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Fix positive integers $k, N$ and let $\omega$ be a Dirichlet character mod $N$.

Let $f\in S_k(N,\phi)$ be a normalized newform (i.e. of weight $k$, level $N$ and character $\phi$) with fourier expansion $\sum_{n\geq 1} a(n)q^n$. In her paper 'Newforms and Functional Equations', Winnie Li showed that if $q$ is a prime dividing $N$ and $\phi$ is not a character mod $N/q$, then $|a(q)|=q^{\frac{k-1}{2}}$. In particular, $a(q)\neq 0$. This is Theorem 3.(ii) of the paper. I should also note that special cases of this were proven earlier Atkin-Lehner, Hecke and Ogg.

I am interested in the Hilbert modular version of this result. More specifically, let $\mathfrak{N}$ be an integral ideal of a totally real number field $K$ of degree $d$ over $\mathbb{Q}$, $\Phi$ be a Hecke character induced by a numerical character mod $\mathfrak{N}$ and $k\in (\mathbb{Z}_+)^d$.

Let $S_k(\mathfrak{N},\Phi)$ be the space of Hilbert modular cusp forms of weight $k$, level $\mathfrak{N}$ and character $\Phi$, viewed adelically to circumvent class number issues (I make no assumptions regarding the class number of $K$). Shimura, in 'The special values of the zeta functions associated to Hilbert modular forms' defined "Fourier coefficients" $C(\mathfrak{m},\textbf{f})$ for cusp forms $\textbf{f}\in S_k(\mathfrak{N},\Phi)$ (this on the bottom of page 649 of the paper).

Let $\textbf{f}\in S_k(\mathfrak{N},\Phi)$ be a normalized newform. I am interested in knowing when $C(\mathfrak{q},\textbf{f})=0$, where $\mathfrak{q}\mid\mathfrak{N}$. In their paper 'Twists of Hilbert Modular Forms', Tom Shemanske and Lynne Walling showed that if the numerical character inducing $\Phi$ is not defined mod $\mathfrak{N}/ \mathfrak{q}$, then this coefficient is nonzero whenever $\mathfrak{q}$ has degree 1 over the rationals or $\mathfrak{q}$ exactly divides $\mathfrak{N}$ (this is part 2 of Theorem 3.3 in the paper). In a remark directly after the Theorem, Tom and Lynne say that the reason for the restrictions on the prime $\mathfrak{q}$ is that this is what is needed to make Ogg's proof in the classical case go through. The 'truth', however, is that the $\mathfrak{q}$-th coefficient should always be nonzero in this case, without any restrictions on the degree of the prime $\mathfrak{q}$. A referee told Tom (who is my advisor) that the representation theory implies this stronger result. Tom lost the referee's report some time ago however, and doesn't know of any reference for this representation theoretic result. Does anyone know of a paper in which this result is proven without any restrictions on the prime $\mathfrak{q}$?

$\textbf{Question}:$ Let $\textbf{f}\in S_k(\mathfrak{N},\Phi)$ be a normalized newform, $\mathfrak{q}$ be a prime dividing $\mathfrak{N}$ and suppose that the numerical character inducing $\Phi$ is not defined mod $\mathfrak{N}\mathfrak{q}^{-1}$. Is it true that $C(\mathfrak{q},\textbf{f})\neq 0$?

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In representation theoretic terms if $f$ is a classical newform of conductor $N$ and $\Pi$ is the associated automorphic representation of $GL_2$, then the Hecke eigenvalue $a_p$ is closely related to the local component $\Pi_p$ at $p$ of $\Pi$. If $p$ does not divide $N$, then $\Pi_p$ is obtained by inducing two characters from the Borel that can be determined from $a_p$. If $p$ does divide $N$ then the picture is (I think) more complicated. However if $\Pi_p$ is a twist by a character $\chi$ of the special representation, then the eigenvalue $a_p$ is related to $\chi$ by a simple formula. –  unknown Oct 22 '10 at 0:07
    
You can read these things in a paper of Casselman in the second of Antwerp volumes "Modular functions of one variable" (pp 119-120 might be useful). I think that a theorem of the type $a_p\neq 0$ can be proved only if one knows a priori that $\pi_p$ is not supercuspidal. –  unknown Oct 22 '10 at 0:13
    
Like unknown above, I am slightly perplexed by the assertions. When $\pi(f)$ is supercuspidal at $v$, I don't see how to prove this result. In fact, I am already rather surprised by the proof in W.Li' article. The results you quote is proved on page 295 but the proof uses a corollary of Ogg. However, I can't see how the hypotheses of the corollary of Ogg are verified. If you are happy to assume that $\pi(f)_{v}$ is not supercuspidal, then this should follow from the description of the Langlands $L$-factor at $v$. –  Olivier Oct 22 '10 at 5:35
    
Olivier - Which of the hypotheses in Ogg's Cor. 1 do you not see as being verified? –  Ben Linowitz Oct 22 '10 at 5:45
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@Olivier: I don't think $\pi$ can be supercuspidal. The condition that the form has conductor $q^t$ at $q$ (here $q$ is a prime ideal and $t\geq1$) and the form also has conductor $q^t$ at $q$ implies that $\pi$ is ramified principal series at $q$, with one unramified and one ramified character, so the result is true. –  Kevin Buzzard Oct 22 '10 at 6:06
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2 Answers

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If I've understood your question correctly, you're right that $C(q,f)\not=0$ always and there is a natural representation-theoretic proof of this result (before I start let me say that I don't know how to get these gothic $q$s and $N$s as in your question, so I am just using usual $q$s and $N$s, but they are ideals of $F$ just like yours). The one thing I am worried about is that I do not know what a "numerical character" is, probably because I think about Hilbert modular forms in a different way to you. For me, a Hilbert modular form is really just an automorphic representation of $GL(2,F)$ with certain properties, and the natural generalisation of the character of a classical modular form in this setting is the the following construction. Take the central character of this representation, which is a character of the ideles of $F$. This character decomposes as the product of a power of the norm character and a finite order character, and this finite order character is the natural generalisation of the character of the form. For me, the theorem is that if the conductor of $f$ is $N$, if $q^t$ is the exact power of $q$ dividing $N$, and if $q^t$ is also the exact power of $q$ dividing the conductor of the finite order character, then $C(q,f)\not=0$, where in this generality I am interpreting that as saying that the local $L$-factor attached to the automorphic representation at $q$ is $(1-c.Norm(q)^{-s})^{-1}$ with $c\not=0$.

So, as the referee states, this statement can be proved purely representation-theoretically. It is also a purely local assertion in fact. The automorphic representation is a tensor product of local representations and the local $L$-factor at $q$ can be computed from knowing $\pi_q$, the factor at $q$. So we are done by the following purely local theorem, where now $K$ is the completion of the totally real field $F$ at the prime $q$, and $\pi$ is $\pi_q$:

Thm) Say $K$ is a finite extension of $\mathbb{Q}_p$, say $\pi$ is a smooth admissible irreducible representation of $GL(2,K)$, say $\pi$ is ramified, has conductor $q^t$ ($q$ a uniformiser of $K$), and say the central character of $\pi$ also has conductor $q^t$. Then $\pi$ is a ramified principal series representation associated to two character, one unramified and one ramified of conductor $q^t$.

The reason the result you want follows is that the $L$-function of $\pi$ is $(1-c.Norm(q)^{-s})^{-1}$ with $c$ equal to the value at a uniformiser of the unramified character. These sorts of assertions (explicit computations of $L$-functions) can all be found in Jacquet-Langlands, a book which changed my life, but I am sure that there are references which are a gazillion times more readable nowadays.

So now all we have to do is to prove the theorem. Well there are probably purely representation-theoretic arguments, but I don't know them [edit: vytas does---see his answer], so I am going to use the following trick: hit everything with local Langlands. This translates the result we want into a question about 2-dimensional representations rather than infinite-dimensional ones, so we'll be in much better shape. To make this part of the argument work you need to have an explicit hold on what local Langlands says for $GL(2)$.

OK so apply local Langlands to $\pi$ and we get a Weil-Deligne representation $(\rho,N)$ of the Weil group of $K$. And we know that the conductor of this representation is $q^t$ and the conductor of its determinant is also $q^t$, and we want to prove that $\rho$ is reducible with one ramified and one unramified character on the diagonal, and that $N=0$. Then we're done.

OK so first I'll show $N=0$. This is because if $N\not=0$ then the definition of a Weil-Deligne representation forces $\rho$ to be $\chi+\chi|.|$ with $|.|$ the norm character. And we now compute conductors. If $\chi$ is unramified then the conductor of $(\rho,N)$ is $q$ but the determinant is unramified, so our hypotheses do not apply (this the situation for elliptic curves with multiplicative reduction, for example; curve has bad reduction but character is unramified at $q$). And if $\chi$ is ramified and has conductor $q^s$ with $s\geq1$ then $\rho$ has conductor $q^{2s}$ so again we can't be here because $s\not=2s$.

It remains to deal with the $N=0$ case. We have a representation $\rho$ with some conductor $q^t$ and its character also has conductor $q^t$---let me drop these $q$s and just talk about conductor $t$ out of laziness. Say first that $\rho$ is the sum of two characters $\sigma_1$ and $\sigma_2$ of conductors $t_1$ and $t_2$. Then the conductor of $\rho$ is $t_1+t_2$ and the conductor of its determinant is at most the max of $t_1$ and $t_2$, so if these are equal then one of the $t_i$ had better be zero, and so the other one had better be non-zero, and this is the case that is really happening.

All that is left now is the case where $\rho$ is irreducible. [Edit: removed incomplete answer and replaced it with complete one]. To do this case one just looks at the definition of the conductor of a representation. It's a sum of the form $\sum_i c_i.\dim(V/V^{G_i})$ where the $c_i$ are rational and the $G_i$ are running through a filtration on the inertia subgroup. The moment some $V^{G_i}$ is zero then you're in trouble, because then the sum contributes 2 to the conductor of $\rho$ and at most 1 to the conductor of its determinant. So each $V^{G_i}$ had better have dimension 1 or 2. In particular there are some inertial invariants. But these form a Galois-stable subspace, so the irreducible case cannot happen and we are finally done!

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I kind of miss the sense of urgency pervasive in the first form of the answer. All this "Will I make it? Will my children be on time for school?" feel was rather enjoyable. That said, the argument about the jumps of the conductor being equal only if $V$ has invariants is nice. –  Olivier Oct 22 '10 at 14:20
    
Sorry Olivier :-) The first version was funnier, but I knew I was doing the wrong thing and couldn't think of the right one in time. When I knew what the right thing was it seemed ridiculous to leave in the kids to school comments. Anyone who really wants to see them can look at previous edits, of course. –  Kevin Buzzard Oct 22 '10 at 15:15
    
Kevin - Thank you so much for your thoughtful answer. You're right that we think about Hilbert modular forms very differently, but it's nice to know that the result I want to use is true, in as much generality as I suspected (so I don't have to restrict the primes dividing the level of my newforms). Now I just need to figure out the best way to cite MO responses. Oh, and finish proving a few things. There's still that as well... –  Ben Linowitz Oct 22 '10 at 15:24
    
Kevin - P.S. To make the Gothic letters I used the \mathfrak command (i.e. \mathfrak{N} ) –  Ben Linowitz Oct 22 '10 at 16:21
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I think I understand what Kevin is saying in the Theorem. Let $\mathfrak o$ be the ring of integers of $K$ and $\mathfrak p$ the maximal ideal. Define the following subgroups of $G=GL(2, K)$:

$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o // \mathfrak p & \mathfrak o \end{pmatrix}$

$I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}$

$Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$

Lemma. Let $\pi$ be a smooth representation of $I$ with a central character, such that $Z_n$ acts trivially, $Z_{n-1}$ does not act trivially and the space of $I_n$-invariants is non-zero. Then the restriction of $\pi$ to $I_n Z_0$ contains a one dimensional subrepresentation of the form $\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where $\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$ is a smooth character of conductor $\mathfrak p^n$.

Proof. Look at the action of the abelian group $Z_0$ on $\pi^{I_n}$.

The pair $(I_nZ_0, \chi)$ is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if $\pi$ is an irreducible smooth representation of $G$, then $Hom_{I_n Z_0}(\chi, \pi)\neq 0$ if and only if $\pi$ is a principal series rep with one character unramified, the other of conductor $\mathfrak p^n$. For this you could look either in the appendix by Heniart to:

http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf

or in the book of Bushnell and Henniart. The main point being that the representation $\chi$ (as a representation of $\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$) shows up in the $U$-coinvariants of $\pi$, where $U$ is unipotent upper (lower?) triangular matrices.

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Thanks Vytas. I was almost sure that there would be an argument which didn't need local Langlands. This sort of argument is important, because it's something that you can hope to actually translate down to a completely elementary argument like the one in Miyake's book. –  Kevin Buzzard Oct 22 '10 at 15:16
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