Inducing a surface area measure on $S^2$ from the Haar measure on $SO(3)$ I'm reading the book "Random Matrices: High Dimensional Phenomena" by G. Blower. There is an example that I've been struggled for a long time. For those who have access to the book, it's the Example 1.2.1 (iii) on p.11.
From my understanding, the author is trying to derive a surface area measure on $S^2$ by a pushforward map. It starts with the fact that every compact group has a unique probability Haar measure on it. So is $\operatorname{SO}(3)=\{U\in M_3(\mathbb{R}):U^TU=I, \det U=1 \}$. It then says that the map $\varphi:SO(3)\rightarrow S^2$ given by $\varphi(U)=Ue_3$, where $e_3=(0,0,1)^T$ is the third canonical basis, induces a measure $\sigma_2$ on $S^2$, normalized to be a probability measure, from the Haar measure on $\operatorname{SO}(3)$.
Also, it mentions that since $\varphi$ is not bijective: $Ve_3=Ue_3$ i.f.f. $V^{-1}U\in\operatorname{SO}(2)\subset\operatorname{SO}(3)$. Thus we have that $\operatorname{SO}(3)/\operatorname{SO}(2)\cong S^2$.
Anyway, it gives the formula for the measure in terms of colatitude $\theta$ and longtitude $\phi$: $\mathrm{d}\sigma_2=\frac{1}{4\pi}\sin\theta\mathrm{d}\theta\mathrm{d}\phi$, which coincides with what I learned from advanced calculus course.
Can anyone tell me how he obtain this area measure? I want to know the procedure (as clear as possible) for computing this $d\sigma$. Thanks!
 A: That's just the volume form of $S^2$ as a Riemannian submanifold of $\mathbb{R}^3$.  You can compute it by taking the parametrization $$\Phi: \theta, \phi \mapsto (\cos(\theta)\sin(\phi), \sin(\theta)\sin(\phi), \cos(\phi))$$ and evaluating $$dx\wedge dy\wedge dz(\Phi_\ast\partial_{\theta}, \Phi_\ast\partial_{\phi}, n(\theta, \phi))$$ where n is the outward unit normal at $\Phi(\theta, \phi)$.  Now since the Euclidean volume form is $SO(3,\mathbb{R})$ invariant, so is this spherical volume form.  In general, if $G$ is compact, the homogeneous space $G/K$ will have a unique $G$ invariant probability measure. The basic idea is that functions on $G/K$ can be lifted to functions on $G$ and then integrated so that a Fubini formula obtains: $$\int_{G/K}\int_Kf(gk)dkd\dot{g}=\int_Gf(g)dg$$ So we've computed the unique $SO(3, \mathbb{R})$-invariant measure on $SO(3)/SO(2)=S^2$, up to normalization. 
Maybe that answers your question?  
If you want more abstraction/generality, you can identify $T_{e_3}(S^2)$ with the subspace $$
 \left( \begin{array}{ccc}
0 & 0 & a \\
0 & 0 & b \\
-a & -b & 0 \end{array} \right)$$ of the Lie algebra of $SO(3,\mathbb{R})$ via the derivative of your map $\varphi$.  Then show that the dot product in $a$ and $b$ gives an $Ad(SO(2, \mathbb{R}))$ invariant inner product, which coincides with the Riemannian metric on $S^2$ induced from $\mathbb{R}^3$.  Then show that your volume form is the one associated to this Riemannian metric.   
A: Employing the Euler parametrization, a generic rotation matrix $R\in SO(3)$ can be expressed as
$$
R(\phi,\theta,\psi)=R_z(\phi)R_y(\theta)R_z(\psi)
$$
with $\phi,\psi\in[0,2\pi)$ while $\theta\in[0,\pi)$, and
$$
R_z(\phi)=\left(
\begin{matrix}
\cos\phi & -\sin\phi & 0\\
\sin\phi & \cos\phi &0\\
0 & 0 & 1
\end{matrix}
\right)
\,,\qquad
R_y(\theta)=\left(
\begin{matrix}
\cos\theta & 0 & \sin\theta\\
0 & 1 & 0\\
-\sin\theta &0 & \cos\theta \\
\end{matrix}
\right)\,.
$$
The Killing metric on $SO(3)$ then reads
$$
ds^2= \frac{1}{2}\mathrm{tr}\big(\dot R^T \dot R\big) dt^2= d\phi^2 + 2\cos\theta\, d\phi\, d\psi + d\theta^2 + d\psi^2\,,
$$
and the (normalized) Haar measure is given by its determinant as
$$
d\mu|_{SO(3)}= \frac{1}{8\pi^2}\,\sin\theta\, d\phi\, d\theta\, d\psi\,.
$$
Now, clearly,
$$
\mathbf n = R(\phi,\theta,\psi)(0,0,1)^T=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)^T\,.
$$
This is the standard embedding of the unit sphere $S^2$ in $\mathbb R^3$ in terms of polar and azimuthal coordinates.
Note that the $SO(2)$ rotation $R_z(\psi)$ in the definition of $R$ does not alter $\mathbf n$, so it parametrizes the kernel of the map between $SO(3)$ and $S^2$, effectively establishing $S^2\simeq SO(3)/SO(2)$.
Consequently, the induced measure is obtained by integrating out $\psi$ from the Haar measure, yielding
$$
d\mu|_{S^2}=\frac{1}{4\pi}\sin\theta\,d\theta d\phi\,.
$$
