Integrable bounded-above quantum modules with finite-dimensional weight spaces are direct sums of simple highest weight modules

Finite rank-one reduction. For homogeneous $x$, choose $N\ge 1$ with $E_i^{N}x=0$ and put $W_x:=\mathrm{span}\{F_i^{(a)}{E}_i^{b}x:0\le b<N,a\ge 0\}$. Local nilpotence of $F_i$ on each of the finitely many $E_i^{b}x$ makes this span finite. The divided commutation formula of [F1] proves $E_i$ preserves it, while $F_i$ and $K_i^{\pm 1}$ preserve it directly. Thus $W_x$ is a finite-dimensional type-one $A_i$-module, with $E_i,F_i$ nilpotent and $K_i$ diagonalizable; its labels lie on the single line through $\mathrm{w}\mathrm{t}(x)$. A general vector has finite label support, so it too lies in a finite sum of such modules.

Finite rank-one strings, proved locally. In a finite-dimensional type-one $A_i$-module, a nonzero $E_i$-highest eigenvector $y$ of integer weight $a$ has a minimal $t\ge 1$ with $F_i^{(t)}y=0$. Applying $E_i$ and [F1] gives $[a-t+1{]}_i{F}_i^{(t-1)}y=0$, hence $t=a+1$ and $a\ge 0$; the positions $F_i^{(n)}y$, $0\le n\le a$, are nonzero and have distinct $K_i$-eigenvalues. Each simple composition factor has such a vector (take an eigencomponent of a last nonzero $E_i$-iterate), whose string is a nonzero submodule and therefore the entire factor. Put $C:=F_i{E}_i+(q_i{K}_i+q_i^{-1}{K}_i^{-1})\mathrm{/}(q_i-q_i^{-1}{)}^2$. The relations of [F1] give $[C,E_i]=[C,F_i]=[C,K_i]=0$, and $C$ acts on a highest string of weight $a$ by $c_a=(q_i^{a+1}+q_i^{-a-1})\mathrm{/}(q_i-q_i^{-1}{)}^2$. These scalars are distinct for $a\ge 0$: equality factors as $(q_i^{a+1}-q_i^{b+1})(1-q_i^{-a-b-2})=0$, forcing $a=b$. Split a finite module into its generalized $C$-eigenspaces. In the $c_a$-block all composition factors are strings of highest weight $a$, so if their number is $k$, the $K_i$-weight spaces $a,a-2,\dots ,-a$ all have dimension $k$ and there are no other weights. Its top space is killed by $E_i$; the maps $F_i^{(n)}$ on that space are injective for $n\le a$ by the first endpoint argument and vanish for $n>a$. Their images therefore exhaust the block's distinct weight spaces. This proves the finite string decomposition without assuming rank-one complete reducibility. Applied to $W_x$, the tops and positions have the chosen labels: on its single root line, the $K_i$-eigenvalue determines the label since $⟨{\alpha }_i,h_i⟩=2$.

Positive spans and restricted-dual integrability. The labels of $U^{\ge 0}x$ lie above one of the finitely many labels of $x$ and below one of the finitely many upper labels ${\lambda }_j$. Each interval contains finitely many labels because the simple roots freely generate $Q$, and each label space is finite-dimensional, so $U^{\ge 0}x$ is finite-dimensional. Fix $\mu ,i$ and an integer $T\ge 0$ with $M_{\mu +t{\alpha }_i}=0$ for all integers $t>T$; such a bound follows from the ${\alpha }_i$ coefficient in each possible expression ${\lambda }_j-\mu -t{\alpha }_i\in Q_{+}$. Put $a=⟨\mu ,h_i⟩$. By step 2.1, any vector on this line is a sum of positions of strings whose top is $\mu +t{\alpha }_i$ with $t\le T$, highest integer $a+2t\ge 0$ and position $0\le n\le a+2t$. The position label $\mu +(t-n){\alpha }_i$ satisfies $-a-T\le t-n\le T$. Thus the line has finite support. Under [F4], $E_i$ raises and $F_i$ lowers a functional's support by ${\alpha }_i$; both sufficiently high powers vanish for a functional at $\mu$. Finite maxima over its finite label support prove that $M^{\ast }$, and the restricted dual of any graded submodule with the same upper bound, is integrable. No global coroot bound is used.

Maximal vectors. Let $F\subseteq M$ be a nonzero finite-dimensional graded $U^{\ge 0}$-stable subspace and choose a maximal label $\mu$ in its finite support. Pick $0\mathrm{\ne }{v}_{\mu }\in F_{\mu }$. Since $F_{\mu +{\alpha }_i}=0$ for each $i$, every $E_i$ kills $v_{\mu }$. Step 2.1 applied in its finite rank-one module gives $⟨\mu ,h_i⟩\ge 0$ for every $i$, so $\mu \in P^{+}$. The cyclic module $L:=U_q{v}_{\mu }$ is integrable and isomorphic to the ordinary-simple $L(\mu )$ by [F3].

The duality retraction. Put $N:=U_{q}F=U_q^{-}F$. Its support lies in the finite union ${\bigcup }_{\nu \in \mathrm{w}\mathrm{t}(F)}(\nu -Q_{+})$; maximality of $\mu$ implies $N_{\mu +{\alpha }_i}=0$ for each $i$ and $N_{\mu }=F_{\mu }$, without requiring the other maximal labels to lie below $\mu$. Choose $\psi \in (N_{\mu }{)}^{\ast }$ with $\psi (v_{\mu })=1$ and extend it by zero on the other labels. Since $\sigma (E_i)$ lowers a label by ${\alpha }_i$, $E_i\psi =0$. By step 3.1, the whole restricted dual $N^{\ast }$ is integrable, so $W:=U_q\psi$ is the integrable highest module $L(\mu )$ by [F3]. Evaluation gives a grading-preserving module map $\mathrm{\Phi }:N\to W_{\mathrm{r}\mathrm{e}\mathrm{s}}^{\ast }$, with $\mathrm{\Phi }(x)(w)=w(x)$: a finite-label vector $x$ defines a finite-label functional, and invariance follows from [F4]. The homogeneous nondegenerate contravariant form on $W$ identifies $W$ with its restricted dual on each finite-dimensional label space. Choose the unique normalized highest-vector isomorphism $W\cong L$ and rescale the composite so that $\pi (v_{\mu })=v_{\mu }$; this is possible since evaluation on $(v_{\mu },\psi )$ is $1$. The grading-preserving module map $\pi :N\to L$ restricts to the identity on the cyclic simple $L$, hence $N=L\oplus \ker \pi$.

Finite induction. For the graded $F$ of steps 3.2 and 4.1 put $G:=(\mathrm{i}\mathrm{d}-\pi )F$. The grading-preserving retraction makes $G$ graded, finite-dimensional and $U^{\ge 0}$-stable, and $U_{q}G=\ker \pi$. Since $v_{\mu }\in F\cap L$ is killed by $\mathrm{i}\mathrm{d}-\pi$, $\dim G<\dim F$. Induction on $\dim F$ therefore decomposes $U_{q}F$ as a finite direct sum of simple highest modules. This also covers an arbitrary finite-dimensional $U^{\ge 0}$-stable generating space $F_0$: its finite graded hull $F$ is stable by [F5], and $U_q{F}_0$ is an ordinary submodule of the finite direct sum $U_{q}F$. Every submodule of a finite direct sum of ordinary simple modules is itself a finite direct sum of simples: induct on the number of summands, splitting off the first summand if it is contained in the submodule, and otherwise inject the submodule into the remaining summands by projection. Each resulting simple is isomorphic to a summand by a nonzero coordinate projection. This argument uses only finite choices.

The general module. For $x\in M$, the finite graded hull of $U^{\ge 0}x$ is finite and stable by step 3.1 and [F5], so step 5.1 places $x$ in a finite sum of ordinary-simple highest modules. Hence $M$ is a sum of simple submodules. By AC choose a maximal independent family in the set of simple submodules; an ordinary simple not contained in its sum has zero intersection with that sum and would enlarge the family, so maximality gives $M={⨁}_{\alpha }{S}_{\alpha }$. Every $S_{\alpha }$ is isomorphic to some $L(\lambda )$ by a nonzero coordinate projection to one of the finite highest-module sums containing a nonzero vector of $S_{\alpha }$. For a nonzero ordinary submodule $V\subseteq M$, choose $0\mathrm{\ne }x\in V$ and a finite simple sum containing $x$. Its intersection with $V$ is a nonzero submodule of that finite sum, so step 5.1 gives a simple highest-module copy inside $V$; the image of its highest vector is a maximal vector whose cyclic submodule is that copy. This proves the concrete clause as an ordinary-module assertion, without assuming an ordinary submodule respects the chosen label grading. The original finite label grading and [F6] retain the datum convention throughout. AC occurs only at the maximal-family step; all finite generated cases were choice-free.