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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.

A scalar field transforming in the Adjoint representation of a Lie algebra transforms by a Lie bracket with the infinitesimal gauge parameter.

For a complex vector space with a nondegenerate symmetric bilinear form $Q$, the map $u\wedge v\mapsto [w\mapsto Q(v,w)u-Q(u,w)v]$ identifies the exterior square of the defining representation with the Adjoint representation of a Lie algebra of $\mathfrak{so}(V)$. Equivariance follows from invariance of $Q$; an orthogonal basis verifies bijectivity. It is irreducible when $\mathfrak{so}(V)$ is simple. In dimension four it splits into the adjoint modules of the two commuting sl2 Lie algebras.

Identify the product with endomorphisms of the defining space. Special unitary group conjugation fixes scalar matrices and preserves the complementary traceless subspace. For $N\ge 2$, the latter is the complex Adjoint representation of a Lie algebra, which is irreducible because ${\mathfrak{sl}}_N$ is simple. With fundamental Dynkin index $1/2$, the trace-Casimir identity gives $C_2(\mathbf{N})=(N^2-1)/(2N)$; the tensor-product Casimir trace identity then gives $C_2(\mathbf{Adj})=N$.

The Jacobi identity is the compatibility condition on a Lie bracketTogether with an Alternating Lie bracket, it defines a Lie algebra. Unlike associativity, it relates three cyclic bracketings rather than equating two bracketings. Equivalently, using antisymmetry of a Lie bracket,Thus the linear map ${\mathrm{ad}}_x:y\mapsto [x,y]$ is a derivation of a Lie algebra: acting on a bracket is the sum of acting on each argument. This makes infinitesimal symmetries act consistently on each other's products. It also gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, the identity that makes the Adjoint representation of a Lie algebra a Lie algebra representation.

For the commutator in an associative algebra, the cancellation is directly visible:Each word occurs once with each sign. This proves the Jacobi identity without a restriction on the characteristic of a field.

For a computed check in the sl2 matrix commutator example,Here the last two terms are individually nonzero over characteristic zero; the identity records their cancellation. In the cross-product Lie algebra, the same cyclic cancellation is the vector triple product identity.

The historical connection is unusually concrete: Lie and Engel's volume I expands nested commutators of differential operators into twelve terms and cancels them cyclically, in Chapter 5, §26, equation (7). See Lie and Engel, §26, in Merker's translation, pp. 105–106.

Let $[e_1,e_2]=e_6$, $[e_1,e_3]=e_4$ and $[e_2,e_3]=e_5$, with $e_4,e_5,e_6$ central. The symmetric tensor $S=e_1\otimes e_5+e_5\otimes e_1+e_3\otimes e_6+e_6\otimes e_3-e_2\otimes e_4-e_4\otimes e_2$ is invariant under the Adjoint representation of a Lie algebra: the derivation action of each of $e_1,e_2,e_3$ cancels in pairs. For $t\ne 0$, $Q=tS+{\sum }_{j=4}^6{a}_j{e}_j\otimes e_j$ has determinant $-t^6$ and inverse metric $g=t^{-1}S(\lambda )-t^{-2}(a_5({\lambda }^1{)}^2+a_4({\lambda }^2{)}^2+a_6({\lambda }^3{)}^2)$ in the dual coframe. Its three hyperbolic planes give the displayed metric signature. On a connected Lie group this supplies a four-parameter family of bi-invariant pseudo-Riemannian metrics. Disconnected components must preserve the tensor separately; the automorphism reversing $e_1,e_4,e_6$ while fixing the other generators reverses $S$ and prevents this extension.

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.

The subspace spanned by $e_4,e_5,e_6$ is the center of a Lie algebra. Each of their tensor squares is therefore invariant. On tensor products the infinitesimal Adjoint representation of a Lie algebra acts by a derivation, $D_x(u\otimes v)=[x,u]\otimes v+u\otimes [x,v]$. Write $u\odot v=u\otimes v+v\otimes u$ and let $S=e_1\odot e_5+e_3\odot e_6-e_2\odot e_4$. The only potentially nonzero checks areThe remaining generators are central, so they act trivially. This proves every claimed tensor invariant.

These tensors are contravariant. To obtain an ordinary metric tensor, take an invertible linear combination and invert it. For arbitrary real $a_4,a_5,a_6$ and $t\ne 0$, setIn the ordered basis $(e_1,e_2,e_3;e_4,e_5,e_6)$ its matrix isSince $K^{T}K=I$, $\det Q=-t^6\ne 0$, andLet ${\lambda }^a$ be the dual left-invariant coframe. The resulting neutral invariant metrics on the free two-step Lie algebra on three generators areHere $({\lambda }^a{)}^2={\lambda }^a\otimes {\lambda }^a$. Infinitesimal invariance of $Q$ implies that its inverse is an adjoint-invariant bilinear form. Its left-translated metric is right-invariant on a connected $G$, giving the desired four-parameter family of bi-invariant metrics. A change of variables removes the central $A$ block from the quadratic form, leaving three off-diagonal two-dimensional blocks. Each has one positive and one negative eigenvalue. Thus the metric signature is $(3,3)$; these are bi-invariant pseudo-Riemannian metrics, not positive-definite ones.

Connectedness, conventionally understood here, matters for the passage from infinitesimal to full group invariance. If $G$ is disconnected, its other components must preserve $Q$ separately. For example the Lie algebra automorphism reversing $e_1,e_4,e_6$ and fixing $e_2,e_3,e_5$ preserves all brackets but sends $S$ to $-S$. In the semidirect product of the simply connected group with this order-two automorphism, none of these nondegenerate $t\ne 0$ tensors is invariant under that component. Thus the Lie-algebra data alone give the stated family on the identity component, rather than an unconditional extension to every disconnected realization.

Let ${\hat{T}}_a={\mathrm{ad}}_{T_a}$ on the Lie algebra itself. The Jacobi identity says, for every $X$,Therefore $[{\hat{T}}_a,{\hat{T}}_b]=c_{ab}^c{\hat{T}}_c$, proving that the displayed structure-constant matrices form the Adjoint representation of a Lie algebra.

The bilinear form ${\kappa }_{ab}=\mathrm{tr}({\hat{T}}_a{\hat{T}}_b)$ is the Killing form. It is symmetric by cyclicity of trace. Its invariance follows by expanding the two commutators:In components this is preciselyEquivalently, the adjoint action preserves the bilinear form, so it is an invariant tensor.

The invariant abelian subalgebra is an abelian ideal of a Lie algebra $\mathfrak{a}$. Assume it is nonzero, as required for the conclusion. For $X\in \mathfrak{a}$ and $Y\in \mathfrak{g}$, the map ${\mathrm{ad}}_X{\mathrm{ad}}_Y$ sends $\mathfrak{g}$ into $\mathfrak{a}$, because it is an ideal. It vanishes on $\mathfrak{a}$, because ${\mathrm{ad}}_Y$ preserves the ideal and ${\mathrm{ad}}_X$ kills it. Relative to $\mathfrak{a}$ and a vector-space complement, this map has zero diagonal blocks, so its trace is zero. ThusThis proves that abelian ideals lie in the radical of the Killing form. Choose a nonzero $X=x^a{T}_a$ in the ideal. Then ${\kappa }_{ba}{x}^a=0$ for every $b$, so the coordinate vector $x$ is a nonzero null eigenvector of the Killing matrix. The nonzero-ideal qualification cannot be dropped: the zero ideal is abelian even in a semisimple algebra with nondegenerate Killing form.

Now suppose $\kappa$ is invertible and let $C_{\kappa }={\kappa }^{bc}{t}_b{t}_c$ in any Lie algebra representation. In matrix notation invariance is $A_a^T\kappa +\kappa A_a=0$, where $(A_a{)}^d{}_b=c_{ab}^d$. Multiplying by the inverse form gives $A_a{\kappa }^{-1}+{\kappa }^{-1}{A}_a^T=0$. HenceNo commutativity of the representation matrices has been assumed. This proves the centrality of the quadratic Casimir operator formed with the inverse Killing form. Schur lemma makes it a scalar on every finite-dimensional complex irreducible representation.

For SU(2), make the normalization explicit. In a Hermitian angular-momentum basis with $[J_a,J_b]=i{ϵ}_{abc}{J}_c$, the structure-constant matrices obeyThus $C_{\kappa }=\frac{1}{2}{\sum }_a{J}_a^2$. In the compact anti-Hermitian basis $T_a=-i{J}_a$, the brackets have real structure constants ${ϵ}_{abc}$ and the Killing form of the SU(2) Lie algebra is instead $-2{\delta }_{ab}$; the inverse contraction gives the same $C_{\kappa }$. This sign change is a basis convention, not a different spectrum.

In the complex highest-weight basis $(H,E^{+},E^{-})$ used earlier, the same Killing form and its inverse areConsequentlyOn a highest vector of weight $m=2j$, the ladder calculation gives $E^{+}{E}^{-}v=mv$ and $E^{-}{E}^{+}v=0$. Its Casimir eigenvalue is thereforeCentrality gives the same value throughout the irreducible multiplet. This is the Killing-normalized SU(2) quadratic Casimir; the familiar angular-momentum operator ${\sum }_a{J}_a^2$ has eigenvalue $j(j+1)$ and is twice this inverse-Killing-form contraction.

In the defining representation $V={\mathbb{C}}^{2n}$, the basis vectors $v_i$ and $v_{\bar{i}}$ have weights ${\epsilon }_i$ and $-{\epsilon }_i$, respectively, each with multiplicity one. Thus its formal character isThe vector $v_1$ is killed by every positive root space, since all positive-root matrices are upper triangular, and is a highest-weight vector. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is irreducible for $n\ge 2$. In terms of the fundamental weights of Dn, its highest weight is ${\omega }_1$ for $n\ge 3$ and ${\omega }_1+{\omega }_2$ for $n=2$.

The flip of the two tensor factors is equivariant, henceUse the invariant symmetric bilinear form $Q(u,v)=u^{T}Jv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the trivial Lie algebra representation. Its inverse-form tensoris invariant and satisfies $C(\tau )=2n$. ConsequentlyThe traceless summand contains the highest-weight vector $v_1\otimes v_1$ of weight $2{\epsilon }_1$. The Weyl complete reducibility theorem states that every finite-dimensional complex semisimple Lie algebra representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2{\epsilon }_1)$. The Weyl dimension formula verifies that it is the entire traceless space: only positive roots involving ${\epsilon }_1$ change their factors, andThis equals $\dim {\mathrm{Sym}}^{2}V-1=n(2n+1)-1$. Thus the traceless symmetric square of the defining even orthogonal representation is irreducible, including $n=2$.

For the other summand, the exterior square of the defining orthogonal representation has the equivariant identificationThe image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the Adjoint representation of a Lie algebra. For $n\ge 3$, the connected Dn Dynkin diagram gives a simple Lie algebra (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is ${\epsilon }_1+{\epsilon }_2$, and $v_1\wedge v_2$ is a highest vector. The tensor-square decomposition of the defining even orthogonal representation is consequentlyThe three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2{\omega }_1$, ${\omega }_2$ and zero for $n\ge 4$; for $n=3$ the middle weight is ${\omega }_2+{\omega }_3$.

For $n=2$, the Chiral decomposition of the complexified so4 Lie algebra splits the adjoint module into its two three-dimensional ideals. ThusThe weights are $2{\omega }_1+2{\omega }_2$, $2{\omega }_1$, $2{\omega }_2$ and zero. Equivalently, labelling highest weights by the two sl2 Lie algebra factors, this is $(2,2)\oplus (2,0)\oplus (0,2)\oplus (0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L({\epsilon }_1)\oplus L(-{\epsilon }_1)$ and $V\otimes V=L(2{\epsilon }_1)\oplus 2L(0)\oplus L(-2{\epsilon }_1)$; there is then no single highest weight for an irreducible natural module.

Apply the Jacobi identity to an arbitrary $z$:Thus the Adjoint representation of a Lie algebra preserves the Lie bracket:

The Adjoint representation of a Lie algebra acts on the Lie algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity givesThus it is a Lie algebra representation. In the basis $T_a$ with $[T_a,T_b]=c^c{}_{ab}{T}_c$, its matrices areThe column index labels the basis vector being acted on; this convention fixes the matrix signs.

Work in finite dimension over characteristic zero, as in the usual real or complex gauge algebras. A simple Lie algebra is nonabelian and has no nonzero proper Lie algebra ideal. A semisimple Lie algebra has no nonzero solvable ideal; equivalently it is a direct sum of simple ideals. The nondegenerate Killing form is $\kappa (X,Y)=\mathrm{tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. Its invariance follows from the representation identity and trace cyclicity:

For the reduction, take a nonzero proper ideal $\mathfrak{a}$, if one exists, and define its Killing-orthogonal complement ${\mathfrak{a}}^{\perp }$. Invariance makes it an ideal: for $y\in {\mathfrak{a}}^{\perp }$, $x\in L$, $z\in \mathfrak{a}$, $\kappa ([x,y],z)=-\kappa (y,[x,z])=0$. Moreover $[\mathfrak{a},{\mathfrak{a}}^{\perp }]=0$, because for $u\in \mathfrak{a}$, $v\in {\mathfrak{a}}^{\perp }$, and any $x\in L$,and the global Killing form is nondegenerate. Their intersection is consequently an abelian ideal: its elements lie in both commuting ideals. Semisimplicity forces that intersection to be zero. The dimension formula for a nondegenerate bilinear form then givesThe summands commute and are semisimple, since a solvable ideal in either summand would also be an ideal of $L$. Repeating on smaller summands terminates in simple ideals. This proves the orthogonal ideal splitting for a nondegenerate Killing form and describes the required reduction; abelian one-dimensional factors do not occur in a semisimple algebra.

If $\kappa (T_a,T_b)=\ell {\delta }_{ab}$ with $\ell \ne 0$, invariance givesso $c_{abc}=-c_{acb}$ after lowering the output index with $\delta$. The bracket already gives $c_{abc}=-c_{bac}$. These two adjacent transpositions generate all permutations, proving that the structure constants of a Lie algebra are completely antisymmetric. This is the antisymmetry of Killing-lowered structure constants; the scalar $\ell$ cancels in the chosen normalization.

For the gauge calculation write $\Lambda ={\lambda }_a{t}_a$, $A_{\mu }=A_{\mu a}{t}_a$, and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The representation generators here obey $[t_a,t_b]=c_{abc}{t}_c$; factors of $i$ and the gauge coupling are absorbed in this connection convention. Expanding the varied derivative givesFor this to equal $\Lambda D_{\mu }\varphi$, choose the gauge transformation in the derivative-plus-connection conventionThen all extra terms cancel, proving covariance of the gauge covariant derivative for arbitrary space-dependent parameters.

Apply both derivatives to a test field. The second derivatives commute and the terms involving one derivative of the field cancel, leaving the Yang-Mills field strengthThus its components areThe already obtained variation can be written $\delta D_{\mu }=[\Lambda ,D_{\mu }]$. The operator Jacobi identity then yieldsorFinally, with fixed space-time metric,The contracted product is symmetric in $a,c$, while $c_{abc}$ is antisymmetric in those indices. Hence the Yang-Mills action density is gauge invariant, with the sign convention derived rather than assumed.

For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

Let $R_a=\mathrm{ad}(T_a)$, where $\mathrm{ad}(X)Y=[X,Y]$. In the given basis, the Adjoint representation of a Lie algebra isThe Jacobi identity gives $[\mathrm{ad}X,\mathrm{ad}Y]=\mathrm{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}{R}_c$, as required for a Lie algebra representation. The Killing form is the symmetric Trace form of a Lie algebra representationCyclicity of the trace proves its invariance directly:In components this is precisely $\boxed{{\kappa }_{dc}{f}^d{}_{ab}+{\kappa }_{bd}{f}^d{}_{ac}=0}$, so the Killing form is an invariant bilinear form on a Lie algebra.

Choose the infinitesimal gauge transformation convention $\delta \varphi =\omega \varphi$, with $\omega (x)={\omega }^a(x)t_a$, and write $A_{\mu }=A_{\mu }^a{t}_a$. A gauge covariant derivative must satisfy $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$ for $U=1+\omega +O({\omega }^2)$; henceEquivalently $\delta A_{\mu }^a=-{\partial }_{\mu }{\omega }^a+f^a{}_{bc}{\omega }^b{A}_{\mu }^c$. Expanding $\delta (D_{\mu }\varphi )$ makes the cancellation transparent:Thus $\boxed{\delta (D_{\mu }\varphi )={\omega }^a{t}_a{D}_{\mu }\varphi }$: it transforms in the same representation as $\varphi$. In adjoint covariant derivative notation, $\delta A_{\mu }=-D_{\mu }^{\mathrm{ad}}\omega$.

The commutator of two gauge covariant derivatives defines the gauge field strength:ThusConjugating the derivative commutator gives $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$, so the gauge field strength transforms homogeneously in the Adjoint representation:Even if the matrices $t_a$ in a particular matter representation are not faithful, the displayed component formula defines the Lie-algebra-valued curvature independently of that representation.