Quotient of semidirect product by one of its factors

Question

Consider for example Poincare' group $$ISO (m) = SO(m) \ltimes R^m$$

I would like to see a formal proof that the coset $ISO(m) / SO(m)$ is diffeomorphic (but not isomorphic, as I initially stated) to $R^m$

Answer 1

"Isomorphic" is the wrong word here, because $SO(m)$ is not normal in $ISO(m)$, and thus the quotient is not a group.

The best way to understand this quotient is in terms of homogeneous spaces. Definition: A homogeneous space is a smooth manifold $M$ equipped with a smooth transitive action of a Lie group $G$ on $M$. There are two fundamental theorems about homogeneous spaces, both of which are proved in my Introduction to Smooth Manifolds (2nd. ed., Theorems 21.17 and 21.18):

Homogeneous Space Construction Theorem: Let $G$ be a Lie group and let $H$ be a closed subgroup of $G$. The left coset space $G/H$ is a topological manifold of dimension equal to $\dim G - \dim H$, and has a unique smooth structure such that the quotient map $G\to G/H$ is a smooth submersion. The left action of $G$ on $H$ given by $g_1\centerdot (g_2H)=(g_1g_2)H$ turns $G/H$ into a homogeneous $G$-space.

Homogeneous Space Characterization Theorem: Let $G$ be a Lie group, let $M$ be a homogeneous $G$-space, and let $p$ be any point of $M$. The isotropy group $G_p$ is a closed subgroup of $G$, and the map $F\colon G/G_p\to M$ defined by $F(gG_p) = g\centerdot p$ is an equivariant diffeomorphism.

Note that $ISO(m)$ acts smoothly and transitively on $\mathbb R^n$ by rigid motions, and the isotropy group of the origin is $SO(m)$. The first theorem implies that $ISO(m)/SO(m)$ is a homogeneous $SO(m)$-space, and the second theorem implies that it is (equivariantly) diffeomorphic to $\mathbb R^n$.