## Notes of Brian Bowditch’s Cambridge lecture 20-04-2017

Bounding genera of singular surfaces

Ultimate motivation: understand the curve complex for non-compact surfaces. But today, only closed surfaces ${\Sigma}$ around.

1. Genus distance

The curve complex ${\mathcal{C}}$ of ${\Sigma}$ is hyperbolic, pseudo-Anosov diffeos act loxodromically, moving vertices at a linear speed.

Say ${\alpha\sim\beta}$ if there is a compact surface ${S}$ with ${\partial S=\alpha_0\cup \beta_0}$ and a map ${S\rightarrow\Sigma}$ mapping ${\alpha_0}$ to ${\alpha}$ and ${\beta_0}$ to ${\beta}$. Let ${A=A(\alpha)}$ denote the equivalence class of ${\alpha}$. There are two cases. Either ${\alpha\sim0}$, i.e. ${\alpha}$ is a separating curve. Or ${\alpha\not\sim 0}$.

Fact. ${A}$ is 3-dense in ${\mathcal{C}}$ (1-dense in the separating case).

Give, ${\alpha,\beta\in A}$, define ${\rho(\alpha,\beta)=}$ minimal genus of surface ${S}$ achieving ${\alpha\sim\beta}$. This is a metric. Morally, this is related to commutator length.

Question. Is this genus distance comparable to the distance in ${\mathcal{C}}$?

2. Pseudo-Anosov distorsion

Example. Let ${\phi}$ be a pseudo-Anosov diffeo such that ${h(\alpha)\sim\alpha}$ (such maps exist). Recall that ${d(\alpha,\phi^n\alpha)\sim n}$. I claim that ${\rho(\alpha,\phi^n\alpha)\sim n}$.

The proof requires 3-manifold topology. Let ${g=\rho(\alpha,\phi^n\alpha)}$, achieved by some surface ${S}$ with a map ${f:S\rightarrow\Sigma}$. If ${\gamma\subset S}$ is an essential simple closed curve, then ${f(\gamma}$ is not null homotopic in ${\Sigma}$ (otherwise, one could surge ${S}$ into a surface of lower genus). Let ${M_\phi}$ be the mapping torus. It is hyperbolic. Let ${M\rightarrow M_\phi}$ be the cyclic cover, diffeomorphic to ${\Sigma\times{\mathbb R}}$, with periodic geometry, let ${\psi}$ be the deck transformation. Let ${\alpha^*}$ be the closed geodesic in ${M}$ freely homotopic to ${\alpha}$. In ${M}$,

$\displaystyle \begin{array}{rcl} d(\alpha^*,\psi^n\alpha^*)\sim n. \end{array}$

Thurston-Bonahon realise the composed map ${F:S\rightarrow\Sigma\rightarrow M}$ by a 1-Lipschitz map from ${S}$ equipped with a hyperbolic structure with concave boundary (it amounts to triangulating ${S}$). Note that Area${(S)\leq 2\pi(2g+1)}$ is linear in ${g}$. Also, the injectivity radius of ${S}$ is bounded from below independently on ${n}$. So the diameter of ${S}$ is linear in ${g}$. Thus

$\displaystyle \begin{array}{rcl} d(\alpha^*,\psi^n\alpha^*)\leq C\,g, \end{array}$

and ${g\geq c\, n}$. However, ${c}$ depends on ${\phi}$ and ${\alpha}$. Note that diameter${(\alpha^*)}$ is bounded.

3. Result

Theorem 1 There is a constant ${L}$, depending only on ${\Sigma}$, such that for all ${\alpha,\beta\in A}$,

$\displaystyle \begin{array}{rcl} d(\alpha,\beta)\leq L\,\rho(\alpha,\beta). \end{array}$

The proof is similar. Let ${f:S\rightarrow\Sigma}$ minimize the genus ${g}$ of ${S}$.

Fact. There exists a complete hyperbolic 3-manifold ${M}$, homeomorphic to ${\Sigma\times{\mathbb R}}$, with arbitrarily short representatives ${\alpha^*}$ and ${\beta^*}$.

It can be taken quasi-Fuchsian. Again, there exists a 1-Lipschitz map ${F:S\rightarrow\Sigma\rightarrow M}$, where ${S}$ is hyperbolic with concave boundary, hence linear area. However, the injectivity radius of ${M}$ is not controlled. The thin part of ${S}$ is mapped to the thin part of ${M}$, a union of solid tori. Let us electrify tubes: change to a metric which is zero on the thin part. Then ${F}$ remains 1-Lipschitz. In the electrified metric, the diameter of ${S}$ is bounded by its area, so the distance between ${\alpha^*}$ and ${\beta^*}$ in the electrified metric is linear in ${g}$.

Fact. ${d(\alpha,\beta)\leq L\,d_{elec}(\alpha^*,\beta^*)}$.

This is a consequence of the Ending Lamination Theorem. One uses the quasi-isometric model of ${M}$. The constant ${L}$ depends only on ${\Sigma}$, but it is not effective.

Question (H. Wilton). Stable commutator length is more natural than commutator length. Would replacing ${\alpha}$ and ${\beta}$ by powers simplify the argument ? I do not see how it could help.