Let $f$ be a modular form of weight $2k$, that is $f$ is a holomorphic on $\mathbb{H}$, $f$ extends holomorphically at infinity and satisfies $$ f\left(\frac{az+b}{cz+d}\right) = (cz+d)^{2k}f(z), \hspace{8mm} \begin{pmatrix} a & b \\ c & d \end{pmatrix}\in SL_2(\mathbb{Z}) $$ Let $\mathcal{M}_k$ denote the space of modular forms and define the Hecke operator $T_m\colon \mathcal{M}_k\to \mathcal{M}_k$ by $$ T_mf(z) = m^{2k-1}\sum_{ad=m, a\geq 1, 0\leq b < d} d^{-2k} f\left(\frac{az+b}{d}\right) $$ If we write $f(z) = \sum_{n\in\mathbb{Z}} c(n)q^n$ with $q = e^{2\pi i z}$, then there is a proof showing $$ T_mf(z) = \sum_{n\in\mathbb{Z}} C(n)q^n, \hspace{8mm} C(n) = \sum_{a\geq 1, a\mid (m,n)} a^{2k-1} c\left(\frac{mn}{a^2}\right) $$ Now define the multiplicative analogue $M_m\colon \mathcal{M}_k\to \mathcal{M}_{k\sigma_1(m)}$ by $$ M_mf(z) = \prod_{ad=m, a\geq 1, 0\leq b < d} m^kd^{-2k} f\left(\frac{az+b}{d}\right) $$ This is well-defined. Now, my question is: Is there an equivalent result concerning the Fourier expansion of $M_mf$ in terms of the Fourier coefficients of $f$? So far what I have is \begin{align*} M_mf(z) &= \prod_{ad=m, a\geq 1, 0\leq b < d} m^kd^{-2k} \left(\sum_{n\in\mathbb{Z}} c(n)e^{2\pi i n\frac{az+b}{d}}\right) \\ &= \prod_{ad=m, a\geq 1} m^kd^{-2k}\prod_{b=0}^{d-1} \left(\sum_{n\in\mathbb{Z}} c(n)e^{2\pi i n\frac{az+b}{d}}\right) \\ &= \prod_{ad=m, a\geq 1} m^kd^{-2k}\sum_{n\in\mathbb{Z}} \left(\sum_{n=n_0+\ldots+n_{s-1}}\prod_{b=0}^{s-1}c(n_b)e^{2\pi i n_b\frac{b}{d}}\right)e^{2\pi i \frac{az}{d}n} \end{align*} where I am not very confident about the last equality. I figure if this expression somehow reduces in a nice way, this naive approach could work, but I have not succeeded in doing so.
If anyone knows if this can be done / has been done, or have any suggestions as to make some progress, let me now!
