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Remove the common Grassmann parameter and write the BRST transformation as the odd derivation $s$:The last two equations immediately give $s^2{\bar{c}}^a=s^2{B}^a=0$. Applying $s$ to the ghost and using the graded product rule givesThe cancellation is the Jacobi identity for the $SU(2)$ structure constants together with anticommutation of the ghost fields. For the gauge field, the component calculation isThe graded product rule givesso the two terms cancel. Thus $s^2$ vanishes on every elementary field.

For two independent Grassmann parameters, ${\delta }_1={\eta }_{1}s$ and ${\delta }_2={\eta }_{2}s$ satisfy ${\delta }_1{\delta }_{2}O={\eta }_1{\eta }_2{s}^{2}O=0$. Since $s$ obeys the graded Leibniz rule, induction extends $s^{2}O=0$ from the generators $A,c,\bar{c},B$ to every polynomial $O(A,c,\bar{c},B)$. Hence the BRST transformations are nilpotent.

In components the BRST transformation isThe last two equations immediately give $s^2\bar{c}=s^{2}B=0$. Applying the graded Leibniz rule to $s^{2}c$ produces a sum of three ghost monomials whose coefficient is the Jacobi identity $f^{e[ab}{f}^{c]de}$, so $s^{2}c=0$. Finally,the derivative terms cancel by anticommutation of the Faddeev-Popov ghost fields, and the remaining terms again cancel by the Jacobi identity. Hence $s^2=0$ on every field: $s$ is a nilpotent Grassmann-odd differential.

For an affine algebraic group $G$, its Lie algebra is the tangent space at the identity,Equivalently, it is the space of left-invariant derivations of the coordinate ring $\mathbb{C}[G]$. The Lie bracket is the commutator of derivations,It is antisymmetric because $[Y,X]=-[X,Y]$. Associativity of composition givesafter all six triple products cancel in pairs, proving the Jacobi identity. This construction is the Lie algebra of an affine algebraic group.

Differentiate ${\mathrm{Ad}}_{e^{tX}}Y=e^{tX}Y{e}^{-tX}$ at zero. This gives the Adjoint representation of a Lie algebraThe Jacobi identity is exactlyso this is a Lie-algebra representation on $\mathfrak{o}(3)$.

Treat the BRST transformation $Q$ as an odd graded derivation. In matrix notation the first three transformations areThenafter substituting $Qc$ and using the Jacobi identity. Similarly, the two terms incancel because the ghost components anticommute and only the Lie-algebra commutator survives. Applying $Q$ once more to $Qc$ gives a sum proportional to $f^d{}_{e[a}{f}^e{}_{bc]}{c}^a{c}^b{c}^c$, which vanishes by Jacobi. Finally,Thus $Q^2=0$ on every field.

It is cleaner to use the action of the charge than to expand every double sum. The commutators from part a give, for $v^{\rho }=x^{\rho },p^{\rho }$, or any ${\alpha }_n^{\rho }$,Thus $M^{\mu \nu }$ acts on every mode in the vector representation of the Lorentz algebra. Apply the Jacobi identity to the action of $[M^{\mu \nu },M^{\kappa \lambda }]$ on every mode. The result agrees with the action ofThere is no extra scalar term: the orbital and oscillator pieces commute with one another, and direct use of their canonical commutation relations gives no central contribution. Hence the displayed operators obey precisely the standard Lorentz algebra.

Bilinearity and alternatingness ofare immediate. For three elements, the $\mathfrak{g}$ component of the Jacobi sum vanishes by the Jacobi identity in $\mathfrak{g}$. In the $V$ component, the coefficient of a vector such as $u$ isbecause the action is a Lie algebra representation; the other terms cancel cyclically in the same way. Hence the bracket satisfies Jacobi and defines the semidirect product of a Lie algebra and a module $\mathfrak{g}⋉V$.

A Lie bracket is bilinear, antisymmetric, and satisfies the Jacobi identity. For a basis $T_a$,The structure constant of a Lie algebra coefficients satisfy $f_{ab}{}^c=-f_{ba}{}^c$ and $f_{ab}{}^d{f}_{dc}{}^e+f_{bc}{}^d{f}_{da}{}^e+f_{ca}{}^d{f}_{db}{}^e=0$.

The curvature of $D=d+A$ isor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$. The covariant exterior derivative and the Jacobi identity give the Bianchi identity

Since the Hodge star operator satisfies ${\ast }^2=1$ on Euclidean two-forms,Using the opposite choice of sign as well givesEquality holds precisely for a self-dual or anti-self-dual Yang-Mills instanton, with the sign selected by that of $k$.

Now assume ${\partial }_0{A}_{\mu }=0$ and write $\Phi =A_0$. ThenThe spatial Bianchi identity and the self-duality equations becomewhere the upper four-dimensional sign gives the first displayed reduced sign under the orientation used here. Consequently

Gauge invariance of the inner product givesFor $w=1-|\Phi {|}^2$,Since $w\to 0$ at infinity, the weak maximum principle for elliptic operators excludes a negative interior minimum. Hence

With $d^{4}x=d{x}^0{d}^{3}x$, direct decomposition givesBecause $D_i{B}_i=0$,so the Pontryagin density reduces to the surface charge $4\pi N$ per unit $x^0$. The conventional three-dimensional energyobeys the Bogomolny bound $E_3\ge 4\pi |N|$, saturated by the Bogomolny-Prasad-Sommerfield monopole equation. The four-dimensional action density per unit $x^0$ is $2E_3$; signs relating $k$ and $N$ depend on the self-duality and orientation convention.

For the hedgehog ansatz for a monopole in the question, direct differentiation with $[T_a,T_b]=-{ϵ}_{abc}{T}_c$ givesOn a sphere of radius $r$,ThereforeThe boundary conditions $f\to 1$ and $r^2\alpha \to 1$ make the integrand equal to $r^{-2}+o(r^{-2})$. Thus

The Adjoint representation of a Lie algebra isIt is linear. The Jacobi identity givesso it is a Lie algebra representation.

A smooth antisymmetric contravariant two-tensor defines a biderivation $\{F,G\}=\omega (dF,dG)$. It is a Poisson bivector when this bracket satisfies the Jacobi identity. Its rank may vary; where it is nondegenerate it inverts to a symplectic form. The coordinate Jacobi condition for a Poisson bivector makes the integrability constraint explicit.

A Poisson manifold is a smooth manifold with a bilinear bracket on smooth functions that is antisymmetric, obeys the Jacobi identity, and is a derivation in each argument. Its Poisson bivector may have nonconstant rank, unlike the inverse of a symplectic form.

For field functionals, a Poisson operator $J$ is an operator whose displayed pairing of variational derivatives defines a Poisson bracket. It must be skew under the integral pairing and satisfy the functional Jacobi identity. A constant skew differential operator, such as $J={\partial }_x$ with periodic or vanishing boundary conditions, gives the simplest examples. The Hamiltonian evolution is $u_t=J(\delta H/\delta u)$.

Two Poisson structures are compatible when every linear combination ${\Pi }_1+\lambda {\Pi }_2$ is a Poisson structure. This family is a Poisson pencil. In terms of the Schouten-Nijenhuis bracket, compatibility means $[{\Pi }_1,{\Pi }_2]=0$, in addition to the two individual Jacobi identities. Constant Poisson bivectors are compatible. Over a real manifold, complex parameter values refer to the complexification of its function algebra.

A Poisson structure is an antisymmetric smooth bivector field whose induced function bracket satisfies the Jacobi identity. In coordinates its condition is the vanishing cyclic sum of ${\omega }^{dc}{\partial }_d{\omega }^{ab}$. Applying Jacobi to coordinate functions proves necessity; expanding arbitrary brackets cancels the second derivatives and proves sufficiency. A function bracket-commuting with every function is a Casimir function of a Poisson manifold.

For the Levi-Civita connection, the curvature form of a connection is the endomorphism-valued two-formIt has tensoriality in all three arguments. Its first Bianchi identity follows from torsion-freeness: the cyclic sum is ${\sum }_{\mathrm{cyc}}({\nabla }_X[Y,Z]-{\nabla }_{[Y,Z]}X)={\sum }_{\mathrm{cyc}}[X,[Y,Z]]=0$ by the Jacobi identity.

For root spaces ${\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }$, the Jacobi identity gives $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subset {\mathfrak{g}}_{\alpha +\beta }$ when $\alpha +\beta \ne 0$. If the bracket is nonzero, that nonzero weight must be a root. When $\beta =-\alpha$, the bracket instead lies in the Cartan subalgebra and can be nonzero; zero is not a root. Thus the common nonzero-bracket implication requires the opposite-root exception.

If $D:\mathfrak{b}\to \mathrm{Der}(\mathfrak{a})$ is a Lie algebra homomorphism into the derivations of a Lie algebra, the semidirect product has underlying vector space $\mathfrak{a}\oplus \mathfrak{b}$ and bracket $[(a,b),(a^{\prime },b^{\prime })]=([a,a^{\prime }]+D_b{a}^{\prime }-D_{b^{\prime }}a,[b,b^{\prime }])$. The homomorphism and derivation identities ensure the Jacobi identity. The first factor is an ideal; the second is a subalgebra whose action records the twisting. The diamond Lie algebra arises by letting a one-dimensional algebra act on a Heisenberg Lie algebra.

// Target: linear-algebra.bigb

An oriented three-dimensional Lie algebra has its antisymmetric structure constants encoded by the displayed matrix, whose symmetric part is $n^{bd}$ and whose antisymmetric part uniquely defines the covector $a_e$. In a volume-normalized basis, the inverse is $C_a{}^b{}_c=\frac{1}{2}{ϵ}_{acd}{n}^{bd}+\frac{1}{2}({\delta }_c^b{a}_a-{\delta }_a^b{a}_c)$. The only independent triple in the Jacobi identity has coefficient $-\frac{1}{2}{n}^{be}{a}_e$, so Jacobi holds exactly when $na=0$. This decomposition is useful in classifying three-dimensional Lie algebras.