Proof that $S^4$ is not a symplectic manifold

Question

I started learning symplectic geometry, but apparently, i forgot some of my differential geometry. I am studying from the book by Aebischer et al.

In it, it is claimed that $S^4$ is not a symplectic manifold, and the proof goes as follows: suppose $\omega$ is a symplectic form on $S^4$. Because of vanishing of De-Rham cohomology, $\omega$ must be of the form $\omega = d(\alpha)$ for some 1-form $\alpha$.

But then, the volume form $\Omega = \omega \wedge \omega$ is also exact, with $d(\alpha \wedge \omega) = \Omega$.

Now, by Stokes theorem,

$\int_{S^4} \Omega = \int_{\partial S^4} \alpha \wedge \omega = 0$.

And the book claims that this is impossible for a volume form.

My two (probably very naive) questions: 1. Why is this intgeral zero? 2. Why can't the integral of a volume form vanish?

Answer 1

1. The boundary of $S^4$ is empty.

2. By definition, the integral of a volume form is greater than zero.

Answer 2

More generally, if $(X^{2n}, \omega)$ is a compact symplectic manifold, then $\omega^n$, as a volume form, represents a generator of top cohomology $H^{2n}(X, \mathbb{R})$, from which it follows that the image of $\omega$ in $H^2(X, \mathbb{R})$ must have the property that its $n^{th}$ cup power can't vanish; in particular, all of the even-dimensional cohomology groups $H^{2k}(X, \mathbb{R})$ contain at least one nontrivial element $\omega^k$ and hence also can't vanish. Conversely, if $X$ is a compact manifold such that any of the even-dimensional cohomology groups $H^{2k}(X, \mathbb{R})$ vanish, $0 \le k \le n$, then $X$ cannot be a symplectic manifold.

This implies, for example, that $S^2$ is the only sphere that can be a symplectic manifold, and also that, for example, $S^1 \times S^3$ cannot be a symplectic manifold. The stronger condition that there is some element of $H^2(X, \mathbb{R})$ whose $n^{th}$ power is nontrivial also shows that, for example, $S^2 \times S^4$ cannot be a symplectic manifold, despite having nontrivial even-dimensional cohomology groups.

From this perspective, arguably the "minimal" examples of compact symplectic manifolds are the complex projective spaces $\mathbb{CP}^n$, which are symplectic when equipped with the imaginary part of their Fubini-Study forms, and whose cohomologies are in fact generated by the powers of their symplectic forms.