Theorem 2 (Cartan's criterion)

Let g be a Lie algebra, M a finite-dimensional vector space, ρ a representation of g on M and β the bilinear form on g associated with ρ. Then ρ(g) is solvable if and only if $\mathscr{D}$g is orthogonal to g with respect to β.

It can obviously be reduced to the case where g is a Lie subalgebra of $\mathfrak{gl}(M)$ and ρ is the identity mapping. If g is solvable, $\mathscr{D}$g is contained in the largest nilpotency ideal of the identity representation of g (Theorem 1) and hence is orthogonal to g with respect to β (§ 4, Proposition 4 (d)). Suppose that $\mathscr{D}$g is orthogonal to g with respect to β. We prove that g is solvable. Let T be the set of $t \in \mathfrak{gl}(M)$ such that $[t, g] \subset \mathscr{D}g$. If $t \in T$ and x, y belong to g, then $[t, x] \in \mathscr{D}g$ and hence

whence by linearity $\mathrm{Tr}(tu) = 0$ for all $u \in \mathscr{D}g$. Also, clearly $\mathscr{D}g \subset T$. Hence (Lemma 3) every element of $\mathscr{D}$g is nilpotent. It follows that $\mathscr{D}$g is nilpotent (§ 4, Corollary 3 to Theorem 1) and hence that g is solvable (no. 3, Corollary 5 to Theorem 1).