Moment map ($\mu :M\to {\mathfrak{g}}^{\ast }$)

A moment map assigns to each element of a Lie algebra a Hamiltonian function for its fundamental vector field, and is equivariant for the coadjoint action. For an abelian Lie group, constant shifts are allowed; the numerical scale also depends on the chosen Lie algebra generators and the normalization of the symplectic form.

On the canonical cotangent bundle $T^{\ast }{\mathbb{R}}^n$, $\omega =i\sum d{z}_j\wedge d{\bar{z}}_j$ and $H=z^{†}z$. The unitary group acts by $z\mapsto Uz$, preserving both. For a Hermitian $A$, the generator $-iA$ has Hamiltonian ${\mu }_A=z^{†}Az$. The Poisson bracket satisfies $\{{\mu }_A,{\mu }_B\}={\mu }_{-i[A,B]}$, with no central term. The central generator has moment map $H$; imaginary antisymmetric matrices give ordinary angular momentum, while real symmetric matrices give quadratic observables mixing position and momentum.

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.

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 semisimple Lie algebra, its nondegenerate Killing form turns a two-cocycle into an operator $D$ through $B(DX,Y)=\kappa (X,Y)$. The cocycle identity makes $D$ a derivation. Choose $Z$ by $B(Z,X)=\mathrm{tr}(D\mathrm{ad}X)$. Using $[D,\mathrm{ad}X]=\mathrm{ad}(DX)$ and cyclicity of trace gives $B([Z,X],Y)=B(DX,Y)$, so $D=\mathrm{ad}Z$. Consequently $\kappa (X,Y)=B(Z,[X,Y])$ is a coboundary. This proves that component Hamiltonians can be normalized to a Lie algebra homomorphism.

For the symplectic form $\omega =idz\wedge d\bar{z}/(1+|z{|}^2{)}^2$, the circle group rotates $z$ by $e^{i\theta }$. Its real generator is $K=iz{\partial }_z-i\bar{z}{\partial }_{\bar{z}}$. The contraction is ${\iota }_K\omega =-dH$, so $H$ is a Hamiltonian function in this sign convention. In the other chart $w=1/z$, $H=1/(1+|w{|}^2)$, proving smoothness at infinity. Its image is $[0,1]$ with the endpoints at the fixed poles. The symplectic area is $2\pi$; rescaling to the round area $4\pi$ would double this moment function.