,G)\text{.}\)

- Show that \(G\) is a group of transformations.
- Determine a minimally invariant set in \((\mathbb{C}^+,G)\) that contains the Euclidean line through \(p\) and \(q\text{.}\)
- Determine a minimally invariant set in \((\mathbb{C}^+,G)\) that contains the perpendicular bisector of segment \(pq\text{.}\)
- For any point \(z \neq p, q\) in \(\mathbb{C}^+\text{,}\) characterize all points in \(\mathbb{C}^+\) congruent to \(z\text{.}\)
- Is \((\mathbb{C}^+,G)\) homogeneous?

Repeat the previous exercise for the set \(G\) consisting of all hyperbolic Möbius transformations that fix \(p\) and \(q\text{.}\)