The highest weight and negative half integral crystal triples are balanced

Simultaneous induction and the base layer. Replace the assertions by the following height-$r$ induction statements: X$(r)$: the triple of $U_A^{-}$ is integrally balanced in every degree of height $\le r$ with bar-invariant global lifts $G_{\mathrm{\infty }}(b)$ forming an $A$-basis and $\mathbb{Z}$-intersection $\mathbb{Z}\mathcal{B}(\mathrm{\infty })$; Y$(r)$: the same for every $V_A(\lambda )$ with lifts $G_{\lambda }(b)$ and intersection $\mathbb{Z}\mathcal{B}(\lambda )$; Z$(r)$: the negative-half membership $G_{\mathrm{\infty }}(b)\in I_{(i,1)}^A$ whenever ${\tilde{e}}_{i}b\mathrm{\ne }0$, for every new degree of height $\le r$; and the derived clause D$(r)$: for the new degrees, the based-quotient compatibility ${\pi }_{\lambda }(G_{\mathrm{\infty }}(b))=0$ if and only if ${\pi }_{\lambda }(b)=0$, and ${\pi }_{\lambda }(G_{\mathrm{\infty }}(b))=G_{\lambda }({\pi }_{\lambda }(b))$ otherwise. For $r=0$ the degree component is $A\cdot 1$ or $A{v}_{\lambda }$, the charts meet in $\mathbb{Z}\cdot 1$ or $\mathbb{Z}{v}_{\lambda }$, and the stated properties hold by inspection. Assume X, Y, Z, D for all heights $<r$ and fix a degree $\alpha$ of height $r\ge 1$.

The string ideals at height $r$ are balanced. The induction hypotheses are exactly the input of [F1] at height $r$, so for every $i\in I$ and $n\ge 1$ the ideals $I_{(i,n)}^A$ and $J_{(i,n)}^A$ are integrally balanced in degree $-\alpha$, respectively $\lambda -\alpha$, with adapted free bases whose balanced $\mathbb{Z}$-intersections map isomorphically onto $\mathbb{Z}(\mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_i^n\mathcal{B}(\mathrm{\infty }))$ and $\mathbb{Z}(\mathcal{B}(\lambda )\cap {\tilde{f}}_i^n\mathcal{B}(\lambda ))$. In particular, for each $b\in \mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_i^n\mathcal{B}(\mathrm{\infty })$ there is an adapted lift $g\in E_{I_{(i,n)}}:=I_{(i,n)}^A\cap \mathcal{L}(\mathrm{\infty })\cap \psi (\mathcal{L}(\mathrm{\infty }))$ whose residue class in $(I_{(i,n)}^A\cap \mathcal{L}(\mathrm{\infty }))\mathrm{/}(I_{(i,n)}^A\cap q\mathcal{L}(\mathrm{\infty }))$ is $b$; these are the only lifts used below.

Every string lift is bar fixed. Let $g\in E_I:=I_{(i,n)}^A\cap \mathcal{L}(\mathrm{\infty })\cap \psi (\mathcal{L}(\mathrm{\infty }))$ be any adapted lift. By [F3] the ideal $I_{(i,n)}^A$ is generated over $A$ by bar-fixed divided-word vectors, so write $g={\sum }_w{a}_{w}w$ with $a_w\in A$ and $\psi (w)=w$; then $g-\psi (g)={\sum }_w(a_w-\psi (a_w))w$, and each antisymmetric Laurent polynomial $a_w-\psi (a_w)$ is divisible in $A$ by $q-q^{-1}$, so $h:=(g-\psi (g))\mathrm{/}(q-q^{-1})\in I_{(i,n)}^A$. Since $g,\psi (g)\in \mathcal{L}(\mathrm{\infty })\cap \psi (\mathcal{L}(\mathrm{\infty }))$ and $(q-q^{-1}{)}^{-1}=-q(1-q^2{)}^{-1}\in q{A}_0$, one gets $h\in q\mathcal{L}(\mathrm{\infty })$; since $(q-q^{-1}{)}^{-1}=q^{-1}(1-q^{-2}{)}^{-1}\in A_{\mathrm{\infty }}$, one gets $h\in \psi (\mathcal{L}(\mathrm{\infty }))$. Thus $h\in E_I$ and $h\in q\mathcal{L}(\mathrm{\infty })$, so its class in $(I_{(i,n)}^A\cap \mathcal{L}(\mathrm{\infty }))\mathrm{/}(I_{(i,n)}^A\cap q\mathcal{L}(\mathrm{\infty }))\cong \mathbb{Z}(\mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_i^n\mathcal{B}(\mathrm{\infty }))$ is zero; by injectivity of the balanced residue map of F2 applied to $I_{(i,n)}^A$, $h=0$ and $g=\psi (g)$. The same argument applies to the module ideals $J_{(i,n)}^A$.

Gluing across colors (Kashiwara’s gluing lemma). Let $b\in \mathcal{B}(\mathrm{\infty }{)}_{-\alpha }$ lie in ${\tilde{f}}_i\mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_j\mathcal{B}(\mathrm{\infty })$, and let $G_i(b),G_j(b)\in \mathcal{L}(\mathrm{\infty })\cap \psi (\mathcal{L}(\mathrm{\infty }))$ be the corresponding adapted lifts of $I_{(i,1)}^A$ and $I_{(j,1)}^A$ from step 2.1. Write $b$ as a lowering word with rightmost color $k$ and choose, using [F7], a dominant integral auxiliary label $\lambda$ in the regular datum with $\lambda (h_k)=0$ and all other coordinates larger than the degree bounds. By the kernel statement of [F7], $\ker {\pi }_{\lambda }$ in degree $\alpha$ equals the image of $U_q^{-}{F}_k$, and ${\pi }_{\lambda }(b)=0$: modulo $q$, projection commutes with every lowering operator, while the rightmost ${\tilde{f}}_k$ kills the highest class because $\lambda (h_k)=0$. For each of the two lifts, $G_i(b)v_{\lambda }$ lies in $J_{(i,1)}^A$ by [F3], lies in $q\mathcal{L}(\lambda )$ because its crystal residue ${\pi }_{\lambda }(b)=0$, and lies in $\psi (\mathcal{L}(\lambda ))$ by bar invariance; the balancedness of $J_{(i,1)}^A$ from step 2.1 forces $G_i(b)v_{\lambda }=0$, and likewise $G_j(b)v_{\lambda }=0$. Hence $Q:=G_i(b)-G_j(b)$ lies in $U_A^{-}\cap U_q^{-}{F}_k$, in $q\mathcal{L}(\mathrm{\infty })$ (both lifts have residue $b$), and in $\psi (\mathcal{L}(\mathrm{\infty }))$. By the reversal [F4], $Q^{\star }$ lies in $U_A^{-}\cap F_k{U}_q^{-}\cap q\mathcal{L}(\mathrm{\infty })\cap \psi (\mathcal{L}(\mathrm{\infty }))$; by [F3], $U_A^{-}\cap F_k{U}_q^{-}=I_{(k,1)}^A$; and then the balancedness of $I_{(k,1)}^A$ from step 2.1 again forces $Q^{\star }=0$, hence $Q=0$.

Global half lifts and spanning. For $\alpha \mathrm{\ne }0$, [F5] supplies for every $b\in \mathcal{B}(\mathrm{\infty }{)}_{-\alpha }$ a color $i$ and a predecessor $b^{\mathrm{\prime }}\in \mathcal{B}(\mathrm{\infty }{)}_{-\alpha +{\alpha }_i}$ with $b={\tilde{f}}_i{b}^{\mathrm{\prime }}$, so $b\in \mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_i\mathcal{B}(\mathrm{\infty })$ and step 2.1 provides an adapted lift in $I_{(i,1)}^A$; define $G_{\mathrm{\infty }}(b)$ to be this element. Step 3.2 shows the definition does not depend on the chosen color, and two lifts for the same color differ by an element of $E_I\cap q\mathcal{L}(\mathrm{\infty })=0$ by step 3.1, so $G_{\mathrm{\infty }}(b)$ is well defined and bar invariant. By [F3], every nonempty divided-word generator in degree $-\alpha$ begins with a positive divided power $F_i^{(l)}$ and therefore belongs to $I_{(i,1)}^A$, hence $U_A^{-}={\sum }_{i\in I}{I}_{(i,1)}^A$ in this degree; by step 2.1 and the lower-height part of the induction, each $I_{(i,1)}^A$ has the free $A$-basis $\{G_{\mathrm{\infty }}(b):b\in \mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_i\mathcal{B}(\mathrm{\infty })\}$. Therefore $U_A^{-}$ in degree $-\alpha$ equals ${\sum }_{b\in \mathcal{B}(\mathrm{\infty }{)}_{-\alpha }}A{G}_{\mathrm{\infty }}(b)$, and X$(<r)$ supplies the lower-height components, so the family $(G_{\mathrm{\infty }}(b){)}_b$ spans $U_A^{-}$ over $A$ in every degree of height $\le r$.

The half triple is integrally balanced (X$(r)$). Put $F:={⨁}_{b\in \mathcal{B}(\mathrm{\infty }{)}_{-\alpha }}\mathbb{Z}{G}_{\mathrm{\infty }}(b)$; it is a finite free $\mathbb{Z}$-module contained in $U_A^{-}\cap \mathcal{L}(\mathrm{\infty })\cap \psi (\mathcal{L}(\mathrm{\infty }))$ because the lifts are bar invariant and lie in the adapted integers of the string ideals. It generates $U_A^{-}$ over $A$ by step 4.1. Its image in $\mathcal{L}(\mathrm{\infty })\mathrm{/}q\mathcal{L}(\mathrm{\infty })$ is ${⨁}_b\mathbb{Z}b$ by [F6], and its image under $\psi$ is another free $\mathbb{Z}$-span of the same rank inside the infinity residue; the two images are injective. Applying the chart generation criterion F2 to $M=U_A^{-}$, $L_0=\mathcal{L}(\mathrm{\infty })$ and $L_{\mathrm{\infty }}=\psi (\mathcal{L}(\mathrm{\infty }))$ gives that the half triple in degree $-\alpha$ is integrally balanced, with $E=F$ and $E\to (U_A^{-}\cap \mathcal{L}(\mathrm{\infty }))\mathrm{/}(U_A^{-}\cap q\mathcal{L}(\mathrm{\infty }))$ an isomorphism onto ${⨁}_b\mathbb{Z}b$; the same holds in every lower degree by X$(<r)$, so X$(r)$ holds.

Adapted membership. An adapted lift of $I_{(i,n)}^A$ and the global lift with the same residue both lie in the now balanced half intersection; its injectivity identifies them. Thus $I_{(i,n)}^A={⨁}_{c\in \mathcal{B}(\mathrm{\infty })\cap {\tilde{f}}_i^n\mathcal{B}(\mathrm{\infty })}A{G}_{\mathrm{\infty }}(c)$, and Z$(r)$ holds. Since $I_{(i,n)}^A$ spans $F_i^n{U}_q^{-}$ by the integral string decomposition, the corresponding global vectors are also a field basis of that string space. The chart extension in F2 makes them an $A_0$-basis of $F_i^n{U}_q^{-}\cap \mathcal{L}(\mathrm{\infty })$.

The exact top-string transition. Suppose ${\tilde{e}}_{i}b=0$ and $n\ge 1$. Write the integral string decomposition $G_{\mathrm{\infty }}(b)={\sum }_{l\ge 0}{F}_i^{(l)}{u}_l$ of [F3]. The string-shift definitions give $u_0=(1-{\tilde{f}}_i{\tilde{e}}_i)G_{\mathrm{\infty }}(b)\in \mathcal{L}(\mathrm{\infty })$ with residue $b$, and $z:=F_i^{(n)}{u}_0={\tilde{f}}_i^n{u}_0\in \mathcal{L}(\mathrm{\infty })$ with residue ${\tilde{f}}_i^{n}b$, by [F5]. All $u_l$ are integral, so the divided-power multiplication formula gives $\phi :=F_i^{(n)}{G}_{\mathrm{\infty }}(b)\in I_{(i,n)}^A$ and $\phi -z={\sum }_{l\ge 1}{\left(\genfrac{}{}{0px}{}{n+l}{n}\right)}_i{F}_i^{(n+l)}{u}_l\in I_{(i,n+1)}^A$. Expand $\phi ={\sum }_{c\in {\tilde{f}}_i^n\mathcal{B}(\mathrm{\infty })}{a}_c{G}_{\mathrm{\infty }}(c)$ with $a_c\in A$. In the quotient $F_i^n{U}_q^{-}\mathrm{/}{F}_i^{n+1}{U}_q^{-}$, step 6.1 identifies the zero chart with the $A_0$-span of the images indexed by $c\in {\tilde{f}}_i^n\mathcal{B}(\mathrm{\infty })\setminus {\tilde{f}}_i^{n+1}\mathcal{B}(\mathrm{\infty })$. Since $\phi$ and $z$ have the same quotient image, those $a_c$ belong to $A\cap A_0=\mathbb{Z}[q]$ and reduce to $1$ at $c={\tilde{f}}_i^{n}b$ and to $0$ otherwise. Both $\phi$ and all global vectors are bar fixed, so uniqueness of their $A$-basis expansion gives $a_c(q)=a_c(q^{-1})$. A bar-fixed polynomial in $\mathbb{Z}[q]$ is constant; hence the indicated coefficients are exactly $1$ and $0$. The remaining terms lie in $I_{(i,n+1)}^A$ by step 6.1, proving (v) without presuming that $\phi$ itself lies in the zero chart.

Killed module images vanish. Fix $\lambda \in P^{+}$ and let $b\in \mathcal{B}(\mathrm{\infty }{)}_{-\alpha }$ have ${\pi }_{\lambda }(b)=0$. By [F5] choose $i$ with ${\tilde{e}}_{i}b\mathrm{\ne }0$; then $b\in {\tilde{f}}_i\mathcal{B}(\mathrm{\infty })$, so $G_{\mathrm{\infty }}(b)\in I_{(i,1)}^A$ by Z$(r)$ and $G_{\mathrm{\infty }}(b)v_{\lambda }\in J_{(i,1)}^A$ by [F3]. Since its residue ${\pi }_{\lambda }(b)$ is zero it lies in $q\mathcal{L}(\lambda )$, and since $G_{\mathrm{\infty }}(b)$ is bar invariant and ${\pi }_{\lambda }$ commutes with $\psi$ it lies in $\psi (\mathcal{L}(\lambda ))$. The balancedness of $J_{(i,1)}^A$ from step 2.1 forces the class of $G_{\mathrm{\infty }}(b)v_{\lambda }$ in its balanced intersection to be zero, so $G_{\mathrm{\infty }}(b)v_{\lambda }=0$; this is the local form of Kashiwara’s killed-image lemma.

The surviving module images. For ${\pi }_{\lambda }(b)\mathrm{\ne }0$, the element $G_{\mathrm{\infty }}(b)v_{\lambda }$ is bar invariant, lies in $\mathcal{L}(\lambda )$ by the lattice projection of [F5], and its residue in $\mathcal{L}(\lambda )\mathrm{/}q\mathcal{L}(\lambda )$ is ${\pi }_{\lambda }(b)$ by the derived clause D$(<r)$ and the operator compatibility of [F5]; the residues ${\pi }_{\lambda }(b)$ over the surviving $b$ biject onto $\mathcal{B}(\lambda )$ by [F5]. These images span $V_A(\lambda )$ over $A$ because $U_A^{-}$ is spanned by the $G_{\mathrm{\infty }}(b)$ and, by [F3] and step 7.2, ${\pi }_{\lambda }$ is surjective with the killed generators contributing zero.

The module triple is integrally balanced (Y$(r)$). Let $F_{\lambda }:={⨁}_{b:{\pi }_{\lambda }(b)\mathrm{\ne }0}\mathbb{Z}{G}_{\mathrm{\infty }}(b)v_{\lambda }$ inside $V_A(\lambda )\cap \mathcal{L}(\lambda )\cap \psi (\mathcal{L}(\lambda ))$. It generates $V_A(\lambda )$ over $A$ by step 8.1 and its image in $\mathcal{L}(\lambda )\mathrm{/}q\mathcal{L}(\lambda )$ is ${⨁}_{\beta \in \mathcal{B}(\lambda )}\mathbb{Z}\beta$ by the bijection of residues, hence is a free generator group; the bar image is the corresponding free span in $\psi (\mathcal{L}(\lambda ))\mathrm{/}{q}^{-1}\psi (\mathcal{L}(\lambda ))$, injective of the same rank. The chart generation criterion F2, applied to $M=V_A(\lambda )$, $L_0=\mathcal{L}(\lambda )$, $L_{\mathrm{\infty }}=\psi (\mathcal{L}(\lambda ))$, gives integral balancedness in degree $\lambda -\alpha$ with $E_{\mathbb{Z}}={⨁}_{{\pi }_{\lambda }(b)\mathrm{\ne }0}\mathbb{Z}{G}_{\mathrm{\infty }}(b)v_{\lambda }$ and residue group $\mathbb{Z}\mathcal{B}(\lambda )$; with the induction hypothesis in lower degrees this is Y$(r)$. Uniqueness of the balanced lift identifies each $G_{\mathrm{\infty }}(b)v_{\lambda }$ with the adapted module lift $G_{\lambda }({\pi }_{\lambda }(b))$, which also proves (iv) in degree $\alpha$.

Saturation and the adapted string bases. The lifts $G_{\lambda }(\beta )$ for $\beta \in \mathcal{B}(\lambda )$ now form an $A$-basis of $V_A(\lambda )$ in every degree of height $\le r$, and by [F3] the ideals $J_{(i,n)}^A$ are generated by the images of the corresponding string ideal generators. By the rational string decomposition and crystal partial inverses, the dimension of $F_i^{n}V(\lambda )$ is the number of surviving positions at least $n$. The adapted basis of $J_{(i,n)}^A$ lies in that field space and has exactly this many independent lifts, so it spans the field space. Its intersection with the full global $A$-lattice therefore equals the $A$-span of the adapted lifts $\{G_{\lambda }(\beta ):\beta \in \mathcal{B}(\lambda )\cap {\tilde{f}}_i^n\mathcal{B}(\lambda )\}$, which is exactly $J_{(i,n)}^A$; this proves the saturation identity and the displayed free decompositions of (iii) for the module, and the half decompositions were proved in step 6.1.

Conclusion. Steps 1.1 and 2.1 set up the induction and produce the balanced string ideals; steps 3.1 and 3.2 prove bar invariance and the gluing of string lifts; steps 4.1 and 5.1 construct the global half basis, its chart injections and X$(r)$; steps 6.1 and 7.1 prove Z$(r)$ and the exact top-string transition (v); steps 7.2, 8.1, 9.1 and 10.1 prove Y$(r)$, the projection rule (iv), the module saturation and the adapted decompositions (iii) for both halves. The ranks in (ii) are the field dimensions computed by [F5] rank by rank, with the classical PBW identification supplied by [F8] and its stated Choice hypothesis; the dimension count is finite in each root degree because only finitely many compositions occur. This closes the induction in every degree and proves (i)-(v); the Choice use is confined to [C1].