0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Rohil Prasad May 22 at 0:34
3
$\begingroup$

I think that the best reference for this question is still the (relatively) old paper by Plante Foliations with measure preserving holonomy Ann. of Math. (2) 102 (1975), no. 2, 327–361, although it is a bit of an overkill for one-dimensional foliations. For instance, by Theorem 4.1 holonomy invariant measures exist for any foliation with a subexponential leaf. There are many other ways, of course.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the reference! $\endgroup$ – Rohil Prasad May 22 at 3:05
1
$\begingroup$

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 $dt/(2n+1)$ on the interval $[-n,+n]$ under the map $t\mapsto\phi^t(*)$. 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 $\phi^t$ (which amounts to a transverse invariant measure). 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: $$\vert(\phi^t_*\mu_n-\mu_n)f\vert=\frac{1}{2n+1}\vert\int_{-n+t}^{n+t}f(\phi^t(s))ds-\int_{-n}^{+n}f(\phi^t(s))ds\vert\le\frac{2t}{2n+1}\Vert f\Vert_\infty$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $\tau$? $\endgroup$ – Rohil Prasad May 25 at 14:24

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.