Weighted circle actions and weighted projective singular quotients

With $z_j=x_j+i{y}_j$, contraction of the weighted fundamental field against ${\omega }_0$ gives ${\iota }_{{\xi }_M}{\omega }_0={\sum }_j{w}_j(y_{j}d{y}_j+x_{j}d{x}_j)=d\left(\frac{1}{2}{\sum }_j{w}_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2\right),$ so $\mu (z)=-\frac{1}{2}{\sum }_j{w}_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2+c$ satisfies the component equation for every constant $c$ by [F2].

The function $\mu$ is invariant under the weighted action because each $\mathrm{\mid }{z}_j\mathrm{\mid }$ is, and the coadjoint action of the circle is trivial; hence $\mu$ is an equivariant moment map.

Since $n\ge 1$ and $c>0$, the level ${\mu }^{-1}(0)$ is the nonempty ellipsoid ${\sum }_j{w}_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2=2c$. For each $j$, it contains the point with $\mathrm{\mid }{z}_j{\mathrm{\mid }}^2=2c\mathrm{/}{w}_j$ and all other coordinates zero. At that point the equation $e^{i{w}_j\theta }{z}_j=z_j$ holds exactly when $e^{i{w}_j\theta }=1$, so the stabilizer is the cyclic group of order $w_j$. If $w_j\ge 2$ this is nontrivial, so the action on this level is not free and the hypotheses of [F3] fail. At every point of the level some coordinate is nonzero, so $d\mu =-{\sum }_j{w}_j(x_{j}d{x}_j+y_{j}d{y}_j)$ is nonzero: the value is regular. Stabilizers are finite since they are intersections of the cyclic groups for the nonzero coordinates. The action is proper because the circle is compact. Thus it is precisely freeness that fails when a weight exceeds one.

Define weighted projective space as $({\mathbb{C}}^n\setminus \{0\})\mathrm{/}{\mathbb{C}}^{\ast }$ for the action $\lambda \cdot z=({\lambda }^{w_j}{z}_j{)}_j$. For each nonzero $z$, the function $f_z(r)={\sum }_j{w}_j{r}^{2w_j}\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$ is continuous and strictly increasing from $0$ to infinity on $r>0$, so has a unique solution $r(z)$ to $f_z(r)=2c$. Its derivative is positive; the implicit function theorem shows $r(z)$ is smooth. Radial normalization $z\mapsto r(z)\cdot z$ meets every complex orbit in exactly one circle orbit, since $\lambda =r{e}^{i\theta }$ uniquely. This normalization and the level inclusion induce mutually inverse continuous quotient maps, identifying the level quotient with weighted projective space.

On the open set $z_j\mathrm{\ne }0$, choose a complex scalar with ${\lambda }^{w_j}{z}_j=1$. The residual ambiguity is precisely ${\zeta }^{w_j}=1$, acting on the remaining coordinates by $u_k\mapsto {\zeta }^{w_k}{u}_k$. Thus the quotient has chart ${\mathbb{C}}^{n-1}\mathrm{/}{\mathit{\mu }}_{w_j}$. Locally on overlaps one chooses a root of the nonzero coordinate; the corresponding coordinate substitutions are holomorphic with holomorphic inverses, and two choices differ by the indicated finite group actions. These charts give the natural, possibly ineffective orbifold structure. At the axis point the whole chart group fixes the origin. At a general point the stabilizer is the subgroup fixing all nonzero coordinates, hence cyclic as in step 3.1. The common ineffective subgroup has order $\gcd (w_1,\dots ,w_n)$: a scalar acts identically exactly when all its $w_j$-th powers equal one. This chart construction is independent of any unproved singular-reduction assertion in [F3].

For $n=1$ the coarse quotient is a point and its orbifold chart retains the group ${\mathit{\mu }}_{w_1}$ acting trivially. If all weights are one, all stabilizers are trivial and the charts are the ordinary projective charts. Nontrivial isotropy need not make the coarse space nonmanifold (already a finite rotation quotient of $\mathbb{C}$ has underlying space $\mathbb{C}$). The example therefore exhibits failure of the free-action hypothesis, not a claim that every nonfree quotient fails to be a manifold. The assumption $n\ge 1$ excludes the empty level at $n=0$ and $c>0$.