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 $({\frak i}\Bbb R)^*\cong \Bbb R$): $$\mu(z) =-\frac{{\frak i}k}{2}|z|^2. $$


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