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Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.

An ideal of a Lie algebra is a vector subspace $I\subseteq \mathfrak{g}$ such that $[\mathfrak{g},I]\subseteq I$. The derived series of a Lie algebra is defined by ${\mathfrak{g}}^{(0)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(r+1)}=[{\mathfrak{g}}^{(r)},{\mathfrak{g}}^{(r)}]$.

Suppose that $I$ is an ideal and $x\in \mathfrak{g}$, $u,v\in I$. The Jacobi identity givesBoth terms on the right lie in $[I,I]$, because $[x,u],[x,v]\in I$. Thus $[I,I]$ is again an ideal. Starting from the ideal $\mathfrak{g}$ and applying this observation inductively proves that every term ${\mathfrak{g}}^{(r)}$ of the derived series is an ideal of $\mathfrak{g}$.

The space $\mathrm{End}(V)$ becomes a Lie algebra under the commutatorA Lie subalgebra $L$ is abelian when $[L,L]=0$; nilpotent when its lower central series of a Lie algebra ${\gamma }_{r+1}=[L,{\gamma }_r]$ reaches zero; and soluble when its derived series of a Lie algebra $L^{(r+1)}=[L^{(r)},L^{(r)}]$ reaches zero.

If $H⊊L$ and $L$ is nilpotent, choose the least $r$ with ${\gamma }_r(L)\subseteq H$. Then ${\gamma }_{r-1}(L)⊈H$, whileThus the normalizer of a Lie subalgebra strictly contains $H$. If $H$ is maximal proper, its normalizer must be all of $L$, so $H$ is an ideal. Solubility is insufficient: in the two-dimensional affine Lie algebra $L=⟨x,y⟩$ with $[x,y]=y$, the maximal subalgebra $\mathbb{C}x$ is not an ideal.

A derivation of a Lie algebra is a linear map $D$ satisfyingIt is inner when $D=\mathrm{ad}z$ for some $z\in L$.

Every nonzero finite-dimensional nilpotent Lie algebra has an outer derivation of a nilpotent Lie algebra. Choose a codimension-one maximal subalgebra $K$; it is an ideal by the result above, and write $L=K\oplus \mathbb{C}z$. The centralizer $C_L(K)$ is nonzero because it contains $Z(L)$. Let $n$ be largest such thatand choose $z_0\in C_L(K)\setminus {\gamma }_{n+1}(L)$. DefineBecause $K$ is an ideal and $z_0$ centralizes $K$, the derivation identity holds on $K\times K$ and on $z\times K$, hence everywhere. If $D=\mathrm{ad}x$, then $D(K)=0$ would put $x$ in $C_L(K)\subseteq {\gamma }_n(L)$, socontrary to the choice of $z_0$. Thus $D$ is outer.

The analogous assertion fails for soluble algebras. In the affine example, a derivation hasand equals $\mathrm{ad}(dx-by)$. Thus every derivation is inner although $L$ is nonzero and soluble.

The one-dimensional quotient $V/W$ is trivial because every one-dimensional representation vanishes on the derived algebra $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$. Choose $v\in V$ mapping to $1\in V/W$. Thenis a $1$-cocycle:We show that it is a coboundary.

Decompose $W$ into generalized eigenspaces of its Casimir element $\Omega$. These are subrepresentations because $\Omega$ is central. On a generalized eigenspace with nonzero eigenvalue, $\Omega$ is invertible. If $(x_i)$ and $(x^i)$ are dual bases of ${\mathfrak{sl}}_2$ for the Killing form, putInvariance of the Killing form and the cocycle identity give the standard Casimir calculationThus on every nonzero generalized eigenspace, $c(x)=x({\Omega }^{-1}u)$.

On the zero generalized eigenspace, every irreducible composition factor has zero Casimir eigenvalue. By part i and the classification of finite-dimensional sl2 representations, each such factor is trivial. In a basis adapted to a composition series, the image of ${\mathfrak{sl}}_2$ is therefore strictly upper triangular and hence solvable. Since ${\mathfrak{sl}}_2$ is simple and non-solvable, that image is zero. The cocycle then vanishes because it kills $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]$.

Combining the generalized eigenspaces gives $w\in W$ such that $c(x)=xw$ for every $x$. Hence $v-w$ is invariant, andis a decomposition into subrepresentations. This proves the codimension-one case of the Weyl complete reducibility theorem.