Geometry with an Introduction to Cosmic Topology

Example 4.1.2. Group of translations.

Let \(\cal T\) denote the collection of all translations of the plane \(\mathbb{C}\text{.}\) In particular, for each \(b \in \mathbb{C}\text{,}\) let \(T_b: \mathbb{C} \to \mathbb{C}\) denote the translation \(T_b(z) = z + b\text{.}\) The set \(\cal T\) consists of all \(T_b\text{,}\) for all \(b \in \mathbb{C}\text{.}\) That is,

Show that \(\cal T\) is a group of transformations.

Example 4.1.4. Finite group of rotations.

Consider the set \(H = \{R_0, R_{\pi/2}, R_{\pi}, R_{3\pi/2}\}\) consisting of four rotations of \(\mathbb{C}\) about the origin (by 0, \(\pi/2\text{,}\) \(\pi\text{,}\) and \(3\pi/2\) radians). We observe first that \(H\) forms a group. Since \(R_0(z)=e^{i0}z = z\text{,}\) \(H\) contains the identity transformation on \(\mathbb{C}\text{.}\) The set also satisfies closure, and the reader can check all possible compositions. For instance, \(R_{3\pi/2} \circ R_{\pi} = R_{\pi/2}\text{.}\) Finally, the inverse of each transformation in \(H\) is again in \(H\text{.}\) Check that \(R_0^{-1} = R_0\text{,}\) \(R_{\pi/2}^{-1} = R_{3\pi/2}\text{,}\) \(R_\pi^{-1} = R_\pi\text{,}\) and \(R_{3\pi/2}^{-1} = R_{\pi/2}\text{.}\) Thus, \(H\) is a group and we may study the geometry \((\mathbb{C},H)\text{.}\) For instance, is the circle \(C\) given by \(|z-i|=.5\) congruent to the circle \(D\) given by \(|z| = .5\text{?}\) Well, is there a transformation in \(H\) that maps \(C\) onto \(D\text{?}\) No! The only four circles congruent to \(C\) are pictured below. These are found by rotating \(C\) about the origin by 0, \(\pi/2\text{,}\) \(\pi\text{,}\) or \(3\pi/2\) radians, the only allowable transformations in this geometry.

Notice also that any point \(z \neq 0\) is congruent to four points: \(z\text{,}\) \(e^{i\pi/2}z\text{,}\) \(e^{i\pi}z\text{,}\) and \(e^{i3\pi/2}z\text{.}\) How many points are congruent to \(z = 0\text{?}\) Are all lines congruent in this geometry? Nope. We are only allowed these few rotations, so we have no way to map the line \(y=x\text{,}\) say, to the line \(y=x+1\text{.}\)

Example 4.1.5. A two-element group.

Consider reflection of \(\mathbb{C}\) across the real axis, given by \(r(z) = \overline{z}\text{.}\) Since \(r \circ r\) is the identity map, the set \(G = \{1,r\}\) is a group of transformations on \(\mathbb{C}\text{,}\) and we may define the geometry \((\mathbb{C},G)\text{.}\) Notice that while \(3+i\) is congruent to \(3-i\) in this geometry, it is not congruent to \(-3+i\text{.}\) Also, the circle \(|z-2i|=1\) is congruent to the circle \(|z+2i|=1\) but not the circle \(|z-3i|=1\text{.}\)

Example 4.1.6. Translational geometry.

Let \(\cal T\) denote the group of translations in Example 4.1.2, and consider the geometry \((\mathbb{C}, {\cal T})\text{.}\) We call this geometry translational geometry. Which figures in Figure 4.1.7 are congruent in this geometry?

Remember, two figures are congruent if we can find a transformation that “moves” one figure on top of the other. Since our allowable moves here are translations, we cannot change the radius of a circle (that's a dilation), and we cannot rotate objects. So, in translational geometry the only figures congruent in Figure 4.1.7 are \(H\) and \(L\text{.}\)

Example 4.1.11. Euclidean geometry.

Euclidean geometry is the geometry \((\mathbb{C}, \cal{E})\text{,}\) where \(\cal{E}\) consists of all transformations of the form \(T(z) = e^{i\theta}z + b\text{,}\) where \(\theta\) is a real number and \(b\) is in \(\mathbb{C}\text{.}\) Note that \(\cal E\) consists of precisely those general linear transformations of the form \(T(z) = az+b\) in which \(|a| = 1\text{.}\) In the exercises, you check that this collection is indeed a group of transformations.

The group \(\cal{E}\) includes rotations and translations, but not dilations. Let's take a look at some familiar properties of objects that should be invariant in Euclidean geometry.

The Euclidean distance between two points \(z_1\) and \(z_2\) is defined to be \(|z_1 - z_2|\text{.}\) To show that this is an invariant function of \((\mathbb{C}, \cal{E})\text{,}\) we need to show that for any \(T\) in the group \(\cal E\text{,}\) the distance between \(z_1\) and \(z_2\) equals the distance between \(T(z_1)\) and \(T(z_2)\text{:}\)

Thus, Euclidean distance is preserved in \((\mathbb{C}, \cal{E})\text{.}\)

Angles are preserved as well. We have already proved that general linear transformations preserve angles (Theorem 3.1.12), and Euclidean transformations are general linear transformations, so angles are preserved in \((\mathbb{C}, \cal{E})\text{.}\)

Example 4.1.13. Homogeneous geometry.

Translational geometry \((\mathbb{C},{\cal T})\) is homogeneous because there is a translation that will map any point of \(\mathbb{C}\) to any other point of \(\mathbb{C}\text{.}\) That is, any two points of \(\mathbb{C}\) are congruent. Of course, without rotations, translational geometry is not isotropic. Here's a formal argument that \((\mathbb{C},{\cal T})\) is homogeneous:

Suppose \(p\) and \(q\) are arbitrary points in \(\mathbb{C}\text{.}\) We must find a translation \(T\) in \(\cal{T}\) such that \(T(p) = q\text{.}\) Let \(w = q - p\text{,}\) and consider the translation \(T_w\) in \(\cal{T}\text{.}\) Then \(T_w(p) = p + w = p + (q - p) = q\text{.}\) Thus \(T_w(p) = q\) and \(p \cong q\text{.}\) Since \(p\) and \(q\) are arbitrary points in \(\mathbb{C}\) it follows that \((\mathbb{C},\cal{T})\) is homogeneous.

Euclidean geometry \((\mathbb{C},\cal{E})\) is homogeneous since it contains all translations, but the geometries of Example 4.1.4 and Example 4.1.5 are not.

Example 4.1.15. Euclidean metric.

The Euclidean metric is defined by \(d(z,w)=|z-w|\text{.}\) We have already shown that \(d\) is preserved under Euclidean transformations, and the first two conditions of being a metric follow directly from the definition of modulus. We establish the triangle inequality by direct computation in the following lemma.