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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?

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  • $\begingroup$ Hint: Write out the first homology group in terms of $L$. $\endgroup$ – rfauffar Jan 14 '15 at 13:35
  • $\begingroup$ @rfauffar The fundamental polygon of a complex torus $\mathbb{C}/L$ is a parallelogram with sides $w_1,w_2$ where $w_1,w_2$ are the linearly indipendent vectors that generate the lattice $L$. So, in terms of $L$, $H^1(X)=\{n w_1+m w_2, \, n,m \in \mathbb{Z} \}$. Is there a relation (possibly homotethic) between the complex numbers $w_i$ and $\int_{\gamma_i} \omega$, for $i=1,2$? I stopped here during the resolution of the exercise and I can't continue... $\endgroup$ – TheWanderer Jan 15 '15 at 9:54
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    $\begingroup$ Yes there is! If you write out the integral by definition and you identify, for instance, $\gamma_1$ with the curve that starts at 0 and goes to $w_1$ via the side of the parallelogram, then $\int_{\gamma_1}dz=w_1$. This way you can identify $L$ with $\Lambda!$ $\endgroup$ – rfauffar Jan 15 '15 at 11:50
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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 $$ \mathrm{Jac}(X)=\mathbb{C}/\Lambda=\mathbb{C}/L=X. $$

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