3 My God I had two $\pi$s! Why didn't someone tell me!

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 $q$: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$is ramified, has conductor$\pi^t$q^t$ ($\pi$ $q$ a uniformiser of $K$, so $\pi$ is my old $q$), K$), and say the central character of$\pi$also has conductor$\pi^t$. q^t$. Then $\pi$ is a ramified principal series representation associated to two character, one unramified and one ramified of conductor $t$.q^t$. The reason the result you want follows is that the$L$-function of$\pi$is$(1-c.Norm(\pi)^{-s})$(1-c.Norm(q)^{-s})^{-1}$ 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 $\pi^t$ q^t$and the conductor of its determinant is also$\pi^t$, q^t$, 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 $\pi$ 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$is ramified and has conductor$\pi^s$q^s$ with $s\geq1$ then $\rho$ has conductor $\pi^{2s}$ 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$with some conductor$\pi^t$q^t$ and its character also has conductor $\pi^t$. 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}$had better have dimension 1 or 2. In particular there are some inertial invariants. But these form a Galois-stable subspace, so the irreducible case cannot happen and we are finally done! 2 finished the proof! All that is left now is the case where$\rho$is irreducible. I'm sure there is a better way to [Edit: removed incomplete answer and replaced it with complete one]. To do this (indeed I think I've known case one in the past) but I have to get just looks at the kids up for school in 10 minutes so I am suddenly under time pressure. I am going to do it by cases, unfortunately. The main case is that$\rho$is induced from a character$\psi$on a quadratic extension$L$definition of$K$. Then the conductor of$\rho$is equal to the norm a representation. It's a sum of the conductor of form$\psi$, times \sum_i c_i.\dim(V/V^{G_i})$where the discriminant of $L/K$. If $L/K$ is unramified then we can compute conductors and so on after making an unramified base change to $L$ and this reduces us to a previous case, because now we c_i$are ramified principal series attached to$\psi$and its conjugate, rational and the preceding argument shows that$\psi$had better be ramified and its conjugate not, which can't happen. If G_i$ are running through a filtration on the other hand inertia subgroup. The moment some $L/K$ V^{G_i}$is ramified zero then you're in trouble, because then the same argument should work but unfortunately I have sum contributes 2 to stop writing. This is a shame because this case is the "heart" conductor of the question. OK so$\rho$and at this point, most 1 to the argument should be regarded as incompleteconductor of its determinant. This is not yet an answerSo each$V^{G_i}$had better have dimension 1 or 2. I'll try to find the time to finish it when I get to workIn particular there are some inertial invariants. If you're not familiar with this language then the whole thing must look quite daunting, but let me stress that this sort of argument is entirely standard and familiar for people who think about But these things in this wayform a Galois-stable subspace, so should not be regarded as profound in any way---rather, it is a "simple calculation" :-/the irreducible case cannot happen and we are finally done! 1 If I've understood your question correctly, you're right that$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$: Thm) Say$K$is a finite extension of$\mathbb{Q}_p$, say$\pi$is a smooth admissible irreducible representation of$GL(2,K)$, say$\pi$is ramified, has conductor$\pi^t$($\pi$a uniformiser of$K$, so$\pi$is my old$q$), and say the central character of$\pi$also has conductor$\pi^t$. Then$\pi$is a ramified principal series representation associated to two character, one unramified and one ramified of conductor$t$. The reason the result you want follows is that the$L$-function of$\pi$is$(1-c.Norm(\pi)^{-s})$with$c$equal to the value at a uniformiser of the unramified character. These sorts of assertions (explicit computations of$L$-functions) can all be found in Jacquet-Langlands, a book which changed my life, but I am sure that there are references which are a gazillion times more readable nowadays. So now all we have to do is to prove the theorem. Well there are probably purely representation-theoretic arguments, but I don't know them, so I am going to use the following trick: hit everything with local Langlands. This translates the result we want into a question about 2-dimensional representations rather than infinite-dimensional ones, so we'll be in much better shape. To make this part of the argument work you need to have an explicit hold on what local Langlands says for$GL(2)$. OK so apply local Langlands to$\pi$and we get a Weil-Deligne representation$(\rho,N)$of the Weil group of$K$. And we know that the conductor of this representation is$\pi^t$and the conductor of its determinant is also$\pi^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)$is$\pi$but the determinant is unramified, so our hypotheses do not apply (this the situation for elliptic curves with multiplicative reduction, for example; curve has bad reduction but character is unramified at$q$). And if$\chi$is ramified and has conductor$\pi^s$with$s\geq1$then$\rho$has conductor$\pi^{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$with some conductor$\pi^t$and its character also has conductor$\pi^t$. Say first that$\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$is irreducible. I'm sure there is a better way to do this (indeed I think I've known one in the past) but I have to get the kids up for school in 10 minutes so I am suddenly under time pressure. I am going to do it by cases, unfortunately. The main case is that$\rho$is induced from a character$\psi$on a quadratic extension$L$of$K$. Then the conductor of$\rho$is equal to the norm of the conductor of$\psi$, times the discriminant of$L/K$. If$L/K$is unramified then we can compute conductors and so on after making an unramified base change to$L$and this reduces us to a previous case, because now we are ramified principal series attached to$\psi$and its conjugate, and the preceding argument shows that$\psi$had better be ramified and its conjugate not, which can't happen. If on the other hand$L/K\$ is ramified then the same argument should work but unfortunately I have to stop writing. This is a shame because this case is the "heart" of the question.

OK so at this point, the argument should be regarded as incomplete. This is not yet an answer. I'll try to find the time to finish it when I get to work. If you're not familiar with this language then the whole thing must look quite daunting, but let me stress that this sort of argument is entirely standard and familiar for people who think about these things in this way, so should not be regarded as profound in any way---rather, it is a "simple calculation" :-/