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Let $Y$ be a hyperbolic manifold that fibers over $S^1$, with fibration $\pi:Y \to S^1$ with fiber $\Sigma$. Thurston states that the monodromy $\phi:\Sigma \to \Sigma$ of this projection is then pseudo-Anosov. Moreover, we can write $Y = N/\mathbb{Z}$ where $N \to \mathbb{R}$ is a hyperbolic manifold fivering over $\mathbb{R}$ with fiber $\Sigma$ and $\mathbb{Z}$ acts on $N$ by isometries. Let $\Phi$ be the generator of the $\mathbb{Z}$-action.

Question. Is there a way of constructing a canonical smooth respresentative of the monodromy $\phi$ of the fiber bundle using the deck transformation $\Phi$?

For instance, you could imagine that there is a projection $N \to \Sigma$ (perhaps with geodesic fibers) where $\Phi$ permutes the fibers of the projection, and where $\Phi$ descends to a diffeomorphism of $\Sigma$ isotopic to $\phi$.

I'm particularly interested in the case of a fibered hyperbolic knot complement.

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    $\begingroup$ Do you know what "pseudo-Anosov" means? $\endgroup$ Mar 23, 2023 at 20:38
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    $\begingroup$ @MoisheKohan Yes. If the answer is obvious to you, feel free to share. $\endgroup$ Mar 23, 2023 at 21:34
  • $\begingroup$ Oh, I missed that you require smoothness. Then you can use a harmonic map or use conformal barycentric extension. $\endgroup$ Mar 24, 2023 at 1:05
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    $\begingroup$ What would "canonical" mean to you? It's a difficult word in this particular context: you may know that the pseudo-Anosov homeomorphism $\phi$ itself is canonical in a topological sense (it is unique up to conjugacy by a homeomorphism isotopic to the identity), but it generally fails to be smooth. $\endgroup$
    – Lee Mosher
    Mar 24, 2023 at 15:48

2 Answers 2

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Here is an answer of sorts; it is not completely canonical though. First of all, you have to pick a conformal or hyperbolic structure on the fiber $\Sigma$. This can be made almost canonical, since there is a unique $\phi$-invariant (real) Teichmuller geodesic. In any case, you get $\Sigma={\mathbb H}^2/\Gamma$, where $\Gamma$ is $\pi_1(\Sigma)$ acting on the hyperbolic plane (isometrically, freely, properly). The map $\phi$ induces a canonical $\Gamma$-equivariant homeomorphism $f: S^1\to S^1$ (of the boundary circle of the hyperbolic plane). Now, use the Douady-Earle extension to extend $f$ to a $\Gamma$-equivariant diffeomorphism $F$ of the hyperbolic plane. Lastly, project $F$ to a diffeomorphism $\phi$ of the surface $\Sigma$.

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  • $\begingroup$ This is an excellent answer. A few follow up questions (1) is the homeomorphism of the ideal boundary of $\mathbb{H}$ determined canonically by the mapping class? Or are you suggesting to first take the usual non-smooth representative of $\phi$ and then to extend its action on the boundary? $\endgroup$ Mar 27, 2023 at 17:50
  • $\begingroup$ (2) Is it possible to make the map area preserving? I'm not sure if this is a property of the Douady-Earle extension. I noticed this paper: arxiv.org/abs/0911.4124 which gives a different extension with this property. $\endgroup$ Mar 27, 2023 at 17:59
  • $\begingroup$ (3) Also any basic references on this material would be amazing :) I'm very much a novice in these things :) Thank you! $\endgroup$ Mar 27, 2023 at 18:00
  • $\begingroup$ Check "Primer on mapping class group" by Farb and Margalit. $\endgroup$ Mar 27, 2023 at 19:29
  • $\begingroup$ I think DE extension is area preserving. $\endgroup$ Mar 27, 2023 at 19:31
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The pseudo-Anosov homeomorphism is a diffeomorphism away from finitely many singular points, so in general will not be a smooth diffeomorphism. However, for certain fibered knots there are not singularities away from the punctures, and hence the map is a smooth diffeomorphism and unique up to smooth conjugacy. This works for the figure 8 knot and more generally certain 2-bridge knots (considered in this paper by Sakata ).

In general though there will be interior singularities of the pseudo-Anosov map of the fiber. Gerber and Katok proved that a pseudo-Anosov map is topologically conjugate to a smooth diffeomorphism. So this gives a kind of canonical smooth diffeomorphism realization, but only up to topological conjugacy (I do not know if diffeomorphism realizations are smoothly conjugate; there are also analytic realizations).

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