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.

On the complex projective line, use the symplectic formand the circle group action $z\mapsto e^{i\theta }z$. Its fundamental vector field is $K=iz{\partial }_z-i\bar{z}{\partial }_{\bar{z}}$. For $H(z)=|z{|}^2/(1+|z{|}^2)$, direct contraction gives ${\iota }_K\omega =-dH$.

Thus $H$ is the moment component in the convention ${\iota }_{X_H}\omega =-dH$. In the plus-sign convention $d\mu ={\iota }_K\omega$ used in Hamiltonian group action, the same generator instead has $\mu =-H$, or $\mu =1-H$ after a permitted additive constant. Reversing the chosen Lie algebra generator reverses the component as well. These choices describe the same rotation with explicitly matched signs.

In the other chart $w=1/z$, $H=1/(1+|w{|}^2)$, so it extends smoothly at infinity. Its range is $[0,1]$, with endpoints at the two fixed poles. In polar coordinates, $\omega =2rdr\wedge d\theta /(1+r^2{)}^2$ has total symplectic area $2\pi$. Scaling to area $4\pi$ doubles the corresponding moment component.