Torus and Moduli space

Let w_1, w_2\in \mathbb{C}, define group G\subset \text{Aut}(\mathbb{C}) generated by two operations

z\longrightarrow z+w_1

z\longrightarrow z+w_2

Assume w_1, w_2 are real linear independent, then \mathbb{C}/G is called a torus. In general we would orient the torus by assuming  \displaystyle \text{Im}\frac{w_2}{w_1}>0.

\mathbf{Proposition:}  \mathbb{C}/G_1 and \mathbb{C}/G_2 are conformally equivalent if and only if G_1 and G_2 are conjugate in \text{Aut}(\mathbb{C}).

\mathbf{Proof:} If G_1 and G_2 are conjugate then it is easy to prove the conclusion. 

 Suppose \phi: \mathbb{C}/G_1\cong \mathbb{C}/G_2 and \pi_i:\mathbb{C}\to \mathbb{C}/G_i. Since the universal cover of \mathbb{C}/G_i is \mathbb{C}, by the lifting lemmam, there exists a unique \tilde{f}: \mathbb{C}\to \mathbb{C} such that the following diagram commutes

From f^{-1}, \exists ! \psi. Then \tilde{f}\psi=\psi\tilde{f}=id, which means \tilde{f}\in \text{Aut}(\mathbb{C}).

Suppose g_1\in G_1, then from \pi_2\circ \tilde{f}=f\circ \pi_1,  \pi_2\tilde{f}g_1(z)=f \pi_1(z)=\pi_2\tilde{f}(z). So there exists \tau_z\in G_2 such that \tilde{f}g_1(z)=\tau_z(\tilde{f}(z)). Actually, the choice of \tau_z does not depend on z. This is because the group G_1 and G_2 act discontinouly on \mathbb{C}, then there is a neighborhood of z such that \tau_z remain the same.

So fix z_0\in\mathbb{C}, \displaystyle S=\{z|\tau_z=\tau_{z_0}\} is an open set. S is also closed, since G_1,G_2 are compatible with the topology of \mathbb{C} and f,\tilde{f} are continuous. So S=\mathbb{C}, which means for a fixed g_1, \exists\, \tau\in G_2 such that \tilde{f}\circ g_1=\tau\circ \tilde{f}. Thus \tilde{f}G_1\tilde{f}^{-1}\subset G_2. For the same sake, we have \tilde{f}^{-1}G_2\tilde{f}\subset G_1. G_1 and G_2 are conjugate in \text{Aut}(\mathbb{C}). \text{Q.E.D. }\hfill \square

Consider \mathbb{C}/G_1\cong \mathbb{C}/G_2. Suppose G_1 is generated by the mobius transformation z\to z+w_1 and z\to z+w_2, G_2 is generated by z\to z+\tau_1 and z\to z+\tau_2. Let us denote this four mobius transformations as w_1, w_2,\tau_1,\tau_2 and \displaystyle w=\frac{w_1}{w_2}, \displaystyle \tau=\frac{\tau_1}{\tau_2}.

Since \text{Aut}(\mathbb{C})=\{az+b|a\neq 0\}, \forall\, f=az+b, fG_1f^{-1}=\langle aw_1,aw_2\rangle. Then \exists\,k,l,m,n\in \mathbb{Z} such that



Then \displaystyle w=\frac{k\tau+l}{m\tau+n}.  Since w and \tau are symmetric, \begin{pmatrix} k &l\\m & n \end{pmatrix} must have inverse in \text{Mat}_{2}(\mathbb{Z}). Thus \det \begin{pmatrix} k &l\\m & n \end{pmatrix}=\pm 1. Since we require \text{Im}\, w and \text{Im}\, \tau be positive, then \det \begin{pmatrix} k &l\\m & n \end{pmatrix}=1.

\mathbf{Thm:} \mathbb{C}/G_1 and \mathbb{C}/G_2 are conformally equivalent if and only if \displaystyle \frac{w_1}{w_2}\sim \frac{\tau_1}{\tau_2} in PSL(2,\mathbb{Z}). The module space of torus is PSL(2,\mathbb{Z}).

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