Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 140 / 6

For $X\in \mathfrak{h}$, the fundamental vector field of the restricted Lie group action is $(iX{)}_M$. If $\varphi$ is the original moment map, thenThis is the defining Hamiltonian action identity, in the convention ${\iota }_{X_H}\omega =dH$.

If equivariance is included in the definition of a moment map, it is also preserved: the dual map $\pi =i^{\ast }$ intertwines the coadjoint actions of $H$, since $i({\mathrm{Ad}}_{h}X)={\mathrm{Ad}}_h(iX)$. Thusis a moment map for the restricted Hamiltonian action of the Lie subgroup $H$.

The diagonal Lie algebra inclusion is $i(a)=(a,\dots ,a)$. Its dual map is characterized byConsequently

Use the paper's Fubini-Study form normalization ${\tilde{\omega }}_{FS}=\frac{i}{2}\partial \overline{\partial }\log (1+|z{|}^2)$, the Hamiltonian vector field convention ${\iota }_{X_H}\omega =dH$, and the Lie algebra generator whose circle action is $e^{i\theta }$. Write $D=|z_0{|}^2+|z_1{|}^2+|z_2{|}^2$. The normalized moment map for the torus action isThe moment map image is the filled triangleIndeed, the three quantities $|z_j{|}^2/D$ are nonnegative and add to one, and any such triple is realized by choosing their square roots as homogeneous coordinates.

For the diagonal circle action, a point in the fixed-point set satisfies $[z_0,t{z}_1,t{z}_2]=[z_0,z_1,z_2]$ for every $t\in S^1$. If $z_0\ne 0$, the projective scaling must be one, forcing $z_1=z_2=0$. If $z_0=0$, all coordinates are scaled together. ThereforeBy part (b), the diagonal moment map isThe factor $1/2$ follows from the specified Fubini-Study form, not the integral normalization in the linked general article. For example, in one affine complex coordinate $w=r{e}^{i\theta }$ the symplectic form is $r(1+r^2{)}^{-2}dr\wedge d\theta$, whose interior product of a differential form with ${\partial }_{\theta }$ is $-r(1+r^2{)}^{-2}dr=d[-r^2/(2(1+r^2))]$. Choosing the generator $e^{2\pi it}$ rescales all moment maps by $2\pi$; reversing the defining sign reverses their signs. The negative level in parts (d) and (e) uses the convention displayed here.

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

The level equation is $|z_1{|}^2+|z_2{|}^2=|z_0{|}^2$. In particular $z_0\ne 0$, so the global affine coordinates on this level are $w_j=z_j/z_0$. They identify it withThe diagonal circle action becomes $(w_1,w_2)\mapsto (t{w}_1,t{w}_2)$ and is free: at least one coordinate is nonzero, so fixing a point forces $t=1$. Since the circle is compact, the Free proper Lie-group action theorem makes the quotient a smooth two-dimensional smooth manifold.

The map $(w_1,w_2)\mapsto [w_1,w_2]$ is the Hopf fibration. It is onto the complex projective line, and two unit representatives determine the same complex line precisely when they differ by a scalar of modulus one, hence by a circle orbit. Local sections, for example $[1,u]\mapsto (1,u)/\sqrt{1+|u{|}^2}$, show that the induced identification is a diffeomorphism. ThereforeThis identifies the symplectic reduction as a smooth manifold; no computation of its reduced symplectic form is needed.