Does there exist a double cover with trivial deck transformation group?

Question

Sorry for the naive question. The following statement at the beginning of Bredon, chapter 4, §20, got me confused:

Let $\pi:X \to Y$ be a two-sheeted covering map. Let $g:X \to X$ be the unique nontrivial deck transformation. [...]

With no additional hypotheses on $X$ and $Y,$ is Bredon claiming that the deck transformation group of $\pi$ is always $\mathbb{Z}/2?$ I thought that it could happen to be the trivial group, but I haven't been able to work out a concrete example yet.

Answer 1

Look at the classification of covering spaces (e.g. Hatcher's Theorem 1.38 and Proposition 1.39): A two-sheeted cover $p : (\tilde X, \tilde x_0) \to(X, x_0)$ is determined by the subgroup $\Gamma = p_*\left[\pi_1(\tilde X, \tilde x_0)\right] \subset \pi_1(X, x_0)$, which is of index two. Now, an index-two subgroup is automatically normal, and then the deck transformation group is isomorphic to the quotient $\pi_1(X, x_0)/\Gamma$, which is of order two.

As Travis remarked in his commentary, the nontrivial deck transformation is quite easy to describe explicitly. This is nothing but a geometric translation of the well-known proof that an index-two subgroup is normal (which I can recall in few words: if $H \subset G$ is of index two, its only non trivial left coset has to be $G \setminus H$, and ditto for the right coset, so we have $gH = Hg$ for every $g \not\in H$. Since it's also true [and obvious] for $g \in H$, $H$ is indeed normal).

Answer 2

A small addendum to the uniqueness statement: Bredon only works with arcwise connected covering spaces. If we do not assume $X$ to be connected consider a map between spaces with discrete topologies $\{a_1,a_2, b_1,b_2\} \to \{a,b\}$, where $a_i\mapsto a, b_i\mapsto b$. This $2$-sheeted covering has $4$ Deck transformations (the deck transformation group is isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$).

This is a special case of a more general construction of $p$-sheeted covering maps from a space with $pn$ points to a space with $n$ points with deck transformation group isomorphic to $\prod_{i=1}^n S_p$. Here $S_p$ denotes the permutation group. We may also replace the points with disjoint unions of any fixed space to get this results in a general situation.

Answer 3

Why an answer to a five year old question? Simply because it is an interesting question and the existing answers do not really take up the question in the title.

We shall show hat the answer is "no" without any assumptions on $X,Y$.

The scope of Bredon's book is limited to Hausdorff, arcwise connected and locally arcwise connected spaces (see also Valentin's answer). Bredon's definition of a covering map is therefore not the most general one (see e.g. here). However, within Bredon's scope PseudoNeo tells the complete story. Valentin shows that for general covering spaces the group of deck transformations may be bigger than $\mathbb Z_2$.

So let us prove the following two theorems for general two-sheeted covering maps $\pi : X \to Y$.

Theorem 1. There exists a unique deck transformation $g : X \to X$ without fixed points. This deck transformation flips the points in each fiber $\pi^{-1}(y)$, $y \in Y$.

Remark. Each deck transformation $\phi$ has the property $\phi \circ \phi = id_X$. Thus all non-trivial elements of the group $\mathcal D(\pi)$ of deck transformations of $\pi$ have order two. If $\phi \ne id_X$, then $D(\phi) =\{id_X,\phi\}$ is a subgroup of $\mathcal D(f)$ such that $D(\phi) \approx \mathbb Z_2$. As Valentin has shown, there may be more than one such $\phi$, i.e. the $\mathcal D(f)$ may have more than one subgroup isomorphic to $\mathbb Z_2$.

Theorem 2. If $X$ is connected, then $id_X$ and $g$ are the only deck transformations. Thus $\mathcal D(\pi) \approx \mathbb Z_2$.

Proof of Theorem 1: For each $x \in X$ there exists a unique $g(x) \in X$ such that $\pi^{-1}(\pi(x)) = \{x,g(x)\}$. This gives us a unique function $g : X \to X$ such that

1. $\pi \circ g = \pi$
2. $g$ has no fixed points.

Clearly, $g$ is a bijection such that $g^{-1} = g$. We shall show that $g$ is continuous (which implies that $g$ is a homeomorphism and thus a deck transformation). Each $y \in Y$ has an open neighborhood $U(y)$ in $Y$ which is evenly covered, that is $\pi^{-1}(U(y)) = V_{+1}(y) \cup V_{-1}(y)$ with disjoint open $V_i(y) \subset X$, $i = \pm 1$, such that the restrictions $\pi_i : V_i(y) \to U(y)$ of $\pi$ are homeomorphisms. Clearly each $V_i(y)$ contains exactly one point of each fiber $\pi^{-1}(y')$, $y' \in U(y)$. In other words, if $x \in V_i(y)$, then $g(x) \in V_{-i}(y)$. For $x \in \pi^{-1}(U(y))$ we therefore have $$g(x) = \begin{cases} \pi_{-1}(\pi_{+1}(x)) & x \in V_{+1}(y) \\ \pi_{+1}(\pi_{-1}(x)) & x \in V_{-1}(y) \end{cases}$$ This shows that $g \mid_{\pi^{-1}(U(y))}$ is continuous. But $X$ is covered by the open sets $\pi^{-1}(U(y))$, $y \in Y$, thus $g$ is continuous.

Proof of Theorem 2: Certainly $id_X$ and $g$ are distinct deck transfomations. Let $\phi : X \to X$ be an arbitrary deck transformation. Clearly $id_X, g,\phi$ are lifts of $\pi : X \to Y$. Since $X$ is connected, it is well-known that two lifts agree if they agree at some point $x \in X$. See e.g. Hatcher Proposition 1.34. Pick any $x \in X$. Then either $\phi(x) = x = id_X(x)$ or $\phi(x) = g(x)$ and we conclude that either $\phi = id_X$ or $\phi = g$.