Moment Maps and Symplectic Reduction — Examples

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.

The zero level of $\mu$ is ${\mu }^{-1}(0)=\{z:\mathrm{\mid }z{\mathrm{\mid }}^2=2c\}$, the sphere of radius $r=\sqrt{2c}>0$.

The circle acts freely on this sphere: if $e^{i\theta }z=z$ with $z\mathrm{\ne }0$ then $e^{i\theta }=1$. The value $0$ is therefore regular by [F3], and the circle is compact, so the action is proper.

By [F2] the reduction $M_0={\mu }^{-1}(0)\mathrm{/}{S}^1$ is a symplectic manifold of dimension $2n-1-1=2n-2$, and its form is the unique form pulled back from ${\iota }^{\ast }{\omega }_0$. The quotient of the sphere by the scalar circle action is the Hopf quotient $S_r^{2n-1}\mathrm{/}{S}^1$, the standard model of ${\mathbb{C}\mathbb{P}}^{n-1}$: scalar multiplication and $z\mapsto \lambda z$ identify the same line, and the quotient is free away from the origin.

In that model the Fubini--Study form is defined exactly by the basic-form property ${\pi }^{\ast }{\omega }_{\mathrm{F}\mathrm{S}}={\iota }^{\ast }{\omega }_0$ on the sphere, so the reduced form is the Fubini--Study form in the normalization fixed by the sphere of radius $r$; replacing $c$ by ${\lambda }^{2}c$ replaces the sphere of radius $r$ by the sphere of radius $\lambda r$ and rescales the reduced form by ${\lambda }^2$. For $n=1$ the sphere is $S^1$, the circle acts transitively, and the quotient is a single point with the zero form.

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 $c>0$.

For $\xi =({\xi }_1,{\xi }_2,{\xi }_3)$ let $\widehat{\xi }=\left(\begin{array}{ccc}0& -{\xi }_3& {\xi }_2\\ {\xi }_3& 0& -{\xi }_1\\ -{\xi }_2& {\xi }_1& 0\end{array}\right)\mathrm{.}$ Then $\widehat{\xi }u=\xi \times u$, and $\xi \mapsto \widehat{\xi }$ is a linear bijection ${\mathbb{R}}^3\to \mathfrak{s}\mathfrak{o}(3)$. Direct expansion of the coordinate cross product gives $[\widehat{\xi },\widehat{\eta }]=\widehat{\xi \times \eta }$ and $-\frac{1}{2}\mathrm{tr}(\widehat{\xi }\widehat{\eta })=\xi \cdot \eta$.

Expanding [F4] gives $p\cdot (\xi \times q)=p_1({\xi }_2{q}_3-{\xi }_3{q}_2)+p_2({\xi }_3{q}_1-{\xi }_1{q}_3)+p_3({\xi }_1{q}_2-{\xi }_2{q}_1)\mathrm{.}$ Grouping these six terms by ${\xi }_i$ gives $\xi \cdot (q\times p)$. The same six-term expansion identifies $w\cdot (u\times v)$ with the determinant whose columns are $w,u,v$. Thus the scalar triple-product identity is proved from the coordinate definition.

For $R\in SO(3)$, for every $w$ the determinant identity in step 1.2 gives $(Rw)\cdot ((R\xi )\times (Ru))=\det (R)\det (w,\xi ,u)=w\cdot (\xi \times u)$. Orthogonality also makes this last expression $(Rw)\cdot R(\xi \times u)$. As $Rw$ ranges over all vectors, nondegeneracy of the dot product gives $R(\xi \times u)=(R\xi )\times (Ru)$, hence ${\mathrm{Ad}}_R\widehat{\xi }=R\widehat{\xi }{R}^{-1}=\widehat{R\xi }$. The trace pairing of step 1.1 identifies $\mathfrak{s}\mathfrak{o}(3{)}^{\ast }$ with ${\mathbb{R}}^3$; under that identification the definition of the coadjoint action gives ${\mathrm{Ad}}_R^{\ast }v=Rv$, since $v\cdot R^{-1}\xi =(Rv)\cdot \xi$.

By [F3], the fundamental field of $\widehat{\xi }$ on ${\mathbb{R}}^3$ is ${\xi }_Q(q)={\frac{d}{dt}\mid }_0\exp (-t\widehat{\xi })q=-\widehat{\xi }q=-\xi \times q\mathrm{.}$ Substituting into [F2] gives ${\mu }^{\xi }(q,p)=-p(-\xi \times q)=p\cdot (\xi \times q)\mathrm{.}$

The scalar triple product identity proved in step 1.2, $p\cdot (\xi \times q)=\xi \cdot (q\times p)$ identifies this with the linear functional $\xi \mapsto \xi \cdot (q\times p)$; under the trace-pairing identification of step 2.1, the covector $\mu (q,p)\in \mathfrak{s}\mathfrak{o}(3{)}^{\ast }$ is therefore the vector $q\times p$.

The resulting map $\mu (q,p)=q\times p$ satisfies the component moment equations by [F2]. The cotangent lift of $q\mapsto Rq$ sends the covector represented by $p$ to the one represented by $Rp$, since $p\cdot R^{-1}v=(Rp)\cdot v$. Thus under the lifted action, $(Rq)\times (Rp)=R(q\times p)$, which is exactly the coadjoint action computed in step 2.1. Together with the component equations this proves equivariance and the moment-map assertion. If $q=0$, $p=0$ or the two vectors are parallel, the formula gives zero without any division or freeness assumption. The countable-choice assumption is exactly [A1].

Freeness makes the stabilizer trivial, so $j_q$ is injective by [F5]. Since $b$ is constant along orbits, $\mathrm{im}{j}_q\subseteq \ker d{b}_q$. Both spaces have dimension $\dim G$ by injectivity and the quotient dimension/submersion assertion in [F2], so they are equal. By [F1], $\mu (q,p)=0$ exactly when $p$ annihilates $\mathrm{im}{j}_q=\ker d{b}_q$.

On vertical fibre variations $\delta p\in T_q^{\ast }Q$, the derivative of $\mu$ is $(d\mu {)}_{(q,p)}(0,\delta p)=-\delta p\circ j_q$. This is surjective onto ${\mathfrak{g}}^{\ast }$: a linear functional on $j_q(\mathfrak{g})$ extends to $T_{q}Q$ by completing a finite basis. Thus $\mu$ is a submersion everywhere, and zero is a regular value. Equivariance in [F1] makes $Z$ invariant, since every linear coadjoint map fixes zero. The assumed free proper lifted action restricts to a free proper action on $Z$: $Z$ is closed as the inverse image of zero, and the action-map preimage of a compact subset of $Z\times Z$ is the same compact preimage as in $T^{\ast }Q\times T^{\ast }Q$. Consequently [F4] supplies $Z\mathrm{/}G$ and its reduced form; its quotient map $\pi :Z\to Z\mathrm{/}G$ is a surjective submersion by [F2].

At $(q,p)\in Z$, surjectivity of $d{b}_q$ and the annihilator description in step 1.1 give a unique $\beta \in T_{b(q)}^{\ast }B$ with $p=(d{b}_q{)}^{\ast }\beta$: define $\beta (v)=p(\tilde{v})$ for any lift, independent of the lift because their difference lies in $\ker d{b}_q$. Define $\mathrm{\Psi }(q,p)=(b(q),\beta )$. It is smooth: in submersion coordinates $b(x,y)=x$, covectors in $Z$ have precisely the form $(p_x,0)$, and $\mathrm{\Psi }(x,y,p_x,0)=(x,p_x)$. Since $b\circ a_g=b$, the cotangent lift transports $(d{b}_q{)}^{\ast }\beta$ to $(d{b}_{gq}{)}^{\ast }\beta$. Thus $\mathrm{\Psi }$ is invariant, onto, and its fibres are exactly the $G$-orbits: representatives of the same point of $B$ differ by the action, and the pullback covector at each representative is unique.

The induced map $\bar{\mathrm{\Psi }}:Z\mathrm{/}G\to T^{\ast }B$ is therefore bijective. It is smooth, since local sections of $\pi$ express it locally as $\mathrm{\Psi }$ composed with a smooth section. To see its inverse is smooth, let $s:U\to Q$ be a local section of $b$ supplied by [F2]. On $T^{\ast }U$, the inverse is $(x,\beta )\mapsto \pi (s(x),(d{b}_{s(x)}{)}^{\ast }\beta )$, a smooth expression which lands in $Z$ by step 1.1. Smoothness into $Z$ also follows from the submersion covector coordinates of step 2.2. These expressions cover the target and agree by uniqueness of the orbit, proving that $\bar{\mathrm{\Psi }}$ is a diffeomorphism.

For $z=(q,p)\in Z$ and $v\in T_{z}Z$, put $\mathrm{\Psi }(z)=(b(q),\beta )$. The correctly typed projection identity is ${\tau }_B\circ \mathrm{\Psi }=b\circ {\tau }_Q\circ \iota$. Therefore $({\mathrm{\Psi }}^{\ast }{\lambda }_B{)}_z(v)=\beta (d{b}_{q}d({\tau }_Q\circ \iota {)}_{z}v)=p(d({\tau }_Q\circ \iota {)}_{z}v)=({\iota }^{\ast }{\lambda }_Q{)}_z(v)\mathrm{.}$ Here $p=(d{b}_q{)}^{\ast }\beta$ by step 2.2, and the tangent vector is a tangent to the level, the domain of $\mathrm{\Psi }$. Applying $-d$ gives ${\mathrm{\Psi }}^{\ast }{\omega }_B={\iota }^{\ast }{\omega }_Q$.

Since $\mathrm{\Psi }=\bar{\mathrm{\Psi }}\circ \pi$, steps 2.1 and 4.1 give ${\pi }^{\ast }({\bar{\mathrm{\Psi }}}^{\ast }{\omega }_B)={\pi }^{\ast }{\omega }^{\mathrm{r}\mathrm{e}\mathrm{d}}$. Pullback by a surjective submersion is injective on forms: at each base point, choose a point above it and lift every finite tuple of tangent vectors by the surjective differential to evaluate the form. Hence ${\bar{\mathrm{\Psi }}}^{\ast }{\omega }_B={\omega }^{\mathrm{r}\mathrm{e}\mathrm{d}}$, proving the canonical symplectomorphism. If $Q$ is empty both spaces are empty. If $G$ is trivial the construction is the identity; if $\dim G=0$ the derivative surjectivity onto its zero-dimensional dual is vacuous and the same descent works. Zero covectors cause no exception. No connection or choice of horizontal distribution enters the map, and the local sections used to prove smoothness do not enter its definition.

For $v=(v_1,v_2,v_3)$ put $\widehat{v}=\left(\begin{array}{ccc}0& -v_3& v_2\\ v_3& 0& -v_1\\ -v_2& v_1& 0\end{array}\right)\mathrm{.}$ Then $\widehat{v}u=v\times u$, and every skew-symmetric $3\times 3$ matrix is uniquely of this form. Direct multiplication using the coordinate cross-product formula gives $[\widehat{v},\widehat{w}]=\widehat{v\times w}$. Moreover, for $R\in SO(3)$ the scalar-triple-product identity and $\det R=1$ give $R(v\times u)=(Rv)\times (Ru)$, so $R\widehat{v}{R}^{-1}=\widehat{Rv}$. Thus the adjoint action is the standard rotation action. Under the dot-product identification $({\mathbb{R}}^3{)}^{\ast }\cong {\mathbb{R}}^3$, orthogonality of $R$ then makes the coadjoint action the same rotation action.

By step 1.1 the coadjoint orbit of $\alpha$ is the set of vectors of the same length, hence the sphere $S_r^2$ of radius $r=\mathrm{\mid }\alpha \mathrm{\mid }$ when $\alpha \mathrm{\ne }0$, and the origin when $\alpha =0$.

With the library's negative-exponential convention for fundamental fields, step 1.1 gives ${\xi }_{\mathcal{O}}(\alpha )=\alpha \times \xi$. The KKS formula [F2] therefore reads ${\omega }_{r,\alpha }({\xi }_{\mathcal{O}}(\alpha ),{\eta }_{\mathcal{O}}(\alpha ))=\alpha \cdot (\xi \times \eta )$. Write $\alpha =r\widehat{\alpha }$. Dilation intertwines the rotation actions, hence $d(D_r^{-1}{)}_{\alpha }{\xi }_{\mathcal{O}}(\alpha )={\xi }_{\mathcal{O}}(\widehat{\alpha })$ and similarly for $\eta$. The vector identity $\widehat{\alpha }\cdot ((\widehat{\alpha }\times \xi )\times (\widehat{\alpha }\times \eta ))=\widehat{\alpha }\cdot (\xi \times \eta )$ and [F3] give ${\omega }_{r,\alpha }({\xi }_{\mathcal{O}}(\alpha ),{\eta }_{\mathcal{O}}(\alpha ))=r{\omega }_{S^2,\widehat{\alpha }}(d(D_r^{-1}{)}_{\alpha }{\xi }_{\mathcal{O}}(\alpha ),d(D_r^{-1}{)}_{\alpha }{\eta }_{\mathcal{O}}(\alpha )),$ which is exactly $D_r^{\ast }{\omega }_r=r{\omega }_{S^2}$.

Consequently the total area of the coadjoint orbit $S_r^2$ is $4\pi r$, and the form is nondegenerate and closed by [F2]; the rotation action is transitive on the sphere and preserves the form, as required of a coadjoint orbit.

By [F2] the inclusion $S_r^2\hookrightarrow {\mathbb{R}}^3$ is an equivariant moment map for the rotation action with this KKS form; explicitly, for $\alpha$ of length $r$ and $\xi \in {\mathbb{R}}^3$, $d⟨\mathrm{\Phi },\xi {⟩}_{\alpha }({\eta }_{\mathcal{O}})=\alpha \cdot (\eta \times \xi )=-{\omega }_{\alpha }({\xi }_{\mathcal{O}},{\eta }_{\mathcal{O}})$, which is the moment equation of the library convention.

For $\xi \in \mathfrak{u}(k)$ and $X\in T_{A}M=M$, using [F3] and $(\xi A{)}^{\ast }=A^{\ast }{\xi }^{\ast }=-A^{\ast }\xi$, $\omega ({\xi }_M(A),X)=\mathrm{Im}\mathrm{tr}((-\xi A{)}^{\ast }X)=\mathrm{Im}\mathrm{tr}(A^{\ast }\xi X)\mathrm{.}$

With $\mu (A)=-\frac{i}{2}A{A}^{\ast }+\lambda iI$ we have $\mu (A{)}^{\ast }=-\mu (A)$, so $\mu (A)\in \mathfrak{u}(k)$ and the component is ${\mu }^{\xi }(A)=⟨\mu (A),\xi ⟩=\mathrm{Re}\mathrm{tr}(\mu (A{)}^{\ast }\xi )$; its derivative in the direction $X$ picks up $\frac{i}{2}(X{A}^{\ast }+A{X}^{\ast })$ and, using $\mathrm{tr}(X^{\ast }\xi A)=-\overline{\mathrm{tr}(A^{\ast }\xi X)}$ together with cyclicity, equals $d{\mu }_A^{\xi }(X)=-\mathrm{Im}\mathrm{tr}(A^{\ast }\xi X)=-\omega ({\xi }_M(A),X)\mathrm{.}$ Hence the component moment equations hold.

Equivariance: $\mu (gA)=-\frac{i}{2}gA{A}^{\ast }{g}^{-1}+\lambda iI=g\mu (A)g^{-1}$, which is the coadjoint action by [F4]; the added central term $\lambda iI$ is fixed. Hence $\mu$ is an equivariant moment map.

The zero level is ${\mu }^{-1}(0)=\{A:A{A}^{\ast }=2\lambda I\}$: it is nonempty because $k\le n$ and contains $A=\sqrt{2\lambda }(I_{k}0)$. If $gA=A$ on this level, then $g=gA{A}^{\ast }(2\lambda {)}^{-1}=A{A}^{\ast }(2\lambda {)}^{-1}=I$, so the action is free; [F5] therefore makes $0$ a regular value. The action is proper because $U(k)$ is compact, so the zero level is an embedded submanifold and reduction applies.

Quotient: two frames $A,A^{\mathrm{\prime }}$ with $A{A}^{\ast }=A^{\mathrm{\prime }}{A}^{\ast }=2\lambda I$ lie in the same $U(k)$-orbit exactly when their rows span the same $k$-plane, so the quotient is the Grassmannian $\mathrm{Gr}(k,n)$ of $k$-planes in ${\mathbb{C}}^n$.

Dimension check: $\dim M=2kn$ and $\dim G=\dim G_0=k^2$ because $0$ is a central coadjoint value, so by [F5] $\dim M_0=2kn-2k^2=2k(n-k)$, the dimension of $\mathrm{Gr}(k,n)$. For the form, let $S_{\lambda }(A)=\sqrt{2\lambda }A$ carry the unit-frame level $A{A}^{\ast }=I$ onto the level $A{A}^{\ast }=2\lambda I$. This map is $U(k)$-equivariant, preserves row spaces, and satisfies $S_{\lambda }^{\ast }\omega =2\lambda \omega$. Pulling the two reduction identities back along $S_{\lambda }$ therefore gives ${\bar{S}}_{\lambda }^{\ast }{\omega }_{\lambda }^{\mathrm{r}\mathrm{e}\mathrm{d}}=2\lambda {\omega }_{1\mathrm{/}2}^{\mathrm{r}\mathrm{e}\mathrm{d}}$ on the common Grassmannian quotient. Thus, with ${\omega }_{\mathrm{Gr}}:={\omega }_{1\mathrm{/}2}^{\mathrm{r}\mathrm{e}\mathrm{d}}$, the reduced form at level $\lambda$ is $2\lambda {\omega }_{\mathrm{Gr}}$, rather than one fixed form for every $\lambda$.

By [F1] each factor contributes the angular momentum $q_i\times p_i$, computed from the tautological moment map with the library's negative fundamental-field convention.

These are specifically the tautological cotangent maps by [F1], so [F3] gives ${\mu }_i(R\cdot (q_i,p_i))={\mathrm{Ad}}_R^{\ast }{\mu }_i(q_i,p_i)$ for every $R\in SO(3)$. Thus each factor meets the equivariance hypothesis of [F2], independently of any covering-group example.