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The scalar circle group action on ${\mathbb{C}}^{n+1}$ with the standard symplectic form has moment map $\mu =-|z{|}^2/2$ in the convention ${\iota }_{X_H}\omega =dH$. Reduction of the radius-$R$ sphere gives Complex projective space and the form ${\omega }_R=(i{R}^2/2)\partial \bar{\partial }\log (1+|w{|}^2)$. Its area on a complex projective line is $\pi R^2$. Thus $R=\sqrt{2}$ yields the Fubini-Study form normalized to projective-line area $2\pi$, while $R=1$ yields half that form. The Hopf fibration identifies the quotient.

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.

At a regular free circle group level of a moment map, the inverse image of a Lagrangian submanifold of the reduced space is a Lagrangian submanifold of the original space. Its restricted symplectic form is the pullback of the reduced form and hence vanishes. Its dimension is one more than that of the reduced Lagrangian, exactly half the ambient dimension. It is a circle bundle over the original Lagrangian.

For a Hamiltonian group action, let $c$ be a regular value of the moment map fixed by the coadjoint action, and assume the action on ${\mu }^{-1}(c)$ is free and proper. The quotient is a symplectic manifold of dimension $\dim M-2\dim G$. Its unique reduced symplectic form is characterized by ${\pi }^{\ast }{\omega }_c={\iota }^{\ast }\omega$, where $\iota$ includes the level set and $\pi$ is the quotient projection. The kernel of the restricted two-form consists precisely of the orbit directions. More general noncentral levels use their stabilizer group; the stated central-level version suffices for circle group reduction.

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.

A symplectic manifold is a smooth even-dimensional manifold $M$ with a differential two-form $\omega$ that is closed, $d\omega =0$, and nondegenerate at every point. Nondegeneracy identifies vector fields with differential one-forms by the interior product $X\mapsto {\iota }_X\omega$. In this solution use the explicit Hamiltonian vector field conventionThis contraction sign differs from the negative-sign convention sometimes used for Hamiltonian vector fields; all subsequent signs are fixed by the displayed choice. In Darboux coordinates with $\omega ={\sum }_{a}d{x}_a\wedge d{y}_a$ it givesThe Poisson bracket is bilinear and antisymmetric, and the ordinary product rule gives $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Thus it remains to check the Jacobi identity for the Poisson bracket.

The Cartan formula for the Lie derivative and closedness of the symplectic form give ${\mathcal{L}}_{X_f}\omega =d({\iota }_{X_f}\omega )+{\iota }_{X_f}d\omega =d^{2}f=0$. Henceand nondegeneracy givesApplying both sides to an arbitrary smooth function $h$ gives $\{f,\{g,h\}\}-\{g,\{f,h\}\}=\{\{f,g\},h\}$, which is exactly the Jacobi identity for the Poisson bracket. Consequently the smooth functions, with their pointwise product and this Poisson bracket, form a Poisson algebra. Constants have zero Hamiltonian vector field. If $f,g$ are components of moment maps, the same displayed relation relates the brackets of their Hamiltonian vector fields to the Poisson bracket of those components. In the opposite contraction convention the corresponding relation is the Hamiltonian Lie algebra homomorphism with the compatible bracket sign.

For the point vortices, remove the collision diagonals from $({\mathbb{R}}^2{)}^N$, since the logarithmic Hamiltonian function is singular there. Set $s_{ab}=(x_a-x_b{)}^2+(y_a-y_b{)}^2>0$ for $a\ne b$. Differentiating each pair contribution givesThe Hamiltonian vector field therefore gives the equations of motionEach contribution is perpendicular to the displacement from vortex $b$ to vortex $a$ and has magnitude $2/|{\mathbf{r}}_a-{\mathbf{r}}_b|$. This is the velocity induced by a point vortex of positive circulation, with the normalization $\Gamma /(2\pi )=2$. Each vortex is advected by the velocity of the others, without a self-induced singular velocity. Pairwise contributions cancel in the total velocity, so ${\sum }_a{x}_a$ and ${\sum }_a{y}_a$ are conserved. For two vortices the separation is constant and they rotate counterclockwise about their midpoint, as expected for identical positive point vortices.

For $L=\frac{1}{2}{\sum }_a(x_a^2+y_a^2)$, compute its Poisson bracket with $H$ directly:The last sum vanishes by pairing $(a,b)$ with $(b,a)$, since their numerators are opposite and denominators equal. Thus $L$ Poisson commutes with $H$ and is conserved. Equivalently, the Hamiltonian function depends only on pairwise distances and is invariant under simultaneous rotations.

The standard positive generator of the simultaneous circle group $SO(2)$ action isIts interior product with the symplectic form isUse the moment map convention $d⟨\mu ,\xi ⟩=-{\iota }_{Y_{\xi }}\omega$ for the usual positive rotation generator. Identifying $\mathfrak{so}(2{)}^{\ast }$ with $\mathbb{R}$, this proves $L$ is the moment map for the standard rotation action. It is invariant under the action, hence equivariant because circle group $SO(2)$ is abelian. Additive constants are possible; the natural normalization is $L=0$ at the origin of the ambient space. On the collision-free space this polynomial is its restriction. With the solution's positive contraction convention, the rotation generator is $Y=-X_L$. Confusing these two signs would reverse the final augmented Hamiltonian.

Let ${\mathcal{R}}_{\phi }$ denote simultaneous counterclockwise rotation through $\phi$, and introduce the co-rotating coordinates by $\mathbf{r}(t)={\mathcal{R}}_{\Omega t}{\mathbf{r}}^{\prime }(t)$. Both the symplectic form and $H$ are rotation-invariant, so pulling back the inertial Hamiltonian vector field leaves $X_H$ unchanged. Differentiating the rotation gives the co-rotating equationThe co-rotating positions are constant precisely when this Hamiltonian vector field vanishes. By nondegeneracy of the symplectic form, that is equivalent toConversely a collision-free critical point of this augmented Hamiltonian function produces a rigidly rotating solution. This is a rotating point-vortex relative equilibrium.

As a sign and scaling check, scaling every position by $s>0$ changes $H$ by $-N(N-1)\log s$. Therefore ${\sum }_a{\mathbf{r}}_a\cdot {\nabla }_{a}H=-N(N-1)$. At a rotating point-vortex relative equilibrium, ${\nabla }_{a}H+\Omega {\mathbf{r}}_a=0$, so $2\Omega L=N(N-1)$. For $N\ge 2$ the angular velocity is positive in this circulation convention. An equilateral three-vortex configuration of circumradius $\rho$ has $L=3{\rho }^2/2$ and $\Omega =2/{\rho }^2$, agreeing directly with the equations of motion.

Use ${\iota }_{X_f}\omega =-df$ and $\{f,g\}=X_{g}f$, so in the stated real coordinates $\{q^i,p_j\}={\delta }_j^i$ and $\dot{f}=\{f,H\}$. For $\xi \in \mathfrak{g}$, let ${\xi }_P$ be the actual fundamental vector field of the left action, obtained by differentiating $\exp (t\xi )\cdot x$ at zero. Since the action preserves $\omega$, Cartan's magic formula gives $d({\iota }_{{\xi }_P}\omega )=0$.

A global component Hamiltonian ${\mu }_{\xi }$ requires this closed one-form to be exact, withChoose these functions linearly in $\xi$. This defines the moment map, with equivariance when the constants have the required normalization. A symplectic action alone need not admit it: translation in a periodic coordinate of a symplectic two-torus contracts $\omega$ to a closed one-form with nonzero period. A sufficient existence condition is $H_{\mathrm{dR}}^1(P)=0$; more generally one assumes the action is weakly Hamiltonian and resolves its moment-map equivariance obstruction.

Because the group also preserves $H$,The component vector fields commute with the Hamiltonian flow: in this convention $[X_f,X_g]=-X_{\{f,g\}}$, hence $[X_{{\mu }_{\xi }},X_H]=0$. These are both conserved observables and infinitesimal symmetries of the dynamics.

For a left action, fundamental vector fields obey $[{\xi }_P,{\eta }_P]=-[\xi ,\eta {]}_P$. Together with the Hamiltonian-field identity, this makeshave zero Hamiltonian vector field. It is therefore constant on each connected component of $P$. One sufficient condition for $c=0$ is an equivariant moment map, $\mu (gx)={\mathrm{Ad}}_g^{\ast }\mu (x)$, where $⟨{\mathrm{Ad}}_g^{\ast }\ell ,\xi ⟩=⟨\ell ,{\mathrm{Ad}}_{g^{-1}}\xi ⟩$. Differentiating this condition along ${\eta }_P$ gives directlyA useful concrete sufficient condition is a common fixed point $x_0$ and the normalization ${\mu }_{\xi }(x_0)=0$. The constant $c$ is zero there because all fundamental fields vanish, and hence zero everywhere on connected $P$. This is fixed-point normalization removes the moment-map cocycle. If the action has an ineffective Lie-algebra kernel, the map to observables can have that kernel; equality of brackets alone is not a claim of a faithful representation.

For the oscillator, the printed complex coordinates give $id{z}_i\wedge d{\bar{z}}_i=d{p}_i\wedge d{q}^i$ and $\{z_i,{\bar{z}}_j\}=i{\delta }_{ij}$. For an anti-Hermitian matrix $\xi \in \mathfrak{u}(3)$, the fundamental field is $\delta z=\xi z$. Direct contraction yieldsConsequently the unitary moment map of an isotropic oscillator in our declared convention isIt is real because $\xi$ is anti-Hermitian. Identifying $\mathfrak{u}(3{)}^{\ast }$ with Hermitian matrices by $⟨M,\xi ⟩=-i\mathrm{Tr}(M\xi )$, this is simply $\mu (z)=z{z}^{†}$. Unitary invariance of $H=z^{†}z$ proves conservation of all nine real matrix components. Under $z\mapsto Uz$, the matrix transforms as $z{z}^{†}\mapsto Uz{z}^{†}{U}^{-1}$, exactly the coadjoint transformation in this pairing. The origin is the required fixed point.

The bracket can also be checked without an equivariance theorem. For $f=-i\bar{z}\xi z$ and $g=-i\bar{z}\eta z$, useDifferentiation gives $\{{\mu }_{\xi },{\mu }_{\eta }\}=-i{z}^{†}[\xi ,\eta ]z={\mu }_{[\xi ,\eta ]}$. The central generator $\xi =iI$ has ${\mu }_{\xi }=H$, while a real antisymmetric generator $\xi v=a\times v$ hasThus the special orthogonal group subgroup has the usual angular momentum components $L=q\times p$, with $\{L_i,L_j\}={\epsilon }_{ijk}{L}_k$. The remaining unitary generators supply the other conserved quadratic observables, with their signs fixed by the stated choice of complex coordinates and Hamiltonian convention.

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.

Let $\mu :P\to \mathbb{R}$ be the moment map for the circle group, assume $c$ is a regular value, and assume the circle action on $N={\mu }^{-1}(c)$ is free. The action is proper because the circle is compact. Then $N$ has dimension $2n-1$ and $P_c=N/SO(2)$ is a smooth manifold of dimension $2n-2$.

If ${\xi }_P$ generates the action, $\omega ({\xi }_P,v)=d\mu (v)=0$ for every $v\in TN$. In fact $(TN{)}^{\omega }$ is exactly the line spanned by ${\xi }_P$, since $TN=\ker d\mu$ is a hyperplane and $\omega$ is nondegenerate. Therefore the kernel of the restricted two-form is precisely the orbit direction. The restriction is invariant and horizontal, so it descends to a unique nondegenerate closed two-form ${\omega }_c$ withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian $H$ is invariant, then $\{\mu ,H\}=0$: its Hamiltonian flow stays on $N$, projects to the quotient and is generated there by the descended function $H_c$. This is a reduced Hamiltonian system in $2n-2$ local variables. At a critical level or a nonfree orbit the quotient can be singular, so the smooth dimension assertion needs the stated hypotheses.

For a mass-$m$ particle in the plane with central force potential $V(r)$, work on $r>0$ and writeExterior differentiation gives $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Rotation translates $\theta$ and has moment map $\ell$. Fix $\ell ={\ell }_0$ and quotient out $\theta$ to obtain the planar rotational symplectic reductionThe reduced equations are $\dot{r}=p_r/m$ and ${\dot{p}}_r={\ell }_0^2/(m{r}^3)-V^{\prime }(r)$. The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from $\dot{\theta }={\ell }_0/(m{r}^2)$. For ${\ell }_0\ne 0$ the origin cannot lie on the level; for ${\ell }_0=0$ the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.

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 Yang-Mills instanton is a localized finite-action Euclidean gauge field with self-dual or anti-self-dual gauge curvature. It is a four-dimensional space-time event, rather than a static three-dimensional finite-energy monopole. Take gauge group $SU(2)$ first, with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, $[T_a,T_b]={ϵ}_{abc}{T}_c$, and $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Writeand fix orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. The positive Euclidean action isVarying $F$ gives $\delta F=D\delta A$. Integration by parts makes stationarity equivalent to $D_{\mu }{F}_{\mu \nu }=0$. The Bianchi identity $DF=0$ follows by expanding $F=dA+A\wedge A$, using $d^2=0$, and cancelling the remaining matrix commutator terms. Consequently $F=\pm \ast F$ implies the Yang-Mills equations: $D\ast F=\pm DF=0$. Self-duality of gauge curvature is an extra first-order condition, not a property of every finite-action gauge connection.

For the usual instanton decay conditions, the field is asymptotically pure gauge, $A\sim h^{-1}dh$ on the sphere at infinity. The boundary map $h:S^3\to SU(2)\simeq S^3$ has an integer degree. With the present trace convention define the instanton number byThe identityturns this integral into a boundary winding integral: for $A=h^{-1}dh$, the three-form in parentheses is $-\frac{1}{3}\mathrm{tr}(h^{-1}dh{)}^3$, giving $k=(24{\pi }^2{)}^{-1}{\int }_{S^3}\mathrm{tr}(h^{-1}dh{)}^3$, with the chosen boundary orientation. The normalized invariant three-form computes the degree of $h$, so $k\in \mathbb{Z}$. Equivalently the gauge bundle extends across conformal infinity to a bundle on $S^4$, and this is its Second Chern number convention. A smooth deformation preserving the boundary class cannot change this integer.

Square completion gives the Yang-Mills instanton Bogomolny bound:Self-dual gauge curvature saturates the bound for $k>0$, and anti-self-dual gauge curvature for $k<0$. These are Yang-Mills action minima within their topological sectors. Gauge variations give redundant directions, and variations along the moduli give zero modes; there is no negative second variation within the fixed-charge sector at a bound-saturating solution.

An explicit charge-one solution is the BPST instanton. Put $y=x-a$, $r^2=y_{\mu }{y}_{\mu }$, $d=r^2+{\rho }^2$, with $\rho >0$. The self-dual 't Hooft symbols are defined by ${\eta }_{ij}^a={ϵ}_{aij}$ and ${\eta }_{i4}^a={\delta }_{ai}$, extended antisymmetrically. ThenFor completeness, differentiation supplies $-4{\eta }_{\mu \nu }^a/d-4K_{\mu \nu }^a/d^2$, where $K_{\mu \nu }^a={\eta }_{\nu \lambda }^a{y}_{\lambda }{y}_{\mu }-{\eta }_{\mu \lambda }^a{y}_{\lambda }{y}_{\nu }$. The matrix commutator term supplies $4C_{\mu \nu }^a/d^2$, and the symbol identityreduces their sum to the stated gauge curvature. Since $\ast {\eta }^a={\eta }^a$, it is self-dual and solves the Yang-Mills equations. The gauge potential is regular at the center in this gauge, and gauge curvature falls as $r^{-4}$ at infinity.

The squared-symbol sum is 12, soThe second integral follows from $d^{4}x=2{\pi }^2{r}^{3}dr$ and the substitution $u=r^2$. ThusReplacing the symbols by their anti-self-dual counterparts reverses the charge and gives an anti-instanton. A compact non-Abelian group with an $SU(2)$ subgroup admits corresponding embedded examples.

The center $a\in {\mathbb{R}}^4$ gives four position moduli and $\rho >0$ gives one size modulus. In a framed instanton moduli space, gauge transformations are required to approach the identity at infinity, so three global $SU(2)$ orientation parameters remain, making eight real parameters for one instanton. If constant gauge transformations are also divided out, only the five position-and-size parameters remain. The size costs no classical Yang-Mills action: under $A_{\mu }(x)\mapsto \lambda A_{\mu }(\lambda x)$, gauge curvature scales by ${\lambda }^2$ and the four-dimensional volume by ${\lambda }^{-4}$. As $\rho \to 0$, gauge curvature concentrates at the center; the limiting object is a singular bubbling boundary of moduli space, not a smooth zero-size solution.

Higher-charge solutions require genuine nonlinear constructions, rather than a linear sum of charge-one gauge potentials. The ADHM construction gives a finite-dimensional description. For $V={\mathbb{C}}^{\ell }$, $W={\mathbb{C}}^N$, take $B_1,B_2\in \mathrm{End}(V)$, $I:W\to V$, $J:V\to W$, satisfyingmodulo the natural group action of $U(\ell )$. Here $\ell$ is the charge magnitude, avoiding confusion with the signed $k$ above. Regularity requires the following matrix to have full row rank everywhere:The complex constraint kills the off-diagonal block of ${\mathcal{D}}_z{\mathcal{D}}_z^{†}$, while the real constraint equates its diagonal blocks. Thus ${\mathcal{D}}_z{\mathcal{D}}_z^{†}=1_2\otimes f^{-1}$. Choose an orthonormal matrix of kernel columns $\Psi$, with ${\Psi }^{†}\Psi =1_N$, and define $A={\Psi }^{†}d\Psi$.

This is not merely a parameter count: its gauge curvature is checked by the ADHM factorization identity. The complementary orthogonal projection is ${\mathcal{D}}_z^{†}(1_2\otimes f){\mathcal{D}}_z$. Differentiating ${\mathcal{D}}_z\Psi =0$ in the identity $F=d{\Psi }^{†}(1-\Psi {\Psi }^{†})\wedge d\Psi$ givesThe space-time differential two-forms appearing here are generated by $d{z}_1\wedge d{\bar{z}}_1-d{z}_2\wedge d{\bar{z}}_2$, $d{z}_1\wedge d{\bar{z}}_2$, and $d{\bar{z}}_1\wedge d{z}_2$. For orientation $1234$ these are anti-self-dual. Hence the construction yields anti-self-dual gauge connections, of charge $k=-\ell$ in the present convention; the reversed construction gives positive charge. Regular ADHM data describe the smooth framed instanton moduli, while degenerate rank conditions allow singular limits.

There are $4{\ell }^2+4\ell N$ real data parameters. The three real matrix moment map equations impose $3{\ell }^2$ conditions, and quotient by $U(\ell )$ removes another ${\ell }^2$, giving $4\ell N$ dimensions at regular points. For $SU(2)$ this is $8\ell$ framed dimensions and $8\ell -3$ unframed dimensions on the irreducible locus. This explains how the one-instanton parameters fit into the higher-charge family.

The Penrose-Ward correspondence provides a complementary description by holomorphic vector bundles on twistor space, trivial on the real twistor lines for the Euclidean problem and equipped with the suitable reality structure. Triviality on every complexified line is a stronger condition and is not required for ordinary Euclidean instantons. Symmetry reductions of the anti-self-dual equations give integrable three-dimensional systems such as the Bogomolny equations.

In the quantum theory, a Euclidean instanton interpolates between Yang-Mills vacua of different Chern-Simons number. Choosing Euclidean time $x^1$ and spatial orientation $234$, its signed charge is the final Chern-Simons number minus the initial one, so it describes quantum tunnelling between winding sectors. Semiclassical contributions are suppressed by $\exp [-8{\pi }^2|k|/g^2]$; a Yang-Mills theta term adds a phase $\exp (i\vartheta k)$. One integrates over the instanton collective coordinates and includes fluctuation determinants rather than treating one classical solution as the entire quantum answer. Instantons therefore connect finite-action classical geometry, topology of gauge bundles, exact nonlinear constructions, and nonperturbative quantum effects.