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Is it true that the conductor of a holomoprhic or a Maass cusp form with trivial nebentypus corresponding to a two-dimensional dihedral representation (over $\mathbb{Q}$ )is non-square-free?

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The conductor could be 1, right? Isn't that squarefree? Are you talking about 2-dimensional representations? Of an arbitrary number field? –  Kevin Buzzard Mar 8 '10 at 12:03
    
A Dirichlet character is monomial and can have prime conductor. I think your question is too terse/ambiguous currently. –  Kevin Buzzard Mar 8 '10 at 13:14
    
There's a weight 1 cuspidal modular form of level 23, whose associated Galois representation is induced from an unramified character of Q(sqrt(-23)). OK so I really give in now ;-) –  Kevin Buzzard Mar 8 '10 at 13:28
    
Sorry. I was too careless in phrasing the question. There are certainly many dihedral forms of prime conductors. I forgot to add the condition that I believe should give the desired conclusion. I have modified the question. –  Idoneal Mar 8 '10 at 17:02
    
Is this a p-adic or a complex representation? I am envisaging the 2-adic Tate module of X_0(32), which is "dihedral" for some values of dihedral, and has "conductor 1" for some values of conductor. But apart from what, read what Emerton said about the Steinberg: it's the only smooth irred rep of GL_2(Q_p) of conductor p and with unramified central character, and it can't show up for global reasons, so if you have trivial det you can't have p dividing the conductor eactly once in a dihedral setting. –  Kevin Buzzard Mar 8 '10 at 20:51
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2 Answers 2

Thinking about Idoneal's question to Emerton about what is being used about the Steinberg, it's not just standard facts about the Steinberg one needs via this approach, but also local-global compatibility. The standard facts about the Steinberg are also easily proved if one uses local Langlands (i.e. works on the Galois side). This made me realise that in fact one can pull off the entire argument on the Galois side! Let's assume you're talking about complex 2-dimensional representations of the Galois group. Emerton has already observed that they must be ramified at some prime $p$, so it suffices to prove that $p^2$ divides the conductor. But now say

$\rho:Gal(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)\to GL_2(\mathbf{C})$

has trivial determinant and conductor $p$. One instantly gets a contradiction: the inertial invariants can't be 0-dimensional because already this would mean the conductor is at least $p^2$, and they can't be 1-dimensional because if $i\in I_p$ has one eigenvalue 1 then the other eigenvalue must be 1 too, and $\rho(i)$ is diagonalisable and hence trivial, so the inertial invariants must be 2-dimensional but now the conductor must be 1. This answers the question completely without recourse to the smooth representation theory side of things and is surely the easiest approach to the question.

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Yes, the Galois side is a slightly more familiar territory for me and this proof is quite clear. Thanks. I shall appreciate very much, nevertheless, if you (or anyone) can suggest some references for the representation theory facts used in the earlier arguments. I looked up Bump's text book but that seems not to be adequate. –  Idoneal Mar 9 '10 at 11:51
    
In some sense I'm not sure that the representation theory facts shed any more light on the situation---they're just a translation of the Galois comments above to the automorphic side. For example, the statement that the Steinberg is (up to unram twist) the only smooth admiss rep of GL_2(Q_p) with unramified central char and conductor p is just a trivial consequence of the local Langlands conjectures and a calculation on the Galois side very analogous to the above (but with Weil-Deligne groups instead of Galois groups, so you're allowed a monodromy operator). Read any intro to LL for GL_n/Q_p! –  Kevin Buzzard Mar 9 '10 at 12:40
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Just to augment Kevin's series of comments: I think that the conductor of the induction of some character $\chi$ over a quadratic field to $\mathbb Q$ would normally equal $D N(C)^2$, where $D$ is the discriminant of the quadratic field, $C$ is the condutor of the character (an ideal in the quadratic field) and $N$ is the norm from the quadratic field to $\mathbb Q$. E.g. in Kevin's $23$ example, one inducing a character of conductor 1 from $\mathbb Q(\sqrt{-23})$, so the conductor is $23$. [Added in response to an edit in the question: This form has nebentypus equal to the Legendre symbol mod 23.]

In the Maass case one should be able to do something similar, by e.g. choosing a prime $p \equiv 1 mod 4$ such that $\mathbb Q(\sqrt{p})$ has non-trivial class group, and then inducing a non-trivial character of conductor 1. [Added in response to an edit in the question: Such examples will have nebentypus equal to the Legendre symbol mod p, I think.]

[Added in response to an edit in the question:] Based on the formula above for the conductor, I think that to have square free conductor one will need to induce a character with trivial conductor, i.e. coming from the (strict) class group. I think that such an induction will always have non-trivial nebentypus, though. (The key point being that if $H/{\mathbb Q}$ is the stict Hilbert class field of the real quadratic field, then this is a generalized dihedral extension.)

Another argument, pointed out me by a colleague, is that if the conductor is square fee and the nebentypus is trivial, then all the local factors of the automorphic representation at primes in the conductor are Steinberg, which is not possible for the induction of a character.

[One more remark:] It seems to me that if we replace $\mathbb Q$ by some well-chosen number field $F$, then it will be possible to find an unramified quadratic extension $E$ of $F$ such that $E$ in turn admits a degree $4$ extension $K$, everywhere unramified, so that $K$ over $F$ (a degree 8 extension) is Galois with the quaternion group as Galois group (as opposed to a dihedral group). I think if we then take the corresponding order 4 ideal class character of $E$ and induce it to $F$, we get a monomial representation of $F$ with trivial determinant whose conductor is equal to one (and in particular, is square-free).

In other words, one is a little bit "lucky" in the $\mathbb Q$-case that Hilbert class fields of real quadratic fields are dihedral over $\mathbb Q$.

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I think this answers my question although I don't understand it fully due to my ignorance about Steinberg representations. Can someone point out a reference for the fact being used about Steinberg representation? As I understand, the crucial fact is the uniqueness of Steinberg representation that Buzzard has mentioned in his last comment. –  Idoneal Mar 9 '10 at 6:16
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