Conductor of characters

Question

Consider characters of $PGL(2)$. The conductor of a local component $\pi_p$ of an automorphic representation is defined as the smallest index such that there is nontrivial fixed vector by the congruence subgroup $$K_0(p^r) = \left\{ M \in GL(2, \mathcal{O}_{F_p}) \ : \ M \equiv \left( \begin{array}{cc} \star & \star \\ & \star \end{array} \right) \mod p^r \right\}$$

In the case of a character of $GL(2)$ of the form $\chi_0 \circ \det$, where $\chi_0$ is a character on the base field $F_p$, is there a natural relation between this notion and the conductor of $\chi_0$, defined as the smallest index $r$ such that $\chi_0$ is trivial on $1 + p^r$?

Answer 1

The lower-right corner is supposed to be a $1$, not a star (unless the central character is trivial, which it isn't in your case unless $\chi_0$ is quadratic).

Regardless, as LSpice notes, these representations do not have a conductor in that sense, unless the conductor of $\chi_0$ is $1$, because they are not invariant under any of these subgroups.

This is not so surprising as the theorem only says that generic representations have a conductor.

Maybe a better thing to try is to tensor your representation with the Steinberg representation, obtaining an infinitesimally equivalent representation which is generic, whose conductor is equal to $\min(c(\chi_0)^2, p)$.