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Let $G$ be a compact Hausdorff group. Let $X$ be a Hausdorff space. Suppose $G$ acts continuously on $X$. Is the orbit space $X/G$ Hausdorff? If not, I would like to know an counter-example.

Remark As my answer to this question shows, if $X$ is a locally compact Hausdorff space, $X/G$ is Hausdorff.

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Makoto Kato
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    This should be useful: http://math.stackexchange.com/questions/50044/quotient-space-of-hausdorff-space – Rankeya Nov 22 '12 at 06:19
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    Yes. This follows from lemma 9 of your answer: the projection $X \to X/G$ is open and the orbit equivalence relation $\Gamma$ is closed in $X \times X$ because $G$ is compact and $X$ is Hausdorff. – commenter Nov 22 '12 at 08:24
  • @commenter Could you explain why $\Gamma$ is closed? – Makoto Kato Nov 22 '12 at 08:35

2 Answers2

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In view of lemma 9 of your answer it suffices to prove that the orbit equivalence relation $\Gamma$ is closed:

Let $G$ be a compact group acting continuously by homeomorphisms on a Hausdorff space $X$. Then the orbit equivalence relation $\Gamma \subset X \times X$ is closed.

Suppose $(x_i, y_i) \to (x,y)$ is a convergent net in $X \times X$ with $(x_i,y_i) \in \Gamma$. Then $x_i = g_i y_i$ for some net $g_i \in G$. Since $G$ is compact, there is a subnet $g_j$ which converges, say $g_j \to g$. Since $y_j \to y$ and $g_j \to g$ we have $x_j = g_j y_j \to gy$. But by assumption $x_j \to x$, so $x = gy$ because $X$ is Hausdorff and hence $(x,y) \in \Gamma$.

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  • Am I wrong or in order to deduce that $x_j=g_jy_j\rightarrow gy$ you also need that $X$ and $G$ are second-countable? – W4cc0 Dec 23 '18 at 14:50
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This is part of an exercise of Munkres' Topology, Exercise 8, Section 31.

Let $X$ be a space; let $G$ be a topological group. An action of $G$ on $X$ is a continuous map $\alpha : G \times X \rightarrow X$ such that, denoting $\alpha(g \times x)$ by $g \cdot x$, one has:

  1. $e \cdot x = x$ for all $x \in X$.
  2. $g_1 \cdot (g_2 \cdot x) = (g_1 \cdot g_2 ) \cdot x$ for all $x \in X$ and $g_1 , g_2 \in G$.

Define $x ∼ g \cdot x$ for all $x$ and $g$; the resulting quotient space is denoted $X/G$ and called the orbit space of the action $\alpha$.

Theorem. Let $G$ be a compact topological group; let $X$ be a topological space; let $\alpha$ be an action of $G$ on $X$. If $X$ is Hausdorff, or regular, or normal, or locally compact, or second-countable, so is $X/G$.

Note that in Munkres, a topological group is $T_1$ by definition and can be proved to be Hausdorff. (Exercise 7(b) of Supplementary Exercises of Chapter 2.)

Very kindly, Munkres gave us another exercise right before this one as follows. This exercise is fairly easy. So I will omit its proof and take it as given.

Let $p : X \rightarrow Y$ be a closed continuous surjective map such that $p^{−1} ( \{ y \} )$ is compact for each $y \in Y$. (Such a map is called a perfect map.)

  1. Show that if $X$ is Hausdorff, then so is $Y$.
  2. Show that if $X$ is regular, then so is $Y$.
  3. Show that if $X$ is locally compact, then so is $Y$.
  4. Show that if $X$ is second-countable, then so is $Y$.

So now we only need to show the quotient map $p: X \rightarrow X/G$ is closed and $p^{−1} ( \{ y \} )$ is compact for each $y \in X/G$ closed. (Continuity and surjectivity are by definition true for a quotient map.)

Suppose $A$ is a closed subset of $X$, $p(A)$ is closed if and only if $G \cdot A$ is closed. The latter is true because $G$ is compact and $A$ is closed. So $p$ is a closed map.

Let $p(x) = y \in X/G$. $p^{−1} ( \{ y \} ) = \{ g \cdot x | g \in G \}$ is compact because $G$ is compact and $\alpha(\cdot, x): G \rightarrow X$ is continuous.

Now we have all the ingredients to reach the conclusion.

Petra Axolotl
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