The integral residue lattices $L_A(\mathrm{\infty })$ and $L_A(\lambda )$

Let $A=\mathbb{Z}[q,q^{-1}]$, with rational discrete valuation rings $A_0$ at $q=0$ and $A_{\mathrm{\infty }}$ at $q=\mathrm{\infty }$. Put $M_A=U_A^{-}$ or $V_A(\lambda )=U_A^{-}{v}_{\lambda }$, $L=\mathcal{L}(\mathrm{\infty })$ or $\mathcal{L}(\lambda )$, and $L^{-}=\psi (L)$, the bar lattice, as in The bar lattice and the divided-power integral lattice of a based module.

The integral residue lattice is $L_{\mathbb{Z}}:=M_A\cap L$, a $\mathbb{Z}[q]$-module, and its bar lattice is $L_{\mathbb{Z}}^{-}:=M_A\cap L^{-}$, a $\mathbb{Z}[q^{-1}]$-module. The balanced intersection is $E_{\mathbb{Z}}:=M_A\cap L\cap L^{-}$, a $\mathbb{Z}$-module. Since $q^{-1}{M}_A=M_A$, one has $M_A\cap qL=q{L}_{\mathbb{Z}}$; consequently the residue map $L_{\mathbb{Z}}\mathrm{/}q{L}_{\mathbb{Z}}\to L\mathrm{/}qL$ is injective and is $\mathbb{Z}$-linear. Its image is an integral subgroup of the rational crystal space, not the entire $\mathbb{Q}$-space. No Laurent-$A$-linearity is asserted on the residue lattice or on the residue map.

The field projection ${\pi }_{\lambda }:U_q^{-}\to V(\lambda )$, ${\pi }_{\lambda }(P)=P{v}_{\lambda }$, restricts to an $A$-linear surjection $U_A^{-}\to V_A(\lambda )$ (The Lusztig divided-power integral form $U_A$), and it sends $L_{\mathbb{Z}}(\mathrm{\infty })$ into $L_{\mathbb{Z}}(\lambda )$ by the already-proved grand-loop lattice projection (The grand loop: existence of the highest weight crystal bases). The module $V_A(\lambda )$ is $U_A$-stable: the divided-power and toral reordering identities of Divided-power multiplication, toral binomials and the $E_i^{(a)}{F}_i^{(b)}$ reordering identity reduce the action on $v_{\lambda }$ to negative divided powers and Gaussian scalars in $A$, with all positive generators killing the highest vector except through the unit, and the residue compatibility of the projection is Projection from $\mathcal{L}(\mathrm{\infty })$ to $\mathcal{L}(\lambda )$ and its compatibility with the Kashiwara operators. No global freeness, residue spanning or balancedness is assumed here. The integral residue subgroup for the negative half is proved below to be $\mathbb{Z}B(\mathrm{\infty })$; the module subgroup $\mathbb{Z}B(\lambda )$ is a consequence of globalization.