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 $f\_\alpha$, $f\_\beta$? 



Adjoint of Atkin-Lehner's U_p
The following is essentially taken from Miyake's "On Automorphic Forms on GL2 and Hecke Operators", with some help interpreting notation by looking at Gelbart's "Automorphic forms on adele groups".


From the point of view of representation theory, you can define Hecke operators at any prime (well, up to a normalizing factor) by the integral operator
$$(T\_pf)(g)=\int\_{K\_p(N)}\omega^{-1}(k)f\Big(g\cdot k\cdot \Big(\matrix{p&0\cr 0&1}\Big)\Big)\ dk$$
where $f$ is your cusp form, $K\_p(N)$ is the p-adic version of $\Gamma\_0(N)$ (it is the largest subgroup under which f is invariant under right translation), and $\omega$ is the nebentypus (central character) of f. This is equivalent to convolution in $L^2(Z\_p\backslash G\_p,\omega)$ with the characteristic function of $K\_p(N) \Big(\matrix{p&0\cr 0&1}\Big) K\_p(N)$.


Now the adjoint of "convolution with a function $\phi(g)$" on a Hilbert space is convolution with $\overline {\phi(g^{-1})}$, which in our case turns out to be 
$$(T\_p^\*f)(g)=\int\_{K\_p(N)}\omega^{-1}(k)f\Big(g\cdot k\cdot \Big(\matrix{p^{-1}&0\cr 0&1}\Big)\Big)\ dk$$


Since it is possible to write the integral operator version of $T\_p$ in classical terms, you should be able do the same with the adjoint . . . 


Miyake proves (he says he is generalizing Hecke and Atkin-Lehner) that the largest subspace of $S\_k(N,\omega)$ on which the algebra generated by all the Hecke operators and their adjoints act commutatively (and normality is the only obstruction to this) is the space of newforms. This implies that whenever there are oldforms, there is at least one prime for which the Hecke operator is not normal. In particular, in your simplest case of prime level $p$, $U\_p$ is not normal. (Though, I guess, since $p^3$ does not divide $p$, it is diagonalizable)


Two papers I came across: Li, "Diagonalizing modular forms" adjusts some $U\_p$ to get a family of Hecke operators at all $p$ which diagonalize the whole space of modular forms. Choie and Kohnen, "Diagonalizing 'bad' Hecke operators, etc" shows that the $U\_p$ are almost always diagonalizable on $S\_k(pN)$, for $N$ square-free, prime to $p$.


To summarize: 1) you can write down the adjoint if you work adelically, 2) normality basically fails, 3) diagonalizability often succeeds.


Disclaimer: I don't spend much time working with classical modular forms: until tonight, I didn't know how the $U\_p$ were defined, and I think this might be the first time I've typed "nebentypus".


UPDATE: Further googling led me to [this page](http://maths.dept.shef.ac.uk/magic/course.php?id=48) of power point slides by Jens Funke. In lecture 19, he claims that the adjoint of $U\_n$ is $n^kV\_n$, where $V\_ng(z)=g(nz)$, though this is left "as an exercise".


