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In my book on lie algebras, it is written the following stuff :

Let $G$ be a Lie group, $H$ a Lie subgroup of G.

$G/H$ is then a manifold.

Is it hard to prove ?

It is a little abstract for me : $G/H$ could not be "smooth" in my vision (we could have like a discrete set of elements, so we could'nt find open sets to define the charts). But it seems it's not the case.

But I don't have any idea to how to prove this.

(I'm a huge beginner in group theory and in Lie groups, I just started to learn this)

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  • $\begingroup$ What's $H$? ${}{}{}$ $\endgroup$
    – user99914
    Apr 6, 2017 at 17:55
  • $\begingroup$ Sorry, H is a lie subgroupe of G, I edit my message $\endgroup$
    – StarBucK
    Apr 6, 2017 at 17:55
  • $\begingroup$ What do you mean by "discrete set of elements"? For example if $G=S^1$ and $H$ is any finite subgroup of $G$ then $G/H\simeq S^1$. Another example, if $G_1, G_2$ are Lie groups then $G_1\times G_2$ is a Lie group, and $H=G_1 \times \{1\}$ is a normal Lie subgroup. And then $(G_1\times G_2)/H\simeq G_2$. $\endgroup$
    – freakish
    Apr 6, 2017 at 21:09

2 Answers 2

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It follows from the Quotient Manifold Theorem:

If $G$ is a Lie group acting smoothly, freely, and properly on a smooth manifold $M$, then the quotient space $M/G$ is a topological manifold with a unique smooth structure such that the quotient map $M\to M/G$ is a smooth submersion.

Note that if $H$ is a closed Lie subgroup of $G$ then $H$ acts smoothly, freely and properly on $G$. Being closed is necessary for the action to be proper, if it is not closed then the quotient is not even $T_1$. Finally the quotient space under this action is equal to the space of cosets. BTW This also implies that $G/H$ is always a manifold, even when $H$ is not normal.

The proof of the Quotient Manifold Theorem (which is not trivial at all) can be found for example in John M. Lee "Introduction to Smooth Manifolds".

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  • $\begingroup$ I think we need to assume that H is closed for the action to be proper? $\endgroup$
    – JKEG
    Dec 15, 2019 at 13:41
  • $\begingroup$ @JKEG well, yes. If $H$ is not closed then $G/H$ is never a manifold, it is not even a $T_1$ space. $\endgroup$
    – freakish
    Dec 15, 2019 at 15:58
  • $\begingroup$ So you need to say: Note that if “H is a closed Lie subgroup...” or is there a reason that is redundant? $\endgroup$
    – JKEG
    Dec 16, 2019 at 21:42
  • $\begingroup$ @JKEG yes, I've updated the answer. $\endgroup$
    – freakish
    Jan 10, 2020 at 12:00
  • $\begingroup$ Thanks - could you clarify $𝑇_1$ space implies? $\endgroup$ Aug 11, 2021 at 16:46
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If $H$ is a subgroup of $G$, not necessarily normal, we can form the set of left cosets, $G/H$ and we have the projection, $p: G \rightarrow G/H$, where $p(g) = gH$. We can give $G/H$ the quotient topology, where a subset $U\subseteq G/H$ is open iff $p^{-1}(U)$ is open in $G$. With this topology, p is continuous, but $G/H$ need not be Hausdorff. However, if $H$ is a closed subgroup, it is Hausdorff. This is the case for a Lie subgroup $H$, and here exists a unique structure of a $C^{\infty}$-manifold on $G/H$ which is compatible with the topological structure. This is not hard to show (but for the details it might be better to look into a book on Lie groups).

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  • $\begingroup$ Can you recommend a book? $\endgroup$
    – SM10
    Aug 12, 2020 at 11:34

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