Does moving a small enough distance in Teichmüller space change the marking?

65 Views Asked by At

Let $S_{g}$ be a genus $g$ closed Riemann surface. The Teichmüller space $\mathcal{T}(S_{g})$ is the set of all pairs $(X,\phi)$ where $X$ is a Riemann surface of genus $g$ and $\phi : S_{g} \rightarrow X$ is a quasiconformal homeorphism, quotient by a certain equivalence relation. We say that two such pairs $(X_{1},\phi_{1})$ and $(X_{2},\phi_{2})$ are equivalent if there exists a conformal map $h : X_{1} \rightarrow X_{2} $ such that $h \circ \phi_{1}$ is homotopic to $\phi_{2}$. We consider the Teichmüller distance $d_{T}$ defined on this space, where $d_{T} ([(X_{1},\phi_{1})],[(X_{2},\phi_{2})])=\frac{1}{2}logK_{h}$ such that $h : X_{1} \rightarrow X_{2}$ is the unique quasiconformal map in the homotopy class of $\phi_{2} \circ \phi_{1}^{-1}$ having the lowest quasiconstant $K_{h}$. Let us consider the point $x=[(S_{g}, id_{S_{g}})] \in \mathcal{T}(S_{g})$. Does there exist a positive real $r$ such that given any point $y=[(X,\phi)] \in B(x;r)$ there exists a pair $(Y,\psi)$ where $Y$ is a Riemann surface with underlying set same as $S_{g}$, $\psi : S_{g} \rightarrow Y$ is a quasiconformal homeomorphism homotopic to $id_{S_{g}}$ and $(X,\phi)$ and $(Y,\psi)$ are in the same equivalence class, i.e. $[(X,\phi)]=[(Y,\psi)]$ ?

Basically, what I am asking is that whether it's true that if we move a small enough distance from a point in Teichmüller space to some other point then the final point has the same making as the starting point.

1

There are 1 best solutions below

0
On BEST ANSWER

Yes, this is true with $r=\infty$, in other words every point $[(X,\phi)] \in \mathcal T(S_g)$ has this property.

For the proof, one defines $Y$ by simply replacing the given Riemann surface structure on $S_g$ with the Riemann surface structure obtained from $X$ by pulling it back to $S_g$ via the homeomorphism $\phi : S_g \to X$, namely the unique conformal structure on the topological surface $S_g$ such that $\phi$ is conformal. The map $\psi = \phi^{-1} \phi$ is thus the identity map on $S_g$, and it is quasiconformal from $S_g$ to $Y$.

If you are not familiar with the pullback structure, it can be defined by choosing an atlas for $X$, and then for each coordinate chart $\chi : U \to \mathbb C$ in that axis (with $U \subset X$ open) one takes the corresponding coordinate chart in $S_g$ to be $\chi \circ \phi : \phi^{-1}(U) \to \mathbb C$.