1

This is a follow up of this question.

Suppose $ G $ is a (Hausdorff) topological group and suppose that $ H \le G $ is a closed subgroup. We define a toplogy on the quotient $ H \backslash G $ by $ U \subseteq H \backslash G $ is open iff $ \pi^{-1}(U) \subseteq G $ is open, where $ \pi : g \rightarrow H \backslash G $ is the quotient map ($ \pi ( g) = H g $). It is easy show that $ \pi $ is an open mapping, and that $ H \backslash G$ is Hausdorff.

We call a subset $ A \subseteq G $ compact modulo $ H $ if the following equivalent definitions hold:

  1. $ \pi(A) \subseteq H \backslash G $ is compact
  2. $ A \subseteq HK $ where $ K $ is compact

I want to show that these definitions are indeed equivalent. I'm stuck in both directions. Here is my attempt:

$ 2 \implies 1 $: suppose $ A \subseteq HK $ where $ K $ is compact, then $ \pi(A) \subseteq \pi (HK) = \pi(K) $. Since $ \pi $ is continuous, we have that $ \pi(K) $ is compact. If we knew that $ \pi(A) $ was closed, we would be done.

$ 1 \implies 2 $: write $ \pi(A) = \{H g_i \mid i\in I\} $ where $ H g_i \ne H g_j $ for $ i \ne j $. Denote $ C = \{ g_i \mid i \in I \} $ and define $ K = \{ g_i \mid H g_i \in A \} $. If we knew that $ \pi \restriction_C : C \rightarrow H \setminus G $ was an open map, then it would be a homeomorphism and therefore $ K = \pi \restriction_C^{-1}( \pi(A) ) = \pi^{-1}( \pi(A) ) \cap C = HA \cap C $ would be compact with $ A \subseteq HK $.

The latter is the proof suggested in the linked question. I find it hard to understand why $ \pi\restriction_C $ is an open map: suppose that $ W \subseteq C $ is open, then $ W = U \cap C $ where $ U \subseteq G $ is open. Now we need to show that $ \pi\restriction_C(W) $ is open, i.e. that $ \pi^{-1} {\pi \restriction_C(W)} = H \cdot W $ is open, which I don't understand why.

Community
  • 1
darkl
  • 352

1 Answers1

2

I’ll suppose that $H \setminus G $ is the space of left cosets of $G$ by $H$. In particular, if $H=\{e\}$ then we can identify the space $H\setminus G$ with $G$ and the map $\pi$ with the identity embedding. This already reveals an error in the claim you are trying to proof: in “1” the set $A$ should be closed, whereas in “2” it is not necessarily. But even this fix doesn’t work, that is “2” does not imply “1” even for closed $A$. As a counterexample consider $G=\Bbb R^2$, $H=\{0\}\times\Bbb R$, $K=[0,1]\times\{0\}$, and $A=\{(x,1/x):0<x\le 1\}$. Then $A$ is closed in $G$ but $\pi(A)=(0,1]$ is not closed in $H\setminus G$. So I propose to assume additionally that $\pi(A)$ is closed in $H\setminus G$. Then you already have the implication $2\Rightarrow 1$. The opposite implication asking for the lifting of $\pi(A)$ to a compact set $K$ in $G$, looks a complicated task and probably can be proved only in particular cases using powerful selection theorems (because the set $C$ which you consider may be descriptively very bad), for which you can ask at MathOverflow.

Alex Ravsky
  • 106,166