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Invariance of the Killing form givesIf $x\in {\mathfrak{g}}^{\perp }$, the right-hand side vanishes for every $y$, so $[a,x]\in {\mathfrak{g}}^{\perp }$. Thus ${\mathfrak{g}}^{\perp }$ is an ideal.

Let $\mathfrak{h}=\mathrm{ad}({\mathfrak{g}}^{\perp })\subseteq \mathfrak{gl}(\mathfrak{g})$. For $x\in [{\mathfrak{g}}^{\perp },{\mathfrak{g}}^{\perp }]$ and $y\in {\mathfrak{g}}^{\perp }$,The Cartan solvability criterion makes $\mathfrak{h}$ solvable. The kernel of the adjoint map on ${\mathfrak{g}}^{\perp }$ lies in its center and is abelian, so ${\mathfrak{g}}^{\perp }$ is itself solvable. This proves the Solvability of the radical of the Killing form.

The Killing form is $\kappa (X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. From part (a),where the middle equality uses cyclicity of the trace.

Each structure constant of a Lie algebra satisfies $f_{ab}{}^d=-f_{ba}{}^d$, so $f_{abc}=f_{ab}{}^d{\kappa }_{dc}$ is antisymmetric in $a,b$. Invariance of the Killing form givesThis cyclic symmetry together with antisymmetry in the first pair implies antisymmetry under every transposition. Thus $f_{abc}$ is totally antisymmetric.

For the trace trilinear form of a Lie algebra representation, the trace of a commutator vanishes:Taking $X=T_b$, $Y=T_a$, $Z=T_c$, $W=T_d$ and expanding each Lie bracket givesSince $f_a{}^{bc}$ is antisymmetric in $b,c$,Raising indices with the inverse Killing form gives $f_a{}^{bc}{f}_{bc}{}^e={\delta }_a{}^e$ in the stated normalization. Using $H(T_e,T_d)=-\mu {\delta }_{ed}$ proves

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

A positive invariant metric is a positive-definite real symmetric invariant bilinear form on a Lie algebra. Each adjoint map is skew-adjoint. Invariance gives $[\mathfrak{g},\mathfrak{g}{]}^{\perp }=\mathfrak{z}(\mathfrak{g})$, so the algebra splits orthogonally into its Abelian center and its derived algebra. The latter has zero center, and $\kappa (X,X)=\mathrm{tr}({\mathrm{ad}}_X^2)=-\Vert {\mathrm{ad}}_X{\Vert }^2$ is strictly negative for nonzero $X$. Thus it is compact semisimple; the whole algebra is compact reductive. Conversely, a compact semisimple summand has metric $-\kappa$ and the center can carry any positive metric. The Killing form alone vanishes on the center.

For a nontrivial finite-dimensional anti-Hermitian Lie algebra representation of a compact Lie algebra that is simple, its Trace form of a Lie algebra representation is a positive multiple of the Killing form. Define $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance makes $T$ commute with the Adjoint representation of a Lie algebra. Symmetry makes $T$ self-adjoint for $-\kappa$, so its eigenspaces are ideals. Simplicity implies $T=\mu I$. The kernel of the representation is an ideal and hence vanishes; therefore $H(X,X)=-\mathrm{Tr}(d(X{)}^{†}d(X))<0$ for nonzero $X$, proving $\mu >0$.

// Target: lie-theory.bigb

The radical of a bilinear form of the Killing form is the ideal of a Lie algebra $\{x:B(x,y)=0\text{ for all }y\}$. It is zero for a complex semisimple Lie algebra. Every element of the center of a Lie algebra lies in this radical.

The usual uniqueness theorem assumes a complex simple Lie algebra, or a real form with simple complexification. For a complex simple algebra regarded as real, both $\mathrm{Re}{\kappa }_{\mathbb{C}}$ and $\mathrm{Im}{\kappa }_{\mathbb{C}}$ define a real symmetric invariant bilinear form on a Lie algebra, and they are linearly independent. The real Killing form is $2\mathrm{Re}{\kappa }_{\mathbb{C}}$. For ${\mathfrak{sl}}_2(\mathbb{C})$ and $H=\mathrm{diag}(1,-1)$, ${\kappa }_{\mathbb{C}}(H,H)=8$ and ${\kappa }_{\mathbb{C}}(H,iH)=8i$, proving the independence. This real algebra is simple: its real complexification is two complex simple ideals exchanged by conjugation, and a real ideal would complexify to a conjugation-stable sum of these ideals, hence either zero or the whole algebra.

For a Cartan subalgebra $H$ of a finite-dimensional complex semisimple Lie algebra $L$, $H$ is abelian and $C_L(H)=H$, so it is maximal among abelian subalgebras. Starting from the nilpotent self-normalizing definition, the generalized-weight decomposition for a nilpotent Lie algebra and the Engel theorem show that the zero generalized weight space of its adjoint action is exactly $H$. Invariance of the Killing form makes $H$ orthogonal to every nonzero generalized weight space. The Lie theorem gives $B([H,H],H)=0$, and nondegeneracy then forces $[H,H]=0$. An element commuting with $H$ normalizes it, hence belongs to $H$.

A real or complex Lie group is semisimple when its Lie algebra is semisimple. For a real group, the Killing form then defines a nondegenerate bi-invariant pseudo-Riemannian metric. Compact semisimple groups have negative-definite Killing form; noncompact examples such as $SL(2,\mathbb{R})$ have indefinite Killing form. A disconnected group can also have a semisimple Lie algebra, so connectedness is a separate hypothesis when integrating infinitesimal actions.

For a semisimple Lie algebra, its nondegenerate Killing form turns a two-cocycle into an operator $D$ through $B(DX,Y)=\kappa (X,Y)$. The cocycle identity makes $D$ a derivation. Choose $Z$ by $B(Z,X)=\mathrm{tr}(D\mathrm{ad}X)$. Using $[D,\mathrm{ad}X]=\mathrm{ad}(DX)$ and cyclicity of trace gives $B([Z,X],Y)=B(DX,Y)$, so $D=\mathrm{ad}Z$. Consequently $\kappa (X,Y)=B(Z,[X,Y])$ is a coboundary. This proves that component Hamiltonians can be normalized to a Lie algebra homomorphism.

A semisimple Lie algebra over a field of characteristic zero is split if it has a Cartan subalgebra whose adjoint action is simultaneously diagonalizable over that field. Its root-space decomposition then needs no extension of scalars. The sl2R Lie algebra is split, with the real diagonal Cartan subalgebra. The real special unitary Lie algebra $\mathfrak{su}(2)$ is not: its adjoint operators are skew-adjoint for the positive definite inner product obtained by negating its Killing form. Thus a split adjoint operator would have to be zero.

Declare a global frame of left-invariant vector fields parallel on a semisimple Lie group. This affine connection is flat and preserves a left-invariant metric tensor because its frame components are constant. If $[e_a,e_c]=C_a{}^b{}_c{e}_b$, its torsion tensor is $T_a{}^b{}_c=-C_a{}^b{}_c$. For fixed $c$, the matrix with entries $T_a{}^b{}_c$ is $\mathrm{ad}{e}_c$. Its trace vanishes because a semisimple Lie algebra is perfect and adjoints of brackets are commutators. The trace of the product for $c,d$ is the Killing form $K_{cd}$, proving both contractions. This flat metric connection differs from the curved torsion-free Levi-Civita connection of the Killing metric.