Killing form ($\kappa (X,Y)$)

The Killing form isIt is symmetric and invariant: $\kappa ([X,Y],Z)=\kappa (X,[Y,Z])$.

For the general linear Lie algebra of all $n\times n$ matrices over a field of characteristic zero, write ${\mathrm{ad}}_X=L_X-R_X$ on the matrix space. The traces of $L_X{L}_Y$ and $R_X{R}_Y$ are $n\mathrm{tr}(XY)$, whereas both mixed traces are $\mathrm{tr}X\mathrm{tr}Y$. Expanding gives the displayed Killing form. Its radical is the scalar matrices because $\kappa (X,Y)=0$ for every $Y$ forces $nX-\mathrm{tr}(X)I=0$. Thus the full general linear algebra is not semisimple, even though its traceless summand is semisimple for $n\ge 2$.

// Target: lie-theory.bigb

Every Lie algebra automorphism $S$ obeys ${\mathrm{ad}}_{SX}=S{\mathrm{ad}}_X{S}^{-1}$. The cyclic property of the trace therefore makes the Killing form invariant under $S$. For a matrix Lie group, $S={\mathrm{Ad}}_g$ is conjugation by $g$, proving the displayed identity even when the group is disconnected. This directly proves gauge invariance of Killing-form contractions of adjoint fields.

For $[e_1,e_2]=a{e}_3$, $[e_2,e_3]=b{e}_1$, and $[e_3,e_1]=c{e}_2$, the Adjoint representation matrices give the displayed Killing form. Cross traces vanish. Nonzero $a,b,c$ make it nondegenerate and hence the algebra semisimple. If the three constants have the same sign it is negative-definite; mixed signs give signature $(2,1)$. In particular $(a,b,c)=(2,2,2)$ is the compact $\mathfrak{su}(2)$ case, whereas $(2,2,-2)$ is the split real traceless two-by-two algebra. This gives a direct compactness criterion from the Killing form in a small explicit example.

For an ideal of a Lie algebra $I$ and nondegenerate Killing form, invariance makes $I^{\perp }$ an ideal. An element of $I\cap I^{\perp }$ commutes with $I$, so the intersection is an abelian ideal. Abelian ideals lie in the radical of the Killing form, forcing the intersection to vanish. Thus $\mathfrak{g}=I\oplus I^{\perp }$ as commuting ideals. Iterating minimal nonzero ideals gives the direct-sum formulation of a semisimple Lie algebra.

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.

A finite-dimensional real Lie algebra has negative-definite Killing form exactly when it is compact semisimple. In the compact direction, average an inner product over a compact group; all adjoint maps are skew, so $\kappa (X,X)=-\Vert {\mathrm{ad}}_X{\Vert }^2$, and semisimplicity removes the center. Conversely, negative-definiteness gives the positive invariant metric $-\kappa$, embeds the faithful adjoint algebra in an orthogonal algebra, and nondegeneracy gives semisimplicity. This is an algebraic compactness criterion, not a claim that every global group with an Abelian compact-type algebra is compact.

If $\mathfrak{a}$ is an abelian ideal of a Lie algebra, then ${\mathrm{ad}}_x{\mathrm{ad}}_y$ maps $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$ whenever $x\in \mathfrak{a}$. Its trace is zero, so $\kappa (\mathfrak{a},\mathfrak{g})=0$. Every nonzero solvable ideal has a last nonzero term in its derived series of a Lie algebra, which is such an abelian ideal. Consequently a nondegenerate Killing form forces the solvable radical to vanish.

The radical ${\mathfrak{g}}^{\perp }=\{x:\kappa (x,\mathfrak{g})=0\}$ of the Killing form is a solvable ideal. Invariance makes it an ideal, and the Cartan solvability criterion applied to its adjoint image proves solvability.

A finite-dimensional complex Lie algebra is semisimple if and only if its Killing form is nondegenerate.

For a finite-dimensional complex Lie algebra $L$, the radical $R$ of its Killing form is a Lie algebra ideal by invariance. For $x,y\in R$, their adjoint actions vanish on $L/R$, so $B_R(x,y)=B_L(x,y)=0$. The Cartan solvability criterion makes the image ${\mathrm{ad}}_R(R)$ solvable; its kernel is the abelian center, so $R$ is solvable. A semisimple Lie algebra has no nonzero solvable ideal, forcing its Killing radical to vanish.

Every invariant bilinear form on a finite-dimensional complex simple Lie algebra is a scalar multiple of its Killing form. A nondegenerate invariant form identifies the algebra with its dual; comparing this identification with the Killing form gives an endomorphism of the irreducible adjoint representation, so Schur lemma makes it scalar.

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 ${\mathfrak{sl}}_n(\mathbb{C})$,