Let $X = S^2 \times \mathbb{R}$ and $G=S^1\times \mathbb{Z}$. Let $G$ act on $X$ as follows - $S^1$ acts on $S^2$ by rotation leaving the north and south poles fixed and acts trivially on $\mathbb{R}$. The generator of $\mathbb{Z}$ reflects $S^2$ about the equator and translates $\mathbb{R}$ along one unit.
I have two questions -
- What does this action do to $X$? More precisely, what is $X/G$?
I can see that the orbit space of the $S^1$ action on $S^2$ is the unit interval $[0,1]$ and the orbit space of $\mathbb{Z}$ action on $\mathbb{R}$ is the circle, but this reflection about equator confuses me. Can I visualize this action by breaking it up like I did or am I going about it wrong? Is there an explicit expression for this action? That is, what would be the image of $(x,r)\in X$ be under $(z,n) \in G$?
- What is $\bar{G}$, the closure of $G$ in Homeo$(X)$ with the compact open topology?
I thought I could show $G$ is closed in Homeo$(X)$ using nets. However since I haven't understood what the action does to $X$ I am having trouble starting this one.
Thank you.
Let $p$ be the projection $X \to \hat{X} = X/S^1$ (we quantify by the action of $S^1 \times \{0\} \subset G$). One can translate the action of $\mathbb Z$ onto $\hat{X}$: $$k\cdot p(x):= p(k\cdot x)$$ (get sure, that this is correct!). But $\hat{X}$ is homeomorphic to $\mathbb{R} \times [0,1]$.Let's denote the action of $\mathbb{Z}$ on the $\hat{X}$ in the coordinates: $$1\cdot(x, t) = (x+1, 1-t)$$ Consider then a square $[0;1]\times[0;1]$ and a map $v$ from the square to itself: $(y, s) \mapsto (y+1\mod(1), 1-s)$. It is well known, that the result of gluing on $v$ is the M$\ddot o$bius strip. But the action of 1 on $\hat{X}$ is just a periodic extending of map $v$ on the whole tape.
It might be not as precise as you wish, but I hope, the idea is clear.