Classical work by Casselman shows that for an irreducible admissible representation $\rho$ of $GL\_2$ over a non-archimedean field $k$, there is a minimal power $n\geq 0$ of the prime ideal $\mathfrak{p}$ such that $\rho$ has a fixed vector under $(\begin{smallmatrix} \* & \* \\ \mathfrak{p}^n & 1+\mathfrak{p}^n\end{smallmatrix})\subset GL\_2(\mathcal{O}\_k)$. Furthermore, the fixed space is 1-dimensional.


Suppose $\rho$ is special or supercuspidal. Then it corresponds by Jacquet-Langlands to an irreducible, admissible $D^\times$-representation $\pi$, where $D$ is the non-split quaternions over $k$.


My question is:
**Is there a compact open subgroup $K\subset D^\times$ such that $\pi^K$ is one-dimensional?**


Edit: I think, when $\rho$ is special $\chi\cdot St$, its JL-transfer is just $\chi\circ Nrd$, so this case is easy.



Conductor of quaternionic representation
I consider a Casselman type of local newform theory on quaternion algebras in [my paper on the basis problem](https://arxiv.org/abs/1804.04234) (sections 2 and 3), which gives you a positive answer to your question half of the time (Cases 1 and 2 below). Here is a brief summary. For simplicity, I'll assume trivial central characters.


Let's say $\pi'$ is the Jacquet-Langlands transfer of $\pi$ to GL(2) and the conductor is $n$. Then $n$ is the minimal positive integer such that $\pi$ is trivial on $U\_D^{n-1}=1+P\_D^{n-1}$, the $(n-1)$-st higher unit group in $D$ (here $P\_D$ is the prime ideal of $D$). Incidentally this is a *normal* subgroup of $D$, so $\pi$ factors through the finite group $D^\times/U\_D^{n-1}$.
Thus if you want an open $K$ such that $\dim \pi^K = 1$ we may as well assume $K$ contains $U\_D^{n-1}$.


**Case 1:** $\pi'$ is special, so $\pi$ is 1-dimensional (factors through the reduced norm as you observed). Then you can just take $K=U\_D^{n-1}$
and $\dim \pi^K = 1$.


**Case 2:** $\pi'$ is supercuspidal of conductor $n=2m+1$. Then you can take $K=O\_E^\times U\_D^{2m}$ where $E/F$ is the unramified quadratic extension,
and $\dim \pi^K = 1$ (see Theorem 3.5 in my paper).


**Case 3:** $\pi'$ is supercuspidal of conductor $n=2m$. For simplicity also assume $\pi$ is minimal (conductor is minimal among twists). Then you can take $K = O\_E^\times U\_D^{2m-1}$ where $E/F$ is any ramified quadratic extension to get $\dim \pi^K = 2$.
Here there is no subgroup that contains a maximal quadratic order which will give you dimension 1 (if you take $E$ unramified, the dimension is 0). But you can pick out a 1-dimensional space by specifying how $\pi$ acts on a unifomizer $\varpi\_D$ (necessarily $\pm 1$).


Subgroups $K$ of the above form are unit groups of Hijikata-Pizer-Shemanske orders and are natural objects to consider for certain reasons. I do not know whether there are other subgroups one can use to get dimension 1 in Case 3, but I know of no suitable natural choices. 


Note by allowing $K$ to be non-compact however (adjoin in $\langle \varpi\_D \rangle$) you can get $\dim \pi^K = 1$ but to me choosing this vector in the 2-dimensional space given in Case 3 is not really the right thing to do for a local newform theory. My feeling is that you would just picking out a 1-dimensional space arbitrarily for the sake of getting a 1-dimensional space, but it would make more sense to pick the vector whose eigenvalue for $\varpi\_D$ matches with an epsilon factor. This however depends upon $\pi$.


