Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 146 (Symplectic Topology)

For a symplectic ball centered at the origin, its symplectic blowup replaces a smaller concentric ball by the disk bundle in ${\mathcal{O}}_{{\mathbb{CP}}^{n-1}}(-1)$. The zero section is the exceptional divisor $E\cong {\mathbb{CP}}^{n-1}$; the blowdown collapses $E$ to the original point and is a symplectomorphism away from $E$. The new symplectic form agrees with ${\omega }_0$ outside the surgery region, while its integral over a line in $E$ specifies the blowup size. In real dimension four, $E\cong {\mathbb{CP}}^1$ and $E^2=-1$.

A Lefschetz pencil on the compact oriented four-manifold $X$ is a mapwith finite base locus $B$, such that orientation-compatible complex coordinates give the local model $f(z_1,z_2)=[z_1:z_2]$ at a base point and $f(z_1,z_2)=z_1^2+z_2^2$ at each critical point. Blowing up all $b=|B|$ base points produces a Lefschetz fibration $\tilde{X}\to S^2$ whose regular fiber is the closed smooth fiber of the pencil.

If that fiber has genus $g$ and there are $m$ critical points, thenEvery blowup raises the Euler characteristic by one, so $\chi (\tilde{X})=\chi (X)+b$. Therefore the Euler characteristic of a Lefschetz pencil is

Now let $X={\mathbb{CP}}^2$ with its Fubini-Study form. A symplectic smooth fiber $F$ represents $dH\in H_2({\mathbb{CP}}^2;\mathbb{Z})$ for some positive integer $d$, because its symplectic area is positive. The Symplectic adjunction formula, using $K_{{\mathbb{CP}}^2}=-3H$, givesand henceThese values are $0,0,1,3,6,10,\dots$. They are all attained: two sufficiently general homogeneous polynomials of degree $d$ generate a pencil of complex plane curves with only the singularities allowed in a Lefschetz pencil, and its smooth fibers are symplectic. Thus the displayed list is exactly the list described by the Genus of a symplectic Lefschetz pencil on the complex projective plane.

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.

Regard $X={\mathbb{CP}}^2$ as a Kähler manifold with its standard structure, letand choose $ϵ\ne 0$. The smooth conicrepresents $2H=[C_1]+[C_2]$. The standard complex structure is compatible with the Fubini-Study form, and $C_1$, $C_2$, and $\Sigma$ are all its complex submanifolds. It therefore supplies the required example.

Take the Hirzebruch surface $X={\mathbb{F}}_2$ with its standard toric Kähler structure. Let $C_1=S$ be its negative section of a Hirzebruch surface, and let $C_2=F$ represent the fiber class of a Hirzebruch surface and meet $S$ once. ThusThe preceding smoothing produces an embedded symplectic $\Sigma$ in class $S+F$.

Suppose a compatible almost complex structure made all three surfaces almost complex. Since $S$ and $\Sigma$ are distinct, Positive intersection of J-holomorphic curves would imply $S\cdot \Sigma \ge 0$. Homologically, however,a contradiction. Hence this example cannot admit such a simultaneous compatible structure.

With the conventionthe time-dependent Hamiltonian vector field is uniquely determined because $\omega$ is nondegenerate. If ${\varphi }_t$ is its local flow, Cartan's magic formula givesThus ${\varphi }_t^{\ast }\omega =\omega$ wherever the flow is defined, which is the Hamiltonian flow preserves the symplectic form property.

Write $v=(a,b)$ and ${\omega }_0=dx\wedge dy$. The linear Hamiltonianhas Hamiltonian vector field $X_{H_0}=v$. Choose a smooth cutoff $\rho$ that is one on $B(r+|v|)$ and zero outside $B(r+|v|+ϵ)$, and put $H=\rho H_0$. Every trajectory beginning in $B(r)$ and following $v$ remains in $B(r+|v|)$ for time $0\le t\le 1$, so the time-one map translates that ball by $v$. Outside $B(r+|v|+ϵ)$ its vector field vanishes, so the map is the identity. This is a compactly supported Hamiltonian translation and hence a compactly supported symplectomorphism.

For the connected-sum construction, choose a Darboux chart about the unique transverse intersection and straighten the two Lagrangian sheets to ${\mathbb{R}}^2$ and $i{\mathbb{R}}^2$ in ${\mathbb{C}}^2$. Remove small disks from the two sheets and join their boundary circles by the standard Lagrangian neckin ${\mathbb{R}}_x^2\oplus {\mathbb{R}}_y^2$, where $(a(t),b(t))$ follows a smooth arc from one positive coordinate ray to the other and agrees with those rays near its ends. Its pullback of ${\sum }_{j}d{x}_j\wedge d{y}_j$ vanishes because $u\cdot u^{\prime }=0$. Gluing this neck to the unchanged surfaces performs Lagrangian surgery. Topologically it is their connected sum, so it gives a Lagrangian connected sum

Finally, choose an immersed circle ${\gamma }_l:S^1\to {\mathbb{R}}^2$ with exactly $l$ transverse double points and no other multiple points; one may add $l$ small figure-eight kinks to an embedded circle. Let $c:S^1\to {\mathbb{R}}^2$ be an embedded circle and defineThis Product Lagrangian immersion satisfies ${\iota }^{\ast }{\omega }_0=0$. If $p$ is a double point of ${\gamma }_l$, its two local branches times $c(S^1)$ meet along $\{p\}\times c(S^1)$. Their tangent spaces intersect precisely in the tangent line to that circle, so the intersection is clean. The $l$ double points therefore give exactly $l$ disjoint clean self-intersection circles.

Near each clean circle, perturb one Lagrangian sheet by the graph of $\delta dh$ in its Weinstein neighborhood, where $h:S^1\to \mathbb{R}$ is a Morse function with one minimum and one maximum. The clean circle is replaced by two transverse double points. Resolve both by Lagrangian surgery. Each resolution attaches one one-handle and lowers the Euler characteristic by two, so resolving all $l$ clean circles changes the Euler characteristic of the original torus from zero toChoose the orientation-reversing neck at one double point; the resulting connected surface is nonorientable, while all the surgeries remove their double points and leave an embedding. Thus the Givental construction of nonorientable Lagrangian surfaces gives a closed connected nonorientable surface of Euler characteristic $-4l$ Lagrangian embedded in $({\mathbb{R}}^4,{\omega }_0)$.