Suppose $H\subset G$ is a subgroup of a topological group $G$, and $Y\subset X$ is a subspace of a topological space $X$. Suppose we are given a continuous group action $\rho : G\times X\rightarrow X$ on $X$, and suppose that $Y$ is $H$ stable, that is $h.y \in Y$ for all $h\in H$ and $y\in Y$. You can form the quotient spaces $H\setminus Y$ and $G \setminus X$, and there is a natural, continuous, in general neither injective nor surjective map $\theta : H\setminus Y\rightarrow G\setminus X$. I am looking for conditions that assure this is a homeomorphism.

You can show easily that $\theta$ is onto $\mathrm{iff}~Y$ intersects all orbits, and one to one $\mathrm{iff} ~ \forall y\in Y, H.y=G.y\cap Y$. So I'll suppose these two conditions.

Under what general conditions on $Y,~X,~H,~G$ and $\rho$ (like $Y$ being a closed subspace (or even compact), $H$ being a closed subgroup (or possibly compact) and/or $\rho$ being a proper group action) is $\theta$ a homeomorphism? All spaces $X$ I have in mind are Hausdorff, but not necessarily locally compact. Also, the groups $G$ I consider are Lie groups, but I am interested in weaker conditions too.

I have found an obvious set of conditions that ensure $\theta$ is a homeomorphism, by ensuring $G\setminus X$ to be Hausdorff, $H$ and $Y$ compact. Then $H\setminus Y$ is compact Hausdorff and $\theta$ must be closed.

My goal with this is to try to find easy proofs for separation of quotients (my immediate goal) by working with smaller spaces and smaller (whenever possible compact) subgroups. For instance, if you take $\mathrm{Fr}(d,X)$ to be the space of all $d$ frames of a finite dimensional vector space $X$, I could show that the quotient space under the natural action of $\mathrm{GL}(d)$ is the same as the one obtained from the orthonormal frames with the action of the orthogonal group $\mathrm{O}(d)$. This is of course a very minor achievement, but I have found it helpful in order to show that the natural action of $\mathrm{GL}(X)$ on the grassmannian of $d$ planes of $X$ is continuous, and it's helped me 'believe' in some proofs involving grassmannians and Stiefel manifolds.

Thank you for your time!

Olivier Bégassat
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  • Note (in case I'm not the only one who didn't know this notation): $G\setminus X$ is another notation for $X/G$, the space of orbits of the group action, i.e. the quotient space of $X$ with respect to the equivalence relation induced by the group action (see http://en.wikipedia.org/wiki/Group_action#Orbits_and_stabilizers). – joriki Jun 09 '11 at 04:59
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    @joriki: To explain a bit more (in case it isn't obvious): If $G$ acts on $X$ from the left then you can write $Gx$ for an orbit. Thus $G\backslash X$ consists of such sets. This notation is particularly convenient if you have two compatible actions: $G$ acts on $X$ from the left and $H$ from the right and $(gx)h = g(xh)$. Then $H$ acts on $G \backslash X$ from the right and $G$ acts on $X/H$ from the left in the obvious way. Think of a pair of subgroups $G$ and $H$ in a larger group $X$, for example. – t.b. Jun 09 '11 at 11:23

1 Answers1


I don't know if my answer will be satisfactory, but it is actually very easy.

A continuous bijection $\varphi$ is a homeomorphism iff it is open and/or closed (openness/closeness of $\varphi$ rephrases continuity of the inverse $\varphi^{-1}$). As written in the question if $Y$ is compact (enough if $H\setminus Y$ is) and $G\setminus X$ is Hausdorff, then $\theta$ is obviously closed.

On the other hand, when $Y$ is open, then it easilly follows that the map $\theta$ is open.

Proof. Denote the quotient map $Y\to H\setminus Y$ by $\pi^Y$ and similarly let $\pi^X:X\to G\setminus X$ be the other obvious projection. We want to check that for all open $U\subseteq H\setminus Y$ the image $\theta(U)$ is open in $G\setminus X$. Since $U=\pi^Y\left({(\pi^Y)}^{-1}(U)\right)$ and $\theta\circ\pi^Y = \pi^X$, we can write $\theta(U)=\pi^X\left({(\pi^Y)}^{-1}(U)\right)$ and it suffices to prove that $\pi^X$ is open.

Consider any open $V\subseteq X$. We need to check if $\pi^X(V)$ is open in the quotient, so by definition of the quotient topology if its inverse image via the quotient map is open in $X$. We have the equality $${(\pi^X)}^{-1}\Big(\pi^X(V)\Big)= G.V \overset{def.}{=} \{g.v\ |\ g\in G,\ v\in V \}$$ and $G.V$ is open as the sum of open sets of the form $g.V$. Thus $\pi^X$ is open and so is $\theta$, which ends the proof.

Summing up: it is very common that a bijective $\theta$ is a homeo: it is enough to assume that $Y$ is compact (and $G\setminus X$ Hausdorff) or open (and nothing more).

From the above proof we can see that a sufficient and necessary condition for openness of $\theta$ (equivalently: for it to be a homeomorphism) is that for any open subset $W\subseteq Y$ (enough to check for W that are $H$-stable) the set $G.W$ is open in $X$.

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