Semisimple groups in characteristic zero are linearly reductive

It suffices to prove the assertion after extending scalars to an algebraic closure $k^{\mathrm{a}}$ of $k$: if every finite-dimensional representation of $G_{k^{\mathrm{a}}}$ is semisimple, then [F1] gives semisimplicity of every finite-dimensional representation of $G$. We therefore assume in the rest of the proof that $k$ is algebraically closed.

For the algebraically closed field $k$ one has $X(G)=0$ by [F3], so by [F2] it suffices to verify condition (c): every simple subrepresentation $W$ of codimension one in a finite-dimensional $V$ is a direct summand. If $V$ is trivial, any finite-dimensional linear complement to $W$ is a $G$-subrepresentation, so the condition holds in every dimension, including zero. Otherwise $\bar{\mathfrak{g}}=\rho (\mathfrak{g})\mathrm{\ne }0$ by [F5], and we may use its Casimir operator.

Let $W$ be a simple subrepresentation of codimension one in a nonzero finite-dimensional $V$ with $\bar{\mathfrak{g}}\mathrm{\ne }0$, and let $c_V$ be the Casimir endomorphism of [F4]. The quotient $V\mathrm{/}W$ is a one-dimensional rational representation, hence trivial by [F3]; therefore $\mathfrak{g}V\subseteq W$. Since $c_V$ is a sum of products $\rho (x)\rho (y)$ with $x,y\in \mathfrak{g}$, it follows that $c_V(V)\subseteq \mathfrak{g}V\subseteq W$.

The restriction $c_V{\mathrm{\mid }}_W$ is a $G$-module endomorphism by [F4]. Since $k$ is algebraically closed and $W$ is nonzero finite-dimensional, it has an eigenvalue $a\in k$. The kernel of $c_V{\mathrm{\mid }}_W-a{\mathrm{id}}_W$ is nonzero and is a $G$-subrepresentation, because that difference is $G$-equivariant. Simplicity of $W$ makes this kernel all of $W$. Hence $c_V{\mathrm{\mid }}_W=a{\mathrm{id}}_W$, by the eigenvalue-kernel argument of Schur's lemma applied directly to the rational $G$-module.

The scalar $a$ is nonzero: $c_V$ maps $V$ into $W$, so $\mathrm{tr}(c_V{\mathrm{\mid }}_V)=\mathrm{tr}(c_V{\mathrm{\mid }}_W)=a\dim W$, while [F4] gives $\mathrm{tr}(c_V{\mathrm{\mid }}_V)={\dim }_k\bar{\mathfrak{g}}\mathrm{\ne }0$; hence $a\mathrm{\ne }0$. Therefore $c_V{\mathrm{\mid }}_W$ is invertible, $\ker c_V$ intersects $W$ trivially, $c_V(V)=W$, and $\dim \ker c_V=\dim V-\dim W=1$. Since $c_V$ is a $G$-module endomorphism by [F4], its kernel is a $G$-submodule, and $V=W\oplus \ker c_V$ exhibits $W$ as a direct summand.

Condition (c) of [F2] holds for every finite-dimensional $V$ by step 5.1, so by [F2] every finite-dimensional rational representation is semisimple; by step 1.1 this descends to the original field.

For an arbitrary rational representation $M$, order by inclusion the sets of simple submodules whose sum is direct. The empty set is such a family, and the union of a chain is such a family because each finite relation occurs in one member of the chain. By [A1] choose a maximal family with sum $S$. If $S\mathrm{\ne }M$, [F7] gives a finite-dimensional submodule $W$ containing a vector outside $S$. By step 6.1, $W$ is a finite direct sum of simple submodules. Since $W$ is not contained in $S$, one such summand $C$ is not contained in $S$; simplicity gives $C\cap S=0$, so adjoining $C$ extends the family, a contradiction. Therefore $M=S$ is a direct sum of simple submodules, establishing linear reductivity.