Minkowski as a Quotient Space

Question

I'm trying to understand how one can obtain $\mathbb{M}^4$ as the quotient space 

$$ISO(3,1)/SO(3,1),$$
and equivalently de Sitter and Anti-de Sitter as $SO(4,1)/SO(3,1)$ and $SO(3,2)/SO(3,1)$, respectively. I.e. we can see these manifolds by taking their symmetry Lie group and quotienting by the normal Lorentz subgroup $SO(3,1)$. 

This construction seems intuitive but I cannot find a decent reference where these steps are taken with a little more care. I would like to see $\mathbb{M}^4$ emerging from explicit group theoretical calculations (I'm not so familiar with group actions, orbits and all that).

Can anyone provide a nice and specific explanation without referencing to general theorems but starting from general grounds?

I'm not interested in deepening the construction coming from topology.. I know it is a general fact that one can construct a manifold  given a transitive  action of a Lie group, as the quotient of the Lie group by the stabilizer of a point.

Thank you!

Answer 1

A detailed discussion of the topic can be found in 5.2.2 of FRE P.G. - Gravity, a Geometrical Course. Volume 2. Black Holes, Cosmology and Introduction to Supergravity. You can find the book on the net.

Comments

Thank you! That reference is clear and easy to follow! I can now understand the constructions for the (A)dS manifolds very easily (that's what the book does basically). I'm still a bit confused regarding Minkowski as $ISO(3, 1)/SO(3, 1)$ though. The book talks about homogeneous group actions that preserve metrics of signature  $m+n$.. but Poincaré group is not homogeneous so how does this change things? 

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@ucci See definition 5.2.1 in the aforementioned book.

A Riemannian or pseudo-Riemannian manifold $M_g$ is said to be
homogeneous if it admits as an isometry the transitive action of a group $G$. A group
acts transitively if any point of the manifold can be reached from any other by means of the group action.

According to this definition, the Minkowski space is a homogeneous space because $G=ISO(3)$ acts transitively, and it is an isometry group.

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@Andrey Feldman Sorry I meant that  it is not homogeneous in the sense that Poincaré is sometimes referred to as the inhomogeneous Lorentz group because of translations.
Yes I get that general fact. I am struggling to adapt the Minkowski case to the construction of the given examples of $\mathbb{H}^\pm$ though. I mean take eq  (5.2.13), what is $ISO(3, 1)/SO(3, 1)$ there? Is it included as a subcase of those two expressions? If not, how to proceed on the same lines to get Minkowski? Sorry if I'm asking to many questions..
Thank you!!

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@ucci $ISO(3,1)$ is a group. It has $SO(3,1)$ as a subgroup. Then one just constructs a coset $ISO(3,1)/SO(3,1)$ by identifying points connected by the action of $SO(3,1)$ in complete analogy with (5.2.13).