Orbit spaces of linear involutions on spheres

Question

Consider the following involutions ($\mathbb{Z}_2$-actions) on the unit $2$-sphere $S^2 \subseteq \mathbb{R}^3$:

1. $(x, y, z) \mapsto (-x,-y,-z)$, the antipodal action; the orbit space is the projective plane $\mathbb{R}P^2$.

2. $(x, y, z) \mapsto (-x, -y, z)$, the rotation around the $z$ axis.

3. $(x,y, z) \mapsto (-x,y,z)$, the reflection along the $x$ axis, with the orbit space the $2$-disk $D^2$.

My first question is: what is the orbit space of the second involution? I've found a post on MathOverflow (https://mathoverflow.net/questions/20836/involutions-of-s2) which states that it is again $S^2$, but why is that the case?

A follow-up question: what happens with the orbit spaces in higher dimensions? In dimension $n$ we have $n+1$ analogous involutions, varying by the number of "minuses". For the antipodal action, the orbit space is $\mathbb{R}P^n$; when there's only one "minus" (analogue of case 3), the orbit space is $D^n$. Can something be said in the remaining cases?

Answer 1

Let's use polar coordinates on the $(x,y)$ plane and catesian on the $z$ axis. Then the application $$\varphi : S^2 \to S^2, \quad (\rho,\theta,z) \mapsto (\rho,2\theta,z)$$ is a ramified covering whose fibers are the orbits of your second involution.

More generally, in higher dimensions as you say you can classify the involution by the spectrum of the corresponding elements in $O^n$. Let's consider for $0 \leq i \leq n$ an involution $z_{i,n+1-i}$ of signature $(i,n+1-i)$ :

$$z_{i,n+1-i} : (x_1, \cdots, x_n, x_{n+1}) \mapsto (x_1, \cdots, x_i, -x_{i+1}, \cdots, -x_{n+1})$$

Then, you have a branched covering

$$\varphi_{i,n+1-i} : S^n \to E$$

where $E$ is the topological quotient space $D^i \times \mathbb{P}^{n-i} / R$, with $R$ being the equivalence relation defined as follows : $$ \forall x\in S^{i-1} \subset D^ i, \forall y_1, y_2 \in \mathbb{P}^{n-i}, \quad (x,y_1) \sim (x,y_2).$$ You can think of $E$ as $D^i \times \mathbb{P}^{n-i}$ that collapses on the border $S^{i-1}$ of $D^i$.

$\varphi_{i,n+1-i}$ is then defined by : $$\varphi_{i,n+1-i} : (x_1, \cdots, x_{n+1}) \mapsto \left(x_1, \cdots, x_i, \pi_{n-i}(x_{i+1}, \cdots, x_{n+})\right),$$ where $\pi_{n-i}$ is the canonical projection $$\pi_{n_i} : \mathbb{R}^{n+1-i}\setminus \{0\} \to \mathbb{P}^{n-i}.$$

Note that the fact that $\pi_{n-i}$ is not defined on $0$ is not a problem as it corresponds to $(x_1, \cdots, x_i)$ being in $S^{i-1}$ so we can just let it undefined and map $(x_1, \cdots, x_i, 0, \cdots, 0)$ to the corresponding point in $E$.

On the other hand, the fibers of $\varphi_{i,n+1-i}$ are exactly the orbits of $z_{i,n+1-i}$, so $E$ is the orbit space.

Now, I understand that this answer may be incomplete because one would like to have a better description or understanding of $E$. While I am unable to give a better general answer, we can already discuss some special cases :

• If $i=0$, then $E$ is the projective space $\mathbb{P}^n$.
• If $i=n$, then $E$ is the closed ball $D^n$.
• If $i=1$, then $E$ is the suspension of $\mathbb{P}^{n-1}.$ That gave us a simple description for $n=2$ because of the exceptional case $\mathbb{P}^1 \sim S^1$, but that is not the case in general.