Symplectic Manifolds, Moser Stability, and Darboux–Weinstein Theory

Apply [F2] to $\omega$. Its normal-form basis consists of $r$ pairs $e_i,f_i$ and $s$ radical vectors, with $\dim V=2r+s$.

By [F1] the radical is zero, so $s=0$. Taking $n=r$ gives the asserted even dimension and the displayed standard symplectic basis. This includes $V=0$, where $n=0$ and the basis is empty.

By [F1], ${\omega }^{\mathrm{♭}}$ restricts to an isomorphism $W^{\omega }\to \mathrm{ann}W$. Finite-dimensional annihilator algebra gives $\dim \mathrm{ann}W=\dim V-\dim W$, proving the dimension identity.

Alternation shows $W\subseteq (W^{\omega }{)}^{\omega }$: if $w\in W$ and $v\in W^{\omega }$, then $\omega (w,v)=-\omega (v,w)=0$. Applying step 1.1 to $W^{\omega }$ gives $\dim (W^{\omega }{)}^{\omega }=\dim W$, so the inclusion is equality. For $W=0$ or $W=V$ the same calculation gives the stated endpoint identities.

If $L=L^{\omega }$, [F1] gives $2\dim L=2n$; [F2] then makes $L$ both isotropic and coisotropic. Thus condition 1 implies conditions 2 and 3.

If condition 2 holds, then $L\subseteq L^{\omega }$ and [F1] gives $\dim L^{\omega }=n=\dim L$, hence equality. If condition 3 holds, the reverse inclusion and the same dimension calculation likewise give equality. Thus conditions 2 and 3 each imply condition 1.

A self-orthogonal $L$ is maximal isotropic: if an isotropic $K$ contains $L$, then $K\subseteq K^{\omega }\subseteq L^{\omega }=L$, hence $K=L$.

Conversely, suppose $L$ is maximal isotropic. If $L\mathrm{\ne }{L}^{\omega }$, choose $v\in L^{\omega }\setminus L$; alternation and $v\in L^{\omega }$ make $L+\mathbb{R}v$ a strictly larger isotropic subspace, which maximality forbids. Hence $L=L^{\omega }$. This argument also covers $n=0$, when $L=0$.

The quotient is defined by [F1]. Replacing $u$ by $u+a$ with $a\in W^{\omega }$, or $v$ by $v+b$ with $b\in W^{\omega }$, does not change $\omega (u,v)$ because $u,v\in W$. Hence $\bar{\omega }$ is well-defined, bilinear, and alternating.

If $[u]$ lies in its radical, then $\omega (u,v)=0$ for every $v\in W$, so $u\in W^{\omega }$ and $[u]=0$. Thus the descended form is nondegenerate. When $W=V$ this recovers $V$; when $W$ is Lagrangian the quotient is the zero symplectic space.

On graph vectors one has $(-{\omega }_V\oplus {\omega }_W)((u,Au),(v,Av))=-{\omega }_V(u,v)+{\omega }_W(Au,Av)$. Thus the graph is isotropic exactly when $A^{\ast }{\omega }_W={\omega }_V$.

If the graph is Lagrangian, [F1] and $\dim \mathrm{graph}A=\dim V$ give $2\dim V=\dim V+\dim W$, hence $\dim V=\dim W$. Step 1.1 also says $A$ preserves the forms, which makes $A$ injective by nondegeneracy; equal dimensions make it an isomorphism.

Conversely, if $A$ is a symplectic isomorphism, step 1.1 makes its graph isotropic and its dimension is half that of $V^{-}\oplus W$, so [F1] makes it Lagrangian. If instead $A$ is a symplectic embedding with $\dim W>\dim V$, step 1.1 still gives isotropy but the half-dimension equality fails. The zero spaces cause no exception.

Fix $p\in M$. If ${\omega }_p$ is nondegenerate, [F2] supplies a basis $e_1,\dots ,e_n,f_1,\dots ,f_n$ with ${\omega }_p={\sum }_i{e}^i\wedge f^i$. Hence ${\omega }_p^n=n!e^1\wedge f^1\wedge \cdots \wedge e^n\wedge f^n\mathrm{\ne }0$.

Conversely, if $0\mathrm{\ne }v$ lies in the radical of ${\omega }_p$, then the graded contraction rule gives ${\iota }_v({\omega }_p^n)=n({\iota }_v{\omega }_p)\wedge {\omega }_p^{n-1}=0$. A nonzero top covector has nonzero contraction by every nonzero vector: extend $v$ to a basis and evaluate on the remaining basis vectors. Therefore ${\omega }_p^n=0$.

Steps 1.1--1.2 prove the equivalence at every $p$, which is exactly [F3]. For $n=0$, ${\omega }^0=1$ and the zero tangent space is nondegenerate, so the same conclusion holds. Closedness is irrelevant to this pointwise equivalence.

By [F1], ${\omega }^n$ is a smooth nowhere-zero top form; division by the positive number $n!$ preserves that property. Thus ${\mu }_{\omega }$ is a volume form.

A nonzero top covector $({\mu }_{\omega }{)}_p\in \det T_p^{\ast }M$ selects the ray of tangent determinants $v\in \det T_{p}M$ on which $({\mu }_{\omega }{)}_p(v)>0$. This ray varies smoothly and therefore defines an orientation by [F2]. For $n=0$, ${\mu }_{\omega }=1$ selects the positive sign at each point, so the boundary case is included.

The form $-\omega$ is closed and has the same radical as $\omega$, so it is symplectic. By [F2], $d\mathrm{\Omega }={\mathrm{pr}}_M^{\ast }d\omega +{\mathrm{pr}}_N^{\ast }d\eta =0$.

Under $T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N$, if $\mathrm{\Omega }((u,v),(u^{\mathrm{\prime }},v^{\mathrm{\prime }}))=0$ for all $(u^{\mathrm{\prime }},v^{\mathrm{\prime }})$, taking $v^{\mathrm{\prime }}=0$ and then $u^{\mathrm{\prime }}=0$ gives $u=0$ and $v=0$ by [F1]. Thus $\mathrm{\Omega }$ is nondegenerate, including when either factor has dimension zero.

For every $p\in L$, [F1] and [F2] give $\dim T_{p}L=n$.

Since $L$ is an embedded constant-dimensional submanifold, its dimension equals the dimension of any tangent space, hence $\dim L=n$. This includes $n=0$.

The expression $p(d\pi \xi )$ in [F1] involves intrinsic maps and their natural pairing, so changing coordinates cannot change its value. It is linear in $\xi$, hence defines a covector at every $(q,p)$.

Write $p={\sum }_i{p}_{i}d{q}^i{\mathrm{\mid }}_q$ and $\xi ={\sum }_i{a}^i{\mathrm{\partial }}_{q^i}+{\sum }_i{b}_i{\mathrm{\partial }}_{p_i}$. Since $d\pi (\xi )={\sum }_i{a}^i{\mathrm{\partial }}_{q^i}$, [F1] gives $\lambda (\xi )={\sum }_i{p}_i{a}^i=({\sum }_i{p}_{i}d{q}^i)(\xi )$. The displayed coefficients are smooth, so $\lambda$ is smooth.

From [F1], $-d\lambda =-{\sum }_{i}d{p}_i\wedge d{q}^i={\sum }_{i}d{q}^i\wedge d{p}_i$. Also $d{\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}=-d^2\lambda =0$.

For $X={\sum }_i(a^i{\mathrm{\partial }}_{q^i}+b_i{\mathrm{\partial }}_{p_i})$, step 1.1 gives ${\iota }_X{\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}={\sum }_i(a^{i}d{p}_i-b_{i}d{q}^i)$. This vanishes only when all $a^i,b_i$ vanish, so the form is nondegenerate. For $n=0$ the same assertion is vacuous. Thus it is symplectic.

The formula for $\widehat{f}$ is smooth with inverse $\widehat{f^{-1}}$, and the projections satisfy ${\pi }_{Q^{\mathrm{\prime }}}\circ \widehat{f}=f\circ {\pi }_Q$. For $\xi \in T_{(q,p)}{T}^{\ast }Q$, the tautological definition gives $({\widehat{f}}^{\ast }{\lambda }_{Q^{\mathrm{\prime }}})(\xi )=((d{f}_q^{-1}{)}^{\ast }p)(d{\pi }_{Q^{\mathrm{\prime }}}d\widehat{f}\xi )=p(d{\pi }_Q\xi )={\lambda }_Q(\xi )$.

By [F1], [F2], and step 1.1, ${\widehat{f}}^{\ast }{\omega }_{Q^{\mathrm{\prime }}}=-{\widehat{f}}^{\ast }d{\lambda }_{Q^{\mathrm{\prime }}}=-d({\widehat{f}}^{\ast }{\lambda }_{Q^{\mathrm{\prime }}})=-d{\lambda }_Q={\omega }_Q$. Hence the diffeomorphism $\widehat{f}$ is symplectic. The case $\dim Q=0$ is included.

Since $\pi \circ s_{\alpha }={\mathrm{id}}_Q$, the tautological formula gives $s_{\alpha }^{\ast }\lambda =\alpha$. Therefore [F1] and [F4] give $s_{\alpha }^{\ast }{\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}=-d\alpha$.

The graph section is an embedding and its image has dimension $n$, half of $\dim T^{\ast }Q=2n$. By [F2]--[F3], it is Lagrangian exactly when the pulled-back symplectic form vanishes. Step 1.1 says this occurs exactly when $d\alpha =0$, proving both directions, including $n=0$.

Nondegeneracy of $\omega$ uniquely defines an invertible $A\in \mathrm{End}(V)$ by $k(u,v)=\omega (u,Av)$. Skew-symmetry of $\omega$ gives $k(u,A^{-1}v)=-k(A^{-1}u,v)$, so $A^{\ast }=-A$. Hence $-A^2=A^{\ast }A$ is positive definite.

Let $P=(-A^2{)}^{1\mathrm{/}2}$ be the positive square root from [F1]. Since $A$ commutes with $-A^2$, it preserves each eigenspace of $-A^2$; on that eigenspace $P$ is multiplication by the positive square root of the eigenvalue, so $P$ commutes with $A$. Define $J=A{P}^{-1}$. Then $J^2=A^2{P}^{-2}=-I$.

Since $P$ is positive definite and commutes with $A$, $\omega (u,Jv)=\omega (u,A{P}^{-1}v)=k(u,P^{-1}v)=k(P^{-1\mathrm{/}2}u,P^{-1\mathrm{/}2}v)\mathrm{.}$ This is symmetric and positive definite, so [F2] makes $J$ compatible. For $V=0$ the same formulas give the unique endomorphism, and all conditions are vacuous.

In each fibre [F1] gives a unique positive-definite self-adjoint square root $S_x=T_x^{1\mathrm{/}2}$. These fibre maps automatically define a bundle endomorphism set-theoretically; it remains to prove local smoothness.

Fix $x_0$ and a smooth orthonormal frame near it, so self-adjoint maps are symmetric matrices. For $\mathrm{\Phi }(R)=R^2$, the derivative at the positive matrix $S_{x_0}$ is $D{\mathrm{\Phi }}_{S_{x_0}}(H)=S_{x_0}H+H{S}_{x_0}$. In an orthonormal eigenbasis of $S_{x_0}$ its $(i,j)$ entry is $(s_i+s_j)H_{ij}$; every $s_i>0$, so this derivative is an isomorphism on symmetric matrices.

Apply [F2] to $R^2-T_x=0$. It produces a unique smooth symmetric solution $R(x)$ near $x_0$ with $R(x_0)=S_{x_0}$. After shrinking, positivity persists; fibrewise uniqueness in [F1] then gives $R(x)=S_x$. Thus $S$ is smooth near every point, and the unique local roots agree on overlaps. In rank zero the unique empty endomorphism supplies the result.

Spend the assumed ${\mathrm{A}\mathrm{C}}_{\omega }$ only through [F1] to choose a smooth Riemannian metric $k$. Define the smooth invertible bundle map $A$ by $k(u,v)=\omega (u,Av)$. Fibrewise skew-symmetry gives $A^{\ast }=-A$, so $-A^2=A^{\ast }A$ is smooth, self-adjoint, and positive definite.

By [F2], $P=(-A^2{)}^{1\mathrm{/}2}$ is a smooth positive bundle endomorphism. Fibrewise, $A$ preserves the eigenspaces of $-A^2$ and $P$ is scalar on each of them, so $P$ commutes with $A$. Hence $J=A{P}^{-1}$ is smooth and $J^2=-I$.

Fibrewise, $\omega (u,Jv)=k(u,P^{-1}v)=k(P^{-1\mathrm{/}2}u,P^{-1\mathrm{/}2}v),$ which is symmetric and positive definite. Thus [F3] proves compatibility. Empty and zero-dimensional manifolds carry the unique such structure.

The first formula is [F1] with initial time zero. If $d{\omega }_t=0$, [F2] turns its parenthesis into $d{\sigma }_t+d({\iota }_{X_t}{\omega }_t)$. Thus the Moser equation ${\iota }_{X_t}{\omega }_t=-{\sigma }_t$ makes the derivative zero.

For nondegenerate ${\omega }_t$, the smooth bundle map ${\omega }_t^{\mathrm{♭}}:TM\to T^{\ast }M$ is invertible, so the unique solution is $X_t=-({\omega }_t^{\mathrm{♭}}{)}^{-1}{\sigma }_t$. Matrix inversion in local coordinates proves joint smoothness in $(t,p)$.

Use [F1] to fix one Riemannian metric. The set of all strongly convex open neighbourhoods is an open cover, so compactness extracts a finite subcover $\mathcal{U}=(U_1,\dots ,U_r)$. Every nonempty finite intersection $U_{i_0\cdots i_s}$ is strongly convex by [F1]. There are only finitely many such intersections; choose one point in each and contract the intersection to it along the unique smoothly endpoint-dependent geodesics. By [F3], these contractions give fixed linear Poincaré homotopy operators $K_{i_0\cdots i_s}$. In local coordinates their coefficients are finite-interval integrals of coefficients of the pulled-back form and the fixed smooth contraction. Differentiation under that compact integral therefore shows directly that each $K_{i_0\cdots i_s}$ preserves smooth dependence on the finite-dimensional parameter.

Use [F2] finitely many times to fix a partition of unity subordinate to $\mathcal{U}$. Start the Čech--de Rham descent with $b_i=K_i(\alpha {\mathrm{\mid }}_{U_i})$, so $d{b}_i=\alpha {\mathrm{\mid }}_{U_i}$. The alternating differences $\delta b$ are closed because $d\delta =\delta d$. On each nonempty double intersection apply its fixed $K_{ij}$ to obtain a primitive; subtracting it makes the next alternating discrepancy closed one degree lower. Repeat. After at most $k$ repetitions the remaining discrepancy is a Čech cocycle of locally constant functions on the finite good cover.

Regard the last cocycle as a vector in the finite-dimensional simplicial cochain complex of the nerve. Because $\alpha$ is globally exact, comparison with any global primitive shows that this cocycle lies in the image of the preceding Čech coboundary. Fix a linear right inverse of that coboundary on its image by choosing bases once. Solve there, then reverse the finite descent. At the final gluing step the fixed partition gives a global $(k-1)$-form $R\alpha$ with $dR\alpha =\alpha$. Thus $R$ is one fixed linear operator on the space of exact $k$-forms; no primitive of an individual input was selected.

Put ${\beta }_p=R{\alpha }_p$. Restriction, the finitely many homotopy integrals, Čech differences, multiplication by fixed partition functions, and the fixed finite-dimensional linear solver all commute with differentiation in the finite-dimensional parameter. Hence ${\beta }_p$ is jointly smooth and $d{\beta }_p={\alpha }_p$. The empty manifold is immediate. The sole nonfinite choice input is the metric supplied under ${\mathrm{A}\mathrm{C}}_{\omega }$; all subsequent selections are finite.

Put ${\alpha }_t={\omega }_t-{\omega }_0$. Constancy of the de Rham class says each ${\alpha }_t$ is exact. Although $[0,1]$ is not a boundaryless parameter manifold, the proof of [F1] constructs one fixed linear primitive operator from a finite good cover, finite spatial homotopy integrals, a finite-dimensional linear solver, and a fixed partition of unity. Apply that same operator pointwise to ${\alpha }_t$. Every one of its finite operations preserves all one-sided time derivatives and joint spatial smoothness at the closed endpoints, so ${\lambda }_t=R{\alpha }_t$ is smooth up to $t=0,1$ and satisfies $d{\lambda }_t={\alpha }_t$. Put ${\sigma }_t={\dot{\lambda }}_t$; differentiating gives $d{\sigma }_t={\dot{\omega }}_t$. By [F2], the equations ${\iota }_{X_t}{\omega }_t=-{\sigma }_t$ have a unique jointly smooth solution $X_t$ up to both endpoints.

We first put the field on a genuinely open time interval without assuming an extension of the forms. Take the step $\tau$ from [F4]. On $0<s<1$ its defining quotient has positive derivative, since $\frac{d}{ds}\log (\beta (s)\mathrm{/}\beta (1-s))=s^{-2}+(1-s{)}^{-2}>0$; hence it maps $(0,1)$ diffeomorphically onto $(0,1)$. Define $Y_s={\tau }^{\mathrm{\prime }}(s)X_{\tau (s)}$ for $0<s<1$ and $Y_s=0$ outside. Every derivative of $\tau$ is flat at $0,1$ by [F4], while all one-sided mixed derivatives of $X_t$ from step 1.1 are continuous on compact $[0,1]\times M$. The product rule therefore shows that $Y_s$ is a smooth time-dependent field on the open interval $\mathbb{R}$. Apply [F3] to $Y$. Fix the Riemannian metric used in [F1]'s construction; $\parallel Y_s\parallel$ is bounded on $[0,1]\times M$. The distance along a trajectory between times $r,s$ is at most its length and at most $C\mathrm{\mid }r-s\mathrm{\mid }$, so a finite-time maximal trajectory is Cauchy. Compactness gives its limit, and [F3] at that interior time extends it. Thus the evolution ${\psi }_s$ exists through $s=1$, with inverse given by reverse evolution. For $0<t<1$ set ${\varphi }_t={\psi }_{{\tau }^{-1}(t)}$, with ${\varphi }_0=\mathrm{id}$ and ${\varphi }_1={\psi }_1$. Changing variables in the coordinate integral equation for $\psi$ shows that, on each short time interval whose trajectory lies in one chart, ${\varphi }_t(p)={\varphi }_{t_0}(p)+{\int }_{t_0}^t{X}_u({\varphi }_u(p))du$ in that chart. The integral equation and the up-to-endpoint smoothness of $X_t$ bootstrap ${\varphi }_t$ and its spatial derivatives to joint smoothness in $(t,p)$ through both endpoints, despite the nonsmooth inverse of $\tau$ there. Each ${\varphi }_t$ is a diffeomorphism, with inverse from the reverse evolution.