TITLE: Adjoint of Atkin-Lehner's U_p
QUESTION [13 upvotes]: 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 $f_\alpha$, $f_\beta$?

REPLY [11 votes]: A part of this question was unanswered: the last one, computing the adjoint of $U_p$ on
the two-dimensional space $V_f$ of old forms of level $\Gamma_0(p)$, attached to a normalized eigenform $f$ of weight $2k$ and level $1$. I observe that the reference to Gross given in the comments deals only with the case of new forms, and the same can be said of the reference given in the update of BR's answer.
In the unlikely event that the answer to this technical question might be of interest for someone else than me, here it is:
In the basis $f(z),f(pz)$ of $V_f$, the matrix of the adjoint of $U_p$ for the Peterson's scalar product is $$U'_p = \left( \matrix{ 0 & - p^{-1} \cr p^{2k} & a_p} \right).$$
Here $a_p$ is the $p$-coefficient of $f$. Note that $U_p$ and its adjoint never commute, which is consistent with BR's answer. 
To get the matrix in the basis $f_\alpha=f(z)-\beta f(pz)$, $f_\beta=f(z)-\alpha f(pz)$,
make a change of basis. The result is unpleasant (to my surprise), so I don't give it.
To prove the result, let us consider the action $g \mapsto w(g)=p^{-k}z^{-2k} g(-1/pz)$ 
of the Arkin-Lehner involution on $V_f$, the normalization being that of the 1970 paper of Atkin and Lehner at Math. annalen).
Lemma 7 (combined with Lemma 9 and Lemma 14) of that paper tells that for all $g$ modular form for $\Gamma_0(p)$ of weight $2k$, the form $p^{1-k} U_p(g)+w(g)$ is of level $\Gamma_0(1)$, hence a multiple of $f$. One easily deduces that $w(f)= x f(z) + p^{k} f(pz)$ for some scalar $x$, $w(f(pz)) = y f(z)$ for some scalar $y$. Using that the action $w$ is an involution, we easily get the values of $x$ and $y$, i.e. the matrix of the action of $w$ in the basis $f(z),f(pz)$:
$$w=\left(\matrix{0 & p^{-k} \cr p^k & 0}\right).$$ The matrix of $U_p$ in the same basis is as is well-known $$U_p=\left(\matrix{a_p & 1 \cr -p^{2k-1} & 0}\right).$$ Since the adjoint of $U_p$ is $w U_p w^{-1}$, one gets the result given above.