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Saturday, July 7, 2018

When is $G\to G/H$ a covering map of topological groups?

The following, or something similar will appear in Monday's exam so I better work on it. Also, here, every neighborhood is declared to be open.

Exercise. Let $G$ be a topological group and $H$ a discrete, closed subgroup of $G$. Then $G\to G/H$ is a covering space.

The proof can be done manually, but we'll use the following proposition.

Proposition (1.40 from Hatcher's book). If an action of $H$ on a space $Y$ satisfies that each $y\in Y$ has a neighborhood $U$ such that $hU\cap U\neq \varnothing$ implies $h=1$ (A strong hypothesis of a properly discontinuous action), then
  • The quotient map $Y\to Y/H$ is a normal covering space.
  • $H$ is the group of deck transformations of $Y\to Y/H$ if $Y$ is path-connected.
  • $H$ is isomorphic to $\pi_1(Y/H)/\pi_1(Y)$ (Where $\pi_1(Y)$ is its identification as a subgroup of $\pi_1(Y/H)$) if $Y$ is path-connected and locally path-connected.
Easy Lemma. Let $G$ be a topological group. Then there for each neighborhood $U$ of $1$ there exists a neighborhood $V\subseteq U$ of $1$ such that $V=V^{-1}$ and $V^2=\{vv'\mid v,v'\in V\}\subseteq U$.
Proof of Easy Lemma. By continuity of the product $\pi$ and the product topology of $G\times G$, there exist open sets $V_1,V_2\subseteq G$ such that $1\in \pi(V_1\times V_2)\subseteq U$. Now take $V=(V_1\cap V_2)\cap (V_1\cap V_2)^{-1}$. Then $1\in V$, $V=V^{-1}$ and $V\subseteq V_1 \subseteq V_1 V_2\subseteq U$. Furthermore $V^2=\pi(V\times V)\subseteq \pi(V_1\times V_2)\subseteq U$.
Proof of Exercise. We only need to prove that the action of $H$ on $G$ by left multiplication is a covering space action. This can be done as follows.

Let $U$ be a neighborhood of $1$ such that $U\cap H=\{1\}$. Take a second neighborhood $V\subseteq U$ of $1$ such that $V=V^{-1}$ and $V^2=\{vv'\mid v,v'\in V\}\subseteq U$. It also holds that $V\cap H=\{1\}$.

Now, for each $g\in G$, take $V_g=Vg$, and let's suppose that $z\in hVg\cap Vg\neq \varnothing$ for some $h\in H$. Then $z=hvg=v'g$ for some $v,v'\in V$. This means that $hv=v'$ or $h=v'v^{-1}\in V^2\cap H$ thus $h=1$.

So, the action of $H$ on $G$ is a covering space action and $G$ is a covering space of $G/H$ via the canonical projection.

There are more consequences: $H$ is the group of deck transformations of this cover, and isomorphic to $\pi_1(G/H)/\pi_1(G)$, given that $G$ is path-connected and locally path-connected.

Easy Example. Take $G=S^1$ and $H=U_n$, the group of the $n$th roots of unity. $G/H$ is just $S^1$ with the projection defined as $z\mapsto z^n$, and $U_n\cong \bZ_n$ is the group of deck transformations.
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