A fact about non-homothetic lattices

Question

Let $\mathbb H$ be the upper half-plane, and let $\Gamma:=SL_2(\Bbb Z)$. Let $\tau, \tau' \in \mathbb H$ such that $\Gamma\tau \neq \Gamma\tau'$, and let $L:=\mathbb Z \oplus \tau\mathbb Z$ and $L':=\mathbb Z \oplus \tau'\mathbb Z$. Then $L$ and $L'$ are two compact Riemann surfaces. Why they are not $\Bbb C$-analytically isomorphic? Why they are diffeomorphic? (I can check that the action of $SL_2(\Bbb R)$ is transitive, but I can not see why the transitivity of the action of $SL_2(\Bbb R)$, implies they are diffeomorphic.)

Answer 1

A complex analytic isomorphism $f:\Bbb{C}/L\to \Bbb{C}/L'$ lifts to a complex analytic isomorphism $\Bbb{C\to C}$ thus it has to be of the form $z\to az+b$ ie. $L' = aL$.

Equivalently $dz$ is an holomorphic 1-form on $\Bbb{C}/L$, the other mermorphic 1-forms are $h(z)dz$ with $h$ meromorphic on $\Bbb{C}/L$, thus $h(z)dz$ is an holomorphic 1-form iff $h$ has no poles nor zeros ie. $h$ is constant. $df(z)$ is also an holomorphic 1-form on $\Bbb{C}/L$, thus $df(z)=adz$ for some $a\ne 0$ and hence $L'=aL$.