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Is it true that $d(r\omega)= r^2 dr\,d\sigma(\omega)$ where $\omega$ is a vector on $S^2$ and $\sigma(\omega)$ is a Lebesgue measure on $S^2$ ?
If yes how would you prove it rigorously ?
I don't know necessarily about differential forms but I don't trust Calculus loose technics either

NotaChoice
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    Your use of $d$ is totally non-standard. Please explain? If $x\in\Bbb R^3$ is a general vector in space, you expect $dx$ all of a sudden to be the volume element? – Ted Shifrin Jun 22 '21 at 00:40
  • Which is also a source of confusion to me, I am taking x to be $r\omega$ and thus integrating over $\mathbb{R}^3$ is transformed to integrating over $r$ in $\mathbb{R}_+$ and over $\omega\in S^2$ while replacing $dx$ by $r^2 dr d\sigma(\omega)$, the integral over $S^2$ is then like a 'projection' integral over its surface. I found this in a proof of the Radon inverse theorem. – NotaChoice Jun 22 '21 at 00:46
  • $f(x)=\frac{1}{(2\pi)^3}\int_{\mathbb{R}^3}\hat{f}(\xi)e^{i\xi \cdot x}d\xi = \frac{1}{(2\pi)^3}\int_{\omega\in S^2}\int_{\rho= 0}^\infty \hat{f}(\rho\omega)e^{i\rho\omega \cdot x}\rho^2 d\rho , d\sigma(\omega)$ , $\hat{f}$ being the Fourier transform of $f$ . – NotaChoice Jun 22 '21 at 00:48
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    If we let $S(r)\subset\Bbb{R}^n$ denote the sphere of radius $r$ centered at the origin, and $\sigma_r$ the induced surface measure, then for any (Lebesgue) measurable $f:S(r)\to [0,\infty]$, we have $\int_{S(r)}f(\xi),d\sigma_r(\xi)= \int_{S(1)}f(r\omega) r^{n-1},d\sigma_1(\omega)$. I'm not sure if this addresses your concern – peek-a-boo Jun 22 '21 at 00:48
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    I suggest you forget your notation and just remember how to integrate in spherical coordinates. That's all that’s going on here. – Ted Shifrin Jun 22 '21 at 00:50
  • well in here it is important, since I want to prove the inversion formula of the Radon transform, Fourier transform involves integration over $\mathbb{R}^3$, the Radon transform involves two parameters $s, \omega$ hence I need the above equality specifically to be true. – NotaChoice Jun 22 '21 at 00:54
  • Thanks peek-a-boo, I think this solves it, but how to justify your equality? with change of variable $d\sigma_r(r\omega)= r^{n-1}d\sigma_1(\omega)$ ? sounds correct since the surface has $n-1$ dimension, but why ? – NotaChoice Jun 22 '21 at 00:58

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The proof is just parametrizing most of the sphere by polar coordinates and then understanding how determinants work. By polar coordinates parametrization, I shall mean a smooth (injective immersion) mapping $\alpha:U\to V_1$ where $U$ is some open set in $\Bbb{R}^{n-1}$, and $V_1$ is a certain open subset of the sphere $S_1$, such that $\sigma_1[S_1\setminus V_1]=0$. For example for the unit circle in the plane, the mapping $\theta\mapsto (\cos\theta,\sin\theta)$ where $0<\theta<2\pi$ covers most of the circle, except for a single point. For the case of unit sphere in $\Bbb{R}^3$, polar coordinates covers the whole sphere except for "half a meridian" (I'm not sure what the right terminology is, but hopefully you get the idea). In general, you can obtain a recursive formula for $\alpha$ which is a parametrization using $n-1$ angles, but that's not relevant for us.

Given the parametrization $\alpha$, we can clearly parametrize the sphere of radius $r$ by defining $\alpha_r(x):= r\cdot \alpha(x)$. So, letting $V_r= \alpha_r[U]$, we have \begin{align} \int_{S_r}f(\xi)\,d\sigma_r(\xi)&=\int_{V_r}f(\xi)\,d\sigma_r(\xi)\\ &:=\int_{U}f(\alpha_r(x))\cdot \sqrt{\left|\det \left\langle \frac{\partial \alpha_r}{\partial x^i}(x), \frac{\partial \alpha_r}{\partial x^j}(x)\right\rangle\right|}\, d\lambda_{n-1}(x)\\ &=\int_{U}f(r\cdot \alpha(x))\cdot\sqrt{\left|\det \left\langle \frac{\partial \alpha}{\partial x^i}(x), \frac{\partial \alpha}{\partial x^j}(x)\right\rangle\right|}r^{n-1}\,d\lambda_{n-1}(x)\\ &:=\int_{V_1}f(r\omega)r^{n-1}\,d\sigma_{1}(\omega)\\ &=\int_{S_1}f(r\omega)r^{n-1}\,d\sigma_1(\omega) \end{align} Here, I used $\lambda_{n-1}$ to denote the usual Lebesuge $(n-1)$-dimensional measure on the open subset $U\subset \Bbb{R}^{n-1}$.

If you're only interested in the $n=3$ case i.e 2-dimensional spheres in $\Bbb{R}^3$, then the explicit formula for the parametrization is $\alpha_r:(0,\frac{\pi}{2})\times (0,2\pi)\to \Bbb{R}^3$, \begin{align} \alpha_r(\theta,\phi)&=(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta) \end{align} If you calculate the determinant above, then you'll see that it equals $r^2\sin\theta$, and surely you must have seen this before.

peek-a-boo
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