= Solution
Suppose $\tau_1=(a\tau_2+b)/(c\tau_2+d)$, with $ad-bc=1$, and put $\alpha=(c\tau_2+d)^{-1}$. The identities
$$
\alpha=a-c\tau_1,\qquad \alpha\tau_2=d\tau_1-b
$$
show that multiplication by $\alpha$ takes one <period lattice> isomorphically to the other: its basis matrix has determinant $ad-bc=1$. If $N\mid c$, the image of $1/N$ is congruent to $a/N$ modulo the target <period lattice>. Moreover $ad\equiv1\pmod N$, so $a$ is a unit modulo $N$. The image therefore generates the same <cyclic subgroup>.
Conversely, a <holomorphic> group isomorphism of <complex tori> lifts to $z\mapsto\alpha z$, with $\alpha\Lambda_{\tau_2}=\Lambda_{\tau_1}$. Indeed its lift fixing zero is additive: its failure to be additive is a continuous lattice-valued function, hence zero. An additive <holomorphic> map is complex linear. Write
$$
\alpha=a-c\tau_1,\qquad \alpha\tau_2=-b+d\tau_1.
$$
The corresponding integral basis matrix has determinant $1$: complex multiplication preserves the real orientation, and both bases $(1,\tau_j)$ are positively oriented. Solving gives $\tau_1=(a\tau_2+b)/(c\tau_2+d)$. Preservation of the <cyclic subgroup> means $\alpha/N\equiv u/N\pmod{\Lambda_{\tau_1}}$ for some unit $u$ modulo $N$. Comparing the two real basis coefficients gives $N\mid c$. Thus the matrix lies in the <Gamma 0 congruence subgroup>, and \b[the classification is]
$$
\boxed{\mathcal E\cong\Gamma_0(N)\backslash\mathbb H.}
$$
This is the noncompact <modular curve> classification; no cusp points have been added.
Back to article page
