Uniqueness: crystal bases of integrable modules decompose into highest weight crystals

Step A: the top isotypic decomposition. By [F1] write $M$ as a direct sum of simple integrable highest weight modules; let $\lambda$ be a maximal weight of $M$ and let $N_0:=U_q(\mathfrak{g})M_{\lambda }$ be the sum of the summands isomorphic to $V(\lambda )$, and $N_1$ the sum of the remaining ones. Each nonzero $w\in M_{\lambda }$ generates a nonzero integrable cyclic highest weight module of weight $\lambda$, isomorphic to $V(\lambda )$ by [F2]; the sum of these cyclic submodules is $N_0$, and since $M_{\lambda }$ has dimension $m$ and each copy of $V(\lambda )$ contributes a one-dimensional $\lambda$-weight space, $N_0\cong V(\lambda {)}^{\oplus m}$ with $m=\dim M_{\lambda }$, while $M=N_0\oplus N_1$ by [F1]. A simple constituent $V(\nu )$ of $N_1$ with $\lambda \in \nu -Q_{+}$ would satisfy $\nu \ge \lambda$, and maximality of $\lambda$ in $\mathrm{w}\mathrm{t}(M)$ gives $\nu =\lambda$, a contradiction; hence $N_1$ has no weight $\lambda$.

Step B: descent for $V(\lambda {)}^{\oplus m}$. Since $M_{\lambda }=N_0{}_{\lambda }$ has dimension $m=\mathrm{\#}{B}_{\lambda }$ by [F3], choose one lift in $L_{\lambda }$ of each element of the finite basis $B_{\lambda }$. Their reduction modulo $q{L}_{\lambda }$ is a basis, so their determinant in any $A_0$-basis of the free module $L_{\lambda }$ is a unit of $A_0$ and they form an $A_0$-basis of $L_{\lambda }$. Each lift generates a copy of $V(\lambda )$ by [F2], these $m$ cyclic submodules are linearly independent and their sum has $\lambda$-weight space of dimension $m$, so they sum directly to $N_0$; fixing the identification $N_0\cong V(\lambda {)}^{\oplus m}$ that sends the chosen lift to the standard top generator $v_{\lambda }^{(j)}$ of the $j$-th copy, one has ${\mathrm{\Phi }}_0(L_{\lambda })={⨁}_j\mathcal{L}(\lambda {)}_{\lambda }^{(j)}$ and $B_{\lambda }$ corresponds to the $m$ standard top residues $b_{\lambda }^{(j)}$. Claim: if $u\in V(\lambda {)}_{\mu }^{\oplus m}$ with $\mu \mathrm{\ne }\lambda$ and ${\tilde{e}}_{i}u\in \mathcal{L}(\lambda {)}^{\oplus m}$ for all $i\in I$, then $u\in \mathcal{L}(\lambda {)}^{\oplus m}$. Indeed, since $\mathcal{L}(\lambda )$ generates $V(\lambda )$ over $\mathbb{Q}(q)$, some $q^{n}u$ lies in $\mathcal{L}(\lambda {)}^{\oplus m}$; take the minimal such $n\ge 0$. If $n>0$, then $b:=q^{n}u+q\mathcal{L}(\lambda {)}^{\oplus m}\mathrm{\ne }0$ is homogeneous of weight $\mu$ and ${\tilde{e}}_i(q^{n}u)=q^n{\tilde{e}}_{i}u\in q^n\mathcal{L}(\lambda {)}^{\oplus m}\subseteq q\mathcal{L}(\lambda {)}^{\oplus m}$ for all $i$, so $b$ is killed by every ${\tilde{e}}_i$; writing $b={\sum }_j{b}^{(j)}$ componentwise, each $b^{(j)}$ is a class in $\mathcal{L}(\lambda {)}_{\mu }\mathrm{/}q\mathcal{L}(\lambda {)}_{\mu }$ killed by every ${\tilde{e}}_i$, so $b^{(j)}=0$ for all $j$ by [F5] (note $\mu \mathrm{\ne }\lambda$), and $b=0$, a contradiction. Hence $n=0$ and $u\in \mathcal{L}(\lambda {)}^{\oplus m}$.

Step C: $L\cap N_0=\mathcal{L}(\lambda {)}^{\oplus m}$. Use the identification of step 2.1 and put $L_0:=L\cap N_0$. It is weight-graded and stable under both Kashiwara operators, since these commute with the inclusion of the graded submodule $N_0$. Its top part equals the standard top lattice. Every standard lattice generator is a lowering word applied to a top lift, hence lies in $L_0$; therefore $\mathcal{L}(\lambda {)}^{\oplus m}\subseteq L_0$, and $L_0$ generates $N_0$ over the field by [F4]. For the reverse inclusion, induct on $\mathrm{ht}(\lambda -\mu )$. The top parts agree; at a lower weight every ${\tilde{e}}_{i}u$, $u\in (L_0{)}_{\mu }$, belongs to the standard lattice by the induction hypothesis. The descent criterion of step 2.1 then places $u$ in that lattice. Weights outside $\lambda -Q_{+}$ have zero $N_0$-space. Thus $L_0=\mathcal{L}(\lambda {)}^{\oplus m}$.

Step D: the splitting of $L$. Put $N_1$ as in step 1.1 and let $u\in L_{\mu }$; write $u=u_0+u_1$ with $u_j\in N_j$. Prove $u_j\in L\cap N_j$ by induction on $\mathrm{ht}(\lambda -\mu )$. At $\mu =\lambda$ one has $u_1=0$ because $N_1$ has no weight $\lambda$ by step 1.1, and $u_0=u\in L\cap N_0$. For $\mu \mathrm{\ne }\lambda$, each ${\tilde{e}}_{i}u\in L_{\mu +{\alpha }_i}$ splits by the induction hypothesis into its $N_0$- and $N_1$-parts, and since the operators preserve the $U_q(\mathfrak{g})$-submodules $N_j$, the $N_0$-part of ${\tilde{e}}_{i}u$ is ${\tilde{e}}_i{u}_0$; the induction hypothesis gives ${\tilde{e}}_i{u}_0\in L\cap N_0=\mathcal{L}(\lambda {)}^{\oplus m}$ for all $i$, and step 2.1 gives $u_0\in \mathcal{L}(\lambda {)}^{\oplus m}\subseteq L$; then $u_1=u-u_0\in L\cap N_1$. Weights outside $\lambda -Q_{+}$ have $N_0$-component zero, so the induction starts there. Hence $L=(L\cap N_0)\oplus (L\cap N_1)$, and both summands are free $A_0$-modules generating $N_0,N_1$ respectively (their $\mathbb{Q}(q)$-spans are $N_0,N_1$ because $\mathbb{Q}(q)\otimes L=M=N_0\oplus N_1$ and the splitting is compatible); both are stable and weight-graded, so they are crystal lattices.

Step E: the residue basis splits. Put $L_1:=L\cap N_1$ and $B_0:=B\cap (L_0\mathrm{/}q{L}_0)$. The standard basis $S:={⨆}_j\mathcal{B}(\lambda {)}^{(j)}$ lies in $B_0$, because each standard vertex is a lowering word from a top residue in $B_{\lambda }$. Since $S$ spans $L_0\mathrm{/}q{L}_0$ and $B$ is linearly independent, $B_0=S$. To show that each remaining vertex lies in $L_1\mathrm{/}q{L}_1$, use the splitting of step 4.1 and induct on $\mathrm{ht}(\lambda -\mu )$. Outside $\lambda -Q_{+}$ the $L_0$-component is zero; at $\mu =\lambda$ every vertex is in $B_0$. At a lower weight write $b=c_0+c_1$, with $c_j\in L_j\mathrm{/}q{L}_j$. If $c_0\mathrm{\ne }0$, [F5] supplies an $i$ with ${\tilde{e}}_i{c}_0\mathrm{\ne }0$. Then ${\tilde{e}}_{i}b$ is a nonzero basis vertex; its nonzero $L_0$-component and the induction hypothesis place it in $B_0$. Partial inversion gives $b={\tilde{f}}_i{\tilde{e}}_{i}b\in B_0$. Hence every $b\in B\setminus B_0$ has $c_0=0$. The direct-sum decomposition and the basis property of $B$ now imply that $B_1:=B\setminus B_0$ is a basis of $L_1\mathrm{/}q{L}_1$. Operators preserve both summands and $B\cup \{0\}$, so both pairs are crystal bases.

Step G: locally finite iteration. Fix upper bounds ${\lambda }_a$ with $\mathrm{w}\mathrm{t}(M)\subseteq {\bigcup }_{a=1}^s({\lambda }_a-Q_{+})$ and define $D(\nu ):=\max \{\mathrm{ht}({\lambda }_a-\nu ):{\lambda }_a-\nu \in Q_{+}\}$. Each level $D(\nu )=n$ is finite, since it is contained in the finite union of root-height-$n$ slices of those cones. If $\tau >\nu$, every cone containing $\tau$ contains $\nu$, and $D(\tau )\le D(\nu )-\mathrm{ht}(\tau -\nu )<D(\nu )$. Thus, in increasing order of $D$, the highest weights of the simple constituents in [F1] are maximal among the remaining weights and may be peeled using steps 3.1-5.1. For a fixed weight $\mu$, only finitely many constituent tops $\nu \ge \mu$ occur: in each cone the nonnegative root coefficients of $\nu -\mu$ and ${\lambda }_a-\nu$ sum to those of ${\lambda }_a-\mu$. Each such top has finite multiplicity, bounded by $\dim M_{\nu }$. Consequently finitely many peelings exhaust every fixed weight space. The module decomposition of [F1] is algebraic, so every vector has finite constituent support; the compatible lattice and residue splittings therefore give $\mathrm{\Phi }(L)={⨁}_{j\in J}\mathcal{L}({\lambda }_j)$ and a crystal bijection $B\to {⨆}_j\mathcal{B}({\lambda }_j)$ under an isomorphism $\mathrm{\Phi }:M\to {⨁}_{j}V({\lambda }_j)$. The family is locally weight-finite and has exactly the constituent multiplicities of [F1]. AC permits the simultaneous top lifts used in these identifications. This proves (i) and (ii) without requiring $J$ finite.

Step H: components. Under $\mathrm{\Phi }$ the connected components of $B$ are exactly the $\mathcal{B}({\lambda }_j)$: each $\mathcal{B}({\lambda }_j)$ is connected because by [F4] every vertex is a lowering word from $b_{{\lambda }_j}$, and distinct copies share no vertices, so the partition into copies is the partition into connected components. Each $\mathcal{B}({\lambda }_j)$ has highest weight element $b_{{\lambda }_j}$ of dominant weight ${\lambda }_j$ and equals $F(b_{{\lambda }_j})$ by [F5], and it has no other highest weight element by the uniqueness clause of [F5]; hence the highest weight elements of $B$ are exactly the $b_{{\lambda }_j}$, each generating its own component, and the crystal graph of $B$ is determined up to isomorphism by $M$. This proves (iii).

Choice audit. The axiom of choice of The Axiom of Choice is used exactly as recorded: once, through [F1], to obtain the decomposition of $M$ into simple integrable highest weight modules and to select bases of the finite-dimensional maximal-vector spaces simultaneously, and for the simultaneous selection of lifts of the finite residue bases in the (possibly infinite) family of top spaces in steps 2.1 and 6.1. Every subsequent lattice and crystal argument is a finite induction on a fixed weight interval and involves no further selection; no incompatible-axiom branch and no global $\mathbb{Z}[q,q^{-1}]$-freeness is used.