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I have a question about the leaves of "Stable lamination" as defined in the paper "LAMINATIONS, TREES, AND IRREDUCIBLE AUTOMORPHISMS OF FREE GROUPS" and the periodic conjugacy classes.

More precisely, let suppose that we have an IWIP outer automorphism $\phi$ (irreducible and $\phi ^{k}$ irreducible for every $k$) of a free group $F_n$, we construct the stable lamination $\Lambda$, and we choose a leaf $l$ of $\Lambda$ in $H$-coordinates for some marked metric graph $H$. Then is it possible for $l$ to contain arbitrarily long segments representing periodic conjugacy classes?

Actually, my original question is if this can be happened the same time for the leaves of the stable and the unstable lamination in $H$ -coordinates. But if it can not be happened for neither of them, that's enough.

Thanks a lot in advance.

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No, this is not possible. The only way for an IWIP (or fully irreducible) outer automorphism $\phi$ to have a periodic conjugacy class is if $\phi$ is represented by a pseudo-Anosov homeomorphism $f : S \to S$ of a compact surface $S$ with connected boundary. The circle $\partial S$, and its iterates, represent the only periodic conjugacy classes. The unstable lamination $\Lambda$ is represented in the interior of $S$ as a geodesic lamination, with respect to any hyperbolic structure on $S$ with totally geodesic boundary. If there existed leaves $\ell$ of $\Lambda$ with arbitrarily long segments staying close to $\partial S$, then for a long enough such segment the geodesic $\ell$ would be forced to cross itself transversely, which is impossible in a geodesic lamination.

Edited to give another proof:

Here also is a way to see this directly without using surface theory. There exists a stable train track representative $f : G \to G$ of some power $\phi^k$ such that $f$ has a unique closed, indivisible Nielsen path (up to reversal of orientation). That Nielsen path, its reversal, and their iterates, represent the unique $\phi$-periodic conjugacy classes. But that Nielsen path has an illegal turn, and no leaf of $\Lambda$ has an illegal turn, so no leaf of $\Lambda$ can even go as much as once around a circuit representing a periodic conjugacy class.

That proves what you ask for in $G$, and the same follows in $H$ coordinates for any marked graph $H$ without much trouble, using the bounded cancellation lemma.

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  • $\begingroup$ Really thanks for the answer, but because I don't know a lot about the the Teichmuller space, is it possible to give me some proof (or hint) using Nielsen paths? $\endgroup$ – user75691 Jun 6 '14 at 17:40
  • $\begingroup$ It is my pleasure. $\endgroup$ – Lee Mosher Jun 6 '14 at 17:52

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