For $G$ a topological group acting continuously on a space $X$, when can we write the space (topologically) as the disjoint union of orbits i.e., when is $$ X\cong\coprod_{O\in X/G}O $$ Further, for $x\in O$, when is the canonical bijection a homeomorphism $$ O\cong G/\mathrm{Stab}\left(x\right) $$ I am looking for the conditions on $X$ and $G$, as well as on the action.
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As given in this answer, that doesn't seem to be the case: https://math.stackexchange.com/questions/47239/totally-disconnected-orbit-stabilizer-theorem – Chetan Vuppulury Mar 27 '21 at 18:05
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1Assume $G$ is Hausdorff, locally compact, and has a countable basis of open sets. If $X$ is also Hausdorff, and the orbit $G.x$ is also locally compact, then the natural continuous bijection $G/G_x \rightarrow G.x$ is a homeomorphism. – D_S Mar 27 '21 at 18:08
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In this case, can $X$ be written as the disjoint union of its orbits? – Chetan Vuppulury Mar 27 '21 at 18:13
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Usually not. That would only be the case when each orbit is both open and closed in $X$. In general, the orbits are neither open nor closed. – D_S Mar 27 '21 at 18:14
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@D_S it is ALWAYS a disjoint union of the orbits - this is a set-theoretic statement having nothing to do with topology. – Igor Rivin Mar 27 '21 at 18:17
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1@IgorRivin OP probably means to ask when that disjoint union has the disjoint union topology. https://en.wikipedia.org/wiki/Disjoint_union_(topology) – D_S Mar 27 '21 at 18:18
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@D_S yes, that is what I mean. So there must be some conditions on the action saying when the orbits are both open and closed. – Chetan Vuppulury Mar 27 '21 at 18:20
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open and closed -> each orbit is a union of connected components of $X.$ Not what one would describe as an. interesting situation. – Igor Rivin Mar 27 '21 at 18:32
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I suggest you read about the Slice Theorem, which is the best general result about the local structure of group actions https://encyclopediaofmath.org/wiki/Slice_theorem – Ruy Mar 27 '21 at 20:14