Circle rotation on complex n-space and its quadratic moment map

For $\xi \in \mathbb{R}$ and $z_j=x_j+i{y}_j$, the curve $t\mapsto e^{-it\xi }z$ has velocity $\iota (\xi {)}_M=\xi {\sum }_j(y_j{\mathrm{\partial }}_{x_j}-x_j{\mathrm{\partial }}_{y_j})$ at $t=0$.

Contracting with ${\omega }_0$ using [F2] gives ${\iota }_{\iota (\xi {)}_M}{\omega }_0=\xi {\sum }_j(y_{j}d{y}_j+x_{j}d{x}_j)=d\left(\frac{\xi }{2}\mathrm{\mid }z{\mathrm{\mid }}^2\right)\mathrm{.}$

Under the dual identification in the Given block, the component of $\mu (z)=-\frac{1}{2}\mathrm{\mid }z{\mathrm{\mid }}^2+c$ at $\iota (\xi )$ is ${\mu }^{\iota (\xi )}(z)=\xi (-\frac{1}{2}\mathrm{\mid }z{\mathrm{\mid }}^2+c)$. Hence step 2.1 gives $d{\mu }^{\iota (\xi )}=-{\iota }_{\iota (\xi {)}_M}{\omega }_0$ for every $\xi$, so the displayed scalar formula defines a genuine $({\mathfrak{s}}^1{)}^{\ast }$-valued moment map.

The map $\mu$ is invariant: $\mathrm{\mid }{e}^{i\theta }z\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$. Since the coadjoint action of $S^1$ is trivial, invariance is equivariance, so $\mu$ is an equivariant moment map. The opposite quadratic $+\frac{1}{2}\mathrm{\mid }z{\mathrm{\mid }}^2$ has differential $+{\iota }_{{\xi }_M}{\omega }_0$ and therefore does not satisfy the library equation.