[proofplan] We prove that every invariant subspace has an invariant complement, and then complete reducibility follows by induction on the dimension of the representation space. Given a short exact sequence of $\mathfrak g$-modules $0 \to W \to V \to Q \to 0$, choose an arbitrary linear section $s: Q \to V$ and measure its failure to be $\mathfrak g$-equivariant by a $1$-cocycle with values in $\operatorname{Hom}(Q,W)$. Whitehead's First Lemma kills this cocycle because $\mathfrak g$ is finite-dimensional semisimple over a field of characteristic $0$ and all modules in sight are finite-dimensional. Correcting the section by the resulting [linear map](/page/Linear%20Map) produces a $\mathfrak g$-equivariant section, whose image is the desired invariant complement. [/proofplan] [step:Reduce complete reducibility to invariant complements] Let $k$ denote the ground field, and let \begin{align*} \rho: \mathfrak g &\to \mathfrak{gl}(V) \end{align*} be a finite-dimensional representation of $\mathfrak g$ on a finite-dimensional $k$-[vector space](/page/Vector%20Space) $V$. We regard $V$ as a $\mathfrak g$-module by writing $xv := \rho(x)(v)$ for $x \in \mathfrak g$ and $v \in V$. It is enough to prove that every $\mathfrak g$-submodule $W \subset V$ has a $\mathfrak g$-submodule complement $U \subset V$, meaning \begin{align*} V = W \oplus U. \end{align*} Indeed, once this is known, induction on $\dim_k V$ gives a decomposition into irreducible submodules: the cases $\dim_k V = 0$ and irreducible $V$ are immediate, while a reducible nonzero $V$ has a proper nonzero submodule $W$, a complement $U$, and both $W$ and $U$ have smaller dimension than $V$. [/step] [step:Encode an arbitrary splitting by a cohomological defect] Let $W \subset V$ be a $\mathfrak g$-submodule. Define the quotient $\mathfrak g$-module \begin{align*} Q := V / W, \end{align*} with quotient map $\pi: V \to Q$ and action $x(v+W) := xv + W$. This action is well-defined because $W$ is $\mathfrak g$-invariant. Choose a $k$-linear section \begin{align*} s: Q &\to V \end{align*} of $\pi$, so $\pi \circ s = \operatorname{id}_Q$. Define the finite-dimensional $\mathfrak g$-module \begin{align*} E := \operatorname{Hom}_k(Q,W), \end{align*} with action \begin{align*} (xA)(q) := x(A(q)) - A(xq) \end{align*} for $x \in \mathfrak g$, $A \in E$, and $q \in Q$. Define \begin{align*} c: \mathfrak g &\to E \\ x &\mapsto c(x), \end{align*} where \begin{align*} c(x)(q) := xs(q) - s(xq). \end{align*} For every $x \in \mathfrak g$ and $q \in Q$, \begin{align*} \pi(c(x)(q)) = xq - xq = 0, \end{align*} so $c(x)(q) \in W$ and $c$ is well-defined as a map into $E$. [guided] The section $s: Q \to V$ is only a linear splitting of the quotient map; it need not respect the $\mathfrak g$-action. The expression \begin{align*} c(x)(q) := xs(q) - s(xq) \end{align*} measures exactly this failure. If $s$ were $\mathfrak g$-equivariant, then $xs(q) = s(xq)$ for all $x \in \mathfrak g$ and $q \in Q$, so $c$ would vanish. We must first check that $c(x)(q)$ really lies in $W$. Since $\pi: V \to Q$ is the quotient map and $\pi \circ s = \operatorname{id}_Q$, we compute \begin{align*} \pi(c(x)(q)) = \pi(xs(q)) - \pi(s(xq)) = x\pi(s(q)) - xq = xq - xq = 0. \end{align*} Thus $c(x)(q) \in \ker \pi = W$. Therefore $c(x)$ is an element of $E = \operatorname{Hom}_k(Q,W)$, and $c: \mathfrak g \to E$ is a well-defined $k$-linear map. [/guided] [/step] [step:Verify that the defect is a $1$-cocycle] We claim that $c$ is a Lie algebra $1$-cocycle, meaning \begin{align*} c([x,y]) = x c(y) - y c(x) \end{align*} in $E$ for all $x,y \in \mathfrak g$. Evaluating both sides at $q \in Q$ gives \begin{align*} (xc(y) - yc(x))(q) &= x\bigl(ys(q) - s(yq)\bigr) - c(y)(xq) - y\bigl(xs(q) - s(xq)\bigr) + c(x)(yq) \\ &= xys(q) - xs(yq) - \bigl(ys(xq) - s(yxq)\bigr) \\ &\quad - yxs(q) + ys(xq) + \bigl(xs(yq) - s(xyq)\bigr) \\ &= [x,y]s(q) - s([x,y]q) \\ &= c([x,y])(q). \end{align*} Thus $c \in Z^1(\mathfrak g,E)$. [guided] The cocycle identity is the algebraic compatibility condition forced by the Lie bracket. We verify it directly from the definitions, using the action of $\mathfrak g$ on $E = \operatorname{Hom}_k(Q,W)$: \begin{align*} (xA)(q) = x(A(q)) - A(xq). \end{align*} For $x,y \in \mathfrak g$ and $q \in Q$, this gives \begin{align*} (xc(y))(q) &= x(c(y)(q)) - c(y)(xq), \\ (yc(x))(q) &= y(c(x)(q)) - c(x)(yq). \end{align*} Substituting $c(y)(q) = ys(q) - s(yq)$ and $c(x)(q) = xs(q) - s(xq)$, we obtain \begin{align*} (xc(y) - yc(x))(q) &= x\bigl(ys(q) - s(yq)\bigr) - \bigl(ys(xq) - s(yxq)\bigr) \\ &\quad - y\bigl(xs(q) - s(xq)\bigr) + \bigl(xs(yq) - s(xyq)\bigr) \\ &= xys(q) - yxs(q) + s(yxq) - s(xyq) \\ &= [x,y]s(q) - s([x,y]q) \\ &= c([x,y])(q). \end{align*} Since this holds for every $q \in Q$, the equality holds in $E$. Hence $c$ is a $1$-cocycle. [/guided] [/step] [step:Apply Whitehead's First Lemma to correct the splitting] The module $E = \operatorname{Hom}_k(Q,W)$ is finite-dimensional because $Q$ and $W$ are finite-dimensional. Since $\mathfrak g$ is finite-dimensional semisimple over a field of characteristic $0$, [Whitehead's First Lemma](/page/Whitehead%20First%20Lemma) applies and gives \begin{align*} H^1(\mathfrak g,E) = 0. \end{align*} Because $c \in Z^1(\mathfrak g,E)$, there exists $A \in E$ such that \begin{align*} c(x) = xA \end{align*} 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*} Here $A(q) \in W \subset V$, so $s_0$ is a well-defined $k$-linear map. Also \begin{align*} \pi(s_0(q)) = \pi(s(q)) - \pi(A(q)) = q, \end{align*} so $s_0$ is still a section of $\pi$. For $x \in \mathfrak g$ and $q \in Q$, \begin{align*} xs_0(q) - s_0(xq) &= xs(q) - xA(q) - s(xq) + A(xq) \\ &= c(x)(q) - \bigl(xA(q) - A(xq)\bigr) \\ &= c(x)(q) - (xA)(q) \\ &= 0. \end{align*} Thus $s_0$ is $\mathfrak g$-equivariant. [guided] Now the hypotheses of Whitehead's First Lemma are exactly the hypotheses of the theorem plus finite-dimensionality of the auxiliary module. The Lie algebra $\mathfrak g$ is finite-dimensional, semisimple, and defined over a field of characteristic $0$ by assumption. The module $E = \operatorname{Hom}_k(Q,W)$ is finite-dimensional because $V$ is finite-dimensional and both $W$ and $Q = V/W$ are finite-dimensional. Therefore [Whitehead's First Lemma](/page/Whitehead%20First%20Lemma) applies to the $\mathfrak g$-module $E$ and yields \begin{align*} H^1(\mathfrak g,E) = 0. \end{align*} The vanishing of $H^1$ means that every $1$-cocycle is a $1$-coboundary. Since we proved $c \in Z^1(\mathfrak g,E)$, there is an element $A \in E$ satisfying \begin{align*} c(x) = xA \end{align*} for all $x \in \mathfrak g$, where the action on $E$ is \begin{align*} (xA)(q) = x(A(q)) - A(xq). \end{align*} We use $A$ to remove the defect of equivariance. Define \begin{align*} s_0: Q &\to V \\ q &\mapsto s(q) - A(q). \end{align*} This is well-defined because $A(q) \in W$ and $W \subset V$. It remains a section because $\pi(A(q)) = 0$: \begin{align*} \pi(s_0(q)) = \pi(s(q)) - \pi(A(q)) = q. \end{align*} Finally, we check equivariance directly: \begin{align*} xs_0(q) - s_0(xq) &= xs(q) - xA(q) - s(xq) + A(xq) \\ &= c(x)(q) - \bigl(xA(q) - A(xq)\bigr) \\ &= c(x)(q) - (xA)(q) \\ &= 0. \end{align*} Thus $xs_0(q) = s_0(xq)$ for all $x \in \mathfrak g$ and $q \in Q$, so $s_0$ is a $\mathfrak g$-module homomorphism. [/guided] [/step] [step:Take the image of the corrected section as the invariant complement] Define \begin{align*} U := s_0(Q) \subset V. \end{align*} Since $s_0$ is $\mathfrak g$-equivariant, $U$ is a $\mathfrak g$-submodule of $V$. Since $\pi \circ s_0 = \operatorname{id}_Q$, the restriction $\pi|_U: U \to Q$ is an isomorphism. Hence $U \cap W = \{0\}$ and every $v \in V$ decomposes as \begin{align*} v = \bigl(v - s_0(\pi(v))\bigr) + s_0(\pi(v)), \end{align*} with $v - s_0(\pi(v)) \in W$ and $s_0(\pi(v)) \in U$. Therefore \begin{align*} V = W \oplus U. \end{align*} By the reduction step, every finite-dimensional representation of $\mathfrak g$ is completely reducible. [/step]