These are my live-TeXed notes (reorganized according to my tastes) for the second part of the course Introduction to Riemann surfaces and Teichmuller theory by Professor Feng Luo, June 28 — July 14, at Math Science Center of Tsinghua University.

TopClassification of Riemann Surfaces

Surfaces (assumed to be connected and orientable) of finite type (i.e., with finitely generated fundamental group) have a complete topological classification due to Mobius ([1, Theorem 1.1]):

Theorem 1 (Topological classification of surfaces)
  • Any closed surface is homeomorphic to the connected sum of a sphere with $g\ge0$ tori, where $g$ is called the genus.
  • Any compact surface is obtained from a closed surface by removing $b\ge0$ disjoint disks, where $b$ is the number of boundary components.
  • Any surface is obtained from a compact surface by removing $p\ge0$ points from the interior, where $p$ is the number of punctures.

Among these, $S^2$, $\mathbb{D}^2$, $S^1\times\mathbb{R}$, $S^1\times S^1$ are the only cases with abelian fundamental group $\pi_1(X)$. However, each single topological class may contain various analytic structures.

Definition 1 Let $\Sigma$ be a topological surface. Denote by $\Mod(\Sigma)$ the set of biholomorphism classes of Riemann surfaces that are homeomorphic to $\Sigma$. This is called Riemann's moduli space.

We can restate the uniformization theorem as follows.

Theorem 2 (Uniformization Theorem) $$\Mod(S^2)=\{[\hat{\mathbb{C}}]\},\quad \Mod(\mathbb{D}^2)=\{[\mathbb{C}], [\mathbb{D}]\}$$

Let $X$ be a Riemann surface with nontrivial fundamental group. From general algebraic topology property, we know that there exists a simply connected surface $\tilde X$ and a covering map $p:\tilde X\rightarrow X$ such that $\Gamma=\pi_1(X)$ acts freely and discontinuously on $\tilde X$: for any $\gamma\in \Gamma$, $\gamma:\tilde X\rightarrow\tilde X$ is a homeomorphism and $p(x)=p(x')$ if and only if $x'{}=\gamma(x)$ for some $\gamma\in\Gamma$; for any $x\in\tilde X$ there exists a neighborhood $U$ of $x$ such that $\gamma(U)\cap U=\varnothing$ for any nontrivial $\gamma$. In particular, for any nontrivial $\gamma$, we have $\gamma(x)\ne x$ for any $x\in\tilde X$. $\tilde X$ is a Riemann surface such that $p$ is analytic. The Uniformization Theorem tells us that $\tilde X$ is either $\hat{\mathbb{C}}$, $\mathbb{C}$ or $\mathbb{D}$.

The main goal of this section is to prove the following classification of Riemann surfaces with simple topology.

Theorem 3 Suppose that $X$ is a Riemann surface such that its fundamental group $\pi_1(X)$ is abelian and nontrivial. Then
  1. $\tilde X\not\cong\hat{\mathbb{C}}$.
  2. If $\tilde X=\mathbb{C}$, then $X\cong \mathbb{C}^*$ or $\mathbb{C}/L$, where $L=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ is a lattice in $\mathbb{C}$.
  3. If $\tilde X=\mathbb{H}$, then $X\cong A_r=\{1<|z|<r\}$, the annulus of module $\log r$, where $1<r\le\infty$. Moreover, $A_{r_1}\cong A_{r_2}$ if and only if $r_1=r_2$.

To prove Theorem 3, we need the following two lemmas.

Lemma 1
  1. For any $g\in \Aut(\tilde X)$, we have $\tilde X/\Gamma\cong \tilde X/g\Gamma g^{-1}$.
  2. Any continuous map $f: \tilde X/\Gamma\rightarrow \tilde Y/\Gamma'$ can be lifted to $\tilde f:\tilde X\rightarrow \tilde Y$ such that $p_Y\circ \tilde f=f\circ p_X$, where $p_X$ and $p_Y$ are the universal covering maps. Furthermore, if $f$ is a homeomorphism then $\tilde f$ is so. Obviously, if $f$ is analytic then $\tilde f$ is so, hence if $f\in\Aut(X)$ then $\tilde f\in\Aut(\tilde X)$.
Proof (a) is direct. For (b), since $\tilde X$ is simply connected and $p_Y:\tilde Y\rightarrow Y$ is a covering map, we choose $\tilde f$ to be the lifting of $f\circ p_X$. Furthermore, suppose $f$ is a homeomorphism with $f^{-1}=g$. Taking $a\in X$, $b=f(a)$ and a lift $\tilde a\in \tilde X$, $\tilde b=\tilde f(\tilde a)$, we get a lifting $\tilde g: (\tilde Y,\tilde b)\rightarrow (\tilde X,\tilde a)$ similarly. Then $$p_X\circ\tilde g\circ\tilde f=g\circ P_Y\circ \tilde f= g\circ f\circ p_X=p_X,$$ so $\tilde g\circ \tilde f$ and $\mathrm{id}_{\tilde X}$ are both the liftings of $\Id_X$ sending $\tilde a$ to $\tilde a$. The uniqueness of the lifting ensures that $\tilde g,\tilde f$ are inverses.
Remark 1 Let $L=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ be a lattice in $\mathbb{C}$ and $g(z)=\lambda z\in \Aut(\mathbb{C})$ where $\lambda\in\mathbb{C}^*$, then $gLg^{-1}=\lambda L$. By lemma 1, we know $\mathbb{C}/L\cong\mathbb{C}/\lambda L$.
Lemma 2 Let $\gamma\in SL(2,\mathbb{R})$, then $\gamma(z)\ne z$ for any $z\in \mathbb{H}$ if and only if all eigenvalues of $\gamma$ are real.
Proof Since $\det \gamma=1$, the eigenvalues are of the form $\lambda,\lambda^{-1}$. The condition that the eigenvalues are real implies that $|a+d|\ge2$. The fixed points condition implies that $(d-a)^2+4bc\ge0$, namely $|a+d|\ge2$, too.

Now we are in a position to prove Theorem 3.

Proof (Proof of Theorem 3) (a) Since every Mobius transformation $\gamma$ has a fixed point in $\hat{\mathbb{C}}$, $\Gamma=\pi_1(X)$ cannot act on $\hat{\mathbb{C}}$ freely.

(b) Suppose $\tilde X=\mathbb{C}$. The automorphism $\gamma=az+b$ has no fixed points, hence $a=1$ and $b\ne0$, namely $\Gamma\subseteq (\mathbb{C},+)$. If $\Gamma=\mathbb{Z}(\gamma)\cong\mathbb{Z}$, then $X=\mathbb{C}/(z\sim z+b)\cong \mathbb{C}^*$ by $p:\mathbb{C}\rightarrow \mathbb{C}^*,\quad z\mapsto e^{2\pi i z/b}$. If $\Gamma=\mathbb{Z}(\gamma_1)\oplus\mathbb{Z}(\gamma_2)\cong\mathbb{Z}^2$ with $\gamma_i=z+\omega_i$, then topologically $X$ is $S^1\times S^1$. We claim that $\omega_1/\omega_2\not\in \mathbb{R}$. If $\omega_1$ is a rational multiple of $\omega_2$, then $n\omega_1+m\omega_2=0$ for $n,m\in \mathbb{Q}$, therefore $\gamma_1^n\gamma_2^m=\mathrm{id}$, which contradicts the fact that $\gamma_1$ and $\gamma_2$ are generators of $\Gamma$. If $\omega_1$ is an irrational multiple of $\omega_2$, then $\{n\omega_1+m\omega_2: n,m\in\mathbb{Z}\}$is dense in the line $\{a\omega_1:a\in\mathbb{R}\}$, so that $\Gamma(0)$ has limit points which contradicts the discontinuity. So $X\cong \mathbb{C}/L$ where $L$ is a lattice. A similar argument shows that $\Gamma$ cannot have more than two generators.

(c) Suppose $\tilde X=\mathbb{H}$.

  1. If $\Gamma=\mathbb{Z}(\gamma)\cong\mathbb{Z}$, with $\gamma\in SL(2,\mathbb{R})$, then every fixed point of $\gamma$ is inside $\mathbb{R}\cup\{\infty\}$ by freeness. After conjugating by some $g\in SL(2,\mathbb{R})$, we may assume that $\gamma$ has the form $\mathrm{diag}(\lambda, \lambda^{-1})$, where $\lambda\in\mathbb{R}^*$ (by Lemma 2) or $\gamma=\pm
1 & 1 \\
0 & 1
 \end{bmatrix}$. If $\gamma=\pm
1 & 1 \\
0 & 1
 \end{bmatrix}$, then $\gamma(z)=z+1$. So $X\cong \mathbb{H}/(z\sim z+1)\cong \{|z|>1\}\cong \mathbb{D}^*=\mathbb{D}-\{0\}$ by $$p:\mathbb{H}\rightarrow \{|z|>1\},\quad z\mapsto e^{-2\pi iz}.$$ If $\gamma=\pm\mathrm{diag}(\sqrt{\lambda},\sqrt{\lambda^{-1}})$, then $\gamma(z)=\lambda z$, where $\lambda\in\mathbb{R}_{>0}$. Then $X=\mathbb{H}/(z\sim \lambda z)$ is biholomorphic to $A_\mu$ where $|\ln \lambda \ln \mu|=2\pi^2$, by composing the logarithm map, the strip-scaling map and the exponential map $$p:\mathbb{H}\rightarrow A_\mu,\quad z\mapsto e^{-i\frac{\ln \mu}{\pi} \ln z}.$$
  2. If $\Gamma=\mathbb{Z}(\gamma_1)\oplus \mathbb{Z}(\gamma_2)\subseteq SL(2,\mathbb{R})$, then $\gamma_i(\mathrm{Fix}(\gamma_j))=\mathrm{Fix}(\gamma_j)$ since $\gamma_i\gamma_j=\gamma_j\gamma_i$. Up to conjugation, we may assume that $\gamma_1(z)=z+1$ or $\gamma_1(z)=\lambda z$ ($\lambda\in \mathbb{R}_{>0}-\{1\}$). If $\gamma_1(z)=z+1$, then $\mathrm{Fix}(\gamma_1)=\infty$ and $\gamma_2(z)=z+b$, so $\Gamma$ cannot act discontinuously; if $\gamma_1(z)=\lambda z$ , then $\mathrm{Fix}(\gamma_1)=\{0,\infty\}$ and $\gamma_2(z)=\mu z$ with $\ln\lambda/\ln\mu\in\mathbb{R}$. If $\ln \lambda/\ln \mu\in\mathbb{Q}$, then $\Gamma\cong\mathbb{Z}$; if $\ln \lambda/\ln \mu\not\in\mathbb{Q}$, then $\Gamma$ cannot act discontinuously.

We claim that no two $A_\mu$'s are not biholomorphically equivalent. If $\mu<\infty$ then $A_\mu\not\cong A_\infty=\mathbb{D}-\{0\}$ by Riemann's theorem on removable singularities. If $1<\mu_1<\mu_2<\infty$, suppose $h:A_{\mu_1}\rightarrow A_{\mu_2}$ is a biholomorphism, then $\ln \lambda_i\ln \mu_i=2\pi^2$. By Lemma 1, $h$ lifts to $\tilde h\in \Aut(\mathbb{H})$ and $\tilde h\circ \gamma_1=\gamma_2^\pm\circ \tilde h$. Comparing the eigenvalues, we know $\lambda_1=\lambda_2$, therefore $\mu_1=\mu_2$. The claim follows.

Remark 2 $\ln \lambda$ in the third case can be viewed as the length of the hyperbolic geodesic (cf. Example 7).
Remark 3 Here is another proof of the claim in the proof using the following beautiful theorem due to Caratheodory.
Theorem 4 (Caratheodory) The Riemann mapping $\mathbb{D}\rightarrow\Omega$ extends to a homeomorphism $\bar{\mathbb{D}}\rightarrow\bar\Omega$.

Now suppose $h:A_{\mu_1}\rightarrow A_{\mu_2}$ is a biholomorphism. Then by Caratheodory's Theorem 4, $h$ extends to a homeomorphism $\bar A_{\mu_1}\rightarrow \bar A_{\mu_2}$. Applying the Schwarz reflection theorem repeatedly, this map extends to a biholomorphism $\mathbb{C}^*\rightarrow\mathbb{C}^*$. Thus it is a scalar, which has to be 1 since $h$ maps $A_{\mu_1}$ to $A_{\mu_2}$. So $\mu_1=\mu_2$.

Corollary 1 $\Mod(S^1\times S^1)=\mathbb{H}/SL(2,\mathbb{Z})$. $\Mod(S^1\times\mathbb{R})\cong
(1,\infty]$ where $\infty$ consists of two points $\mathbb{D}-\{0\}$ and $\mathbb{C}^*$.
Proof The second part of the corollary follows directly from Theorem 3 and the first part follows Theorem 3 along with the following lemma.
Lemma 3 If $\eta,\eta'\in\mathbb{H}$ then $\mathbb{C}/\mathbb{Z}+\mathbb{Z}\eta\cong\mathbb{C}/\mathbb{Z}+\mathbb{Z}\eta'$ if and only if $\eta, \eta'$ are equivalent under the action of $SL(2,\mathbb{Z})$.
Proof $: by change of lattice basis and Remark 1. $: composing $h$ by an automorphism of $\mathbb{C}/L$, we may assume $h[0]=[0]$. Let $\tilde h:\mathbb{C}\rightarrow \mathbb{C}$ be the lifting of $h$ such that $\tilde h(0)=0$ using Lemma 1. Hence $\tilde h(z)=\lambda z, \lambda\ne0$, which implies that $\lambda L=L'$ and $\eta, \eta'$ are equivalent under the action of $GL(2,\mathbb{Z})$. Moreover, $\eta, \eta'\in\mathbb{H}$ ensures that they are actually equivalent under $SL(2,\mathbb{Z})$.

By now, we have completely described Riemann's moduli spaces$\Mod(\Sigma)$ for all topological surfaces $\Sigma$ with abelian fundamental groups.

Remark 4 Suppose that $g\ge2$ and $\Sigma_g$ is a genus $g$ closed Riemann surface. Riemann computed that $\dim_\mathbb{R}\Mod(\Sigma_g)=6g-6$. To see that $\dim_\mathrm{R}(\Mod(\Sigma_g))\le6g-6$ intuitively, let $[X]\in \Mod(\Sigma_g)$, then $X\cong\mathbb{H}/\pi_1(X)$, where the fundamental group $\pi_1(X)$ is generated by $A_i, B_i\in SL(2,\mathbb{R})$ ($i=1,\ldots, g$) subject to $\prod A_iB_iA_i^{-1}B_i^{-1}=\mathrm{Id}$. The total number of degrees of freedom is $3\cdot 2g-3$. Since $\mathbb{H}/\Gamma\cong\mathbb{H}/g\Gamma g^{-1}$ for $g\in SL(2,\mathbb{R})$, the total dimension is no more than $6g-6$. More generally, let $\Sigma$ be a topological surface of finite type (homeomorphic to a compact space) such that $\pi_1(\Sigma)$ is generated by $n$ elements, then $\dim_\mathbb{R}(\Mod(\Sigma))\le3n-3$.

TopQuasi-conformal maps

We fix the following conventions in this section: $\Omega, \tilde \Omega$ are open in $\mathbb{C}$. $f:\Omega\rightarrow\tilde\Omega$ is a $C^1$- smooth function. Denote $z=x+iy$, $f(z)=u(x,y)+i v(x,y)$, $u_x=\frac{\partial}{\partial x}u$, $\frac{\partial}{\partial \bar z}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$ and so on. The following proposition can be thought of the chain rule with respect the variables $z $ and $\bar z$.

Proposition 1 Let $f: \Omega\rightarrow\tilde\Omega$ and $g:\tilde\Omega\rightarrow\mathbb{C}$ be $C^1$-smooth, $w=f(z)$. Then
  1. $\frac{\partial}{\partial z}(g(f(z)))=\frac{\partial g}{\partial w}\frac{\partial w}{\partial z}+\frac{\partial g}{\partial \bar w}\frac{\partial \bar w}{\partial z}$.
  2. $\frac{\partial}{\partial \bar z}(g(f(z)))=\frac{\partial g}{\partial w}\frac{\partial w}{\partial \bar z}+\frac{\partial g}{\partial \bar w}\frac{\partial \bar w}{\partial \bar z}$.
Lemma 4 The derivative $D(f): \mathbb{C}\rightarrow\mathbb{C}$ of $f$ is given by $D(f)(w)=f_z\cdot w+f_{\bar z}\cdot \bar w$.
Proof By definition and Proposition 1, we have $$D(f)(w)=\frac{d}{dt}\Big\vert_{t=0}(f(z+tw))=f_z\frac{d(z+tw)}{dt}+f_{\bar z} \frac{d(\overline{z+tw})}{dt}=f_z\cdot w+f_{\bar z}\cdot \bar w.$$ The lemma follows.

It is easy to see that every $\mathbb{R}$-linear map $A:\mathbb{C}\rightarrow\mathbb{C}$ is given by $A(z)=az+b\bar z$, where $a,b\in\mathbb{C}$ satisfies $A(1)=a+b$ and $A(i)=(a-b)i$. Moreover, $\det(az+b\bar z)=|a|^2-|b|^2$. If $|a|\ne|b|$, then $A^{-1}(z)=\frac{\bar a z-b\bar z}{|a|^2-|b|^2}$. It follows that the Jacobian of $f$ is equal to $|f_z|^2-|f_{\bar z}|^2$. So $f$ preserves orientation if and only if $|f_z|> |f_{\bar z}|$.

Geometrically, $az+b\bar z$ sends circles to ellipses. The following easy lemma describes which of the radii of a circle are sent to the axes of the corresponding ellipse.

Lemma 5 For $A(z)=az+b\bar z$ (assume that $|a|>|b|$), we have $$(|a|-|b|)|z|\le |A(z)|\le (|a|+|b|)|z|.$$ The second equality holds if and only if $az/b\bar z\in\mathbb{R}_{>0}$, if and only if $\Arg(z)=\frac{1}{2}\Arg(b/a)$. The first equality holds if and only if $az/b\bar z\in\mathbb{R}_{<0}$, if and only if $\Arg(z)=\frac{1}{2}\Arg(b/a)+\frac{\pi}{2}$.
Definition 2 The dilatation of $A$, denoted by $D_A$, is the ratio of axes of the ellipse $A\{|z|=1\}$, which is equal to $\frac{|a|+|b|}{|a|-|b|}$.

Note that $A^{-1}$ will switch the major axis and the minor axis. Also, it is easy to see that $D_A=D_{A^{-1}}$ and $D_{B\circ A}\le D_B\cdot D_A$.

Definition 3 The Beltrami coefficient is defined to be $\mu_A=b/a=A_{\bar z }/A_z\in\mathbb{C}$. We have $\Arg(\mu_A)=2\Arg(\text{minor axis})$ and $|\mu_A|=|b|/|a|=\frac{D_A-1}{D_A+1}$.
Definition 4 Assume $f$ preserves orientation. Define $D_f$ to be the dilatation of $D(f)$ and $\mu_f$ to be the Beltrami coefficient of $D(f)$. Then $D_{f\circ g}\le D_f\cdot D_g$.
Lemma 6 Suppose $g:\tilde\Omega\rightarrow \mathbb{C}$ is $C^1$-smooth, then $$\mu_{g\circ f}=\frac{\mu_f+\gamma_f\mu_g}{1+\gamma_f\overline{\mu_f}\mu_g}$$ where $\gamma_f=\overline{f_z}/f_z$.
Proof By definition, $$\mu_{g\circ f}=\frac{(g\circ f)_{\bar z}}{(g\circ f)_{z}}=\frac{g_w f_{\bar z}+g_{\bar w} (\bar f)_{\bar z}}{g_wf_z+g_{\bar w}(\bar f)_z}=\frac{\mu_f+\mu_g\overline{f_z}/f_z}{1+\mu_g\overline{f}_z/f_z}=\frac{\mu_f+\gamma_f\mu_g}{1+\gamma_f\overline{\mu_f}\mu_g}.$$ The lemma follows.
Corollary 2 If $g_{\bar w}=0$, then $g$ is conformal and $\mu_{g\circ f}=\mu_f$. If $f_{\bar z}=0 $, then $f$ is conformal and $\mu_{g\circ f}=\gamma_f\mu_g$.
Definition 5 Say $f:\Omega\rightarrow\tilde\Omega$ is a quasi-conformal map if $f$ is an orientation-preserving homeomorphism, $C^1$-smooth outside a discrete set $P $ and there exists $K>0$ such that $D_f(z)\le K$ for any $z\in\Omega- P$. Denote $K_f=\sup\{D_f(z): z\in\Omega-P\}$, then $K_f<\infty$.
Remark 5 $K_f=1$ if and only if $f$ is analytic. Hence the ratio of the two axes is conformally invariant.

The concept of quasi-conformal maps can be generalized to Riemann surfaces using local coordinates. A natural question is to find the minimum dilatation quasi-conformal maps (in given a homotopy class) between Riemann surfaces with different analytic structures. This is the content of Teichmuller theory and the main theorem is the following.

Theorem 5 (Teichmuller Theorem) Suppose $X,Y$ are compact Riemann surfaces of genus $g\ge2$ and $h:X\rightarrow Y$ is a given homeomorphism. Then there exists a unique quasi-conformal map $f:X\rightarrow Y$ such that $f$ is homotopy equivalent to $h$ and $D_f$ has the smallest dilatation among all such $f$. Moreover, $D_f(z)$ is a constant independent of $z $.

This theory is motivated by the classical Grotzsch problem.

Example 1 (Grotzsch problem) Let $R_i$ be rectangles with vertices $A_i, B_i, C_i, D_i$ and size $a_i\times b_i$, $i=1,2$. Consider all $C^1$-smooth diffeomorphism $f:R_1\rightarrow R_2$ such that $f(A_1)=A_2, \ldots, f(D_1)=D_2$. We have:
Theorem 6 $D_f\ge D_g=\max\{\frac{a_2/b_2}{a_1/b_1},\frac{a_1/b_1}{a_2/b_2}\}$ where $g$ is the affine map.
Proof For any $y\in[0, b_1]$, let $\gamma_y$ be the horizontal line segment $\{(x,y): 0\le x\le a_1\}$, then $$a_2\le\text{length} (f(\gamma_y))=\int_0^{a_1}|f_x|dx=\int_0^{a_1}|f_z+f_{\bar z}|dx.$$ So using the triangle inequality, we obtain $$a_2b_1\le\int_0^{b_1}\int_0^{a_1}(|f_z|+|f_{\bar z}|)dxdy.$$ Now the Jacobian $J_f=|f_z|^2-|f_{\bar z}|^2$, so 
  a_2b_1&\le\iint_{R_1}\sqrt{\frac{|f_z|+|f_{\bar z}|}{|f_z|-|f_{\bar z}|}}\sqrt{J_f}dxdy \\
  &\le\left(\iint_{R_1}\frac{|f_z|+|f_{\bar z}|}{|f_z|-|f_{\bar z}|}dxdy\right)^{1/2}\text{area}(R_2)^{1/2}\\
by Cauchy's inequality. The equality holds if and only if $f$ is affine: $f_z/f_{\bar z}\in \mathbb{R}>0$ is a constant and $J_f$ is a constant, therefore $|f_z|$ and $|f_{\bar z}|$ are constants. The theorem now follows.
Example 2 Let $A_i=\{1<|z|<e^{r_i}\}$ be an annulus of module $r_i$ ($i=1,2$). Then $A_i\cong B_i/(z\sim z+2\pi\sqrt{-1})$ by $z\mapsto e^{z}$, where $B_i=\{z\in\mathbb{C}: 0<\Re z<r_i\}$. So if $f:R_1=[0, r_1]\times [0, 2\pi]\rightarrow R_2=f(R_1)$ is a $C^1$-smooth diffeomorphism, using the same proof as in Theorem 6, we know that $K_f\ge r_2/r_1$ and the equality holds if and only if the lift $\tilde f:B_1\rightarrow B_2$ is affine given by $(x,y)\mapsto (x, ky+c)$.
Question How can we generalize the solution to the Grotzsch problem to arbitrary Riemann surfaces?

Teichmuller used quadratic differentials to solve this problem.

TopQuadratic differentials

Definition 6 A holomorphic 1-form $\eta$ on $X=\Omega/\Gamma$ is an analytic function $h:\Omega\rightarrow \mathbb{C}$ such that $h(\gamma(z))\gamma'(z)=h(z)$ for any $\gamma\in\Gamma$. Similarly, a quadratic differential $\eta$ on $X=\Omega/\Gamma$ is an analytic function $h:\Omega\rightarrow \mathbb{C}$ such that $h(\gamma(z))\gamma'(z)^2=h(z)$ for any $\gamma\in\Gamma$. More generally, for a Riemann surface $X$ with analytic charts $\{U_\alpha, z_\alpha: \alpha\in I\}$, a holomorphic 1-form on $X$ is a collection of analytic functions $g_\alpha: z_\alpha(U_\alpha)\rightarrow\mathbb{C}$ such that $g_\alpha dz_\alpha= g_\beta dz_\beta$, i.e., $g_\alpha=g_\beta(\phi_{\beta\alpha}(z))\phi_{\beta\alpha}'(z)$. Similarly, a quadratic differential on $X$ is a collection of analytic functions such that $g_\alpha=g_\beta(\phi_{\beta\alpha}(z))\phi_{\beta\alpha}'(z)^2$.
Example 3 There are no holomorphic 1-forms on $\hat{\mathbb{C}}$ ($H_1(\hat{\mathbb{C}},\mathbb{Z})=0$). All holomorphic 1-forms and quadratic differentials on $\mathbb{C}/L$ are of the forms $kdz$ and $kdz^2$ where $k\in \mathbb{C}$ ($\dim_{\mathbb{R}} H_1(\mathbb{C}/L,\mathbb{R})=2$).
Remark 6 The following theorem is classical.
Theorem 7 (Riemann-Roch) Let $X$ be a compact Riemann surface of genus $g\ge2$, then the vector space $\QD(X)$ of all quadratic differentials on $X$ is of complex dimension $3g-3$ (and real dimension $6g-6$).

Roughly speaking, we will see a correspondence between $\QD(\Sigma_g)$ and $\Mod(\Sigma_g)$, which explains the significance of quadratic differentials in Teichmuller's construction (cf. Remark 4and Theorem 14).

A holomorphic 1-form at $p\in X$ gives an $\mathbb{R}$-linear map from the tangent space $T_p X$ to $\mathbb{C}$. Similarly, a quadratic differential $h(z)dz^2$ at $p\in X$ gives a quadratic form $$B:v=[\alpha]\mapsto h(\alpha(t))\alpha'(t)^2|_{t=0}$$ from $T_p X$ to $\mathbb{C}$. A quadratic differential provides the following geometric data.

  1. A horizontal direction in $T_p X$, namely $v\in T_p X$ such that $B(v)>0$. So $\Arg(v)=-\frac{1}{2}\Arg(h(p))$.
  2. A vertical direction in $T_p X$, namely $v\in T_p X$ such that $B(v)<0$. So $\Arg(v)=-\frac{1}{2}\Arg(h(p))+\frac{\pi}{2}$.
  3. A flat Riemannian metric $|hdz^2|$ near $p$ with $h(p)\ne0$, given by $v\mapsto |B(v)|=|h(p)||v|^2$. This metric is conformal to the standard Euclidean metric, i.e., angles in $hdz^2$ are the same as angles on the Riemann surface $X$.
  4. $|hdz^2|$ is locally isometric to the Euclidean space $\mathbb{E}^2$. Indeed, given $|g^2(z)dz^2|$, define $w=\int_0^z g(t)dt$, then $w(0)=0$ and $w'(0)=g(0)\ne0$ and $|g^2(z)dz^2|=|dw^2|$.
Theorem 8 (Teichmuller) Suppose $\eta$ is a nonzero quadratic differential on a Riemann surface $X$, then for any $p\in X$, there exists an analytic coordinate $(U,z)$ at $p$ with $z(p)=0$ such that $\eta$ in $(U,z)$ is given by $z^mdz^2$ where $m\in\mathbb{Z}_{\ge0}$. We call this analytic coordinate the natural coordinate for $\eta$.
Proof We may work locally and assume that $X\subseteq \mathbb{C}$, $p=0$ and $w=h(z)dz^2$ where $h:X\rightarrow\mathbb{C}$ is analytic. If $h(0)\ne0$, we have just proved that $\eta=dw^2$ where $w=\int_0^z \sqrt{h(t)}dt$. Now suppose that $h(0)=0$. Let $h(z)=z^mg(z)$ where $g$ is analytic and $g(0)\ne0$. Our goal is to write $z^mg(z)dz^2$ as $w(z)^m(dw(z))^2$ where $w(z)$ is analytic, $w(0)=0$ and $w'(0)\ne0$. Namely, we would like to solve the equation 
  w^m(z)w'(z)^2=z^mg(z). \tag{1}
Formally, it is equivalent to $$\frac{1}{m/2+1}(w^{m/2+1}(z))'{}=z^{m/2}\sqrt{g(z)},$$ or $$w(z)=\left((m/2+1)\int_0^z t^{m/2}\sqrt{g(t)}dt\right)^{1/(m/2+1)}.$$ We need to justify that this is well-defined. Suppose $\sqrt{g(z)}=\sum_{n\ge0} a_nz^n$ where $a_0\ne0$. Then $$\int_0^zt^{m/2}\sqrt{g(t)}dt=z^{m/2+1}\sum_{n\ge0}\frac{a_nz^n}{n+m/2+1}.$$ Hence $$w(z)=z\left((m/2+1)\sum_{n\ge0}\frac{a_nz^n}{n+\frac{m}{2}+1}\right)^{1/(m/2+1)}$$ is single-valued and satisfies Equation 1.
Definition 7 For a nonzero quadratic differential $\eta$, we denote $\Sing(\eta)=\{p:\eta_p=0\}$.
Remark 7 Let $\eta\not\equiv0$ be a quadratic differential on a compact Riemann surface $X$, then we obtain a flat Riemannian metric $|\eta|$ on $X-\Sing(\eta)$, so the area of $\eta$ is well-defined: $$\text{area}(\eta)=\int_{X-\Sing(\eta)} dxdy=\int_X dA_\eta=\int_X|\eta|.$$ We also denote it by $||\eta||_{L^1}$, then $\QD(X)$ is a $L^1$-normed vector space.
Example 4 The metric $|z^mdz^2|$ near $z=0$ is a cone with cone angle $\pi (m+2)$. The cone minus the vertex has a flat metric. Suppose the cone angle is $\alpha$, then the pullback metric of $dw^2$ via $w(z)=z^{\alpha/2\pi}$ is $(\alpha/2\pi)^2(z^{\alpha/2\pi-1})^2dz^2=kz^\beta dz^2$. For example, the metric $|zdz^2|$ is obtained by gluing three half planes along the origin.

TopTeichmuller Uniqueness Theorem

Let $X$ be a Riemann surface, $\eta\not\equiv0$ a quadratic differential on $X$ and $K=\frac{1+k}{1-k}\ge1$, where $0\le k<1$. Then there exists a new Riemann surface $X_{\eta, K}$ such that the underlying set of $X$ and $X_{\eta, K}$ are the same, and there exists a quadratic differential $\tilde\eta$ on $X_{\eta, K}$ such that $\mathrm{Id}: X\rightarrow X_{\eta, K}$ is $K$-quasi-conformal. Let us construct the analytic charts for $X_{\eta, K}$ as follows. Pick any point $p\in X$.

Type I $\eta_p\ne0$. By Theorem 8, there exists a natural coordinate $(U,z)$ of $\eta$ at $p$ with $z(p)=0$ and $\eta=dz^2$. Define a new analytic chart $(U,\tilde z)$ for $X_{\eta, K}$ by $\tilde z=\frac{z+k\bar z}{1-k}$, in other words, $\tilde x=Kx$, $\tilde y=y$.

Note that if another $w=w(z)$ is analytic with $w(0)=0$ and $w'(0)\ne0$ such that $dz^2=dw^2=w'(z)^2dz^2$, then $w'(z)^2=1$, which implies that $w(z)=\pm z$. So if $(U_1, z_1)$ and $(U_2, z_2)$ are two natural coordinates of Type I for $\eta$, the the transition function $\tilde z_1\tilde z_2^{-1}$ between them is analytic: we know that $z_1=\pm z_2+c$ where $c$ is a constant, therefore $\tilde z_2=\pm \tilde z_1+c'$, so $\tilde z_1\tilde z_1^{-1}(t)=\pm t+c'$. Hence $d\tilde z_2=\pm\tilde dz_1$, $d\tilde z_2^2=d\tilde z_1^2=\tilde \eta$ is a well-defined quadratic differential on $X_{\eta, K}-\Sing(\eta)$.

Type II $\eta_p=0$. Choose a natural coordinate $(U,z)$ at $p$ such that $\eta=z^mdz^2$ by Theorem 8. Define a chart for $X_{\eta, K}$ by $$\tilde z=\left(\frac{z^{m/2+1}+k\bar z^{m/2+1}}{1-k}\right)^{2/(m+2)}.$$ This is well-defined since the right hand side is $z\left(\frac{1+k(\bar z/z)^{m/2+1}}{1-k}\right)^{2/(m+2)}$ and $\Re(1+k(\bar z/z)^{m/2+1})\ge 1-k>0$, so we can just choose the principal branch. Moreover, $z\mapsto\tilde z$ is a homeomorphism $\mathbb{C}\rightarrow\mathbb{C}$. Indeed, $\tilde z^{m/2+1}=(z^{m/2+1}+k\bar z^{m/2+1})/(1-k)$, so $$z=\left(\frac{\tilde z^{m/2+1}-k\bar{\tilde z}^{m/2+1}}{1-k}\right)^{2/(m+2)}.$$

We claim that the transition function $\tilde z_2\tilde z_1^{-1}$ for a Type I chart $(U_1,\tilde z_1)$ and a Type II chart $(U_2, \tilde z_2)$ is analytic. Suppose $\eta=z_2^mdz_2^2$ at $p$. We may assume that $p\not\in U_1\cap U_2$. Then we find that $$z_2^mdz_2^2=d\left(z_2^{m/2+1}/(m/2+1)\right)^2$$ in $U_1\cap U_2$, so $z_1=\pm (z_2^{m/2+1}/(m/2+1)+c)$. Hence $$\tilde z_1=\pm1/(m/2+1)\left(\frac{z_2^{m/2+1}+k\bar z_2^{m/2+1}}{1-k}\right)+c'{}=\pm 1/(m/2+1)\tilde z_2^{m/2+1}+c'$$ is analytic and $d\tilde z_1^2=\tilde z_2^md(\tilde z_2)^2$.

Remark 8 Teichmuller proved that all analytic structures on a topological surface can be obtained via this construction.

The following uniqueness theorem can be regarded as the general answer to the Grotzsch problem.

Theorem 9 (Teichmuller Uniqueness Theorem) Let $X$ be a compact Riemann surface with a nonzero quadratic differential $\eta$. Let $X_{\eta, K}$ be the new Riemann surface. If $f: X\rightarrow X_{\eta, K}$ is a quasi-conformal map homotopic to the identity map, then $K_f\ge K_{\mathrm{Id}}=K$ and the equality holds if and only if $f=\mathrm{Id}$.

Denote $Y=X_{\eta, K}$ for simplicity.

Lemma 7 (Key Lemma) Suppose $\alpha:[0,1]\rightarrow Y$ is an embedded horizontal arc in $Y-\Sing(\tilde\eta)$ and $\tilde\alpha:[0,1]\rightarrow Y$ such that $\alpha$, $\tilde\alpha$ are homotopic relative to $\{0,1\}$. Then the length of $\alpha$ is no greater than the length of $\tilde\alpha$ in $|\tilde\eta|$.

The problem occurs due to the presence of singularities (negative curvatures). To prove this Key Lemma, we need the following version of Gauss-Bonnet Theorem.

Theorem 10 (Gauss-Bonnet) Let $P $ be a polyhedral surface obtained by isometric gluing of Euclidean triangles. For a vertex $v $ of $P $, define its curvature 
    2\pi-\text{cone angle at } v, & v \text{ interior}, \\
    \pi-\text{sum of angles at } v, & v \text{ boundary}.
Then $\sum_{v} K_v=2\pi\chi(P)$.
Proof (Proof of Key Lemma 7) Let us minimize the length of $\tilde\alpha$. We may assume that $\tilde\alpha$ is a piece-wise geodesic path homotopic to $\alpha$ such that the vertices of $\tilde\alpha$ are in $\Sing(\tilde\eta)$. Moreover, any vertex other than $\alpha(0)$ and $\alpha(1)$ has cone angle at least $\pi$ (otherwise we can decrease the length), so $K_v\le0$ for these vertices. Now by the Gauss-Bonnet Theorem, we get $(\pi-\theta_1)+(\pi-\theta_1)>\sum_{v} K_v=2\pi$, impossible unless $\tilde\alpha=\alpha$.
Lemma 8 Let $f:X\rightarrow Y$ be a $K$-quasi-conformal map homotopic to the identity map. Then there exists a constant $M=M(f)$ such that for any horizontal arc $\alpha$ in $X-\Sing(\eta)$, we have $$\ell_{\tilde\eta}(f(\alpha))\ge K\cdot\ell_\eta(\alpha)-M.$$
Proof Let $H:X\times I\rightarrow Y$ be a $C^1$-smooth homotopy between $\mathrm{Id}$ and $f$. Let $M=\frac{1}{2}\max\{\ell_{\tilde\eta}(H(p,\cdot)): p\in X\}$. Using Key Lemma 7, since $\alpha$ is also horizontal in $\tilde\eta$ and $\ell_{\tilde\eta}(\alpha)=K\ell_{\eta}(\alpha)$, we know that $$K\cdot\ell_\eta(\alpha)\le\ell_{\tilde\eta}(H(0,\cdot))+\ell_{\tilde\eta}(f(\alpha))+\ell_{\tilde\eta}(H(1,\cdot))\le\ell_{\tilde\eta}(f(\alpha))+M.$$ This completes the proof.
Lemma 9 Let $f_x$ be the partial derivative with respect to the natural coordinate $z=x+iy$ in $\eta$, $\tilde\eta$. Let $|f_x|$ be measured in $|\eta|,|\tilde\eta|$. Then $$\int_X |f_x|dA_\eta\ge K\cdot\text{area}(\eta).$$
Proof Let $X'$ be $X$ minus all horizontal lines through $\Sing(\eta)$ (several branches at one singular point), then $X'$ has full measure in $X$. For any $p\in X'$ and $L>0$, define $\alpha_L$ be the horizontal line centered at $p$ of length $2L$. Then $\alpha_L$ is simple since the horizontal direction is unique outside $\Sing(\eta)$. We choose the arc length parametrization, namely $|\alpha_L'(t)|_\eta=1$, so $\alpha_L(t)=(x_0+t, y_0)$, where $p=(x_0,y_0)$ under the natural coordinate. Using Key Lemma 7, we know that $$\delta(p, L)=\ell_Y(f(\alpha_L))\le 2KL-M,$$ where $M=M(f)$ is a constant. But by the chain rule $$\delta(p,L)=\int_{-L}^L|f_x(\alpha_L(t))|dt, $$ we know that 
  \int_{X'}\delta(p,L)dA(p)&=\int_{X'}\int_{-L}^L |f_x(p+t)|dt d A(p)\\
  &=\int_{-L}^L\left(\int_{X'}|f_x(p+t)|dA(p)\right)dt \\
  &\qquad ((x,y)\mapsto(x+t,y) \text{ preserves the measure }dxdy) \\
So $$2L\int_X|f_x|dA\ge (2KL-M)\cdot\text{area}(\eta).$$ The lemma then follows by letting $L\rightarrow \infty$.
Proof (Proof of Theorem 9) By Lemma 9 and Cauchy's inequality, we have $$K\cdot\text{area}(\eta)\le\int_X |f_x|dx\le\sqrt{K_f\text{area}(\eta)\text{area}(\tilde \eta)}=\sqrt{K_f K}\text{area}(\eta)$$ since $\frac{\text{area}(\tilde \eta)}{\text{area}(\eta)}=K$ when $X$ is compact. The theorem follows.

TopTeichmuller spaces

Let us return to the study of Riemann's moduli space $\Mod(\Sigma)$ consisting of all Riemann surfaces $X$ homeomorphic to a fixed topological surface $\Sigma$. We observe that it is the equivalent to study all complex structures on $\Sigma$.

Definition 8 Let $$\mathcal{A}=\{\mathcal{C}: \text{ complex structures on }  \Sigma\}.$$ Let $h:\Sigma\rightarrow\Sigma$ be a homeomorphism, then we can pull $\mathcal{C}$ back to another complex structure $\mathcal{C}_h$ on $\Sigma$. Define $$\Home^+(\Sigma)=\{\text{orientation preserving self-homeomorphisms of }\Sigma\}.$$ Then the Riemann moduli space $\Mod(\Sigma)=\mathcal{A}/\Home^+(\Sigma)$.
Definition 9 We say two complex structures $\mathcal{C}_1$ and $\mathcal{C}_2$ are Teichmuller equivalent if there is a biholomorphism $h: \mathcal{C}_1\rightarrow\mathcal{C}_2$ such that $h$ is homotopic to the identity map. The set of all Teichmuller classes is called the Teichmuller space, denoted by $\Teich(\Sigma)$. Define $$\Home_0(\Sigma)=\{h\in\Home^+(\Sigma): h\simeq \Id\}.$$ Then $\Teich(\Sigma)$ is the same as $\mathcal{A}/\Home_0(\Sigma)$.
Definition 10 The group $\MCG=\Home^+(\Sigma)/\Home_0(\Sigma)$ is called the mapping class group of $\Sigma$. Then $\Mod(\Sigma)=\Teich(\Sigma)/\MCG$.
Example 5 For $\Sigma=S^1\times S^1$, then $\Teich(S^1\times S^1)=\mathbb{H}$ and $\MCG=SL(2,\mathbb{Z})$.

There is a subtle problem concerning "markings" on the Teichmuller space since we fix the underlying set $\Sigma$.

Example 6 What is the space of all congruence classes of triangles in the plane? Let $\Delta$ be a fixed triangle. Then $$\Teich(\Delta)=\{(\ell_1,\ell_2,\ell_3)\in\mathbb{R}_{>0}^3: \ell_i+\ell_j>\ell_k\}=\{\text{iso. classes of marked triangles}\},$$ which is isomorphic to $\mathbb{R}^3$. But the Riemann moduli space $\Mod(\Delta)=\Teich(\Delta)/S_3$ is isomorphic to $\{(x,y,z):z\ge0\}$.

Instead of fixing the underlying set $\Sigma$, we may consider the equivalence classes of marked Riemann surfaces.

Definition 11 Fix a topological surface $\Sigma$. A marked Riemann surface $X$ by $\Sigma$ is a pair $(X,h)$ where $h:\Sigma\rightarrow X$ is a homeomorphism. $(X_1, h_1)$ and $(X_, h_2)$ is Teichmuller equivalent if there exists a biholomorphism $\phi:X_1\rightarrow X_2$ such that $\phi\circ h_1\simeq h_2$ are homotopic. So the Teichmuller space $\Teich(\Sigma)$ is the set of all Teichmuller equivalent classes of $(X, h)$.
Remark 9 If $h_1, h_2:\Sigma\rightarrow X$ are homotopic, then $(X, h_1)$ and $(X, h_2)$ are Teichmuller equivalent, so we may replace $h$ by its homotopy class.
Remark 10 If $\chi(\Sigma)<0$, then $h_1\simeq h_2:\Sigma\rightarrow X$ if and only if $(h_1)_*=\alpha(h_2)_*\alpha^{-1}:\pi_1(\Sigma)\rightarrow\pi_1(\Sigma)$, where $\alpha\in\pi_1(\Sigma)$. This can be proved by looking at the universal covering $\mathbb{H}$. So a marking on the Riemann surface $\Sigma$ can be regarded as a choice of generators of the fundamental group $\pi_1(\Sigma)$ under the equivalence of conjugation.

Our goal is to show that $\Teich(\Sigma_g)$ is isomorphic to the Fricke space $$\mathcal{T}_g\subseteq \Hom(\pi_1(\Sigma_g), PSL(2,\mathbb{R}))/\text{conjugation}.$$ Fricke showed that $\mathcal{T}_g$ can be viewed as a subset in a $6g-6$-dimensional manifold $PSL(2,\mathbb{R})^{2g-2}$, and then one can prove that $\mathcal{T}_g$ is actually homeomorphic to $\mathbb{R}^{6g-6}$ (cf. Uniformization Theorem 14).

TopHyperbolic geometry

This subsection is a crash course on the basics of hyperbolic geometry. Define a Riemannian metric $ds^2=\frac{dx^2+dy^2}{y^2}$ on $\mathbb{H}$.

Example 7 The $y$-axis is a geodesic in $\mathbb{H}$. Let $\gamma(t)$ be a path in $\mathbb{H}$ with staring point $x(0)=0$. Then 
  \ell(\gamma(t))&=\int_{\alpha}^\beta\frac{\sqrt{x'(t)^2+y'(t)^2}}{y(t)} dt
  \ge \int_\alpha^\beta\frac{|y'(t)|}{y(t)}dt \\
  &\ge \left| \int_\alpha^\beta \frac{y'(t)}{y(t)}dt\right|
  =\ln \beta/\alpha,
and equality holds if and only $x(t)=x'(t)=0$ and $y'(t)\ge0$, if and only if $\alpha(t)$ is along the $y$-axis.

Note that $(x,y)\mapsto (x+a,y)$ is an isometry of $ds^2$. Thus we know that all vertical lines are geodesics. Note that $(x,y)\mapsto (\lambda x,\lambda y)$ and $z\mapsto-1/z$ are also isometries. Since all Mobius transformations are composition of these three types of isometries, we know that $PSL(2,\mathbb{R})\subseteq \Iso^+(\mathbb{H}, ds^2)$. Actually one can show that $PSL(2,\mathbb{R})= \Iso^+(\mathbb{H}, ds^2)$. Using the transformation $z\mapsto-1/z$ and the translations, it follows that all geodesics are either vertical lines or semicircles perpendicular to the $x$-axis.

Remark 11 $(\mathbb{H},ds^2)$ has constant Gaussian curvature $K$ since $PSL(2,\mathbb{R})$ acts on $\mathbb{H}$ transitively. Indeed, for a Riemannian metric $h(z)^2|dz^2|$, the Gaussian curvature is given by $K=-\frac{4}{h}\frac{\partial^2\log h}{\partial z\partial \bar z}$. A computation then shows that $K=-1$.
Example 8 The distance $$d(i\alpha,i\beta)=\ln\beta/\alpha=\ln(i\beta/i\alpha)=\ln(i\alpha,i\beta,\infty, 0).$$ Since the cross ratio is invariant under a Mobius transformation, we have $$d(z,w)=\ln(z,w,w',z'),$$ where $z', w'$ are the intersection points of the semicircle through $z,w$ with the $x$-axis.

Let $X$ be a compact Riemann surface of genus $g\ge2$. The uniformization theorem tells us that $X=\mathbb{H}/\Gamma$. The metric $ds^2$ is invariant under $\Gamma$, so it induces a Riemannian metric $ds^2_X$ on $X$ (thus such an $X$ is called a hyperbolic surface) and now it makes sense to talk about lengths on $X$.

Lemma 10 Suppose $\mathbb{H}/\Gamma$ is a closed hyperbolic surface, then for $\gamma\in\Gamma-\{\Id\}$, we have $|\tr(\gamma)|>2$, or equivalently, $\gamma$ has two distinct real eigenvalues (which are called hyperbolic).
Proof By the proof of Lemma 2, we know that $|\tr(\gamma)|\ge2$. If $|\tr(\gamma)|=2$, after conjugation, we may assume that $\gamma(z)=z+1$. Since $X$ is compact, we can find finitely many small balls $B_i$ covering $X$ (so $B_i$'s are contractible). By Lebesgue's Lemma, there exists $\varepsilon>0$ such that any loop of length less than $\varepsilon$ in $X$ is in some $B_i$ (hence homotopic to a point). But for $\gamma(z)=z+1$, consider the image of $t\mapsto(t, 2/\varepsilon)$, whose length is $\varepsilon/2<\varepsilon$. It is homotopically nontrivial, a contradiction.

TopUniformization theorem

We are now going to talk about the remaining two main theorems, namely the uniformization theorem and the measurable Riemann mapping theorem. Teichmuller asserted that the Teichmuller space $\Teich(\Sigma)$ is isomorphic to $\mathbb{R}^{6g-6}$, but this result was not rigorously proved until the work of Ahlfors and Bers in 1950s. We will utilize some corollaries of the measurable Riemann mapping theorem (MRM) to prove the uniformization theorem in this section and discuss the MRM in the last section.

Let $\Sigma_g$ be a closed surface of genus $g\ge2$. Let $\tilde X$ be the universal covering of $X$, then there exists a biholomorphism $g:\tilde X\rightarrow\mathbb{H}$ such that $g\pi_1(X)g^{-1}=\Gamma\subseteq PSL(2,\mathbb{R})$ acts freely and discontinuously on $\mathbb{H}$. Moreover, $g$ induces a biholomorphism on the quotient $h_1:X\rightarrow\mathbb{H}/\Gamma$. In summary, we get a map 
  \Psi:\Teich(\Sigma_g)&\rightarrow \mathcal{T}_g\subseteq\Hom(\pi_1(\Sigma_g),PSL(2,\mathbb{R}))/\text{conjugation}, \\ [(X,h)]&\mapsto (\phi:\gamma\mapsto g h_1(\gamma) g^{-1}).
The image $\mathcal{T}_g$ is called the Frick space of $\Sigma_g$. By Remark 10 , the map $\Psi$ is injective. So we can identify $\Teich(\Sigma_g)$ with $\mathcal{T}_g$.

Theorem 11 (Fricke) $\mathcal{T}_g$ is a subset of the $6g-6$ dimensional smooth manifold $PSL(2,\mathbb{R})^{2g-2}$.
Remark 12 $\mathcal{T}_g$ has a natural topology induced from $PSL(2,\mathbb{R})^{2g-2}$, hence we can use $\Psi$ to define a topology on $\Teich(\Sigma_g)$ such that $\Psi$ is a homeomorphism.
Proof Let $\pi_1(\Sigma_g)=\langle a_1, b_1,\ldots, a_g, b_g: [a_1, b_1]\cdots[a_g,b_g]=1\rangle$. Let $\rho: \pi_1(\Sigma_g)\rightarrow PSL(2,\mathbb{R})$ be a discrete faithful representation and $A_i=\rho(a_1)$, $B_i=\rho(b_i)$. We get 
A_1B_1A_1^{-1}B_1^{-1}=[A_g, B_g]^{-1}\cdots [A_2, B_2]^{-1}.  \tag{2}
After conjugation, we may assume that $A_1=\pm\diag(\lambda, \lambda^{-1})$, where $\lambda>1$ by Lemma 10. Suppose $B_1=\pm
  a & b \\
  c & d
\end{bmatrix}$. By the proof of Theorem 3, we know $\mathrm{Fix}(\rho(a_1))\cap\mathrm{Fix}(\rho(b_1))=\varnothing$ in $\mathbb{H}\cup \mathbb{R}\cup\infty$, since $\langle a_1, b_1\rangle$ is nonabelian. But $\mathrm{Fix}(\rho(a_1))=\{0,\infty\}$, so $c\ne0$ (consider $\infty$) and $b\ne0$ (consider 0).

Now $\diag(\mu, \mu^{-1})$ commutes with $A_1$, so after conjugation we get $$
  \mu & 0\\
  0 & \mu^{-1}
  a & b \\
  c & d
  \mu & 0 \\
  0 & \mu^{-1}
  a & \mu^2b \\
  \mu^{-2}c & d

We may assume $c=\pm1$ for $B_1$. If we can solve uniquely $A_1, B_1$ from Equation 2, then $\mathcal{T}_g\subseteq \{(A_2, B_2, \cdots, B_g)\in PSL(2,\mathbb{R})^{2g-2}\}$. Indeed, we compute that $$A_1B_1A_1^{-1}B_1^{-1}=
  ad-\lambda^2 bc & ab(\lambda^2-1) \\
  (\lambda^{-2}-1)cd & ad-\lambda^{-2}bc
\end{bmatrix}.$$ and one can show that $\lambda$, $b$, $a$, $d$ can be solved successively using the fact that $c=\pm1$, $b\ne0$.

By the Riemann-Roch Theorem 7, $\QD(X)$ is a $6g-6$ dimensional $\mathbb{C}$-vector space and has a natural $L^1$-norm. Let $\QD_1(X)=\{\mu\in \QD(X): ||\mu||_{L^1}<1\}$. Denote $X_\eta=X_{\eta, k}$, where $k=||\eta||_{L^1}$. Define the Teichmuller map $$\Phi: \QD_1(X)\rightarrow \Teich(\Sigma_g),\quad \eta\mapsto [X_\eta]=[(X_\eta,\Id)].$$

Now the Teichmuller Uniqueness Theorem 9 can be restated as follows.

Theorem 12 (Teichmuller Uniqueness Theorem) $\Phi$ is injective.
Proof If $\eta_1, \eta_2\in \QD_1(X)$ and $[(X_{\eta_1},\Id_1)]=[(X_{\eta_2},\Id_2)]$, then there exists a biholomorphism $\phi: X_{\eta_1}\rightarrow X_{\eta_2}$ such that $\phi\circ\Id_1$ and $\Id_2$ are homotopic. By the Teichmuller Uniqueness Theorem 9, we know that $$k_{\phi\circ\Id_1}=k_{\Id_2},$$ therefore $\phi=\Id$. It follows from the construction of $X_{\eta,k}$ that $\eta_1=\eta_2$.

Combining Theorem 11 and Theorem 12, we obtain the following picture: $$\xymatrix{\QD_1(X) \ar@{^{(}->}[r]^-\Phi & \Teich (\Sigma_g) \ar[r]^-\Psi_-{\sim}  & \mathcal{T}_g\subseteq M^{6g-6}.}$$

The next two lemmas will be proved using MRM in the next section.

Lemma 11 $\Psi\circ\Phi: \QD_1(X)\rightarrow \mathcal{T}_g$ is continuous and proper.
Lemma 12 $\mathcal{T}_g$ is connected.

Now the uniformization theorem will follow easily from the above two lemmas and Brouwer's famous invariance of domain theorem.

Theorem 13 (Brouwer's Invariance of Domain) Let $U$ be an open set in $\mathbb{R}^N$ and $f:U\rightarrow M^N$ be a one-to-one continuous map into a manifold of the same dimension, then $f(U)$ is open in $M^N$ and $f: U\rightarrow f(U)$ is a homeomorphism.
Theorem 14 (Uniformization Theorem) $\Psi\circ\Phi: \QD_1(X)\rightarrow \mathcal{T}_g$ is a homeomorphism. In particular, the Teichmuller space $\Teich(\Sigma_g)$ and the Fricke space $\mathcal{T}_g$ are homeomorphic to $\mathbb{R}^{6g-6}$.
Proof By the continuity in Lemma 11 and Invariance of Domain Theorem 13, we know that the image of $\Psi\circ\Phi$ is open in $\mathcal{T}_g$. By the properness in Lemma 11and the fact that the target $\mathcal{T}_g$ is a locally compact Hausdorff space, we find that $\Psi\circ\Phi$ is a closed map and thus the image of $\Psi\circ\Phi$ is also closed in $\mathcal{T}_g$. But $\mathcal{T}_g$ is connected by Lemma 12, hence the image of $\Psi\circ\Phi$ is the whole of $\mathcal{T}_g$, which completes the proof.

TopMeasurable Riemann Mapping Theorem

As promised in the last section, we shall discuss the Measurable Riemann Mapping Theorem and utilize it to finish the proof of the uniformization theorem.

Let $f:\Omega\rightarrow \mathbb{C}$ be a quasi-conformal map. By definition, the supremum norm of the Beltrami coefficients $||\mu_f||_{L^\infty}=k<1$. The following inverse question asks about the existence of such quasi-conformal maps.

Question (Beltrami Equation) For $\mu\in L^\infty(\Omega)$ with $||\mu||_{L^\infty}<1$, is there any quasi-conformal map $f$ such that $f_{\bar z}=\mu f_z$?

The Measurable Riemann Mapping Theorem gives affirmative answers to this question.

Theorem 15 (Measurable Riemann Mapping Theorem) Let $\mu\in L^\infty(\hat{\mathbb{C}})$, $||\mu||_{L^\infty}<1$, then there exists a unique quasi-conformal map $f:\hat{\mathbb{C}}\rightarrow\hat{\mathbb{C}}$ satisfying the normalization $f(0)=0$,$f(1)=1$ and $f(\infty)=\infty$ such that $f_{\bar z}=\mu f_z$.
Remark 13 The $f$ in question is absolutely continuous, and $f$ is smooth whenever $\mu$ is so.

This deep theorem was proved by Lavrentev for $C^0$-functions $\mu$ and by Morrey for measurable functions $\mu$.

MRM can be also generalized to Riemann surfaces. From Sullivan's geometric point of view, the MRM simply says the following: place an ellipse on each tangent space on a Riemann surface, then there exists a unique quasi-conformal map sending this ellipse field to a circle field. In other words, for a pair $(X, \lambda)$, we need to construct a new Riemann surface $X_\lambda$ satisfying the above geometric condition.

Lemma 13 Suppose $X, Y$ are Riemann surfaces and $f: X\rightarrow Y$ is a quasi-conformal map. Then the Beltrami differential (of $f$) $\mu_f=\frac{f_{\bar z}}{f_z} \frac{d\bar z}{dz}$ is well-defined on $X$.
Proof Suppose $g(z)=\alpha(f(\beta(z)))$, where $\alpha, \beta$ are analytic. Let $w=\beta(z)$. By the chain rule, we have $$g_z=\alpha' f_w \beta', \quad g_{\bar z}=\alpha' f_{\bar w} \bar \beta'.$$ Then $$\frac{g_{\bar z}}{g_z}=\frac{f_{\bar w}}{f_w}\frac{\bar \beta'}{\beta'}.$$ So $\mu_f$ is well-defined.

Now let $\Beltr(X)$ be the vector space of all Beltrami differentials on $X$. It has a natural $L^\infty$-norm given by $|\lambda(z)\frac{d\bar z}{dz}(p)|=|\lambda(p)|$. We have a pairing $$\QD(X)\times \Beltr(X)\rightarrow \mathbb{C},\quad \left(\phi(z)dz^2,\lambda(z)\frac{d\bar z}{dz}\right)\mapsto \int_X \phi(z)\lambda(z) dzd\bar z.$$ So $\Beltr(X)$ can be viewed as the dual of $\QD(X)$ in some sense.

Looking at the universal covering and using the normal MRM Theorem 15, one can prove the following MRM on Riemann Surfaces.

Theorem 16 (MRM on Riemann Surfaces) Let $X$ be a Riemann surface and $\mu$ be a Beltrami differential on $X$ with $||\mu||_{L^\infty}=k<1$. Then there exists a Riemann surface $Y$ and a quasi-conformal map $f$ such that $\mu_f=\mu$ and $f$ is unique up to biholomorphism.
Remark 14 Moreover, Ahlfors-Bers showed that the solution $f^\mu$ depends analytically on $\mu$. This will imply that $\Mod(\Sigma_g)$ is a complex manifold (see [2]).
Remark 15 We can interpret the Teichmuller map $\Phi: \QD_1(X)\rightarrow\Teich(\sigma_G)$ in terms of Beltrami differentials. For a quadratic differential $\eta=h(z)dz^2$, the associated $X_{\eta,K}$ at $z $ is constructed by stretching by a factor of $K$ along the horizontal direction $-\frac{1}{2}\Arg(h(z))$. On the other hand, we know that $|\eta|/\eta$ transforms as $|dz^2|/dz^2=d\bar z/dz$ and thus it is a Beltrami differential. The quasi-conformal map $f^\mu$ associated to the Beltrami differential $\mu=k|\eta|/\eta$ in Theorem 16satisfies $f^\mu_{\bar z}/f^\mu_z=k |h(z)|/h(z)$, hence it locally behaves as stretching by a factor of $K=\frac{1+k}{1-k}$ along the horizontal direction $-\frac{1}{2}\Arg(h(z))$ too. In summary, the Teichmuller map can be obtained by taking $f^\mu: X\rightarrow Y$ of the Beltrami differential $\mu=||\eta||_{L^1}|\eta|/\eta$.

Finally, here comes the proof of Lemma 11 and Lemma 12, which in turn completes the proof of Theorem 14.

Proof (Proof of Lemma 11) By Remark 15, $\Phi: \QD_1(X)\rightarrow \Teich(\Sigma_g)$ is the composition of the continuous MRM given by Theorem 16 and the continuous map $\eta\mapsto||\eta||_{L^1}|\eta|/\eta$, hence continuous. If it is not proper, suppose $\eta_{n}$ satisfies $||\eta_n||_{L^1}\rightarrow 1$ ($K_n\rightarrow\infty$) and $(X_n=\Phi(\eta_n),\Id_n)$ converges to $(X_\infty, \phi)$. Then we get quasi-conformal maps $f_n: X_\infty\rightarrow X_n$ such that $f_n\circ\phi\simeq \Id_n$. But $K_n=K_{\Id_n}\le K_{f_n\circ \phi}\le K_{f_n}K_\phi\le K'K_\phi$ for some constant $K'$ as $X_n$ converges to $X_\infty$, which contradicts the assumption that $K_n\rightarrow\infty$.
Proof (Proof of Lemma 12) We use MRM to prove that $\mathcal{T}_g$ is connected. Namely, we need to show that there is a continuous process from one $\mathbb{H}/\Gamma_0$ to another $\mathbb{H}/\Gamma_1=\tilde f^\mu\circ \Gamma_0\circ \tilde (f^\mu)^{-1}$ such that every matrix representation of $\Gamma_t$ is discrete and faithful. This can be done by using $\Gamma_t=\tilde f^{t\mu}\circ \Gamma_0\circ (\tilde f^{t\mu})^{-1}$.


[1]Benson Farb and Dan Margalit, A Primer on Mapping Class Groups,

[2]Lars V. Ahlfors, Lectures on Quasiconformal Mappings (University Lecture Series), American Mathematical Society, 2006.