TITLE: Why spheres are not symplectic manifolds?
QUESTION [11 upvotes]: Reading some books on diferential geometry, a found that $S^{2n}$ (with $ n>1$) are not symplectic manifolds. They say it's because the de Rham cohomology of this spheres are R, but I do not understand this argument. It will be helpfull if anyone can explain me this fact (a silly explanation)...thx.

REPLY [20 votes]: Compact symplectic manifolds must have nontrivial $H^2$.
A symplectic manifold $M^{2n}$, by definition, possesses a closed, nondegenerate 2-form $\omega$. Because $\omega$ is closed (i.e., $d\omega = 0$), $\omega$ represents a de Rham cohomology class $[\omega] \in H^2(M)$.
The nondegeneracy condition implies $\omega^n$ (the wedge product of $\omega$ with itself $n$ times) is a volume form. This means that if $M$ is compact, $\int_M \omega^n$ is nonzero, and $[\omega]^n$ is nonzero in $H^{2n}(M)$, meaning $[\omega]$ must be nonzero in $H^2(M)$. In particular, $H^2(M)$ must be nonzero, which is not the case for $S^{2n}$ for $n>1$. ($H^\ast (S^k) = \mathbb{R} \text{ if } \ast = 0 \text{ or } k$ and is zero otherwise.)