3
$\begingroup$

I'm lacking ideas how to attack the following problem:

Given $G$ a topological group, and $H$ a discrete subgroup, I have to show that that $G \rightarrow G/H$ is a regular covering space.

Could someone give me a pointer?

Thank you

Stefan

1 Answers 1

3

I think you may want $G$ to be Hausdorff (or maybe even something stronger), [also see t.b.'s comment] I think a counterexample for what you stated above is given by $G=\mathbb R$ with the countable complement topology and $H =\mathbb Z$. [does not turn $G$ into a topological group it seems!]

Let $\pi: G\to G/H$ be the quotient map.

HINT 1: The quotient map is open.

HINT 2: Let $gH$ be any point in $G/H$. Can you find a neighbourhood $U$ of $g$ in $G$, such that

$\pi|_U: U \to \pi(U)$

is a homeomorphism? Note that we only need $\pi|_U$ to be a bijection because of hint 1.

For such $U$: What is $U\cap Uh$ for $h\ne e$, where $h\in H$?

Conclude that $\pi(U)$ is an evenly covered neighbourhood of $gH$. So $\pi: G\to G/H$ is a covering map.

To show that the covering is regular, we must prove that the group of deck transformations acts transitively on fibers. Can you guess what the group of deck transformations is?

HINT 3: \pi(g) = \pi(g') if and only if g' = gh for some $h\in H$.

  • 4
    Re: first sentence. Exercise: $T_0$ implies all the separation axioms $T_1$, $T_2$, $T_3$ and $T_{3\frac12}$ for topological groups.2011-12-13