A compact free affine plane quotient comes from a rank-two lattice

Every $g\in G$ is a biholomorphism of $\mathbb{C}$, so $g(z)=az+b$ with $a,b\in \mathbb{C}$, $a\mathrm{\ne }0$.

$G$ is nontrivial: if $G=\{e\}$ then the quotient map $q$ is a bijection, so [F8] makes $q$ a homeomorphism and $\mathbb{C}\mathrm{/}G\cong \mathbb{C}$. The discs $D(0,n)$ for $n\ge 1$ cover $\mathbb{C}$, while any finitely many of them, $D(0,n_1),\dots ,D(0,n_k)$, are all contained in $D(0,N)$ with $N:={\max }_{1\le i\le k}{n}_i$, a proper subset of $\mathbb{C}$; so no finite subfamily covers $\mathbb{C}$, and $\mathbb{C}$ is not compact. This contradicts the hypothesis that $\mathbb{C}\mathrm{/}G$ is compact.

Suppose $a\mathrm{\ne }1$ for some $g\in G$ of the form $g(z)=az+b$. Then $z_0:=b\mathrm{/}(1-a)$ satisfies $g(z_0)=z_0$, so freeness forces $g=e$; but the identity map is $z\mapsto 1\cdot z+0$, whose linear coefficient is $a=1$, a contradiction. Hence $a=1$ for every $g\in G$, that is, every element of $G$ is a translation $z\mapsto z+\lambda$, and the identity corresponds to $\lambda =0$ while every nonidentity element corresponds to a nonzero $\lambda$.

Composition of translations adds vectors and inverses subtract them, so $\mathrm{\Lambda }=\{\lambda \in \mathbb{C}:(z\mapsto z+\lambda )\in G\}$ is a subgroup of $(\mathbb{C},+)$ containing $0$, and $G=\{t_{\lambda }:\lambda \in \mathrm{\Lambda }\}$ where $t_{\lambda }(z)=z+\lambda$; in particular $g\cdot z=z+\lambda$ for $g=t_{\lambda }$. By step 1.2 this subgroup is nontrivial, so $\mathrm{\Lambda }\mathrm{\ne }\{0\}$.

The closed unit disc $K:=\overline{D(0,1)}\subseteq \mathbb{C}$ is compact. Proper discontinuity applied to $K$ makes $S:=\{g\in G:g\cdot K\cap K\mathrm{\ne }\mathrm{\varnothing }\}$ finite. If $\lambda \in \mathrm{\Lambda }$ with $\mathrm{\mid }\lambda \mathrm{\mid }\le 2$, then $z:=\lambda \mathrm{/}2$ has $\mathrm{\mid }z\mathrm{\mid }\le 1$ and $\mathrm{\mid }z-\lambda \mathrm{\mid }=\mathrm{\mid }-\lambda \mathrm{/}2\mathrm{\mid }\le 1$, so $z\in K$ and $z-\lambda \in K$, whence $z=(z-\lambda )+\lambda \in (K+\lambda )\cap K=t_{\lambda }\cdot K\cap K\mathrm{\ne }\mathrm{\varnothing }$ and $t_{\lambda }\in S$. Therefore $\mathrm{\Lambda }\cap \overline{D(0,2)}$ is contained in the set of translation vectors of the finite set $S$, so it is finite.

The quotient map $q$ is open: for open $V\subseteq \mathbb{C}$ one has $q^{-1}(q(V))={\bigcup }_{\lambda \in \mathrm{\Lambda }}(V+\lambda )$, a union of open sets, hence open in $\mathbb{C}$, so $q(V)$ is open in $\mathbb{C}\mathrm{/}G$ by [F8].

The finite set $\mathrm{\Lambda }\cap \overline{D(0,2)}$ is either $\{0\}$ or contains an element of positive modulus; in the first case put $\rho :=1$, and in the second case put $\rho :=\min \{\mathrm{\mid }\lambda \mathrm{\mid }:\lambda \in \mathrm{\Lambda },0<\mathrm{\mid }\lambda \mathrm{\mid }\le 2\}>0$, a minimum of a finite nonempty set of positive real numbers. In both cases $\mathrm{\Lambda }\cap D(0,\rho )=\{0\}$. Since $\mathrm{\Lambda }$ is an additive subgroup, every ${\lambda }_0\in \mathrm{\Lambda }$ then satisfies $\mathrm{\Lambda }\cap D({\lambda }_0,\rho )=\{{\lambda }_0\}$: if $\lambda \in \mathrm{\Lambda }$ with $\mathrm{\mid }\lambda -{\lambda }_0\mathrm{\mid }<\rho$, then $\lambda -{\lambda }_0\in \mathrm{\Lambda }\cap D(0,\rho )=\{0\}$. Thus distinct elements of $\mathrm{\Lambda }$ have distance at least $\rho$, and $\mathrm{\Lambda }$ is discrete. Every compact subset of $\mathbb{C}$ is covered by finitely many discs of radius $\rho \mathrm{/}3$, each meeting $\mathrm{\Lambda }$ in at most one point, so it meets $\mathrm{\Lambda }$ in a finite set. In particular, if $\mathrm{\Lambda }\mathrm{\ne }\{0\}$ and $w\in \mathrm{\Lambda }\setminus \{0\}$, the set $\mathrm{\Lambda }\cap \overline{D(0,\mathrm{\mid }w\mathrm{\mid })}$ is finite and nonempty, and its element of least positive modulus is an element of $\mathrm{\Lambda }\setminus \{0\}$ of least modulus overall.

By step 5.1 and step 3.1 choose $v\in \mathrm{\Lambda }\setminus \{0\}$ of least modulus. Let $L:=\mathbb{R}v$, let $P:\mathbb{C}\to L^{\perp }$ be the orthogonal projection onto the real line perpendicular to $v$, and let $H:=P(\mathrm{\Lambda })$, a subgroup of $(L^{\perp },+)$, which we identify with $(\mathbb{R},+)$.

A discrete subgroup $\mathrm{\Gamma }$ of $(\mathbb{R},+)$ is either $\{0\}$ or of the form $\mathbb{Z}\gamma$ for some $\gamma \mathrm{\ne }0$: if $\mathrm{\Gamma }\mathrm{\ne }\{0\}$ and $x_1\in \mathrm{\Gamma }\setminus \{0\}$, then $\mathrm{\Gamma }\cap [-\mathrm{\mid }{x}_1\mathrm{\mid },\mathrm{\mid }{x}_1\mathrm{\mid }]$ is finite by discreteness, so there is $\gamma \in \mathrm{\Gamma }\setminus \{0\}$ of least modulus, and for arbitrary $x\in \mathrm{\Gamma }$ Euclidean division gives $m\in \mathbb{Z}$ with $\mathrm{\mid }x-m\gamma \mathrm{\mid }\le \mathrm{\mid }\gamma \mathrm{\mid }\mathrm{/}2<\mathrm{\mid }\gamma \mathrm{\mid }$; the element $x-m\gamma \in \mathrm{\Gamma }$ therefore vanishes and $x=m\gamma$.

The quotient map $q$ is a covering map. Let $\epsilon :=\inf \{\mathrm{\mid }\lambda \mathrm{\mid }:\lambda \in \mathrm{\Lambda }\setminus \{0\}\}>0$, where positivity holds by step 5.1. For $z\in \mathbb{C}$ put $D_z:=D(z,\epsilon \mathrm{/}3)$ and $U_z:=q(D_z)$. If $w,w^{\mathrm{\prime }}\in D_z$ with $q(w)=q(w^{\mathrm{\prime }})$ then $w-w^{\mathrm{\prime }}\in \mathrm{\Lambda }$ and $\mathrm{\mid }w-w^{\mathrm{\prime }}\mathrm{\mid }<2\epsilon \mathrm{/}3<\epsilon$, so $w=w^{\mathrm{\prime }}$; thus $q{\mathrm{\mid }}_{D_z}$ is a bijection onto $U_z$, and it is a homeomorphism because for open $A\subseteq D_z$ the set $q(A)$ is open in $\mathbb{C}\mathrm{/}G$ by step 4.2, hence open in $U_z$. Moreover $q^{-1}(U_z)={\bigcup }_{\lambda \in \mathrm{\Lambda }}D(z+\lambda ,\epsilon \mathrm{/}3)$: a point of $D(z+\lambda ,\epsilon \mathrm{/}3)$ has the form $u+\lambda$ with $u\in D_z$ and maps to $q(u)\in U_z$, and conversely $q(w)\in U_z$ means $w=u+\lambda$ with $u\in D_z$, $\lambda \in \mathrm{\Lambda }$. These discs are pairwise disjoint, because $D(z+\lambda ,\epsilon \mathrm{/}3)\cap D(z+{\lambda }^{\mathrm{\prime }},\epsilon \mathrm{/}3)\mathrm{\ne }\mathrm{\varnothing }$ with $\lambda \mathrm{\ne }{\lambda }^{\mathrm{\prime }}$ would give $\mathrm{\mid }\lambda -{\lambda }^{\mathrm{\prime }}\mathrm{\mid }<2\epsilon \mathrm{/}3<\epsilon$; and on the disc with index $\lambda$ the map $q$ equals $q{\mathrm{\mid }}_{D_z}\circ t_{-\lambda }$, a homeomorphism onto $U_z$. Hence $U_z$ is an evenly covered neighbourhood of $q(z)$ and $q$ is a covering map.

The subgroup $H$ is discrete. If it were not, then taking $\sigma =\mathrm{\mid }v\mathrm{\mid }\mathrm{/}2$ there would be $0\mathrm{\ne }h\in H$ with $\mathrm{\mid }h\mathrm{\mid }<\mathrm{\mid }v\mathrm{\mid }\mathrm{/}2$; choose $\lambda \in \mathrm{\Lambda }$ with $P(\lambda )=h$ and write $\lambda =tv+h$ with $t\in \mathbb{R}$, choose $m\in \mathbb{Z}$ with $\mathrm{\mid }t-m\mathrm{\mid }\le 1\mathrm{/}2$, and set ${\lambda }^{\mathrm{\prime }}:=\lambda -mv\in \mathrm{\Lambda }$. Then ${\lambda }^{\mathrm{\prime }}=(t-m)v+h\mathrm{\ne }0$ because $h\mathrm{\ne }0$, while $\mathrm{\mid }{\lambda }^{\mathrm{\prime }}{\mathrm{\mid }}^2\le \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}4+\mathrm{\mid }h{\mathrm{\mid }}^2<\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2$, so $\mathrm{\mid }{\lambda }^{\mathrm{\prime }}\mathrm{\mid }<\mathrm{\mid }v\mathrm{\mid }$, contradicting the minimality of $\mathrm{\mid }v\mathrm{\mid }$ among nonzero elements of $\mathrm{\Lambda }$. Hence $H\cap D(0,\mathrm{\mid }v\mathrm{\mid }\mathrm{/}2)=\{0\}$, and the finite-minimum argument of step 5.1 applied inside the line $L^{\perp }\cong \mathbb{R}$ shows that $H$ is discrete.

The maps ${\phi }_z:=(q{\mathrm{\mid }}_{D_z}{)}^{-1}:U_z\to D_z\subseteq \mathbb{C}$ of step 6.3 are homeomorphisms onto open subsets of $\mathbb{C}$, and the sets $U_z$ cover $\mathbb{C}\mathrm{/}G$. If $W:=U_z\cap U_{z^{\mathrm{\prime }}}\mathrm{\ne }\mathrm{\varnothing }$, then on each connected component $C_0$ of ${\phi }_z(W)$ the difference $u\mapsto {\phi }_{z^{\mathrm{\prime }}}(q(u))-u$ is a continuous map into the discrete group $\mathrm{\Lambda }$, hence is constant. Thus each transition map is a translation on each component of its domain, so these charts define a compatible holomorphic atlas; in each chart the local expression of $q$ is the identity. The quotient is second countable: since $q$ is open by step 4.2, the images of a countable basis of $\mathbb{C}$ form a basis of $\mathbb{C}\mathrm{/}G$. Hausdorffness is established after the lattice structure is determined.

$H\mathrm{\ne }\{0\}$: suppose $H=\{0\}$, so that $\mathrm{\Lambda }\subseteq L$. Choose a nonzero $\mathbb{R}$-linear functional $\mathrm{\ell }:\mathbb{C}\to \mathbb{R}$ vanishing on $L$. Since $\mathrm{\ell }(\lambda )=0$ for every $\lambda \in \mathrm{\Lambda }$, the formula $\bar{\mathrm{\ell }}([z]):=\mathrm{\ell }(z)$ defines a map $\bar{\mathrm{\ell }}:\mathbb{C}\mathrm{/}G\to \mathbb{R}$; it is surjective because $\mathrm{\ell }$ is a surjective linear map, and it is continuous: for every open interval $(a,b)\subseteq \mathbb{R}$ its preimage satisfies $q^{-1}({\bar{\mathrm{\ell }}}^{-1}(a,b))={\mathrm{\ell }}^{-1}(a,b)$, which is open in $\mathbb{C}$, so ${\bar{\mathrm{\ell }}}^{-1}(a,b)$ is open in $\mathbb{C}\mathrm{/}G$ by [F8]. Then $\mathbb{R}=\bar{\mathrm{\ell }}(\mathbb{C}\mathrm{/}G)$ is a continuous image of the compact space $\mathbb{C}\mathrm{/}G$, hence compact. But the open cover $\{(-n,n):n\ge 1\}$ of $\mathbb{R}$ has no finite subcover, since a finite subcover is contained in $(-N,N)$ for the largest index $N$ and omits $N+1$; this contradiction shows $H\mathrm{\ne }\{0\}$.

Choose $w\in \mathrm{\Lambda }$ with $P(w)=h$, where $h\mathrm{\ne }0$ generates $H=\mathbb{Z}h$ as in step 6.2, and let $m\in \mathbb{Z}$ with $P(\lambda )=mh$ for a given $\lambda \in \mathrm{\Lambda }$. Then $\lambda -mw\in \mathrm{\Lambda }\cap \ker P=\mathrm{\Lambda }\cap L$, a discrete subgroup of the line $L$ containing $v$; applying step 6.2 inside $L\cong \mathbb{R}$ and using the minimality of $\mathrm{\mid }v\mathrm{\mid }$ gives $\mathrm{\Lambda }\cap L=\mathbb{Z}v$, so $\lambda =mw+nv$ for some $n\in \mathbb{Z}$. Hence $\mathrm{\Lambda }=\mathbb{Z}v+\mathbb{Z}w$. Finally $v$ and $w$ are linearly independent over $\mathbb{R}$: if $tv+sw=0$ with $s\mathrm{\ne }0$, then projecting onto $L^{\perp }$ gives $0=P(tv+sw)=tP(v)+sP(w)=sP(w)=sh\mathrm{\ne }0$, a contradiction; hence $s=0$, and then $tv=0$ with $v\mathrm{\ne }0$ gives $t=0$. So the only real relation is the trivial one.

The deck group of the covering $q$ is exactly $\{t_{\lambda }:\lambda \in \mathrm{\Lambda }\}$. Each $t_{\lambda }$ with $\lambda \in \mathrm{\Lambda }$ satisfies $q\circ t_{\lambda }=q$, so it is a deck transformation. Conversely, if $h$ is a deck transformation, then $q(h(0))=q(0)$, so $h(0)\in q^{-1}(q(0))=\mathrm{\Lambda }$, and the translation $t_{h(0)}$ is a deck transformation with $t_{h(0)}(0)=h(0)$; the total space $\mathbb{C}$ is connected, so $h=t_{h(0)}$ by [F5].

Let $T:{\mathbb{R}}^2\to \mathbb{C}$ be the $\mathbb{R}$-linear isomorphism $T(s,t):=sv+tw$ of step 9.1; it carries ${\mathbb{Z}}^2$ bijectively onto $\mathrm{\Lambda }$, so the formula $\bar{T}([(s,t)]):=q(T(s,t))$ is well defined, and it is a bijection $\bar{T}:{\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2\to \mathbb{C}\mathrm{/}G$ because $T$ and $q$ are onto and $T(x)-T(y)\in \mathrm{\Lambda }$ forces $x-y\in {\mathbb{Z}}^2$; it is continuous because $\bar{T}$ composed with the quotient map is the continuous map $q\circ T$. The set $\mathrm{\Lambda }$ is closed in $\mathbb{C}$: if $x$ were a limit of points of $\mathrm{\Lambda }$ outside $\mathrm{\Lambda }$, two nearby points $\lambda ,{\lambda }^{\mathrm{\prime }}\in \mathrm{\Lambda }$ would give $0<\mathrm{\mid }\lambda -{\lambda }^{\mathrm{\prime }}\mathrm{\mid }<\rho$ with the $\rho$ of step 5.1, contradicting $\mathrm{\Lambda }\cap D(0,\rho )=\{0\}$. Hence for distinct classes $[z]\mathrm{\ne }[z^{\mathrm{\prime }}]$ the number $\delta :=\mathrm{dist}(z-z^{\mathrm{\prime }},\mathrm{\Lambda })$ is positive, and the open sets $q(D(z,\delta \mathrm{/}2))$, $q(D(z^{\mathrm{\prime }},\delta \mathrm{/}2))$ are disjoint: a common class would give $u\in D(z,\delta \mathrm{/}2)$, $u^{\mathrm{\prime }}\in D(z^{\mathrm{\prime }},\delta \mathrm{/}2)$ with $u-u^{\mathrm{\prime }}\in \mathrm{\Lambda }$, forcing $\mathrm{dist}(z-z^{\mathrm{\prime }},\mathrm{\Lambda })<\delta$. So $\mathbb{C}\mathrm{/}G$ is Hausdorff. The quotient ${\mathbb{R}}^2\mathrm{/}{\mathbb{Z}}^2$ is compact, being a continuous image of the compact square $[0,1{]}^2$. A continuous bijection from a compact space onto a Hausdorff space is a homeomorphism, so $\bar{T}$ is a homeomorphism and $\mathbb{C}\mathrm{/}G$ is compact and Hausdorff; together with step 7.2's second-countable topology and holomorphic atlas, this makes $\mathbb{C}\mathrm{/}G$ a compact Riemann surface.

The standard torus $S^1\times S^1$ is the connected sum of one copy of itself, hence is ${\mathrm{\#}}_1{T}^2$ in the notation of [F10], and step 10.2 identifies the compact Riemann surface $\mathbb{C}\mathrm{/}G$ with it. By the definition of genus [F10] and the uniqueness of the homeomorphism type in the topological classification, the genus of $\mathbb{C}\mathrm{/}G$ is $1$. Steps 2.1, 9.1, 10.1 and 10.2 therefore exhibit $\mathbb{C}\mathrm{/}G$ as the complex torus $\mathbb{C}\mathrm{/}(\mathbb{Z}v+\mathbb{Z}w)$ whose deck group of translations is the rank-two lattice $\mathrm{\Lambda }=\mathbb{Z}v+\mathbb{Z}w\cong G$. The only use of the Axiom of Choice is through the genus definition [F10], which assumes it; every choice made in the argument itself was finite.