# Minkowski as a Quotient Space

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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!

reshown Sep 6, 2017

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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.

answered Sep 2, 2017 by (904 points)

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?

@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.

@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!!

@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).

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