The genus of a compact Riemann surface determines its uniformization type

Setup. By [F2] the surface $X$ is biholomorphic to $M\mathrm{/}G$, where $M$ is its universal-covering type, exactly one of $\widehat{\mathbb{C}},\mathbb{C},\mathbb{D}$, and $G$ is the deck group of the holomorphic universal covering $p:M\to X$, acting freely and properly discontinuously by holomorphic automorphisms; by [F3], $G\cong {\pi }_1(X,x_0)$; and by [F1], $X\cong {\mathrm{\#}}_g{T}^2$ for the unique genus $g=g(X)$.

The sphere model forces $g=0$. Suppose $M=\widehat{\mathbb{C}}$. Every element of $G$ is a biholomorphic automorphism of $\widehat{\mathbb{C}}$, hence a Möbius transformation by [F6], and every nonidentity Möbius transformation has a fixed point in $\widehat{\mathbb{C}}$ by [F6]; since $G$ acts freely we get $G=\{e\}$, and then [F2] makes the covering $p$ a biholomorphism, so $X\cong \widehat{\mathbb{C}}$ as Riemann surfaces. By [F5], $\widehat{\mathbb{C}}\cong S^2={\mathrm{\#}}_0{T}^2$, so $X$ is homeomorphic to ${\mathrm{\#}}_0{T}^2$ and the uniqueness in [F1] gives $g=0$. Thus $M=\widehat{\mathbb{C}}$ implies $g=0$.

The plane model forces $g=1$. Suppose $M=\mathbb{C}$. Then $G$ is a group of biholomorphisms of $\mathbb{C}$ acting freely and properly discontinuously (step 1.1) whose quotient $\mathbb{C}\mathrm{/}G\cong X$ is compact, so [F7] applies: $G$ is a rank-two lattice and $\mathbb{C}\mathrm{/}G$ is a complex torus of genus one. Since $X$ is biholomorphic, hence homeomorphic, to $\mathbb{C}\mathrm{/}G$, the uniqueness of the genus in [F1] gives $g=1$. Thus $M=\mathbb{C}$ implies $g=1$.

Cayley conjugation of the disc case. Suppose $M=\mathbb{D}$. The map $\gamma (z):=(z-i)\mathrm{/}(z+i)$ is a Möbius transformation [F6], hence a biholomorphic self-map of $\widehat{\mathbb{C}}$ [F6]; it is holomorphic on $\mathbb{H}$ with ${\gamma }^{\mathrm{\prime }}(z)=2i\mathrm{/}(z+i{)}^2\mathrm{\ne }0$, and $\mathrm{\mid }\gamma (z){\mathrm{\mid }}^2=\mathrm{\mid }z-i{\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }z+i{\mathrm{\mid }}^2$ with $\mathrm{\mid }z-i{\mathrm{\mid }}^2<\mathrm{\mid }z+i{\mathrm{\mid }}^2$ exactly when $\mathrm{Im}z>0$, so $\gamma$ maps $\mathbb{H}$ bijectively onto $\mathbb{D}$ and restricts to a biholomorphism $\gamma :\mathbb{H}\to \mathbb{D}$ whose inverse is also Möbius. Conjugation by the homeomorphism $\gamma$ carries the free properly discontinuous action of $G$ on $\mathbb{D}$ to a free properly discontinuous action of $G^{\mathbb{H}}:={\gamma }^{-1}G\gamma$ on $\mathbb{H}$ by holomorphic automorphisms, and $G^{\mathbb{H}}\cong G$; by [F6] each element of $G^{\mathbb{H}}$ is a real Möbius map $z\mapsto (az+b)\mathrm{/}(cz+d)$ with $ad-bc>0$.

Nonidentity elements of $G^{\mathbb{H}}$ have their fixed points on $\mathbb{R}\cup \{\mathrm{\infty }\}$. Let $g_0\in G^{\mathbb{H}}$ be nonidentity. By [F6] it is a real Möbius map and has one or two fixed points in $\widehat{\mathbb{C}}$. If a fixed point $w$ satisfied $\mathrm{Im}w\mathrm{\ne }0$, then because the coefficients of $g_0$ are real the conjugate $\bar{w}$ is also a fixed point, and one of $w,\bar{w}$ lies in $\mathbb{H}$, contradicting the freeness of the action of $G^{\mathbb{H}}$ on $\mathbb{H}$ (step 2.3). Hence every fixed point of $g_0$ lies in $\mathbb{R}\cup \{\mathrm{\infty }\}$, and there are one or two of them.

Hyperbolic normal form and its centralizer. Suppose $g_0\in G^{\mathbb{H}}$ is nonidentity with two fixed points in $\mathbb{R}\cup \{\mathrm{\infty }\}$ (step 3.1). Relabel them so either $v=\mathrm{\infty }$ with $u\in \mathbb{R}$, or $u,v\in \mathbb{R}$ with $u<v$. If $v=\mathrm{\infty }$, take $T(z)=z-u$; otherwise take $T(z)=(z-u)\mathrm{/}(v-z)$. These are real Mobius maps with positive determinant, hence automorphisms of $\mathbb{H}$ by [F6], and they carry the fixed points to $0$ and $\mathrm{\infty }$. Then $h:=T{g}_0{T}^{-1}$ fixes $0$ and $\mathrm{\infty }$, so $h(z)=\lambda z$ for some $\lambda \mathrm{\ne }1$; preserving $\mathbb{H}$ forces $\lambda >0$. The centralizer of $h$ in $\mathrm{Aut}(\mathbb{H})$ is exactly $\{z\mapsto kz:k>0\}$. Indeed, write a commuting automorphism as $u_0(z)=(az+b)\mathrm{/}(cz+d)$ with real coefficients and positive determinant [F6]. The identity $u_0(\lambda z)=\lambda u_0(z)$ gives $ac=bc=bd=0$ after cross-multiplication, since $\lambda >0$ and $\lambda \mathrm{\ne }1$. If $c\mathrm{\ne }0$, then $a=b=0$, contradicting $ad-bc\mathrm{\ne }0$; hence $c=0$. The determinant then forces $a,d\mathrm{\ne }0$, and $bd=0$ forces $b=0$, leaving $u_0(z)=(a\mathrm{/}d)z$ with $a\mathrm{/}d>0$.

Parabolic normal form and its centralizer. Suppose $g_0\in G^{\mathbb{H}}$ is nonidentity with exactly one fixed point $u\in \mathbb{R}\cup \{\mathrm{\infty }\}$ (step 3.1). Choose $T\in \mathrm{Aut}(\mathbb{H})$ carrying $u$ to $\mathrm{\infty }$: take $T=\mathrm{id}$ if $u=\mathrm{\infty }$, and $T(z)=-1\mathrm{/}(z-u)$ if $u\in \mathbb{R}$. The latter is a real Mobius map with determinant $1$ and hence an automorphism of $\mathbb{H}$ [F6]. Then $h:=T{g}_0{T}^{-1}$ fixes $\mathrm{\infty }$ and no other point, so $h(z)=\alpha z+\beta$ with real $\alpha >0$; the second fixed point $\beta \mathrm{/}(1-\alpha )$ would be finite if $\alpha \mathrm{\ne }1$, so $\alpha =1$ and $\beta \mathrm{\ne }0$. If $\beta <0$, replace $h$ by $h^{-1}$ and $\beta$ by $-\beta$, so $\beta >0$. Conjugating by the positive dilation $z\mapsto z\mathrm{/}\beta$ and composing it with $T$ normalizes the translation to $h(z)=z+1$. The centralizer of $h$ in $\mathrm{Aut}(\mathbb{H})$ is exactly $\{z\mapsto z+t:t\in \mathbb{R}\}$: if $u_0$ commutes with $h$, then $u_0(\mathrm{\infty })$ is a fixed point of $h$, hence is $\mathrm{\infty }$; so $u_0(z)=\alpha z+\gamma$ with real $\alpha >0$ [F6], and commutation gives $\alpha =1$.

An abelian group in a centralizer is cyclic. Assume $G$ is abelian, as will be the case when the genus is $1$. If $G$ is trivial the conclusion holds; otherwise choose a nonidentity element $g_0\in G^{\mathbb{H}}$. In the hyperbolic or parabolic case of steps 4.1 and 4.2, conjugation carries $G^{\mathbb{H}}$ to a free properly discontinuous group $H$ containing the normalized element $h$. Since $G$ is abelian, $H$ is abelian, so every element of $H$ commutes with $h$ and lies in the centralizer computed in those steps. Under the identification of that centralizer with $(\mathbb{R},+)$ (directly for translations and by $k\mapsto \log k$ for positive dilations), $H$ corresponds to a subgroup $\mathrm{\Gamma }\le \mathbb{R}$. If $\mathrm{\Gamma }$ were not discrete, the choice allowed by the Axiom of Choice would give nonzero parameters ${\gamma }_n\to 0$. After passing to a distinct subsequence, the corresponding automorphisms converge uniformly to the identity on a compact neighbourhood $K$ of a point of $\mathbb{H}$, so $h_{{\gamma }_n}(K)\cap K\mathrm{\ne }\mathrm{\varnothing }$ for infinitely many distinct elements, contradicting proper discontinuity. Hence $\mathrm{\Gamma }$ is discrete. A discrete subgroup of $\mathbb{R}$ is trivial or infinite cyclic, so $G\cong H$ is trivial or infinite cyclic.

The disc model forces $g\ge 2$. Suppose $M=\mathbb{D}$, so $X\cong \mathbb{D}\mathrm{/}G$ with $G$ acting freely and properly discontinuously by automorphisms of $\mathbb{D}$ (step 1.1). If $g=0$, then $X\cong S^2$ by [F1], so ${\pi }_1(X)=0$ by [F4] and therefore $G\cong {\pi }_1(X)$ is trivial by [F3]; then [F2] makes the covering $p$ a biholomorphism, so $X\cong \mathbb{D}$, which is impossible because $X$ is compact and $\mathbb{D}$ is not compact by [F9]. If $g=1$, then $X\cong T^2$ by [F1], so ${\pi }_1(X)\cong {\mathbb{Z}}^2$ by [F4] and $G\cong {\mathbb{Z}}^2$ by [F3]. In particular $G$ is abelian and nontrivial, so step 5.1 makes it infinite cyclic, contradicting $G\cong {\mathbb{Z}}^2$. Hence $M=\mathbb{D}$ implies $g\mathrm{\notin }\{0,1\}$, that is, $g\ge 2$.

Elimination and conclusion. Exactly one model $M$ occurs for $X$ by [F2], so the three cases of steps 2.1, 2.2 and 6.1 are exhaustive and mutually exclusive. If $g=0$, then $M\mathrm{\ne }\mathbb{C}$ by step 2.2 and $M\mathrm{\ne }\mathbb{D}$ by step 6.1, so $M=\widehat{\mathbb{C}}$ and $X$ has spherical type. If $g=1$, then $M\mathrm{\ne }\widehat{\mathbb{C}}$ by step 2.1 and $M\mathrm{\ne }\mathbb{D}$ by step 6.1, so $M=\mathbb{C}$ and $X$ has parabolic type. If $g\ge 2$, then $M\mathrm{\ne }\widehat{\mathbb{C}}$ by step 2.1 and $M\mathrm{\ne }\mathbb{C}$ by step 2.2, so $M=\mathbb{D}$ and $X$ has hyperbolic type. The Axiom of Choice [A1] enters through [F1], [F2] and [F7] and licenses the countable sequence selected in step 5.1; all other selections are finite.