symplectic manifolds

Question

I know sometimes they use advanced methods to prove a given 4-manifold is not symplectic. for instance by Seiberg-Witten theory. But for a manifold to be symplectic we just need to check that there is an element in the second cohomology which is closed and when we cup product with itself gives us a nonzero element. Isn't this easy to check by hand? for instance I guess connected sum of two copies of $CP^2$ is not symplectic, can't we check this by hand without using any S-W theory?

Answer 1

The simplest obstruction to a symplectic structure on a given closed manifold $M^{2n}$ is, as you say, the de Rham algebra $H^*_{dR}(M)$. If there is a symplectic structure on $M$, there must be an element $[\omega] \in H^2_{dR}(M)$ such that $[\omega^n]$ is a nonvanishing element in $H^{2n}_{dR}(M)$. This rules out that, say, spheres of dimension greater than two are symplectic.

But this is far from sufficient. We're not just looking for a closed 2-form whose highest exterior power is not exact, we're looking for a nondegenerate closed 2-form - something whose highest power is a volume form. That is to say, the form $\omega^n$ is nowhere zero, not just cohomologically nonzero. There's no reason to believe the existence of an $[\omega]$ as above is sufficient.

There are better obstructions than the cohomology ring alone. For instance, if I have a symplectic manifold $M$, it had better admit an almost-complex structure. Supposing $M$ is a 4-manifold, this existence question is just another homological criterion:

A closed 4-manifold $M$ admits an almost complex structure $J$ if and only if there exists $h \in H^2(M;\Bbb Z)$ such that $h^2 = 3\sigma(M) + 2\chi(M)$ and $h \equiv w_2(M) \mod 2$. Here $\sigma$ is the signature, $\chi$ is the Euler characteristic, and $w_2$ is the second Whitney class. Now we can verify that $\#_2 \Bbb{CP}^2$ does not support a symplectic structure by writing $H^2(M;\Bbb Z)$ as $\Bbb Z \oplus \Bbb Z$; $w_2 = (1,1)$; and so we must have $h^2 = 14$. But you can check by hand that $14$ cannot be written as the sum of two odd squares (or the sum of two squares, period). Using this theorem, one can actually show that a simply connected, closed 4-manifold supports an almost complex structure iff $b_2^+(M)$ is odd.

And then there are even better obstructions, though they're harder to use. As an example, $\#_3 \Bbb{CP}^2$ has a perfectly nice cohomology ring - three generators in degree 2 that all square to a positive form in top degree, and does support an almost complex structure. But its Seiberg-Witten invariants vanish because it can be written as a connected sum of a manifold with $b_2^+>0$ and another with $b_2^+ > 1$, so it cannot support a symplectic structure (as Taubes has proved that symplectic manifolds have nonvanishing Seiberg-Witten invariants when $b_2^+ \geq 3$.)

Answer 2

On Mike Miller's suggestion, I've turned my comments into an answer:

From a more analytic perspective, you're asking for a global solution $\omega \in \Omega^2(M)$ to the differential equation $d\omega = 0$ on $M^{2n}$ which also satisfies $\omega^n \neq 0$. In general, I don't think it's easy to know when a differential equation has global solutions or not.

Whether there exist non-degenerate 2-forms (not necessarily closed) is a purely topological question; a complete characterization of such manifolds can be given in terms of characteristic classes. But it's not clear which of those also support a non-degenerate 2-form which solves $d\omega = 0$. (Just as it is not clear, for example, which smooth manifolds admit integrable complex structures.)