Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 (Symplectic Topology)

Let $J_0$ be the standard complex structure on ${\mathbb{R}}^{2n}$, let ${\omega }_0$ be the standard symplectic form, and let $g_0$ be the Euclidean inner product. They satisfyThe two-out-of-three property for unitary structures says that a real linear map preserving any two of these structures preserves the third. In terms of the corresponding matrix groups,

For example, if $A$ preserves $J_0$ and ${\omega }_0$, thenso $A$ preserves $g_0$. If it preserves $J_0$ and $g_0$, the second displayed identity shows that it preserves ${\omega }_0$. Finally, $J_0$ is uniquely determined by $g_0(J_{0}u,v)={\omega }_0(u,v)$, so preservation of $g_0$ and ${\omega }_0$ implies $A{J}_0=J_{0}A$. The common intersection is therefore the unitary group.

The unitary group acts on the Lagrangian Grassmannian by $A\cdot L=A(L)$. Every Lagrangian subspace has an orthonormal basis, and adjoining its $J_0$-image gives a unitary basis, so this action is transitive. The stabilizer of the standard real subspace ${\mathbb{R}}^n\subset {\mathbb{C}}^n$ consists exactly of real unitary matrices, namely $O(n)$. Henceis a continuous bijection from a compact space to a Hausdorff space and is therefore a homeomorphism.

For $n=1$, every line in ${\mathbb{R}}^2$ is Lagrangian. ThusConcretely, the line making angle $\theta$ with the real axis corresponds to $e^{2i\theta }$.

On the quotient from part (b), the Maslov mapis well defined because $\det (Q{)}^2=1$ for $Q\in O(n)$. Consider the loop of Lagrangian subspacesIts endpoints agree as unoriented subspaces, and $\mu \circ \ell (t)=e^{2\pi it}$ has degree one. If $\eta$ is the generator of $H^1(S^1;\mathbb{Z})$, thenConsequently ${\mu }^{\ast }\eta \ne 0$, provingThe same integer is the Maslov index of $\ell$.

Any symplectic form on $S^2$ orients its tangent bundle. Choose a compatible complex structure and inner product; this reduces the structure group to $SO(2)$. Since every line in an oriented symplectic plane is Lagrangian,is an oriented circle bundle.

If a transition function of $T{S}^2$ rotates vectors through an angle $\alpha$, its action on unoriented lines rotates the coordinate $e^{2i\theta }$ through $2\alpha$. The Euler class of the Lagrangian-line bundle of an oriented plane bundle therefore givesBy the Poincaré-Hopf theorem,for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal{L}Gr(T{S}^2)$ is $\pm 4$, which is nonzero. A smoothly trivial oriented circle bundle has zero Euler class, so $\mathcal{L}Gr(T{S}^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.

Take the toruswith the translation-invariant symplectic formThe coordinate vector fields give a global symplectic frame of $TM$, so $TM$ is symplectically trivial. Passing fiberwise to the Lagrangian Grassmannian bundle giveswhich is a smooth trivialization over the compact symplectic manifold $T^{2n}$.

The Weinstein neighborhood theorem states that if $L$ is a compact Lagrangian submanifold of $(M,\omega )$, then neighborhoods of $L$ in $M$ and of the zero section in its cotangent bundle $T^{\ast }L$ are symplectomorphic. The symplectomorphism restricts to the identity on $L$, and the canonical form on $T^{\ast }L$ is taken with the sign matching the chosen convention.

Take $L=S^2$. For any Lagrangian embedding of $S^2$ into a symplectic four-manifold, the symplectic form identifies the normal bundle with $T^{\ast }{S}^2$. The self-intersection formula and the Euler characteristic giveup to the harmless orientation sign. If a smooth isotopy displaced $L$, its final image would represent the same homology class but have intersection number zero with $L$, a contradiction. This is the smooth non-displaceability from self-intersection.

For a compact ambient example, equip $S^2\times S^2$ with $\omega \oplus (-\omega )$. Its diagonalis Lagrangian because the two summands cancel on $T\Delta$, and the preceding argument shows that it is not smoothly displaceable.

Let $M=S^2$ with an area form and let $L$ be an equator dividing the sphere into two open hemispheres of equal area. Every curve in a symplectic surface is Lagrangian. A small normal push moves $L$ to a nearby latitude, so it is displaceable by a smooth isotopy.

Suppose a symplectic isotopy had final image $L^{\prime }$ disjoint from $L$. The curve $L^{\prime }$ must lie in one hemisphere. Of the two discs bounded by $L^{\prime }$, the one contained in that hemisphere has area strictly below half the total area, and the other has area strictly above half. On the other hand, a symplectomorphism maps the original two hemispheres to the two discs bounded by $L^{\prime }$ and preserves their areas, so both would have half the total area. This contradiction is the symplectic non-displaceability of an area bisector.

LetFor any sufficiently small nonzero $\delta$, translationis a symplectic isotopy with ${\varphi }_1(L)\cap L=\varnothing$. Choosing $|\delta |$ arbitrarily small makes the isotopy arbitrarily small in every $C^k$ norm.

This isotopy has nonzero flux homomorphism, represented by $\delta dx$. In contrast, every Hamiltonian isotopy has zero flux. If a Hamiltonian image $L^{\prime }$ were disjoint from $L$, the two homologous essential circles would bound an annulus $A$, and evaluation of the flux on $[L]$ would equal the signed symplectic areawhich is nonzero. This contradicts vanishing Hamiltonian flux, so $L$ is not Hamiltonian displaceable.

Again take a symplectic two-sphere, but let $L$ be a small latitude bounding a cap of area strictly below half the total area. A rotation carrying that cap to a disjoint cap carries $L$ to a disjoint Lagrangian circle. Rotations of the symplectic sphere are flows of Hamiltonian vector fields; for rotation about an axis, a height function is a Hamiltonian function. Thus this rotation is a Hamiltonian isotopy, and $L$ is Hamiltonian displaceable.

Moser's trick states that if $M$ is compact and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms whose de Rham cohomology class is independent of $t$, then there is an isotopy ${\varphi }_t$ withSince $[{\dot{\omega }}_t]=0$, choose a smooth family of one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy of ${\omega }_t$ uniquely determines a vector field $X_t$ byCompactness makes its flow ${\varphi }_t$ exist for the whole interval. Cartan's magic formula and $d{\omega }_t=0$ givewhich proves the theorem.

Smooth degree-$d$ hypersurfaces form the complement of the discriminant in the projective space of degree-$d$ homogeneous polynomials. This complement is path connected, so $X$ and $X^{\prime }$ lie in a smooth one-parameter family. The Ehresmann fibration theorem identifies the fibers smoothly. Under such an identification, the restrictions of the Fubini-Study form form a family ${\omega }_t$ whose cohomology class is the fixed restricted hyperplane class. Moser's trick therefore gives the symplectic equivalence of smooth projective hypersurfaces.

It remains to construct the finite subgroup for one convenient hypersurface. On the Fermat hypersurfacethe group $({\mu }_d{)}^{n+1}$ acts by diagonal coordinate multiplication. It preserves both $X_F$ and the Fubini-Study form. The kernel of its projective action is the diagonal subgroup ${\mu }_d$, so the effective Fermat hypersurface diagonal symmetry group isConjugating this action by a symplectomorphism $X_F\to X$ gives the required subgroup of $\mathrm{Symp}(X,{\omega }_{FS}{|}_X)$.

The symplectic neighborhood theorem says that a neighborhood of a compact symplectic submanifold is determined, up to symplectomorphism, by the restricted symplectic form and its symplectic normal bundle. More precisely, if $f:S_0\to S_1$ is a symplectomorphism and an isomorphism of symplectic normal bundles covers $f$, then that bundle isomorphism extends to a symplectomorphism between neighborhoods of $S_0$ and $S_1$.

Let $Q\subset {\mathbb{CP}}^2$ be a smooth complex conic. Its homology class is $2H$, so its self-intersection number isRescale the Fubini-Study form so that $Q$ and $C$ have the same symplectic area. Their normal bundles have opposite Euler numbers, $+4$ and $-4$, so the symplectic sum can be formed along $Q$ and $C$.

Concretely, remove tubular neighborhoods $\nu (C)$ and $\nu (Q)$ and glue the boundaries by a fiber-reversing bundle map. The symplectic neighborhood theorem supplies the standard models needed for the gluing, and the symplectic-sum construction supplies a symplectic form onThe second piece is a rational homology ball. Thus this operation replaces the neighborhood of the $-4$ sphere by that rational ball and is the symplectic rational blowdown of a minus-four sphere.

Let $F_0$ and $F_1$ be general homogeneous cubic forms with base locus $B$ of nine points. The incidence varietyis the blowup of ${\mathbb{CP}}^2$ at $B$. Projection onto the second factor is the elliptic fibration of the rational elliptic surfaceand its fibers are connected plane cubics. Over any base point $p\in B$, the exceptional curve $E_p=\{p\}\times {\mathbb{CP}}^1$ maps isomorphically to the base, so it is a holomorphic section.

Write $H$ for the pullback of a line and $E_1,\dots ,E_9$ for the exceptional classes. The fiber class and the canonical class of the rational elliptic surface areThe degree of $\pi {|}_C$ is $C\cdot F=d$. The Adjunction formula now yieldsTherefore the genus formula for a multisection of the rational elliptic surface is