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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.

An adjoint scalar has adjoint covariant derivative $D_{\mu }\varphi ={\partial }_{\mu }\varphi +[A_{\mu },\varphi ]$. A symmetric invariant bilinear form on a Lie algebra contracts the gauge field strength and scalar kinetic terms to give gauge invariance. On a compact real form the positive internal metric is $-\kappa$, because the Killing form itself is negative definite; complex scalar components require the associated Hermitian pairing. The Killing form is degenerate on Abelian factors, where another invariant metric is needed for a nondegenerate kinetic term.

Every Adjoint representation operator raises the lower central series of a Lie algebra filtration. In a basis adapted to that finite filtration, all adjoint operators are strictly triangular. Their products have zero trace, so the Killing form vanishes. Consequently its radical of the Killing form is the whole algebra. For strictly upper triangular matrices the filtration can be seen directly by the gap between the column and row indices.

// Target: algebra.bigb

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

In the ordered basis $(J,E_1,E_2)$ of the Euclidean motion Lie algebra in two dimensions, the Killing form has matrix $\mathrm{diag}(-2,0,0)$. Its radical is the translation ideal, and the algebra is solvable rather than semisimple.

In the anti-Hermitian basis $T^a=-i{\sigma }^a/2$, the SU(2) Lie algebra has $[T^a,T^b]={\epsilon }^{abc}{T}^c$ and Killing form $\kappa (T^a,T^b)=-2{\delta }^{ab}$. This is negative definite; the positive internal form for compact Yang-Mills theory is its negative.

For a real compact semisimple algebra, $B=-\kappa$ is a positive internal metric. With signature $(+---)$ and connection $D_{\mu }={\partial }_{\mu }+A_{\mu }$, the healthy Lagrangian is $\mathcal{L}=-B(F_{\mu \nu },F^{\mu \nu })/(4g^2)=\kappa (F_{\mu \nu },F^{\mu \nu })/(4g^2)$. Gauge invariance follows from Killing-form invariance and ${\delta }_{X}F=ϵ[X,F]$. The physical Hamiltonian density is $[B(E_i,E_i)+B(B_i,B_i)]/(2g^2)$ after imposing the Gauss law constraint in gauge theory and treating the spatial boundary flux appropriately. An Abelian factor needs a separate positive invariant metric because its Killing form vanishes.

Lowering and raising structure constants with the Killing form gives $c_a{}^{mn}{c}_{mn}{}^r=-{\delta }_a{}^r$. Since $c_a{}^{mn}$ is antisymmetric in $m,n$, only the commutator part of the cubic trace tensor of a Lie algebra representation contributes, proving $c_a{}^{mn}{B}_{mnb}=-H_{ab}/2$. In an adapted compact basis with ${\kappa }_{ab}=-{\delta }_{ab}$ and $H_{ab}=-\mu {\delta }_{ab}$, the resulting sign is positive: $+\mu {\delta }_{ab}/2$. Defining the lowered structure constants with a positive Euclidean metric instead would reverse this sign.

// Target: lie-theory.bigb

With Hermitian angular-momentum generators $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the Killing form is $2{\delta }_{ab}$ and its inverse contraction is $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. Equivalently the anti-Hermitian basis $T_a=-i{J}_a$ has Killing form $-2{\delta }_{ab}$ and gives the same operator. In the complex highest-weight basis, $C_{\kappa }=H^2/8+(E^{+}{E}^{-}+E^{-}{E}^{+})/4$. It acts on spin $j$ by $j(j+1)/2$, so it differs in normalization from the usual angular-momentum quadratic Casimir operator.

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.

For ambient quadratic form $Q=-x_0^2-x_4^2+x_1^2+x_2^2$, associate $M=\left(\begin{array}{cc}{x}_4+x_1& x_2+x_0\\ x_2-x_0& x_4-x_1\end{array}\right)$. Its determinant is $-Q$, so the unit Anti-de Sitter spacetime hyperboloid is the special linear group of real two-by-two matrices. On its tangent space at the identity, the induced bilinear form is $\frac{1}{2}\mathrm{tr}(UV)$. The Killing form is $4\mathrm{tr}(UV)$, and left and right matrix multiplication preserve the ambient form. Consequently the induced bi-invariant pseudo-Riemannian metric is $B/8$. This model has periodic time; its causally unwrapped version is the universal covering Lie group.

For a complex semisimple Lie algebra, the Cartan subalgebra is orthogonal under the Killing form to every nonzero root space. A vector in the Cartan subalgebra orthogonal to that whole subalgebra is therefore orthogonal to the entire Lie algebra and must vanish. On the Cartan subalgebra the form is $\kappa (h,h^{\prime })={\sum }_{\alpha \in \Phi }\alpha (h)\alpha (h^{\prime })$.

For a root of a root system of a complex semisimple Lie algebra, define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$ using the Killing form on its Cartan subalgebra. Then $\alpha (t_{\alpha })\ne 0$. Otherwise vectors $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$ would generate a solvable Lie algebra with $[x,y]=t_{\alpha }$ central. The Lie theorem would make $\mathrm{ad}{t}_{\alpha }$ nilpotent, while the root-space decomposition makes it diagonalizable, forcing $t_{\alpha }$ into the zero center of a Lie algebra.

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 displayed matrices describe the orientation-preserving component $SE(2)$ of the full Euclidean group $E(2)=O(2)⋉{\mathbb{R}}^2$. Both have the same Lie algebra, so this naming convention does not affect the calculation. Choose the translation generators $P_1,P_2$ and rotation generator $J$ in the Lie algebra of this Matrix Lie group asTaking matrix commutators givesAll remaining brackets follow from antisymmetry. The corresponding left-invariant vector fields in coordinates $(x,y,\psi )$ areTheir Lie brackets of vector fields obey the same relations. This also fixes which sign of the rotation generator is being used.

There is an important qualification to the printed connection prescription. For an arbitrary vector field $Z$ and smooth function $f$, its literal right-hand side obeysAn affine connection instead requires ${\nabla }_{L_i}(fZ)=L_i(f)Z+f{\nabla }_{L_i}Z$. Thus the formula for arbitrary $Z$ defines a connection only when $\lambda =1$. For example, at $\psi =0$, take $i=1$, $f=x$ and $Z=L_1$: the erroneous formula gives $\lambda L_1$, whereas the Leibniz rule requires $L_1$.

The intended one-parameter family is the bracket connection on a Lie group: prescribe the rule for left-invariant $Z$, then extend it by the Leibniz rule. If $[L_i,L_j]=c_{ij}{}^k{L}_k$, this meansExtend linearly over functions in the first slot. This is a well-defined affine connection for every real $\lambda$. The following Ricci tensor calculation applies to this intended family; under the literal arbitrary-$Z$ reading only its $\lambda =1$ member exists.

Use the curvature conventionFor left-invariant vector fields the structure constants of a Lie algebra are constant. The Jacobi identity givesWriting $A=\lambda (\lambda -1)$, the only nonzero curvature actions, apart from antisymmetry in the first two slots, areFor instance, $[[J,P_1],J]=[P_2,J]=P_1$. Contracting the first and output slots gives $\mathrm{Ric}(L_3,L_3)=-2A$, while every other component vanishes. Equivalently, $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, soThis is the Killing form. Here ${\mathrm{ad}}_J$ rotates the two-dimensional translation space and vanishes on $J$, so $B(J,J)=-2$; translation generators give all other components zero. ThereforeSince the coframe component dual to $L_3$ is $d\psi$, the same Ricci tensor is $2\lambda (1-\lambda )d\psi \otimes d\psi$ in these coordinates. Reversing the curvature convention reverses this sign.

At both $\lambda =0$ and $\lambda =1$, the full curvature vanishes. At $\lambda =0$, the left-invariant vector fields form a global parallel frame. At $\lambda =1$, every right-invariant vector field is parallel because left- and right-invariant vector fields commute; the connection is flat in that global parallel frame. Flatness does not imply zero torsion: for left-invariant arguments the torsion tensor isThus the two endpoint connections are flat, with opposite nonzero torsion. For comparison, $\lambda =1/2$ is torsion-free but has nonzero curvature, with ${\mathrm{Ric}}_{33}=1/2$. No metric has been specified, so these connections should not automatically be identified with a Levi-Civita connection.

The Adjoint representation of a Lie algebra acts on the algebra itself: ${\mathrm{ad}}_X(Y)=[X,Y]$. In the basis $T_b$, its matrices areFor any algebra element $X$, the Jacobi identity givesTherefore the $D\times D$ matrices satisfy the same Lie bracket relations:This construction does not require that the algebra be simple or that the adjoint action be faithful.

For the field calculation, write $A_{\mu }=A_{\mu }^a{T}_a$ and $\lambda ={\lambda }^a{T}_a$. In the derivative-plus-connection convention the Yang-Mills field strength isThe sign of the parameter here is the negative of the matter-field parameter in the usual gauge transformation in the derivative-plus-connection convention. Work with the stated $\lambda$ throughout. Expanding the variation givesThe second derivatives cancel by commutativity of partial derivatives, and all remaining derivatives of $\lambda$ cancel by antisymmetry of the Lie bracket. The Jacobi identity makes the last two terms $[[A_{\mu },A_{\nu }],\lambda ]$. HenceUnlike the connection, the curvature has no inhomogeneous derivative term: it transforms in the Adjoint representation.

The quadratic form ${\kappa }_{ab}=\mathrm{tr}(T_a^{\mathrm{ad}}{T}_b^{\mathrm{ad}})$ is the Killing form. Its symmetry follows from cyclicity of the matrix trace. Using the just-proved adjoint commutators,ThusEquivalently, each adjoint generator is skew with respect to this invariant bilinear form. This invariance holds even when the Killing form is degenerate; no inverse form is needed here.

The spacetime metric used to raise Lorentz indices is unaffected by the internal gauge transformation. Since $\delta F=-{\mathrm{ad}}_{\lambda }F$, invariance givesTherefore the Killing-form contraction of the field strengths is gauge invariant, as required for the internal contraction in a Yang-Mills action.

Use the index convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$. For the left- and right-invariant coframes, the Maurer-Cartan equations areTheir dual invariant vector fields satisfyFor example, evaluating $d{\lambda }^b$ on two dual fields gives $-{\lambda }^b([L_a,L_c])$, since its invariant coefficient functions are constant. The opposite right-invariant bracket gives the other sign. The mixed bracket vanishes because left and right translations commute.

For a semisimple Lie algebra, the Killing form $B(X,Y)=\mathrm{Tr}(\mathrm{ad}X\mathrm{ad}Y)$ is nondegenerate. It is adjoint invariant: using $\mathrm{ad}[Z,X]=[\mathrm{ad}Z,\mathrm{ad}X]$ and cyclicity of trace givesGroup conjugation preserves the trace as well, so this is invariant under the full group adjoint action. Translation supplies a bi-invariant pseudo-Riemannian metric.

Lower the indicated structure index by $C_{abc}=B_{be}{C}_a{}^e{}_c$. ThenThis changes sign under exchange of $a,c$ by the bracket and under exchange of $a,b$ by invariance. Hence $C_{abc}=C_{[abc]}$.

Let $\eta (X,Y,Z)=B(X,[Y,Z])$ at the identity and extend it as a left-invariant differential form. Its value is adjoint invariant, so the extended form is also right-invariant. The three-form in the question is $-\eta$ with the index-lowering convention just specified, an irrelevant overall sign for these properties. To prove closedness rather than only invariance, apply the exterior derivative on four left-invariant fields. Its six bracket terms combine by invariance intoThe final equality is the Jacobi identity. Therefore the Cartan three-formis both bi-invariant and closed.

Define the raw Killing form byFor a finite-dimensional real or complex Lie algebra in characteristic zero, it is nondegenerate exactly when the algebra is a semisimple Lie algebra. For a real compact semisimple algebra it is negative definite. Consequently the positive compact Killing metric is $g=-K$: for example on $\mathfrak{su}(2)$ with $[e_i,e_j]={\epsilon }_{ijk}{e}_k$, $K_{ij}=-2{\delta }_{ij}$ and $-K$ gives a Riemannian metric on $SU(2)$. An overall minus sign in the term “Killing metric” is common here; raw trace $K$ itself is negative definite in this example.

For a Lorentzian metric use $\mathfrak{sl}(2,\mathbb{R})$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. Direct traces givewhose metric signature is $(2,1)$. For a non-Abelian degenerate example, the three-dimensional Heisenberg Lie algebra has $[x,y]=z$ central. Every product of two adjoint maps is zero, so $K=0$.

For any adjoint-invariant nondegenerate metric on a Lie group, the Koszul formula on left-invariant fields simplifies toTake $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. The Jacobi identity givesTaking the trace of $Z\mapsto R(Z,X)Y=-\mathrm{ad}Y\mathrm{ad}X(Z)/4$ yieldsThus the metric $g=K$ is a Killing-form Einstein metric with Einstein constant $-1/4$. With $g=-K$ the same connection has constant $+1/4$. This proves the Einstein manifold assertion for every semisimple group, including products when the unscaled Killing form is used on all factors.

There is also a different, flat metric-preserving connection. Declare the global left-invariant frame parallel:Its curvature is zero because applying two derivatives to coefficient functions and subtracting their commutator gives zero. Its metric derivative vanishes since $g_{ab}$ is constant in this frame. Its torsion tensor on invariant fields isFor each fixed $c$, antisymmetry gives $T_a{}^b{}_c=C_c{}^b{}_a$, so this is the matrix of $\mathrm{ad}{e}_c$. The torsion contractions of a left-parallel Lie-group connection therefore areThe zero trace follows because a semisimple algebra is perfect: every element is a sum of brackets, and the adjoint of a bracket is a matrix commutator with zero trace. If the metric convention is $g=-K$, the second contraction is $-g_{cd}$. These flat translation connections on a Lie group are not the torsion-free Levi-Civita connection; nonzero torsion permits flatness alongside the curved Killing metric.