Existence of harmonic symplectic structure on a symplectic manifold (which is cohomologue to the initial symplectic structure)

Question

Assume that $(M, \omega)$ is a symplectic manifold which is equiped with a Riemannian metric.

Is there a symplectic structure $\omega '$ which is a harmonic $2$-form? Can one choose such a $\omega'$ such that it would be de Rham cohomolgue to the initial form $\omega$

Answer 1

A difficulty with your question is that there need not be any relation between the given symplectic form $\omega$ and the given Riemannian metric $g$; consequently, it is far from clear that (some symplectic form cohomologous to) the symplectic form $\omega$ can be $g$-harmonic, or that there exists a $g$-harmonic symplectic form for that matter.

Let's assume first that $\omega$ and $g$ are 'related', namely that they are adapted to each other i.e. there is an almost complex structure $J$ such that $\omega(-, J-) = g(-,-)$ and $\omega(J-,J-) = \omega(-,-)$. Then it turns out that $\omega$ is co-closed for $g$, hence $g$-harmonic; see for instance Theorem 1 in Delanoë's Analyzing non-degenerate 2-forms with riemannian metrics. It is well-known that the set of metrics adapted to a given symplectic form is non-empty, hence given $\omega$, there is always several choices of $g$ for which $\omega$ is harmonic. (In dimension 2, a simple calculation shows that a symplectic form is $g$-harmonic if and only if it is adapted to $g$, and your questions are thus always answered positively.)

If we drop the condition that $\omega$ and $g$ be adapted, then it is possible (in dimension 4 and higher) that the answers to your questions be negative. To be more specific, in dimension 4, given a symplectic form $\omega$, there is always a choice of $g$ for which the $g$-harmonic form cohomologous to $\omega$ is not symplectic; If moreover $b_2 = 1$ (e.g. $M = \mathbb{C}P^2$), then for the same $g$, no harmonic form is symplectic. (By taking cartesian products, I suspect one can get counter-examples in higher dimensions too).

Here is a proof of the above claims. The Luttinger-Simpson(-Perutz-Taubes) theorem implies that any 4-manifold which admits a symplectic structure admits cohomologous nonsymplectic 'nearly symplectic' forms; see Theorem 1 of this paper of Taubes for a more precise statement. Now, according to Proposition 1 in this paper of Auroux-Donaldson-Katzarkov, any nearly symplectic form is $g$-harmonic for some metric $g$. The combination of these two results and the uniqueness of the harmonic form in its cohomology class yield the claims.

Answer 2

Hodge's theorem states that on a closed Riemannian manifold, any closed form has a unique harmonic representative. So the answer to your question is yes, and moreover it is unique, provided this harmonic representative is nondegenerate (as a bilinar form). But that is not always the case (e.g. when $\omega$ is exact, $\omega' = 0$). I don't know whether there is a way to predict that the harmonic representative of $\omega$ will be degenerate or not.