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

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.