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In studying topology, I encountered this problem:

Let $S$ be a topological space and let $G$ be a topological group acting continuously on $S$ (group action as $G \times S \to S$ map is continuous). Define a relation $\sim$ in the following way: $x \sim y \iff \exists g \in G, y = g.x$ and give $S/ \sim$ quotient topology. Prove that quotient map $\pi: S \to S/ \sim$ is open.

Basically, the proof consists in showing $\pi^{-1}(\pi(V)) = \cup_{g \in G} g.V$, for $V$ open in $S$.

After that, to gain some intuition I've tried to think of several good examples of this, and I was satisfied only with one example I thought of: $O(3)$ acting on $\mathbb R^3$.

Do you know any other good examples, possibly geometric in nature?

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  • $\begingroup$ Ooops, I'll fix that. $\endgroup$ – ante.ceperic Jul 31 '14 at 14:48
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    $\begingroup$ Similarly, the circle group acting on the plane. Or: the additive group $\mathbb R$ acting on the plane, by $a*(x,y)=(x+a,y)$ $\endgroup$ – Lubin Jul 31 '14 at 15:01
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Usually the quotient is written as $G \setminus S$ (or $S ~/ ~ G$ if $G$ acts on the right).

A simple example is the quotient of the $n$-sphere $S^n$ by the "antipodal" action of $\mathbb{Z}/2\mathbb{Z}$. The quotient identifies with the projective space $\mathbb{R} \mathbb{P}^n$. It follows that the projection $S^n \to \mathbb{R} \mathbb{P}^n$ is open. This makes it rather easy to construct a manifold structure on $\mathbb{R} \mathbb{P}^n$, for example.

Another typical example is the "translation" action of $\mathbb{Z}$ on $\mathbb{R}$. The quotient $\mathbb{R}/\mathbb{Z}$ identifies with the circle $S^1$.

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There are a lot of these. i.e. Let $G\subseteq GL_n(\Bbb R)$ such that all $g\in G$ restricted to a non-trivial subspace $V$ of $\Bbb R^n$ is an automorphism of $V$ (it should be easy to come up with a specific $G$). Then $G$ acts smoothly and non-transitively on $\Bbb R^n-\{0\}$ .

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  • $\begingroup$ @900sit-upsaday $\Bbb R^n$ is not affine over $\Bbb C^n$. What I said about irreducible components may be true over non-algebraically closed fields, but I couldn't find a source. $\endgroup$ – PVAL-inactive Aug 3 '14 at 15:49
  • $\begingroup$ @900sit-upsaday I thought about it a bit and found a proof that only utilizes algebraic geometry over $\Bbb C$, and updated that as my answer. $\endgroup$ – PVAL-inactive Aug 3 '14 at 16:32

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