Return to Answer

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 $E$ 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 $\mathbb Q$.

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 $\mathbb Q$.

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.

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 $E$ 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 $\mathbb Q$.

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.]

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.