Weyl's complete reducibility theorem

We first work over an algebraically closed field. Every one-dimensional $\mathfrak{g}$-module is trivial: its representation kills $[\mathfrak{g},\mathfrak{g}]$, which equals $\mathfrak{g}$ by [L5].

Suppose $V\mathrm{/}W$ is one-dimensional and $W$ is simple. Replace $\mathfrak{g}$ by its image $\mathfrak{h}$ in $\mathfrak{g}\mathfrak{l}(V)$; the kernel is an ideal and [L8] makes the quotient $\mathfrak{h}$ semisimple. If $\mathfrak{h}=0$ the action is trivial and every line complementary to $W$ is invariant, so assume $\mathfrak{h}\mathrm{\ne }0$. By [L9] its trace form on $V$ is invariant and symmetric, so [L10] makes its radical an ideal; [L6] makes that radical solvable, hence zero. Form its Casimir $C$ as in [L7]. It kills the trivial quotient $V\mathrm{/}W$, while [L2] and [L3] say that it acts on $W$ by a scalar $a$. Moreover ${\mathrm{tr}}_V(C)=\dim \mathfrak{h}$, by summing the dual-basis identities. Characteristic zero makes this nonzero, so $a\mathrm{\ne }0$. Consequently $\ker C$ is an invariant line complementary to $W$. This, together with $W=0$, is the induction base.

Retain $\dim (V\mathrm{/}W)=1$ but allow arbitrary $W$, and assume the assertion for submodules of smaller dimension. If $W$ is not simple, choose a nonzero proper submodule $W^{\mathrm{\prime }}$. By the induction hypothesis, $W\mathrm{/}{W}^{\mathrm{\prime }}$ has a complement $V^{\mathrm{\prime }}\mathrm{/}{W}^{\mathrm{\prime }}$ in $V\mathrm{/}{W}^{\mathrm{\prime }}$. Now $V^{\mathrm{\prime }}\mathrm{/}{W}^{\mathrm{\prime }}$ is one-dimensional, and another induction application splits $V^{\mathrm{\prime }}=W^{\mathrm{\prime }}\oplus L$. The line $L$ is disjoint from $W$ and complements it in $V$.

For general $W\subseteq V$, give ${\mathrm{Hom}}_k(V,W)$ the action $(xf)(v)=xf(v)-f(xv)$. Let $V_1$ consist of maps whose restriction to $W$ is a scalar multiple of the identity, and $W_1$ of maps vanishing on $W$. They are submodules and $V_1\mathrm{/}{W}_1$ is one-dimensional. Step 3.1 gives an invariant line $L=kf$ complementary to $W_1$. By step 1.1, $L$ is trivial, so $f$ intertwines the action. Its restriction to $W$ is a nonzero scalar—otherwise $f\in W_1$—and after rescaling it is the identity. Therefore $\ker f$ is an invariant complement to $W$.

For an arbitrary characteristic-zero ground field, choose bases of $\mathfrak{g}$ and $V$ adapted to $W$, and let $k_0$ be the subfield generated over $\mathbb{Q}$ by their finitely many structure and action coefficients. Nondegeneracy of the Killing matrix shows that the resulting $k_0$-form of $\mathfrak{g}$ is semisimple. Embed the finitely generated field $k_0$ in $\mathbb{C}$. Steps 1.1–4.1 over $\mathbb{C}$ produce an equivariant projection $P:V_{\mathbb{C}}\to W_{\mathbb{C}}$ restricting to the identity. The conditions $P{\mathrm{\mid }}_W=1$ and $P\rho (x)=\rho (x)P$ for the chosen finite bases form a finite linear system over $k_0$. Row reduction shows that consistency over $\mathbb{C}$ already gives a $k_0$-solution; extending it to the original field gives an invariant kernel complementary to $W$. Since $W$ was arbitrary, every submodule has an invariant complement. Induction on $\dim V$ now gives the direct-sum condition in [L1]: for $V\mathrm{\ne }0$, choose a nonzero submodule $S$ of least dimension, which is irreducible; if $S\mathrm{\ne }V$, split $V=S\oplus T$ and decompose the smaller module $T$ by induction. For $V=0$ the empty direct sum is [L1]'s convention. Every descent, embedding, and row reduction uses only finite data, so no choice principle is used.