The topology of the Deck group of a covering map

Question

I have never been interested in this before, and I have become interested in to find some answers and my teaching on the fundamental group has led me in this direction. Neither, I don't know if it suits for MO. So, please down-vote or vote-to-close after you provided some references.

I think if $X$ is semi-locally simply connected, path connnected, and locally path connected then for the universal cover $p:\widetilde{X}\to X$, its group of Deck transformations, say $\mathrm{Deck}(p)$ acts properly discontinuously on $X$, hence it is equipped with the discrete topology. This then would imply that any covering over $X$ must have discrete group of Deck transformations. Indeed, this does not imply that it is finite, infinite or even countable. So, it could be $\mathbb{R}^\delta$ that is $\mathbb{R}$ equipped with the discrete topology. Is this conclusion correct?

So, if I wish to find a covering whose group of Deck transformations has a non-discrete topology then either $X$, the base space of my covering map, should not be semi-locally simply connected ( like the infinite earring) or $X$ should not be either path connected nor locally path connected. I wonder if there is a place that I can look for examples of such coverings where the topology of the Deck group is determined. I know that in Munkres's book there are some statements/exercises about this. But, anything more recent or some survey articles on this?

Answer 1

I reiterate from my comments that a covering map is a discrete-sort-of-thing by definition, no matter what space you're looking at. If you are interested in non-discrete deck transformations or situations where such things are useful, there is a large literature on generalizations of covering space theory (see my answer to this question). I don't know of a reference for the following claim but we can put together a short proof with some basic facts.

Claim: If $p:\widetilde{X}\to X$ is a covering map with $\widetilde{X}$ connected, then the subspace topology on $Deck(p)$ inherited from $Homeo(\widetilde{X})$ is discrete.

Proof. First, $Homeo(\widetilde{X})$ may not be a topological group, but it is a quasitopological group and is therefore homogeneous. So it suffices to show that the trivial subgroup $\{id\}$ is open in $Deck(p)$. Now, let's recall perhaps the strongest version of the uniqueness of lifts: If $Y$ is connected and $f,g:Y\to\widetilde{X}$ are maps such that $p\circ f=p\circ g$ and such that $f(y)=g(y)$ for at least one point $y\in Y$, then $f=g$. Applying this to our situation, we see that if $f,g\in Deck(p)$ agree on at least one point, then $f=g$.

Fix a point $\tilde{x}\in\widetilde{X}$ and let $U$ be an open neighborhood of $p(\tilde{x})$ that is evenly covered by $p$. Write $p^{-1}(U)=\coprod_{j\in J}V_{j}$ where $V_{j}$ is open and is mapped homeomorphically onto $U$ by $p$. We have $\tilde{x}\in V_{k}$ for some $k\in J$. Now $\mathcal{V}=\{f\in Deck(p)\mid f(\{\tilde{x}\})\subseteq V_{k}\}$ is an open neighborhood of $id$ in $Deck(p)$ (recall subbasic sets in the compact-open topology). If $f\in \mathcal{V}$, then $f(\{\tilde{x}\})\subseteq V_{k}\cap p^{-1}(p(\tilde{x}))=\{\tilde{x}\}$. Therefore, $f$ and $id$ agree at a point and must be equal. We conclude that $\{id\}=\mathcal{V}$ is open in $Deck(p)$. $\square$