Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 16 / 5 / c

The Marsden-Weinstein theorem says that for a Hamiltonian group action, if $c$ is a regular value fixed by the coadjoint action and $G$ acts freely and properly on ${\mu }^{-1}(c)$, thenis a symplectic manifold with a unique form characterized by ${\pi }^{\ast }{\omega }_c={\iota }^{\ast }\omega$, where $\iota$ includes the level set and $\pi$ is the quotient projection. Its dimension is $\dim M-2\dim G$.

Equip ${\mathbb{C}}^{n+1}$ with the standard symplectic formLet the circle group act by $e^{i\theta }\cdot z=e^{i\theta }z$. Its generator is $X={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$, andConsequently the moment map in our convention is $\mu (z)=-|z{|}^2/2$. It is invariant, hence equivariant because the circle group is abelian. The level ${\mu }^{-1}(-1)$ is the sphere of radius $\sqrt{2}$; it is regular, the circle group acts freely there, and properness follows from compactness of the group. The quotient is Complex projective space, by $z\mapsto [z]$, a scaled Hopf fibration. The Marsden-Weinstein theorem therefore produces a reduced symplectic form on ${\mathbb{CP}}^n$.

To identify it rather than only assert its existence, use the primitiveOn the affine chart choose the local section $s(w)=\sqrt{2}(1,w)/\sqrt{S}$ of the quotient. Direct substitution givesDifferentiating, using $\bar{\partial }\partial =-\partial \bar{\partial }$, givesThis is the Fubini-Study form from circle reduction. The unit sphere instead produces half this form; our radius $\sqrt{2}$ is exactly what gives the $2\pi$ line-area normalization used above.