# Conductor of quaternionic representation

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.

• A somewhat related question is answered in the first paragraph of the proof of Theorem 3.1 of this paper: doi.org/10.1090/S0002-9939-02-06918-6. This cites an earlier paper of Koch and Zink (in German). – Peter Humphries Dec 17 '18 at 22:22

I consider a Casselman type of local newform theory on quaternion algebras in my paper on the basis problem (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 $$\pi$$.