Subjects: Algebraic Topology
Links: Covering Maps, The Monodromy Action of Covering Maps, Automorphism Group,

Def: Suppose $q:E\to X$ is a covering map. An automorphism of $q$ is a covering isomorphism from $q$ to itself, that is a homeomorphism $\phi :E\to E$ such that $q\circ \phi =q$:

\usepackage{tikz-cd}
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\begin{tikzcd}[row sep=2cm, column sep=2cm]
E \arrow[rr, "\varphi"] \arrow[dr, "q"] && E \arrow[dl, "q"'] \\
& X
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Covering automorphisms are also variously known as deck transformations or covering transformations.

Let $\text{Deck}(E/X)={\text{Aut}}_q(E)$ denote the set of all automorphisms of the covering $q:E\to X$. Thus ${\text{Aut}}_q(E)$ is a group, called the automorphism group of the covering, or the covering group. It acts on $E$ in a natural way, and the definition of covering map isomorphisms implies that each orbit is a subset of a single fiber.

Properties of the Covering Group: Let $q:E\to X$ be a covering map.

• If two automorphisms of $q$ agree at one point, they are identical,
• Given $x\in X$, each covering automorphism restricts to a ${\pi }_1(X,x)$-automorphism of the fiber $q^{-1}\{x\}$ (with respect to the monodromy action).
• For any evenly covered open subset $U\subseteq X$, each covering automorphism permutes the components of $q^{-1}[U]$.
• The group ${\text{Aut}}_q(E)$ acts freely on $E$ by homeomorphism.

Examples:

• For the covering map $\epsilon :\mathbb{R}\to {\mathbb{S}}^1$, the integral translations $x\mapsto x+k$ for $k\in \mathbb{Z}$ are automorphisms. We can prove that ${\text{Aut}}_{\epsilon }(\mathbb{R})\cong \mathbb{Z}$. We can generalise this further to ${\epsilon }^n:{\mathbb{R}}^n\to {\mathbb{T}}^n$, then ${\text{Aut}}_{{\epsilon }^n}({\mathbb{R}}^n)\cong {\mathbb{Z}}^n$.
• If $q:{\mathbb{S}}^n\to {\mathbb{RP}}^n$ is the natural covering map, then the antipodal map $\alpha :{\mathbb{S}}^n\to {\mathbb{S}}^n$ defined by $\alpha (x)=-x$ is an automorphism. The covering automorphism group is the two element $\{\text{Id},\alpha \}\cong \mathbb{Z}/2\mathbb{Z}$.

Orbit Criterion for Covering Automorphisms: Let $q:E\to X$ be a covering map. If $e_1,e_2\in E$ are two points in the same fiber $q^{-1}\{x\}$, there exists a covering automorphism taking $e_1$ to $e_2$ iff the induced subgroups $q_{\ast }[{\pi }_1(E,e_1)]=q_{\ast }[{\pi }_1(E,e_2)]\le {\pi }_1(X,x)$.

Normal Coverings Have Transitive Automorphism Groups: If $q:E\to X$ is a covering map, then ${\text{Aut}}_q(E)$ acts transitively on each fiber iff $q$ is a normal covering.

Prop: Suppose $q_1:E\to X_1$ and $q_2:E\to X_2$ are normal coverings. There exists a covering $X_1\to X_2$ making the diagram commute iff ${\text{Aut}}_{q_1}(E)\le {\text{Aut}}_{q_2}(E)$.

Th: Suppose $q:E\to X$ is a covering map and $x\in X$. The restriction map $\phi \mapsto \phi {|}_{q^{-1}\{x\}}$ is a group isomorphism between ${\text{Aut}}_q(E)$ and the group ${\text{Aut}}_{{\pi }_1(X,x)}(q^{-1}\{x\})$ of ${\pi }_1(X,x)$-automorphism of $q^{-1}\{x\}$.

Covering Group Structure Theorem: Suppose $q:E\to X$ is a covering map, $e\in E$, and $x=q(e)$. Let $G={\pi }_1(X,x)$ and $H = $q_{\ast }[{\pi }_1(E,e)]\le {\pi }_1(X,x)$. For each path class $\gamma \in N_G(H)$ (the normalizer of $H$ in $G$), there is a unique covering automorphism ${\phi }_{\gamma }\in {\text{Aut}}_q(E)$ that satisfies ${\phi }_{\gamma }(e)=e\cdot \gamma$. The map $\phi \mapsto {\phi }_{\gamma }$ is a surjective homomorphism from $N_G(H)$ to ${\text{Aut}}_q(E)$ with kernel equal to $H$, so it descends to an isomorphism from $N_G(H)$ to ${\text{Aut}}_q(E)$:

Normal Case: If $q:E\to X$ is a normal covering, then for any $x\in X$ and any $e\in q^{-1}\{x\}$, the map $\gamma \mapsto {\phi }_{\gamma }$ of the theorem above induces an isomorphism from ${\pi }_1(X,x)/q_{\ast }[{\pi }_1(E,x)]$ to ${\text{Aut}}_q(E)$.

Simply Connected Case: If $q:E\to X$ is a covering map and $E$ is simply connected, then the automorphism group of the covering map is isomorphic to the fundamental group of $X$. In fact, for any $x\in X$ and $e\in q^{-1}\{x\}$, the map $\gamma \mapsto {\phi }_{\gamma }$ of the theorem above is an isomorphism from ${\pi }_1(X,x)$ to ${\text{Aut}}_q(E)$.

Prop: If we wanted to give a topology to ${\text{Aut}}_q(E)$ such that its action on $E$ is continuous, then the only topology that works is the discrete topology.

Classification Theorem

Prop: Suppose $q:E\to X$ is a covering map. Let $E/{\text{Aut}}_q(E)$ be the orbit space, and let $\pi :E\to E/{\text{Aut}}_q(E)$ be the quotient map. Then there is $q^{\prime }:E/{\text{Aut}}_q(E)\to X$ such that $q^{\prime }\circ \pi =q$.

Classification Theorem: Let $X$ be a topological space that has a universal covering space, and let $x_0\in X$ be any base point. there is a one-to-one correspondence between isomorphism classes of coverings of $X$ and conjugacy classes of subgroups of ${\pi }_1(X,x_0)$. The correspondence associates each covering $\hat{q}:\hat{E}\to X$ with the conjugacy class of its induced subgroup.

This gives us a Galois correspondence between the covering spaces and subgroups of the fundamental group, given that the original space has a universal covering space. This is akin to having a normal and separable field extension.

Cor: Suppose $q:E\to X$ is the universal covering space of $X$, and $x_0\in X$ is any base point. Given a subgroup $H\le {\pi }_1(X,x_0)$, let $\hat{H}\le {\text{Aut}}_q(E)$ be the subgroup corresponding to $H$ under the isomorphism between ${\text{Aut}}_q(E)$ and ${\pi }_1(X,x_0)$. Then $q$ descents to a continuous map $\hat{q}:E/\hat{H}\to X$; which is a covering space whose induced subgroup is $H$, .i.e, $H={\hat{q}}_{\ast }[{\pi }_1(E/\hat{H},{\hat{e}}_0)]$ for some ${\hat{e}}_0\in {\hat{q}}^{-1}\{x_0\}$.

Classification of Torus Coverings: Every covering of ${\mathbb{T}}^2$ is isomorphic to precisely one of the following:

• The universal covering ${\epsilon }^2:{\mathbb{R}}^2\to {\mathbb{T}}^2$.
• A covering $q:{\mathbb{S}}^1\times \mathbb{R}\to {\mathbb{T}}^2$ by $q(z,y)=(z^a\epsilon (y{)}^b,z^b\epsilon (y{)}^{-a})$, where $(a,b)$ is a pair of integers with $a\ge 0$ and $b>0$ if $a=0$.
• A covering $q:{\mathbb{T}}^2\to {\mathbb{T}}^2$ by $q(z,w)=(z^a{w}^b,w^c)$ where $(a,b,c)$ are integers with $a>b\ge 0$ and $c>0$.