I am trying to write an element of $SU(2)$ with rational elements. These are typically written in the form:
$$ \left( \begin{array}{cc} x + iy & z + iw \\ -z + iw & x - iy \end{array}\right) \in SU(2, \mathbb{Q}) $$
And this should check out that this $2 \times 2$ matrix has determinant equal to $1$. I've calculated:
$$ \det = ||x+iy||^2 + ||z + iw||^2 = x^2 + y^2 + z^2 + w^2 = 1$$
In order to find four numbers that when squared, sum to another square.
$$ \left[ \begin{array}{cc} x + iy & z + iw \\ -z + iw & x - iy \end{array}\right]\in \text{GL}_2(\mathbb{Z}[i]) \mapsto \left[ \begin{array}{cc|cc} x & -y & z & -w \\ y & x & w & z \\ \hline -w & -w & x & y \\ w & -z & -y & x \end{array} \right] \in \text{GL}_4(\mathbb{Z}) $$
This can't be right. This map is not a homomorphism. In particular the determinants are different. The second one is:
$$ \det = \big( x^4 + y^4 + z^4 + w^4 \big) + (\dots)$$
and there are no cancellations. This is slightly disconcerting. I think this is becomes an attempt to write a "map" of some kind from: $$ \mathbb{H} \to SU(2) \to SO(4) $$
We also know that the special unitary group is a $2:1$ cover of the $SO(3)$ lie group, so there must be a $3 \times 3$ matrix element as well.
I could try to write the thing using quaternions, I don't have the multiplications off hand:
$$ 1 \mapsto \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right]$$ The elements $i,j,k \in \mathbb{H}$ could also be represented a quaternions.
$$ 1 \mapsto \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{array} \right]$$ This is looking close to what I have. Maybe it's no longer reasonable to expect the determinants to be the same?