Question on Teichmuller metric as infimum of the dilatations

Question

I'm learning Teichmuller theory from the book "Primer on Mapping class group". There is something I can not understand, I hope someone could clarify it.

Let $\mathcal{T}_g$ be the Teichmuller space of closed Riemann surfaces of genus $g\ge 2$.
The authors say that, given two marked Riemann surfaces $(X_1,f_1:S_g\rightarrow X_1),(X_2,f_2:S_g\rightarrow X_2)\in \mathcal{T}_g$, their Teichmuller distance (which I write as $d_{\mathcal{T}}(X_1,X_2)$) is defined as the infimum of the dilatations of the quasi-conformal maps homotopic to the change of markings $f_2\circ f_1^{-1}:X_1\rightarrow X_2$. This fact suggest that the dilatation of the quasi-conformal maps should be calculated in the coordinates of the conformal atlases of $X_1$ and $X_2$. Then, by the two Teichmuller's theorems (existence and unicity of Teichmuller map) we get that this infimum is realized by the Teichmuller map between $X_1$ and $X_2$. The proof of the Teichmuller's unicity theorem relies on the fact that the Teichmuller map is affine in the natural coordinates of some quadratic differentials $q_1$ (on $X_1$) and $q_2$ (on $X_2$) and that its horizontal stretch factor is $\sqrt{K}$ and its vertical stretch factor is $1/\sqrt{K}$ ($K>1$) and so the dilatation is $K$.

So we get that $d_{\mathcal{T}}(X_1,X_2)$ equals the dilatation of the Teichmuller map (for the quadratic differentials $q_1$ and $q_2$) computed in the natural coordinates of $q_1$ and $q_2$.

This is my question: Why is the dilatation of the Teichmuller map computed in the natural coordinates of $q_1$ and $q_2$ and not in the coordinates of the conformal atlases of $X_1$ and $X_2$? Is it the same thing? If yes, why?

Thank you.

Answer 1

Let $K$ denote the coefficient of quasiconformality for maps between domains in the complex plane; it satisfies the following general inequality: $$ K(f\circ g)\le K(f) K(g). $$ Furthermore, a homeomorphism $f$ is conformal if and only if $K(f)=1$. It follows that if $f$ is conformal then $K(f\circ g)\le K(g)$ for any $g$. By composing with $f^{-1}$ we conclude that $K(f\circ g)=K(g)$ for every conformal $f$ and quasiconformal $g$. The same works for precompositions with conformal maps. From this, you conclude that for any map $f: X\to Y$ between Riemann surfaces, the coefficient of quasiconformality of $\varphi\circ f\circ \psi^{-1}$ is independent of the choice of conformal charts $\varphi, \psi$ on $X$ and $Y$. This staff will be covered in any book on Teichmuller theory.