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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?

TheWanderer
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  • Hint: Write out the first homology group in terms of $L$. – rfauffar Jan 14 '15 at 13:35
  • @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... – TheWanderer Jan 15 '15 at 09:54
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    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 $\Lambda!$ – rfauffar Jan 15 '15 at 11:50

1 Answers1

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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 $$ \mathrm{Jac}(X)=\mathbb{C}/\Lambda=\mathbb{C}/L=X. $$

Zé Pequeno
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  • I believe that the integrations should take place over homology classes (and not cohomology classes). But it appears that the standard notation for the first cohomology group is being used here to denote the first homology group. – Dan Asimov Nov 06 '23 at 02:59