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Let $B = \left(\frac{-1,-11}{\mathbb{Q}}\right)$ be a choice of quaternion algebra ramifying at $11$ and consider the maximal order $\mathfrak{O}=\mathbb{Z}\oplus\mathbb{Z}i\oplus\mathbb{Z}\frac{1+j}{2}\oplus\mathbb{Z}\frac{i+k}{2}$.

Let $g = \begin{bmatrix} 1 & (1+3i)j\\ 0 & j\end{bmatrix}$.

I need to get my hands on all $2\times2$ matrices $A = \begin{bmatrix} \alpha & \beta\\ \gamma & \delta\end{bmatrix}$ with entries in $B$ such that:

1) $N(\alpha)+N(\beta) = 1$.

2) $N(\alpha) = N(\delta)$ and $N(\beta) = N(\gamma)$.

3) $\alpha\bar{\gamma}+\beta\bar{\delta} = 0$

4) $gAg^{-1} \in$ GL$_2(\mathfrak{O})$

A priori I know that there are exactly $48$ such matrices but have only managed to find $4$ of them by guessing a huge amount of matrices of the form:

  1. $\begin{bmatrix} \alpha & 0 \\ 0 & \delta\end{bmatrix}$ where the entries are units in $\mathfrak{O} = \{\pm1, \pm i\}$.

  2. $\begin{bmatrix} 0 & \beta \\ \gamma & 0\end{bmatrix}$ wth similar entries.

  3. $\begin{bmatrix} j^{-1}a & j^{-1}b\\ j^{-1}c & j^{-1}d\end{bmatrix}$ where $a,b,c,d\in\mathfrak{O}$ have norms between $1$ and $10$, $N(a)+N(b)=11$ and $N(a) = N(d), N(b) = N(c)$.

Between case 1 and case 3 I find $4$ matrices. Looking at a calculation done by Ibukiyama for ramified prime $2$ (in the longer version of "On Symplectic Euler factors of genus two") I expected the majority of the matrices to be of these forms and are stuck at making other guesses.

Has anyone got any ideas off the top of their heads what form the other $44$ matrices may take? Is there even an algorithm that can help?

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