If $\Gamma=\Gamma_1(N)$, or $\Gamma=\Gamma_0(N)$, the Hecke operator $[\Gamma diag(1,l) \Gamma]$ for $l$ a prime (acting on the space of cusp forms of level $\Gamma$ and some weight $k$) is in general denoted $T_l$ when $l$ does not divide $N$, but $U_l$ otherwise. It is well-known that the Hecke operators $T_l$ are normal (that is commute with their adjoint) for the Petersson's inner product, but not the $U_l$. Or are they ?

Is there any value of the level $N$, the weight $k$, and a prime $l$ dividing $N$ such that $U_l$ is normal ?

Of course, this would imply that $U_l$ is diagonalizable, but this is conjectured (known if $k=2$) to happen if $l^3$ does not divide $N$ (cf. Coleman-Edixhoven, Math. Annalen, 1998).

This question is surely easy, but my problem is that I can't really compute the adjoint of $U_l$. Of course, it is the Hecke operator $\Gamma diag(l,1) \Gamma$ and it's not hard to write this double class as a union of simple class, but the formulas I get look awful.

For example consider this simplest case. Let $f$ be form of weight $k$ and level $1$, normalized eigenform for all $T_l$. Choose a prime $p$. As it is well known, the space of form of level $\Gamma_0(p)$ with same eigenvalues as $f$ for all the $T_l$ with $l \neq p$ is two dimensional, generated by $f(z)$ and $f(pz)$, or preferably generated by $f_\alpha(z) = f(z) - \beta f(pz)$ and $f_\beta(z) = f(z)-\alpha f(pz)$ where $\alpha$ and $\beta$ are the two roots of $X^2-a_pX + p^{k-1}$, with $T_p f = a_p f$ (I suppose $\alpha \neq \beta$, which is always conjectured and often known). The interest of this basis is that $U_p$ is diagonal in it: $U_p f_\alpha = \alpha f_\alpha$ and $U_p f_\beta = \beta f_\beta$

Can you compute the matrix of the adjoint of $U_p$ in the basis $f_\alpha$, $f_\beta$?