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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 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.

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 $\pi R^2$.

On the affine chart $[1:z_1:\cdots :z_n]$ of ${\mathbb{CP}}^n$, the normalized Fubini-Study form isIts integral over a projective line is $\pi$. The Darboux theorem says that every point of a symplectic $2n$-manifold has local coordinates in which the form is ${\sum }_{j}d{x}_j\wedge d{y}_j$.

To form the blowup ${\tilde{X}}_{\lambda }$, choose a symplectic embedding of the closed standard ball $B^{2n}(\lambda )$ centered at $p$, remove its interior, and collapse each characteristic Hopf circle of its boundary $S^{2n-1}$ to a point. The boundary becomes the exceptional divisor $E\cong {\mathbb{CP}}^{n-1}$, and the reduced form extends the old form outside the ball with every projective line in $E$ having area $\pi {\lambda }^2$. This is the symplectic blowup of size $\lambda$. In real dimension four its volume isThus, whenever two different sizes are allowed, different $\lambda$ give different total symplectic volumes and hence nonsymplectomorphic blowups.

The punctured area-$\pi$ sphere ${\mathbb{CP}}^1\setminus \{\infty \}$ is symplectomorphic to the open unit disc. Consequentlyfor every $r<1$. The bound is sharp. Given a hypothetical larger ball, choose a compatible almost complex structure agreeing with the pushed-forward standard structure on the ball. A standard J-holomorphic-curve result supplies a sphere in one ruling class through the ball center. Its total area is $\pi$, while monotonicity inside the ball requires at least $\pi r^2$. Hence $r\le 1$. Equivalently, the Gromov width of the monotone product of projective lines is

An almost complex structure $J$ on $(M,{\omega }_M)$ is a compatible almost complex structure when $J^2=-I$, ${\omega }_M(Ju,Jv)={\omega }_M(u,v)$, andis a positive-definite inner product. To prove existence, choose any Riemannian metric $h$ and define $A$ by ${\omega }_M(u,v)=h(Au,v)$. The metric construction of a compatible almost complex structureis smooth and compatible, so the space is nonempty.

Identify the compact symplectic manifold $N$ with its image under the symplectic embedding. Along $N$ there is a symplectic splittingChoose compatible almost complex structures on both summands and take their direct sum. Its associated metric makes the two summands orthogonal. Extend this metric from the closed submanifold $N$ to all of $M$ using a partition of unity, and apply the metric construction again. Along $N$ it recovers the prescribed direct sum, so the resulting global compatible $J$ satisfies $J(TN)=TN$. This is the relative extension of a compatible almost complex structure.

For the two compact complex curves, use the supplied holomorphic coordinates at their transverse intersection. There $C_1\cup C_2$ is $\{xy=0\}$. Replace it in a small ball by the complex annulus $\{xy=ϵ\}$ and use a cutoff in a surrounding annulus to rejoin the unchanged curves. For sufficiently small nonzero $ϵ$, the result is an embedded symplectic surface $\Sigma$. This local replacement is a bordism between the old and new cycles, soThis is the symplectic smoothing of a positive transverse node.

A symplectic ball is a Euclidean ball in $({\mathbb{R}}^{2n},{\omega }_0)$, or its image under a symplectic embedding.

A symplectic capacity is monotone under symplectic embeddings, scales by $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and has positive unit-ball capacity and finite unit-cylinder capacity. It is invariant under symplectomorphisms. A normalized symplectic capacity additionally assigns $\pi$ to both unit domains.