# (p.684) B Quotients and universal covering spaces

# (p.684) B Quotients and universal covering spaces

The purpose of the present appendix is to give a formal definition of some of the concepts used in Chapter 3. In particular, we define the notion of simple connectedness and of a universal covering space. However, we do not define what a quotient space is. The reader interested in a formal discussion of this notion is referred to, e.g., [15, Chapter 4]. Moreover, we assume the reader to be familiar with the concept of a manifold.

# B.1 Simple connectedness

Let us begin by defining the concept of a simply connected manifold.

Definition B.1

Let

Mbe a manifold. AloopinMis a continuous function $\mathrm{\gamma}:[a,b]\to M$ such that $\mathrm{\gamma}(a)=\mathrm{\gamma}(b)$. A loop $\mathrm{\gamma}:[a,b]\to M$ is said to becontractibleif there is a continuous function $F:[a,b]\times [0,1]\to M$ such that $F(t,0)=\mathrm{\gamma}(t)$, $F(a,s)=\mathrm{\gamma}(a)$, $F(b,s)=\mathrm{\gamma}(b)$ and $F(t,1)=\mathrm{\gamma}(a)$ for all $(t,s)\in [a,b]\times [0,1]$. The manifoldMis said to besimply connectedif every loop inMis contractible.

Example B.2

The spheres ${\mathbb{S}}^{n}$ for $n\ge 2$ and ${\mathbb{R}}^{n}$ are simply connected manifolds. However, ${\mathbb{S}}^{1}$, ${\mathbb{T}}^{n}$ and ${\mathbb{R}}^{2}\setminus \{0\}$ are not.

Due to Theorem 3.2, it would be desirable to only consider simply connected manifolds when trying to characterise topology via geometry. However, such a restriction would not be reasonable. On the other hand, it turns out that given any manifold, there is a canonically associated simply connected manifold, called the universal covering space:

Theorem B.3

Given a manifold

M, there is a manifold $\tilde{M}$, called theuniversal covering spaceofM, and a map $\mathrm{\pi}:\tilde{M}\to M$, called theuniversal covering map(or universal covering projection), such that

• $\tilde{M}$ is simply connected,

• given $p\in M$, there is a connected open neighbourhood

Uofpsuch that ${\mathrm{\pi}}^{-1}(U)={\cup}_{\mathrm{\alpha}\in A}{U}_{\mathrm{\alpha}}$, whereAis some index set and ${U}_{\mathrm{\alpha}}$ and ${U}_{\beta}$ are disjoint if $\mathrm{\alpha}\ne \beta $,• $\mathrm{\pi}$, restricted to ${U}_{\mathrm{\alpha}}$, yields a homeomorphism from ${U}_{\mathrm{\alpha}}$ to

U.

(p.685) Remark B.4

If ${g}_{ij}$ is a Riemannian metric on

M, then the above construction yields a natural Riemannian metric on $\tilde{M}$, say ${\tilde{g}}_{ij}$; in fact, ${\tilde{g}}_{ij}=({\mathrm{\pi}}^{\ast}g{)}_{ij}$. Moreover, ifMis a closed manifold, then $(\tilde{M},{\tilde{g}}_{ij})$ is geodesically complete.

Example B.5

Thinking of ${\mathbb{S}}^{1}$ as the unit circle in the complex plane, there is a map ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}:\mathbb{R}\to {\mathbb{S}}^{1}$ defined by ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}(t)={e}^{2\mathrm{\pi}it}$. Let us argue that the existence of this map demonstrates that $\mathbb{R}$ is the universal covering space of ${\mathbb{S}}^{1}$ and that ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}$ is the universal covering projection. To start with, it is intuitively clear that $\mathbb{R}$ is simply connected. Let $p\in {\mathbb{S}}^{1}$ and

Ube any connected open neighbourhood ofpwhich does not coincide with the entire circle. Then ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}^{-1}(U)$ is a union of disjoint sets, say ${U}_{i}$, $i\in \mathbb{Z}$. Moreover, restricting ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}$ to one of these sets, say ${U}_{i}$, we obtain a homeomorphism from ${U}_{i}$ toU. As a consequence, $\mathbb{R}$ is the universal covering space of ${\mathbb{S}}^{1}$ and ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}$ is the universal covering projection; cf. Figure 3.2 for an illustration.

Given a manifold, the universal covering space is unique up to diffeomorphism. Say now, for the sake of argument, that it is possible to classify simply connected manifolds. The question then arises how this knowledge can be used to obtain information concerning manifolds that are not simply connected. In other words, given a simply connected manifold *M*, how does one determine the manifolds that have *M* as a universal covering space? This question leads us to the concept of a deck transformation.

Definition B.6

Let $\tilde{M}$ be the universal covering space of

Mand let $\mathrm{\pi}$ be the covering projection. Adeck transformationis then a diffeomorphism $\varphi $ from $\tilde{M}$ to itself such that $\mathrm{\pi}\circ \varphi =\mathrm{\pi}$.

Remark B.7

If there is a Riemannian metric on

M, then the deck transformations are isometries of the corresponding metric on the universal covering space.

Example B.8

Consider Example B.5. If $n\in \mathbb{Z}$ and ${\varphi}_{n}(x)=x+n$, then, clearly, ${\mathrm{\pi}}_{{\mathbb{S}}^{1}}\circ {\varphi}_{n}={\mathrm{\pi}}_{{\mathbb{S}}^{1}}$. Thus ${\varphi}_{n}$ is a deck transformation if $n\in \mathbb{Z}$. In fact, in the case of Example B.5, all the deck transformations are of this form.

It turns out that the set of deck transformations, say Γ, form a group. Moreover, this group clearly acts in a natural way on the universal covering space. Furthermore, if *g* is a Riemannian metric on *M*, then Γ is a subgroup of the isometry group of the corresponding metric on the universal covering space. Finally, the group action has certain properties; it is free and properly discontinuous. The formal definition of these concepts is as follows; cf., e.g., [98, Definition 6, p. 188].

Definition B.9

A group Γ of diffeomorphisms of a manifold

Mis said to act freely and properly discontinuously provided

1. each $p\in M$ has a neighbourhood

Usuch that if $\varphi (U)$ meetsUfor $\varphi \in \mathrm{\Gamma}$, then $\varphi $ is the identity,2. points $p,q\in M$ not in the same orbit have neighbourhoods

UandVsuch that for every $\varphi \in \mathrm{\Gamma}$, $\varphi (U)$ andVare disjoint.

Remark B.10

For convenience, most of the time we simply say that the group acts nicely.

From our perspective, the main point of the definition is that if Γ acts freely and properly discontinuously on a simply connected manifold, then the quotient $M/\mathrm{\Gamma}$ is also a manifold; cf., e.g., [98, Proposition 7, p. 188]. Moreover, *M* is the universal covering space of the quotient and the natural projection $\mathrm{\pi}:M\to M/\mathrm{\Gamma}$ is the universal covering projection. As a consequence of the above considerations, there are two natural perspectives.

(p.686) Perspective 1

Let

Mbe a manifold. Then there is a universal covering space $\tilde{M}$ and a group of diffeomorphisms Γ (acting freely and properly discontinuously on $\tilde{M}$) such that $M=\tilde{M}/\mathrm{\Gamma}$. Moreover, given a Riemannian metric onM, there is a corresponding metric on $\tilde{M}$ and Γ is a group of isometries.

Perspective 2

Let $\tilde{M}$ be a simply connected manifold. If Γ is a group of diffeomorphisms of $\tilde{M}$ acting freely and properly discontinuously, then $M=\tilde{M}/\mathrm{\Gamma}$ is a manifold. Moreover, the natural projection is a covering projection making $\tilde{M}$ into the universal covering space of

M. If $\tilde{M}$ is endowed with a Riemannian metric and Γ is a subgroup of the isometry group, we, in addition, obtain a metric onMwhich is locally isometric to that of $\tilde{M}$.

Conclusion

If we are in a position to classify the simply connected manifolds and the collection of free and properly discontinuous group actions on these manifolds, we are able to classify the non-simply connected manifolds as well.