## Tag Archives: torus

### 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

$aw_1=k\tau_1+l\tau_2$

$aw_2=m\tau_1+n\tau_2$

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 $PSL(2,\mathbb{Z})$.