moment map

Definition

A smooth function $\mu :M\to {\mathfrak{g}}^{\ast }$ is called a moment map if

• ${\xi }_M=X_{⟨\mu ,\xi ⟩}$ (i.e. ${\iota }_{{\xi }_M}\omega =d⟨\mu ,\xi ⟩$) for all $\xi \in \mathfrak{g}$
• $\mu$ is $G$-equivariant

Where

• ${\xi }_M$ is the fundamental vector field of $\xi \in \mathfrak{g}$ of the Lie algebra of the Lie group $G$.
• $⟨\mu (p),\xi ⟩=\mu (p)(\xi )$ which evaluates $\xi$ in $\mu (p)\in {\mathfrak{g}}^{\ast }$

Motivation

The goal is to form a systematic way to choose the Hamiltonian functions needed to make a Hamiltonian action.

So, we are looking for some map (this is sometimes called the comoment map)

$$\begin{array}{rl}\tilde{\mu }:\mathfrak{g}& ⟶C^{\infty }(M)\\ \xi & ⟼H_{\xi }\end{array}$$

such that

${\iota }_{{\xi }_M}\omega =d{H}_{\xi }.$

We know that we need this map to be linear on $\mathfrak{g}$, so instead of defining the map itself, we can map into ${\mathfrak{g}}^{\ast }$ which guarantees linearity.

Consider a map

$\mu :M\to {\mathfrak{g}}^{\ast }$

then using the pairing $⟨f,\xi ⟩:-f(\xi )$ for all $f\in {\mathfrak{g}}^{\ast }$ and $\xi \in \mathfrak{g}$ get a map

$⟨\mu (-),\xi ⟩:M⟶\mathbb{R}.$

So the goal is to define $\mu$ such that

$$\begin{array}{rl}{\mu }^{\ast }:\mathfrak{g}& ⟶C^{\infty }(M)\\ \xi & ⟼⟨\mu (-),\xi ⟩\end{array}$$

is a Lie algebra homomorphism.

Equivariance

What does equivariance mean here? Usually, equivariance is taken to mean $\varphi (g\cdot p)=g\cdot \varphi (p)$ So in this context, we consider the coadjoint action of $G$ on ${\mathfrak{g}}^{\ast }$:

$$\begin{array}{rl}\mu (g\cdot p)& =g\cdot \mu (p)\\ & ={\text{Ad}}_{g^{-1}}^{\ast }\mu (p)\end{array}$$

Why require equivariance? At first glance, this would seem like a somewhat arbitrary condition.

Equivariance of moment map

Theorem: A map $\mu :M\to {\mathfrak{g}}^{\ast }$ is equivariant if and only if ${\mu }^{\ast }:\mathfrak{g}\to C^{\infty }(M)$ is a Lie algebra homomorphism (using the definitions of ${\mu }^{\ast }$ given $\mu$ above.)

Proof:todo