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

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 $\omega$

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.

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.

• Thank you very much for your answer and your attention to my question. The main motivation for the question was the Hodge theorem, but the nondegeneracy is the main question – Ali Taghavi Sep 19 '18 at 8:15
• I think that the unique ness is up to coboundary. For example in $\mathbb{R}^2$ let $let \omega= e^x dx\wedge dy$. In this example $\omega$ is exact but one can put $\omega'=dx\wedge dy$ a nondegenerate harmonic 2-form. – Ali Taghavi Sep 19 '18 at 13:21
• @Seub The "counterexample" you provide is irrelevant - a symplectic form on a closed manifold is never exact. So the nondegeneracy question remains open. – Amitai Yuval Sep 19 '18 at 13:56
• @AmitaiYuval I think the symplectic form on a COMPACT manifold is never exact, right? – Ali Taghavi Sep 19 '18 at 14:21
• @AliTaghavi Yes. When I say a "closed manifold", I mean compact with no boundary. – Amitai Yuval Sep 19 '18 at 14:37