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The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

Fix the current sign convention by defining the localized variation as ${\delta }_{ϵ}S=-\int d^{d}x({\partial }_{\mu }{ϵ}^a)j_a^{\mu }$. For a first-derivative Lagrangian invariant without a boundary term under constant parameters, this means $j_a^{\mu }=-\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )t_a\varphi$; include the usual improvement term if the constant variation is a total derivative. On solutions, arbitrary compactly supported parameters imply ${\partial }_{\mu }{j}_a^{\mu }=0$. The Noether charge is $Q_a=\int d^{d-1}x{j}_a^0$ and is conserved if the spatial flux vanishes. This sign convention matches the printed Ward identity; reversing the current also reverses the corresponding generator convention.

Assume an invariant regulated functional measure, invariant vacuum boundary conditions and no quantum anomaly. Changing variables in the normalized path integral gives $0=⟨{\delta }_{ϵ}X⟩+i⟨X{\delta }_{ϵ}S⟩$. Integration by parts then yieldsFor a product of scalar fields, the local Ward identity contact terms areAway from the insertions this is current conservation. Time-ordering or the distributional functional identity supplies the contact terms.

For the gauge theory write $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$ and use the left-acting BRST differential $s$, with ${\delta }_{ϵ}=ϵs$. It obeys the graded Leibniz rule andAssume the structure constants satisfy the Jacobi identity and the dot product is invariant; antisymmetry alone would not be enough. Since $c$ and $D_{\mu }c$ are odd, their Lie brackets are symmetric in these two arguments. ConsequentlyAlso $s{F}_{\mu \nu }=D_{\mu }{D}_{\nu }c-D_{\nu }{D}_{\mu }c=[F_{\mu \nu },c]$. The graded Jacobi identity gives $s[c,c]=0$, so $s^{2}c=0$; the other three fields have zero second variation immediately. For independent odd parameters, ${\delta }_{{ϵ}^{\prime }}{\delta }_{ϵ}=-{ϵ}^{\prime }ϵs^2=0$. These are off-shell BRST nilpotence identities because the Nakanishi-Lautrup field $b$ is retained.

The Yang-Mills theory variation is proportional to $F^{\mu \nu }\cdot [F_{\mu \nu },c]=0$. The gauge-fixing variation is $({\partial }^{\mu }b)\cdot D_{\mu }c$, while the ghost variation is its negative; the $b^2$ term does not vary. Equivalently these terms are $s\Psi$ for the gauge-fixing fermion $\Psi =({\partial }^{\mu }\bar{c})\cdot A_{\mu }+\xi \bar{c}\cdot b/2$. BRST nilpotence makes this expression invariant. Ghost-number scaling also leaves every term invariant: the ghost and antighost factors carry opposite weights.

Here are the two explicit currents in the same sign convention as the Ward identity. Localizing the even ghost parameter gives coefficient $({\partial }_{\mu }\theta )[\bar{c}\cdot D^{\mu }c-({\partial }^{\mu }\bar{c})\cdot c]$. Localizing the odd parameter, keeping it on the left, gives coefficientThe last sign comes from moving the odd parameter through the odd antighost derivative. Since the current was defined as minus this coefficient,These are the ghost-number Noether current and the BRST current in derivative-b gauge fixing. If the opposite Noether sign is used, both displayed currents acquire an overall minus sign. Integrating the gauge-fixing term by parts changes the Noether representative by the associated boundary improvement; mixing the two Lagrangian conventions without that improvement gives incorrect signs.

With no BRST anomaly, the conserved odd BRST charge has $Q_B^2=0$. The BRST cohomology identifies closed states $Q_B|\psi ⟩=0$ modulo exact states $Q_B|\chi ⟩$. Nilpotence puts every exact state in the closed space. In the usual indefinite gauge-fixed state space, a Hermitian BRST charge makes exact states orthogonal to closed states; the standard no-ghost/positivity assumptions then give a physical inner product on the quotient. The physical sector is its ghost-number-zero component,The ghost number assigns $+1$ to $c$, $-1$ to $\bar{c}$ and zero to gauge and auxiliary fields. Gauge-invariant observables and the chosen vacuum have ghost number zero; unphysical ghost excitations are removed in BRST pairs. Thus physical representatives are expected to satisfy $Q_G|\psi ⟩=0$. This zero-grading selection is part of the physical-state prescription, not a consequence of nilpotence alone.

Expand $T=T^{\mu }{}_{\nu }{\partial }_{\mu }\otimes d{x}^{\nu }$. Using the tensor Lie derivative of functions, vectors and covectors and its Leibniz rule givesThe contravariant slot has a minus sign and the covariant slot a plus sign.

For the commutator identity for Lie derivatives, put $D=[{\mathcal{L}}_X,{\mathcal{L}}_Y]-{\mathcal{L}}_{[X,Y]}$. The commutator of two tensor derivations is itself a tensor derivation, and $D$ commutes with tensor contractions. On a function, $Df=XYf-YXf-[X,Y]f=0$. On a vector field $Z$,by the Jacobi identity, which follows here by expanding the commutators of the operators $X,Y,Z$ acting on functions. For a type $(1,1)$ tensor, $T(Z)$ is a vector field, and contraction compatibility givesAs this holds for every $Z$, $DT=0$. ThereforeThe derivation argument also establishes the identity for arbitrary tensor types by applying it to covector–vector pairings and then to tensor products.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.

Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.

Let $P=({\omega }^{ab})$ be a smooth antisymmetric matrix field. It defines a Poisson bivector throughThis Poisson bracket is bilinear, antisymmetric and a derivation in each argument. It defines a Poisson manifold when it also satisfies the Jacobi identity. Apply that identity in the form $\{\{f,g\},h\}+\{\{g,h\},f\}+\{\{h,f\},g\}=0$ to the coordinate functions. Since $\{x^a,x^b\}={\omega }^{ab}$, it givesThis is the coordinate Jacobi condition for a Poisson bivector. It is also sufficient: in the Jacobi identity for arbitrary functions the terms containing second derivatives cancel by antisymmetry, leaving $J^{abc}({\partial }_{a}f)({\partial }_{b}g)({\partial }_{c}h)$. Thus the coordinate condition captures the whole obstruction.

Now suppose the Poisson bivector is nondegenerate. Write $S=P^{-1}$, so $S_{ab}{\omega }^{bc}={\delta }_a^c$, and define the 2-formAn overall minus sign in identifying the symplectic form depends on the convention for Hamiltonian vector fields; it does not affect the closure argument. Differentiating the inverse matrix givesContract the coordinate Jacobi identity with $S_{ia}{S}_{jb}{S}_{kc}$. Antisymmetry gives ${\omega }^{dc}{S}_{kc}=-{\delta }_k^d$, and similarly for the other terms, henceThe cyclic expression is precisely the coefficient of the exterior derivative $d\Omega$. ThereforeSince $S$ is antisymmetric and nondegenerate, $\Omega$ is a symplectic form. This proves the closure of the inverse of a nondegenerate Poisson bivector. The matrix entries ${\omega }^{ab}$ are scalar functions; the codomain in the source's matrix description should be read as the space of antisymmetric matrices, rather than a vector-valued individual entry.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.

Choose Lie-algebra generators with $[T_b,T_c]=f^a{}_{bc}{T}_a$ and an invariant orthonormal color metric. Since the coupling is placed outside the gauge kinetic term, it is absorbed into $A_{\mu }$ in the connection convention. The gauge field strength and adjoint covariant derivative areThe structure constants are antisymmetric in $b,c$ and obey the Jacobi identity. These definitions give $[D_{\mu },D_{\nu }]v=[F_{\mu \nu },v]$ on an adjoint-valued field. The field $c$ is a Grassmann-odd Faddeev-Popov ghost field; the Lie bracket in $D_{\mu }c$ contracts its color components in the same way as for any adjoint field.

Put the anticommuting parameter on the left and write $\delta \Phi =ϵs\Phi$. Then $s$ is a left-acting BRST differential: it is odd and satisfies the graded product rule $s(UV)=(sU)V+(-1{)}^{|U|}U(sV)$. The Faddeev-Popov ghost field $c$ and antighost field $b$ are odd, whereas $A_{\mu }$ and the Nakanishi-Lautrup field $h$ are even. Use $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $sb=h$, $sh=0$.

A total derivative in the Lagrangian variation must be included in the Noether current. For constant $ϵ$, the ghost covariant derivative has $s(D_{\mu }c)=0$, by the Jacobi identity and the odd statistics of $c$. The Yang-Mills term is invariant, and the remaining variation isThe term proportional to $\xi$ does not vary because $sh=0$.

To obtain the signs without an ambiguity about fermionic canonical momenta, now allow $ϵ=ϵ(x)$. In particular,The coefficients of ${\partial }_{\mu }ϵ$ in the Yang-Mills, ghost, and multiplier terms are respectivelyThe positive sign of the last ghost expression results from moving ${\partial }_{\mu }ϵ$ past the odd ${\partial }^{\mu }{b}^a$. Subtracting the total-derivative term $K^{\mu }$ therefore gives the Yang-Mills BRST Noether currentIndeed, the full localized variation is $\delta \mathcal{L}=({\partial }_{\mu }ϵ)j^{\mu }+{\partial }_{\mu }(ϵK^{\mu })$. The Noether theorem then gives ${\partial }_{\mu }{j}^{\mu }=0$ on the field equations. With spatial boundary terms vanishing, the corresponding BRST charge in four spacetime dimensions isIt is odd and has ghost number one. Overall generator phases depend on the convention relating this Noether charge to quantum commutators; one may use $s\mathcal{O}=i[Q_{\mathrm{BRST}},\mathcal{O}\}$. The displayed current fixes the classical Noether normalization for the left-parameter convention.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.