Conductor of monomial forms with trivial nebentypus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:06:12Z http://mathoverflow.net/feeds/question/17461 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17461/conductor-of-monomial-forms-with-trivial-nebentypus Conductor of monomial forms with trivial nebentypus Idoneal 2010-03-08T11:16:18Z 2010-03-09T10:26:00Z <p>Is it true that the conductor of a holomoprhic or a Maass cusp form <em>with trivial nebentypus</em> corresponding to a two-dimensional dihedral representation (over $\mathbb{Q}$ )is non-square-free?</p> http://mathoverflow.net/questions/17461/conductor-of-monomial-forms-with-trivial-nebentypus/17488#17488 Answer by Emerton for Conductor of monomial forms with trivial nebentypus Emerton 2010-03-08T16:27:26Z 2010-03-09T05:28:13Z <p>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.]</p> <p>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.]</p> <p>[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.)</p> <p>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.</p> <p>[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). </p> <p>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 $\mathbb Q$. </p> http://mathoverflow.net/questions/17461/conductor-of-monomial-forms-with-trivial-nebentypus/17593#17593 Answer by Kevin Buzzard for Conductor of monomial forms with trivial nebentypus Kevin Buzzard 2010-03-09T10:26:00Z 2010-03-09T10:26:00Z <p>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</p> <p><code>$\rho:Gal(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)\to GL_2(\mathbf{C})$</code></p> <p>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.</p>