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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.