3
$\begingroup$

Let $\pi: M \to N$ be a smooth submersion from a bounded open subset of $\mathbb{R}^m$ onto $ N \subset \mathbb{R}^n$, $m \geq n$. Further, let $M$ be given a probability measure $\mu$. Then the map induces a probability measure $\nu$ on $N$. The goal is to compute the latter's density with respect to Lebesgue measure, $\frac{d\nu}{d\lambda}$.

Consider the derivative matrix of $\pi$, $d\pi$, of dimension $m \times n$. One can take the generalized Jacobian at $x \in M$ defined by \begin{equation} J_x:= \sqrt{\det (d\pi_x^t d\pi_x)}, \end{equation} where $d\pi_x^t$ denotes the transpose of the matrix $d \pi_x$. Note that $\pi$ being a submersion ensures that $J_x$ is nonzero, and that $\pi^{-1}(y)$ is always a smooth sub-manifold (see Guillemin and Pollack).

Intuitively, a large $J_x$ means a small neighborhood around $x$ will get mapped onto a large neighborhood around $\pi(x)$, hence the density $\frac{d \nu}{d \lambda}$ at $\pi(x)$ should be inversely related to $J_x$. Also larger $\frac{d\mu}{d\lambda}(x)$ obviously gives larger $\frac{d \mu}{d\lambda}$. This reasoning leads to the following conjecture:

\begin{equation} \frac{d \nu}{d\lambda}(y) = \int_{\pi^{-1}(y)} J_x^{-1} \frac{d\mu}{d\lambda}(x) \lambda_{m-n}(dx), \end{equation} where $\lambda_{m-n}$ is the Lebesgue-induced measure on the $(m-n)$-dimensional sub-manifold $\pi^{-1}(y)$, coinciding with the Hausdorff fractal measure.

My question is, is there any textbook/paper that deals with the "obvious" equality above? Or is it an easy consequence of standard results such as the change of variable formula?

$\endgroup$
1
$\begingroup$

Answering my own question: Corollary 1.4 of this online note seems to address exactly the problem: http://www3.nd.edu/~lnicolae/Coarea.pdf

As expected it is intimately related to the co-area formula (Theorem 1.3). The latter is proved using Fubini's theorem locally, partition of unity, and implicit function theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.