Adjoint representation of a Lie algebra ($\mathrm{ad}$)

The adjoint representation is ${\mathrm{ad}}_x(y)=[x,y]$. For a semisimple Lie algebra, its nonzero weights are the roots and its zero-weight space is the Cartan subalgebra.

The complex Adjoint representation of SU(3) is the action by conjugation on traceless $3\times 3$ matrices. With $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, its highest-weight vector is $E_{13}$ of Dynkin labels $(1,1)$. The six off-diagonal matrix units have weights $(2,-1),(-2,1),(-1,2),(1,-2),(1,1),(-1,-1)$, and the two-dimensional Cartan subalgebra has weight zero. In coordinates $(h_1,(h_1+2h_2)/\sqrt{3})$, the weight diagram is a regular hexagon together with a central point of weight multiplicity two. This representation describes both the gluon colour multiplet and the approximate flavour baryon octet, under different physical SU(3) symmetries.

For traceless matrices $T,S$ transforming by conjugation under SU(3), the trace $\mathrm{tr}(TS)$ supplies a singlet, the commutator $[T,S]$ an antisymmetric octet, and the traceless anticommutator a symmetric octet. Symmetrizing the two upper and two lower indices and removing traces supplies the 27. Contracting an antisymmetric lower pair with the invariant alternating tensor and symmetrizing the remaining three upper indices supplies the 10, and the conjugate construction supplies the conjugate 10. These nonzero equivariant maps exhaust the dimensions: the symmetric part has $36=1+8+27$, the antisymmetric part $28=8+10+10$.

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

Every invariant subspace of the Adjoint representation of a Lie algebra is an ideal. Hence the adjoint representation of a simple Lie algebra is irreducible.