Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 55 / 4

Fix the convention ${\iota }_{X_f}\omega =-df$ for a Hamiltonian vector field. Nondegeneracy of the symplectic form gives one unique smooth $X_f$ for each smooth real-valued function $f$. Define the Poisson bracket byIn coordinates with $\omega =\sum d{q}^i\wedge d{p}_i$, this is $\sum (f_{q^i}{g}_{p_i}-f_{p_i}{g}_{q^i})$. Thus the sign convention is explicit and agrees with the usual coordinate Poisson bracket. By Cartan's magic formula, ${\mathcal{L}}_{X_f}\omega =d{\iota }_{X_f}\omega +{\iota }_{X_f}d\omega =0$. The contraction–Lie derivative commutator identity now givesHence the Hamiltonian Lie algebra homomorphism is $\boxed{\Phi (f)=X_f,[X_f,X_g]=X_{\{f,g\}}}$. It is linear and onto the space of Hamiltonian vector fields by definition. For completeness, the Jacobi identity for the bracket on functions follows from closure of $\omega$: evaluating $d\omega =0$ on $X_f,X_g,X_h$ and using the displayed commutator relation gives the cyclic Jacobi sum zero. Thus this really is a map of Lie algebras, not just a bracket-preserving notation.

Its kernel is determined by nondegeneracy:On a connected manifold these are precisely the real constants; on a disconnected manifold the constant may differ on each component. Therefore the quotient by these functions is isomorphic to the Lie algebra of Hamiltonian vector fields. The printed “homeomorphis” is interpreted as homomorphism: the map before taking this quotient is not an isomorphism because of its nontrivial kernel. The alternative convention ${\iota }_{X_f}\omega =df$ requires a corresponding bracket sign change to retain this homomorphism.

Differentiate the circle group action at angle zero. The resulting real vector field isIn real coordinates $z=x+iy$, this is $K=-y{\partial }_x+x{\partial }_y$; the complex coefficients are conjugates, so it is indeed real. Its integral curves satisfy $\dot{z}=iz$ and hence $z(t)=e^{it}z(0)$, with period $2\pi$ away from the fixed points.

The affine coordinate does not cover the whole complex projective line, so check the other chart. With $w=1/z$, the same vector field is $K=-iw{\partial }_w+i\bar{w}{\partial }_{\bar{w}}$, smooth at $w=0$. It vanishes at $z=0$ and $w=0$, the two fixed poles. It therefore generates the stated global rotation of $S^2$, not merely a flow in one chart.

Set $s=|z{|}^2$. Contraction of the symplectic form with the rotation vector field givesWith the convention fixed in the unheaded solution, ${\iota }_K\omega =-dH$, so the Hamiltonian function isIts derivative is $dH=ds/(1+s{)}^2$. In the chart $w=1/z$, $H=1/(1+|w{|}^2)+C$, so it extends smoothly across infinity; the additive constant is the kernel freedom already identified. Thus $K=X_H$ globally, proving that the action is a Hamiltonian group action. With $C=0$, the moment map for rotation of the complex projective line takes values in $[0,1]$, reaching its endpoints at the two fixed poles.

One can check the normalization in polar coordinates: $\omega =2r(1+r^2{)}^{-2}dr\wedge d\varphi$ has total symplectic area $2\pi$, not $4\pi$, and $H=r^2/(1+r^2)$. Equivalently, in a polar angle with $r=\tan (\vartheta /2)$, $H=(1-\cos \vartheta )/2$. This explains the half-height factor that would be lost by replacing the printed form with the unscaled round-sphere area form. Under the opposite convention ${\iota }_{X_H}\omega =dH$, the Hamiltonian is the negative of this function, up to a constant.