Momentum map

One Parameter group of Diffeomorphism

Let $M$ be a manifold and $X$ a complete vector field on $M$. Let ${\rho }_t:M\to M$, $t\in \mathbb{R}$, be the family of diffeomorphisms generated by $X$. For each $p\in M$, ${\rho }_t(p)$, $t\in \mathbb{R}$, is by definition the unique integral curve of $X$ passing through $p$ at time $0$, i.e., ${\rho }_t(p)$ satisfies $\{\begin{array}{ll}{\rho }_0(p)& =p\\ & \\ \frac{d{\rho }_t(p)}{dt}& =X({\rho }_t(p)).\end{array}$ We have that ${\rho }_t\circ {\rho }_s={\rho }_{t+s}$. The map $\mathbb{R}\to \text{Diff}(M)$, $t\to {\rho }_t$, is a group homomorphism. The family $\{{\rho }_t|t\in \mathbb{R}\}$ is then called a one-parameter group of diffeomorphisms of $M$ and denoted ${\rho }_t=\exp tX$.

Action

An action of a Lie group $G$ on $M$ is a group homomorphism $\chi :G\to \text{Diff}(M).$ $g\to {\chi }_g$ The evaluation map associated with an action $\chi$ is ${\text{ev}}_{\chi }:M\times G\to \text{Diff}(M)$ $(p,g)\to {\chi }_g(p).$ The action $\chi$ is smooth if ${\text{ev}}_{\chi }$is a smooth map.

Every complete vector field gives rise to a smooth action of $\mathbb{R}$ on $M$. Conversely, every smooth action of $\mathbb{R}$ on $M$ is defined by a complete vector field.

Symplectic Action

The action $\chi$ is a symplectic action if $\chi :G\to \text{Sympl}(M,\omega )\subset \text{Diff}(M)$

Hamiltonian Action of $S^1$ or $\mathbb{R}$

A symplectic action $\psi$ of $S^1$ or $\mathbb{R}$ on $(M,\omega )$ is hamiltonian if the vector field generated by $\psi$ is hamiltonian. Equivalently, an action $\psi$ of $S^1$ or $\mathbb{R}$ on $(M,\omega )$ is hamiltonian if there is $H:M\to \mathbb{R}$ with $dH={\iota }_X\omega$, where $X$ is the vector field generated by $\psi$.

Coadjoint Representation

Let $⟨\cdot ,\cdot ⟩$ be the natural pairing between ${\mathfrak{g}}^{\ast }$ and $\mathfrak{g}$: \begin{align*} \langle \cdot, \cdot \rangle : \mathfrak{g}^* \times \mathfrak{g} &\longrightarrow \mathbb{R} \\ (\xi, X) &\longmapsto \langle \xi, X \rangle = \xi(X) \ . \end{align*} Given $\xi \in {\mathfrak{g}}^{\ast }$, we define ${\mathrm{Ad}}_g^{\ast }\xi$ by $⟨{\mathrm{Ad}}_g^{\ast }\xi ,X⟩=⟨\xi ,{\mathrm{Ad}}_{g^{-1}}X⟩,\text{for any }X\in \mathfrak{g}.$ The collection of maps ${\mathrm{Ad}}_g^{\ast }$ forms the coadjoint representation (or coadjoint action) of $G$ on ${\mathfrak{g}}^{\ast }$: \begin{align*} \mathrm{Ad}^* : G &\longrightarrow \mathrm{GL}(\mathfrak{g}^*) \\ g &\longmapsto \mathrm{Ad}^*_g \ . \end{align*}

Hamiltonian Action and Momentum Map

Let $(M,\omega )$ be a symplectic manifold, $G$ be a Lie group with $\psi :G\to \text{Sympl}(M,\omega )$, a symplectic action and $\mathfrak{g}$ is the Lie algebra of $G$, ${\mathfrak{g}}^{\ast }$ is the dual vector space of $\mathfrak{g}$. The action $\psi$ is a hamiltonian action if there exists a map $\mu :M⟶{\mathfrak{g}}^{\ast }$ satisfying:

1. For each $X\in \mathfrak{g}$, let ${\mu }^X:M\to \mathbb{R},{\mu }^X(p):=⟨\mu (p),X⟩$, be the component of $\mu$ along $X$, $X^{\#}$ be the vector field on $M$ generated by the one-parameter subgroup$\{\exp tX\mid t\in \mathbb{R}\}\subseteq G$. Then $d{\mu }^X={\iota }_{X^{\#}}\omega$ i.e., ${\mu }^X$ is a hamiltonian function for the vector field $X^{\#}$.
2. $\mu$ is equivariant with respect to the given action $\psi$ of $G$ on $M$ and the coadjoint action ${\mathrm{Ad}}^{\ast }$ of $G$ on ${\mathfrak{g}}^{\ast }$: $\mu \circ {\psi }_g={\mathrm{Ad}}_g^{\ast }\circ \mu ,\text{for all }g\in G.$ The vector $(M,\omega ,G,\mu )$ is then called a hamiltonian $G$-space and $\mu$ is a moment map

Remark

For a given Lie group action $\psi$, we can’t describe it with single vector field or flow when the Lie group is more than one dimensional, so we consider the action given by it’s Lie algebra instead, with all these actions being “Hamiltonian”, and the $\psi$ well-represented by these actions (equivariant), we define momentum map.

Proposition

For connected Lie groups, hamiltonian actions can be equivalently defined in terms of a comoment map ${\mu }^{\ast }:\mathfrak{g}⟶C^{\infty }(M),$ with the two conditions rephrased as:

1. ${\mu }^{\ast }(X):={\mu }^X$ is a hamiltonian function for the vector field $X^{\#}$,
2. ${\mu }^{\ast }$ is a Lie algebra homomorphism: ${\mu }^{\ast }[X,Y]=\{{\mu }^{\ast }(X),{\mu }^{\ast }(Y)\}$ where $\{\cdot ,\cdot \}$ is the Poisson bracket on $C^{\infty }(M)$.