# Quotient Manifolds

Quotient spaces are fundamental objects of study in mathematics, but they nonetheless feel strange or unclear the first few (dozen?  hundred?) times they are used.  Math majors at St. Mary’s first encounter them in their Foundations of Mathematics class but come to exclusively associate them with topics from their Abstract Algebra sequence, most notably factor groups and factor rings.  In this post I will review the basics of this concept and how it is useful to geometry.

## Equivalence Relations and Quotient Spaces

Let $S$ be a set equipped with an equivalence relation $\sim$.  Note that we needn’t impose any additional structure on $S$ for what follows — $S$ is not assumed to be a group, a ring, a manifold, a vector space, or even a topological space, its just a set. The equivalence class of an element $s \in S$ is the subset

$\displaystyle [s] = \{x \in S : s\sim x\}.$

The three defining properties of an equivalence relation imply that the original set, $S$, can be partitioned into disjoint equivalence classes.  In set notation this means that

$\displaystyle S = \bigcup_{s\in S} [s]$

where $[s] \cap [x] \neq \emptyset \iff [s]=[x]$.  Viewing $S$ in this way offers us a simplified version of the original set; rather than examining individual elements $s \in S$ we examine these equivalence class subsets.  Indeed, one forms the set of equivalence classes and notates thusly

$\displaystyle \overline{S} = S/\sim \hspace{0.2cm} = \{[s] : s \in S\}.$

It is this object, $\overline{S}$, that one refers to as a quotient space.  This new set naturally feels a bit strange since, technically, it is a set of subsets (that is $\overline{S} \subset \mathcal{P}(S)$).  However, after some practice and examples, the strangeness wears off.  One gets used to treating the equivalence classes $[s]$ as individual elements, performing group operations between them (when $S = G$ is a group and $[e] = H$ is a normal subgroup) or, in our case, thinking about them as points in a manifold. Given an arbitrary subset $A \subseteq S$, one can construct an equivalence relation $\sim_A$ defined by

$\displaystyle s \sim_A x \iff s, x \in A \text{ or } s=x.$

This equivalence relation declares two elements of $S$ to be “the same” if they both lie in $A$, otherwise they are declared to only be equivalent to themselves.  The associated quotient space, then, views all elements of $A$ as one single element, effectively collapsing the subset to a single-point, and it views all other elements as distinct.  This kind of quotient space is so important we give it its own notation:

$\displaystyle \frac{S}{A} = \{[s] : s \in S\} \text{ where } [s] = \{x \in S : s\sim_Ax\}.$

The quotient space $S/A$ is often described as being obtained from $S$ by “killing $A$.”  This notation is also used in a slightly different way, and we will need to clear this up soon.  For now, though, here are  few examples.

Example 1. Let $S = \mathbb{R}$ and let $A = \mathbb{R}^*$.  The quotient space $\mathbb{R}/\mathbb{R}^*$ consists of two points:

$\displaystyle \frac{\mathbb{R}}{\mathbb{R}^*} = \{[1],[0]\}.$

In the quotient, all of the non-zero real numbers are “collapsed” into the single equivalence class $[1] = [\pi] = [-1/2]$, and the element $0 \in \mathbb{R}$ is placed inside of its own, single-element equivalence class $[0] = \{0\}$.

Example 2. Let $S = S^n$ be the unit sphere in $\mathbb{R}^n$ and define the equivalence relation

$\displaystyle \bold{x} \sim \bold{y} \iff \bold{x} = \pm\bold{y}$.

The set of equivalence classes, $\overline{S}$, contains infinitely many elements.  One can write down these elements using points from $S^n$ that lie in only one “hemisphere.”  For instance, when $n = 2$, the points $\bold{x} = (x_1, x_2, x_3)$ that lie in the southern hemisphere (where $x_3 < 0$) are enough to describe most of the elements in $S/\sim$.  The only ones we are missing are the points along the boundary circle where $x_3 = 0$, and, as before, many of these points are identified into the same equivalence class, too.  To the right you can see an attempt at visualizing this collapsed space.   The points labeled $a$ and $b$ are identified to similarly labeled points, ones that are located at “opposite” positions.  This means that were you a small bug crawling on the outside of this surface, once you walked through point $a$ on the blue edge you would pop out at the point $a$ on the red edge.  Although they are difficult (if not impossible) to visualize, higher dimensional versions of these objects can be described or thought of in similar ways.

Interestingly enough, these quotient spaces coincide with the real projective spaces mentioned in my post on differentiable manifolds.  Since every line through the origin intersects the unit sphere in exactly two points — say $\pm\bold{x}$ — we can think of these lines as being represented by the equivalence classes $[\bold{x}] = \{\bold{x}, -\bold{x}\}$.  Some other interesting facts?  Every even-dimensional instance of this example is non-orientable, but the odd-dimensional ones are orientable.  Lots of interesting art work has been based off of these spaces, including famous sculptures of Boy’s surface.

Example 3.  The finite cyclic groups, $\mathbb{Z}_n$, arise via the equivalence relation

$\displaystyle a\sim b \iff a-b = k\,n \text{ for some } k \in \mathbb{Z}$.

Example 4.  Every calculus student is actually familiar with equivalence classes of continuous functions $f(x)$.  Let us define two such functions to be equivalent, $f(x) \sim g(x)$, if $f(x) - g(x) = c$ for some constant $c \in \mathbb{R}$.  Then one of the fundamental theorems of calculus stipulates

$\displaystyle \int \! f(x) \, dx = [F(x)]$

where $F'(x) = f(x)$.

Example 5.  The set of rational numbers, $\mathbb{Q}$, is usually defined or constructed as a quotient space of $\mathbb{Z}\times\mathbb{Z}$ where two pairs of integers are said to be equivalent if

$(a,b) \sim (c, d) \iff ad = bc$

and then one extends the operations of addition and multiplication to these equivalence classes.  An important note deserves to be made here: when operations or functions are defined in terms of equivalence class inputs, it is quite common to use a representative from (or, more accurately, an element in) an abstract equivalence class.  As a result, one is left with checking that the given definition is, in fact, well-defined; that is, that it does not depend on the element chosen but only on the class from which it was chosen.  For instance, when we define addition between two rational numbers we assert

$\displaystyle [(a,b)] + [(c, d)] = [(ad+cb, bd)].$

One is left to check that if $[(x,y)]=[(a,b)]$ and $[(z,w)] = [(c,d)]$, then $[(ad+cb, bd)] = [(xw+zy, yw)]$.

Similarly, one way to define the set of real numbers, $\mathbb{R}$, is as a set of equivalence classes of Cauchy sequences of rational numbers.  We define two such sequences $\{a_n\}$ and $\{b_n\}$ to be equivalent if

$\displaystyle \{a_n\}\sim\{b_n\} \iff \lim_{n\to\infty} \left(a_n-b_n\right) = 0$.

## Discontinuous Group Actions

Every quotient space comes equipped with a natural function called the projection map, $\pi:S\to\overline{S}$.  It is defined by sending every element $s \in S$ to its equivalence class:

$\displaystyle \pi(s) = [s].$

Really, this is just a function-theoretic way of reminding ourselves how $\overline{S}$ is constructed from $S$.  It tells us that we “collect equivalent points in $S$ and treat them as single elements in $\overline{S}$.”  Note that since every equivalence class in $\overline{S}$ contains at least one element from $S$, this function is always surjective.

The projection function captures the essential information about our quotient space, and so when we want to equip $\overline{S}$ with additional structure, we prefer to do so with respect to this function. That is, if we start not just with a set $S$ but with a manifold $S = M$, then, given an equivalence relation $\sim$, we would like to equip the resulting quotient space $M/\sim$ with a manifold structure, too, but we would like to do so in a way that uses or acknowledges the manifold structure of the source, $M$.  Since the projection function $\pi$ is the link between these two objects, one typically equips the quotient space with a manifold structure so that $\pi$ becomes a nice mapping.

In general, we will want $\pi$ to become a differentiable mapping, but we will not want $\pi$ to be injective — after all, this would then imply that $\pi$ is a diffeomorphism and so the “new” quotient manifold $M/\sim$ won’t be all that new. First thing’s first, though: we must understand the main or basic ways in which a manifold gives rise to quotient manifold.  Instead of using a subset $A \subset M$ or an arbitrarily defined equivalence relation, we use an an abstract group, $G$.  However the group $G$ is originally given or written down, when its elements $g \in G$ can be made to act as diffeomorphisms $g:M\to M$, then we say that $G$ acts on $M$.  The group structure of $G$ should interact well with the differentiable structure of $M$, and so we also require that group multiplication correspond to diffeomorphism composition.  That is, we stipulate

$\displaystyle \left(g_1*g_2\right)(p) = g_1\left(g_2(p)\right)$

for all $g_1, g_2 \in G$ and $p \in M$.  It is also not surprising that we require the identity element $e \in G$ to act as the identity diffeomorphism on $M$.

We next form an equivalence relation, $\sim$, on $M$ by declaring two points $p, q \in M$ to be equivalent if

$p \sim q \iff g(p) = q \text{ for some } g \in G$

where $g\neq e$. These equivalence classes contain all points that are confused or mixed together under the action of $G$ so that $[p] = [q]$ if and only if you can move $p$ to $q$ using some mapping $g \in G$. The notation for this quotient space is usually $M/G$, which looks an awful lot like our previous notation for the quotient $S/A$, only in this context $G$ is not required to be a subset of $M$.  Moreover, even when $G$ is a subset of $M$, the equivalence classes being formed in $M/G$ are different than those in the $S/A$ setting.

Take, for example, the case when $M = \mathbb{R}$ and $G = \mathbb{Z}$ (under addition).  The quotient space $\mathbb{R}/\mathbb{Z}$ in this setting is diffeomorphic to the one-dimensional sphere:

$\displaystyle \mathbb{R}/\mathbb{Z} = \bigcup_{x\in[0,1)}\,[x] \cong S^1$

Here we have let $g\in G = \mathbb{Z}$ act on $\mathbb{R}$ by addition: $g(x) = x+g$.  The equivalence class of $\sqrt{2}$, for instance, contain all of the integer translates of $\sqrt{2}$: $[\sqrt{2}] = \{\dots, \sqrt{2}-1, \sqrt{2}, \sqrt{2}+1, \sqrt{2}+2, \dots\}$.  Points in the new manifold $\mathbb{R}/\mathbb{Z}$ may be represented using points from the original, source manifold, but they now have an interesting property; two points that differ by an integer amount are actually the same.  This means that $[0]=[1]=[-10]$ and $[1/2] = [9/2]$.  Indeed, ever point in this quotient space can be uniquely represented as a point $[x]$ where $x \in [0,1)$.  Observe that this quotient space is, in fact, the same one obtained by starting with the set $S = [0,1]$, say, and then collapsing the subset $A = \{0, 1\}$.

On the other hand, if we form the quotient space of $\mathbb{R}/\mathbb{Z}$ treating $\mathbb{Z}$ as a set, we obtain a very different object, one in which every integer is collapsed into a single element, $[1]$, but where all non-integers are in individual, single-point equivalence classes.  In particular, $[1/2] \neq [3/2]$ in this second setting.

From here on out, whenever we use the notation $M/G$ we will mean that $G$ acts on $M$, and the equivalence relation is given in terms of the group action.  In purely group theoretic terms we say that $G$ acts on $M$ if there exists a group homomorphism $\Phi : G \to \text{Diff}(M)$ where $\text{Diff}(M)$ denotes the group of all diffeomorphisms from $M$ to itself (with the operation of composition).  While this phrasing includes all of the aforementioned properties for free (group elements act as diffeomorphisms, group multiplication corresponds to function composition, and the identity element acts as the identity function), it is still not enough to guarantee that the new set $M/G$ inherits a meaningful differentiable structure from $M$.  To ensure that this happens, we also require the following

• For every $p \in M$ there exists a neighborhood $p \in U \subseteq M$ such that $g(U)\cap U = \emptyset$ for all $g \neq e$

In other words, on potentially very small scales, the group action of $G$ on $M$ moves points to new points (unless you’re using the identity element).  Stated this way, its not so clear why such a condition is needed or desired, nor is it necessarily clear why a group action with this property is called a proper discontinuous group action.  The name aside (at least for the moment), it turns out this property is essential, as demonstrated by the following

Theorem: Let $G$ act on a manifold $M$.  Then the projection map $\pi:M\to M/G$ is a local diffeomorphism if and only if $G$ acts properly discontinuously.

As mentioned in our textbook, $\pi$ being a local diffeomorhism means that for every point $p \in M$ there exists a neighborhood $p \in U \subseteq M$ on which $\pi:U\to\pi(U)$ is a diffeomorphism.  Since the projection mapping is known to be (globally) surjective, the key issue is local injectivity.   And this is where our definition of a properly discontinuous group action comes in: $\pi:M\to M/G$ is injective near $p$ if and only if there are no $G$-equivalent points near $p$ (except for $p$ itself, of course). The differentiability of $\pi$ is a routine check that Do Carmo explains quite well, and so I will omit it from this discussion.  The take home message is this:  Building a manifold structure on $M/G$ that respects or in some way “comes from” the manifold structure on $M$ is achieved by building a manifold structure on $M/G$ that makes $\pi$ a local diffeomorphism.

## Examples of Quotient Manifolds

Example 1. Real projective spaces are, of course, examples of quotient manifolds.  In fact, they are examples in (at least) two ways.  First, as introduced by Do Carmo, they arise as quotient manifolds of $M = \mathbb{R}^{n+1}$ under the group action of “rescaling,” i.e. $G = \mathbb{R}^* = \{\lambda \in \mathbb{R} : \lambda \neq 0\}$ and $\lambda \in G$ acts on $\bold{x} \in M$ by $\lambda(\bold{x}) = \lambda\bold{x}$.

The second way in which projective spaces arise as quotient manifolds involves $n$-dimensional spheres $M = S^n$.  The group acting on this space is then given by $G = \{\text{id}, -\text{id}\} \cong \mathbb{Z}_2$ so that two elements $\bold{x}, \bold{y} \in S^n$ are declared to be equivalent if they are equal or if one is the negative of the other. This example is important for a number of reasons, including the fact that the same quotient manifold can be described in multiple ways.

Example 2.  Let $M = \mathbb{R}^n$ and let $G = \mathbb{Z}^n$ act on $M$ by addition: $g(\bold{x}) = \bold{x} + g$.  It is trivial to verify that this is, indeed, a group action.  Moreover, it is not difficult to confirm that the action is properly discontinuous: given any $g \neq e$ in $\mathbb{Z}^n$ and any $\bold{x} \in \mathbb{R}^n$ one can let $U$ be a ball about $\bold{x}$ of radius $1/2$.  It then follows that $g(U) \cap U = \emptyset$.  After all, $g(U)$ will translate the ball $U$ to a new ball centered at a point of at least unit distance from $\bold{x}$. This example generalizes the “rolled up plane” picture we often see for the $2$-torus, $T^2$.  Indeed, these manifolds are referred to as $n$-tori:

$\displaystyle T^n = \mathbb{R}^n/\mathbb{Z}^n$

Finally, it is not difficult to verify that $T^n$ is diffeomorphic to the $n$-fold product of $S^1$.  Indeed, one can use the function

$\displaystyle f(\,[\bold{x}]\,) = \left(\cos(2\pi\,x_1), \sin(2\pi\,x_1), \cos(2\pi\,x_2), \sin(2\pi\,x_2)\, \dots, \cos(2\pi x_n), \sin(2\pi x_n)\right).$

There are, of course, a number of properties to check.  First and foremost, though, is the well-definedness of $f$.  The given formula appears to depend on how the input $[\bold{x}] = [(x_1, x_2, \dots, x_n)]$ is represented since the output $f(\,[\bold{x}]\,)$ contains expressions that involve the components $x_i$.  One must show that if $[bold{x}] = [\bold{y}]$, then replacing the $x_i$‘s with $y_i$‘s does not change the function’s output.

Second, one should verify that $f(\,[\bold{x}]\,) \in S^1\times S^1\times\cdots\times S^1$.  This is much easier than it may first seem since

$\displaystyle (\cos(2\pi \, x_i), \sin(2\pi \, x_i)) \in S^1 \text{ for all } i$.

Lastly, we need to confirm that $f$ is a diffeomorphism — invertible and differentiable.

Example 3.  Let $M = \text{SL}_n(\mathbb{R})$ and let $G = \{\text{id}, -\text{id}\} \cong \mathbb{Z}^2$.  It is trivial that $G$ acts properly discontinuously, and the resulting quotient manifold is called the projective special linear group

$\text{PSL}_n(\mathbb{R}) = \text{SL}_n(\mathbb{R})/\{\text{id}, -\text{id}\}.$

As happened with our real projective spaces in example 1, there is an alternate way to define this space:

$\text{PSL}_n(\mathbb{R}) = \text{GL}_n(\mathbb{R})/G$

where $G = \mathbb{R}^*$ acts by non-zero multiplication.

Example 4.  The Grassmannians $\text{G}(k,n)$ mentioned in the previous post can be constructed as quotient manifolds.  There is a good, thorough, but topologically-detailed description of this here (on pages 1-3) where one starts with the so-called Stiefel Manifold, $\text{St}(k, n)$.  This space is defined as a set of $n\times k$ matrices:

$\displaystyle \text{St}(k,n) = \left\{ n \times k \text{ matrices with rank } k \right\} \subset \mathbb{R}^{n\,k}.$

This set of matrices is in correspondence with the set of $k$-dimensional subspaces of $\mathbb{R}^n$, a fact one can argue by recalling that every such space is determined by a basis of $k$ vectors which can then be arranged as the columns of a matrix $A \in \text{St}(k,n)$.  Just like $\text{GL}_n(\mathbb{R})$ is an open set of $\mathbb{R}^{n^2}$, so too is $\text{St}(k,n)$ an open subset of $\mathbb{R}^{n\,k}$ (and so, in particular, has dimension $nk$).  We can then define a proper, discontinuous action on $\text{St}(k,n)$ by $G = \text{GL}_k(\mathbb{R})$ as follows: $g \in G$ acts on $A \in \text{St}(k,n)$ by

$\displaystyle g(A) = Ag \in \text{St}(k,n)$.

We then have

$\displaystyle \text{G}(k, n) = \text{St}(k,n)/\text{GL}_k(\mathbb{R})$

which corroborates our intuition that each $k$-dimensional subspace $W \subset \mathbb{R}^n$ is determined by an equivalence class of basis vectors, $\mathcal{B} = \{\bold{w}_1, \dots, \bold{w}_k\}$.  Here one basis, $\mathcal{B}$, is equivalent to another basis, $\mathcal{B}'$, if both bases span the same space $W$; this happens precisely when the two bases differ by an invertible, linear action (i.e. by an action of $g \in \text{GL}_k(\mathbb{R})$).