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Let $X$ be a topological space and $G$ a group acting on $X$. Do we have this property:

$\operatorname{orb}(x)=\operatorname{orb}(y)\iff\operatorname{stab}(x)\sim \operatorname{stab}(y)\qquad ?$ where $\operatorname{orb}(x)$ is the orbit of $x$, $\operatorname{stab}(x)=\{g\in G\mid gx=x\}$, and the symbol $\sim$ means conjugate to.

One way is obvious: if $\operatorname{orb}(x)=\operatorname{orb}(y)$, then $x=gy$ for some $g\in G$, so $\operatorname{stab}(x)=\operatorname{stab}(gy)=g\operatorname{stab}(y)g^{-1}.$ But the other way is not obvious to me: if $\operatorname{stab}(x)\sim \operatorname{stab}(y)$, then $\exists k\in G$ such that for all $g\in \operatorname{stab}(x)$, $ \exists h \in \operatorname{stab}(y)$ such that $g=khk^{-1}$. Now since $gx=x$ and $hy=y$, then $khk^{-1}x=x$ so $hk^{-1}x=k^{-1}x$ hence $h\in \operatorname{stab}(k^{-1}x)$, but I can't go any further... Thanks for your help.

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    This approach seems quite optimistic. Forgetting about the topology for a second, if you examine any action of say the integers on an uncountable set, there will be uncountably many orbits but only countably many conjugacy classes of stabilizers. More generally, you are trying to capture global dynamical behavior by a local, algebraic criterion, which is a massive loss of information.2011-12-04

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The answer is no.

Let the trivial group $G=\{Id\}$ act on $X$ containing at least two different points $x$ and $y$. Then $\operatorname{orb}(x) \neq \operatorname{orb}(y)$ but $\operatorname{Stab}(x)=\operatorname{Stab}(y)$.

More generally, invariant points (i.e. with stabilizer equal to $G$) are not in the same orbit but have the same stabilizer.

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    I doubt that it is the right way to go. even for "nice" actions that allow to define new objects, such as projective spaces (action of $\mathbb{R}$ by scalar multiplication) all nonzero elements have the same stabilizer {1} but need not be in the same orbit. And once again, this is a very simple and nice behaved action (and yet non trivial)...2011-12-04