Prove that hyperspherical coordinates are a diffeomorphism, derive Jacobian

The explicit form for the transformation into hyperspherical coordinates is

$$x_1 = r\sin\theta_1 \sin\theta_2 \dotsb \sin \theta_{n-1} \\ x_2 = r\sin\theta_1 \sin\theta_2 \dotsb \cos \theta_{n-1} \\ x_3 = r\sin\theta_1 \dotsb \cos \theta_{n-2}\\ \vdots \\ x_{n} = r \cos\theta_1$$

for $0 \leq \theta_i \leq \pi \;\;(1\leq i \leq n-2)$ and $0\leq \theta_{n-1} \leq 2\pi$. It has Jacobian $r^{n-1} \sin^{n-2}\theta_1 \sin^{n-3}\theta_2 \dotsb \sin{\theta_{n-2}}.$

I wonder if someone could provide me with a reference for an intuitive explanation as to why this indeed is a diffeomorphism from $\mathbb{R}^n\setminus\{0\} \to \mathbb{R}^n \setminus\{0\}$, and why this is the Jacobian. Or perhaps someone could indicate the idea of a proof. Thanks as always

I will sketch the proof for the Jacobian by induction on $n$, thus in a different fashion compared to What is the metric tensor on the n-sphere (hypersphere)? (which perhaps is smoother, though).

For $n=2$, we have the transformation law $A_2$ $$ x^1 = \rho\cos\phi\\ x^2 = \rho\sin\phi. $$ Hence $$ \frac{\partial(x^1,x^2)}{\partial(\rho,\phi)}= \left( \begin{array}[cc] \ \cos\phi & -\rho \sin\phi\\ \sin\phi & \rho \cos\phi \end{array} \right) $$ and the Jacobian is $J_2=\rho(\cos^2\phi+\sin^2\phi)=\rho$, i.e. $dx^1 dx^2=\rho\, d\rho d\phi$.

The idea of induction works as follows: for $n=3$, instead of using the transformation $A_{3}$, given by $$ x^1 = \rho \cos\phi \sin\theta\\ x^2 = \rho \sin\phi \sin\theta\\ x^3 = \rho \cos\theta, $$ directly, one uses the two combined transformations $A_{23}$, given by $$ z^1 = \rho \cos\phi\\ z^2 = \rho \sin\phi\\ \theta=\theta, $$ with Jacobian $J_2=\rho$, and, letting $|z|\equiv\sqrt{(z^1)^2+(z^2)^2}=\rho$, $A_{3z}$, given by $$ x^1 = z^1 \sin\theta\\ x^2 = z^2 \sin\theta\\ x^3 = |z| \cos\theta. $$ Now, $$ \frac{\partial(x^1,x^2,x^3)}{\partial(z^1,z^2,\theta)}= \left( \begin{array}[ccc] \ \sin\theta & 0 & z^1\cos\theta\\ 0 & \sin\theta & z^2\cos\theta\\ z^1|z|^{-1}\cos\theta & z^2 |z|^{-1}\cos\theta & -|z|\sin\theta \end{array} \right) $$ which gives, expanding with respect to the last line, $$ J_{3z}= z^1|z|^{-1}\cos\theta(z^1\cos\theta\sin\theta)+ z^2 |z|^{-1}\cos\theta(z^2\sin\theta\cos\theta)+ |z|\sin\theta \sin^2\theta\\ =|z|\sin\theta $$ so that, being $A_3=A_{3z}\circ A_{23}$, $$ J_3 = J_2 \cdot J_{3z} = \rho |z|\sin\theta = \rho^2 \sin\theta, $$ i.e. $dx^1dx^2dx^3 = \rho^2\sin\theta\, d\rho d\theta d\phi$.

Let $A_n$ be $$ x^1 = \rho \cos\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\ x^2 = \rho \sin\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\ x^3 = \rho \cos\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3} \sin\theta_{n-2}\\ \ldots\\ x^{n-1}=\rho \cos\theta_{n-3}\sin\theta_{n-2}\\ x^n = \rho \cos\theta_{n-2}; $$ the transformation itself is built recursively: at the $n$th step, multiply the $n-1$ old coordinates by sine of the new angle and add a new coordinate $x^n$ as $\rho$ times the cosine of the new angle. In the same spirit of the above calculation, let $A_{(n-1)n}$ be $$ z^1 = \rho \cos\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\ z^2 = \rho \sin\phi \sin\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\ z^3 = \rho \cos\theta_1 \sin\theta_2 \ldots \sin\theta_{n-3}\\ \ldots\\ z^{n-1}=\rho \cos\theta_{n-3}\\ \theta_{n-2} = \theta_{n-2} $$ which, by induction, will have Jacobian $J_{n-1} = \rho^{n-2}\sin\theta_{1}\sin^2\theta_{2}\ldots\sin^{n-3}\theta_{n-3}$. Let $A_{nz}$ be $$ x^1 = z^1 \sin\theta_{n-2}\\ x^2 = z^2 \sin\theta_{n-2}\\ \ldots\\ x^{n-1} = z^{n-1} \sin\theta_{n-2}\\ x^n = |z| \cos\theta_{n-2}. $$ The Jacobi matrix reads $$ \frac{\partial(x^1,x^2,\ldots,x^{n-1},x^n)}{\partial(z^1,z^2,\ldots,z^{n-1},\theta_{n-2})} = \left( \begin{array}[ccccc] \ \sin\theta_{n-2} & 0 & \ldots & 0 & z^1\cos\theta_{n-2}\\ 0 & \sin\theta_{n-2} & \ldots & 0 & z^2\cos\theta_{n-2}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \ldots & \sin\theta_{n-2} & z^{n-1}\cos\theta_{n-2}\\ z^1|z|^{-1}\cos\theta_{n-2} & z^2|z|^{-1}\cos\theta_{n-2} & \ldots & z^{n-1}|z|^{-1} & -|z|\sin\theta_{n-2} \end{array} \right), $$

where $$|z|\equiv \sqrt{\sum_{k=1}^{n-1}(z^k)^2 }=\rho.$$ Again, the Jacobian for $A_{nz}$ is easily expressed expanding with respect to the last row $$ J_{nz} = z^1|z|^{-1}\cos\theta_{n-2} z^1\cos\theta_{n-2} \sin^{n-2}\theta_{n-2}\\ + z^2 |z|^{-1}\cos\theta_{n-2} z^2\cos\theta_{n-2}\sin^{n-2}\theta_{n-2}\\ + \ldots\\ + z^{n-1}|z|^{-1}\cos\theta_{n-2} z^{n-1}\cos\theta_{n-2}\sin^{n-2}\theta_{n-2}\\ + |z| \sin^2\theta_{n-2}\sin^{n-2}\theta_{n-2}\\ = |z| \sin^{n-2}\theta_{n-2}. $$ Finally $A_{n} = A_{nz}\circ A_{(n-1)n}$ and hence $$ J_n = J_{n-1}\cdot J_{nz} = \left( \rho^{n-2} \prod_{k=0}^{n-3}\sin^k\theta_{k}\right) |z| \sin^{n-2}\theta_{n-2}\\ =\rho^{n-1} \prod_{k=0}^{n-2} \sin^k\theta_k, $$ where, for notational convenience $\theta_0\equiv \pi/2$. This proves the Jacobian formula for any $n$.

If I recall correctly, any transformation with nonsingular Jacobi matrix gives rise to a local diffeomorphism, therefore our derivation above proves that this change of coordinates is good, except at the “north poles” and “south poles” $\theta_k =0,\pi$. To see that this indeed covers the whole $\mathbb{R}^n$, with the exception of such singularities, one can work out explicitly the inversion formulas, expressing the hyperspherical coordinates in terms of $x^1,\ldots,x^n$. This can be done, again, recursively and I think it is nicely given on https://en.wikipedia.org/wiki/N-sphere.

I hope it helps!

This Jacobian is calculated on the first pages of the book "Linear equations of mathematical physics" written by S.G.Mikhlin.