TITLE: Miranda's Exercise on the Jacobian of a complex torus
QUESTION [5 upvotes]: I have to prove that the Jacobian of a complex torus $X=\mathbb{C}/L$ is isomorphic to $X$ by explicity showing that the subgroups of periods $\Lambda \subset \mathbb{C}$ is a lattice which is homotethic to the defining lattice $L$ for $X$, i.e. there is a nonzero complex number $\mu$ such that $\mu \Lambda=L$.
My idea, that I can't formalize, is the following:
the first homology group of the torus is the free group of rank $2$, i.e. $Z^2$, so the set $\Lambda =\{ \int_c \omega, \, \, c \in H_1(X,\mathbb{Z}) \}$ is of the form $\{n_1 \int_{\gamma_1} \omega+n_2 \int_{\gamma_2} \omega, \, \, n_1, n_2 \in \mathbb{Z} \}$. This implies that $\Lambda$ is a lattice and $Jac(X)=\mathbb{C}/\Lambda$ is a complex torus.
Now I have to prove that there is a nonzero complex number $\mu$ such that $\mu \Lambda=L$. How can I solve this exercise?

REPLY [7 votes]: We consider the lattice $L=\{m\cdot z_1+n\cdot z_2:m,n\in\mathbb{Z}\}$ for some $\mathbb{R}$-linearly independent $z_{1},z_{2}\in\mathbb{C}$. Let $\pi:\mathbb{C}\rightarrow X$ be the natural projection, where $X=\mathbb{C}/L$. 
The genus $g$ of $X$ is $1$, so
$$
1=g=\dim_\mathbb{C}\Omega^1(X). 
$$
It follows that $\mathrm{d}z$ generates $\Omega^1(X)$ as a $\mathbb{C}$-vector space. Therefore, the subgroup of periods $\Lambda\subseteq\mathbb{C}$ is
$$
\Lambda=\left\{\int_{[c]}\mathrm{d}z:[c]\in H^1(X,\mathbb{Z})\right\}.
$$
On the other hand, if we denote
$$
\gamma_1:[0,1]\rightarrow \mathbb{C}, t\mapsto \lambda\cdot z_1,
$$
$$
\gamma_2:[0,1]\rightarrow \mathbb{C}, t\mapsto \lambda\cdot z_2,
$$
we have that the classes of $a=\pi\circ\gamma_1$ and $b=\pi\circ\gamma_2$ generate the group $H^1(X,\mathbb{Z})$. It follows that
$$
\Lambda=\left\{ m\cdot\int_{[a]}\mathrm{d}z+n\cdot\int_{[b]}\mathrm{d}z:m,n\in\mathbb{Z}\right\}.
$$
Now, since the integral of a form along the push forward of a path is the integral of the pull back of the form along the path, it follows that
$$
\int_{[a]}\mathrm{d}z=\int_{\gamma_1}\mathrm{d}z=\int_0^1z_1=z_1,
$$
$$
\int_{[b]}\mathrm{d}z=\int_{\gamma_2}\mathrm{d}z=\int_0^1z_2=z_2.
$$
Hence, $\Lambda=L$ and therefore
$$
\mathrm{Jac}(X)=\mathbb{C}/\Lambda=\mathbb{C}/L=X.
$$