Why spheres are not symplectic manifolds?

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.

Solution 1:

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.)

Solution 2:

Suppose $(X, \omega)$ is a closed symplectic manifold of dimension $2n$. If $\omega = d\alpha$, then $$\int_X\omega^n = \int_X d(\alpha\wedge \omega^{n-1}) = \int_{\partial X}\alpha\wedge \omega^{n-1} = 0$$ but this is absurd as $\omega^n$ is a volume form. Therefore, a symplectic form on a closed manifold is not exact, so it defines a non-zero element of $H^2_{\text{dR}}(X)$. As $H^2_{\text{dR}}(S^{2n}) = 0$ for $n > 1$, $S^{2n}$ is not symplectic.

A similar argument can be used to show that $\omega^k$ is not exact for $k = 1, \dots, n$, so in fact $H^{2k}_{\text{dR}}(X) \neq 0$ for a closed symplectic manifold $X$.