Balanced triples over the two charts of ${\mathbb{P}}^1$ and the globalization map

Let $V$ be a finite-dimensional vector space over $\mathbb{Q}(q)$, let $M\subseteq V$ be an $A=\mathbb{Z}[q,q^{-1}]$-submodule spanning $V$, and let $L_0,L_{\mathrm{\infty }}$ be free full rational chart lattices over $A_0,A_{\mathrm{\infty }}$ (as for a based module in The bar lattice and the divided-power integral lattice of a based module). Put $L_{\mathbb{Z}}:=M\cap L_0,L_{\mathbb{Z}}^{-}:=M\cap L_{\mathrm{\infty }},E_{\mathbb{Z}}:=M\cap L_0\cap L_{\mathrm{\infty }}\mathrm{.}$ The integral triple $(M,L_0,L_{\mathrm{\infty }})$ is balanced if the canonical $\mathbb{Z}$-linear map $E_{\mathbb{Z}}⟶L_{\mathbb{Z}}\mathrm{/}q{L}_{\mathbb{Z}}=(M\cap L_0)\mathrm{/}(M\cap q{L}_0)$ is an isomorphism. Its inverse is the integral globalization map on the integral residue subgroup. Rational balancedness is the corresponding isomorphism $\mathbb{Q}{\otimes }_{\mathbb{Z}}{E}_{\mathbb{Z}}\to L_0\mathrm{/}q{L}_0$, after extending the integral subgroup by $\mathbb{Q}$. In the applications of this page the integral subgroup is $\mathbb{Z}B$ for the lower global family $B$, and its globalization gives the lower lifts $G(b)$.

For a balanced triple, the two-chart lemma of this page proves $M\cong A{\otimes }_{\mathbb{Z}}{E}_{\mathbb{Z}}$, $L_{\mathbb{Z}}\cong \mathbb{Z}[q]{\otimes }_{\mathbb{Z}}{E}_{\mathbb{Z}}$ and $L_{\mathbb{Z}}^{-}\cong \mathbb{Z}[q^{-1}]{\otimes }_{\mathbb{Z}}{E}_{\mathbb{Z}}$. If the triple is balanced and $E_{\mathbb{Z}}$ has a finite $\mathbb{Z}$-basis $B_E$ whose rational residue classes are a basis of $L_0\mathrm{/}q{L}_0$, these lifts give bases of $M,L_0,L_{\mathrm{\infty }}$ over $A,A_0,A_{\mathrm{\infty }}$ respectively. Thus $A$-freeness of $M$ is a conclusion from integral balancedness together with a proved free integral residue basis; it is not automatic for an arbitrary not-finitely-generated $\mathbb{Z}$-submodule. All definitions apply componentwise to the finite root or label spaces.