Hamiltonian Action of $S^1$ on $\mathbb{C}^n$

Question

Given the n-dimensional complex space, regarded as a symplectic manifold when equipped with the usual symplectic form $\sum_i r_i dr_i \wedge d\theta_i$, we consider the action of $S^1$ defined by multiplication: $$ z\longmapsto t z $$ were $t$ denotes an element of $C^1$.

My question is: what is the vector field induced by the one parameter subgroup $e^{uX}$, were $u\in R$ and $X\in \mathfrak{g}$ the Lie algebra of $S^1$, in $\mathbb{C}^n$ itself?

The answer is trivial according to the book I'm following but I just can't figure it out. Namely this is the vector field which I would intuitively think of as rotations: $$ \partial/\partial \theta_1 \ldots \partial/\partial \theta_n$$

Answer 1

I assume that you are referring to the fundamental vector field $V_X$ determined by $X\in Lie(S^1)=i\mathbb{R}.$ In this case, we have $$V_X(z)=\frac{d}{dt}\vert_{t=0}(e^{tX}z)=Xz$$ if we regard the tangent space to $z\in\mathbb{C}^n$ as $\mathbb{C}^n$ itself.

Answer 2

I'm restricting the reasoning to $\mathbb{C}$, then generalisation should be partially straightforward.

We have $S^1=\{e^{i\varphi}|\ \varphi\in\mathbb{R}\}$ and therefore its Lie algebra $\mathfrak{g} = \{i\varphi|\ \varphi\in\mathbb{R}\}$. The defined multiplicative action is therefore: $$ e^{i\varphi}\longmapsto \psi_{e^{i\varphi}}=multiplication\ by\ e^{i\varphi} $$ or explicitly $$ \psi_{e^{i\varphi}}(z) = e^{i\varphi}z.$$

The vector field $X^\#$ defined by the one-parameter subgroup of $S^1$ is the one defined by $S^1$ itself: $$ X^\# = \frac{\mathrm{d}}{\mathrm{d}\varphi}\Big|_{\varphi = 0} \psi_{e^{i\varphi}}(z) = \frac{\mathrm{d}}{\mathrm{d}\varphi}\Big|_{\varphi = 0} e^{i\varphi}z = iz =i\ re^{i\vartheta} $$ where $\theta = arg(z)$ and $r=|z|$. Now we notice that: $$ i\ re^{i\vartheta} = re^{i(\vartheta+\pi/2)} = -r\sin \vartheta + i r\cos\vartheta\simeq -r\sin\vartheta \frac{\partial}{\partial x}+ r\sin\vartheta\frac{\partial}{\partial y}. $$ But $$ \frac{\partial}{\partial \vartheta} = \frac{\partial x}{\partial \vartheta}\frac{\partial}{\partial x} + \frac{\partial y}{\partial \vartheta}\frac{\partial}{\partial y}= -r\sin\theta \frac{\partial}{\partial x}+r\sin\vartheta\frac{\partial}{\partial y}. $$ Hence $X^\#=\partial/\partial\vartheta$ and the statement is proved for $n=1$.

In general for n dimensions we just get to consider the action of $\psi_{e^{i\varphi}}$ on each coordinate $\mathbb{C}$ plane, getting therefore: $$ \partial/\partial\vartheta_1+\ldots+\partial/\partial\vartheta_n. $$