Jacobian of a transformation on a Manifold

Question

Let $a,b\geq{0}$. I am trying to find the Jacobian of the following transformation

\begin{equation} \begin{aligned} \mathbb{R}^3\times\mathbb{S}^2&\longmapsto\mathbb{R}^3\times\mathbb{R}^3\\ (v,w)&\longmapsto (p,q), \end{aligned} \end{equation} where \begin{equation} \begin{aligned} p&=\frac{v}{2}+\sqrt{av^2+b}\,\,w,\,\,\, and\\ q&=\frac{v}{2}-\sqrt{av^2+b}\,\,w\,. \end{aligned} \end{equation} Well, the fact that $w\in\mathbb{S}^2$ yields to a constraint satisfied by $p$ and $q$, as $$(p-q)^2-a(p+q)^2=b,$$ and thus the final and initial spaces have the same dimension.

I computed the Jacobian of the same transformation, but starting from $\mathbb{R}^3\times\mathbb{R}^3$, and it was $(av^2+b)^{\frac{3}{2}}$.

Could anyone please give me an idea on how to proceed? I'm very grateful.

Answer 1

You cannot take the Jacobian without considering the constraint. Otherwise the Jacobian at a point of a manifold would be the Jacobian of the surrounding flat space.

I'm not sure what can be assumed from the question, but here is an idea of how to proceed:

• Parametrize the sphere using your favorite mapping, say $w=(\cos\theta\cos\phi,\cos\theta\sin\phi,\sin\theta)$. Then $f$ becomes a function $\Omega\subset\mathbb{R}^5\to\mathbb{R}^6$. The five tangent vectors $f_v$, $f_\theta$, $f_\phi$ live in $\mathbb{R}^6$.
• Form the first fundamental form $5\times5$ matrix $g_{ij}=f_i\cdot f_j$, where $i,j$ range over the five vectors. Finally, the 'Jacobian' is $\sqrt{\det[g]}$.

Using the given function, I calculate (without double checking) $\sqrt{\det[g]}=\sqrt{\frac{1}{2}(av^2+b)(4a^2v^2+av^2+b)}\cos\theta$.