Let $F$ be a compact oriented surface with a foliation $\cal F$ with $k$-prong singularities only (or, if it helps, assume that $\cal F$ admits an invariant measure). Is it true then there exists a pseudo-Anosov $\phi$ such that for every loop or an arc (with endpoints in $\partial F$) $\gamma$, $\phi^n(\gamma)$ is transversal to $\cal F$ for large $n$?

Here is a rough argument why it is probably true: The $MCG(F)$ action on ${\cal PMF}(F)$ is minimal, cf. Fathi-Laudenbach-Poenaru, Thurston's Work on Surfaces, for closed $F$. (A reference for bounded $F$ would be useful although the original proof seems to work). This implies that the unstable foliations $\cal F^u_\phi$ of pseudo-Anosovs $\phi$ are dense in ${\cal PMF}(F)$. So it is reasonable to expect that there is $\phi$ with all angles between $\cal F^u_\phi$ and $\cal F$ greater than some $\alpha>0.$ (But I don't have a rigorous argument for that.) Now by the work of Thurston (cf. the book above), the angles between $\phi^n(\gamma)$ and $\cal F^u_\phi$ converge to $0$ as $n\to \infty.$ Consequently, $\phi^n(\gamma)$ is transverse to $\cal F$ for large $n$.

EDIT: To address the comment below, let us just assume that $\cal F$ admits an invariant measure of full support.

  • $\begingroup$ Its false for every $\cal F$ that has a Reeb component. en.wikipedia.org/wiki/Reeb_foliation#/media/… And even when $\cal F$ has a Reeb component it will always have an invariant transverse measure. $\endgroup$ – Lee Mosher Mar 6 '18 at 14:55
  • $\begingroup$ @LeeMosher: When you say $\cal F$ has an invariant transverse measure even with a Reeb component, do you mean a measure of full support? Wouldn't an arc connecting the two boundaries of the Reeb annulus intersect each leaf infinitely many times and, hence, have an infinite transverse length? $\endgroup$ – Adam Mar 6 '18 at 21:35
  • $\begingroup$ As you say, the support of the measure is disjoint from the interior of the Reeb component. $\endgroup$ – Lee Mosher Mar 7 '18 at 0:17
  • $\begingroup$ Speaking of that, is there some result asserting that every $\cal F$ has an invariant measure with partial, but "reasonably" large support? $\endgroup$ – Adam Mar 7 '18 at 2:30
  • $\begingroup$ I assume that the power $n$ is allowed to vary with the loop or arc $\gamma$? If we have that, and if the foliation $\mathcal{F}$ has a measure of full support, then we should win. $\endgroup$ – Sam Nead Mar 7 '18 at 8:36

Given a full support measured foliation $\mathcal F$ and given a pseudo-Anosov $\phi$, what you want will work as long as $\mathcal F^u_\phi$ can be isotoped to be transverse to $\mathcal F$.

The trouble is, this is not always possible. For general choice of $\mathcal F$ and $\phi$, there are a few things one can say.

First, one can say that $\mathcal F$ and $\mathcal F^u_\phi$ can be altered in their Whitehead equivalence classes so that they become transverse.

Second, measured foliations themselves are annoying because of Whitehead equivalence, but you can do away with that annoyance in several ways. The most standard way is to work with measured geodesic laminations: the measured geodesic laminations associated to $\mathcal F$ and $\mathcal F^u_\phi$ will always be transverse, unless they are equal.

Another somewhat less satisfactory way is to work with the "partial measured foliations" that are used in one or two places in FLP. Every measured foliation $\mathcal F$ on $S$ is represented by a partial measured foliation $\mathcal F'$ which is canonical up to isotopy (slice apart all saddle connections, producing a partial measured foliation which has only annulus components and arational components; and then normalize the singularities on the arational components, with no interior saddle connections and with all boundary singularities being 3-pronged). Then one can say that $\mathcal F'$ and $\mathcal F^u_\phi$ will be almost transverse after a homotopy of $\mathcal F^u_\phi$; nonetheless there may still be violations of transversality at singularities, and notice that I wrote "homotopy" instead of "isotopy", because nearly parallel leaf segments of $\mathcal F^u_\phi$ may be forced to coincide in order to achieve transversality with $\mathcal F$.

But to return to your actual question, given $\mathcal F$ does there exist $\phi$ so that $\mathcal F^u_\phi$ can be isotoped to be transverse to $\mathcal F$? The answer is still no, because one can construct a measured foliation $\mathcal F$ which is not transversely recurrent: there is a simple closed curve $\sigma$ which is a concatenation of saddle connections of $\mathcal F$ such that there does not exist a simple closed curve transverse to $\mathcal F$ and having nonempty intersection with $\sigma$. (The reference for this construction, I think, is the Penner-Harer book "Combinatorics of Train Tracks").

Here's the construction of $\mathcal F$. Choose an essential, separating simple closed curve $c \subset S$, breaking $S$ into a union of two essential subsurfaces $S=S_1 \cup S_2$ with $c = S_1 \cap S_2$. On the surface $S_i$, choose an arational measured foliation $\mathcal F_i$ which is orientable and hence transversely orientable (we are assuming $S$ is oriented). We may, at first, assume that $\mathcal F_i$ has canonical singularities: no interior saddle connections; all boundary singularities 3-pronged. But now alter $\mathcal F_i$ as follows. Pick a boundary saddle connection $\alpha_i$ for $\mathcal F_i$, and alter $\mathcal F_i$ by Whitehead equivalence, collapsing the arc $c - \alpha_i$ to a point $p_i$. Now the only boundary singularity of $\mathcal F_i$ is $p_i$. However, $p_1$ and $p_2$ might not be the same point, but if not then $c$ is a union of two saddle connections with endpoints $p_1,p_2$. Collapse one of those saddle connections to get the final example $\mathcal F$.

Now one can prove that there is no simple closed curve $\gamma$ transverse to $\mathcal F$ that has nontrivial intersection with $c$. If $\gamma$ existed, write it as a concatenation $$\gamma = \gamma_1 * ... * \gamma_{2K} $$ where each $\gamma_{2k+1}$ is contained in $S_1$ transverse to $\mathcal F_1$, and each $\gamma_{2k}$ is contained in $S_2$ transverse to $\mathcal F_2$. The curve $\gamma_1$ has endpoints on $c$ transverse to $\mathcal F$ at each endpoint. At the initial point the orientation on $\gamma_1$ points inwards on $c = \partial S_1$, and at the terminal point it points outward. However, that violates the fact that $\mathcal F_1$ is transversely oriented.

  • $\begingroup$ Many thanks for this very comprehensive answer! A quick follow-up: Why $\cal F$ being not trasversely recurrent implies that no $\cal F_\phi^u$ is transversal to $\cal F$? $\endgroup$ – Adam Mar 14 '18 at 14:03
  • $\begingroup$ Because if $\mathcal{F}^u_\phi$ were transverse to $\mathcal{F}$ then a long leaf segment of $\mathcal{F}^u_\phi$ which intersects every leaf of $\mathcal{F}$ could be approximated by a closed curve transverse to and intersecting every leaf of $\mathcal F$, which directly violates the property of transverse recurrence. $\endgroup$ – Lee Mosher Mar 14 '18 at 14:12

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.