Theorem 4

Let $g$ be a Lie algebra, $r$ its radical, $\rho$ a finite-dimensional representation of $g$, $g' = \rho(g)$ and $r' = \rho(r)$. Then the following conditions are equivalent: (a) $\rho$ is semi-simple; (b) $g'$ is reductive and its centre consists of semi-simple endomorphisms; (c) $r'$ consists of semi-simple endomorphisms; (d) the restriction of $\rho$ to $r$ is semi-simple.

(a) $\Rightarrow$ (b): if $\rho$ is semi-simple, $g'$ is reductive (Proposition 5); the associative algebra generated by 1 and $g'$ is semi-simple (\emph{Algebra}, Chapter VIII, § 5, no. 1, Proposition 3), hence its centre is semi-simple (\emph{loc. cit.}, § 5, no. 4, Proposition 12) and hence the elements of this centre are semi-simple (\emph{loc. cit.}, § 9, no. 1, Proposition 2).

(b) $\Rightarrow$ (c): if $g'$ is reductive, its centre is equal to its radical, that is $r'$, whence the implication (b) $\Rightarrow$ (c).

(c) $\Rightarrow$ (d): suppose that $r'$ consists of semi-simple endomorphisms. As $[g', r']$ consists of nilpotent endomorphisms (no. 4, Proposition 6), $[g', r'] = \{0\}$. Then the implication (c) $\Rightarrow$ (d) follows from \emph{Algebra}, Chapter VIII, § 9, no. 2, Theorem 1.

(d) $\Rightarrow$ (a): let $s$ be the nilpotent radical of $g$ and $\rho'$ the restriction of $\rho$ to $r$. The elements of $\rho(s)$ are nilpotent and hence $s$ is contained in the largest nilpotency ideal of $\rho'$. As $\rho'$ is semi-simple, $\rho'(s) = \{0\}$ and $g'$ is reductive (Corollary to Proposition 6), so that $g' = a' \times r'$ with $a'$ semi-simple (Proposition 5). Let $A'$ (resp. $R'$) be the associative algebra generated by 1 and $a'$ (resp. $r'$). It is semi-simple (\emph{Algebra}, Chapter VIII, § 5, no. 1, Proposition 3), hence $A' \otimes_K R'$ is semi-simple (\emph{loc. cit.}, § 7, no. 6, Corollary 4 to Theorem 3) and hence the associative algebra generated by 1 and $g'$, which is a quotient of $A' \otimes_K R'$, is semi-simple, which proves that $\rho$ is semi-simple.