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Let $f\in S_k(\Gamma_0(N))$ be a cusp form for $N>1$. Consider the following operators acting on $f$ via the natural action of $GL_2^{+}(\mathbb{R})$ :

$$ W_N=\begin{pmatrix} 0 & -1\\ N & 0 \end{pmatrix}$$

$$ U_q=\sum\limits_{i=0}^{q-1}\begin{pmatrix} q & i\\ 0 & q \end{pmatrix}$$ for prime $q\mid N$.

Does $U_qW_Nf=W_NU_qf$?

Remark : The action for $\gamma=\begin{pmatrix} a & b\\ c & d\end{pmatrix}\in GL_2^{+}(\mathbb{R})$ is given by :

$$\gamma f(z)= (\operatorname{det}(\gamma))^{k/2}(cz+d)^{-k} f\left(\dfrac{az+b}{cz+d}\right).$$

Any help is deeply appreciated.

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EDIT. In the answer below, $U_q$ refers to the usual Hecke operator given on Fourier expansions by $\sum_{n \geq 1} a_n x^n \mapsto \sum_{n \geq 1} a_{qn} x^n$. The operator $U_q$ in the OP is given by $\sum_{n \geq 1} a_n x^n \mapsto q \sum_{n \geq 1} a_{qn} x^{qn}$. As explained in the comments this does not preserve the space of forms of level $N$.

If $f$ is a newform in $S_k(\Gamma_0(N))$ then $f$ is an eigenfunction for both $U_q$ and $W_N$. But in general $U_q$ and $W_N$ do not commute. You can find an example in Shimura, "Introduction to the arithmetic theory of automorphic functions", Remark 3.59. There he constructs eigenfunctions $f$ for $U_q$ such that $W_N f$ is not an eigenfunction for $U_q$, hence $W_N U_q f \neq U_q W_N f$.

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  • $\begingroup$ Simply take $q=N=2$ and $f(\tau)=\Delta(2\tau)$ with $\Delta$ the usual weight $12$ cusp form of level $1$. $\endgroup$
    – Henri Cohen
    Commented Jun 18, 2022 at 12:14
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    $\begingroup$ I am not sure but have you observed that the stated operator U_q is different from the usual Hecke operator for prime q dividing N. $\endgroup$
    – Akash Yadav
    Commented Jun 18, 2022 at 12:35
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    $\begingroup$ I see, your notation is not standard (usually $U_p(\sum a_n q^n) = \sum a_{pn} q^n$). I think Henri Cohen's example works. $\endgroup$
    – François Brunault
    Commented Jun 18, 2022 at 13:21
  • $\begingroup$ Just a final question, the operators still commute for a newform right? $\endgroup$
    – Akash Yadav
    Commented Jun 19, 2022 at 10:18
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    $\begingroup$ @AkashYadav If I understood your definitions correctly, your operator $U_q$ doesn't preserve the space of forms of level $N$. For example let $f=\sum_{n \geq 1} a_n x^n$ be a newform of level $N$, then $U_q f = q \sum_{n \geq 1} a_{qn} x^{qn} = q a_q f(qz)$ since the Fourier coefficients of $f$ are completely multiplicative. So $U_q f$ has level $Nq$, and applying $W_N$ doesn't make sense. $\endgroup$
    – François Brunault
    Commented Jun 19, 2022 at 23:25

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