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In the matrix expression, $t^a={\lambda }^a/2$ denotes a fundamental generator, whereas the component adjoint covariant derivative uses adjoint generators. With $\mathrm{Tr}(t^d{t}^h)={\delta }^{dh}/2$,Using antisymmetry to write $f^{acd}=f^{dac}$ and $f^{bed}=f^{dbe}$ gives exactly the component mass term. This is the gauge-boson mass matrix from an adjoint scalar.

For ${\Phi }_0=v\mathrm{diag}(1,1,-2)$, the Gell-Mann matrices $t^1,t^2,t^3,t^8$ commute with ${\Phi }_0$. Their gauge bosons are therefore massless. The generators $t^4,t^5,t^6,t^7$ connect the first two eigenspaces to the third; the eigenvalue difference is $3v$. Directly taking the matrix trace giveswith all remaining entries zero. ThusThe factor $1/2$ is the canonical real-vector mass term normalization. The unbroken Lie algebra is $\mathfrak{su}(2)\oplus \mathfrak{u}(1)$, giving the usual description SU(2) × U(1). More precisely, the centralizer of the vacuum expectation value inside $SU(3)$ isThis global quotient refines the adjoint-Higgs breaking of SU(3) to SU(2) and U(1) without changing the local gauge boson spectrum.

The Jacobi identity for the Poisson bracket impliesFor first-class constraints with linearly independent differentials this gives $f_{[ij}{}^l{f}_{k]l}{}^m=0$, the Jacobi identity for the structure constants of the constraint algebra. Independence is an implicit assumption: for constraints obeying identities or vanishing identically, only the contracted identity follows, and arbitrary coefficients multiplying such constraints need not satisfy a Lie algebra identity.

With $G={ϵ}^i{\phi }_i$ and $\delta F=\{F,G\}$, the canonical variables transform asTo fix the sign convention, vary the phase-space action directly:Thus action invariance requires $\delta {\lambda }^i={\dot{ϵ}}^i+{ϵ}^j{\lambda }^k{f}_{jk}{}^i$. For $F^i={\lambda }^i-{\bar{\lambda }}^i$, the Faddeev-Popov determinant is that ofUsing anticommuting Faddeev-Popov ghost fields, the invariant-convention result isThe original PDF has a sign error in the stated multiplier transformation; the TeX transcription has the consistent plus sign. Keeping the PDF's displayed minus sign mechanically would instead give $i\int dt{b}_i({\dot{c}}^i-c^j{\bar{\lambda }}^k{f}_{jk}{}^i)$. That expression exponentiates the determinant of the printed transformation, but that transformation does not preserve the stated action with the canonical convention above. It cannot be used as the invariant result without changing another convention consistently.

As for the particle, constant multiplier moduli and any residual zero mode in field theory must be handled separately; the constant gauge fixing is understood locally on the gauge orbit.

A simple Lie algebra is a nonabelian Lie algebra whose only ideals of a Lie algebra are $0$ and the whole algebra.

Use the standard basis $e,f,h$ of the sl2 Lie algebra, withLet $I\ne 0$ be an ideal and choose $0\ne x=ae+bf+ch\in I$. Since $I$ is invariant under the Adjoint representation, it is invariant under the linear operator $\mathrm{ad}h$. The three basis vectors are eigenvectors of $\mathrm{ad}h$ with distinct eigenvalues $2,-2,0$. Applying the corresponding polynomial spectral projections to $x$ shows that $I$ contains at least one nonzero multiple of $e$, $f$, or $h$.

If $e\in I$, then $[f,e]=-h\in I$ and $[f,h]=2f\in I$; the cases $f\in I$ and $h\in I$ are identical after taking brackets with the other basis vectors. Hence $e,f,h\in I$, so $I={\mathfrak{sl}}_2(\mathbb{C})$. Therefore ${\mathfrak{sl}}_2(\mathbb{C})$ is simple.

The Killing form of a finite-dimensional Lie algebra $\mathfrak{g}$ is the symmetric bilinear formThe cyclicity of the trace makes it invariant:Consequently its radical of a bilinear form $R=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal, since $x\in R$ implies $\kappa ([y,x],z)=\kappa (x,[z,y])=0$ for all $y,z\in \mathfrak{g}$.

If $\mathfrak{g}$ is simple, then $R$ is either $0$ or $\mathfrak{g}$. In the second case the Killing form vanishes identically, so the stated solvability criterion makes $\mathfrak{g}$ a solvable Lie algebra. A nonabelian simple Lie algebra cannot be solvable: its first derived algebra is a nonzero ideal and hence equals $\mathfrak{g}$, after which the derived series never reaches zero. Thus $R=0$, and the Killing form of a simple Lie algebra is nondegenerate.

A Lie group is a group and smooth manifold whose multiplication and inversion are smooth. A Lie algebra is a vector space with a bilinear alternating bracket satisfying the Jacobi identity. For a Matrix Lie group $G\subset GL(m,\mathbb{C})$, a logarithmic chart of a matrix Lie group near $g$ sends $h$ to $\log (g^{-1}h)\in \mathfrak{g}=T_{I}G$; the matrix exponential is its local inverse, and the Baker--Campbell--Hausdorff formula makes the local group operations smooth.

For $B_1,B_2\in \mathfrak{g}$, consider the group commutatorIts logarithm takes values in the vector space $\mathfrak{g}$, and expansion at $(0,0)$ givesThus $\mathfrak{g}$ is closed under the commutator. Bilinearity, alternation, and the Jacobi identity follow from matrix multiplication, so this proves that the Lie algebra of a matrix Lie group has

A principal bundle $\pi :P\to M$ with structure group $G$ is a smooth fiber bundle with a free right $G$-action, each fiber a single orbit, and equivariant local trivializations ${\pi }^{-1}(U)\cong U\times G$.

For the right action of the unitary group on $GL(n,\mathbb{C})$, defineThe matrix $g{g}^{\ast }$ is a positive-definite matrix and a Hermitian matrix, and $q(gu)=q(g)$ for $u\in U(n)$. The polar decomposition of an invertible complex matrix gives the unique factorizationConsequentlyis a diffeomorphism. If $W\subset H(n)$ is a neighborhood of zero and $N=q^{-1}(W)$, then $N$ is an open neighborhood of $I$, it is a union of complete orbits, andMoreover, two matrices have the same value of $q$ exactly when they differ by right multiplication by a unitary matrix. Thus $q$ induces the smooth identificationunder which $\pi$ is projection $H(n)\times U(n)\to H(n)$. This is the general linear group modulo the unitary group, and $\pi$ is in fact a globally trivial principal $U(n)$-bundle.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

Let $G$ be a simple Lie group, let $T^a$ be Hermitian matrices for a finite-dimensional unitary representation $R$, and normalizeA matter field $\psi$ transforms locally as ${\psi }^{\prime }(x)=U(x)\psi (x)$, where $U(x)=e^{ig{ϵ}^a(x)T^a}$. An ordinary derivative of $\psi$ does not transform covariantly because it differentiates $U$. Introduce a gauge field $A_{\mu }=A_{\mu }^a{T}^a$ and the gauge covariant derivativeDemanding $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$ determines the Yang-Mills gauge transformationTo first order in $ϵ$,

The gauge field strength is defined by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$:or, in components,Covariance of the commutator givesThe commutator term distinguishes Yang-Mills theory from an Abelian gauge theory and produces cubic and quartic gauge-boson interactions.

For a Dirac field of mass $m$ in $R$, the Lagrangian isEquivalently, the gauge term is proportional to $-\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$. The cyclic property of the trace and $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$ make it invariant. Unitarity gives ${\bar{\psi }}^{\prime }=\bar{\psi }{U}^{-1}$, while $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$, so both the matter kinetic term and mass term are invariant. A complex scalar $\varphi$ in a unitary representation may instead be coupled throughprovided the scalar potential $V$ is $G$-invariant.

The simplicity assumption means that the Lie algebra $\mathfrak{g}$ is nonabelian and has no proper nonzero ideal of a Lie algebra. Its Adjoint representation is therefore irreducible, and every invariant symmetric bilinear form is proportional to the Killing form. Consequently the pure gauge kinetic term has one overall gauge coupling for a simple factor. The theory has no independent Abelian gauge direction; if the gauge algebra were a direct sum of simple and Abelian ideals, each factor could instead carry its own coupling. A simple group may still have a discrete center, but this does not add a gauge boson because gauge bosons are indexed by the Lie algebra.

Because the bilinear form from part (ii) is invariant, $\rho (SU(2))\subseteq O(3)$. The mapis continuous. Since $SU(2)$ is connected and $\det \rho (I_2)=1$, it is identically one. Hence

To find the kernel, suppose $U$ commutes with every traceless skew-Hermitian matrix. Commutation withforces $U$ to be diagonal. Commutation withthen forces its two diagonal entries to agree. Thus $U=z{I}_2$. Since $U\in SU(2)$,and therefore

It remains to prove surjectivity. The derived representation at the identity isIf this vanishes for every $A$ in the SU(2) Lie algebra, the same commutant calculation makes $X$ scalar; tracelessness then gives $X=0$. Thus $d{\rho }_{I_2}$ is injective. Its domain and codomain are both three-dimensional real Lie algebras, so it is an isomorphismThe inverse function theorem now shows that the image of $\rho$ contains a neighbourhood of the identity in $SO(3)$. Therefore the image is an open subgroup. The special orthogonal group $SO(3)$ is connected, since every rotation can be deformed continuously to the identity by reducing its rotation angle. A connected topological group has no proper open subgroup, soTogether with the kernel calculation, this is the Adjoint double cover from SU(2) to SO(3).

The space $\mathrm{End}(V)$ becomes a Lie algebra under the commutatorA Lie subalgebra $L$ is abelian when $[L,L]=0$; nilpotent when its lower central series of a Lie algebra ${\gamma }_{r+1}=[L,{\gamma }_r]$ reaches zero; and soluble when its derived series of a Lie algebra $L^{(r+1)}=[L^{(r)},L^{(r)}]$ reaches zero.

If $H⊊L$ and $L$ is nilpotent, choose the least $r$ with ${\gamma }_r(L)\subseteq H$. Then ${\gamma }_{r-1}(L)⊈H$, whileThus the normalizer of a Lie subalgebra strictly contains $H$. If $H$ is maximal proper, its normalizer must be all of $L$, so $H$ is an ideal. Solubility is insufficient: in the two-dimensional affine Lie algebra $L=⟨x,y⟩$ with $[x,y]=y$, the maximal subalgebra $\mathbb{C}x$ is not an ideal.

A derivation of a Lie algebra is a linear map $D$ satisfyingIt is inner when $D=\mathrm{ad}z$ for some $z\in L$.

Every nonzero finite-dimensional nilpotent Lie algebra has an outer derivation of a nilpotent Lie algebra. Choose a codimension-one maximal subalgebra $K$; it is an ideal by the result above, and write $L=K\oplus \mathbb{C}z$. The centralizer $C_L(K)$ is nonzero because it contains $Z(L)$. Let $n$ be largest such thatand choose $z_0\in C_L(K)\setminus {\gamma }_{n+1}(L)$. DefineBecause $K$ is an ideal and $z_0$ centralizes $K$, the derivation identity holds on $K\times K$ and on $z\times K$, hence everywhere. If $D=\mathrm{ad}x$, then $D(K)=0$ would put $x$ in $C_L(K)\subseteq {\gamma }_n(L)$, socontrary to the choice of $z_0$. Thus $D$ is outer.

The analogous assertion fails for soluble algebras. In the affine example, a derivation hasand equals $\mathrm{ad}(dx-by)$. Thus every derivation is inner although $L$ is nonzero and soluble.

As a vector space, the Lie algebra of $G$ is $\mathfrak{g}=T_{e}G$. For $\xi \in \mathfrak{g}$, definewhere $L_g$ is Left translation on a Lie group. This vector field is smooth and left-invariant because $(d{L}_h{)}_g{l}_{\xi }(g)=l_{\xi }(hg)$. The assignment $\xi \mapsto l_{\xi }$ is linear and injective by evaluation at $e$. Conversely, every left-invariant vector field $V$ satisfies $V(g)=(d{L}_g{)}_{e}V(e)$, so it equals $l_{V(e)}$. Hence the map is an isomorphism.

The Lie bracket of vector fields is bilinear, alternating, satisfies the Jacobi identity, and is natural under diffeomorphisms. Therefore the bracket of two left-invariant vector fields is left-invariant. DefineThenand the inherited bilinearity, alternation, and Jacobi identity make $T_{e}G$ a Lie algebra.

The special orthogonal group in three dimensions isDifferentiating $A(t{)}^{T}A(t)=I$ at the identity shows that its Lie algebra is the space of skew-symmetric matrices. A convenient basis iswith $[J_a,J_b]={ϵ}_{abc}{J}_c$.

The infinitesimal action of $\exp (t{J}_a)$ on a point $x\in {\mathbb{R}}^3$ is $J_{a}x=e_a\times x$. It therefore generates the vector fieldFundamental vector fields for this left action form an antihomomorphism with the stated convention, and direct differentiation givesso their span is closed under the Lie bracket of vector fields.

The brackets $\{x,y\}=z$, $\{y,z\}=x$, and $\{z,x\}=y$ define the rotational Lie-Poisson structure on R3. With the convention that a Hamiltonian vector field acts by $X_H(f)=\{f,H\}$,Hence the required Hamiltonians are simplyThe quadratic functionsatisfies $\{F,x_a\}=0$ for all $a$, so it is a Casimir function of a Poisson manifold. Its nonzero regular level sets $F=R^2$ are spheres. The Poisson tensor has rank two there and is tangent to each level set, so it inverts to a symplectic form; each sphere is a symplectic leaf. Rotations preserve both $F$ and the alternating tensor ${ϵ}_{abc}$, hence preserve the restricted Poisson tensor and its inverse symplectic form. The $SO(3)$ action therefore restricts to a symplectic action on every sphere $S_R^2\subset {\mathbb{R}}^3$.

For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.

A finite-dimensional Lie algebra $L$ is semisimple when its solvable radical is zero, equivalently when it has no nonzero solvable ideals.

We prove the Weyl complete reducibility theorem. Induct on the dimension of a finite-dimensional $L$-module $V$. It is enough first to split a submodule $W$ of codimension one. The one-dimensional quotient is trivial because a semisimple Lie algebra is perfect. By induction, $W$ is a direct sum of irreducible modules. A Casimir element formed using the Killing form commutes with the $L$-action, acts as zero on every trivial summand, and acts by a nonzero scalar on every nontrivial irreducible summand. Its image is therefore the sum $W_1$ of the nontrivial summands, while its kernel contains the trivial summands $W_0$ and maps onto $V/W$. ThusInside $\ker \Omega$, choose a lift $v$ of a basis of $V/W$. For $x\in L$, $xv\in W_0$, and $L$ acts trivially on $W_0$. Hence $[x,y]v=0$ for all $x,y\in L$. Since $L=[L,L]$, actually $xv=0$ for every $x$, so $\mathbb{C}v$ is the required invariant complement.

This codimension-one case implies the general case. For an arbitrary submodule $W\subseteq V$, letwith the natural Hom representation. The maps vanishing on $W$ form an $L$-submodule $X_0$ of codimension one. Splitting $X_0$ supplies an $L$-equivariant $f$ with $f{|}_W=I_W$. Thenso every invariant subspace has an invariant complement and every finite-dimensional representation is completely reducible.

For the requested example, embed ${\mathfrak{sl}}_2$ as the upper-left $2\times 2$ block in the special linear Lie algebra ${\mathfrak{sl}}_3$. Under the restricted Adjoint representation,Here $V_n$ is the irreducible ${\mathfrak{sl}}_2$-module of highest weight $n$. The first summand is the three-dimensional adjoint module, the second is trivial, and the last two are the two-dimensional defining module and its dual, which are isomorphic. This explicit direct sum demonstrates complete reducibility.

For $\xi$ in the Lie algebra $\mathfrak{g}$, set ${\Phi }^t=R_{\exp (t\xi )}$. The one-parameter subgroup law gives the flow law, andso this is the global flow of the left-invariant vector field $l_{\xi }$.

For $f(g)=R_g^{\ast }\alpha$, every tangent vector at $g$ is the initial velocity of $g\exp (t\xi )$ for some $\xi$. Since $R_{g\exp (t\xi )}=R_{\exp (t\xi )}\circ R_g$,Thus all ${\mathcal{L}}_{l_{\xi }}\alpha$ vanish exactly when every derivative of $f$ vanishes. This is equivalent to $f$ being a locally constant function.

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.

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

The first term $g{A}_{\mu }{g}^{-1}$ lies in $\mathfrak{g}$ because the adjoint action preserves the Lie algebra. For the second, fix $x$ and consider the group curve $s\mapsto g(x+s{e}_{\mu })g(x{)}^{-1}$ through the identity. Its tangent at zero is $({\partial }_{\mu }g)g^{-1}$, so this is also in $\mathfrak{g}$. Since a Lie algebra is a vector space, $A_{\mu }^{\prime }\in \mathfrak{g}$.

The Coleman–Mandula theorem says that, under its assumptions on a nontrivial analytic relativistic S-matrix and the particle spectrum, every continuous bosonic symmetry algebra is a direct sum of the Poincare algebra and an internal symmetry algebra. Supersymmetry becomes possible by weakening the assumption that the symmetry algebra is an ordinary Lie algebra: a Z2-graded Lie superalgebra admits odd generators whose bracket is an anticommutator. The Haag–Łopuszański–Sohnius theorem then classifies the allowed extension and leads to the Super-Poincaré algebra.

The special unitary group is $SU(2)=\{U\in M_2(\mathbb{C}):U^{†}U=I,\det U=1\}$. Differentiating these equations at the identity shows that its Lie algebra isFor $T_i=-i{\sigma }_i/2$, the Pauli-matrix identity $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$ givesso the structure constants in this basis are $c^k{}_{ij}={ϵ}_{ijk}$.