\in {\text{Aut}}(F_A).$$ The element $(\phi_B)_*(g)$ is a generic element of ${\text{Aut}}(F_{A})$, namely $\prod_{j=0}^n \rho_j^{h_j} $. We turn our attention to describing the dual group $G^T$ to $G$. If we push the element $g^T:= (e^{2\pi i s_i/d})_i \in {\text{Aut}}(F_{dI})$ down via the map $\phi_{B^T}$, then we get the action $$(\phi_{B^T})_*(g^T) =(e^{\frac{2\pi i }{d} \sum_{i=0}^n s_i b_{ij}})_j= \prod_{i=1}^n (\rho^T_i)^{s_i} .$$

In other words, we have (surjective) group homomorphisms $$\begin{aligned}
(\phi_B)_*&: {\text{Aut}}(F_{dI}) \rightarrow {\text{Aut}}(F_A);\text{ and }\\
(\phi_{B^T})_*&: {\text{Aut}}(F_{dI}) \rightarrow {\text{Aut}}(F_{A^T}).
\end{aligned}$$

This gives us a new interpretation of the choice of groups $G$ and $G^T$: both are pushforwards of subgroups of ${\text{Aut}}(F_{dI})$ via the Shioda maps $\phi_B$ and $\phi_{B^T}$, respectively.

Reinterpretation of the Dual Group
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We now reformulate the relationship between the groups $G$ and $G^T$ via a bilinear pairing. Consider the map $$\langle , \rangle_B: {\mathbb{Z}}^{n+1} \times {\mathbb{Z}}^{n+1} \longrightarrow {\mathbb{Z}}$$ where $\langle \mathbf{s}, \mathbf{h}\rangle_B := \mathbf{s}^T B \mathbf{h}$. Choose a subgroup $G\subset {\text{Aut}}(F_A)$, so that $J_{F_A} \subseteq G$. Then set $H : = ((\phi_B)_*)^{-1}(G)$. Note that the map $(h_j)_j \mapsto (e^{2\pi i h_j / d})_j$ induces a natural, surjective group homomorphism $${\mathbb{Z}}^{n+1} \stackrel{pr}{\rightarrow} {\text{Aut}}(F_{dI}).$$ Take $\tilde H$ to be the inverse image $\tilde H:= pr^{-1}(H)$ of $H$ under this map. We 