Example: the sphere as a homogeneous space

For $n\geq1$ , consider the standard action of $SO(n{+}1)$ on $\mathbb R^{n+1}$ , hence on the unit sphere

S^n=\{x\in\mathbb R^{n+1}:\|x\|=1\}.

This is a smooth Lie group action.

Transitivity and stabilizer

Fix the “north pole” $p=e_{n+1}=(0,\dots,0,1)$ . For any $x\in S^n$ , there exists $g\in SO(n{+}1)$ with $g\cdot p=x$ (choose an orthonormal basis sending $e_{n+1}$ to $x$ ), so the action is transitive.

The stabilizer of $p$ consists of rotations preserving the orthogonal complement $p^\perp\cong\mathbb R^n$ , hence

\mathrm{Stab}(p)\cong SO(n),

as an embedded Lie subgroup (see stabilizer).

Identification with a coset space

Define

\Phi: SO(n{+}1)/SO(n) \longrightarrow S^n,\qquad \Phi(g\,SO(n)):=g\cdot p.

This is well-defined because elements of $SO(n)$ fix $p$ . It is bijective by transitivity and the stabilizer description, and it is a diffeomorphism once we use the standard smooth structure on the coset space $SO(n{+}1)/SO(n)$ (which exists because $SO(n)$ is closed; compare the Closed Subgroup Theorem). Thus

S^n \cong SO(n{+}1)/SO(n)

as a homogeneous space.

Dimension check

Using $\dim SO(m)=\frac{m(m-1)}{2}$ ,

\dim\bigl(SO(n{+}1)/SO(n)\bigr) =\frac{(n{+}1)n}{2}-\frac{n(n-1)}{2}=n=\dim S^n,

consistent with the identification.

Special case. For $n=2$ , this reads $S^2\cong SO(3)/SO(2)$ , connecting directly to the $SO(3)$ example.