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Let $G$ be a Lie group, $H < G$ a Lie subgroup. Let $X \subset G$. Under what sufficient conditions on $X$ is it true that the natural mapping $X/H \to G/H$ is a homeomorphism? And what conditions are necessary? (obvious: $X$ must intersect with every equivalence class)

In other words, I'm trying to learn how to compute quotients by subgroups actions. Sometimes we can take $X$ as a complete set of representatives. For example, if $G = \mathbb{R}^2$, $H = \mathbb{R}$, then we can take the graph of any continuous function. Or if $G = S^3$, $H = \{a + bi ~|~ a^2 + b^2 = 1\} \cong S^1$, then we can take as $X$ the subset of all elements of the form $a + bj + ck$ (thus $S^3/S^1 = S ^2$). Sometimes this is impossible: for example, if $G = S^1$, $H = C_n$, then the best we can take is an arc (in which it remains to identify more extreme points), and any complete set of representatives does not have some important good properties for this (unknown to me).

I'm not interested in complete sets of representatives , I just illustrated the question on them: I'm interested in any ways of calculating the quotient.

Update (01.02.2022, 22:34). In the second example, $X$ is not a complete sets of representatives: it is necessary to identify the opposite points of the sphere and identify the whole circle $a = 0$, after which the two-dimensional sphere $S^2$ will again be obtained.

Arshak Aivazian
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  • If $X$ meets every equivalence class, then this map $X/H\to G/H$ is a continuous bijection. Thus if $H$ is closed and $X/H$ is compact then it is a homeomorphism, right? In particular such thing happens whenever $G$ is compact and $X$ is closed (additionally to intersecting each equivalence class). – freakish Jan 31 '22 at 12:06
  • Another property is that if $X/H$ is a manifold of the same dimension as $G/H$, then this will also hold. Continuous bijections between such manifolds are homeomorphisms. My previous comment covers your compact examples, while this one covers your $G=\mathbb{R}^2$, $H=\mathbb{R}$ example. – freakish Jan 31 '22 at 12:11
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    Also, you need to be more precise about what $X/H$ means. Since $H$ does not necessarily act on $X$. In what I wrote I assume that $X/H$ is defined as the quotient of $X$ via $x\sim y$ iff $xH=yH$ relationship. – freakish Jan 31 '22 at 12:16
  • I meant the natural restriction of the equivalence relation from $G$ to $X$, of course, yes, exactly what you wrote. – Arshak Aivazian Feb 01 '22 at 05:23
  • @freakish Could you suggest why the closure of $H$ and the compactness of $X/H$ imply that our mapping is a homeomorphism? It's probably very simple, sorry. – Arshak Aivazian Feb 01 '22 at 06:11
  • If $H$ is closed then $G/H$ is Hausdorff. If $X/H$ is compact then we can apply the fact that a continuous bijection from compact to Hausdorff is a homeomorphism. – freakish Feb 01 '22 at 07:20
  • @freakish And indeed, thank you! – Arshak Aivazian Feb 01 '22 at 08:03
  • @freakish You can summarize your comments in an answer and I'll accept it because in my case $G$ is compact and the sufficient condition you suggested ("$X$ is closed") is incredibly broad and very easy to check. However, my question is of independent interest, so any additional information is welcome. – Arshak Aivazian Feb 01 '22 at 08:10

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Throught I will assume (just like you did) that $X$ meets every equivalence class. Next since $H$ does not necessarily act on $X$, I will define $X/H$ as the quotient via $x\sim y$ iff $xH=yH$ relationship. Then we have an induced continuous bijection

$$f:X/H\to G/H$$ $$f([x]):=xH$$

So when is that a homeomorphism? There are at least two important cases that cover all your examples:

  1. If $X/H$ is a manifold (which implicitly implies that $H$ is closed) of the same dimension as $G/H$ is, then this map is a homeomorphism. In general a continuous bijection between manifolds of the same dimension is a homeomorphism. Which is a consequence of the invariance of domain.
  2. If $H$ is closed then $G/H$ is Hausdorff. If additionaly $X/H$ is compact then we can apply the fact that a continuous bijection from compact to Hausdorff is a homeomorphism. In particular this holds always when $G$ is compact and both $X$ and $H$ are closed.

Both cover all your examples and I'm not aware of other interesting conditions for that to happen.

freakish
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