Crystal lattice polarizations are perfect at $q=0$

Residue orthonormality. Induct on $\mathrm{ht}(\alpha )$. For residues $b,b^{\mathrm{\prime }}\in \mathcal{B}(\lambda {)}_{\lambda -\alpha }$ choose $i$ with ${\tilde{e}}_{i}b\mathrm{\ne }0$; then $b={\tilde{f}}_i{\tilde{e}}_{i}b$ by clause C(r) and J(r), so the adjunction of [F1] gives $⟨b,b^{\mathrm{\prime }}{⟩}_0=⟨{\tilde{e}}_{i}b,{\tilde{e}}_i{b}^{\mathrm{\prime }}{⟩}_0$. If ${\tilde{e}}_i{b}^{\mathrm{\prime }}=0$ this is $0$; otherwise ${\tilde{e}}_i{b}^{\mathrm{\prime }}$ lies at lower height and, by the lower-height orthonormality and the injectivity of ${\tilde{e}}_i$ on the surviving string positions, the pairing is ${\delta }_{b,b^{\mathrm{\prime }}}$. The top vector has norm $1$, and since all $a_{ii}=2$ the operator $Q_i$ of [F2] is the identity, so no norm larger than $1$ occurs.

Perfection and self-duality. By step 1.1 the Gram matrix of the finite component in the residue basis is congruent to the identity modulo $q$; by [F3] its determinant is a unit of $A_0$ and the matrix is invertible, so the pairing on the finite component is perfect and the dual lattice with respect to the field form equals the lattice itself: ${\mathcal{L}}^{\vee }=\mathcal{L}$.

The negative half. For $U_q^{-}$ use the corrected degree/colour comparison of [F2] with a sufficiently large regularised dominant weight: ${\pi }_{\lambda }$ is a lattice isomorphism on the component and a bijection on its crystal residue sets, and $C_{\nu }\equiv 1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q$; hence the Gram residue of the half lattice is the identity as well, and the same finite-matrix argument gives $A_0$-valuedness, perfection and $\mathcal{L}(\mathrm{\infty }{)}^{\vee }=\mathcal{L}(\mathrm{\infty })$ on each finite component.

The norm criterion. Suppose $⟨x,x⟩\in A_0$ but $x$ has a pole in the lattice. Choose the minimal $N>0$ with $q^{N}x\in \mathcal{L}$ and let $\bar{x}\mathrm{\ne }0$ be its residue. In the orthonormal residue basis of step 1.1, $⟨\bar{x},\bar{x}⟩$ is a sum of squares of rational coordinates, hence a positive rational; on the other hand $q^{2N}⟨x,x⟩\in q{A}_0$ has residue $0$, a contradiction. Therefore $\mathcal{L}=\{x:⟨x,x⟩\in A_0\}$ inside the field scalar extension.

Choice. The only choice used is the inherited hypothesis of the negative-half field form, already declared in [F2]; the finite Gram computations and the norm criterion use no further choice.