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Let $G$ be a group of homeomorphisms of a topological space $X$. The action of $G$ on $X$ is said to be discontinuous at a point $x \in X$ if

  1. $G_x :=$ the stabilizer of $x$, is finite.

  2. $x$ has an open neighbourhood $U$ such that $gU \cap U = \emptyset$ for all $g \notin G_x$

Let $H$ be the normal subgroup of $G$ generated by all those elements of $G$ that have fixed points.

I need to show that the action of $G/H$ on $X/H$ is free and discontinuous

For any $gH \in G/H$ and $[x]_H \in X/H$ I can define $gH \cdot [x]_H = [g\cdot x]_H$ where $[x]_H$ is the class of all points of $X$ in the $H$ - orbit of $x$. I have proven that this is a well defined action (independent of the choice of $x$ or $g$). Also I have proven that this action is free.

This is my attempt to prove the action is discontinuous -

Let $\pi : X \rightarrow X/H$ be the orbit map.

  1. for $[x]_H \in X/H$ the stabilizer is just $\{H\}$ which is finite.

  2. Let $x\in \pi^{-1}([x]_H)$. Then there exists an open neighbourhood $U$ of $x$ in $X$ such that $gU \cap U = \emptyset$ for all $g \notin G_x$

    My guess is that $\pi (U)$ is the required neighbourhood of $[x]_H$

    Let $gH$ be a non identity element of $G/H$. Then I need to show that $gH \pi(U) \cap \pi (U) = \emptyset$. Suppose there is a $[y]_H$ in this intersection. Then there will be a $y_1 \in U$ such that $[y]_H = [gy_1]_H$. This will give me that $[y]_G =[y_1]_G$ as $G$ - orbits. How do I get a contradiction to $gU \cap U = \emptyset$ when $g \notin G_x$? Or more precisely where do I get a $g \notin G_x$ from? It is possible I'm missing something really simple here but I have no idea how to proceed.

Thanks!

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The following assumes that $G$ acts discontinuously on $X$, but you assumed that at the beginning of the second point of your attempt.

First, we choose an open set $U \subseteq X$ such that $g U \cap U = \emptyset$ for all $g \notin G_x$.

So you have to prove that $gH \pi(U) \cap \pi(U) = \emptyset$. Let $x_1 \in U$, i.e. $[x_1]_H \in \pi(U)$. If $[x_1]_H \in gH \pi(U)$, then there exists an $x_2 \in U$ such that $[x_1]_H = [g(x_2)]_H$, i. e. for an $h \in H$, $x_1 = h(g(x_2))$. We define $f := h \circ g \in G$. If $f$ is in the stabiliser of $x$, then $gH$ stabilises $[x]_H$ and there is nothing to prove. Otherwise, $x_2 \in U \cap fU$, and hence $U \cap fU \neq \emptyset$, contradiction.

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