The length is bounded

Question

Let $\Sigma$ be a surface of finite type. Let $\mathcal{S}$ be the set of non-trivial isotopy classes of simple closed curves on $\Sigma$. One denotes by $l_x(\alpha)$ the infimal length of curves in the class of $\alpha$ in the metric $x$. The metric $x$ can be considered to be a point in the Teichmüller space $\mathcal{T}$ of $\Sigma$ and hence a hyperbolic metric, the length will be realized on a closed geodesic. Thurston introduced the following asymmetric metric on $\mathcal{T}$

$$ L(x, y)=\log \sup _{\alpha \in \mathcal{S}} \frac{l_y(\alpha)}{l_x(\alpha)} $$

We consider a (Hölder) continuous map $\omega \mapsto f(\omega)$ where $f(\omega)$ are homeomorphisms of $\Sigma$ (or more generally semi-contractions of $\mathcal{T}$ ). You can think that $f:\Omega \to \text{Homes}^+ (\Sigma)$, where $\Omega$ is a compact metric space.

Fix a base point $x_0 \in \mathcal{T}$, is is true that there is $0<C<1$ such that for every $\omega$ $$l_{f(\omega) x_0}(\alpha) \geq C l_{x_0}(\alpha) \quad \forall \alpha \in \mathcal{S}?$$

I don't expect that to be true in the above general concept, but I was wondering if it could be true under some minimal conditions (or generic conditions).

I was trying to use this result that said \begin{aligned} &L\left(x_0, y\right)=\log \sup _{\alpha \in \mathcal{S}} \frac{l_y(\alpha)}{l_{x_0}(\alpha)} \asymp \log \max _{\alpha \in \mu} \frac{l_y(\alpha)}{l_{x_0}(\alpha)}\\ &\text { up to an additive error. } \end{aligned}

Answer 1

I am not sure how well you know the Teichmuller theory, but the basic thing to understand is that $\mathcal T$ is not the space of hyperbolic metrics on $\Sigma$: It is the space of pairs $(\sigma, [\phi])$, where $\sigma$ is a hyperbolic metric on $\Sigma$ and $[\phi]$ is the homotopy (same as isotopy) class of an orientation-preserving homeomorphism $\phi: \Sigma\to \Sigma$. A homeomorphism $h: \Sigma\to \Sigma$ acts on $\mathcal T$ by the precomposition $(\sigma, [\phi])\mapsto (\sigma, [\phi\circ h])$ (or $\phi\circ h^{-1}$, depending on whom you ask, but this does not matter for our purpose; this is the difference between left and right actions). Now, the basic topological fact is that if $\sigma$ has injectivity radius $\ge \epsilon>0$ (and such $\epsilon$ exists for each metric on a compact surface), and $d(h_1, h_2)<\epsilon$ for the uniform metric $d$ on $Homeo(\Sigma)$ defined by $\sigma$, then $[h_1]=[h_2]$. Thus, given any compact subset $K\subset Homeo(\Sigma)$, the image of $K$ in the mapping class group of $\Sigma$ is finite. Given this, it follows that the image of the composition of your map $f: \Omega\to Homeo(\Sigma)$ with the orbit map $x_0=(\sigma_0, Id)\mapsto (\sigma_0, [f(\omega)])$ is a finite subset of $\mathcal T$. From this, the existence of a constant $C$ you are asking about is immediate since every homotopy class of homeomorphisms $(\Sigma, \sigma_1)\to (\Sigma, \sigma_2)$ (where $\sigma_i$ are Riemannian metrics on $\Sigma$) has a bi-Lipschitz representative.