[proofplan] We prove the equivalent splitting statement: every short exact sequence of finite-dimensional $\mathfrak g$-modules splits. Given a vector-space section of the quotient map, its failure to commute with the $\mathfrak g$-action is a $1$-cocycle with values in a Hom-module. Whitehead's First Lemma, proved by the Casimir operator for semisimple Lie algebras in characteristic zero, makes this cocycle a coboundary, so the section can be corrected to a $\mathfrak g$-module section. Once every submodule has an invariant complement, induction on dimension gives a decomposition into irreducible submodules. [/proofplan] [step:Correct an arbitrary linear section to a $\mathfrak g$-module section] Let \begin{align*} 0 \longrightarrow W \xrightarrow{\iota} V \xrightarrow{\pi} Q \longrightarrow 0 \end{align*} be a short exact sequence of finite-dimensional $\mathfrak g$-modules over $F$. We prove that this sequence splits as a sequence of $\mathfrak g$-modules. Choose an $F$-linear section \begin{align*} s: Q &\to V \end{align*} of $\pi$, so that $\pi \circ s = \operatorname{id}_Q$. Define the defect map \begin{align*} c: \mathfrak g &\to \operatorname{Hom}_F(Q,W) \end{align*} by \begin{align*} c(x)(q) := x \cdot s(q) - s(x \cdot q) \end{align*} for $x \in \mathfrak g$ and $q \in Q$. This expression lies in $W$ because \begin{align*} \pi(x \cdot s(q) - s(x \cdot q)) = x \cdot \pi(s(q)) - \pi(s(x \cdot q)) = x \cdot q - x \cdot q = 0. \end{align*} Give $\operatorname{Hom}_F(Q,W)$ its standard $\mathfrak g$-module structure: \begin{align*} (x \cdot T)(q) := x \cdot T(q) - T(x \cdot q) \end{align*} for $x \in \mathfrak g$, $T \in \operatorname{Hom}_F(Q,W)$, and $q \in Q$. We compute, for $x,y \in \mathfrak g$ and $q \in Q$, \begin{align*} (x \cdot c(y))(q) - (y \cdot c(x))(q) &= x \cdot \bigl(y \cdot s(q) - s(y \cdot q)\bigr) - c(y)(x \cdot q) \\ &\quad - y \cdot \bigl(x \cdot s(q) - s(x \cdot q)\bigr) + c(x)(y \cdot q) \\ &= x \cdot y \cdot s(q) - y \cdot x \cdot s(q) - s(x \cdot y \cdot q) + s(y \cdot x \cdot q) \\ &= [x,y] \cdot s(q) - s([x,y] \cdot q) \\ &= c([x,y])(q). \end{align*} Thus $c$ is a $1$-cocycle of $\mathfrak g$ with values in the finite-dimensional $\mathfrak g$-module $\operatorname{Hom}_F(Q,W)$. By Whitehead's First Lemma (citing a result not yet in the wiki: Whitehead's First Lemma), since $\mathfrak g$ is finite-dimensional and semisimple over a field of characteristic $0$, every $1$-cocycle with values in a finite-dimensional $\mathfrak g$-module is a coboundary. Hence there exists an $F$-[linear map](/page/Linear%20Map) \begin{align*} a: Q &\to W \end{align*} such that \begin{align*} c(x) = x \cdot a \end{align*} for every $x \in \mathfrak g$. Define \begin{align*} s_0: Q &\to V, \\ q &\mapsto s(q) - a(q). \end{align*} Since $a(q) \in W = \ker \pi$, we have \begin{align*} \pi(s_0(q)) = \pi(s(q)) - \pi(a(q)) = q. \end{align*} Therefore $\pi \circ s_0 = \operatorname{id}_Q$. Finally, for $x \in \mathfrak g$ and $q \in Q$, \begin{align*} x \cdot s_0(q) - s_0(x \cdot q) &= x \cdot s(q) - x \cdot a(q) - s(x \cdot q) + a(x \cdot q) \\ &= c(x)(q) - \bigl(x \cdot a(q) - a(x \cdot q)\bigr) \\ &= c(x)(q) - (x \cdot a)(q) \\ &= 0. \end{align*} Thus $s_0$ is a $\mathfrak g$-module homomorphism. Its image $s_0(Q)$ is a $\mathfrak g$-submodule of $V$, and exactness gives \begin{align*} V = W \oplus s_0(Q) \end{align*} as $\mathfrak g$-modules. [guided] The goal is to replace an arbitrary vector-space complement by an invariant complement. Start with a short exact sequence \begin{align*} 0 \longrightarrow W \xrightarrow{\iota} V \xrightarrow{\pi} Q \longrightarrow 0 \end{align*} of finite-dimensional $\mathfrak g$-modules. Since this is exact as a sequence of vector spaces, choose an $F$-linear section \begin{align*} s: Q \to V \end{align*} with $\pi \circ s = \operatorname{id}_Q$. The section $s$ need not respect the $\mathfrak g$-action, so we measure its failure by defining \begin{align*} c: \mathfrak g \to \operatorname{Hom}_F(Q,W), \qquad c(x)(q) := x \cdot s(q) - s(x \cdot q). \end{align*} This really has values in $W$: applying $\pi$ gives \begin{align*} \pi(x \cdot s(q) - s(x \cdot q)) = x \cdot \pi(s(q)) - \pi(s(x \cdot q)) = x \cdot q - x \cdot q = 0, \end{align*} so the defect lies in $\ker \pi = W$. We now view $\operatorname{Hom}_F(Q,W)$ as a $\mathfrak g$-module by \begin{align*} (x \cdot T)(q) := x \cdot T(q) - T(x \cdot q). \end{align*} With this action, the defect $c$ satisfies the cocycle identity. Indeed, for $x,y \in \mathfrak g$ and $q \in Q$, \begin{align*} (x \cdot c(y))(q) - (y \cdot c(x))(q) &= x \cdot \bigl(y \cdot s(q) - s(y \cdot q)\bigr) - c(y)(x \cdot q) \\ &\quad - y \cdot \bigl(x \cdot s(q) - s(x \cdot q)\bigr) + c(x)(y \cdot q) \\ &= x \cdot y \cdot s(q) - y \cdot x \cdot s(q) - s(x \cdot y \cdot q) + s(y \cdot x \cdot q) \\ &= [x,y] \cdot s(q) - s([x,y] \cdot q) \\ &= c([x,y])(q). \end{align*} This is exactly the condition that $c$ is a $1$-cocycle. Whitehead's First Lemma applies because $\mathfrak g$ is finite-dimensional and semisimple, the field has characteristic $0$, and $\operatorname{Hom}_F(Q,W)$ is finite-dimensional. It says that this cocycle is a coboundary, so there is an $F$-linear map \begin{align*} a: Q \to W \end{align*} such that $c(x) = x \cdot a$ for every $x \in \mathfrak g$. Define the corrected section \begin{align*} s_0: Q &\to V, \\ q &\mapsto s(q) - a(q). \end{align*} Since $a(q)$ lies in $W=\ker \pi$, the corrected map is still a section: \begin{align*} \pi(s_0(q)) = \pi(s(q)) - \pi(a(q)) = q. \end{align*} The correction was chosen precisely to cancel the defect: \begin{align*} x \cdot s_0(q) - s_0(x \cdot q) &= x \cdot s(q) - x \cdot a(q) - s(x \cdot q) + a(x \cdot q) \\ &= c(x)(q) - \bigl(x \cdot a(q) - a(x \cdot q)\bigr) \\ &= c(x)(q) - (x \cdot a)(q) \\ &= 0. \end{align*} Thus $s_0$ is a $\mathfrak g$-module homomorphism. Its image is a $\mathfrak g$-[stable complement](/theorems/2279) to $W$, and hence \begin{align*} V = W \oplus s_0(Q) \end{align*} as $\mathfrak g$-modules. [/guided] [/step] [step:Split every submodule of a finite-dimensional module] Let $V$ be a finite-dimensional $\mathfrak g$-module, and let $W \subset V$ be a $\mathfrak g$-submodule. The quotient [vector space](/page/Vector%20Space) $V/W$ carries the quotient $\mathfrak g$-module structure \begin{align*} x \cdot (v+W) := x \cdot v + W \end{align*} for $x \in \mathfrak g$ and $v \in V$. The sequence \begin{align*} 0 \longrightarrow W \longrightarrow V \longrightarrow V/W \longrightarrow 0 \end{align*} is therefore a short exact sequence of finite-dimensional $\mathfrak g$-modules. By the previous step, it splits, so there exists a $\mathfrak g$-submodule $U \subset V$ such that \begin{align*} V = W \oplus U. \end{align*} [/step] [step:Induct on dimension to decompose into irreducible submodules] We prove complete reducibility by induction on $\dim_F V$. If $\dim_F V = 0$, then $V=0$ is the empty direct sum of irreducible submodules. Assume $\dim_F V > 0$ and that every finite-dimensional $\mathfrak g$-module of smaller dimension is completely reducible. If $V$ is irreducible, then $V$ is already a direct sum of one irreducible submodule. Otherwise, choose a nonzero proper $\mathfrak g$-submodule $W \subsetneq V$. By the previous step, there is a $\mathfrak g$-submodule $U \subset V$ such that \begin{align*} V = W \oplus U. \end{align*} Both $W$ and $U$ have strictly smaller dimension than $V$. By the induction hypothesis, there exist irreducible $\mathfrak g$-submodules $W_1,\dots,W_r \subset W$ and $U_1,\dots,U_m \subset U$ such that \begin{align*} W = W_1 \oplus \cdots \oplus W_r, \qquad U = U_1 \oplus \cdots \oplus U_m. \end{align*} Therefore \begin{align*} V = W_1 \oplus \cdots \oplus W_r \oplus U_1 \oplus \cdots \oplus U_m, \end{align*} a direct sum of irreducible $\mathfrak g$-submodules. This completes the induction and proves that every finite-dimensional $\mathfrak g$-module is completely reducible. [/step]