I think I understand what Kevin is saying in the Theorem. Let $\mathfrak o$ be the ring of integers of $K$ and $\mathfrak p$ the maximal ideal. Define the following subgroups of $G=GL(2, K)$:
$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o // \mathfrak p & \mathfrak o \end{pmatrix}$
$I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}$
$Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$
Lemma. Let $\pi$ be a smooth representation of $I$, I$with a central character, such that$Z_n$acts trivially,$Z_{n-1}$does not act trivially and the space of$I_n$-invariants is non-zero. Then the restriction of$\pi$to$I_n Z_0$contains a one dimensional subrepresentation of the form$\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where$\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$is a smooth character of conductor$\mathfrak p^n$. Proof. Look at the action of the abelian group$Z_0$on$\pi^{I_n}$. The pair$(I_nZ_0, \chi)$is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if$\pi$is an irreducible smooth representation of$G$, then$Hom_{I_n Z_0}(\chi, \pi)\neq 0$if and only if$\pi$is a principal series rep with one character unramified, the other of conductor$\mathfrak p^n$. For this you could look either in the appendix by Heniart to: http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf or in the book of Bushnell and Henniart. The main point being that the representation$\chi$(as a representation of$\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$) shows up in the$U$-coinvariants of$\pi$, where$U$is unipotent upper (lower?) triangular matrices. 1 I think I understand what Kevin is saying in the Theorem. Let$\mathfrak o$be the ring of integers of$K$and$\mathfrak p$the maximal ideal. Define the following subgroups of$G=GL(2, K)$:$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o // \mathfrak p & \mathfrak o \end{pmatrix}I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$Lemma. Let$\pi$be a smooth representation of$I$, such that$Z_n$acts trivially,$Z_{n-1}$does not act trivially and the space of$I_n$-invariants is non-zero. Then the restriction of$\pi$to$I_n Z_0$contains a one dimensional subrepresentation of the form$\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where$\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$is a smooth character of conductor$\mathfrak p^n$. Proof. Look at the action of the abelian group$Z_0$on$\pi^{I_n}$. The pair$(I_nZ_0, \chi)$is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if$\pi$is an irreducible smooth representation of$G$, then$Hom_{I_n Z_0}(\chi, \pi)\neq 0$if and only if$\pi$is a principal series rep with one character unramified, the other of conductor$\mathfrak p^n$. For this you could look either in the appendix by Heniart to: http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf or in the book of Bushnell and Henniart. The main point being that the representation$\chi$(as a representation of$\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$) shows up in the$U$-coinvariants of$\pi$, where$U\$ is unipotent upper (lower?) triangular matrices.