Symplectic Manifolds, Moser Stability, and Darboux–Weinstein Theory — Examples

For $v={\sum }_i(a^i{\mathrm{\partial }}_{q^i}+b_i{\mathrm{\partial }}_{p_i})$, contraction gives ${\iota }_v{\omega }_0={\sum }_i(a^{i}d{p}_i-b_{i}d{q}^i)$.

This covector vanishes only when every $a^i$ and $b_i$ is zero, so [F1] proves nondegeneracy. Equivalently, ${\omega }_0^n=n!d{q}^1\wedge d{p}_1\wedge \cdots \wedge d{q}^n\wedge d{p}_n\mathrm{\ne }0$. For $n=0$ the zero space is included.

Direct pairing gives $⟨e_1{⟩}^{\omega }=⟨e_1,e_2,f_2⟩$, so $⟨e_1⟩$ is isotropic but not Lagrangian. Taking orthogonals reverses this equality, so $⟨e_1,e_2,f_2⟩$ is coisotropic but not Lagrangian.

Also $⟨e_1,f_1{⟩}^{\omega }=⟨e_2,f_2⟩$, making $⟨e_1,f_1⟩$ symplectic, while $⟨e_1,e_2{⟩}^{\omega }=⟨e_1,e_2⟩$, making the latter Lagrangian.

The four displayed coordinate subspaces therefore exhibit, respectively, a proper isotropic, a proper coisotropic, a proper symplectic, and a Lagrangian subspace.

On overlaps, angular coordinates differ by a locally constant multiple of $2\pi$, so $d{\theta }^{\mathrm{\prime }}=d\theta$ and the fibre coefficient is $p^{\mathrm{\prime }}=p$. Thus the local products glue to $S^1\times \mathbb{R}$, and $pd\theta$ is global.

Applying [F1] and [F2] gives $\lambda =pd\theta$ and ${\omega }_{\mathrm{c}\mathrm{a}\mathrm{n}}=-d(pd\theta )=d\theta \wedge dp$, the standard area form on the cylinder.

Since $d(df)=0$, [F1] makes $\mathrm{graph}(df)$ Lagrangian. The same argument applies to any closed one-form, without asserting it is exact.

On $S^1$, the global angular form $d\theta$ is closed but not exact because its integral around the positively oriented circle is $2\pi$, whereas an exact form integrates to zero. Its graph is the section $p=1$ in $T^{\ast }{S}^1$, so [F1] supplies the promised closed-nonexact example.

A tangent vector to the diagonal is $(v,v)$. On two such vectors, the product form gives $-\omega (v,w)+\omega (v,w)=0$, so the diagonal is isotropic.

If $\dim M=2n$, then $\dim {\mathrm{\Delta }}_M=2n$ and $\dim (M^{-}\times M)=4n$. Thus [F2] upgrades isotropy to Lagrangianity, including $n=0$.

Direct substitution gives $J^2(q,p)=(-q,-p)$. For $u=(a,b)$ and $v=(c,d)$, ${\omega }_0(u,Jv)={\omega }_0((a,b),(-d,c))=a\cdot c+b\cdot d$.

The last expression is the Euclidean inner product, hence symmetric and positive definite. By [F1], $J$ is compatible; the same calculation gives ${\omega }_0(Ju,Jv)={\omega }_0(u,v)$.

Compactness makes both top forms compactly supported. Their difference has integral zero, so [F1] makes it exact and therefore $[{\omega }_0]=[{\omega }_1]$.

Relative to any fixed positive area form, write ${\omega }_i=f_i\mu$ with $f_i>0$. Then ${\omega }_t=(1-t){\omega }_0+t{\omega }_1=((1-t)f_0+t{f}_1)\mu$ stays positive and hence symplectic, and step 1.1 makes its class constant.

Apply [F2] to obtain ${\varphi }_t^{\ast }{\omega }_t={\omega }_0$; in particular ${\varphi }_1^{\ast }{\omega }_1={\omega }_0$. The connected nonempty hypothesis is exactly what [F1] uses.

At the chosen point, $(Q,P)=(0,0)$. Differentiation under the integral gives $dP=P_{x}dx+f(x,y)dy$, and therefore $dQ\wedge dP=dx\wedge (P_{x}dx+fdy)=fdx\wedge dy=\omega$.

The Jacobian determinant of $(x,y)\mapsto (Q,P)$ is $P_y=f>0$, so the inverse function theorem makes $(Q,P)$ a coordinate system after shrinking. Thus it realizes the Darboux conclusion in [F1] explicitly.

On the zero section every $p_i$ is constant zero, so the pullback of [F1] vanishes. On the fibre over fixed $q$, every $q^i$ is constant, so the restriction again vanishes. Both submanifolds are therefore isotropic.

Each has dimension $n$, while $T^{\ast }Q$ has dimension $2n$. By [F2], both are Lagrangian, including the rank-zero case.

Its square is ${\eta }^2=2e^{x_1}d{x}_1\wedge d{y}_1\wedge d{x}_2\wedge d{y}_2$, a nowhere-zero top form, so $\eta$ is nondegenerate.

Exterior differentiation gives $d\eta =e^{x_1}d{x}_1\wedge d{x}_2\wedge d{y}_2\mathrm{\ne }0$. Hence [F1] shows that $\eta$ is not symplectic. Products with standard symplectic factors give the same phenomenon in every even dimension at least four.

Both forms are closed and positive, hence symplectic. By [F2], their cohomology classes are distinct because their integrals are $A>0$ and $2A$.

If $f^{\ast }(2\omega )=\omega$, then $f$ is orientation preserving and change of variables gives $A={\int }_{S^2}{f}^{\ast }(2\omega )=2A$, impossible. Thus [F1] fails for every diffeomorphism, exhibiting the global cohomology obstruction.

Pointwise $A$ preserves $\omega$ because its reciprocal scalings on $(x_1,y_1)$ preserve $d{x}_1\wedge d{y}_1$ and it fixes the other pair. Thus [F1] makes $J$ compatible. Explicitly, with $a=e^{-2x_2}$, $J{\mathrm{\partial }}_{x_1}=a{\mathrm{\partial }}_{y_1}$, $J{\mathrm{\partial }}_{y_1}=-a^{-1}{\mathrm{\partial }}_{x_1}$, and $J$ is standard on the second pair.

For $X={\mathrm{\partial }}_{x_1}$ and $Y={\mathrm{\partial }}_{x_2}$, all coordinate brackets vanish, while $[a{\mathrm{\partial }}_{y_1},{\mathrm{\partial }}_{x_2}]=-a^{\mathrm{\prime }}{\mathrm{\partial }}_{y_1}$. Hence $N_J(X,Y)=J(a^{\mathrm{\prime }}{\mathrm{\partial }}_{y_1})=-(a^{\mathrm{\prime }}\mathrm{/}a){\mathrm{\partial }}_{x_1}=2{\mathrm{\partial }}_{x_1}\mathrm{\ne }0$. By [F2], $J$ is not integrable.