Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 16 / 2

A symplectic manifold is a smooth manifold $M$ equipped with a differential two-form $\omega$ such thatThe first condition says it is a closed differential form; the second says it is pointwise nondegenerate. A nondegenerate skew-symmetric matrix has even size, so the dimension is $2n$. Equivalently, ${\omega }^n$ is a nowhere-zero top-degree differential form. It supplies an orientation and the volume form ${\omega }^n/n!$. These are the defining conditions on the symplectic form.

Odd-dimensional spheres cannot carry a symplectic form, by even-dimensionality. For $S^{2m}$ with $m\ge 2$, the second de Rham cohomology group vanishes. Any hypothetical symplectic form would therefore be $\omega =d\alpha$. The symplectic cohomology obstruction givesby the Generalized Stokes theorem. But the symplectic orientation makes ${\omega }^m$ a positive volume form, whose integral on this nonempty compact manifold is positive. This is a contradiction.

The oriented area form on $S^2$ is nondegenerate and automatically closed, since a two-dimensional manifold has no nonzero three-forms. Thus among positive-dimensional spheres, exactly $S^2$ admits a symplectic structure:If zero-dimensional symplectic manifolds are admitted, $S^0$ also qualifies: its zero two-form is nondegenerate on the zero tangent spaces. This is a convention-dependent additional case, not another positive-dimensional example.

There are no symplectic structures on the Möbius strip. A symplectic form on a surface would be a nowhere-vanishing two-form, hence would provide an orientation. The Möbius strip is nonorientable: transport around its core reverses a transverse direction and therefore reverses any local orientation. This is incompatible with such a two-form. The argument applies whether the boundary is included or only the interior is considered.

A symplectic vector field satisfies ${\mathcal{L}}_X\omega =0$, so its local flow consists of symplectomorphisms. By Cartan's magic formula and $d\omega =0$, this is equivalent to the differential one-form ${\iota }_X\omega$ being closed. In the convention fixed above, a Hamiltonian vector field satisfies ${\iota }_X\omega =dH$ for a globally defined smooth function $H$. It is therefore a symplectic vector field, but the converse requires this closed one-form to be exact.

On the torus with $\omega =dx\wedge dy$,Thus ${\partial }_x$ is a symplectic vector field. The closed differential one-form $dy$ is not exact on the torus: its integral on the loop $\gamma (s)=[(0,s)]$, $0\le s\le 1$, equals one, while the integral of an exact one-form around any closed loop is zero. The vector field is symplectic but not Hamiltonian. The local candidate $H=y$ does not descend to a single-valued function on ${\mathbb{R}}^2/{\mathbb{Z}}^2$.

The Hamiltonian flow of $H$ solvesCompactness and the absence of a boundary make this smooth vector field complete, so the flow exists for all real $t$. The defining equation and Cartan's magic formula giveTherefore the pullback of a differential form differentiation rule yieldsso $({\varphi }_H^t{)}^{\ast }\omega =\omega$. Pullback commutes with the wedge product of differential forms, and consequentlyThus the flow preserves the symplectic volume, in fact the entire symplectic form.