## Section7.7Quotient Spaces

We may bend a sheet of paper and join its left and right edges together to obtain a cylinder. If we let $\mathbb{I}^2 = \{(x,y) \in \mathbb{R}^2 ~|~ 0 \leq x \leq 1, 0 \leq y \leq 1\}$ represent our square piece of paper, and $C = \{(x,y,z) \in \mathbb{R}^3 ~|~ x^2 + y^2 = 1, 0 \leq z \leq 1 \}$ represent a cylinder, then the map

\begin{equation*} p:\mathbb{I}^2 \to C ~~\text{by}~~p((x,y))= (\cos(2\pi x), \sin(2\pi x), y) \end{equation*}

models this gluing process.

This map tries very hard to be a homeomorphism. The map is continuous, onto, and it is almost one-to-one with a continuous inverse. It fails in this endeavor only where we join the left and right edges: the points $(0,y)$ and $(1,y)$ in $\mathbb{I}^2$ both get sent by $p$ to the point $(1,0,y)\text{.}$ But $p$ is nice enough to induce a homeomorphism between the cylinder and a modified version of the domain $\mathbb{I}^2\text{,}$ obtained by “dividing out” of $\mathbb{I}^2$ the mapping redundancies so that the result is one-to-one. The new version of $\mathbb{I}^2$ is called a quotient space. We develop quotient spaces in this section because all surfaces and candidate three-dimensional universes can be viewed as quotient spaces. We need the notion of an equivalence relation on a set. To get this, we need the notion of a relation.

A relation on a set $\boldsymbol S$ is a subset $R$ of $S \times S\text{.}$ In other words, a relation $R$ consists of a set of ordered pairs of the form $(a,b)$ where $a$ and $b$ are in $S\text{.}$ If $(a,b)$ is an element in the relation $R\text{,}$ we may write $a R b\text{.}$ It is common to describe equivalence relations, which we define shortly, with the symbol $\sim$ instead of $R\text{.}$ So, when you see $a \sim b$ this means the ordered pair $(a,b)$ is in the relation $\sim\text{,}$ which is a subset of $S \times S\text{.}$

###### Definition7.7.2.

An equivalence relation on a set $A$ is a relation $\sim$ that satisfies these three conditions:

1. Reflexivity: $x \sim x$ for all $x \in A$
2. Symmetry: If $x \sim y$ then $y \sim x$
3. Transitivity: If $x \sim y$ and $y \sim z$ then $x \sim z\text{.}$

For any element $a \in A\text{,}$ the equivalence class of $\boldsymbol{a}$, denoted $[a]\text{,}$ is the subset of all elements in $A$ that are related to $a$ by $\sim\text{.}$ That is,

\begin{equation*} [a] = \{x \in A ~|~ x \sim a\}. \end{equation*}
###### Example7.7.3.An equivalence relation.

Define $z \sim w$ in $\mathbb{C}$ if and only if Re$(z) - ~\text{Re}(w)$ is an integer and Im$(z) = ~\text{Im}(w)\text{.}$ For instance, $(-1.6 + 4i) \sim (2.4 + 4i)$ since the difference of the real parts (-1.6 - 2.4 = -4) is an integer and the imaginary parts are equal. To show $\sim$ is an equivalence relation, we check the three requirements.

1. Reflexivity: Given $z = a + bi\text{,}$ it follows that $z \sim z$ because $a - a = 0$ is an integer and $b = b\text{.}$
2. Symmetry: Suppose $z \sim w\text{.}$ Then Re$(z) - ~\text{Re}(w) = k$ for some integer $k$ and Im$(z) = ~\text{Im}(w)\text{.}$ It follows that Re$(w) - ~\text{Re}(z) = -k$ is an integer and Im$(w) = ~\text{Im}(z)\text{.}$ In other words, $w \sim z\text{.}$
3. Transitivity: Suppose $z \sim w$ and $w \sim v\text{.}$ We must show $z \sim v\text{.}$ Since $z \sim w\text{,}$ Re$(z) - ~\text{Re}(w) = k$ for some integer $k\text{,}$ and since $w \sim v\text{,}$ Re$(w) - ~\text{Re}(v) = l$ for some integer $l\text{.}$ Notice that
\begin{align*} k+l \amp =[\text{Re}(z)-~\text{Re}(w)]+[\text{Re}(w) - ~\text{Re}(v)]\\ \amp = \text{Re}(z) - ~\text{Re}(v)\text{.} \end{align*}
So, Re$(z) - ~\text{Re}(v)$ is an integer. Furthermore, we have Im$(z) =$ Im$(w) =$ Im$(v)\text{.}$ Thus, $z \sim v\text{.}$

The equivalence class of a point $z = a+bi$ consists of all points $w = c + bi$ where $a-c$ is an integer. In other words, $c = a+n$ for some integer $n\text{,}$ so $w = z + n$ and we may express the equivalence class as $[z] = \{ z + n ~|~ n \in \mathbb{Z}\}\text{.}$

A partition of a set $A$ consists of a collection of non-empty subsets of $A$ that are mutually disjoint and have union equal to $A\text{.}$ An equivalence relation on a set $A$ serves to partition $A$ by the equivalence classes. Indeed, each equivalence class is non-empty since each element is related to itself, and the union of all equivalence classes is all of $A$ by the same reason. That equivalence classes are mutually disjoint follows from the following lemma.

Suppose there is some element $c$ that is in both $[a]$ and $[b]\text{.}$ We show $[a] = [b]$ by arguing that each set is a subset of the other.

That $[a]$ is a subset of $[b]\text{:}$ Suppose $x$ is in $[a]\text{.}$ We must show that $x$ is in $[b]$ as well. Since $x$ is in $[a]\text{,}$ $x \sim a\text{.}$ Since $c$ is in $[a]$ and in $[b]\text{,}$ $c \sim a$ and $c \sim b\text{.}$ We may use these facts, along with transitivity and symmetry of the relation, to see that $x \sim a \sim c \sim b\text{.}$ That is, $x$ is in $[b]\text{.}$ Therefore, everything in $[a]$ is also in $[b]\text{.}$

We may repeat the argument above to show that $[b]$ is a subset $[a]\text{.}$ Thus, if $[a]$ and $[b]$ have any element in common, then they are entirely equal sets, and this completes the proof.

In light of Lemma 7.7.4, an equivalence relation on a set provides a natural way to divide its elements into subsets that have no points in common. An equivalence relation on $A\text{,}$ then, determines a new set whose elements are the distinct equivalence classes.

###### Definition7.7.5.

If $\sim$ is an equivalence relation on a set $A\text{,}$ the quotient set of $A$ by $\sim$ is

\begin{equation*} A/_\sim = \{[a] ~|~ a \in A\}\text{.} \end{equation*}

We will be interested in quotients of three spaces: the Euclidean plane $\mathbb{C}\text{,}$ the hyperbolic plane $\mathbb{D}\text{,}$ and the sphere $\mathbb{S}^2\text{.}$ If we build a quotient set from one of these spaces, we will call a region of the space a fundamental domain of the quotient set if it contains a representative of each equivalence class of the quotient and at most one representative in its interior.

### SubsectionOrbit Spaces

We may construct a natural quotient set from a geometry $(X,G)\text{.}$

The group structure of $G$ defines an equivalence relation $\sim_G$ on $X$ as follows: For $x, y \in X\text{,}$ let

\begin{equation*} x \sim_G y ~\text{if and only if}~ T(x) = y ~\text{for some}~T \in G\text{.} \end{equation*}

Indeed, for each $x \in X\text{,}$ $x \sim_G x$ because the group $G$ must contain the identity transformation, so the relation is reflexive. Next, if $x \sim_G y$ then $T(x) = y$ for some $T$ in $G\text{.}$ But the group contains inverses, so $T^{-1}$ is in $G$ and $T^{-1}(y) = x\text{.}$ Thus $y \sim_G x\text{,}$ and so $\sim_G$ is symmetric. Third, transitivity of the relation follows from the fact that the composition of two maps in $G$ is again in $G\text{.}$

Given geometry $(X,G)$ we let $X/G$ denote the quotient set determined by the equivalence relation $\sim_G\text{.}$ In this setting we call the equivalence class of a point $x$ in $X\text{,}$ the orbit of $x\text{.}$ So, the orbit of $x$ consists of all points in the space $X$ to which $x$ can be mapped under transformations of the group $G\text{:}$

\begin{equation*} [x] = \{ y \in X ~|~ T(x) = y~\text{for some}~T \in G\}\text{.} \end{equation*}

Put another way, the orbit of $x$ is the set of points in $X$ congruent to $x$ in the geometry $(X,G)\text{.}$

Note that if the geometry $G$ is homogeneous, then any two points in $X$ are congruent and, for any $x \in X\text{,}$ the orbit of $x$ is all of $X\text{.}$ In this case the quotient set $X/G$ consists of a single point, which is not so interesting. We typically want to consider orbit spaces $X/G$ in which $G$ is a “small” group of transformations.

We say that a group of transformations $G$ of $X$ is a group of homeomorphisms of $X$ if each transformation in $G$ is continuous. In this case, we call $X/G$ an orbit space. If $X$ has a metric, we say that a group of transformations of $X$ is a group of isometries if each transformation of the group preserves distance between points.

###### Example7.7.6.Building a topological cylinder.

Consider the horizontal translation $T_1(z) = z + 1$ of $\mathbb{C}\text{.}$ This transformation is a (Euclidean) isometry of $\mathbb{C}$ and it generates a group of isometries of $\mathbb{C}$ as follows. Put $T_1$ and $T_1^{-1}$ in the group, along with any number of compositions of these transformations. Fortunately, any number of compositions of these two maps results in an isometry that is easy to write down. Any finite composition of copies of $T_1$ and $T_1^{-1}$ indicates a series of instructions for a point $z\text{:}$ at each step in the long composition $z$ moves either one unit to the left if we apply $T_1^{-1}$ or one unit to the right if we apply $T_1\text{.}$ In the end, $z$ has moved horizontally by some integer amount. That is, any such composition can be written as $T_n(z) = z + n$ for some integer $n\text{.}$ We let $\langle T_1 \rangle$ denote the group generated by $T_1\text{,}$ and we have

\begin{equation*} \langle T_1 \rangle = \{T_n(z) = z + n ~|~ n \in \mathbb{Z}\}\text{.} \end{equation*}

The orbit of a point $p$ under this group of isometries is

\begin{equation*} [p] = \{p + n ~|~ n \in \mathbb{Z}\}\text{.} \end{equation*}

A fundamental domain for the orbit space $\mathbb{C}/\langle T_1 \rangle$ is the vertical strip consisting of all points $z$ with $0 \leq \text{Re}(z) \leq 1\text{,}$ as in the following figure. Every point in $\mathbb{C}$ is related to a point in this shaded vertical strip. Furthermore, no two points in the interior of the strip are related. By passing to the quotient, we are essentially “rolling” up the plane in to an infinitely tall cylinder. The rolling up process is described by the map $p:\mathbb{C} \to \mathbb{C}/\langle T_1 \rangle$ given by $p(z) = [z]\text{.}$

###### Example7.7.7.A quotient space from rotations.

The rotation $R_{\frac{\pi}{2}}$ of $\mathbb{C}$ by $\pi/2$ about the origin generates a group of isometries of $\mathbb{C}$ consisting of four transformations. We generate the group as before, by considering all possible compositions of $R_{\frac{\pi}{2}}$ and $R_{\frac{\pi}{2}}^{-1}\text{.}$ This group turns out to be finite: Any combination of these rotations produces a rotation by 0, $\pi/2\text{,}$ $\pi\text{,}$ or $3\pi/2$ radians, giving us

\begin{equation*} \langle R_{\frac{\pi}{2}} \rangle = \{1, R_{\frac{\pi}{2}}, R_\pi, R_{\frac{3\pi}{2}}\} \text{.} \end{equation*}

The orbit of the point 0 is simply $\{0\}$ because each transformation in the group fixes 0, but the orbit of any other point in $\mathbb{C}$ is a four-element set. For instance, the orbit of $1$ is $[1] = \{i,-1,-i, 1\}\text{.}$

It turns out that every surface can be viewed as a quotient space of the form $M/G\text{,}$ where $M$ is either the Euclidean plane $\mathbb{C}\text{,}$ the hyperbolic plane $\mathbb{D}\text{,}$ or the sphere $\mathbb{S}^2\text{,}$ and $G$ is a group of isometries in Euclidean geometry, hyperbolic geometry, or elliptic geometry, respectively. In topology terminology, the space $M$ is called a universal covering space of the orbit space $M/G\text{.}$

###### Example7.7.8.$H_1$ as quotient of $\mathbb{C}$.

Suppose $a$ and $b$ are positive real numbers. Let $\langle T_a, T_{bi}\rangle$ be the group of homeomorphisms generated by the horizontal translation $T_a(z) = z + a$ and the vertical translation $T_{bi}(z) = z + bi\text{.}$

This group contains all possible compositions of these two transformations and their inverses. Thus, the orbit of a point $z$ consists of all complex numbers to which $z$ can be sent by moving $z$ horizontally by some integer multiple of $a$ units, and vertically by some integer multiple of $b$ units. An arbitrary transformation in $\Gamma = \langle T_a, T_{bi}\rangle$ has the form

\begin{equation*} T(z) = z + (ma + nbi) \end{equation*}

where $m$ and $n$ are integers. A fundamental domain for the orbit space consists of the rectangle with corners $0, a, a + bi, bi\text{.}$ The resulting quotient space is homeomorphic to the torus. Notice that points on the boundary of this rectangle are identified in pairs. In fact, the fundamental domain, with its boundary point redundancies, corresponds precisely to our polygonal surface representation of the torus.

If the space $M$ has a metric and our group of homeomorphisms is sufficiently nice, then the resulting orbit space inherits a metric from the universal covering space $M\text{.}$ To be sufficiently nice, we first need our homeomorphisms to be isometries. The group of isometries must also be fixed-point free and properly discontinuous. The group $G$ is fixed-point free if each isometry in $G$ (other than the identity map) has no fixed points. The group $G$ is properly discontinuous if every $x$ in $X$ has an open 2-ball $U_x$ about it whose images under all isometries in $G$ are pairwise disjoint. The interested reader is encouraged to see [10] or [9] for more detail. If our group $G$ is a fixed-point free, properly discontinuous group of isometries, then the resulting orbit space inherits a metric from $M\text{.}$

Consider the quotient space in Example 7.7.7. The group here is a group of isometries, since rotations preserve Euclidean distance, but it is not fixed-point free. All maps in $G$ have fixed points (rotation about the origin fixes 0). This prevents the quotient space from inheriting the geometry of its mother space. Indeed, a circle centered at [0] with radius $r$ would have circumference $\frac{2\pi r}{4}\text{,}$ which doesn't correspond to Euclidean geometry.

The group of isometries in the torus example is fixed-point free and properly discontinuous, so the following formula for the distance between two points $[u]$ and $[v]$ in the orbit space $\mathbb{C}/\langle T_a, T_{bi}\rangle$ is well-defined:

\begin{equation*} d([u],[v]) = \text{min}\{|z-w| ~|~ z \in [u], w \in [v]\}\text{.} \end{equation*}

Figure 7.7.9 depicts two points in the shaded fundamental domain, $[u]$ and $[v]\text{.}$ The distance between them equals the Euclidean distance in $\mathbb{C}$ of the shortest path between any point in equivalence class $[u]$ and any point in equivalence class $[v]\text{.}$ There are many such nearest pairs, and one such pair is labeled in Figure 7.7.9 where $z$ is in $[u]$ and $w$ is in $[v]\text{.}$ Also drawn in the figure is a solid line (in two parts) that corresponds to the shortest path one would take within the fundamental domain to proceed from $[u]$ to $[v]\text{.}$ This path marks the shortest route a ship in the video game from Chapter 1 could take to get from $[u]$ to $[v]\text{.}$

###### Example7.7.10.$\mathbb{P}^2$ as quotient of $\mathbb{S}^2$.

Let $T_a: \mathbb{S}^2 \to \mathbb{S}^2$ be the antipodal map $T_a(P) = -P\text{.}$ This map is an isometry that sends each point on $\mathbb{S}^2$ to the point diametrically opposed to it, so it is fixed-point free. Since $T_a^{-1} = T_a\text{,}$ the group generated by this map consists of just 2 elements: $T_a$ and the identity map. The quotient space $\mathbb{S}^2/\langle T_a\rangle$ is the projective plane.

###### Example7.7.11.$H_2$ as quotient of $\mathbb{D}^2$.

We may build a regular octagon in the hyperbolic plane whose interior angles equal $\pi/4$ radians. We may also find a hyperbolic transformation that takes an edge of this octagon to another edge. Labelling the edges as in the following diagram, let $T_{a}$ be the hyperbolic isometry taking one $a$ edge to the other, being careful to respect the edge orientations. We construct such a map by composing two hyperbolic reflections about hyperbolic lines: the hyperbolic line containing the first $a$ edge, and the hyperbolic line $m$ that bisects the $b$ edge between the $a$ edges. Since the two lines of reflection do not intersect, the resulting map in $\cal H$ is a translation in ${\cal H}$ and has no fixed points in $\mathbb{D}$ (the fixed points are ideal points).

Define $T_{b},T_{c}\text{,}$ and $T_{d}$ similarly and consider the group of isometries of $\mathbb{D}$ generated by these four maps. This group is a fixed-point free, properly discontinuous group of isometries of $\mathbb{D}\text{,}$ so the resulting quotient space inherits hyperbolic geometry.

The distance between two points $[u]$ and $[v]$ in the quotient space is given by

\begin{equation*} d_H([u],[v]) = ~\text{min}\{d_H(z,w) ~|~ z \in [u], w \in [v]\}\text{.} \end{equation*}

Geodesics in the quotient space are determined by geodesics in the hyperbolic plane $\mathbb{D}\text{.}$

Topologically, the quotient space is homeomorphic to $H_2\text{,}$ and the octagon pictured above serves as a fundamental domain of the quotient space. Moving copies of this octagon by isometries in the group produces a tiling of $\mathbb{D}$ by this octagon. Each copy of the octagon would serve equally well as a fundamental domain for the quotient space. Figure 7.7.12 displays a portion of this tiling, including a geodesic triangle in the fundamental domain, and images of it in neighboring octagons.

All surfaces $H_g$ for $g \geq 2$ and $C_g$ for $g \geq 3$ can be viewed as quotients of $\mathbb{D}$ by following the procedure in the previous example.

Start with a perfectly sized polygon in $\mathbb{D}\text{.}$ The polygon must have corner angle sum equal to $2\pi$ radians, and the edges that get identified must have equal length so that an isometry can take one to the other. (In every example so far, we have used regular polygons in which all sides have the same length, but asymmetric polygons will also work.) Next, for each pair of oriented edges to be identified, find a hyperbolic isometry that maps one onto the other (respecting the orientation of the edges). The group generated by these isometries creates a quotient space homeomorphic to the space represented by the polygon, and it inherits hyperbolic geometry. Note also that the initial polygon can be moved by the isometries in the group to tile all of $\mathbb{D}$ without gaps or overlaps.

### SubsectionDirichlet Domain

We end this section with a discussion of the Dirichlet domain, which is an important tool in the investigation of the shape of the universe. Suppose we live in a surface described as a quotient $M/G$ where $M$ is either $\mathbb{C}\text{,}$ $\mathbb{D}\text{,}$ or $\mathbb{S}^2\text{,}$ and $G$ is a fixed-point free and properly discontinuous group of isometries of the space. For each point $x$ in $M$ define the Dirichlet domain with basepoint $x$ to consist of all points $y$ in $M$ such that

\begin{equation*} d(x,y) \leq d(x,T(y)) \end{equation*}

for all $T$ in $G\text{,}$ where it is understood that $d(x,y)$ is Euclidean distance, hyperbolic distance, or elliptic distance, depending on whether $M$ is $\mathbb{C}\text{,}$ $\mathbb{D}\text{,}$ or $\mathbb{S}^2\text{,}$ respectively.

At each basepoint $x$ in $M\text{,}$ the Dirichlet domain is itself a fundamental domain for the surface $M/G\text{,}$ and it represents the fundamental domain that a two-dimensional inhabitant might build from his or her local perspective. It is a polygon in $M$ (whose edges are lines in the local geometry) consisting of all points $y$ that are as close to $x$ or closer to $x$ than any of its image points $T(y)$ under transformations in $G\text{.}$

We may visualize a Dirichlet domain with basepoint $x$ as follows. Consider a small circle in $M$ centered at $x\text{.}$ Construct a circle of equal radius about all points in the orbit of $x\text{.}$ Then, begin inflating the circle (and all of its images). Eventually the circles will touch one another, and as the circles continue to expand let them press into each other so that they form a geodesic boundary edge. When the circle has filled the entire surface, it will have formed a polygon with edges identified in pairs. This polygon is the Dirichlet domain.

At any basepoint in the torus of Example 7.7.8 the Dirichlet domain will be a rectangle identical in proportions to the fundamental domain. In general, however, the shape of a Dirichlet domain may be different than the polygon on which the surface was built, and the shape of the Dirichlet domain may vary from point to point, which is rather cool.

###### Example7.7.14.Dirichlet domains in a Klein bottle.

Consider the surface constructed from the hexagon in Figure 7.7.13, which appeared in Levin's paper on cosmic topology [23]. Assume the hexagon is placed in $\mathbb{C}$ with its six corners at the points 0, 1, 2, $2 + i\text{,}$ $1+i$ and $i\text{.}$

This polygonal surface represents a cell division of a surface with three edges, two vertices, and one face. The Euler characteristic is thus 0, so the surface is either the torus or Klein bottle. In fact, it is a Klein bottle because it contains a Möbius strip. We may tile the Euclidean plane with copies of this hexagon using the transformations $T(z) = z + 2i$ (vertical translation) and $r(z) = \overline{z}+(1+2i)$ (a transformation that reflects a point about the horizontal axis $y = 1$ and then translates to the right by one unit). The following figure shows the shaded fundamental domain $A$ and its images under various combinations of $T$ and $r\text{.}$

If $\Gamma$ is the group of transformations generated by $T$ and $r\text{,}$ the quotient space $\mathbb{C}/\Gamma$ is the Klein bottle, and its geometry is Euclidean, inherited from the Euclidean plane $\mathbb{C}\text{.}$

It turns out that the Dirichlet domain at a basepoint in this space can vary in shape from point to point. Exercise 7.7.4 investigates the shape of the Dirichlet domain at different points.

### ExercisesExercises

###### 1.

Show that the Dirichlet domain at any point of the torus in Example 7.7.8 is an $a$ by $b$ rectangle by completing the following parts.

1. Construct an $a$ by $b$ rectangle to be the fundamental domain, and place eight copies of this rectangle around the fundamental domain as in Figure 7.7.9. Then plot a point $x$ in the fundamental domain, and plot its image in each of the copies.
2. For each image $x^\prime$ of $x\text{,}$ construct the perpendicular bisector of the segment $xx^\prime\text{.}$ The eight perpendicular bisectors enclose the Dirichlet domain based at $x\text{.}$ Prove that the Dirichlet domain is also an $a$ by $b$ rectangle.
###### 2.

Construct $C_3$ as a quotient of $\mathbb{D}$ by a group of isometries of $\mathbb{D}\text{.}$ Be as explicit as possible when defining the group of isometries.

###### 3.

Explain why the $g$-holed torus $H_g$ can be viewed as a quotient of $\mathbb{D}$ by hyperbolic isometries for any $g \geq 2\text{.}$

###### 4.

Consulting Example 7.7.14, show that the Dirichlet domain at any point $z$ on the line $\text{Im}(z)=1/2\text{,}$ such as the one in Figure 7.7.15, is a square. Show that the Dirichlet domain at any point $z$ on the line $\text{Im}(z)=0$ is a rectangle.