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Let $X = V / \Lambda$ and X' = V' / \Lambda' be two complex tori, of dimensions $n$ and n'. If f : X \to X' be a homomorphism, then $f$ is induced by a linear map F : V \to V' which satisfies F(\Lambda) \subset \Lambda'. I want to know how we find $F$ given a morphism between the lattices $\Lambda$ and \Lambda'.

More explicitly, I'm trying to see the map $F$ induced by an isomorphism of the lattices $\Lambda$ and \Lambda'. For example, if $X$ and X' are both one-dimensional (i.e. marked elliptic curves), then $X$ and X' are isomorphic if and only if there is a $\gamma$ in $SL_2(\mathbb Z)$ such that \gamma \Lambda = \Lambda'.

Write $\gamma = \begin{pmatrix}a & b \\ c & d \end{pmatrix}$. We can find $\tau$ and \tau' such that $\Lambda = \mathbb Z \oplus \tau \mathbb Z$ and \Lambda' = \mathbb Z \oplus \tau' \mathbb Z, and in this case \tau' = \frac{a \tau + b}{c \tau + d}. I have some lecture notes ("Lectures on moduli spaces of elliptic curves", R. Hain) which say the linear map which induces the isomorphism X \to X' in this case is $L : z \mapsto \frac 1{c\tau + d} z$, without going into detail on how this is calculated.

Question: How do we find the map $L$ given $\gamma$?

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    @Giacomo: in this case rank$\Lambda=2\dim V$, so your comment doesn't quite apply2011-05-05

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For your first paragraph: Choose a basis of $\Lambda$ (e.g. $1,\tau$ in elliptic curve case), in this basis choose a subset which is a (complex) basis of $V$ (e.g. $1$ in elliptic curve case), and see how that (complex) basis is mapped by $F$ - it determines a unique linear map V\to V'.

The matrix $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ has a bit different meaning. The lattice $\mathbb{Z}\oplus\tau\mathbb{Z}$ has a basis $\tau,1$, and $\gamma$ gives you another basis $a\tau+b$, $c\tau+d$ of the same lattice. We then look for a linear transformation $z\mapsto kz$ that would map $c\tau+d$ to $1$, so $k=1/(c\tau+d)$ ($a\tau+b$ is mapped to \tau':=(a\tau+b)/(c\tau+d)).