Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 17 / 5

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.

Let $\lambda =\frac{1}{2}{\sum }_j(x_{j}d{y}_j-y_{j}d{x}_j)$ on ${\mathbb{C}}^3$, so $d\lambda ={\omega }_{st}$. On a projective line, work in the affine coordinate $w=\rho e^{i\theta }$ and lift it to the unit three-sphere byThis sphere lies in a complex two-plane in ${\mathbb{C}}^3$. The defining reduction identity for the Fubini-Study form gives $\Omega =s^{\ast }{\omega }_{st}$ on this chart. Differentiating the lift, or using the fact that real radial rescaling contributes no imaginary part to $\bar{z}dz$, givesThe missing point at infinity has zero area, soThis normalization is important: it differs from conventions in which a projective line has area $1$ or $2\pi$.

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 $r_1^4+r_2^4\le 1$.