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A symplectomorphism of a closed symplectic surface carries the complementary components of a separating simple closed curve to those of its image, preserving their unordered symplectic areas. On the unit sphere a latitude at height $h$ has complementary areas $2\pi (1-h)$ and $2\pi (1+h)$. Thus the equator cannot be sent to any nonequatorial latitude.

Translation by a differential one-form $\beta$ is $(x,\xi )\mapsto (x,\xi +{\beta }_x)$. It changes the canonical one-form on a cotangent bundle by ${\pi }^{\ast }\beta$. A closed differential form gives a symplectomorphism; an exact differential form gives a Hamiltonian diffeomorphism when compact support is not required.

A locally uniform limit of symplectomorphisms that is a smooth diffeomorphism is symplectic. By capacity preservation under uniform limits and linear capacity rigidity, its derivative is symplectic or anti-symplectic. Repeat this argument after taking the product of every map with the identity on a symplectic plane. An anti-symplectic derivative on the original factor and a symplectic derivative on the added factor would give neither sign on the product, a contradiction. The argument uses symplectic capacities in the original and stabilized dimensions.

If ${\iota }_{X_{H_t}}\omega =-d{H}_t$ and ${\varphi }_t$ is its flow, then Cartan's magic formula givesThus every defined time slice of a Hamiltonian isotopy is a symplectomorphism.

The displayed distance uses the oscillation length of a compactly supported Hamiltonian isotopy. On a closed manifold one may normalize each Hamiltonian to have zero mean. Additive time-dependent constants do not change the flow or its oscillation. Pullback by a symplectomorphism preserves extrema, so this distance is bi-invariant. Nondegeneracy requires an additional geometric argument; it does not follow merely from the definition as an infimum of path lengths.

Use the Hamiltonian vector field convention ${\iota }_{X_{H_t}}\omega =-d{H}_t$. A smooth time-dependent Hamiltonian function $H:[0,1]\times M\to \mathbb{R}$ with flow defined over the whole interval generates the Hamiltonian isotopy determined byThe group of Hamiltonian diffeomorphisms isThese maps are symplectomorphisms: by Cartan's magic formula, ${\mathcal{L}}_{X_{H_t}}\omega =d{\iota }_{X_{H_t}}\omega =0$. Thus $d({\varphi }_H^t{)}^{\ast }\omega /dt=0$.

The term “group” is justified explicitly. If ${\varphi }_t,{\psi }_t$ are generated by $H_t,K_t$, their product path is generated by $H_t+K_t\circ {\varphi }_t^{-1}$, since the pushforward of $X_{K_t}$ by a symplectomorphism is $X_{K_t\circ {\varphi }_t^{-1}}$. The inverse path is generated by $-H_t\circ {\varphi }_t$. The zero Hamiltonian gives the identity.

On a closed symplectic manifold, all these flows exist on the finite time interval. On an open manifold one often means the compactly supported Hamiltonian diffeomorphism group ${\mathrm{Ham}}_c$, using Hamiltonians with compact support in $[0,1]\times M$. We use that subgroup for the finite-energy definitions on Euclidean space below. The unrestricted complete-flow group is needed for the global cotangent translations in Question 4; those translations are generally not compactly supported.

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.

Let ${\Phi }_t$ be the local flow of a vector field $V$. The Lie derivative of a differential form is ${\mathcal{L}}_V\eta ={\frac{d}{dt}|}_{t=0}{\Phi }_t^{\ast }\eta$. We prove Cartan's magic formulaBoth sides are degree-zero derivations of an algebra on the exterior algebra of differential forms. They agree on a function $f$, since ${\iota }_{V}df=V(f)$. They also agree on $df$: pullback commutes with the exterior derivative, so ${\mathcal{L}}_{V}df=d(Vf)$, while $(d{\iota }_V+{\iota }_{V}d)df=d(Vf)$. Locally every differential form is a sum of products of functions and coordinate differentials, proving the identity in every degree.

For a compactly supported Hamiltonian vector field $V_H$, take the convention ${\iota }_{V_H}\omega =-dH$. Since $d\omega =0$, Cartan's magic formula gives ${\mathcal{L}}_{V_H}\omega =0$, so its complete flow of a vector field consists of symplectomorphisms. Given $p$ and $v\in T_{p}M$, a bump function lets us choose a compactly supported $H$ with $d{H}_p=-{\iota }_v{\omega }_p$, and hence $V_H(p)=v$. Choose such fields for a basis of $T_{p}M$. Their successive small-time flows give a map with invertible derivative at the origin. The inverse function theorem makes the orbit of $p$ open. Every orbit is open; because $M$ is connected, there is only one orbit. Thus the compactly supported symplectomorphisms act transitively on points, by Hamiltonian transitivity on a connected symplectic manifold.

On the unit sphere, the standard symplectic area is $4\pi$. The equator has complementary disks of areas $2\pi ,2\pi$, whereas the latitude at height $1/2$ has complementary disks of areas $\pi ,3\pi$: the cap above height $h$ has area $2\pi (1-h)$. A symplectomorphism must preserve the unordered pair of complementary areas. Therefore no such symplectomorphism exists, by complementary areas obstruct symplectic equivalence of separating curves.

The symplectic neighborhood theorem says that a symplectomorphism between compact symplectic submanifolds, together with a symplectic vector bundle isomorphism of their symplectic normal bundles, extends to a symplectomorphism of neighborhoods. To outline the proof, use tubular neighborhoods to extend the bundle identification smoothly. The two pulled-back symplectic forms agree as bilinear forms along the zero section. The relative Poincaré lemma writes their difference as $d\alpha$, with $\alpha$ vanishing there. Their linear interpolation ${\omega }_t$ remains nondegenerate on a sufficiently small neighborhood. Solvingand integrating $V_t$ gives the relative Moser theorem; the flow fixes the submanifold and carries one form to the other.

For the symplectic fiber sum along a square-zero surface, the self-intersection is the Euler number of the oriented rank-two normal bundle. Thus each normal bundle is trivial. Rescale one ambient symplectic form if necessary to equalize the areas of the two copies of $C$. The Moser theorem for the surface then supplies a base identification preserving their area forms. The symplectic neighborhood theorem identifies each neighborhood with a product carryingRemove small disk neighborhoods and glue collars with the base identification andfor a suitable positive constant $a$. Both signs reverse, so the normal two-forms match; the base forms already match. They define a closed nondegenerate two-form across the neck, agreeing with the ambient forms elsewhere. This constructs the symplectic sum. Its construction uses the area normalization and the chosen normal-bundle gluing.

For a smooth counterexample, take $X=Y={\mathbb{CP}}^2$, and in a small four-ball in each choose an unknotted sphere bounding a three-ball. These spheres are null-homologous and have self-intersection zero. Their standard smooth fiber sum isIndeed, the complement of the standard $S^2\times D^2$ in $S^4$ is $S^1\times D^3$; doubling these complements gives $S^1\times S^3$, and the two ambient projective-plane summands remain as connected sums. This is the smooth fiber sum along unknotted null-homologous spheres.

In the induced orientation, both summands in $Z={\mathbb{CP}}^2\#({\mathbb{CP}}^2\#(S^1\times S^3))$ have positive positive index of the intersection form. The symplectic connected-sum obstruction therefore rules out a symplectic form: connected-sum vanishing of the Seiberg–Witten invariant of a four-manifold contradicts Taubes nonvanishing theorem. In the opposite orientation the intersection form is negative definite, also ruling out a symplectic form, whose class has positive square. Thus the smooth fiber sum need not be symplectic.

A symplectic capacity assigns $c(X,\omega )\in [0,\infty ]$ to symplectic manifolds in a fixed dimension, is monotone under symplectic embeddings, satisfies $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and is nontrivial in the sense thatThe second domain is the standard symplectic cylinder. A normalized symplectic capacity assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under symplectomorphisms and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.

Let ${\varphi }_j$ be symplectomorphisms converging locally uniformly to a smooth diffeomorphism $\varphi$. Work in Darboux charts. For a sufficiently small solid ellipsoid $E$ centered at a chosen point and $0<a<1$, uniform convergence and degree of a continuous mapping give, for large $j$,For the second inclusion, ${\varphi }^{-1}{\varphi }_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline{E}$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^{2}c(E)\le c(\varphi (E))\le a^{-2}c(E)$. Letting $a$ tend to one proves capacity preservation under uniform limits.

At the chosen point, write $T=D\varphi$. The rescaled maps $(\varphi (tz)-\varphi (0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all solid ellipsoids; so does $T^{-1}$.

We prove the required linear capacity rigidity explicitly. Write the standard symplectic form as $\Omega$. Suppose $\Omega (u,v)=1$ andComplete $u,v$ to a symplectic basis, and complete $T^{T}u/\lambda$, with the appropriately signed $T^{T}v/\lambda$, to another symplectic basis. If their basis matrices are $P,P^{\prime }$, then $A=(P^{\prime }{)}^{-1}{T}^{T}P$ has first two columns $\lambda e_1,\pm \lambda f_1$. Hence $A^T$ sends the unit ball into the symplectic cylinder of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T{)}^k$ sends the unit ball into the cylinder of radius ${\lambda }^k$. Monotonicity givesfor every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.

A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thusfor every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an integral domain; consequently one factor vanishes identically. Therefore $T$ is either symplectic or anti-symplectic.

To eliminate the negative sign, repeat the argument for ${\varphi }_j\times {\mathrm{id}}_{{\mathbb{R}}^2}$, which are symplectomorphisms for $\omega \oplus {\omega }_{{\mathbb{R}}^2}$. If ${\varphi }^{\ast }\omega =-\omega$ at the chosen point, the limit pulls this product form back to $-\omega \oplus {\omega }_{{\mathbb{R}}^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus ${\varphi }^{\ast }\omega =\omega$ everywhere. We have proved the Eliashberg–Gromov rigidity theorem: the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms.

Finally put $E=W^{\perp }$, with perpendicularity taken for the standard Euclidean inner product. Its dimension is two, so being nonisotropic means that the restricted symplectic form is nondegenerate. For the standard compatible complex structure $J$, $W^{\omega }=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega (v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a symplectic subspace and admits a symplectic basis extended to the whole space.

In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a symplectic cylinder of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling giveas in capacity of a bounded set thickened by a symplectic subspace. For a normalized symplectic capacity, the two bounds simplify to $\pi r^2$ and $\pi R^2$.

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.

Yes. Choose a global differential one-form $\beta$ with $d\beta =\sigma$, and use the cotangent fiber translationIt is a diffeomorphism with inverse $T_{-\beta }$, and $\pi \circ T_{\beta }=\pi$. Directly from the Liouville one-form,The sign convention $\omega =-d\alpha$ therefore givesThus $T_{\beta }:(T^{\ast }X,\omega )\to (T^{\ast }X,{\omega }_{\sigma })$ is the required symplectomorphism. Equivalently, $T_{-\beta }^{\ast }\omega ={\omega }_{\sigma }$ gives the inverse direction. This translation equivalence of exact twisted cotangent bundles is global even when $X$ is not compact; no completeness argument for a Moser theorem flow is needed.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

First observe that the Liouville one-form is zero as an ambient covector exactly on the zero section. Indeed, ${\alpha }_{(x,p)}=p\circ d\pi$, and $d\pi$ is surjective. Since $dg$ is invertible, $g^{\ast }\alpha =\alpha$ therefore implies that $g$ maps the zero section onto itself. It induces a diffeomorphism $f:X\to X$ defined by $g(x,0)=(f(x),0)$.

The map $g$ is a symplectomorphism, since it preserves $-d\alpha$. Let $E={\sum }_i{p}_i{\partial }_{p_i}$ be the vertical Liouville vector field. With the chosen sign convention,Preservation of both $\omega$ and $\alpha$ forces $g_{\ast }E=E$ because $\omega$ is nondegenerate. Its complete cotangent fiber-dilation flow isUniqueness of integral curves gives $g\circ D_s=D_s\circ g$. If $g(x,p)=(y,P)$, let $s\to -\infty$. Continuity and this commutation yieldThus $y=f(x)$ for every $p$: the whole cotangent fiber over $x$ maps into the fiber over $f(x)$. This step proves that $g$ covers $f$; it was not assumed.

Now write $g(x,p)=(f(x),P(x,p))$. For any $u\in T_{x}X$, choose a tangent vector to $T^{\ast }X$ projecting to $u$. Evaluating $g^{\ast }\alpha =\alpha$ on it givesSince $d{f}_x$ is invertible, $P(x,p)=p\circ (d{f}_x{)}^{-1}$. Thereforeand $f$ is unique. This rigidity of cotangent Liouville-form preservation uses smoothness at the zero section; it follows from the canonical form and its dilation dynamics on the entire cotangent bundle.

For any positive-dimensional $X$, choose a smooth function $h$ with $dh$ not identically zero. A nonconstant smooth bump in a coordinate chart supplies one on an arbitrary smooth manifold. Translate by this exact differential form:The cotangent fiber translation identities giveThus a nontrivial exact fiber translation is a symplectomorphism that fails to preserve the Liouville one-form. For example, on $T^{\ast }\mathbb{R}$ one can take $T(q,p)=(q,p+1)$: it preserves $dq\wedge dp$ but changes $pdq$ to $(p+1)dq$.

If dimension zero is admitted, $T^{\ast }X=X$ and $\alpha =0$, so every diffeomorphism preserves $\alpha$ and this requested distinction is impossible. The example therefore applies in positive dimension, as required for a nonzero differential one-form.