Does incompressible in the smooth sense imply incompressible in the measure theoretic sense?

Question

A smooth map $T$ is called incompressible if it has zero divergence and a measurable map $T$ is incompressible if $T^{-1}(B) \subset B$ implies $\mu(T^{-1}(B))=\mu(B)$ where $\mu$ is some probability measure. Are these two versions of incompressibility equivalent in $R^n$?

Answer 1

These two notions are not equivalent, but there is a connection. To avoid confusion, I'll refer to the two cases as divergence-free vector fields and measure-preserving maps respectively.

As stated, any constant map $\mathbb{R}^n\to\mathbb{R}^n$ is divergence-free as a vector field, but clearly not measure preserving. It's easy to come up with lots of other linear counterexamples, some of which are invertible.

Instead, given a smooth vector field $V:\mathbb{R}^n\to\mathbb{R}^n$, we can (at least locally) define the flow of $V$ to be a family of maps $\Theta_V:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n$ which displaces each point along the integral curves of $V$. In other words, it is defined as the solution to the following initial value problem: $$ \frac{\partial}{\partial t}\Theta_V(t,x)=V(\Theta_V(t,x)),\ \ \ \Theta_V(0,x)=x $$ One can show with a bit of calculus that if $V$ is incompressible, then its flow $\Theta_V$ preserves the Lebesgue measure on $\mathbb{R}^n$ (i.e. each map $\Theta(t,\_):\mathbb{R}^n\to\mathbb{R}^n$ is measure preserving). This will not be the case for other measures on $\mathbb{R}^n$ (though some will give rise to a similar notion of divergence), and not every measure-preserving map is the flow of a vector field.

This notion generalizes beyond $\mathbb{R}^n$. For instance, a sufficiently nice measure on a smooth $n$-manifold $M$ can be described (locally, up to sign) by a nonvanishing differential $n$-form $\omega$. The divergence of a vector field can then be defined in terms of a Lie derivative of $\omega$. $$ \operatorname{div}(X)\omega=\mathcal{L}_X\omega $$ And once again we have that the flow of a divergence-free vector field is measure-preserving. This notion applies, for instance, to other measures on $\mathbb{R}^n$ with smooth nonvanishing density functions and to Riemannian manifolds and their canonical measures.