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.



Fricke involution and Atkin operator
**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 $W\_N U\_q f \neq U\_q W\_N f$.


