Proposition 6

Let g be a Lie algebra, r its radical and s its nilpotent radical. (a) $s = [g, r] = \mathscr{D}g \cap r$. (b) s is the intersection of the orthogonals of g with respect to the bilinear forms associated with the finite-dimensional representations of g.

Clearly $[g, r] \subset \mathscr{D}g \cap r$. Now $\mathscr{D}g \cap r = s$ by Theorem 1 of § 5, no. 3. Let $g' = g/[g, r]$ and f be the canonical homomorphism of g onto $g'$; then $f(r)$ is the radical $r'$ of $g'$ (Corollary 3 to Proposition 2, no. 2), hence $[g', r'] = \{0\}$ and $r'$ is the centre of $g'$; therefore (Proposition 5) $g'$ has a finite-dimensional faithful semi-simple representation, whence $s \subset [g, r]$. This proves (a).

Let t be the intersection of the orthogonals of g with respect to the bilinear forms associated with the finite-dimensional representations of g. Then $s \subset t$ (§ 4, no. 3, Proposition 4 (d)). On the other hand, $g/s$ has a finite-dimensional faithful semi-simple representation and hence (Proposition 5) a finite-dimensional representation $\rho$ such that the associated bilinear form is non-degenerate; considered as a representation of g, $\rho$ has an associated bilinear form $\beta$ on g and the orthogonal of g with respect to $\beta$ is s, whence $t \subset s$. Hence $t = s$.

Even if $s \neq \{0\}$ there may exist non-degenerate symmetric bilinear forms on g (Exercise 18 (c)). Such forms, of course, are not associated with any representation of g.