# Infinitesimal generator of a smooth $S^1$-action over $\mathbb C$

Question: Consider the symplectic manifold $$(\mathbb C, \omega_0 = \frac{i}{2} {\rm d}z \wedge {\rm d}\bar z)$$ and a smooth $$\Bbb S^1$$-action over $$\mathbb C$$ given by $$(t,z) \mapsto t^kz$$ for some fixed $$k \in \mathbb Z$$, with moment map $$\mu: \mathbb C \to \frak {g}^* \cong i\mathbb R$$ given by $$\mu (z) = -\frac{i}{2}k|z|^2$$ What is the infinitesimal generator of this action $$(X^\#)_{z_0} = \displaystyle \frac{d}{dt}\bigg|_{t=0} \exp(tX)\cdot z_0$$?

Attempt: I tried to use polar coordinates $$z = r e^{i\theta}$$. Then the symplectic 2-form is $$\omega_0 = r {\rm d}r \wedge {\rm d}\theta$$ and the moment map should be $$\mu (re^{i\theta}) = -\frac{i}{2}kr^2$$ if we consider the global flow $$\exp : \mathfrak g = \text{Lie} (S^1) \cong i\mathbb R \to S^1$$, $$\exp (is) = e^{is}$$ then

\begin{align}(X^\#)_{z_0} &= \frac{d}{dt}\bigg|_{t=0} \exp(tX)\cdot z_0 = \frac{d}{dt}\bigg|_{t=0} e^{its}\cdot z_0 \\&= \frac{d}{dt}\bigg|_{t=0} e^{itks} z_0 = iksz_0 \\&= iksr_0 e^{i\theta_0} = -ksr_0 \sin \theta_0 + i ksr_0 \cos \theta_0\end{align}

I can't see how this is a vector field written in coordinates $$\frac{\partial}{\partial r}$$ and $$\frac{\partial}{\partial \theta}$$. Any help?

I'll write $$u$$ for elements of $$\Bbb S^1$$, so we don't get confused with the same letter $$t$$ both for an element of $$\Bbb S^1$$ and for parameters of curves. This way, we have $$\Bbb S^1\circlearrowright \Bbb C$$ given by $$u\cdot z\doteq u^kz$$.
The action is symplectic because if $$w=u^kz$$, we have $${\rm d}w\wedge {\rm d}\overline{w} = (u^k {\rm d}z)\wedge (\overline{u^k}{\rm d}\overline{z})= |u|^{2k}{\rm d}z\wedge {\rm d}\overline{z}={\rm d}z\wedge {\rm d}\overline{z},$$since $$u\in \Bbb S^1$$ means $$|u|=1$$.
To compute the infinitesimal generator, let $${\frak i}a\in T_1 (\Bbb S^1)={\frak i}\Bbb R$$ (i.e., $$a\in \Bbb R$$) and consider $$\alpha:\Bbb R\to \Bbb S^1$$ given by $$\alpha(t)=e^{{\frak i}at}$$. Then $$\alpha(0)=1$$ and $$\alpha'(0)={\frak i}a$$, so that we have$$(({\frak i}a)^\#)_z = \frac{\rm d}{{\rm d}t}\bigg|_{t=0} \alpha(t)\cdot z = \frac{\rm d}{{\rm d}t}\bigg|_{t=0} e^{{\frak i}atk}z ={\frak i}kaz,$$which corresponds under $$\Bbb C\cong T_z\Bbb C$$ to $$(({\frak i}a)^\#)_z = {\frak i}kaz\frac{\partial}{\partial z} -{\frak i}ak\overline{z}\frac{\partial}{\partial\overline{z}}.$$Now \begin{align}\iota_{({\frak i}a)^\#}\omega_0 &=\omega_0 (({\frak i}a)^\#,\cdot) =\frac{\frak i}{2} \begin{vmatrix} {\frak i}kaz & {\rm d}z \\ -{\frak i}ka\overline{z} & {\rm d}\overline{z}\end{vmatrix}\\ &=\frac{1}{2} (-kaz \,{\rm d}\overline{z} - ka\overline{z}\,{\rm d}z) = \frac{-ka}{2} {\rm d}(|z|^2) \\ &= {\rm d}\left(-\frac{ka|z|^2}{2}\right) = -{\rm d}\left(-\frac{{\frak i}k({\frak i}a)|z|^2}{2}\right).\end{align}The comoment map is $$\hat{\mu}:{\frak i}\Bbb R\to \mathscr{C}^\infty (\Bbb C)$$ satisfying the relation $$\iota_{({\frak i}a)^\#}\omega_0 = -{\rm d}\mu^{{\frak i}a}$$, right? So we've proved that $$\mu^{{\frak i}a}(z) =-\frac{{\frak i}k({\frak i}a)|z|^2}{2}.$$And the moment map is $$\mu:\Bbb C \to ({\frak i}\Bbb R)^*$$ given by duality and identified with its action on $$1$$ (which corresponds to $$\frak -i$$ under $$({\frak i}\Bbb R)^*\cong \Bbb R$$): $$\mu(z) =-\frac{{\frak i}k}{2}|z|^2.$$