Let $\Gamma(\mathbb{Z}) = \operatorname{SL}_2(\mathbb{Z}) \times \operatorname{SL}_3(\mathbb{Z}) \times \operatorname{SL}_3(\mathbb{Z})$. Then $\Gamma$ acts on a pair $(A,B)$ of $3 \times 3$ matrices via the following action: for $U = \left(\begin{smallmatrix} u_1 & u_2 \\ u_3 & u_4 \end{smallmatrix} \right)$ and $S,T \in \operatorname{SL}_3(\mathbb{Z})$, we have that $(U,S,T)$ acts on $(A,B)$ by
$$\displaystyle (U,S,T) \cdot (A,B) = (u_1 SAT^t + u_2 SBT^t, u_3 SAT^t + u_4 SBT^t)$$
where $T^t$ refers to the transpose of $T$.
Bhargava proved in his Higher composition laws II that the $\Gamma(\mathbb{Z})$-orbits of pairs of integral matrices $(A,B)$ parametrize ideal classes of cubic orders. In particular, the associated cubic order is given by the (class of) binary cubic form $F$ given as
$$F(x,y) = \det(Ax - By).$$
An ideal class in a number field also has a norm form, which is always a decomposable form (i.e., it splits into linear factors over an algebraically closed field). Thus there should be some way of converting the pair $(A,B)$ into a ternary norm form. The obvious construction of taking a determinantal form, analogous to how $F$ is defined, doesn't seem to work as a determinantal form in three variables is in general not decomposable.
How does one canonically give the norm form for the ideal class associated to $(A,B)$ in terms of the coefficients of $(A,B)$?
