The biholomorphic structure of Torus We know that the tours  can be viewed as $\mathbb C/\Lambda$, where $\Lambda$ is the lattice in the complex plane.
Question: Why the Biholomorphic Mapping Group of Torus is isomorphic to $SL_2(\mathbb Z)$, i.e. two tori $\mathbb C/\Lambda$, $\mathbb C/\Lambda'$ are biholomorphic to each other, iff $\exists f\in SL_2(\mathbb Z)$ s.t. $f(\Lambda)=\Lambda'$.
Thanks.
 A: Let lattice $\Lambda$ be generated by two complex numbers $f_1$ and $f_2$. We need $\Lambda \cong \mathbb{Z}^2$, so $f_2/f_1$ can't be a real number, because in this case $\Lambda \cong \mathbb{Z}$. By dividing over $f_1$ we can assume without loss of generality that $f_1=1$ and $f_2$ is in the upper half plane (it is not real, if it is in the lower half-plane we just choose another generator in the upper half plane). So we parameterized all curves by points in the upper half-plane. 
We can choose a different basis of $\Lambda$ of the same form $\{1,f_2'\}$, it is related to the given basis by an element of $GL_2(\mathbb{R})$. And the torus $\mathbb{C}/\Lambda$ does not depend on which basis we choose. 
The lattices $\Lambda$ and $\Lambda'$ give the same curve if and only if there is a non-zero complex number $t$ s.t. $\Lambda=t\Lambda'$. So, the center $GL_2(\mathbb{R})$ acts trivially and we get isomorphic curves if and only if $\Lambda' = f(\Lambda)$, where $f \in PGL_2(\mathbb{R})=PSL_2(\mathbb{R})$, where $f$ changes the basis.
