Weight $2$ Eisenstein series transformation

Question

Let $$G_2(\tau) = \sum_{c \in \mathbb{Z}} \sum_{d \in \mathbb{Z}}\frac{1}{(c\tau + d)^2}$$ be the weight $2$ Eisenstein series for $\tau \in \mathbb{H}$. I'm trying to do an exercise from Diamond and Shurman aimed at establishing the transformation law $$(c\tau + d)^{-2}G_2\left(\frac{a\tau + b}{c\tau + d}\right) = G_2(\tau) - \frac{2\pi i c}{c\tau + d} \quad \textrm{ for }\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \textrm{SL}_2(\mathbb{Z}).$$ The exercise is to show that if the above holds for $\gamma_1,\gamma_2 \in \textrm{SL}_2(\mathbb{Z})$, then it holds for the product $\gamma_1\gamma_2$. Define $j(\gamma) = c\tau + d$ for $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \textrm{SL}_2(\mathbb{Z})$. Let $c,c'$ be the bottom left entires of $\gamma_1$ and $\gamma_2$, respectively. Using the identity $j(\gamma_1\gamma_2,\tau) = j(\gamma_1,\gamma_2\tau)j(\gamma_2,\tau)$, we see that \begin{equation*} \begin{aligned} \frac{G_2(\gamma_1\gamma_2\tau)j(\gamma_2,\tau)^2}{j(\gamma_1\gamma_2,\tau)^2} &= \frac{G_2(\gamma_1(\gamma_2\tau))}{j(\gamma_1,\gamma_2\tau)^2} \\ &= G_2(\gamma_2\tau) - \frac{2\pi i c}{j(\gamma_1,\gamma_2\tau)} \\ &= j(\gamma_2,\tau)^2\left(G_2(\tau) - \frac{2\pi i c'}{j(\gamma_2,\tau)}\right) - \frac{2\pi i c}{j(\gamma_1,\gamma_2\tau)}. \end{aligned} \end{equation*} Dividing by $j(\gamma_1\gamma_2,\tau)^2$ and using the aforementioned identity again, we get \begin{equation*} \begin{aligned}\frac{G_2(\gamma_1\gamma_2\tau)}{j(\gamma_1\gamma_2,\tau)^2} &= G_2(\tau) - \frac{2\pi i c'}{j(\gamma_2,\tau)} - \frac{2\pi ic}{j(\gamma_2,\tau)^2j(\gamma_1,\gamma_2\tau)} \\ &= G_2(\tau) - \frac{2\pi i c'}{j(\gamma_2,\tau)} - \frac{2\pi ic}{j(\gamma_2,\tau)j(\gamma_1\gamma_2,\tau)}. \end{aligned} \end{equation*} For the transformation law to hold, we would need $$\frac{c'}{j(\gamma_2,\tau)} + \frac{c}{j(\gamma_2,\tau)j(\gamma_1\gamma_2,\tau)} = \frac{C}{j(\gamma_1\gamma_2,\tau)},$$ where $C$ is the bottom left entry of $\gamma_1\gamma_2$. But the left-hand side is equal to $$\frac{j(\gamma_1\gamma_2,\tau)c' + c}{j(\gamma_2,\tau)j(\gamma_1\gamma_2,\tau)},$$ so this would say that $$\frac{j(\gamma_1\gamma_2,\tau)c' + c}{j(\gamma_2,\tau)} = C$$ is constant. How can this possibly be? I cannot find where I went wrong in this tedious calculation and it's unbelievably frustrating.

Answer 1

There's nothing wrong: $$Cj(\gamma_2,\tau)-c'j(\gamma_1\gamma_2,\tau)\\=(ca'+dc')(c'\tau+d')-c'\big((ca'+dc')\tau+(cb'+dd')\big)\\=(ca'+dc')d'-c'(cb'+dd')=c(a'd'-b'c')=c.$$