I need some help for the proof of the uniformization theorem (Silverman's Advanced Topics ...).
If we have $G_{4}(\Lambda_{1})=G_{4}(\Lambda_{2}) $ and $ G_{6}(\Lambda_{1})=G_{6}(\Lambda_{2})$ (with $\Lambda_{1},\Lambda_{2}$ two lattices and $G_{n}$: Einsenstein serie).
Why we have $\Lambda_{1}=\Lambda_{2}$ ?
On
It might be easiest to first note that the $j$-invariants $j(\Lambda_1)=j(\Lambda_2)$ are equal and use the theorem that the $j$-invariant defines an injective map from the space of lattice modulo homothety to the affine line. Thus the equality $j(\Lambda_1)=j(\Lambda_2)$ implies that $\Lambda_1=c\Lambda_2$ for some $c\in\mathbb C^*$. Next use the fact that $G_{2k}(c\Lambda)=c^{-2k}G_{2k}(\Lambda)$ and your assumption that $G_4(\Lambda_1)=G_4(\Lambda_2)$ and $G_6(\Lambda_1)=G_6(\Lambda_2)$ to conclude that $c^2=1$, provided that both $G_4$ and $G_6$ values are non-zero. Hence $c=\pm1$, which gives the desired result, since clearly $-\Lambda=\Lambda$. Finally, if one of $G_4$ or $G_6$ is zero, you only get $c^4=1$ or $c^6=1$, but in each case one can show that the lattice have CM by the appropriate root of unity.
There is a nice analytic proof of this. The Weierstrass function $\wp(z)$ associated to $\Lambda$ satisfies a differential equation with coefficients derived from $G_4(\Lambda)$ and $G_6(\Lambda)$. It is the unique even function with principal part $1/z^2$ satisfying this. It has poles at the points of $\Lambda$. So $G_4(\Lambda)$ and $G_6(\Lambda)$ determine $\wp(z)$ which determines $\Lambda$.