Moment Maps and Symplectic Reduction

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$.

Let $G$ be a finite-dimensional real Lie group with Lie algebra $\mathfrak{g}=T_{e}G$ and dual ${\mathfrak{g}}^{\ast }=\mathcal{L}(\mathfrak{g},\mathbb{R})$ (Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$). For $g\in G$ the coadjoint map is the linear map

${\mathrm{Ad}}_g^{\ast }:{\mathfrak{g}}^{\ast }⟶{\mathfrak{g}}^{\ast },{\mathrm{Ad}}_g^{\ast }\alpha :=\alpha \circ {\mathrm{Ad}}_{g^{-1}},$

so that $⟨{\mathrm{Ad}}_g^{\ast }\alpha ,\xi ⟩=⟨\alpha ,{\mathrm{Ad}}_{g^{-1}}\xi ⟩$ for every $\xi \in \mathfrak{g}$. The coadjoint action of $G$ on ${\mathfrak{g}}^{\ast }$ is

$G\times {\mathfrak{g}}^{\ast }⟶{\mathfrak{g}}^{\ast },(g,\alpha )⟼g\cdot \alpha :={\mathrm{Ad}}_g^{\ast }\alpha \mathrm{.}$

The family ${\mathrm{Ad}}^{\ast }:G\to \mathrm{GL}({\mathfrak{g}}^{\ast })$, $g\mapsto {\mathrm{Ad}}_g^{\ast }$, is the coadjoint representation. The coadjoint orbit of $\alpha \in {\mathfrak{g}}^{\ast }$ and its coadjoint stabilizer are the orbit and stabilizer, in the sense of Orbits, stabilizers, and orbit maps of smooth actions, of this action:

$G\cdot \alpha =\{{\mathrm{Ad}}_g^{\ast }\alpha :g\in G\},G_{\alpha }=\{g\in G:{\mathrm{Ad}}_g^{\ast }\alpha =\alpha \}\mathrm{.}$

The maps ${\mathrm{Ad}}_g^{\ast }$ are invertible, with $({\mathrm{Ad}}_g^{\ast }{)}^{-1}={\mathrm{Ad}}_{g^{-1}}^{\ast }$: composing ${\mathrm{Ad}}_g^{\ast }{\mathrm{Ad}}_h^{\ast }$ gives $\alpha \mapsto \alpha \circ {\mathrm{Ad}}_{h^{-1}}\circ {\mathrm{Ad}}_{g^{-1}}$, and ${\mathrm{Ad}}_{h^{-1}}{\mathrm{Ad}}_{g^{-1}}={\mathrm{Ad}}_{h^{-1}{g}^{-1}}$ because $\mathrm{Ad}$ is a group homomorphism (Adjoint is a smooth Lie-group representation), which identifies the composite with ${\mathrm{Ad}}_{gh}^{\ast }$. Hence

${\mathrm{Ad}}_{gh}^{\ast }={\mathrm{Ad}}_g^{\ast }\circ {\mathrm{Ad}}_h^{\ast },{\mathrm{Ad}}_e^{\ast }={\mathrm{id}}_{{\mathfrak{g}}^{\ast }},$

so the coadjoint action is a left action; the inverse $g^{-1}$ in the definition is exactly what makes it left rather than right. The action is jointly smooth. Indeed, in a fixed basis of $\mathfrak{g}$ and its dual, the matrix of ${\mathrm{Ad}}_g^{\ast }$ is the transpose of the matrix of ${\mathrm{Ad}}_{g^{-1}}$; the matrix entries of $g\mapsto {\mathrm{Ad}}_{g^{-1}}$ are smooth because $\mathrm{Ad}$ is a smooth representation and inversion in $G$ is smooth (Lie group), and the coordinates of $(g,\alpha )\mapsto \alpha \circ {\mathrm{Ad}}_{g^{-1}}$ are these smooth matrix entries paired with the coordinates of $\alpha$ (Conjugation and the adjoint representation of a Lie group). Thus the coadjoint action is a smooth left action in the sense of Smooth left actions of Lie groups, and the orbit and stabilizer above are those of a smooth action.

Differentiating the curve $t\mapsto {\exp }_G(-t\xi )\cdot \alpha$ at $t=0$ gives the infinitesimal formula

${\xi }_{{\mathfrak{g}}^{\ast }}(\alpha )(\eta )={\frac{d}{dt}\mid }_0⟨{\exp }_G(-t\xi )\cdot \alpha ,\eta ⟩=\alpha ([\xi ,\eta ]),\eta \in \mathfrak{g},$

for the fundamental vector field ${\xi }_{{\mathfrak{g}}^{\ast }}(\alpha )=\frac{d}{dt}{\mathrm{\mid }}_0{\exp }_G(-t\xi )\cdot \alpha$ of Fundamental vector fields for a left action: the derivative of ${\mathrm{Ad}}_{{\exp }_G(t\xi )}=e^{t{\mathrm{ad}}_{\xi }}$ is ${\mathrm{ad}}_{\xi }$ (Adjoint exponential identity), and ${\mathrm{ad}}_{\xi }\eta =[\xi ,\eta ]$. Equivalently $(\xi \cdot \alpha )(\eta )=-\alpha ([\eta ,\xi ])$ if one writes the infinitesimal coadjoint action, but the formulation above is the one used in this library. A definition of the dual spaces, of the adjoint representation, of orbits and of the $\exp (-t\xi )$ convention, but of no further structure, is involved; the coadjoint action applies verbatim to disconnected $G$, to $\alpha =0$, whose orbit is the singleton $\{0\}$, and to abelian $G$, where it is trivial.

Here ${\mathrm{A}\mathrm{C}}_{\omega }$ is countable choice and is used only through the supplied fundamental-vector-field convention and the adjoint-exponential identity; no further choice is made in this definition.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $G$ be a finite-dimensional real Lie group with Lie algebra $\mathfrak{g}=T_{e}G$, let $(M,\omega )$ be a symplectic manifold (Symplectic form and symplectic manifold), and let $G\times M\to M$, $(g,p)\mapsto g\cdot p$, be a smooth left action (Smooth left actions of Lie groups). For $\xi \in \mathfrak{g}$ let ${\xi }_M$ be its fundamental vector field in the library convention

${\xi }_M(p)={\frac{d}{dt}\mid }_0{\exp }_G(-t\xi )\cdot p(p\in M)$

(Fundamental vector fields for a left action). The action is symplectic when

$g^{\ast }\omega =\omega \text{for every }g\in G,$

that is, when every $p\mapsto g\cdot p$ is a symplectomorphism. It is Hamiltonian when it is symplectic and there is a smooth map

$\mu :M⟶{\mathfrak{g}}^{\ast }$

to the algebraic dual ${\mathfrak{g}}^{\ast }=\mathcal{L}(\mathfrak{g},\mathbb{R})$ (Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$) such that

$d⟨\mu ,\xi ⟩=-{\iota }_{{\xi }_M}\omega \text{for every }\xi \in \mathfrak{g},$

and $\mu$ is equivariant for the given action on $M$ and the coadjoint action on ${\mathfrak{g}}^{\ast }$ (The coadjoint representation, action and orbits):

$\mu (g\cdot p)=g\cdot \mu (p)(g\in G,p\in M)\mathrm{.}$

Such a $\mu$ is an equivariant moment map for the action, and $(M,\omega ,G,\mu )$ is a Hamiltonian $G$-space.

Two conventions are load-bearing. First, the minus sign in the definition of ${\xi }_M$ enters through the exponential ${\exp }_G(-t\xi )$, not through the moment equation, and $d⟨\mu ,\xi ⟩=-{\iota }_{{\xi }_M}\omega$ is the identity used throughout this page. In the convention that generates $\xi$ by ${\exp }_G(t\xi )$ the same equation reads $d⟨\mu ,\xi ⟩={\iota }_{{\xi }^{\mathrm{\#}}}\omega$, so a source written that way is translated by ${\xi }^{\mathrm{\#}}=-{\xi }_M$ rather than by changing the sign of $\mu$. Second, the coadjoint action is the left action $g\cdot \alpha =\alpha \circ {\mathrm{Ad}}_{g^{-1}}$; with the opposite convention equivariance would be replaced by its inverse.

The map $\mu$ is required to be smooth but not to be a submersion, the action is not required to be free, proper, transitive, or to preserve any additional structure, and $G$ and $M$ may be disconnected; those hypotheses enter only in the theorems that use them. For $G$ with $\mathfrak{g}=0$, in particular for $G$ discrete, a Hamiltonian action is exactly a symplectic action and $\mu$ is the constant map to the zero-dimensional dual. Here ${\mathrm{A}\mathrm{C}}_{\omega }$ is countable choice; it is used exactly through the supplied fundamental-vector-field construction, which itself invokes countable choice, and no further choice is made in this definition.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let a smooth left action of $G$ on a symplectic manifold $(M,\omega )$ be given, with fundamental fields ${\xi }_M$ as in Symplectic and Hamiltonian Lie-group actions. Let $\mu :M\to {\mathfrak{g}}^{\ast }$ be a smooth map to the dual ${\mathfrak{g}}^{\ast }=\mathcal{L}(\mathfrak{g},\mathbb{R})$ (Linear functionals and the algebraic dual $V^{\ast }=\mathcal{L}(V,F)$). Its components are the smooth functions

${\mu }^{\xi }:M⟶\mathbb{R},{\mu }^{\xi }(p):=⟨\mu (p),\xi ⟩(\xi \in \mathfrak{g}),$

and they depend linearly on $\xi$, because evaluation of a fixed covector is linear. The map $\mu$ is an infinitesimal moment map for the action when the component moment equations

$d{\mu }^{\xi }=-{\iota }_{{\xi }_M}\omega \text{hold for every }\xi \in \mathfrak{g}$

hold; equivalently, when the map $\mathfrak{g}\to C^{\mathrm{\infty }}(M)$, $\xi \mapsto {\mu }^{\xi }$ is linear and each ${\mu }^{\xi }$ is a Hamiltonian function for the vector field $-{\xi }_M$. The infinitesimal moment map is an equivariant moment map when in addition

$\mu (g\cdot p)=g\cdot \mu (p)(g\in G,p\in M)$

for the coadjoint action $g\cdot \alpha =\alpha \circ {\mathrm{Ad}}_{g^{-1}}$ (The coadjoint representation, action and orbits), and a Hamiltonian action is a symplectic action that admits an equivariant moment map.

The distinction between the two notions is deliberate and is used by the nonequivariance lemma later on this page: an infinitesimal moment map is required only to satisfy the differential equations, while equivariance is an additional group-theoretic condition that can genuinely fail. The nonequivariance defect of an infinitesimal moment map is the alternating bilinear map of functions

$c(\xi ,\eta ):=\{{\mu }^{\xi },{\mu }^{\eta }\}-{\mu }^{[\xi ,\eta ]},\xi ,\eta \in \mathfrak{g},$

where $\{,\}$ is the Poisson bracket of the symplectic form. Equivariance of $\mu$ is not built into the definition of an infinitesimal moment map, and no item on this page assumes it unless it is stated. The countable-choice assumption is inherited from Symplectic and Hamiltonian Lie-group actions and is used only there; no choice is made here, and $\mu$ may be replaced by $\mu +c$ for any constant $c\in {\mathfrak{g}}^{\ast }$ without changing any component differential.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let a smooth left action of $G$ on a symplectic manifold $(M,\omega )$ be symplectic. Then every fundamental vector field ${\xi }_M$ of the action has vanishing Lie derivative on $\omega$:

${\mathcal{L}}_{{\xi }_M}\omega =0.$

Consequently each ${\xi }_M$ is a symplectic vector field, and the flow of ${\xi }_M$ consists of symplectomorphisms.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let $\mu :M\to {\mathfrak{g}}^{\ast }$ be an infinitesimal moment map for a symplectic action, so that $d{\mu }^{\xi }=-{\iota }_{{\xi }_M}\omega$ for every $\xi \in \mathfrak{g}$. Then for every $\xi$ the Hamiltonian vector field of the component ${\mu }^{\xi }$ is the negative of the fundamental field:

$X_{{\mu }^{\xi }}=-{\xi }_M\mathrm{.}$

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Let a symplectic left action of $G$ on a symplectic manifold $(M,\omega )$ be given, and let $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfy the component moment equations $d{\mu }^{\xi }=-{\iota }_{{\xi }_M}\omega$ for every $\xi \in \mathfrak{g}$.

1. If $\mu$ is coadjoint equivariant, then $\{{\mu }^{\xi },{\mu }^{\eta }\}={\mu }^{[\xi ,\eta ]}$ on all of $M$ for all $\xi ,\eta \in \mathfrak{g}$.
2. Conversely, if $G$ is connected and $\{{\mu }^{\xi },{\mu }^{\eta }\}={\mu }^{[\xi ,\eta ]}$ on all of $M$ for all $\xi ,\eta$, then $\mu$ is coadjoint equivariant.

Thus, for connected $G$ and connected $M$, coadjoint equivariance of a map satisfying the component moment equations is equivalent to the moment-map Poisson bracket identity. For a general group the bracket identity is equivalent to equivariance under the identity component $G^0$, and equivariance under all of $G$ requires in addition equivariance under one representative of each coset of $G\mathrm{/}{G}^0$. Connectivity of $M$ is not used by either implication; it is used only when the bracket identity is to be checked at a single point, because the nonequivariance defect is then constant on $M$ by the companion lemma below.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ and let $M$ be connected. Let a symplectic left action of $G$ on $(M,\omega )$ be given and let $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfy the component moment equations $d{\mu }^{\xi }=-{\iota }_{{\xi }_M}\omega$ for every $\xi \in \mathfrak{g}$. Then the nonequivariance defect

$c(\xi ,\eta ):=\{{\mu }^{\xi },{\mu }^{\eta }\}-{\mu }^{[\xi ,\eta ]},\xi ,\eta \in \mathfrak{g},$

is a constant function on $M$ for each pair $(\xi ,\eta )$, depends bilinearly and alternatingly on $(\xi ,\eta )$, and is a Chevalley--Eilenberg two-cocycle with trivial coefficients:

$c([\xi ,\eta ],\zeta )+c([\eta ,\zeta ],\xi )+c([\zeta ,\xi ],\eta )=0\text{for all }\xi ,\eta ,\zeta \in \mathfrak{g}\mathrm{.}$

Consequently, if $G$ is connected, $\mu$ is coadjoint equivariant if and only if $c=0$. For a general group the identity $c=0$ is equivalent to equivariance under the identity component $G^0$, and equivariance under all of $G$ requires in addition equivariance under one representative of each coset of $G\mathrm{/}{G}^0$ (For connected groups, equivariance is equivalent to the moment-map Poisson bracket identity). In every case the identity $c=0$ need only be verified at one point of the connected manifold $M$, because $c$ is constant there by the first part.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ and let $M$ be connected. Fix a symplectic left action of $G$ on $(M,\omega )$. If ${\mu }_1,{\mu }_2:M\to {\mathfrak{g}}^{\ast }$ are two equivariant moment maps for this action, then

${\mu }_1-{\mu }_2=\delta$

for a constant $\delta \in ({\mathfrak{g}}^{\ast }{)}^G$, the space of coadjoint-fixed covectors. Conversely, for every equivariant moment map $\mu$ and every $\delta \in ({\mathfrak{g}}^{\ast }{)}^G$, the translate $\mu +\delta$ is again an equivariant moment map. Hence the set of equivariant moment maps for a fixed action is either empty or an affine space under $({\mathfrak{g}}^{\ast }{)}^G$.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ and let $M$ be connected. Let $\mathfrak{g}$ be a finite-dimensional real semisimple Lie algebra, and let a Hamiltonian action of a Lie group $G$ with Lie algebra $\mathfrak{g}$ on $(M,\omega )$ be given. Then there is at most one equivariant moment map for the action: if one equivariant moment map exists, it is the unique one.

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$, connected $G$, connected $M$, and suppose an infinitesimal moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ is supplied for the action: that is, each closed one-form $-{\iota }_{{\xi }_M}\omega$ has a chosen Hamiltonian function ${\mu }^{\xi }$, depending linearly on $\xi$. If $\mathfrak{g}$ is finite-dimensional real semisimple, then there is a covector $b\in {\mathfrak{g}}^{\ast }$ such that

$\mu -b:M⟶{\mathfrak{g}}^{\ast },(\mu -b{)}^{\xi }={\mu }^{\xi }-b(\xi ),$

is a coadjoint-equivariant moment map. Thus constants can be added to a supplied infinitesimal moment map to make it equivariant. The argument assumes the linear choice of Hamiltonians and does not prove that the component one-forms ${\iota }_{{\xi }_M}\omega$ are exact; it removes only the obstruction to equivariance.