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Let $\mu :P\to \mathbb{R}$ be the moment map for the circle group, assume $c$ is a regular value, and assume the circle action on $N={\mu }^{-1}(c)$ is free. The action is proper because the circle is compact. Then $N$ has dimension $2n-1$ and $P_c=N/SO(2)$ is a smooth manifold of dimension $2n-2$.

If ${\xi }_P$ generates the action, $\omega ({\xi }_P,v)=d\mu (v)=0$ for every $v\in TN$. In fact $(TN{)}^{\omega }$ is exactly the line spanned by ${\xi }_P$, since $TN=\ker d\mu$ is a hyperplane and $\omega$ is nondegenerate. Therefore the kernel of the restricted two-form is precisely the orbit direction. The restriction is invariant and horizontal, so it descends to a unique nondegenerate closed two-form ${\omega }_c$ withThis proves the Marsden-Weinstein theorem in the circle case. If the Hamiltonian $H$ is invariant, then $\{\mu ,H\}=0$: its Hamiltonian flow stays on $N$, projects to the quotient and is generated there by the descended function $H_c$. This is a reduced Hamiltonian system in $2n-2$ local variables. At a critical level or a nonfree orbit the quotient can be singular, so the smooth dimension assertion needs the stated hypotheses.

For a mass-$m$ particle in the plane with central force potential $V(r)$, work on $r>0$ and writeExterior differentiation gives $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Rotation translates $\theta$ and has moment map $\ell$. Fix $\ell ={\ell }_0$ and quotient out $\theta$ to obtain the planar rotational symplectic reductionThe reduced equations are $\dot{r}=p_r/m$ and ${\dot{p}}_r={\ell }_0^2/(m{r}^3)-V^{\prime }(r)$. The centrifugal term arises from the conserved angular momentum; it is not an extra force imposed by hand. Reconstruct the angle from $\dot{\theta }={\ell }_0/(m{r}^2)$. For ${\ell }_0\ne 0$ the origin cannot lie on the level; for ${\ell }_0=0$ the same local reduction holds away from the origin, while the full phase-space origin is a singular fixed orbit.

The Marsden-Weinstein theorem needs a Hamiltonian action, not merely an action by symplectomorphisms. Assume an equivariant moment map $\mu :P\to {\mathfrak{g}}^{\ast }$ exists, choose a coadjoint-fixed regular value $a$ (zero is the usual choice), and assume the action on $C={\mu }^{-1}(a)$ is free and proper. Regularity makes $C$ a submanifold of codimension $g=\dim G$. Equivariance and invariance of $a$ make $C$ invariant under $G$, and freeness and properness give a smooth quotient $P^{\prime }=C/G$. Thus

Let $\iota :C\hookrightarrow P$ and $\pi :C\to P^{\prime }$. For $v\in T_{p}C$ and $\xi \in \mathfrak{g}$,Moreover the conormal of $C$ is spanned by the independent $d{\mu }_{\xi }$, so nondegeneracy of $\omega$ gives $(T_{p}C{)}^{\omega }=\{{\xi }_P(p)\}$. The coadjoint-fixed value makes these orbit directions tangent to $C$. Hence the kernel of ${\iota }^{\ast }\omega$ is exactly the orbit tangent space. The restricted form is invariant and horizontal, so descends uniquely to a two-form ${\omega }^{\prime }$ withIt is closed because $\omega$ is closed. If a quotient tangent vector annihilates ${\omega }^{\prime }$, any lift lies in the kernel just identified and is vertical; the quotient vector is therefore zero. This proves nondegeneracy and completes symplectic reduction.

If the action is not free, singular strata or orbifold phenomena may occur. At a general noncentral regular value one quotients by its coadjoint stabilizer $G_a$, not by all of $G$; the regular dimension is $2n-g-\dim G_a$. These are genuine hypotheses behind the stated dimension formula.

Both proposed examples can be given explicitly. For symplectic reduction of an isotropic oscillator, take two modes withIts Hamiltonian flow is $z_j\mapsto e^{-it}{z}_j$, a circle action with moment map $H$. For $E>0$, $H^{-1}(E)$ is the sphere $S^3$ of radius $\sqrt{2E}$, and the action is free. The Hopf fibration identifies its quotient with ${\mathbb{CP}}^1\cong S^2$. On the chart $z_1\ne 0$, use $w=z_2/z_1$ and choose a section with $z_1=\sqrt{2E/(1+|w{|}^2)}$ real positive. Pulling back $\omega$ to this section givesWriting $w=x+iy$ gives $2Edx\wedge dy/(1+x^2+y^2{)}^2$, whose integral is $2\pi E$. This is the Fubini-Study form scaled by $E$ in the convention of area $2\pi$ at unit scale. The reduced dimension is $4-2=2$. At $E=0$, the circle fixes the origin and the smooth regular-level hypotheses fail.

For a unit-mass free particle in the plane, use the rotation circle action with angular-momentum moment map $\ell =x{p}_y-y{p}_x$. Choose ${\ell }_0\ne 0$ so the entire level avoids $r=0$ and the action is free. Polar canonical coordinates haveFixing $\ell ={\ell }_0$ and quotienting the rotation angle leavesThis planar rotational symplectic reduction replaces a planar free trajectory by radial motion with the centrifugal potential. A nonzero moment value is allowed because the circle is abelian and every value is coadjoint-fixed.

The Marsden-Weinstein theorem says that for a Hamiltonian group action, if $c$ is a regular value fixed by the coadjoint action and $G$ acts freely and properly on ${\mu }^{-1}(c)$, thenis a symplectic manifold with a unique form characterized by ${\pi }^{\ast }{\omega }_c={\iota }^{\ast }\omega$, where $\iota$ includes the level set and $\pi$ is the quotient projection. Its dimension is $\dim M-2\dim G$.

Equip ${\mathbb{C}}^{n+1}$ with the standard symplectic formLet the circle group act by $e^{i\theta }\cdot z=e^{i\theta }z$. Its generator is $X={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$, andConsequently the moment map in our convention is $\mu (z)=-|z{|}^2/2$. It is invariant, hence equivariant because the circle group is abelian. The level ${\mu }^{-1}(-1)$ is the sphere of radius $\sqrt{2}$; it is regular, the circle group acts freely there, and properness follows from compactness of the group. The quotient is Complex projective space, by $z\mapsto [z]$, a scaled Hopf fibration. The Marsden-Weinstein theorem therefore produces a reduced symplectic form on ${\mathbb{CP}}^n$.

To identify it rather than only assert its existence, use the primitiveOn the affine chart choose the local section $s(w)=\sqrt{2}(1,w)/\sqrt{S}$ of the quotient. Direct substitution givesDifferentiating, using $\bar{\partial }\partial =-\partial \bar{\partial }$, givesThis is the Fubini-Study form from circle reduction. The unit sphere instead produces half this form; our radius $\sqrt{2}$ is exactly what gives the $2\pi$ line-area normalization used above.