# Linear map induced by homomorphism of complex tori

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 $\gamma$?

-
If $\mbox{rank}\,\Lambda = \mbox{dim}\,V$ any morphism of lattices $\Lambda \to \Lambda'$ extends uniquely to a linear map $F: V \to V'$. This should be your $F$. –  Giacomo d'Antonio May 5 '11 at 8:52
@Giacomo: in this case rank$\Lambda=2\dim V$, so your comment doesn't quite apply –  user8268 May 5 '11 at 8:58
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 $\tau':=(a\tau+b)/(c\tau+d))$.