Is the spherical measure $\sigma^n$ a pushforward with the Lebesgue measure $\lambda^n$?

Question

Let $\sigma^n$ denote the spherical measure on $\mathbb S^n \subset \mathbb R^{n+1}$ and let $\lambda^n$ denotes the $n$-dimensional Lebesgue measure on $\mathbb R^n$.

We have the following relationships between $\sigma^n$ and $\lambda^{n+1}$ :

$$\sigma^n(A) = \dfrac{1}{\lambda^{n+1}(B(0,1))}\lambda^{n+1}(\{tx : t \in [0,1], x \in A\})$$

Since $\mathbb S^n$ is locally diffeomorphic to $\mathbb R^n$, I was wondering if there was some relationships between $\sigma^n$ and $\lambda^n$ ?

The problem is that we cannot cover $\mathbb S^n$ with only one chart. To avoid this complication, let's restrict ourselves to an open subset of $\mathbb S^n$ where covering with only one chart is possible, say the half-sphere $H^n = \{x \in \mathbb S^n : x_{n+1} > 0\}$ with the following chart $$y \mapsto \varphi(y) = (y_1,...,y_n,\sqrt{1-|y|^2})$$ between $B(0,1) \subset \mathbb R^n$ and $H^n$.

Intuitively, for a measurable set $A \subset H^n$, I would expect something like

$$\int_A f(x) d\sigma^n(x) = \int_{\varphi^{-1}(A)} f(\varphi(y)) |\det D_{\varphi}(x)| d\lambda^n$$

i.e. the restriction of $\sigma^n$ on $H^n$ is the pushforward measure $\varphi \# |\det D_{\varphi}(x)| \lambda^n$.

Can this be made rigorous ? Any reference is welcomed.

Answer 1

The surface measure $\sigma^n$ is the measure on $S^{n}$ induced by the pullback Riemannian metric $\iota^*g$, where $g=\sum_{i=1}^{n+1}dx^i\otimes dx^i$ is the standard Riemannian metric on $\Bbb{R}^{n+1}$ and $\iota:S^{n}\to\Bbb{R}^{n+1}$ is the canonical inclusion. I've written several answers about measures on manifolds, the most recent is Surface Measure and Gauss-Green Theorem. Here is an which talk about change of variables on manifolds Change of Variables in Integration on Manifolds (idk if this is particularly useful for you, but it may be a good read idk).

Anyway, in your case, you have the parametrization $\varphi:B_1(0)\subset\Bbb{R}^n\to H^n\subset\Bbb{R}^{n+1}$ (note by the way typically we use the term 'chart' for the inverse map because it goes from the manifold to the Cartesian space). The fact that the domain and target of $\varphi$ are of different dimensions should already be a warning to you that the Frechet derivative $D\varphi(x)$ is a linear map $\Bbb{R}^n\to\Bbb{R}^{n+1}$, so it makes no sense to take its determinant. Even if you meant the tangent mapping $T\varphi_x:T_x[B_1(0)]\cong \Bbb{R}^n\to T_{\varphi(x)}H^n=T_{\varphi(x)}S^n$, the domain and target are completely different vector spaces (yes they have the same dimension, but they're different spaces) so the determinant is not well-defined, (unless you specify bases for the domain and target and take the determinant of the corresponding matrix).

The correct expression is that we have to take the square root of the determinant of the Gram matrix. i.e for each $y\in B_1(0)$, consider the (symmetric) matrix with entries being the inner product of the tangent vectors associated to the parametrization $\varphi$, i.e the entries are $G_{ij}(y):=\left\langle \frac{\partial \varphi}{\partial y^i}(y),\frac{\partial \varphi}{\partial y^j}(y)\right\rangle$. Now, for any Lebesgue measurable set $A\subset H^n$, and any non-negative Lebesgue-measurable function $f:H^n\to [0,\infty]$, we have (by construction of $\sigma^n$) \begin{align} \int_Af(\eta)\,d\sigma^n(\eta)&=\int_{\varphi^{-1}(A)}f(\varphi(y))\sqrt{|\det (G(y))|}\, d\lambda^n(y). \end{align} So, yes $\sigma^n$ restricted to $H^n$ is the pushforward measure $\varphi_*\bigg(\sqrt{|\det G|}\,\,\lambda^n\bigg)$, where of course $\lambda^n$ is restricted to the unit ball $B_1(0)\subset\Bbb{R}^n$.

More generally (following the notation of my first linked answer), if $(M,g)$ is any pseudo-Riemannian (second-countable) manifold, and $\lambda_g$ is the associated Riemann-Lebesgue volume measure, then given any chart $\alpha:U\subset M\to \alpha[U]\subset\Bbb{R}^n$, we can define for each $p\in \alpha[U]$, the matrix \begin{align} G(p)&=\left[g_{\alpha^{-1}(p)}\left(\frac{\partial}{\partial x^i}\bigg|_{\alpha^{-1}(p)},\frac{\partial}{\partial x^j}\bigg|_{\alpha^{-1}(p)}\right)\right]_{i,j\in \{1,\dots, n\}} \end{align} (the matrix of inner products wrt $g$ of the coordinate-chart induced tangent vectors). Then, we have $\lambda_g= (\alpha^{-1})_*\bigg(\sqrt{|\det G|}\,\lambda^n\bigg)$, where $\lambda_g$ is restricted to $U$ and $\lambda^n$ to $\alpha[U]$ (again, this is just because of how $\lambda_g$ is defined).