I am reading an article where we have $$\iint_{S^2\times\mathbb{R}^3}f(w,v)\,dw\,dv$$ where $w\in S^2$ and $v\in\mathbb{R}^3$ are variable vectors. A change of variable was done $(w,v)\mapsto(p,q)$ where $$p=(v\cdot w)w$$ and $$q=v-p=v-(v\cdot w)w.$$
That is, $p$ is the projection of $v$ on $w$, and $q$ is the projection of $p$ on $w^\perp$, where $w^\perp$ is the rotation of $w$ in the plane $\{w,v\}$ by $\pi/2$. Then $p\in\mathbb{R}^3$ and $q\in\{p\}^\perp$.
They computed the Jacobian to be $$\,dw\,dv=\frac{2}{p^2\sin(p,p+q)}\,dp\,dq.$$ Does anyone have an idea how to compute this Jacobian explicitly?
The article can be found here
The part I am asking about is in page 12 proposition 5.
It has to be related to spherical coordinate, but since $\omega$ is a vector I don't understand how they compute the Jacobian.
I'll use the paper's notation. First, rearrange the claim as $dpdq=\tfrac12p^2\sin(\omega,\,V_\ast)d\omega dV_\ast$. We actually don't need a matrix of partial derivatives to prove this, any more than you need them to prove the more famous $dxdy=rdrd\theta$; in both cases, we can make a geometric argument, by comparing two coordinate systems' infinitesimal area elements.
In our example, the left-hand side is such an element. The right-hand side is half the area of an infinitesimal parallelogram with angle $(\omega,\,V_\ast)$ between sides of length $|p|d\omega,\,|p|dV_\ast$. Don't let this factor of $\tfrac12$ surprise you; it's a consequence of the degeneracy in $p=\pm|p|\omega$. The paper itself makes essentialy this argument, between Eqs. (36) & (37).