Complete reducibility of integrable kac moody o modules

Work first in one Casimir summand $C$ of F1. If $v\in C_{\lambda }$ is nonzero and singular, meaning killed by ${\mathfrak{n}}^{+}$, its cyclic module $M$ has support in $\lambda -Q^{+}$ and top $\mathbb{C}v$ by F6. A proper submodule misses that top, since containing $v$ would generate all of $M$. If a proper nonzero submodule existed, F5 gives it a maximal support weight $\mu$: take a maximal element in the finite upper set above any of its weights. Every vector there is singular. Thus $\mu <\lambda$ and both are primitive weights of $C$, contradicting F1. Hence $M$ is simple, and F2 and F3 identify it with $L_A(\lambda )$ for dominant integral $\lambda$. In this simple module the only singular vectors lie in its top line: any other nonzero singular vector would generate it by simplicity, and F6 would force its original top to be at or below a strictly lower weight, impossible.

For each eigenvalue $a$ and weight $\lambda$, let $C=V^{(a)}$ and $C_{\lambda }^0=\{v\in C_{\lambda }:{\mathfrak{n}}^{+}v=0\}$. It is finite dimensional by F5. Its set of finite ordered bases is nonempty, with the empty basis for zero dimension. The pairs $(a,\lambda )$ form a set. Apply F4 to choose a basis $B_{a,\lambda }$ for every such space. This is the specified use of AC. For fixed $a$, each basis vector $b$ generates a simple submodule $M_b$ by 1.1. There is a natural module map from their algebraic external direct sum to $C$, sending each copy of $M_b$ to its given submodule.

This map is injective. Otherwise a nonzero kernel vector has finite summand support, so its kernel inside the corresponding finite external direct sum $D$ is nonzero. The module $D$ belongs to $\mathcal{O}$: its finite-dimensional weight spaces and finite-cone bounds are finite sums of those of the summands. By F5 its kernel has a maximal support weight and a nonzero singular vector $z$ there. Projecting $z$ to each simple summand gives singular vectors; by 1.1 each component is a scalar multiple of that summand's chosen top vector, and only summands with the weight of $z$ can contribute. Its image in $C$ is thus a linear relation among distinct members of $B_{a,\lambda }$. Their independence makes every scalar zero, contradicting $z\mathrm{\ne }0$. Hence the sum $S={\sum }_b{M}_b$ is direct.

Suppose $C\mathrm{/}S\mathrm{\ne }0$. By F5 it has a maximal support weight $\mu$ and a nonzero highest vector there. The quotient weight-space description supplies a lift $v\in C_{\mu }$, which is primitive in $C$ and does not belong to $S$. For every $i$, $e_{i}v\in S$. If one is nonzero, its finite expression in the direct sum from 3.1 has a nonzero component of weight $\mu +{\alpha }_i$ in some $M_b$ of highest weight $\lambda$. F6 gives $\lambda \ge \mu +{\alpha }_i>\mu$. The weights $\lambda$ and $\mu$ are both primitive in $C$, contrary to F1. Thus every $e_{i}v=0$, hence their generated algebra ${\mathfrak{n}}^{+}$ kills $v$. It lies in $C_{\mu }^0$, whose chosen basis is contained in $S$, so $v\in S$, again a contradiction. Therefore $C=S$.

Reassemble the Casimir direct sum from F1. Steps 1.1–4.1 give a direct sum of dominant simple highest-weight modules in each component, so their combined algebraic sum is the stated decomposition of $V$. Each vector has finitely many Casimir components and finitely many simple components within each. Zero singular spaces contribute empty bases and no summands; the zero module gives the empty sum. One-dimensional singular spaces require just one generator, and zero dominant labels cause no exception. No unproved splitting theorem is used: injectivity and spanning were proved separately. All choices beyond the single family of bases in 2.1 concern finitely many elements or individual witnesses.