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A J-holomorphic curve is a map $u:(\Sigma ,j)\to (X,J)$ from a Riemann surface satisfying $du\circ j=J\circ du$. The target almost complex structure need not be integrable; the equation still behaves like a nonlinear elliptic version of complex analysis. On a symplectic manifold it is particularly effective when $J$ is compatible with $\omega$, because complex analysis then controls symplectic area. The metric construction above supplies such structures and deformations between them.

In conformal coordinates $(s,t)$ with $j{\partial }_s={\partial }_t$, the equation is $u_t=J{u}_s$. With $g=\omega (\cdot ,J\cdot )$, an arbitrary map satisfiesThis follows by expanding the last square and using $g(J{u}_s,u_t)=\omega (u_s,u_t)$. Integrating proves the Energy identity for a J-holomorphic curve. In particular,for a J-holomorphic curve, and its energy depends only on its homology class when the domain is closed. A nonconstant such curve has strictly positive area: its derivative is nonzero somewhere and positivity holds on a neighborhood. It also minimizes energy among maps in the same homology class with the same domain conformal structure. Thus analytic bounds can be obtained from topological information.

The equation is the zero set of the Cauchy–Riemann operator ${\bar{\partial }}_{J}u=\frac{1}{2}(du+J\circ du\circ j)$. Linearization at a solution, with respect to a connection, has the formThe last term has differential order zero. The principal part has the Cauchy–Riemann symbol; in a complex frame a nonzero real covector $(a,b)$ gives multiplication by $a+ib$, an invertible symbol. This establishes ellipticity. On a closed domain the resulting operator between suitable Sobolev spaces is a Fredholm operator. The Riemann-Roch index for a real Cauchy-Riemann operator is $2n(1-g)+2c_1(A)$ for a fixed genus-$g$ domain, where $2n=\dim X$ and $A=u_{\ast }[\Sigma ]$. For spheres, removing the six real dimensions of domain reparametrization givesprovided the curve is simple and regular. Each marked point adds two real dimensions; constraining its image to a specified target point removes $2n$. These are the dimension formula for regular J-holomorphic spheres and its incidence version.

Here regular means that $D_u$ is surjective. Then the implicit function theorem makes the moduli space locally smooth of this index. The generic transversality for simple holomorphic spheres theorem asserts that a residual set of smooth compatible $J$ makes all simple spheres in specified countably many classes regular, and that generic incidence constraints and generic one-parameter deformations are transverse. It does not assert regularity for all multiple covers. This distinction matters whenever one tries to count curves. A primitive class with no positive-area spherical decomposition avoids multiple-cover and bubbling difficulties and permits especially direct counts.

The other central ingredient is Gromov compactness for spheres. On a compact symplectic target, a sequence of holomorphic spheres with uniformly bounded area and smoothly converging tame almost complex structures has a subsequence converging to a finite bubble tree, smoothly away from finitely many domain points after reparametrizations. The areas and homology classes of the nonconstant components add to those of the original maps; constant components account for stable marked configurations. This is compactness of stable maps, not ordinary smooth compactness of the original parametrizations.

The proof mechanism explains why bubbling is indispensable. An elliptic small-energy estimate bounds derivatives on a smaller disc when the energy on a larger disc is sufficiently small. Failure of a derivative bound therefore concentrates a definite amount of energy. Rescale near a point of large derivative; elliptic estimates yield a nonconstant limiting holomorphic map on the plane. Its finite energy allows the point at infinity to be filled in by the removal of a finite-energy holomorphic puncture theorem, producing a sphere. Repeating extracts further spheres. Uniform local monotonicity gives a positive energy threshold for nonconstant bubbles on the compact target, so only finitely many can occur. Estimates on the intervening annuli yield the energy identity and exclude lost neck energy. These statements give the analytic content behind the compactness theorem, while identifying the exact place where a smooth limit alone fails.

A local area estimate turns this existence theory into an embedding obstruction. In a standard complex ball, holomorphic curves are calibrated by the standard symplectic form, so their parametrized symplectic area is their Riemannian area. They are stationary minimal surfaces. The minimal-surface monotonicity formula says that $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing wherever the curve has no boundary in the ball. At a point of multiplicity $m$, its limiting density is $m\pi$. Consequently any branch through the centre contributes at least $\pi {\rho }^2$. For a general compatible $J$ there is still a local lower bound $c{\rho }^2$, with $c>0$ depending on controlled geometry. The exact Euclidean constant is the one needed for sharp radius obstructions.

Here is a global existence argument that feeds this estimate. Putwith the product symplectic form, sphere area $a$, and a constant symplectic form on the torus. Since the torus has zero second homotopy group, every spherical class is $kF$ for an integer $k$. A nonconstant holomorphic sphere in this class has area $ka>0$, so $k\ge 1$. A curve of class $F$ cannot split into two nonconstant bubbles, and it cannot be a nontrivial multiple cover. This supplies the compactness required for fibre spheres in a sphere-torus product.

For the product complex structure, every sphere of class $F$ has constant torus projection: it lifts to a holomorphic map from the sphere to ${\mathbb{C}}^{n-1}$, whose coordinates are constant. Its sphere projection has degree one. Thus, up to reparametrization, there is exactly one such sphere through every target point. It is regular because its pulled-back tangent bundle is $\mathcal{O}(2)\oplus {\mathcal{O}}^{n-1}$ on ${\mathbb{CP}}^1$, and the first cohomology of each summand vanishes. Its linearized Dolbeault operator therefore has zero cokernel.

Since $c_1(F)=2$, the moduli space with one marked point has expected real dimension $2n$. Imposing passage through a fixed point gives dimension zero. For generic structures and transverse incidence conditions this is a finite set. Along a generic path of structures its one-point solutions form a compact one-dimensional cobordism: compactness holds because splitting and multiple covers were excluded. Its boundary contains the endpoint solutions. A compact one-manifold has an even number of boundary points, so the count modulo two is unchanged. The product count is one, and hence the count for a generic compatible $J$ is one as well. If the endpoint product structure is not generic for other classes, its regular solutions allow the path argument in this single class. Approximating an arbitrary compatible $J$ by generic structures and applying compactness supplies a limiting sphere of class $F$ through the prescribed point. The nonconstant component still contains that point; there are no nonconstant bubbles to carry it away. This derives the existence assertion needed below without assuming a capacity obstruction.

Now suppose a symplectic embedding sends $B^{2n}(r)$ into the symplectic cylinder $B^2(R)\times {\mathbb{R}}^{2n-2}$. Choose $0<\rho <{\rho }^{\prime }<r$. The image of the closed radius-${\rho }^{\prime }$ ball is compact. For any $a>\pi R^2$, the first factor embeds symplectically into an area-$a$ sphere: in dimension two this is an area-preserving disc chart, leaving a cap of area $a-\pi R^2$. Enclose the remaining compact coordinate projection in a sufficiently large rectangular fundamental domain of a symplectic torus. The compact ball image thereby embeds into $S^2(a)\times T^{2n-2}$.

Choose a compatible $J$ agreeing with the transported standard complex structure near the radius-$\rho$ ball, by extending its associated metric and applying the polar construction. The sphere of class $F$ through the centre has area $a$. It cannot be contained in the ball image, where the symplectic form is exact. Euclidean monotonicity therefore gives $a\ge \pi {\rho }^2$. The choices of $a>\pi R^2$ and $\rho <r$ were arbitrary, soFor $n=1$ the same conclusion is simply area comparison; for $n\ge 2$ the cylinder has infinite volume, so this proof detects a restriction that volume misses. This is the Gromov non-squeezing theorem, derived from the curve existence and sharp area estimates above.

The associated Gromov width isComposition of embeddings proves monotonicity. Rescaling coordinates in a ball proves $c_G(U,c\omega )=c{c}_G(U,\omega )$ for $c>0$. Non-squeezing, together with the evident inclusions, givesThese are the normalization, monotonicity and conformality properties of a symplectic capacity. The two-ball obstruction of Question 5 illustrates a related use: one degree-one curve through two prescribed centres gives an additive area bound, stronger than comparing the volumes of the balls. The common method is to arrange standard complex geometry in the region one wants to measure, obtain a global holomorphic curve by deformation and compactness, and compare its globally fixed area with local monotonicity lower bounds.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

A symplectic manifold is a smooth even-dimensional manifold $X^{2n}$ with a closed, nondegenerate differential two-form $\omega$. A Lagrangian submanifold $L\subset X$ has dimension $n$ and $\omega {|}_{TL}=0$; equivalently $T_{x}L=(T_{x}L{)}^{\omega }$ at every point.

On $T^{\ast }M$, let $\pi :T^{\ast }M\to M$ and define the canonical one-form on a cotangent bundle by ${\theta }_{(q,p)}(v)=p(d\pi (v))$. In local coordinates $\theta ={\sum }_i{p}_{i}d{q}_i$, so our sign convention givesThe intrinsic definition of $\theta$ makes this form globally defined. It is closed because $d^2=0$, and its displayed coordinate matrix is invertible. The zero section pulls $\theta$, and therefore ${\omega }_0$, back to zero and has half the total dimension. It is a Lagrangian submanifold.

We prove the Weinstein neighborhood theorem, rather than merely invoking it. Along a Lagrangian submanifold $L\subset (X,\omega )$, choose a compatible almost complex structure $J$; its existence is proved by the metric construction in Question 5. Then $JL$ at the tangent-bundle level means $J(TL)$, andIndeed, $Jv\in TL$ with $v\in TL$ would imply $0=\omega (v,Jv)$, forcing $v=0$. Also $\omega (Ju,Jv)=\omega (u,v)=0$ for $u,v\in TL$, so $J(TL)$ is a Lagrangian complement. Identify a normal vector $n\in J(TL)$ with the covector $\eta (v)=\omega (v,n)$. This is a vector-bundle isomorphism with $T^{\ast }L$, and under itwhich is exactly the form ${\omega }_0$ along the zero section. The tubular neighborhood theorem therefore supplies a smooth map $F$ from a neighborhood of that section to a neighborhood of $L$, restricting to the identity and with this derivative along $L$.

Write ${\omega }_1=F^{\ast }\omega$ and $\delta ={\omega }_1-{\omega }_0$. This closed differential form vanishes as an entire bilinear form along the zero section. The relative Poincaré lemma gives $\delta =d\beta$ with $\beta$ zero along the section. Explicitly, if $r_s(q,p)=(q,sp)$ and $R=\sum p_i{\partial }_{p_i}$, useThe homotopy identity gives $d\beta =\delta -r_0^{\ast }\delta =\delta$; the vanishing at the section ensures smoothness of this integral. The forms ${\omega }_t={\omega }_0+t\delta$ are nondegenerate on a sufficiently small neighborhood of the section for all $0\le t\le 1$. Solve uniquelyThe vector field $Y_t$ vanishes on the section. After another neighborhood shrink its flow ${\varphi }_t$ exists up to time one and fixes the section. For noncompact $L$, the sizes of these neighborhoods may vary with the base point; no compactness of $L$ is required for the local conclusion. By Cartan's magic formula,Thus $(F\circ {\varphi }_1{)}^{\ast }\omega ={\omega }_0$. This establishes the universal neighborhood model, with the Lagrangian submanifold identified with the zero section.

For the section $s_{\alpha }(q)=(q,{\alpha }_q)$, the canonical differential one-form satisfies $s_{\alpha }^{\ast }\theta =\alpha$. HenceIts dimension is already half that of $T^{\ast }M$, so vanishing of this pullback is sufficient as well as necessary.

Suppose now that a smooth open six-dimensional region were a fiber bundle with Lagrangian submanifolds diffeomorphic to $S^3$ as its fibres. Fix one compact fibre $L$. Smooth local triviality and compactness imply that the other sufficiently nearby fibres are $C^1$-close embeddings of $L$. In the Weinstein neighborhood, projection to $L$ restricts on each such fibre to a local diffeomorphism close to the identity. It has degree one, and compactness makes it a covering map, so it is a diffeomorphism. Each nearby fibre is therefore a graph of a closed differential one-form $\alpha$ on $S^3$. Since $H^1(S^3;\mathbb{R})=0$, this is $df$ for a smooth real function $f$. A maximum of $f$ exists by compactness, and there $df=0$. The graph meets the original zero section, contradicting disjointness of different fibres. No such Lagrangian three-sphere fibration exists. More generally this is the compact first-cohomology obstruction to a Lagrangian fibration.

For a symplectic vector space $(V,\omega )$ and a vector subspace $W$, the symplectic orthogonal complement isNondegenerate bilinear form identifies $V$ with its dual space, so $\dim W^{\omega }=\dim V-\dim W$. If $W$ is a symplectic subspace, $W\cap W^{\omega }=0$, and hence $V=W\oplus W^{\omega }$; the restriction to $W^{\omega }$ is also nondegenerate.

Here is an inductive construction of a symplectic basis. Choose $e_1\ne 0$. Nondegenerate bilinear form supplies $f_1$ with $\omega (e_1,f_1)=1$. The plane $P=\mathrm{span}(e_1,f_1)$ is a symplectic subspace, and its symplectic orthogonal complement gives $V=P\oplus P^{\omega }$. Repeat on $P^{\omega }$. The process ends in dimension zero, since each step removes two dimensions and a one-dimensional alternating form is degenerate. We obtain $e_1,f_1,\dots ,e_n,f_n$ withThe linear map taking the coordinate vectors of ${\mathbb{R}}^{2n}$ to this basis pulls $\omega$ back to ${\sum }_{j}d{x}_j\wedge d{y}_j$, proving the asserted normal form.

Now split $V=C\oplus D$ with $D=C^{\omega }$, and choose symplectic bases on both planes. Parametrize ${\Gamma }_A$ by $v\mapsto (Av,v)$. Orthogonality removes the cross terms, and a two-dimensional alternating form transforms by the determinant, soConsequently the correct nondegenerate bilinear form criterion isThe strict inequality printed in the PDF requires an extra orientation condition: it says that the symplectic orientation on the graph agrees with the orientation transported from $D$. It is not equivalent to being a symplectic subspace. For example, $A=\mathrm{diag}(-2,1)$ has determinant $-2$ and gives a nondegenerate restriction $-{\omega }_D$. This is the orientation criterion for a graph of symplectic planes.

This same map supplies the requested obstruction. Orient $C$ and ${\Gamma }_A$ by their restricted symplectic forms. If $(d_1,d_2)$ is a positive basis of $D$, then $(A{d}_1+d_1,A{d}_2+d_2)$ is negatively oriented in ${\Gamma }_A$. The coordinate transformation from a basis of $C$ followed by these graph vectors to a basis of $C\oplus D$ has determinant $1$. Reversing the graph basis to give its positive orientation therefore makes the oriented direct sum $C\oplus {\Gamma }_A$ negative relative to the ambient symplectic orientation.

Two distinct complex lines in standard ${\mathbb{C}}^2$ have the opposite behaviour: complex bases of the two lines give a complex-linear isomorphism $\mathbb{C}\oplus \mathbb{C}\to {\mathbb{C}}^2$, whose real determinant is the positive number $|{\det }_{\mathbb{C}}{|}^2$. Their complex orientations are their symplectic orientations. A symplectic linear map preserves both the ambient and each plane's symplectic orientation, so it cannot change this sign. Thus the pair with $A=\mathrm{diag}(-2,1)$ cannot be the image of a pair of complex lines. Interchanging the planes does not change the sign, since both dimensions are even.