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A compatible triple consists of a symplectic form, a compatible almost complex structure, and the Riemannian metric $g(u,v)=\omega (u,Jv)$. Equivalently $g(Ju,v)=\omega (u,v)$.

The metric construction of a compatible almost complex structure is continuous in the auxiliary Riemannian metric, and recovers $J$ when applied to $g_J=\omega (\cdot ,J\cdot )$. Interpolating $g_J$ linearly to a fixed auxiliary metric therefore contracts the space of compatible almost complex structures. In particular it is nonempty and path connected.

For a smooth map between Riemannian manifolds, its Dirichlet energy is half the integral of the squared norm of its differential, using the source and target metrics. When the source is two-dimensional, a conformal change of its metric leaves this energy unchanged. For a J-holomorphic curve into a symplectic manifold with a compatible almost complex structure, the Energy identity for a J-holomorphic curve identifies it with the pulled-back symplectic area. The norm of a differential here is the full tensor norm, summing its squared values on an orthonormal source frame.

In a product symplectic manifold $S^2(a)\times T^{2n-2}$, every compatible almost complex structure admits a J-holomorphic sphere in class $F$ through each point. Here $S^2(a)$ has area $a$, and the torus carries a constant symplectic form. All spherical classes are integer multiples of $F$, so a curve in class $F$ cannot split into positive-area nonconstant spheres. For the product complex structure the fibre through a point is unique and regular. Its one-point count modulo two is one; generic transversality and the compact one-dimensional parameterized moduli spaces preserve this count. Approximation and Gromov compactness for spheres remove the genericity requirement. This gives the area-$a$ curve used in the proof of the Gromov non-squeezing theorem.

For a residual set of smooth compatible almost complex structures, all simple J-holomorphic spheres in a specified countable set of spherical classes are regular. The corresponding evaluation maps at marked points are transverse to generic incidence conditions; generic paths give parameterized transversality. The universal linearization can vary the target structure near an injective point of the curve, which removes its cokernel. This assertion concerns simple curves and does not ensure regularity of arbitrary multiple covers.

A Lagrangian boundary condition for a J-holomorphic curve requires specified boundary components of the source to map into specified Lagrangian submanifolds. For the strip, this is $u(s,0)\in L_0$ and $u(s,1)\in L_1$. For a compatible almost complex structure these target submanifolds are totally real. The condition accompanies the interior equation $u_s+J{u}_t=0$.

Choose a reference path between Lagrangian submanifolds $L_0,L_1$ in a symplectic manifold. With a connecting homotopy strip oriented by $ds\wedge dt$, define $\mathcal{A}(\gamma )=-\int u^{\ast }\omega$. If the path space is connected and $H_2(M,L_0\cup L_1)=0$, relative symplectic-area independence makes this single-valued up to a global constant. Its derivative is $d\mathcal{A}(\xi )=\int \omega (\dot{\gamma },\xi )dt$, so its critical points are the constant paths in $L_0\cap L_1$. For a compatible almost complex structure, its formal downward $L^2$ flow is the J-holomorphic curve equation $u_s+J{u}_t=0$.

A rank-$n$ vector subbundle of a rank-$2n$ symplectic vector bundle is Lagrangian when each fibre is a Lagrangian subspace. If it exists, a compatible almost complex structure supplies a complementary Lagrangian subbundle by applying $J$. Existence is a global condition: the area-oriented tangent bundle of the two-sphere has no real line subbundle, by triviality of real line bundles over the sphere and the Hairy ball theorem.

Choose an auxiliary Riemannian metric $h$ and define the skew-adjoint bundle map $A$ by $\omega (u,v)=h(Au,v)$. Thenis a compatible almost complex structure. The positive square root is defined fiberwise by the spectral theorem for normal operators, and its smooth dependence on $A$ makes $J$ smooth.

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.

DefineEvery projective point with $z_3\ne 0$ has a unique unit representative whose third coordinate is positive real, so $F$ is a diffeomorphism. More explicitly, in the affine coordinate $u=(z_1/z_3,z_2/z_3)$ its inverse is $w=u/\sqrt{1+|u{|}^2}$. Lift $F$ to the unit sphere by $\tilde{F}(w)=(w,\sqrt{1-|w{|}^2})$. The last coordinate is real, so its contribution $d{x}_3\wedge d{y}_3$ pulls back to zero. The reduction identity therefore yieldsThis proves the symplectic ball chart in complex projective space directly, including the precise unit radius.

For the final unheaded request, embed the two disjoint balls in this chart. Fix arbitrary $0<{\rho }_i<r_i$, and choose intermediate radii ${\rho }_i<{\rho }_i^{\prime }<r_i$. On the images of the balls of radius ${\rho }_i^{\prime }$, transport the standard complex structure by the given symplectic embeddings. It is compatible with $\Omega$. Their smaller closed neighborhoods are compact and disjoint. Choose a global Riemannian metric agreeing there with the associated compatible metrics, using a partition of unity outside these neighborhoods, and apply the metric construction. The resulting global compatible almost complex structure $J$ agrees with the transported standard structure on each ball of radius ${\rho }_i$ and on a slightly larger neighborhood. This extends metrics, rather than averaging almost complex structures.

Let $p_1,p_2$ be the centres. They are distinct. The degree-one J-holomorphic curve supplied in the question has total area $\pi$, because its homology class is the projective line class. We use the sharp Euclidean version of the Monotonicity theorem for a J-holomorphic curve: a nonconstant holomorphic curve through the centre of a standard complex ball, with no boundary in the ball and counted with its parametrization multiplicities, has area in the radius-$\rho$ ball at least $\pi {\rho }^2$. This includes branched points: their density is a positive integer $m$, and the stronger lower bound is $m\pi {\rho }^2$. One way to see the constant is that complex curves are calibrated minimal surfaces; their stationary area ratio $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing and its limit at the centre is $m\pi$.

The closed degree-one curve cannot stay entirely inside either chart ball: there the form is exact, so Stokes theorem would give zero total area, contradicting its positive area. Thus the branch through each centre crosses the surrounding spheres, and the stated monotonicity bound applies to each restriction. The two ball images are disjoint, so their area contributions add without overlap. It follows thatLetting both smaller radii increase to their original radii givesThis two-ball packing obstruction in the projective plane is sharper than volume alone: a volume comparison would only give $r_1^4+r_2^4\le 1$.

A compatible almost complex structure is a smooth endomorphism $J:TX\to TX$ satisfying $J^2=-I$, preserving $\omega$, and making $g_J(u,v)=\omega (u,Jv)$ a positive-definite inner product. Preservation and $J^2=-I$ make this bilinear form symmetric.

To construct one, choose any Riemannian metric $h$ and define $A$ by $h(Au,v)=\omega (u,v)$. The endomorphism $A$ is skew-adjoint and invertible. Consequently $-A^2$ is positive definite and self-adjoint. Let $B=(-A^2{)}^{1/2}$ be its positive square root and setThe spectral theorem gives $B$ fibrewise, and the positive-square-root operation is smooth on positive-definite matrices, so this defines a smooth endomorphism. Since $A$ commutes with $B$, $J^2=-I$. Also $J^{\ast }=-J$, so $J^{\ast }J=I$. Commutation then givesThis is the metric construction of a compatible almost complex structure, proving nonemptiness.

If we start with $h=g_J$ for an already compatible $J$, then $h(Ju,v)=\omega (u,v)$, so $A=J$, $B=I$, and the construction recovers $J$. For two compatible structures, linearly interpolate their metrics,and apply the construction to $h_t$. This is a continuous smooth-in-$t$ path with endpoints $J_0,J_1$. Thus the space is nonempty and path connected. Interpolating instead to a single fixed auxiliary metric actually gives the contractibility of compatible almost complex structures.

An almost complex structure $J$ is compatible with $\omega$ if $J^2=-I$, $\omega (Ju,Jv)=\omega (u,v)$, and $g_J(u,v)=\omega (u,Jv)$ is a positive-definite inner product. To construct one, choose a Riemannian metric $h$ and define $A$ by $h(Au,v)=\omega (u,v)$. The map $A$ is invertible and skew-adjoint, so $-A^2$ is positive definite. Its positive square root $P=(-A^2{)}^{1/2}$ commutes with $A$, andThe same identities show that $J$ preserves $\omega$. The positive square root depends smoothly on $A$, giving a smooth compatible almost complex structure.

For a compatible $J$, applying this metric construction of a compatible almost complex structure to $g_J$ returns $J$. Given $J_0,J_1$, apply it to $(1-t)g_{J_0}+t{g}_{J_1}$ to obtain a path joining them. In fact interpolation to any fixed auxiliary metric gives a contraction. Thus the space is nonempty and connected, and even contractible, by contractibility of compatible almost complex structures.

Regularity of a J-holomorphic curve $u$ means that its linearized Cauchy–Riemann operator $D_u$ is surjective. Differentiating in coordinates givesup to the harmless factor $1/2$ in the convention for ${\bar{\partial }}_J$. For constant $u$, the last term vanishes, $J(u)$ is constant, and the pulled-back tangent bundle is ${\mathbb{CP}}^1\times T_{u}M$. Identifying $T_{u}M$ with ${\mathbb{C}}^n$, this is a direct sum of $n$ copies of the Dolbeault operator on the trivial line bundle. Its cokernel is $H^{0,1}({\mathbb{CP}}^1,\mathcal{O})=0$, by Dolbeault cohomology of the projective line. Hence constant maps are regular for every compatible $J$.

The Energy identity for a J-holomorphic curve says that the energy is ${\int }_{{\mathbb{CP}}^1}{u}^{\ast }\omega$. If the homology class is zero, this integral is zero, forcing $du=0$. Thus every zero-class sphere is constant and regular.

In real dimension four, the Fredholm index of the parametrized sphere operator is $4+2c_1(A)$. At a regular nonconstant sphere, quotienting by the six-dimensional Möbius transformation group gives a local moduli space, or an orbifold for finite stabilizers, of real dimensionThis is the dimension formula for regular J-holomorphic spheres.

For a simple J-holomorphic sphere in class $A$, the adjunction inequality for a simple J-holomorphic sphere gives $c_1(A)\le A^2+2$. If $A^2\le -2$, the displayed moduli dimension is negative, so such a regular simple J-holomorphic sphere cannot exist. To cover nonsimple maps as well, interpret regularity here as regularity for all nonconstant maps under discussion. A degree-$m$ cover, $m\ge 2$, of a simple J-holomorphic sphere in class $B$ has $A=mB$, with $B^2\le -1$ and $c_1(B)\le 1$. Its local family of rational covering maps of the projective line modulo domain reparametrization has real dimension $4m-4$, while the predicted dimension isIt therefore cannot be regular. This proves there are no regular spheres in a class with square at most minus two, by negative-square exclusion under full sphere regularity. Regularity only for simple curves would not suffice: multiple covers of an exceptional minus-one sphere are the counterexample.

An almost complex structure is a smooth bundle endomorphism $J:TM\to TM$ with $J^2=-I$. It is an $\omega$-compatible almost complex structure whenThese conditions make $g_J(v,w)=\omega (v,Jw)$ a Riemannian metric. In particular it is symmetric: invariance and $J^2=-I$ give $\omega (Jv,w)=-\omega (v,Jw)$, and skew-symmetry then gives $g_J(w,v)=g_J(v,w)$. Positivity is the second condition. Moreover $J$ is an isometry for $g_J$. Compatibility does not require that the almost complex structure be integrable.

No nonconstant J-holomorphic curve with this closed connected source exists. The standard symplectic form on ${\mathbb{R}}^{2n}$ is exact; for instanceThe Energy identity for a J-holomorphic curve and the Generalized Stokes theorem giveThe target metric from the compatible almost complex structure is positive definite, so the nonnegative continuous energy density must vanish everywhere. Thus $du=0$. Connectedness of $\Sigma$ makes $u$ constant. This argument uses exactness and compatibility, and works even when $J$ is nonintegrable; it does not require the ordinary holomorphic maximum principle on the target.

Fix a reference path ${\gamma }_{\ast }$ from $L_0$ to $L_1$. By path connectedness, a path $\gamma$ can be joined to it by a smooth homotopy strip $u:[0,1{]}_s\times [0,1{]}_t\to M$ withSmooth homotopies may be used by the relative smoothing theorem; piecewise smooth homotopies give the same integrals. Orient the square by $ds\wedge dt$ and define the Lagrangian path-area functional byHere $\mathcal{A}$ denotes the functional called $A$ in the question. The minus sign is the convention for which downward gradient flow is $J$-holomorphic on the standard oriented strip.

For relative symplectic-area independence, glue two choices, with one orientation reversed, along their common reference and final paths. The resulting oriented surface is a relative $2$-cycle $Z$ in $(M,L_0\cup L_1)$. Since $H_2(M,L_0\cup L_1)=0$, it bounds a relative $3$-chain: as an absolute chain, $Z=\partial B+C$ for a chain $C$ supported in $L_0\cup L_1$. The symplectic form is closed, and it restricts to zero on each Lagrangian submanifold. Using piecewise smooth relative chains, Stokes theorem givesEquivalently, integration of the closed symplectic form defines the zero pairing on the zero relative homology group. The two strips therefore define the same value of $\mathcal{A}$.

Choosing a different reference path changes every value by the negative area of one fixed connecting strip, independent of $\gamma$. Changing the initially assigned reference value also adds a constant. Path connectedness ensures that this is one global constant, rather than an independent constant on each component.

For the first variation, let $\xi (t)$ be a variation vector field along $\gamma$, with $\xi (0)\in T_{\gamma (0)}{L}_0$ and $\xi (1)\in T_{\gamma (1)}{L}_1$. Differentiate the defining strip integral. Stokes theorem, or the variation formula for the integral of a closed differential form, reduces it to the final edge; the side edges give zero because their variations and tangent vectors lie in the Lagrangian submanifolds. With the chosen sign,If this vanishes for every admissible $\xi$, it in particular vanishes for every field supported in $(0,1)$. Nondegeneracy of the symplectic form, together with the fundamental lemma of the calculus of variations, forces $\dot{\gamma }(t)=0$ in the interior and hence everywhere by smoothness. Conversely, a constant path makes the displayed integral zero. The constant must lie in both endpoint submanifolds. ThusNo transverse-intersection assumption is required; the critical set can be empty or have positive dimension.

A compatible almost complex structure is a smooth bundle map $J:TM\to TM$ such thatThese conditions makea Riemannian metric. Its symmetry follows from the $J$-invariance and skew symmetry of $\omega$, and its positivity is the final compatibility condition. It also satisfies $g(Jv,w)=\omega (v,w)$.

Use the formal $L^2$ metric on the path space,The variation formula becomesThus a formal downward flow $s\mapsto u(s,\cdot )$ satisfiesPut the standard complex structure $j$ on the strip, with $j{\partial }_s={\partial }_t$. The J-holomorphic curve equation $du\circ j=J\circ du$ says $u_t=J{u}_s$, which is equivalent to the displayed flow equation because $J^2=-I$. The converse is identical: a smooth J-holomorphic curve on the strip with the specified Lagrangian boundary conditions gives a formal downward trajectory of $\mathcal{A}$.

As a sign check, along such a strip,The drop in $\mathcal{A}$ therefore equals the positive symplectic area swept out by the strip. The word formal matters: the $L^2$ calculation gives the interior equation and its Lagrangian boundary conditions; it does not assert that arbitrary smooth initial paths produce a well-posed ordinary flow on the smooth path space. In particular, the expression $J\dot{\gamma }$ need not satisfy the endpoint tangent restrictions for an arbitrary initial path. The claimed correspondence concerns smooth solutions of the strip equation.

A compatible almost complex structure is a smooth bundle map $J:TM\to TM$ such that, for all tangent vectors $u,v$,It determines the Riemannian metricIndeed, the invariance of the symplectic form under $J$ and its antisymmetry imply $\omega (v,Ju)=\omega (u,Jv)$, so $g$ is symmetric; the positivity condition makes it positive definite. A compatible triple consists of this symplectic form, this compatible almost complex structure, and this Riemannian metric. Useful equivalent identities are $\omega (u,v)=g(Ju,v)$ and $g(Ju,Jv)=g(u,v)$.

At a point $p$, write $W=T_{p}L$ and take the orthogonal complement with respect to the Riemannian metric $g$. The compatible triple identity givesIf $L$ is a Lagrangian submanifold, then $\omega (u,v)=0$ for $u,v\in W$, so $JW\subset W^{\perp }$. Both have dimension $\frac{1}{2}\dim M$, hence equality.

Conversely, if $JW=W^{\perp }$, equality of dimensions gives $\dim W=\frac{1}{2}\dim M$, and $g(Ju,v)=0$ for $u,v\in W$ gives $\omega {|}_W=0$. ThereforeA positive-definite inner product has $W\cap W^{\perp }=\{0\}$. Hence $W\cap JW=\{0\}$, proving that every Lagrangian submanifold is a totally real submanifold for a compatible almost complex structure.

Use ${\mathbb{C}}^2$ with the standard symplectic form ${\omega }_0=d{x}_1\wedge d{y}_1+d{x}_2\wedge d{y}_2$ and the compatible almost complex structure given by multiplication by $i$. Consider the embedded submanifoldIn coordinates $(x_1,y_1,x_2,y_2)$, its tangent space is $W=\{(a,b,b,0):a,b\in \mathbb{R}\}$. Applying $J$ gives $J(a,b,b,0)=(-b,a,0,b)$. If this is again in $W$, its fourth coordinate forces $b=0$, and its second and third coordinates then force $a=0$. Thus $W\cap JW=0$ and ${\dim }_{\mathbb{R}}L=2$: $L$ is a totally real submanifold.

However, the pullback of a differential form under the parametrization isThus this totally real submanifold is not a Lagrangian submanifold.