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Suppose that $X$ is a topological space, and $T$ is a topological group which continuously acts on $X$ on the right. We call the pair $(X,T)$ a (right) transformation group. Now suppose that $\equiv$ is an “invariant” equivalence relation on $X$ in the sense that $\equiv$ is an arbitrary equivalence relation such that $x.t\equiv y.t$ whenever $x\equiv y$ for all $t\in T$ and for all $x,y\in X$.

It is easy to see that $(X/\equiv, T)$ is a transformation group with the action of $T$ defined by $[x].t=[x.t]$ where we have assumed $T$ to be given the discrete topology.

However, why is $(X/\equiv,T)$ a transformation group with the original topology of $T$ if the equivalence relation is open or if $X$ is compact and Hausdorff and $\equiv$ is a closed subset of $X\times X$? Even the theorems needed to see why this is true suffices for me to try proving it.

  • I don't think I will be able to answer the question but in order to clarify. Do you mean that, with the notations of your question, $(X/\equiv,T)$ is not a trsanformation group in general if we consider $T$ with its topology but if the relation is open then the pair is? – Dog_69 Mar 11 '19 at 19:21
  • @Dog_69 yes exactly –  Mar 11 '19 at 19:49
  • and do you have any example (counterexample) of that fact? I mean, a tranformstion group and an equivalent relation satisfying the hypothesis and such that the quiotient is not a transformation group. I say this in your related question I haven't seen why you need to consider the discrete topology. Furthermore, the group is well-behaved with respect to the equivalence relation, so it seems strange the necessity of some extra hypothesis, as openess, but... it may be, of course. – Dog_69 Mar 11 '19 at 20:00
  • @Dog_69 I don’t have any example. In fact I’m just new in this subject and the book is a classic apparently without many examples and without proofs of such assertions –  Mar 11 '19 at 20:51
  • I’ve started reading (reviewing and seriously learning) topology so that I don’t face such problems any more @Dog_69 –  Mar 11 '19 at 20:52
  • Which book are you reading? – Dog_69 Mar 11 '19 at 21:41
  • Here you have other references. I don't know any of them but the second is Pontrjagin. – Dog_69 Mar 11 '19 at 21:53
  • @Dog_69 thanks those books must clarify some vague points in my book –  Mar 11 '19 at 22:07
  • Yes, probably. But having a look at internet, I have seen that topological transformation groups seem a very specialised topic, and that not every book about topological groups cover them. For example Bourbaki devotes a chapter to study topological groups and there is no specific result about transformation groups. I suggest you to look on the internet for specific topological transformation group if you haven't done it yet. It appears some books that look fine. – Dog_69 Mar 11 '19 at 23:32
  • Previously I searched it and I found two relevant books, one by Tammo Dieck but it seemed so algebraic which is not of my interest and the other by Kobayashi which is of my interest but still advanced to me @Dog_69 –  Mar 11 '19 at 23:41
  • My current book is fine but my topological background is not strong so I have started reading topology seriously –  Mar 11 '19 at 23:43

1 Answers1

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I will use the notation that a subset of $X/\equiv$ is denoted by $[V]$ where the set $V$ is $\{x\in X\mid [x]\in[V]\}.$ I found this more convenient than referring to sets of equivalence classes. Also, to me "neighborhood of $x$" means "set whose interior contains $x$" - a neighborhood doesn't have to be open.

We need to show that for each $(x,t)\in X\times T$ and each neighborhood $[U]$ of $[x]\cdot t,$ there is a neighborhood $[V]$ of $[x]$ and a neighborhood $S$ of $t$ such that $[V]\cdot S\subseteq [U].$

Suppose the equivalence relation is open. Using continuity of the action map there are neighborhoods $V'$ of $x$ and $S$ of $t$ such that $V'\cdot S\subseteq U.$ Set $V=\bigcup_{v\in V'}\{x\mid x\equiv v\}.$ Then $V$ is open so $[V]$ is open. And $V\cdot S\subseteq U$ as required.

Suppose the equivalence relation is closed and $X$ is compact Hausdorff. The set $U\cdot t^{-1}$ is open, because $y\mapsto y\cdot t^{-1}$ is a homeomorphism. And $U\cdot t^{-1}$ is a union of $\equiv$ equivalence classes. So $[U\cdot t^{-1}]$ is open. $X/\equiv$ is compact Hausdorff - see Question about quotient of a compact Hausdorff space for example. Since compact Hausdorff spaces are regular, there is a closed neighborhood $[K]$ of $[x]$ with $[K]\subseteq [U]\cdot t^{-1} = [U\cdot t^{-1}].$ For each $k\in K$ there are open sets $K_k\subseteq X$ and $T_k\subseteq T$ with $1\in T_k$ and $K_k\cdot T_k\subseteq U\cdot t^{-1},$ by continuity of the action map. The set $K$ is a closed subset of $X,$ so it is compact. There is therefore a finite set $K'$ such that $\bigcup_{k\in K'} K_k$ covers $K.$ The set $T'=\bigcap_{k\in K}T_k$ is an open neighborhood of $1\in T$ and satisfies $K\cdot T'\subseteq U\cdot t^{-1},$ i.e. $[K]\cdot (T't)\subseteq [U].$

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