Exercise on Complex Tori

Question

I can't solve this exercise: Miranda Algebraic Curves and Riemann Surfaces pag. 43 K:

Recall that a lattice $L \subset \mathbb{C}$ is an additive subgroup generated (over $\mathbb{Z}$) by two complex numbers $\omega_1$ and $\omega_2$ which are lineraly indipendent over $\mathbb{R}$. Thus $L=\{m\omega_1 + n\omega_2, m,n \in \mathbb{Z} \}$.

1. Suppose that $L \subset L'$ are two lattices in $\mathbb{C}$.Show that the natural map from $\mathbb{C}/L$ to $\mathbb{C}/L'$ is holomorphic, and is biholomorphic if and only if $L=L'$.
2. Let $L$ be a lattice in $\mathbb{C}$ and let $\alpha$ be a nonzero complex number. Show that $\alpha L$ is a lattice in $\mathbb{C}$ and that the map $$\phi: \mathbb{C}/L \mapsto \mathbb{C}/\alpha L $$ sending the coset $z+L$ to $\alpha z + \alpha L$ is a well defined biholomorphic map.
3. Show that every torus $\mathbb{C}/L$ is isomorphic to a torus which has the form $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$, where $\tau$ is a complex number with strictly positive imaginary part.

According to Miranda's convention $\mathcal{A}=\{D(z_i,\epsilon),\pi ^{-1} \mid _{D(z_i,\epsilon)}\}_{i=1}^{n}$ is the complex atlas on a torus. I proved the second statement in this way:

Let $\pi^{-1}$ and $\pi_\alpha ^{-1}$ the homeomorphisms defining the complex atlases on $\mathbb{C}/L$ and $\mathbb{C}/\alpha L$, so $\pi_\alpha ^{-1} \phi \pi (z)=\alpha z + l_{\alpha,z}$ where $l_{\alpha,z} \in \alpha L$. For the functions $$\pi_\alpha ^{-1} \phi \pi (z)-\alpha z$$ is continuous and $\alpha L$ is discrete, $l_{\alpha,z}$ is indipendent from $z$, thus it is a constant. $\phi$ is injective and trivially surjective, so this is a holomorphic bijection, that's to say a biholomorphism.

But I can't solve the first and the third part... Can you help me, please?

Answer 1

For the first part, show that $\pi_{L'}\circ \pi_L^{-1}$ is well-defined. It is then clear that it is holomorphic.

If $L \neq L'$, then $\pi_L(\omega) \neq \pi_L(0)$ for every $\omega \in L' \setminus L$, and both points are mapped to $\pi_{L'}(0)$, so it is not biholomorphic. If $L = L'$, it is the identity, hence obviously biholomorphic.

For the third part, pick a basis $\{\omega_1,\,\omega_2\}$ of the lattice. Since $\omega_1$ and $\omega_2$ are linearly independent over $\mathbb{R}$, we have $\Im (\omega_2/\omega_1) \neq 0$. If the imaginary part of the quotient is negative, renumber the basis vectors or replace $\omega_2$ with $-\omega_2$. Then $\tau = \frac{\omega_2}{\omega_1}$ lies in the upper half-plane, and the lattice $\frac{1}{\omega_1}L$ is generated by $1$ and $\tau$.