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An affine algebraic group is a linearly reductive algebraic group if every finite-dimensional rational representation is completely reducible; equivalently the functor of invariants is exact. It is a geometrically reductive algebraic group if, whenever $0\ne v\in V^G$, there is a homogeneous invariant polynomial $F\in k[V{]}^G$ of positive degree with $F(v)\ne 0$. Linear reductivity permits degree one; geometric reductivity allows higher degree.

Over the algebraically closed field, an algebraic torus is $T\cong ({\mathbb{G}}_m{)}^r$. Its coordinate ring is the Laurent polynomial ring. A rational action writes the coaction of a vector as a finite sum $\delta (v)={\sum }_{\chi }{v}_{\chi }\otimes t^{\chi }$. The coaction identity and linear independence of Laurent monomials give $\delta (v_{\chi })=v_{\chi }\otimes t^{\chi }$ and $v={\sum }_{\chi }{v}_{\chi }$. ThusEach weight space has scalar action, so any vector-space complement within each weight space is invariant. This proves complete reducibility, and hence tori are linearly reductive in every characteristic. The argument uses characters themselves, not their Lie algebra differentials, which could coincide modulo $p$.

We now prove geometric reductivity of $G=S{L}_2(k)$ by determinant separation for positive-characteristic SL2. Fix a rational $G$-module $V$ and a nonzero fixed vector $v$. For the diagonal torus, choose a weight-zero functional $l\in V^{\ast }$ with $l(v)=1$, using the torus decomposition just proved. Define matrix coefficientsThese are regular functions on $G$, and $\Phi (hw)(g)=\Phi (w)(h^{-1}g)$ makes $\Phi$ equivariant for left translations. Moreover $\Phi (w)(gt)=\Phi (w)(g)$ because $l$ is torus-invariant. Thus $\Phi$ maps into $k[G{]}^T$ for the right torus action, and $\Phi (v)=1$.

Write a matrix's two columns as $a=(a_0,a_1)$ and $b=(b_0,b_1)$, with $\Delta =a_0{b}_1-a_1{b}_0=1$. Right torus multiplication scales these columns by $t$ and $t^{-1}$. Every right-torus-invariant regular function has a weight-zero polynomial lift, hence a sum of polynomials of bidegrees $(j,j)$ in $(a,b)$. To see that a weight-zero lift exists, take any lift and retain its zero-weight component; the ideal $(\Delta -1)$ is torus-stable and the quotient map preserves weights. Multiply each piece by ${\Delta }^{n-j}$ to put the entire sum in bidegree $(n,n)$ without changing its restriction to $G$.

Let $P_n$ be the space of degree-$n$ polynomials in one column. Restriction gives an injective equivariant map $P_n\otimes P_n\to k[G{]}^T$. Indeed if a bidegree-$(n,n)$ polynomial vanishes on determinant-one matrices, scale the first column of any invertible matrix to make its determinant one. Homogeneity then makes the polynomial vanish on every invertible matrix, a dense open subset, so it is zero. Denote the image by $R_n$. These spaces increase because multiplication by $\Delta$ realizes $R_{n-1}\subset R_n$, and their union is $k[G{]}^T$. The finite-dimensional image $\Phi (V)$ therefore lies in one $R_n$, and in any sufficiently larger filtration piece.

Choose $q=p^s$ with $q-1$ at least that bound, and set $n=q-1$. The identity $(1+z{)}^q=1+z^q$ in characteristic $p$ givesThe invariant tensortherefore defines an invertible equivariant linear map $D:P_n^{\ast }\to P_n$: its matrix is anti-diagonal with all entries nonzero. Interpret any tensor $B\in P_n\otimes P_n$ as a map $P_n^{\ast }\to P_n$ and send it to $B{D}^{-1}\in \mathrm{End}(P_n)$. Equivariance of $D$ makes this identification intertwine the diagonal tensor action with conjugation on endomorphisms. Crucially, $D$ restricts to the constant function $1$ on $G$, and is sent to the identity endomorphism.

Under the resulting equivariant linear map $V\to \mathrm{End}(P_n)$, the vector $v$ is sent to $I$. Since $\dim P_n=n+1=q$, the endomorphism determinant is a homogeneous polynomial of positive degree $q$, invariant under conjugation. Pull it back to $V$. The resulting polynomial satisfiesThis constructs the required invariant for every fixed nonzero vector and proves $S{L}_2(k)$ is geometrically reductive, including characteristic two. No averaging by a group order or division by $q$ is used; those operations would fail in the field's characteristic.

For a finite-dimensional Lie algebra, the Killing form is the symmetric bilinear formIn the root-space decomposition of this semisimple Lie algebra, $\mathrm{ad}H$ acts as zero on the abelian Cartan subalgebra and as multiplication by $\alpha (H)$ on ${\mathfrak{g}}_{\alpha }$. For $H,K\in \mathfrak{h}$, their product thus acts as multiplication by $\alpha (H)\alpha (K)$ on each root space. Every root space here is one-dimensional. Taking the trace over the two-dimensional Cartan subalgebra and the eight root spaces provesFor a general weight decomposition the trace would include the corresponding weight multiplicities; the root multiplicities are one in the semisimple setting here.

A complex semisimple Lie algebra has zero centre, so its Adjoint representation is faithful. Its absolute Jordan decomposition is the unique expression $x=x_s+x_n$ with $[x_s,x_n]=0$, $\mathrm{ad}{x}_s$ diagonalizable and $\mathrm{ad}{x}_n$ nilpotent. Existence follows by applying part (ii) to the adjoint embedding, and uniqueness follows from uniqueness of the matrix Jordan–Chevalley decomposition. The representation-compatibility theorem states that for every finite-dimensional Lie algebra representation $\rho$, $\rho (x_s)$ and $\rho (x_n)$ are respectively the semisimple and nilpotent parts of $\rho (x)$. Thus these parts are intrinsic to the Lie algebra, independent of a chosen matrix embedding, and their semisimplicity and nilpotence are preserved in every finite-dimensional representation.

We prove closure using complete reducibility of semisimple Lie algebra representations, rather than assuming that a Lie subalgebra is closed under polynomials in its elements. Write the matrix Jordan–Chevalley decomposition as $X=S+N$. On $\mathrm{End}(V)$, the commutator operator $\mathrm{ad}S$ is diagonalizable: if $S$ has eigenspaces $V_{\lambda }$, then its eigenvalue on $\mathrm{Hom}(V_{\lambda },V_{\mu })$ is $\mu -\lambda$. The operator $\mathrm{ad}N$ is nilpotent: if $N^r=0$, thenThe two operators commute because $[S,N]=0$. Uniqueness of the Jordan–Chevalley decomposition therefore says that $\mathrm{ad}S$ is the semisimple part of $\mathrm{ad}X$. Since a matrix's semisimple part is a polynomial in that matrix, and $\mathfrak{g}$ is invariant under $\mathrm{ad}X$, it follows thatNow regard $\mathrm{End}(V)$ as a Lie algebra representation of $\mathfrak{g}$ under commutators. By complete reducibility of semisimple Lie algebra representations, there is an invariant complement $M$ with $\mathrm{End}(V)=\mathfrak{g}\oplus M$. Write $S=Y+C$, with $Y\in \mathfrak{g}$ and $C\in M$. For $A\in \mathfrak{g}$, the normalization property gives $[C,A]=[S,A]-[Y,A]\in \mathfrak{g}$, while invariance of $M$ gives $[C,A]\in M$. Thus $[C,A]=0$: $C$ commutes with the whole Lie algebra.

Decompose $V={⨁}_j{V}_j$ into irreducible $\mathfrak{g}$-modules. Each $V_j$ is preserved by $X$, hence by its polynomial parts $S,N$, and also by $Y$; therefore it is preserved by $C$. By the Schur lemma, $C{|}_{V_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every element is a sum of Lie brackets and has trace zero on every finite-dimensional representation. Consequently $\mathrm{tr}(X{|}_{V_j})=\mathrm{tr}(Y{|}_{V_j})=0$. Also $N{|}_{V_j}$ is nilpotent, so it has trace zero; henceThus every $c_j$ is zero, $C=0$, and $S=Y\in \mathfrak{g}$. Finally $N=X-S\in \mathfrak{g}$. We have proved

Use the same nonsemisimple matrix from part (i) and the one-dimensional Lie subalgebraIt is a Lie subalgebra because $[aX,bX]=0$ for all scalars $a,b$. Its Jordan–Chevalley decomposition has $X_s=I_2$ and $X_n=X-I_2\ne 0$. If $I_2=cX$, comparing off-diagonal entries forces $c=0$, which contradicts the diagonal entries. If $X_n=cX$, taking traces gives $0=2c$, again forcing $c=0$ although $X_n\ne 0$. ThusThe semisimplicity assumption on the Lie algebra in part (ii) is therefore essential.

For a finite-dimensional Lie algebra, write $A_X=\mathrm{ad}X$ for its Adjoint representation. The Killing form is the symmetric bilinear formThe Jacobi identity gives $A_{[X,Y]}=[A_X,A_Y]$. Using the cyclic property of the trace,Thus it is an invariant bilinear form on a Lie algebra. If $[T_a,T_b]=f_{ab}{}^c{T}_c$, the adjoint matrix has entries $(A_{T_a}{)}^c{}_b=f_{ab}{}^c$. Its trace product yields the structure constants of a Lie algebra formulaA semisimple Lie algebra has zero solvable radical, equivalently no nonzero solvable ideal. In characteristic zero, the Cartan criterion for semisimplicity characterizes it by nondegeneracy of its Killing form. A real compact Lie algebra is the Lie algebra of a compact Lie group, equivalently one admitting a positive-definite invariant inner product. For a real semisimple algebra, the compactness criterion from the Killing form says compactness is equivalent to negative-definiteness of $\kappa$. An abelian compact-type algebra has zero Killing form, so this last characterization specifically concerns the semisimple case.

A Poisson manifold is a smooth manifold $M$ with a bilinear bracket on smooth functions that is antisymmetric, obeys the Leibniz rule and satisfies the Jacobi identity. Equivalently, a Poisson tensor $\pi$ gives $\{f,g\}=\pi (df,dg)$. Its rank need not be constant or maximal. Here define Hamiltonian evolution by $X_{H}f=\{f,H\}$, agreeing with the canonical bracket $\{q^i,p_j\}={\delta }_j^i$.

A smooth Lie group action sends each $x$ along its orbit $Gx$, with stabilizer $G_x$. If the action is free and proper, the orbit space is a smooth quotient manifold and the projection is a submersion. Without these hypotheses the quotient can have singular orbit-type strata, or even fail to be Hausdorff. For an action preserving the Poisson bracket, invariant functions form a Poisson subalgebra. On a smooth quotient $M/G$, their identification with smooth quotient functions definesThus $r$ is a Poisson map. The quotient is generally Poisson, not necessarily symplectic: different reduced momentum sectors can be different leaves. An arbitrary smooth group action alone does not guarantee this bracket descent.

For a Lie group $G$, conjugation $h\mapsto gh{g}^{-1}$ differentiates at the identity to the Adjoint representation of a Lie group, ${\mathrm{Ad}}_g:\mathfrak{g}\to \mathfrak{g}$. Its infinitesimal version is ${\mathrm{ad}}_{\xi }\eta =[\xi ,\eta ]$. The coadjoint representation on the dual space is defined byThe inverse in the first formula makes this a left action. These definitions distinguish the representation on infinitesimal generators from its dual action on momenta.

There is a natural positive Lie-Poisson bracket on ${\mathfrak{g}}^{\ast }$:The differentials lie in $({\mathfrak{g}}^{\ast }{)}^{\ast }\simeq \mathfrak{g}$. In a basis with $[e_a,e_b]=c_{ab}{}^c{e}_c$, linear coordinates satisfy $\{{\ell }_a,{\ell }_b{\}}_{+}=c_{ab}{}^c{\ell }_c$. The bracket extends by the chain rule and Leibniz rule to smooth functions. Its coordinate Jacobi condition reduces precisely to the Jacobi identity for $c_{ab}{}^c$, proving that it is Poisson. Reversing its overall sign also gives a Poisson bracket and is useful for body-frame conventions below.

On a general Poisson manifold, the vectors $X_f$ span a distribution. The Jacobi identity closes this family under commutators, and its orbits under Hamiltonian flows form the symplectic foliation. On each leaf the induced two-form satisfiesIt is well-defined after quotienting out differential covectors giving zero Hamiltonian vector field, is nondegenerate on the leaf, and is closed by the Jacobi identity. Leaf dimensions can vary, so ordinary constant-rank foliation terminology must be qualified. A Hamiltonian trajectory remains in one leaf; a Casimir function of a Poisson manifold is constant along all such trajectories, but its level set can contain more than one leaf.

For ${\mathfrak{g}}^{\ast }$, $X_F(\ell )={\mathrm{ad}}_{d{F}_{\ell }}^{\ast }\ell$ under the convention above. Hence the connected coadjoint orbits are exactly its symplectic leaves for connected $G$; for disconnected $G$, take connected orbit components. The orbit two-form is the Kirillov–Kostant–Souriau symplectic form:If a generator is in the stabilizer, its pairing with every bracket vanishes, proving independence of the choice of tangent generator. Conversely, vanishing pairing with all tangent generators puts it in that stabilizer, proving nondegeneracy. The cyclic derivative relation giving $d\omega =0$ is the Lie algebra Jacobi identity. This explains the symplectic structure of the orbit rather than just naming it.

Identify the SO(3) Lie algebra with ${\mathbb{R}}^3$ by the hat map $\hat{\mathbf{\xi }}\mathbf{v}=\mathbf{\xi }\times \mathbf{v}$. Its Lie bracket becomes the vector cross product, and the invariant Euclidean pairing identifies the dual with angular-momentum vectors $\mathbf{M}$. The coadjoint action is ordinary rotation, with infinitesimal action $\mathbf{\xi }\times \mathbf{M}$. Therefore its nonzero orbits are spheres $|\mathbf{M}|=r$; the origin is a zero-dimensional leaf. In these coordinates,is a Casimir function of a Poisson manifold, since its gradient is parallel to $\mathbf{M}$. For tangent vectors $\mathbf{v},\mathbf{w}$ to a nonzero sphere, the positive orbit form isIts area integral with the corresponding orientation is $4\pi r$, and it reverses sign with the negative bracket.

For a torque-free rigid body, choose body angular momentum $\mathbf{M}$ and principal moments $I_1,I_2,I_3>0$. Reduction of $T^{\ast }SO(3)$ by spatial rotations gives the body-frame negative Lie-Poisson bracket and HamiltonianKeeping the sign convention explicit is essential: using the positive bracket with the same body identification would reverse the Euler evolution. The negative bracket giveswith cyclic expressions for the other two components. These are the Euler equations for a torque-free rigid body. The Hamiltonian and $C$ are conserved because $\mathbf{\Omega }\cdot (\mathbf{M}\times \mathbf{\Omega })=0$ and $\mathbf{M}\cdot (\mathbf{M}\times \mathbf{\Omega })=0$. Trajectories therefore lie on intersections of energy ellipsoids with momentum spheres, the latter being the symplectic leaves.

The body's orientation is reconstructed from $\dot{R}=R\hat{\mathbf{\Omega }}$. Its spatial momentum $\mathbf{J}=R\mathbf{M}$ is constant, since $\dot{\mathbf{J}}=R(\mathbf{\Omega }\times \mathbf{M}+\mathbf{M}\times \mathbf{\Omega })=0$. Thus the body momentum need not be constant even though the spatial conserved momentum is. Poisson reduction retains the dynamics and its momentum leaves while eliminating orientation variables; the unreduced motion is recovered by reconstruction.

For a Lie group $G$, Left translation on a Lie group and right translation on a Lie group are the diffeomorphisms $L_g(h)=gh$ and $R_g(h)=hg$. Their differentials identify the tangent space at the identity with every tangent space. Translating a basis $e_a$ of the Lie algebra gives global left-invariant vector fields $E_a(g)=(d{L}_g{)}_e{e}_a$ and right-invariant vector fields $F_a(g)=(d{R}_g{)}_e{e}_a$. Left and right translations commute. If $[e_a,e_b]=c^c{}_{ab}{e}_c$, thenThe opposite sign for the right-invariant frame is essential.

Define an affine connection ${\nabla }^{-}$ by declaring the $E_a$ parallel. Explicitly, ${\nabla }_X^{-}(f^a{E}_a)=X(f^a)E_a$. Its curvature form of a connection vanishes, since successive derivatives of the component functions satisfy the vector-field commutator identity. Its torsion tensor isSimilarly, declaring the $F_a$ parallel gives another flat affine connection ${\nabla }^{+}$, with $T^{+}(F_a,F_b)=-[F_a,F_b]$. In a left-invariant frame the latter connection obeysTo see this, express $E_a=F_{{\mathrm{Ad}}_g{e}_a}$. Along $g\exp (t{e}_b)$, differentiation of ${\mathrm{Ad}}_{g\exp (t{e}_b)}{e}_a$ gives ${\mathrm{Ad}}_g[e_b,e_a]$, producing the displayed bracket. Consequently $T^{+}(X,Y)=[X,Y]$ for left-invariant $X,Y$. These are the flat translation connections on a Lie group. They have opposite torsion; if $G$ is Abelian, both torsion tensors vanish and the two connections coincide. The phrase “with torsion” does not mean nonzero torsion for Abelian groups.

The average ${\nabla }^0=({\nabla }^{-}+{\nabla }^{+})/2$ is again an affine connection and satisfies ${\nabla }_X^{0}Y=\frac{1}{2}[X,Y]$ for left-invariant fields. Its torsion tensor is zero. With the curvature convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity givesDefine the Ricci tensor by $\mathrm{Ric}(X,Y)=\mathrm{Tr}(Z\mapsto R(Z,X)Y)$. Since $[[Z,X],Y]={\mathrm{ad}}_Y{\mathrm{ad}}_{X}Z$, this giveswhere $K(X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$ is the Killing form. Thus the canonical torsion-free connection on a Lie group is generally curved. Its exact flatness criterion is $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$: Abelian and two-step nilpotent groups are exceptions. A vanishing Ricci tensor alone does not prove flatness.

For a semisimple Lie algebra, the Killing form is nondegenerate. Its adjoint invariance gives $K([X,Y],Z)+K(Y,[X,Z])=0$, so translating it defines a bi-invariant pseudo-Riemannian metric $g_K$. For left-invariant fields,A metric-compatible torsion-free connection is uniquely the Levi-Civita connection; therefore ${\nabla }^0$ is its Levi-Civita connection. Moreover,The special feature is a canonical nondegenerate metric and an Einstein manifold structure. The metric need not be positive definite: noncompact semisimple groups generally require indefinite signature. For compact semisimple groups, using $-K$ gives a bi-invariant Riemannian metric with $\mathrm{Ric}=\frac{1}{4}{g}_{-K}$ instead.

Identify the ambient real four-dimensional vector space with $2\times 2$ real matrices byFor $A,B\in SL(2,\mathbb{R})$, the linear action $X\mapsto AX{B}^{-1}$ preserves this quadratic form. It gives a group homomorphism to the special orthogonal group of signature $(2,2)$; its determinant on matrix space is $(\det A{)}^2(\det B{)}^{-2}=1$. The group $SL(2,\mathbb{R})$ is connected: an oriented Gram–Schmidt factorization writes every element as a rotation times an upper triangular matrix with positive diagonal and determinant one, and each factor varies in a connected space. Thus the image lies in the identity component $S{O}_0(2,2)$.

If $AX{B}^{-1}=X$ for every $X$, taking $X=I$ gives $A=B$. A matrix commuting with every real $2\times 2$ matrix is scalar, and its determinant-one condition then gives $A=B=\pm I$. The kernel is therefore exactly $\{(I,I),(-I,-I)\}$. Infinitesimally the action is $X\mapsto aX-Xb$, with $a,b$ traceless. If this vanishes for every $X$, then $a=b$ and it is scalar and traceless, hence zero. The differential is injective; source and target Lie algebras each have dimension six, so it is an isomorphism. The image is therefore an open subgroup of the connected target, which must be the entire target. ConsequentlyThe finite quotient is by the diagonal central subgroup, not by a sign in just one factor. The connected-component subscript is a genuine qualification of the source's displayed formula: the full $SO(2,2)$ is disconnected and cannot be isomorphic to a quotient of the connected product.

Let the anti-de Sitter radius be $\ell >0$. Its quadric is $-u^2-v^2+x^2+y^2=-{\ell }^2$, or $\det X={\ell }^2$. Thus the map $X\mapsto g=X/\ell$ identifies this quadric smoothly and globally with $SL(2,\mathbb{R})$. This is the group model of three-dimensional anti-de Sitter spacetime. The induced Lorentzian metric is bi-invariant, since the two matrix multiplication actions preserve the ambient quadratic form. At the identity, tangent matrices $a,b$ are traceless andIndeed, for a traceless matrix $a$, $\det a=-\frac{1}{2}\mathrm{Tr}(a^2)$, and polarizing $-\det (\ell a)$ gives this metric. For ${\mathfrak{sl}}_2(\mathbb{R})$ the Killing form is $K(a,b)=4\mathrm{Tr}(ab)$: in the basis $H=\mathrm{diag}(1,-1)$, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, the brackets are $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$, giving $K(H,H)=8$, $K(E,F)=4$ and the remaining independent pairings zero. Therefore $g_{\ell }={\ell }^{2}K/8$.

The canonical torsion-free connection on a Lie group is the Levi-Civita connection of this bi-invariant pseudo-Riemannian metric. Its Ricci tensor, already calculated from the double bracket, isSubstitution into the vacuum Einstein field equations givesThus the group metric solves the three-dimensional vacuum equations with negative cosmological constant.

The antipodal map $X\mapsto -X$ is $g\mapsto -g$, multiplication by the central element $-I$. Hence the antipodal quotient is $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}$, with the same local curvature and cosmological constant. To see the global time issue, useThe induced metric isOn the quadric $t$ has period $2\pi$; curves with fixed $\rho ,\phi$ are closed timelike curves. The antipodal identification is $(t,\phi )\sim (t+\pi ,\phi +\pi )$, and does not eliminate this problem. At $\rho =0$, even the half-period time shift closes after the antipodal quotient.

If global time is not identified, take the universal cover, with $t\in \mathbb{R}$. The topology of $SL(2,\mathbb{R})$ is $S^1\times {\mathbb{R}}^2$, as follows from the same rotation–triangular decomposition, so the universal cover is diffeomorphic to ${\mathbb{R}}^3$ and carries the lifted group structure $\widetilde{SL(2,\mathbb{R})}$. The metric and Einstein equation are unchanged locally. Here $t$ is a global time function with timelike gradient, excluding closed causal curves. Unwrapped anti-de Sitter spacetime is the universal covering group, rather than the original embedded quadric or its antipodal quotient; retaining the antipodal time-shift identification would wrap time again.

A left-invariant vector field $X$ on a Lie group satisfiesor equivalently $(L_g{)}_{\ast }X=X$ for every $g$. For $\xi \in T_{e}G$, defineThis is smooth because multiplication is smooth and its differential depends smoothly on the base points. The identity $L_g\circ L_h=L_{gh}$ gives $(d{L}_g{)}_h{X}_h^{\xi }=X_{gh}^{\xi }$, so $X^{\xi }$ is a left-invariant vector field. Conversely, left invariance forces any such field to have this form with $\xi =X_e$. Evaluation at $e$ and the construction above are inverse linear maps. HenceThe Lie bracket of vector fields is their commutator as derivations on smooth functions: $[X,Y]f=X(Yf)-Y(Xf)$. It is another vector field because the cross terms cancel in the Leibniz rule for its action on a product. The bracket is preserved by a diffeomorphism: conjugating the derivation operators by pullback of a smooth function preserves their commutator. Thereforefor left-invariant $X,Y$. These fields are consequently closed under the bracket.

Transport the bracket to the tangent space byBilinearity and antisymmetry come from the commutator. The Jacobi identity follows by expanding the three nested commutators of derivation operators: their six terms cancel in pairs. Thus $T_{e}G$ is a Lie algebra, and the displayed correspondence is a Lie algebra isomorphism onto the algebra of left-invariant vector fields.

The correspondence in part (a) respects Lie brackets. If the Lie algebra is abelian, any two left-invariant vector fields satisfy $[X,Y]=0$ everywhere, not just at the identity. Part (b) therefore gives $d\omega (X,Y)=0$ for every such pair.

At a point $g$, the map $(d{L}_g{)}_e:T_{e}G\to T_{g}G$ is an isomorphism. Thus every tangent vector at $g$ is the value of a left-invariant vector field. The two-form $d\omega$ consequently vanishes on every pair of tangent vectors at every point, provingNo connectedness assumption on the Lie group is needed: the argument uses its left translations and its abelian Lie algebra.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus

In the quark model, three spin-$1/2$ quarks therefore have total quark spin $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state baryons with orbital angular momentum $L=0$, total $J=S$: the nucleon has $J=1/2$ and the Delta baryon has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric three-quark colour singlet, the spatial wavefunction and flavor symmetry must also satisfy the Pauli constraint on three-quark flavour multiplets; orbital excitations can give other total angular momenta.

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.