Let \(X\) be a space; let \(G\) be a topological group. An action of \(G\) on \(X\) is a continuous map \(\alpha: G \times X \to X\) such that, denoting \(\alpha(g \times x)\) by \(g \cdot x\), one has:
\(e \cdot x = x\) for all \(x \in X\).
\(g_1 \cdot (g_2 \cdot x) = (g_1 \cdot g_2) \cdot x\) for all \(x \in X\) and \(g_1, g_2 \in G\).
Define \(x \sim g \cdot x\) 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 \(X/G\).
[Hint: See Exercise 13 of §26.]
Solution
Let \(p:X\to X/G\) be the orbit projection.
For each \(g\in G\), the map \(x\mapsto g\cdot x\) is a homeomorphism. Hence for every open \(U\subset X\), \[ p^{-1}(p(U))=G\cdot U=\bigcup_{g\in G}gU \] is open. Thus \(p\) is an open quotient map.
If \(C\subset X\) is closed, then \(G\cdot C\) is closed. Indeed, the map \[ \Phi:G\times X\to G\times X,\qquad \Phi(g,x)=(g,g\cdot x) \] is a homeomorphism, and \[ G\cdot C=\pi_X(\Phi(G\times C)). \] Since \(G\) is compact, the projection \(G\times X\to X\) is closed. Therefore \(G\cdot C\) is closed. Since \[ p^{-1}(p(C))=G\cdot C, \] the quotient criterion implies \(p(C)\) is closed. Thus \(p\) is closed. Its fibers are the orbits \(G\cdot x\), continuous images of compact \(G\), hence compact. Therefore \(p\) is a perfect map.
The preceding perfect-map exercise now gives immediately:
- if \(X\) is Hausdorff, then \(X/G\) is Hausdorff;
- if \(X\) is regular, then \(X/G\) is regular;
- if \(X\) is locally compact, then \(X/G\) is locally compact;
- if \(X\) is second countable, then \(X/G\) is second countable.
It remains only normality. Let \(A,B\subset X/G\) be disjoint closed sets. Their inverse images \(p^{-1}(A)\) and \(p^{-1}(B)\) are disjoint closed subsets of normal \(X\), so choose disjoint open sets \(U,V\subset X\) containing them. Since \(p\) is closed, the sets \[ U'= (X/G)\setminus p(X\setminus U),\qquad V'= (X/G)\setminus p(X\setminus V) \] are open. They contain \(A\) and \(B\), respectively, and satisfy \[ p^{-1}(U')\subset U,\qquad p^{-1}(V')\subset V. \] Hence \(U'\cap V'=\varnothing\). Thus \(X/G\) is normal.