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Suppose a symplectic Lie group action on a connected symplectic manifold has global component Hamiltonians and a common fixed point $p_0$. Choose their constants linearly so ${\mu }_{\xi }(p_0)=0$. The difference $c(\xi ,\eta )=\{{\mu }_{\xi },{\mu }_{\eta }\}-{\mu }_{[\xi ,\eta ]}$ has zero Hamiltonian vector field because fundamental fields and Hamiltonian fields have compatible Lie-bracket conventions; it is therefore constant. At $p_0$, all fundamental fields vanish, so both terms in $c$ vanish. Thus $c=0$ everywhere and the Poisson brackets of the components reproduce the group Lie algebra. For a connected group this also yields coadjoint equivariance. A linear unitary action on a vector space has the origin as such a fixed point.

For a Lie group action and $X$ in its Lie algebra, the fundamental vector field is $X_M(p)={\frac{d}{dt}|}_{t=0}\exp (tX)\cdot p$.

For a symplectic left Lie group action, use actual fundamental fields ${\xi }_P$ and ${\iota }_{X_f}\omega =df$. They satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. Cartan's magic formula then gives ${\iota }_{[\xi ,\eta {]}_P}\omega =d[\omega ({\xi }_P,{\eta }_P)]$. Thus every commutator generator has a globally exact contraction. A semisimple Lie algebra is perfect, so every generator has a global Hamiltonian. The separate semisimple moment-map equivariance result removes the constant two-cocycle obstruction and gives an equivariant moment map for a connected group.

A Hamiltonian group action is a Lie group action on a symplectic manifold with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$. In the sign convention ${\iota }_{X_H}\omega =dH$, each component satisfies $d⟨\mu ,X⟩={\iota }_{X_M}\omega$, where $X_M$ is the fundamental vector field. The opposite Hamiltonian vector field convention reverses the sign.

For a smooth left Lie group action, define the fundamental vector field by $X^{\#}(p)={\frac{d}{ds}|}_0\exp (sX)\cdot p$. With the Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, this map is an anti-homomorphism. The map $X\mapsto v_X=-X^{\#}$ is a Lie algebra representation. A right action with the positive exponential has the opposite homomorphism convention, so specifying action side and bracket convention prevents a sign ambiguity.

For a weakly Hamiltonian left Lie group action on a connected symplectic manifold, choose ${\iota }_{X_f}\omega =df$. Fundamental vector fields and Hamiltonian vector fields are both antihomomorphisms, so the displayed difference is constant. The Jacobi identity for the Poisson bracket makes it a two-cocycle. Changing ${\mu }_X$ by $b(X)$ changes the difference by $-b([X,Y])$. Thus its class in Lie algebra cohomology obstructs choosing an equivariant moment map. For a connected group, vanishing of this class gives infinitesimal and group equivariance.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

Write ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ for the fundamental vector field of a left Lie group action, and use ${\iota }_{X_f}\omega =df$ throughout. Such fundamental fields satisfy $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. A moment map has linear components ${\mu }_{\xi }=⟨\mu ,\xi ⟩$ satisfyingFor the action to be Hamiltonian, these components can be chosen equivariantly, or, infinitesimally,This identifies the span of the component Hamiltonians with the Lie algebra image, rather than identifying the entire infinite-dimensional Poisson algebra $C^{\infty }(P)$ with $\mathfrak{g}$. For an actual isomorphism with $\mathfrak{g}$, injectivity is also needed; a faithful infinitesimal action supplies it.

Preservation of $\omega$ alone gives $d{\iota }_{{\xi }_P}\omega =0$, not necessarily exactness. For example, translation generated by ${\partial }_q$ on the symplectic two-torus has contraction $dp$, whose period is nonzero. That action has no globally defined component Hamiltonian. Once the contractions are exact, choose Hamiltonians linearly in $\xi$. On each connected component of $P$, the discrepancyis constant: apply $[X_f,X_g]=-X_{\{f,g\}}$ and the fundamental-field bracket. The Jacobi identity makes $\kappa$ a Lie algebra two-cocycle. Replacing ${\mu }_{\xi }$ by ${\mu }_{\xi }+b(\xi )$ changes it to $\kappa (\xi ,\eta )-b([\xi ,\eta ])$. Thus the exactness condition and vanishing of this moment-map equivariance obstruction are precisely the required infinitesimal conditions.

For a semisimple Lie algebra, both hold. First Cartan's magic formula givesSemisimplicity implies $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so every generator is a sum of commutators and has a globally exact contraction. This proves Hamiltonian existence for a semisimple symplectic action.

Next use the nondegenerate Killing form to define $D$ by $B(D\xi ,\eta )=\kappa (\xi ,\eta )$. Antisymmetry makes $D$ skew for $B$. The cocycle identity and invariance of $B$ imply $D[\xi ,\eta ]=[D\xi ,\eta ]+[\xi ,D\eta ]$, so $D$ is a derivation. Choose $Z$ byThen, using cyclicity of trace and $[D,\mathrm{ad}\xi ]=\mathrm{ad}(D\xi )$,Hence $D=\mathrm{ad}Z$ and $\kappa (\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes the cocycle and proves semisimple moment-map equivariance without merely invoking a cohomology theorem. For a connected group, integrate the infinitesimal equivariance along its one-parameter subgroups to obtain group equivariance. A disconnected group can require an additional discrete-equivariance check.

Use ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. For an actual infinitesimal left Lie group action, define ${\xi }_P(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$. Its fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ with this convention.

Existence and its global obstruction. Preservation of the symplectic form gives ${\mathcal{L}}_{{\xi }_P}\omega =0$. By Cartan's magic formula, ${\iota }_{{\xi }_P}\omega$ is closed. Locally it equals $d{\mu }_{\xi }$; globally this requires its class in $H_{\mathrm{dR}}^1(P)$ to vanish. If it does for every $\xi$, choose potentials linearly in $\xi$ and defineThis gives the component Hamiltonians of a weakly Hamiltonian action. The assumption in the question does not alone guarantee a global moment map: on $T^2$ with $\omega =dx\wedge dy$, translation in $x$ preserves $\omega$, but its contraction is $dy$, whose integral around the $y$ circle is nonzero. It is not an exact one-form. Thus no global real-valued component Hamiltonian exists for that circle action. The intended construction is valid locally, or globally with this exactness condition, for example if $H_{\mathrm{dR}}^1(P)=0$.

Matching the Lie algebra. On each connected component, the moment-map equivariance obstruction is the constantIndeed, the identities $[X_f,X_g]=-X_{\{f,g\}}$ and $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$ show that its differential vanishes. The Jacobi identity for the Poisson bracket saysChanging ${\mu }_{\xi }$ by a linear constant $b(\xi )$ changes $c$ to $c-b([\xi ,\eta ])$. Consequently the necessary and sufficient infinitesimal condition is the cocycle must be a coboundary, so that the constants can be chosen withEquivalently its class in $H^2(\mathfrak{g},\mathbb{R})$ vanishes. For a connected group this infinitesimal equivariance integrates to equivariance under its coadjoint action. A disconnected group can impose additional conditions. Here “coincides” means representing the specified Lie brackets; it does not guarantee injectivity when the group action has ineffective generators.

Proof using the Killing form. A nondegenerate Killing form $B$ makes $\mathfrak{g}$ semisimple. First, it is perfect: the $B$-orthogonal complement of $[\mathfrak{g},\mathfrak{g}]$ is the center by invariance, and the center is zero by nondegeneracy. For symplectic fields,by Cartan's magic formula and the antihomomorphism sign. Every commutator therefore has an exact contraction. Perfectness supplies component Hamiltonians for every generator, even when $P$ is not simply connected: this is Hamiltonian existence for a semisimple symplectic action.

It remains to eliminate the cocycle, not merely to name an existence theorem. Define a linear operator $D$ on $\mathfrak{g}$ by $B(D\xi ,\eta )=c(\xi ,\eta )$. Antisymmetry of $c$ makes $D$ skew for $B$. Combining its cocycle identity with invariance of $B$ givesChoose $Z\in \mathfrak{g}$ by $B(Z,\xi )=\mathrm{tr}(D{\mathrm{ad}}_{\xi })$, possible because $B$ is nondegenerate. Since $D$ is a derivation, $[D,{\mathrm{ad}}_{\xi }]={\mathrm{ad}}_{D\xi }$. Cyclicity of the matrix trace then givesThus $D={\mathrm{ad}}_Z$, and $c(\xi ,\eta )=B(Z,[\xi ,\eta ])$. Taking $b(\xi )=B(Z,\xi )$ removes $c$, proving the required condition. This applies to semisimple Lie groups, including $SU(n)$ for $n\ge 2$, $SO(3)$ and $SL(n,\mathbb{R})$ for $n\ge 2$. It is nondegeneracy, not positive definiteness, that matters. Groups with a nonzero central Lie algebra, such as $U(3)$, do not satisfy this Killing-form criterion, although particular actions can still have equivariant moment maps.

The isotropic oscillator. On $T^{\ast }{\mathbb{R}}^3$, putThe action $z\mapsto Uz$, $U\in U(3)$, preserves the Hermitian norm and the symplectic form, and therefore preserves $H$. A Hermitian matrix $A$ parametrizes the Lie algebra generator $\xi =-iA$. Its unitary moment map of an isotropic oscillator isSince $\{z_j,{\bar{z}}_k\}=-i{\delta }_{jk}$, direct differentiation yieldsThe matrix $-i[A,B]$ is Hermitian and corresponds precisely to $[{\xi }_A,{\xi }_B]$; no central defect remains. The full moment map can be identified, using the pairing with Hermitian matrices, with the rank-one matrix $z{z}^{†}$. The central generator $A=I$ gives ${\mu }_I=H$, whose flow is $z(t)=e^{-it}z(0)$. Globally $U(3)=(SU(3)\times U(1))/{\mathbb{Z}}_3$, so the “$U(1)$ factor” refers to this central circle, rather than a literal direct-product decomposition.

The real orthogonal subgroup acts geometrically by $(x,p)\mapsto (Rx,Rp)$, the cotangent lift of a diffeomorphism of ${\mathbb{R}}^3$. If $K$ is real antisymmetric and $A=iK$, then ${\mu }_A=-x^{T}Kp$. For $Kv=\Omega \times v$, this is $\Omega \cdot (x\times p)$, so its three components are the usual angular momentum $L_i={ϵ}_{ijk}{x}_j{p}_k$.

The remaining six independent Hermitian directions can be taken as real symmetric matrices. Their moment functions spanThe diagonal ones are the individual oscillator energies; their sum is $H$. The off-diagonal ones correlate two oscillator modes. Their flows mix positions and momenta and are not cotangent lifts of the obvious spatial rotation action: for a real symmetric $A$, $\dot{x}=Ap$ and $\dot{p}=-Ax$, rather than $\dot{x}=Kx$, $\dot{p}=Kp$. All nine quadratic functions commute with $H$, since the entire unitary group action preserves the oscillator Hamiltonian.

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.

A Hamiltonian group action is a smooth Lie group action of $G$ on $(M,\omega )$ by symplectomorphisms, together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$. For $\xi \in \mathfrak{g}$, let ${\xi }_M(p)={\frac{d}{dt}|}_0\exp (t\xi )\cdot p$ be its fundamental vector field. In our sign convention the defining identities areHere the left coadjoint action means $({\mathrm{Ad}}_g^{\ast }\nu )(\xi )=\nu ({\mathrm{Ad}}_{g^{-1}}\xi )$. Each component ${\mu }^{\xi }$ is therefore a Hamiltonian function for the corresponding infinitesimal action. Equivariance is part of the definition; merely requiring each infinitesimal generator to be a Hamiltonian vector field is the weaker condition of a weakly Hamiltonian action. Reversing the defining sign of Hamiltonian vector fields reverses the moment map sign as well.

The real vector space of symmetric matrices has dimension $n(n+1)/2$. On its invertible part the derivative of the determinant isAt a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The regular level set theorem therefore yieldsThis applies to every signature component, not only positive-definite matrices.

The special linear congruence action on symmetric matrices preserves symmetry because $(AB{A}^T{)}^T=AB{A}^T$, and preserves determinant because $\det (AB{A}^T)=(\det A{)}^2\det B=1$. The identity acts trivially andThese verify a smooth left Lie group action of the special linear group. By Sylvester's law of inertia, it also preserves the numbers of positive and negative eigenvalues.

For $n=2$, parametrize the symmetric matrices byThus $M$ is exactly the two-sheeted hyperboloidThe two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the special linear congruence action on symmetric matrices. They are individually transitive: on the positive sheet $B=A{A}^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-A{A}^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.

Differentiating the left Lie group action along $\exp (sX)$ gives the fundamental vector fieldFor the three matrices in the question, their coordinate components areA sign convention matters here. With the usual Lie bracket of vector fields $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_{X}z$, the bracket has coefficient $(R_Y{R}_X-R_X{R}_Y)z$. To obtain a Lie algebra representation rather than an anti-representation, use the infinitesimal left-action sign conventionAn explicit answer is thereforeThese are tangent to the two-sheeted hyperboloid: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,They can be written on either sheet using $t=\pm \sqrt{1+x^2+y^2}$:Here the omitted ${\partial }_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.

Direct matrix multiplication givesDirect differentiation of the displayed vector fields gives exactlyFor instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the sl2 Lie algebra. The three fields are linearly independent over constant real coefficients: if $a{v}_1+b{v}_2+c{v}_3$ vanishes on a sheet, its ${\partial }_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its ${\partial }_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is faithful. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.

A principal bundle separates the geometry of internal symmetry from a chosen local gauge potential. Let $\pi :P\to X$ be a principal $G$-bundle with a free right Lie group action. Each fiber is a copy of $G$, but generally no single identification works globally. A principal connection tells us how to compare fibers over nearby base points. Its physical interpretation is a prescription for parallel transport of internal states.

For $\xi \in \mathfrak{g}$, the vertical fundamental vector field is ${\xi }_p^{\#}={\frac{d}{ds}|}_{0}p\exp (s\xi )$. A principal connection is a Lie algebra-valued one-form $\mathcal{A}$ on $P$ satisfyingEquivalently, the horizontal distribution of a principal connection $H_p=\ker {\mathcal{A}}_p$ complements the vertical tangent space and is preserved by right translation. Given a curve in $X$ and an initial point in its fiber, there is a unique horizontal lift. For a closed curve its endpoint differs from its start by a group element: the holonomy of a connection. Thus a principal connection contains both infinitesimal and global transport information.

Choose local sections $s_{\alpha }:U_{\alpha }\to P$. Their local principal connection forms $A_{\alpha }=s_{\alpha }^{\ast }\mathcal{A}$ are the usual gauge potentials. If $s_{\beta }=s_{\alpha }{g}_{\alpha \beta }$ on an overlap, equivariance and reproduction of vertical generators giveThis inhomogeneous law means that a gauge potential is not an ordinary globally defined tensor. Its local formulas glue to a global principal connection. The same law describes changing a local section by a gauge function. For example, to impose temporal gauge locally, solve ${\partial }_{t}g=-A_{t}g$ so the transformed time component vanishes.

The curvature of a principal connection isUnlike the principal connection form, its curvature is horizontal and equivariant. Consequently $F_{\beta }=g_{\alpha \beta }^{-1}{F}_{\alpha }{g}_{\alpha \beta }$, and the gauge curvature is globally a two-form on $X$ with values in the adjoint bundle $\mathrm{ad}P=P{\times }_{\mathrm{Ad}}\mathfrak{g}$. The curvature measures the failure of horizontal directions to close under brackets: for horizontal lifts $U,V$, $\mathcal{F}(U,V)=-\mathcal{A}([U,V])$. A flat principal connection has an integrable horizontal distribution, although nontrivial global holonomy of a connection can remain around noncontractible curves.

The covariant exterior derivative on adjoint bundle-valued forms is locally $D_A=d+[A,\cdot ]$. Expanding $F=dA+A\wedge A$ and using $d^2=0$ yields the Bianchi identityFor a representation $\rho$ of $G$, the same principal connection induces transport in the corresponding associated vector bundle; locally its covariant derivative is $d+{\rho }_{\ast }(A)$. This explains why charged matter and the gauge curvature use the same gauge potential.

Now put an oriented Riemannian metric on a four-dimensional base and take $G=SU(n)$. Use anti-Hermitian matrices with positive pairing $-\mathrm{tr}(XY)$, as in the preceding solution. The Yang-Mills action isIts gauge invariance follows from curvature conjugation and invariance of the trace. Under a compactly supported variation $a=\delta A$, $\delta F=D_{A}a$. Integration by parts therefore givesThis is a second-order equation for the gauge potential. The self-dual Yang-Mills equations $F=\ast F$ and the Anti-self-dual Yang-Mills equations $F=-\ast F$ are first-order equations. Either implies the full Yang-Mills equations immediately, since $D_A\ast F=\pm D_{A}F=0$ by the Bianchi identity. This is self-duality implies Yang-Mills equations.

A Yang-Mills instanton is a smooth finite-action Euclidean solution with self-dual or anti-self-dual gauge curvature. Its defining first-order condition depends on the principal connection, the metric and the orientation. Since the Hodge star operator on two-forms is unchanged under $g\mapsto e^{2f}g$ in four dimensions, self-duality and the Yang-Mills action are conformally invariant. In particular Euclidean solutions can be studied through conformal compactification, with appropriate behavior at infinity.

The global topology is encoded by Chern-Weil theory. Since $\mathrm{tr}F=0$, the Second Chern form and Second Chern number areon a compact oriented four-manifold. The Bianchi identity makes $c_2(A)$ closed. More explicitly, under a connection variation,Thus the integrated Second Chern number is independent of the principal connection on a fixed bundle, with the usual fixed-boundary condition on a noncompact base. The space of connections is affine, so integrating this variation along a straight path also proves that their characteristic forms differ by an exact form.

Locally the same characteristic form has a Chern-Simons 3-form primitive:On ${\mathbb{R}}^4$ the underlying bundle is trivial, but a finite-action instanton with the standard extendible behavior at infinity can define a nontrivial bundle after adding the point at infinity. The transition map on an equatorial $S^3$ takes values in $SU(n)$; for $n\ge 2$, its homotopy class lies in ${\pi }_3(SU(n))\cong \mathbb{Z}$. The boundary integral of the Chern-Simons 3-form computes that integer with the chosen sign convention. This reconciles a local matrix-valued gauge potential on ${\mathbb{R}}^4$ with nonzero global Second Chern number on $S^4$.

Finally the Hodge splitting of Euclidean two-forms produces the Yang-Mills instanton Bogomolny boundEquality holds exactly when the opposite-duality component of the gauge curvature vanishes. Thus a Yang-Mills instanton is an absolute action minimum in its fixed topological sector, not merely a stationary solution. In this trace convention self-dual instantons have $k<0$ and anti-self-dual instantons have $k>0$. In the zero sector, a self-dual or anti-self-dual finite-action solution has $S=0$ and hence $F=0$.

Different principal connections related by bundle gauge transformations describe the same physical configuration. Fixing $k$ and taking the quotient of instanton solutions by this gauge equivalence of principal connections gives the instanton moduli space. Near an anti-self-dual solution, write a variation as $a\in {\Omega }^1(X,\mathrm{ad}P)$. The linearized instanton equation and infinitesimal gauge transformations areA local gauge condition $D_A^{\ast }a=0$ removes this redundancy. The combined operator $D_A^{\ast }\oplus P_{+}{D}_A$ is elliptic: for nonzero covector $\xi$, its principal symbol $a\mapsto ({\iota }_{{\xi }^{♯}}a,P_{+}(\xi \wedge a))$ has zero kernel. Choose $\xi =dt$; the first component removes $a_t$, and the self-dual projection of $dt\wedge a$ removes the three remaining components. This connects the instanton moduli space to geometric analysis, while the characteristic number and transport law explain its topological and physical meaning.

For $X\in \mathfrak{h}$, the fundamental vector field of the restricted Lie group action is $(iX{)}_M$. If $\varphi$ is the original moment map, thenThis is the defining Hamiltonian action identity, in the convention ${\iota }_{X_H}\omega =dH$.

If equivariance is included in the definition of a moment map, it is also preserved: the dual map $\pi =i^{\ast }$ intertwines the coadjoint actions of $H$, since $i({\mathrm{Ad}}_{h}X)={\mathrm{Ad}}_h(iX)$. Thusis a moment map for the restricted Hamiltonian action of the Lie subgroup $H$.

Let $V_i=\partial V/\partial {\varphi }_i$ and $M_{ij}^2={\partial }_i{\partial }_{j}V{|}_{{\varphi }_0}$ be the scalar mass matrix, which is the Hessian matrix of the scalar potential at the vacuum. Invariance under the infinitesimal Lie group action givesDifferentiate with respect to ${\varphi }_j$ and evaluate at the vacuum expectation value ${\varphi }_0$. Since $V_i({\varphi }_0)=0$ at a minimum,Thus every tangent vector $i{t}^a{\varphi }_0$ generated by a broken Lie algebra generator is a zero eigenvalue eigenvector of the scalar mass matrix. The unbroken generators are precisely those in the stabilizer Lie algebra $\mathfrak{h}$ and give the zero tangent vector; independent broken directions span $\mathfrak{g}/\mathfrak{h}$. Hence Goldstone theorem givesmassless scalar modes at the classical level.

A symplectic Lie group action is weakly Hamiltonian when every fundamental vector field is a Hamiltonian vector field. A Hamiltonian group action additionally supplies Hamiltonians as components of an equivariant moment map. Constants in the component Hamiltonians do not affect their vector fields, but the ability to choose them with the required equivariance is an additional global condition.