For $\delta\in\Gamma(1)$, right multiplication by $\delta$ permutes $\Gamma_\infty\backslash\Gamma(1)$. The <automorphy factor> identity
$$
j(\gamma\delta,\tau)
=j(\gamma,\delta\tau)j(\delta,\tau)
$$
therefore gives
$$
E_{k,s}(\delta\tau)
=j(\delta,\tau)^kE_{k,s}(\tau).
$$
The <modular form> $f$ obeys the same weight-$k$ transformation law, while
$$
\operatorname{Im}(\delta\tau)^k
=\frac{\operatorname{Im}(\tau)^k}{|j(\delta,\tau)|^{2k}}.
$$
Consequently
$$
\begin{aligned}
&f(\delta\tau)\overline{E_{k,s}(\delta\tau)}
\operatorname{Im}(\delta\tau)^k\\
&\quad=
j(\delta,\tau)^k\overline{j(\delta,\tau)}^k
|j(\delta,\tau)|^{-2k}
f(\tau)\overline{E_{k,s}(\tau)}\operatorname{Im}(\tau)^k\\
&\quad=
f(\tau)\overline{E_{k,s}(\tau)}\operatorname{Im}(\tau)^k.
\end{aligned}
$$
Thus the product is invariant under the weight-zero action of $\Gamma(1)$, as described by the <invariant product with a weight-k real-analytic Eisenstein series>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 4 / b / Solution

If $f$ is a weight-$k$ modular form, then
$$
f(\tau)\overline{E_{k,s}(\tau)}(\operatorname{Im}\tau)^k
$$
is modular invariant. The factors transform by $j(\gamma,\tau)^k$, $\overline{j(\gamma,\tau)}^k$, and $|j(\gamma,\tau)|^{-2k}$.


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