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)

  • $\begingroup$ 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


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".

  • $\begingroup$ 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

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.