Computing the group of covering translations of a covering space

Question

I'm trying to understand covering spaces and covering projections. I'm looking at the following question:

Consider the space $$Y = \{ (y,z) \in (\mathbb{C}\backslash \ \{0\}) \times S^1 | \frac{y}{|y|} =z^2 \in S^1\}$$ Describe the fibre and compute the group of covering translations of the map: $$q:Y \rightarrow (\mathbb{C}\backslash \ \{0\}); (y,z) \mapsto y$$

So first of all, I'm struggling to visualise what $Y$ is. In particular, what is the intuitive meaning of the condition $\frac{y}{|y|} =z^2 \in S^1$ that each point in $Y$ must satisfy? By setting $\frac{y}{|y|} = \exp(i\theta) = \exp(i2\pi \psi)$, I didn't really get anywhere.

As for the fibre, it should be just the preimage of $q$ of a point in $\mathbb{C}\backslash \ \{0\}$ but what does that look like. At first I thought it was a copy of a circle, but I think that's incorrect.

Finally, how does one "compute" the group of covering translations? I know that it is the group of homeomorphisms of the covering space $h: Y \rightarrow Y$, such that $qh = h$, but not sure how to apply this property to actually show what the group is.

Answer 1

Let $\mathbb C^* = \mathbb C \setminus \{0\}$. For $w \in S^1$ let $F(w) = \{z \in S^1 \mid z^2 = w\}$. This set has exactly two elements.

1. The fibre over $y$ is $$q^{-1}(y) = \{y\} \times F(\frac{y}{|y|})$$ and therefore has two elements.

2. Clearly $h : \mathbb C^* \to (0,\infty) \times S^1 , h(y) = (\lvert y \rvert , \frac{y}{|y|})$, is a homeomorphism whose inverse is given by $h^{-1}(t,w) = tw$. Then also $H = h \times id_{S^1} : \mathbb C^* \times S^1 \to (0,\infty) \times S^1 \times S^1$ is a homeomorphism. Let $X = \{(z^2,z) \in S^1 \times S^1 \mid z \in S^1\}$. We have $$H(Y) = \{(h(y),z) \mid y \in \mathbb C^*, z \in S^1,\frac{y}{|y|} = z^2 \} \\= \{(h(h^{-1}(t,w)),z) \mid (t,w) \in (0,\infty) \times S^1 , z \in S^1,\frac{h^{-1}(t,w)}{|h^{-1}(t,w)|} = z^2 \} \\= \{(t,w,z) \in (0,\infty) \times S^1 \times S^1 \mid w = z^2\} = (0,\infty) \times X .$$ But $X$ is the graph of the map $s : S^1 \to S^1, s(z) = z^2$. Thus $X \approx S^1$ via $\pi : X \to S^1, \pi(w,z) = z$. The inverse of $\pi$ is given by $\pi^{-1}(z) = (z^2,z) = (s(z),z)$. Hence $\Pi = id_{(0,\infty)} \times \pi : (0,\infty) \times X \to (0,\infty) \times S^1$ is a homeomorphism and thus $$\phi = h^{-1} \circ \Pi \circ H : Y \to \mathbb C^*$$ is a homeomorphism. Note that $$\phi(y,z) = \lvert y \rvert z = \frac{y}{z}\quad, \quad \phi^{-1}(y) = \left(\frac{y^2}{|y|}, \frac{y}{|y|}\right) .$$ This nicely describes $Y$.

Now consider the map $$\psi = h \circ q \circ (\Pi \circ H)^{-1} : (0,\infty) \times S^1\to \mathbb (0,\infty) \times S^1 .$$ It is easy to verify that $\psi(t,w) = (t,s(w))$, i.e. $$\psi = id_{(0,\infty)} \times s .$$ The map $s$ is well-known to be a double (i.e. two-sheeted) covering, thus also $\psi$ is a double covering. Therefore $$q = h^{-1} \circ \psi \circ \Pi \circ H$$ is a double covering.

It is then easy to see that $q$ admits exactly two covering translations: The identity $id_Y$ and the map $\tau$ flipping the points in each fibre $q^{-1}(y)$ which is explicitly given by $$\tau(y,z) = (y,-z) .$$ In fact, let $g$ be any covering translation for $q$. The maps $id_Y, \tau, g$ are lifts of $q$. Since $Y$ is connected, it is well-known that two lifts agree if they agree at some point of $(y,z) \in Y$. But either $g(y,z) = (y,z) = id_Y(y,z)$ or $g(y,z) = (y,-z) = \tau(y,z)$, i.e. either $g = id_Y$ or $g = \tau$. Thus the group of covering translations is $\mathbb Z_2$.

See also https://math.stackexchange.com/q/3796928.