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Use the degree-one convention implicit in this enveloping-algebra assertion: a degree-one almost commutative algebra has $F_{0}R=k$, a finite-dimensional $F_{1}R$ containing $1$ which generates $R$, and commutative associated graded ring. Taking the leading symbol of a commutator givesin particular $[F_{1}R,F_{1}R]\subseteq F_{1}R$. Thus $\mathfrak{g}=F_{1}R$, equipped with $[u,v]=uv-vu$, is a finite-dimensional Lie algebra. Bilinearity and antisymmetry are immediate, and expanding the three nested commutators proves the Jacobi identity.

Its inclusion into the associative algebra $R$ is a Lie algebra homomorphism. The universal enveloping algebra is $T(\mathfrak{g})$ modulo the relations $u\otimes v-v\otimes u-[u,v]$. Sending the tensor generators to their elements of $R$ kills these relations and gives an algebra homomorphism $U(\mathfrak{g})\to R$. It is surjective because $F_{1}R$ generates $R$. ThereforeThe vector $1\in F_{1}R$ is retained as a central Lie algebra generator and maps to the associative identity; its corresponding relation in the kernel is that generator minus $1_{U(\mathfrak{g})}$. One should not first replace $F_{1}R$ by $F_{1}R/k1$, since commutators of generators can be nonzero scalars.

Terminology is important here. Merely requiring a commutative finitely generated associated graded ring for an arbitrary weighted filtration is a broader definition of almost commutative algebra; it does not supply the finite-dimensional bracket-closed first step used in this argument. The degree-one assumptions make the requested enveloping presentation precise. The Noetherianity in part (b) also follows directly under the broader finite-generation convention.

This part proves $(c)\Rightarrow (a)$. Write the Weyl algebra generators as $x_1,\dots ,x_n,D_1,\dots ,D_n$, with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ form a basis over any field: one may construct the algebra by successively adjoining the commuting derivations $D_i$ to $k[x_1,\dots ,x_n]$, with multiplication $D_{i}f=f{D}_i+\partial f/\partial x_i$, obtaining unique ordered forms. This also makes $A_n(k)$ infinite-dimensional for $n\ge 1$.

Assume $\mathrm{char}k=0$ and let $I$ be a nonzero two-sided ideal. Choose a nonzero element $a=\sum c_{\alpha \beta }{x}^{\alpha }{D}^{\beta }\in I$, and a nonzero term of maximal total degree, say $({\alpha }_0,{\beta }_0)$. The relations giveApply $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ to $a$. Every lower-degree term vanishes. A term of the same degree survives only if all its exponents dominate $({\alpha }_0,{\beta }_0)$, which forces equality. Thus the resulting element of $I$ isThe factorials are nonzero in characteristic zero. This is a nonzero scalar, so $1\in I$ and $I=A_n(k)$. Therefore $A_n(k)$ is simple in characteristic zero. This scalar extraction by iterated Weyl commutators supplies the full simplicity proof.

Taking $n=e_i$ gives ${\sigma }_i^2=I$. Taking $n=(e_i+e_j)/\sqrt{2}$ for $i\ne j$ and expanding the square gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$. Thus the anticommutator and commutator together implySince $n\cdot \sigma$ is Hermitian and $\theta$ is real, the exponent is anti-Hermitian. Its exponential has inverse equal to its adjoint and is therefore unitary. Splitting the exponential series into even and odd powers, using $(n\cdot \sigma {)}^2=I$, givesPut $N=n\cdot \sigma ,M=m\cdot \sigma$ with $m\perp n$. Then $NM=i(n\times m)\cdot \sigma$, $MN=-NM$, and $NMN=-M$. Expanding $(c-iNs)M(c+iNs)$ gives $(c^2-s^2)M-2icsNM$. ConsequentlyThe spin-half angular momentum is $S_i=\hslash {\sigma }_i/2$. The commutators then become $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, and $(n\cdot S{)}^2={\hslash }^{2}I/4$ expresses the two possible directional spin values $\pm \hslash /2$. Also $S^2=3{\hslash }^{2}I/4$, appropriate to spin one half. The exponential represents a spatial rotation on spin states.

Choose $n=(0,1,0)$ and initially ${\sigma }_3|\chi ⟩=|\chi ⟩$. The conjugation formula sends ${\sigma }_3$ to ${\sigma }_3\cos \theta +{\sigma }_1\sin \theta$. ThereforeThus $U|\chi ⟩$ is spin up along $(\sin \theta ,0,\cos \theta )$, as required.

In the position representation, $\hat{p}=-i\hslash \nabla$. The orbital angular momentum operator and its squared magnitude areFor example, with $D=x\cdot \nabla$, the latter is $-{\hslash }^2[r^2\Delta -D^2-D]$. Operator products here act on smooth wavefunctions in a common invariant domain.

To calculate the angular momentum commutation relations, use $[x_a,p_b]=i\hslash {\delta }_{ab}$ and the vanishing position-position and momentum-momentum commutators. Expanding the products givesMultiplication by ${ϵ}_{iab}{ϵ}_{jcd}$ then givesFinally, the product commutator rule yieldsThe final cancellation pairs the antisymmetric Levi-Civita symbol with an expression symmetric in $j,k$. Thus

Work in finite dimension over $\mathbb{C}$; the characteristic-zero real version follows by complexification. The Cartan solvability criterion isIts semisimplicity consequence, also often called Cartan's criterion, is $\mathfrak{g}$ is semisimple if and only if its Killing form is nondegenerate. We prove both forms.

The auxiliary results used are as follows. The Lie theorem simultaneously triangularizes every finite-dimensional representation of a complex solvable Lie algebra. The Engel theorem says that a matrix Lie algebra consisting entirely of nilpotent endomorphisms has a common annihilated nonzero vector and a basis making all its matrices strictly upper triangular; in particular it is a nilpotent Lie algebra. Nilpotent means that the lower central series terminates, which implies that the derived series terminates as well. Finally, the Additive Jordan decomposition writes any complex endomorphism uniquely as $x=x_s+x_n$, where $x_s$ is diagonalizable, $x_n$ is nilpotent and they commute. On a generalized eigenspace $V_{\lambda }$, the two parts are $\lambda I$ and $x-\lambda I$, and both are polynomials in $x$. On endomorphisms, $\mathrm{ad}{x}_s$ is diagonalizable, $\mathrm{ad}{x}_n$ is nilpotent, and these are the Jordan parts of $\mathrm{ad}x$: the first assertion follows by splitting into maps between eigenspaces, and the second by expanding repeated commutators with the nilpotent $x_n$.

First prove the linear trace version: if $L\subseteq \mathfrak{gl}(V)$ andthen $L$ is solvable. Fix $x\in [L,L]$ and let $V=⨁V_{\lambda }$ be its generalized eigenspace decomposition. Define $y$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. We do not assume that $y$, $x_s$ or $x_n$ belongs to $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}x$ has the form $(\lambda -\mu )I+N$ with $N$ nilpotent, whereas $\mathrm{ad}y$ is multiplication by $\overline{\lambda -\mu }$. Choose a polynomial $q$ with $q(d)=\bar{d}$ for every occurring eigenvalue difference $d$, and with all its derivatives of positive order up to the relevant nilpotency index zero at $d$. Such a polynomial exists by Hermite interpolation, or the Chinese remainder theorem for the pairwise coprime powers of $(t-d)$. At $d=0$ its value is zero, so $q(0)=0$. Expanding $q(dI+N)$ now givesBecause $x\in L$ and $[L,L]$ is an ideal, every positive power of $\mathrm{ad}x$ sends $L$ into $[L,L]$. The zero constant term therefore gives $[y,L]\subseteq [L,L]$.

Express $x$ as a sum of commutators ${\sum }_j[a_j,b_j]$ with $a_j,b_j\in L$. Cyclicity of trace and the hypothesis implyBut on $V_{\lambda }$, $xy$ has trace $\dim (V_{\lambda })|\lambda {|}^2$, since the nilpotent part has trace zero. Henceforcing every eigenvalue of $x$ to vanish. Thus every member of the derived algebra $[L,L]$ is a nilpotent endomorphism. By the Engel theorem, $[L,L]$ is nilpotent and therefore solvable; its derived series terminates, and adjoining the initial term $L$ shows that $L$ is solvable.

Conversely, if $L$ is solvable, the Lie theorem makes it upper triangular. Commutators of upper-triangular matrices have zero diagonal, and multiplying such a commutator by an upper-triangular matrix still has zero diagonal. Therefore the trace condition holds. Apply this equivalence to $L=\mathrm{ad}\mathfrak{g}$. Its trace pairing is the Killing form, and solvability of its image is equivalent to solvability of $\mathfrak{g}$: the kernel is its abelian centre, so a terminating derived series in the quotient terminates after at most one more step in $\mathfrak{g}$. This proves the solvability criterion.

For the semisimplicity criterion, let $R$ be the radical of the Killing form. Invariance makes $R$ a Lie algebra ideal. For $x,y\in R$, their adjoint actions on $\mathfrak{g}/R$ vanish, so computing trace in a basis adapted to $R$ gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$. The solvability criterion applied to $R$ shows it is a solvable ideal. Thus a semisimple Lie algebra has $R=0$ and a nondegenerate Killing form.

Conversely, any nonzero solvable ideal has a last nonzero term $A$ in its derived series; this is an abelian ideal of $\mathfrak{g}$. For $x\in A$, $y\in \mathfrak{g}$, the map $\mathrm{ad}x\mathrm{ad}y$ sends $\mathfrak{g}$ into $A$ and vanishes on $A$, so has trace zero. Hence $A\subseteq R$. A nondegenerate Killing form therefore rules out nonzero solvable ideals, exactly the definition of semisimplicity.

For classification, nondegeneracy of the Killing form provides the duality and root-space pairings used to extract a reduced crystallographic root system from a semisimple Lie algebra. Classification then reduces to the finite Dynkin diagrams and reconstruction of the Lie algebra from its root data.

Take curves $g(s),h(t)$ through the identity with tangent vectors $X,Y$. Form the local group commutator $c(s,t)=g(s)h(t)g(s{)}^{-1}h(t{)}^{-1}$. In a smooth coordinate chart at the identity, its leading mixed term defines the Lie bracket:for a matrix group, where the identity's constant term has zero derivative. More generally take this derivative in the chart with the identity sent to zero. Since the first derivatives vanish on the coordinate axes, changing charts changes the mixed coefficient only by the tangent-space coordinate transformation, so the definition is intrinsic.

For the displayed nilpotent matrices choose $g(s)=I+sX$ and $h(t)=I+tY$, which lie in ${\mathrm{SL}}_2$. Their inverses are $I-sX$ and $I-tY$. Direct multiplication givesIts mixed derivative is thereforeInterchanging the two curves giveswhose mixed derivative is $-H$. Thus $[X,Y]=-[Y,X]$ directly from the curve definition. The general matrix expansion likewise gives the commutator $XY-YX$.

The displayed identity is valid for a highest-weight vector $v$, satisfying $Xv=0$ and $Hv=kv$, not for arbitrary $v\in V$. For example, in the defining module with $k=1$, take $v=y$: then $XYv=0$ but the proposed right side for $n=1$ is $v\ne 0$.

For a highest-weight vector, the sl2 triple relations give $H{Y}^{j}v=(k-2j)Y^{j}v$. The operator identitycomes from expanding a commutator with a product. Apply it to $v$, using $Xv=0$. Each term becomes $(k-2(n-1-j))Y^{n-1}v$, so the sum isThis proves the intended sl2 highest-weight lowering formula. The identity valid for an arbitrary vector is instead

On the basis $v,Yv,\dots ,Y^{k}v$ of the irreducible module, $XY$ has eigenvalue $(j+1)(k-j)$ on $Y^{j}v$, for $0\le j\le k$. Summing gives the trace of a raising-lowering product in an irreducible sl2 module:The last equality follows by the formulas for the sums of the first $k$ integers and their squares.

Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.

Use Minkowski spacetime with $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and units $\hslash =c=1$. The transformation rotates the two components in their internal plane. In particular,Thus both terms in the Lagrangian density are invariant: this is an internal rotation symmetry of two real scalar fields, rather than a transformation of spacetime. The equal masses are essential to this symmetry.

The Noether theorem gives the Noether currentIndeed, the cross terms in its divergence cancel, and the two Klein-Gordon equations implyFor fields decaying sufficiently at spatial infinity, there is no outward current flux. The conserved Noether charge is therefore

In canonical quantization, the canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, with equal-time canonical commutation relationsConsequently $Q=\int d^{3}x({\pi }_1{\varphi }_2-{\pi }_2{\varphi }_1)$. Operators of different components commute, so this expression is Hermitian without an ordering correction. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$, $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $a_{k\mathbf{p}}=a_{\mathbf{p}}^k$. Substituting the oscillator expansions and using the Dirac delta function from the spatial integral givesThe terms with two annihilation operators cancel after $\mathbf{p}\mapsto -\mathbf{p}$; the terms with two creation operators cancel in the same way. In the remaining terms useThere is no commutator constant between the distinct components. The result is already in normal ordering:

For the charged oscillator basis of a scalar doublet, computeIt follows that $[Q,b_{\pm \mathbf{p}}^{†}]=\pm b_{\pm \mathbf{p}}^{†}$. The Fock vacuum satisfies $Q|0⟩=0$, so $b_{+\mathbf{p}}^{†}|0⟩$ has charge $+1$, and $b_{-\mathbf{p}}^{†}|0⟩$ has charge $-1$. To obtain a normalized one-particle state, replace the sharp momentum by ${\int }_{\mathbf{p}}f(\mathbf{p})b_{\pm \mathbf{p}}^{†}|0⟩$, where ${\int }_{\mathbf{p}}|f(\mathbf{p}){|}^2=1$. The charge is unchanged. Equivalently,The generator convention is $\delta {\varphi }_k=i\alpha [Q,{\varphi }_k]$, which reproduces the original rotation signs.

For stability of a two-scalar quartic potential, the relevant potential energy density isAlong either axis boundedness requires $\lambda \ge 0$; along ${\varphi }_1={\varphi }_2$ it requires $\lambda +\mu \ge 0$. Conversely the displayed sum is nonnegative whenever both conditions hold. With $m^2>0$ the origin is then the unique global minimum, including the equality cases. Hence the stable-vacuum conditions areStrict positivity of the quartic term away from the origin would instead require $\lambda >0$ and $\mu >-\lambda$. That stronger condition is unnecessary for a positive mass term. If $m=0$, the same non-strict conditions give boundedness, but the equality cases have flat directions and do not give an isolated classical vacuum.

Finally, the interaction changes the variation of the potential energy byThe interacting Euler-Lagrange equations consequently giveThus the same Noether charge is conserved for all field configurations precisely when $\mu =\lambda$. Then $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the internal rotation symmetry of two real scalar fields; imposing stability additionally requires $\lambda \ge 0$.

The anti-Hermitian operators represent the compact real form; its complexification also contains the Hermitian $H_i$ and the raising operators and lowering operators $E_{\pm }^r$, with $(E_{+}^r{)}^{†}=E_{-}^r$. Define the roots in the two-dimensional Cartan plane byEach has squared length one. For each $r=1,2,3$, setThe displayed commutators give $[J_3^{(r)},J_{\pm }^{(r)}]=\pm J_{\pm }^{(r)}$ and $[J_{+}^{(r)},J_{-}^{(r)}]=2J_3^{(r)}$. Thus these are the three root SU(2) subalgebras of SU(3).

The commuting Hermitian operators $H_1,H_2$ can be simultaneously diagonalized. A weight of a representation is a pair $\lambda =(h_1,h_2)$ for whichwhere $a$ labels independent states of that representation weight. The weight multiplicity is the dimension of this simultaneous eigenspace. Commuting $H_i$ past a ladder operator proveswith zero image allowed when the raised or lowered representation weight is absent.

Choose ${\alpha }_1,{\alpha }_2$ as positive simple roots. A highest-weight vector is a nonzero simultaneous eigenvector annihilated by $E_{+}^1,E_{+}^2,E_{+}^3$; the third condition follows from the first two and their commutator. In an irreducible representation, its highest weight $\Lambda$ has multiplicity one and repeated lowering generates the representation. The three root SU(2) subalgebras of SU(3) give nonnegative integer Dynkin labelsEquivalently, with ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$,These are the SU(3) highest-weight coordinates in the normalization used here.

The Weyl group is generated by the reflections $s_r(\lambda )=\lambda -2(\lambda \cdot {\alpha }_r){\alpha }_r$. The SU(2) Lie algebra root-string symmetry makes these reflections preserve both representation weights and their weight multiplicities. They generate the symmetries of an equilateral triangle: rotations through $120^{\circ }$ and three reflections. The outer polygon is the convex hull of the Weyl group orbit of $\Lambda$. When $p,q>0$ it is a hexagon with alternating side lengths $p,q$ in root-step units; when exactly one is zero it is a triangle; when both vanish it is a point. The representation conjugate to $(p,q)$ has labels $(q,p)$ and negative representation weights. For $p=q$, this gives an additional inversion symmetry and a regular hexagon with $60^{\circ }$ rotations. A general unequal-sided hexagon need not have that extra symmetry.

The shell multiplicities in an SU(3) weight diagram can be stated precisely. Let $P_{a,b}$ be the convex hull of the Weyl group orbit of $a{\omega }_1+b{\omega }_2$. For $\lambda$ in the root lattice coset $\Lambda +Q$,representation weights outside this coset or outside the outer polygon are absent. This states that boundary representation weights have multiplicity one, each inward shell adds one, and the innermost triangular region has constant multiplicity $\min (p,q)+1$. When $p=q$ the innermost region is a single point. All representation weights of a triangular diagram have multiplicity one. Counting the diagram gives $\dim (p,q)=(p+1)(q+1)(p+q+2)/2$. The examples below display both triangular orientations, a regular hexagon, an unequal-sided hexagon, a central multiplicity three, and the trivial diagram.

Take a local gauge transformation $g(x)$ acting on the adjoint field by ${\Phi }^{\prime }=g\Phi g^{-1}$. The connection convention $D_{\mu }={\partial }_{\mu }+[A_{\mu },\cdot ]$ is compatible withIndeed, differentiating ${\Phi }^{\prime }$ gives the extra commutator $[({\partial }_{\mu }g)g^{-1},{\Phi }^{\prime }]$, exactly cancelled by the inhomogeneous term in the transformed gauge potential.

For a direct proof of the Yang-Mills gauge transformation of curvature, put ${\Omega }_{\mu }=({\partial }_{\mu }g)g^{-1}$ and $B_{\mu }=g{A}_{\mu }{g}^{-1}$. Differentiation givesSubstituting $A_{\mu }^{\prime }=B_{\mu }-{\Omega }_{\mu }$ into the Yang-Mills field strength cancels all terms involving $\Omega$ and leavesThis direct calculation also applies when the adjoint action has a kernel.

Expand $D_{\mu }{F}_{\nu \rho }+D_{\nu }{F}_{\rho \mu }+D_{\rho }{F}_{\mu \nu }$. The mixed second derivatives of $A$ cancel in pairs. The first-derivative terms from differentiating $[A_{\nu },A_{\rho }]$ cancel the terms from $[A_{\mu },{\partial }_{\nu }{A}_{\rho }-{\partial }_{\rho }{A}_{\nu }]$. The remaining sum isby the Jacobi identity. Thus the gauge-theory Bianchi identity is

For the variation of the Yang-Mills action, write $a_{\mu }=\delta A_{\mu }$. Differentiating the curvature gives $\delta F_{\mu \nu }=D_{\mu }{a}_{\nu }-D_{\nu }{a}_{\mu }$. Invariance of the Killing form impliesThis is invariant integration by parts for a gauge covariant derivative. Using symmetry of the Killing form, antisymmetry of $F$, and compactly supported variations,Since the Killing form of a semisimple Lie algebra is nondegenerate and the variations are arbitrary, the Yang-Mills equations are

For the constant Yang-Mills theta term, symmetry under exchanging the two antisymmetric index pairs givesup to the boundary variation of the Yang-Mills theta term. Contracting the gauge-theory Bianchi identity with ${ϵ}^{\mu \nu \rho \sigma }$ makes the last integrand vanish. Therefore a constant theta term does not alter the bulk Yang-Mills equations: $D_{\mu }{F}^{\mu \nu }=0$. Its variation is a boundary term, so this conclusion uses compact support or boundary conditions that remove that term.

Finally apply $D^{\rho }$ to the gauge-theory Bianchi identity, obtainingThe adjoint covariant derivative obeys $[D_{\alpha },D_{\beta }]X=[F_{\alpha \beta },X]$. Commute $D^{\rho }$ past the other derivative. The differentiated divergences vanish by the Yang-Mills equations, leavingIn the last line the second commutator needs its sign tracked carefully: directly,Thus the simplified sum gives the covariant wave equation for the Yang-Mills field strengthAll spacetime derivatives and index manipulations here use the fixed flat metric.