On affine Riemann surfaces

Proof. \(S\) is a connected smooth \(1\) dimensional complex manifold. Our argument constructs global coordinates on the universal covering space of \(S\). We begin. For every \((z,w) \in S\) the complex tangent space to \(S\) at \((z,w)\) is \(\ker dF (z,w)\), where \[(0,0) \ne dF (z,w) =du_{|S}(z,w) + \mathrm{i}\, dv_{|S}(z,w) = (du + \mathrm{i} \, dv)_{|S}(z,w)\] for every \((z,w) \in S\). Thus the nonzero vector field \(X_F = (X_u + \mathrm{i}\, X_v)_{|S}\) spans the complex tangent space of \(S\) at each point of \(S\). Because \(X_F\) is nonzero on \(S\), the real vector fields \({X_u}_{|S}\) and \({X_{v}}_{|S}\) are linearly independent at each point of \(S\). To see this we argue as follows. Suppose that the real vector fields \(X_{u_{|S}}\) and \(X_{v_{|S}}\) are linearly dependent at some point \((z,w) \in S\). Then \(\text{span}_{\mathbb{R}} \{ X_{u_{|S}}(z,w), X_{v_{|S}}(z,w) \}\) has real dimension \(1\). Thus \((X_{u_{|S}} + \mathrm{i}\, X_{v_{|S}})(z,w)\) does not span the complex tangent space to \(S\) at \((z,w)\), which is a contradiction.

Consider the \(2\)-form \({\Omega }_{|S}\) on \(S\). Since \(\Omega\) is closed, it follows that \({\Omega }_{|S}\) is closed. Because \({X_u}_{|S}\) and \({X_v}_{|S}\) are linearly independent vector fields on \(S\) and \(\Omega\) is nondegenerate on \({\mathbb{R}}^4\), it follows that \({\Omega }_{|S}\) is nondegenerate on \(\text{span}_{\mathbb{R} } \{ {X_u}_{|S}(z,w),\) \({X_v}_{|S}(z,w) \}\) for every \((z,w) \in S\). To see this from \((X_u + \mathrm{i}\, X_v) \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $} \Omega = dF\) and the fact that \(dF \ne (0,0)\) on \(S\) we get \(\Omega (X_u , X_v) \ne 0\) on \(S\). Hence \({\Omega }_{|S}\) is a symplectic form on \(S\).

Let \(M\) be the universal covering space of \(S\) with covering mapping \(\rho : M \rightarrow S\). Because \(\rho\) is a local diffeomorphism, the \(2\)-form \(\omega = {\rho }^{\ast} ({\Omega }_{|S})\) on \(M\) is symplectic. Consider the smooth functions \(U = {\rho }^{\ast }(u_{|S})\) and \(V = {\rho }^{\ast }(v_{|S})\) on \((M, \omega )\). The corresponding Hamiltonian vector fields \(X_U\) and \(X_V\) on \((M, \omega )\) are given by \(dU = X_U \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}\omega\) and \(dV = X_V \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}\omega\). Since \[\begin{aligned} X_U \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}\omega & = dU = d\, ({\rho }^{\ast}u_{|S}) = {\rho }^{\ast }(du_{|S}) = {\rho }^{\ast }(X_{u_{|S}} \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}{\Omega }_{|S} ) \\ & = {\rho }^{\ast}(X_{u_{|S}}) \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}{\rho }^{\ast }({\Omega }_{|S}) = {\rho }^{\ast }(X_{u_{|S}}) \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}\omega , \end{aligned}\] it follows that \(X_U = {\rho }^{\ast }(X_{u_{|S}})\), because \(\omega\) is nondegenerate. Similarly, \(X_V = {\rho }^{\ast }(X_{v_{|S}})\). Since \(\rho\) is a local diffeomorphism and the vector fields \(X_{u_{|S}}\) and \(X_{v_{|S}}\) are linearly independent at each point of \(S\), the vector fields \(X_U\) and \(X_V\) are linearly independent at each point of \(M\). Thus the \(1\)-forms \(dU\) and \(dV\) on \(M\) are linearly independent at each point of \(M\), because \(\omega\) is nondegenerate. So the vector fields \(\frac{\partial }{\partial U}\) and \(\frac{\partial }{\partial V}\) are linearly independent at each point of \(M\).

Consider the nonzero \(2\)-form \(\varpi = dV \wedge dU\) on \(M\). Since \(M\) is \(2\)-dimensional, the de Rham cohomology group of \(2\)-forms on \(M\) has dimension \(1\). Thus \(\varpi = a \omega\) for some nonzero real number \(a\).¹ Because \(\{ \frac{\partial }{\partial U}, \frac{\partial }{\partial V} \}\) is a basis of the tangent space of \(M\) at each point of \(M\), we may write \(X_U = A \frac{\partial }{\partial U} + B\frac{\partial }{\partial V}\). Then \[dU = X_U \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}\omega = \frac{1}{a} X_U \mbox{$\, \rule{8pt}{.5pt}\rule{.5pt}{6pt}\, \, $}\varpi = \frac{1}{a}(B dU – A dV),\] which implies \(X_U = a \frac{\partial }{\partial V}\). A similiar argument shows that \(X_V = -a \frac{\partial }{\partial U}\).

The pair of functions \((U,V)\) are coordinates on \(M\), since the vector fields \(\frac{1}{a} X_U = \frac{\partial }{\partial V}\) and \(-\frac{1}{a}X_V = \frac{\partial }{\partial U}\) are linearly independent at each point of \(M\) and commute. This latter assertion follows because \[\begin{aligned} \{ u, v \} & = L_{X_v}u = L_{X_{\mathrm{Im}\, F}}(\mathrm{Re}\, F) = L_{\frac{1}{2\mathrm{i}}(X_{F – \mathrm{i}\, F})}\mbox{${\scriptstyle \frac{1}{2}\, }$}(F +\mathrm{i}\, F) \\ & = \frac{1}{4i}[ L_{X_F}F + \mathrm{i} \, L_{X_F}F – \mathrm{i}\, L_{X_F}F + L_{X_F}F ] = 0 \end{aligned}\] implies \([X_v, X_u] = X_{\{ u,v \} } = 0\). From \[T\rho \, [X_U,X_V] = [X_u|S, X_v|S ] \circ \rho = {[X_u, X_v]}_{|S} \circ \rho = 0,\] we get \([X_U,X_V] =0\), because \(\rho\) is a local diffeomorphism. Hence \([\frac{\partial }{\partial U}, \frac{\partial }{\partial V}] =0\). Thus we may identify \(M\) with \({\mathbb{R} }^2\). ◻

Proof. The flow of \(X_{U + \mathrm{i}\, V}\) on \(\mathbb{C}\) is \(U(t) + \mathrm{i}\, V(t) = \big( U(0) + \mathrm{i}at \big) +\big( \mathrm{i}V(0) -at \big)\), since \(X_U = a \frac{\partial }{\partial V}\) and \(X_V = -a \frac{\partial }{\partial U}\). Hence an integral curve of \(X_{U + \mathrm{i}\, V}\) starting at \(U(0) + \mathrm{i}\, V(0)\) is \(t \mapsto \big( U(0) + \mathrm{i}\, V(0) \big) + a( -t + \mathrm{i}t)\), which is a straight line in \(\mathbb{C}\). Thus the flow of \(X_{U + \mathrm{i} V}\) is linear. Since \[\begin{aligned} T\rho X_{U + \mathrm{i}\, V} & = T\rho (X_U + \mathrm{i}\, X_V) = T\rho X_U + \mathrm{i}\, T\rho X_V \\ & = X_{u_{|S}} \circ \rho + \mathrm{i}\, X_{v_{|S}} \circ \rho = X_{(u+\mathrm{i}v)_{|S}} \circ \rho = {X_{F}}_{|S} \circ \rho , \end{aligned}\] the image of the flow of \(X_{U+ \mathrm{i}\, V}\) under the covering map \(\rho\) is the flow of \(X_F\). ◻

Proof. It suffices to show that for every \((z,w) \in S\) \[T_{(z,w)} \delta \, X_F(z,w) = \alpha \frac{\partial }{\partial \zeta } {\rule[-10pt]{.5pt}{22pt}}_{\zeta = \delta(z,w)}. \label{eq-ex-four} \tag{4}\] This we do as follows. Using the definition of the map \({\pi }_{|S}\) and the vector field \((X_F)_{|S} = 2w \frac{\partial }{\partial z} – 6w^5\frac{\partial }{\partial w}\), for every \((z,w) \in S\) we get \[T_{(z,w)}{\pi }_{|S} \, X_F(z,w) = T_{(z,w)}{\pi }_{|S}(2w \frac{\partial }{\partial z} – 6w^5\frac{\partial }{\partial w}) = 2w \frac{\partial }{\partial z} .\] By definition of the function \(f\) (2) we have \(df = \frac{1}{2w} dz\), which implies \(T_zf \big( 2w \frac{\partial }{\partial z} \big)\) \(= \frac{\partial }{\partial \zeta }\). Thus for every \((z,w) \in S\) \[\begin{aligned} T_{(z,w)}\delta \, X_F(z,w) &= \alpha T_zf \Big( T_{(z,w)}{\pi }_{|S} \big( X_F(z,w) \big) \Big) = \alpha \frac{\partial }{\partial \zeta }, \end{aligned}\] which establishes Eq. (4). ◻

Proof. This follows from Eq. (4), which shows that the tangent map of \(\delta\) is injective at each point of \(S\). ◻

Proof of Proposition 2. Let \(U + \mathrm{i}V = {\rho }^{\ast }(\mathrm{Re}\, F) + \mathrm{i}\, {\rho }^{\ast }(\mathrm{Im}\, F)\). By Proposition 1, \(U+\mathrm{i}\, V\) is a coordinate on \(\mathbb{C}\). Define the diffeomorphism \[\lambda : \mathbb{C} \rightarrow \mathbb{C}: \zeta \mapsto U + \mathrm{i}\, V\] by requiring \({\lambda }_{\ast }\big( \alpha \frac{\partial }{\partial \zeta } \big) = X_U + \mathrm{i}\, X_V\), that is, set \(U = \lambda (\mathrm{Re}\,\zeta )\) and \(V = \lambda (\mathrm{Im}\, \zeta )\). By construction we have \(\alpha \frac{\partial }{\partial \zeta } = {\lambda}^{\ast }{\rho }^{\ast } \big( (X_F)_{|S} \big)\), see the proof of Proposition 1. By Eq. (4) we have \(\alpha \frac{\partial }{\partial \zeta } = {\delta }_{\ast }\big( (X_F)_{|S} \big)\). Thus \({\delta }_{\ast } = {\lambda}^{\ast }{\rho }^{\ast }\), which implies \(\rho \circ \lambda \circ \delta = {\mathrm{id}}_S\). To see this suppose that \(\rho \circ \lambda \circ \delta \ne {\mathrm{id}}_S\). Then \({\delta }^{\ast } \circ (\rho \circ \lambda )^{\ast } \ne \mathrm{id}_{TS}\). Hence \({\lambda }^{\ast }{\rho }^{\ast } \ne {\delta }_{\ast }\), which is a contradiction.

Let \[R: \mathbb{C} \rightarrow \mathbb{C} : z \mapsto {\mathrm{e}}^{2\pi \mathrm{i}/6}z. \label{eq-ex-five} \tag{5}\] Then \(f(Rz) = Rf(z)\), where \(f\) is the function defined in (2). To see this we compute. \[\begin{aligned} f(Rz) & = \int^{Rz}_0 \frac{d\xi}{2w(\xi)}, \, \, \mbox{where $w(\xi ) = \sqrt{1-{\xi }^6}$} \\ & = \int^z_0 \frac{R dz}{2 w(z)}, \, \, \mbox{using $\xi = Rz$ and $w(Rz) = w(z)$} \\ & = Rf(z). \end{aligned}\] Thus up to a dialation the image under \(f\) (2) of the closed equilateral triangle \[T' = \{ z = r' {\mathrm{e}}^{\mathrm{i} {\theta}' } \in \mathbb{C} \, \rule[-4pt]{.5pt}{13pt}\, \, 0 \le r' \le 1 \, \, \& \, \, 0 \le {\theta}' \le 2\pi /6 \}\] with vertex at the origin and one edge of length \(1\) along the real axis is the equilateral triangle \[T = f(T') = \{ \zeta = r {\mathrm{e}}^{\mathrm{i} \theta } \in \mathbb{C} \, \rule[-4pt]{.5pt}{13pt}\, \, 0 \le r \le C \, \, \& \, \, 2\pi /6 \le \theta \le 4\pi /6 \} = CR(T'),\] where \(C=\int^1_0 \frac{dz}{\sqrt {1-z^6}}\). Hence \(f\) maps a regular hexagon into another. In particular, it sends the closed regular hexagon \(H'\) with center at the origin \(O\) and edge length \(1\) onto the regular hexagon \(H\) with center at \(O\) and edge length \(C\). Since \(H'\) is simply connected and is contained in the unit disk \(\{ | z | \le 1 \}\), the complex square root \(\sqrt{1-z^6}\) is single valued for all \(z \in H'\). Thus \(H'\) is the image under \({\pi }_{|S}\) of a domain \(\mathcal{D} \subseteq S\), which is contained in some sheet \(S_{{\ell }'}\) of the covering map \(\rho\) of \(S\).

Let \[U: \mathbb{C} \rightarrow \mathbb{C} : z \mapsto \overline{z}. \label{eq-ex-six} \tag{6}\] The regular hexagon \(H\) is invariant under the action of the group \(\mathcal{G}\), generated be the rotation \(R\) and the reflection \(RU\) in the diagonal of \(H\), which is an edge of the triangle \(T\) with the orgin as an end point that is not the real axis. The map \(\delta\) (3) intertwines the action of the group \(G'\) generated by \(\mathcal{R}\) and \(\mathcal{RU}\) on \(S\) with the action of the \(\mathcal{G}\) on \(H\). Thus the domain \(\mathcal{D}\) contains a fundamental domain of the action of the covering group \(G\) on \({\mathbb{R} }^2\).

Let \(\mathcal{T}\) be the abelian group generated by the translations \[{\tau }_k : \mathbb{C} \rightarrow \mathbb{C} : z \mapsto z + u_k, \, \, \mbox{for $k=0,1, \ldots , 5$.}\] Here \(u_k = \sqrt{3}C\, {\mathrm{e}}^{2\pi \mathrm{i}(1/12+ k/6)}\), which is perpendicular to an edge of the equilateral triangle \(R^k(T)\) that lies on the boundary of the hexagon \(H\). The action of \(\mathcal{T}\) on \(\mathbb{C}\) has fundamental domain \(H\). To see this recall that in [2] it is shown that \[\bigcup_{n \ge 0}\hspace{-6pt} \bigcup_{\hspace{10pt}{\ell }_1 + \cdots + {\ell}_k =n} \hspace{-15pt} {\tau }^{{\ell }_1}_1 { \circ } \cdots { \circ } {\tau }^{{\ell }_k}_k(K) = \mathbb{C},\] where \(K\) is the closed stellated hexagon formed by placing an equilateral triangle of edge length \(C\) on each bounding edge of \(H\). But \[K = H \cup \bigcup^5_{k=0} {\tau }_k(R^{(4+k) \! \bmod 6}T).\] So \(H\) is the fundamental domain of the \(\mathcal{T}\) action on \(\mathbb{C}\). Because applying an element of \(G'\) to the domain \(\mathcal{D} \subseteq S\) gives a domain whose boundary has a nonempty intersection with the boundary of \(\mathcal{D}\), it follows that under the mapping \(\delta\) (3) the corresponding element of the group of motions in \(\mathbb{C}\) sends the hexagon \(H\) to a hexagon which has an edge in common with \(H\). Thus this group of motions is the group \(\mathcal{T}\). Because the mapping \(\delta\) intertwines the \(G'\) action on \(S\) with the \(\mathcal{T}\) action on \(\mathbb{C}\) and sends the domain \(\mathcal{D} \subseteq S_{{\ell }'}\) diffeomorphically onto \(H\), it follows that \(\mathcal{D}\) is a fundamental domain for the action of \(G'\) on \(S\). Consider \(\lambda (H)\), which is a regular hexagon with center at the origin, since the coordinate change \(\lambda\) maps straight lines to straight lines. From proposition 2.1 we deduce that \(\lambda (H)\) is a fundamental domain for the action of the covering group \(G\) on \(\mathbb{C} = {\mathbb{R} }^2\) of the affine Riemann surface \(S\). Hence \(S = {\mathbb{R} }^2/\mathcal{T}\). ◻