Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 / 1

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 $T^{2n}$.