Proposition 5

Let $g$ be a Lie algebra and $r$ its radical. The following conditions are equivalent: (a) $g$ is reductive. (b) $\mathscr{D}g$ is semi-simple. (c) $g$ is the product of a semi-simple algebra and a commutative algebra. (d) $g$ has a finite-dimensional representation such that the associated bilinear form is non-degenerate. (e) $g$ has a faithful semi-simple finite-dimensional representation. (f) The nilpotent radical of $g$ is zero. (g) $r$ is the centre of $g$.

(a) $\Rightarrow$ (b): if the adjoint representation of $g$ is semi-simple, $g$ is a direct sum of minimal non-zero ideals $a_i$ and hence $g$ is isomorphic to the product of the $a_i$; and $a_i$ has no ideals other than $\{0\}$ and $a_i$ and hence is simple or commutative of dimension 1. Therefore $\mathscr{D}g$ is equal to the product of those $a_i$ which are simple and hence is semi-simple.

(b) $\Rightarrow$ (c): if $\mathscr{D}g$ is semi-simple, $g$ is isomorphic to the product of $\mathscr{D}g$ by a Lie algebra $h$ (no. 1, Corollary 1 to Proposition 1); $h$ is isomorphic to $g/\mathscr{D}g$ and hence is commutative.

(c) $\Rightarrow$ (d): let $g_1$ and $g_2$ be two Lie algebras, $\rho_i$ a finite-dimensional representation of $g_i$ and $\beta_i$ the bilinear form on $g_i$ associated with $\rho_i$ ($i = 1, 2$); $\rho_1$ and $\rho_2$ can be considered as representations of $g = g_1 \times g_2$; let $\rho$ be their direct sum. Clearly the bilinear form on $g$ associated with $\rho$ is the direct sum of $\beta_1$ and $\beta_2$ and hence is non-degenerate if $\beta_1$ and $\beta_2$ are non-degenerate. Then to prove the implication (c) $\Rightarrow$ (d) it suffices to consider the 2 following cases: (1) $g$ is semi-simple; then the adjoint representation admits as associated form the Killing form, which is non-degenerate; (2) $g = K$; then the identity representation of $g$ on $K$ has an associated bilinear form which is non-degenerate.

(d) $\Rightarrow$ (e): let $\rho$ be a finite-dimensional representation of $g$ and $\beta$ the associated bilinear form; by Proposition 4 of § 4, no. 3, there exists a finite-dimensional semi-simple representation $\sigma$ of $g$ such that the kernel $n$ of $\sigma$ is orthogonal to $g$ with respect to $\beta$. If $\beta$ is non-degenerate, then $n = \{0\}$ and hence $\sigma$ is faithful.

(e) $\Rightarrow$ (f): this is obvious.

(f) $\Rightarrow$ (g): if the nilpotent radical of $g$ is zero, $\mathscr{D}g \cap r$ is zero ($§ 5$, no. 3, Theorem 1); as $[g, r] \subset \mathscr{D}g \cap r$, $r$ is the centre of $g$.

(g) ⇒ (a): if r is the centre of g, the adjoint representation of g is identified with a representation of g/r, which is a semi-simple Lie algebra (§ 5, no. 2, Proposition 3); this representation is therefore semi-simple (Theorem 2).