Canonically representing the monodromy of a hyperbolic manifold fibered over $S^1$

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.

• Do you know what "pseudo-Anosov" means? Mar 23, 2023 at 20:38
• @MoisheKohan Yes. If the answer is obvious to you, feel free to share. Mar 23, 2023 at 21:34
• Oh, I missed that you require smoothness. Then you can use a harmonic map or use conformal barycentric extension. Mar 24, 2023 at 1:05
• 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. Mar 24, 2023 at 15:48

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 $$\Sigma$$.
• 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? Mar 27, 2023 at 17:50