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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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Is there some hypothesis you've left out? This is certainly falsifiable (as written) by considering a trivial group action in which every group element fixes every point. –  user83827 Dec 4 '11 at 16:13
    
At least $G$ needs to act faithfully. –  Qiaochu Yuan Dec 4 '11 at 18:11
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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. –  user83827 Dec 4 '11 at 18:37

1 Answer 1

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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please what are the most general conditions we put on $X$ or on $G$ or on the action of $G$ on $X$ to have this property.. thanks alot. –  palio Dec 4 '11 at 16:23
    
well as I said a first necessary condition would be that there are no invariant points. but I think this is far from sufficient. may I ask why you're asking yourself this question ? –  Glougloubarbaki Dec 4 '11 at 16:26
    
i have a set $F=\{(x,y)\in X \times X \;| \; orb(x)=orb(y)\}$ i want to simplify the expression of $F$ in terms of stabilizer subgroups –  palio Dec 4 '11 at 16:31
    
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)... –  Glougloubarbaki Dec 4 '11 at 17:05

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