Why is the coset space between a Lie group and its subgroup a manifold?

Question

If $G$ is a Lie group and $H$ its subgroup, then we can define an equivalence relation $g \sim g'$ where $g'=gh$ for $h \in H$ and $g,g' \in G$. In the book I'm learning from (Geometry, Topology and Physics by Mikio Nakahara) it is then stated that the coset space $G/H$ is a manifold (but not necessarily a Lie group) with:

$ \mathrm{dim} \; G/H = \mathrm{dim} \; G - \mathrm{dim} \; H \tag{1} $

where $\mathrm{dim} \; X$ denotes the dimensions of some manifold $X$.

Why is it true that $G/H$ is a manifold? And why does it have dimensions given by equation $(1)$?

Comments

This result not only holds in Lie groups but to all groups. If G is a group & H is its subgroup then G/H, i.e. set of cosets, forms a group only when H is normal in G. So this result follows from there. 

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The comment above is nonsense. The question isn't asking when G/H is a group, but why it is a manifold.

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@giri_iitg You have misinterpreted the question, you've just shown that G/H is a group when H is normal in G, but not that G/H is a manifold...  

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I would like to delete this question, because it is not the best question for the site, but I also would like for you to get an answer, so I will answer in this comment. The collection of cosets is a manifold because you can put coordinates on it, all that a manifold means is a space that you can parametrize by coordinates. If you have the set of cosets, you can choose representatives in some way for the coset and use the coordinates from the group on the coset space as your local coordinates. The number of independent coordinates at any one point you can figure out by knowing the tangent space--- it is the total number of coordinates for the full group, minus the coordinates for the subgroup at that point. You should consider a special example, for instance SU(3) with U(1) giving you a 2-d sphere. The question is difficult to answer because it is just a matter of definitions.

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@Ron Maimon I don't understand why you want this question to be closed. It *is* graduate-level.  

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@RonMaimon thanks for your reply. The book I'm reading doesn't always make clear whether the author makes a statement that is too difficult to prove, or too trivial to waist any extra words on it. I guess my question falls in the former category.

Also, I hope these kind of questions are allowed on this site, because I think they are one of the big "unique selling points" compared with physics SE, but maybe we should have this discussion somewhere else.

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@Hunter I agree they should be allowed here, it is certianly a graduate-level question. 

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I agree now that they should be allowed, it is difficult to find answers to not-so-deep questions. But common grad-student confusions get upvoted a lot, because there are a lot of confused grad-students in the world, and only a few people interested in specific detailed research questions, so my worry is that the grad-student confusions will suck attention away from research questions. The question is NOT completely trivial in generalization, because when you quotient a manifold by a group you can get orbifold points at fixed points of the transformation. But for the case of a continuous subgroup of a big lie group you can see that there are no fixed points, that the space always has a local tangent space, because there are always the same number of dimensions you throw away at each tangent space, the ones corresponding to the directions of multiplying by the subgroup. The problem is that the question is not exactly trivial, but it is tough to answer, because the difficulty is getting a working intuition for what it means to have a manifold, what it means to do the quotient construction and so on. This is also why you got a detailed math-y answer.

Answer 1

Well, the statement is true when $H$ is a Lie subgroup of $G$, not a simple subgroup. It is enough, to this end, that $H$ is a closed subgroup of $G$.

Concerning the proof, it is technical, but the general idea can be grasped from the following heuristic reasoning.  $G$ acts transitively on $M:= G/H$ by representation $G \ni g \mapsto f_g$ where $f_g : M \to M$ is defined as:

$f_g :  M \ni  [g'] \mapsto [g'g]$

You can easily see that $f_g= id$ if and only if $g \in H$. Therefore, if you fix $p \in M$ you can reach every other point $q$ by means of a suitable $f_g$ using  as degrees of freedom the $n = dim G$ coordinates determining $g$. This can be done completely disregarding the $m= dim H$ coordinates defining elements in $H$, since they give rise to trivial actions in $M$ which leave $p$ fixed. This means that, to define a point $q$ on $M$ (starting form a fixed on $p$) you need to fix $n-m$ coordinates. Thus $M$ has dimension $n-m = dim G - dim H$.

Comments

Thanks for your answer! I'm not (yet?) allowed to vote on your answer, but it does make sense (although I'll need to go through it in more detail tomorrow when I've time).

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@Hunter you are now allowed to vote.   

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@Dimension10 Thanks for letting me know, I have indeed voted.