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Handel's Theorem(Entropy and semi-conjugacy in dimension two, 1987): let $M$ denote a closed surface. Let $\vartheta$ be a pseudo-Anosov (orientation-presrv.) homeomorphism of $M$ and $g$ be an (orientation-presrv.) homeomorphism of $M$ isotopic to $\vartheta$. If the topological entropy of $\vartheta$ equals the topological entropy of $g$, then $\vartheta$ is semi-conjugate to $g$.

Is there any similar statement that holds for surfaces with boundary ?

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The definition of pseudo-Anosov homeomorphisms on a surface-with-boundary is not as well established as it is on a closed surface. Your question exhibits one reason for this: however you might want to formulate it, this semiconjugacy statement is false on a surface with boundary.

Here's why. Let $S$ be the surface, and list its boundary components as $\partial S = b_1 \cup\cdots\cup b_K$. Let $S/\partial$ be the quotient surface obtained from $S$ by collapsing each $b_k$ to a point denoted $p_k$, and let $P = \{p_1,...,p_K\}$. Whatever it might mean for $g : S \to S$ to be pseudo-Anosov, let us agree that one consequence is that the induced homeomorphism-of-pairs $g / \partial : (S/\partial,P) \to (S/\partial,P)$ satisfies the definition of a pseudo-Anosov homeomorphism on a closed surface, except that a point of $P$ is allowed to have a one-pronged singularity in its unstable measured foliation (and, thus also, in its stable measured foliation). In the setting of the mapping class group of $S/\partial$ relative to $P$, Handel's theorem is true: if $g / \partial$ is pseudo-Anosov, and if $\vartheta : (S/\partial,P) \to (S/\partial,P)$ is isotopic to $g / \partial$ relative to $P$, and if $g$ and $\vartheta$ have the same topological entropy, then there is a semiconjugacy from $\vartheta$ to $g$.

But what then happens back in $S$ itself? The problem is that the restriction of $g$ to $\partial S$ does not contribute to the entropy, and anything goes there. To simplify matters, let me assume that $\partial S = b_1$ is connected. In the definition of pseudo-Anosov for a surface-with-boundary that is presented in [FLP] Exposé Eleven, the restriction $g \mid b_1$ has rational rotation number in $\mathbb{R} / \mathbb{Z}$, because $g \mid b_1$ has a finite invariant set, namely the set of singularities. However, one can isotope $g$ to $g'$, without raising the entropy, so that $g' \mid b_1$ has any rotation number whatsoever. But if two circle homeomorphisms have different rotation numbers, then there is no semiconjugacy in either direction.

Perhaps this can be repaired by imposing artificial conditions on $g \mid \partial S$ which enforce a particular topological behavior on (and near) $\partial S$...

Added: Regarding Handel's theorem in the case of 1-pronged singularities, Handel's proof uses global shadowing in the universal covering space, so that argument has to be tailored to this situation.

It's never been written down, I'm pretty sure, but here's roughly speaking what one might try. Take the singular Euclidean metric associated to the stable and unstable measured foliations on the surface $S/\partial$. Remove the set $P$; now the metric is incomplete. Lift to the universal cover $\widetilde{S/\partial \!- \!P}$, you get an incomplete metric on which the finite rank free group $F = \pi_1(S/\partial-P)$ acts by deck transformations. Take the metric completion. Let me denote that metric space $\widehat S$. The action of $F$ extends to $\widehat S$, although the action is no longer properly discontinuous, because each point added to make the metric completion has infinite cyclic stabilizer; the quotient $\widehat S / F$ is naturally identified with $S / \partial$. One can think of $S/\partial$ as a "developable 2-complex of groups" in the sense of Haefliger, where the points of $P$ are labelled by infinite cyclic groups, and $\widehat S$ is the "universal covering space" of this 2-complex of groups (this theory of complexes of groups is a higher dimensional generalization of the theory of graphs of groups). The global shadowing arguments can now be carried out in $\widehat S$, and the results can be transported back down to $S/\partial$ itself.

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  • $\begingroup$ Thank you very much for this answer! I understand now what can go wrong with boundary. Just an extra precision: you said Handel's Theorem holds for pseudo-Anosov homeo. relative to a nonempty set. Does this follow immediately from Handel's original proof ? $\endgroup$ May 24 '18 at 13:32
  • $\begingroup$ I don't think that Handel's original proof can be applied directly, instead one has to make the proof work in a broader setting. I added some more stuff about how one might try to do that. $\endgroup$
    – Lee Mosher
    May 24 '18 at 14:10
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This question is addressed in detail in Isotopy Stability of Dynamics on Surfaces, Contemp. Math., 246, 17-46, 1999 or arXiv:math/9904160 [math.DS].

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