# Transverse invariant measures to vector fields

Given a smooth, non-vanishing vector field on a compact manifold, when does the 1-dimensional foliation given by its integral curves admit a transverse invariant measure?

I've seen examples of higher-dimensional foliations not admitting transverse invariant measures, but I'd imagine the same question is much easier to address one way of the other in the one-dimensional case.

• Something that may work but I'm not sure about it enough to write an answer: Construct a Birkhoff section (likely disconnected) with a measure on it. Then the measure of a transversal is just defined to be the area of the parts of the Birkhoff section that the transversal first hits when we run it along the flow of the vector field. – Rohil Prasad May 22 at 0:34

As R W says, the answer is in Plante's paper, but the special case of $$1$$-dimensional foliations is actually simple, and well-known: let $$M$$ be the manifold, $$*\in M$$ be a basepoint, $$(\phi^t)$$ be the flow (assumed tangential to the boundary of $$M$$, if any). For every integer $$n\ge 0$$, let $$\mu_n$$ be the image in $$M$$ of the probability measure $$dt/(2n)$$ on the interval $$[-n,+n]$$ under the map $$F:R\to M:t\mapsto\phi^t(*)$$ Since the space of Borelian probability measures on $$M$$ is compact for the weak topology, the sequence $$(\mu_n)$$ has a subsequence weakly converging to a Borelian probability measure $$\mu$$ on $$M$$. Claim: $$\mu$$ is invariant by every $$\phi^t$$. Indeed, for a fixed $$t$$, the sequence of measures $$\phi^t_*\mu_n-\mu_n$$ goes to $$0$$ in the norm topology, since for any continuous real function $$f$$ on $$M$$, one has: $$\vert(\phi^t_*\mu_n-\mu_n)f\vert=\frac{1}{2n}\vert\int_{-n+t}^{n+t}f(\phi^s(*))ds-\int_{-n}^{+n}f(\phi^s(*))ds\vert\le\frac{t}{n}\Vert f\Vert_\infty$$ Finally, the $$R$$-invariant measure $$\mu$$ amounts to a transverse invariant measure $$\nu$$ such that locally (in every local flow box $$B\cong D^{n-1}\times I$$) one has $$\mu=\nu\otimes dt$$. The measure $$\nu(D)$$ of every small transverse disk $$D$$ is the limit of some subsequence of the sequence $$\frac{1}{2n}{\sharp(F^{-1}(D)\cap[-n,+n])}$$ In this construction, we have used the choice axiom (hidden in the compacity of the measures space); in particular, the construction of the subsequence is non-constructive.
• I understand the construction, but I might need this spelled out a bit more explicitly for me. This is certainly an invariant measure, but how does it induce a transverse invariant measure? i.e. given some small transversal $\tau$, what would the total measure of $\tau$ be? Would it just be the limit as $n \to \infty$ of $\frac{1}{2n+1}$ times the number of intersection points of the trajectory with $\tau$? – Rohil Prasad May 25 at 14:24