2
$\begingroup$

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 $\omega$

$\endgroup$
2
+50
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Ali Taghavi Sep 19 '18 at 8:15
  • $\begingroup$ 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. $\endgroup$ – Ali Taghavi Sep 19 '18 at 13:21
  • 1
    $\begingroup$ @Seub The "counterexample" you provide is irrelevant - a symplectic form on a closed manifold is never exact. So the nondegeneracy question remains open. $\endgroup$ – Amitai Yuval Sep 19 '18 at 13:56
  • $\begingroup$ @AmitaiYuval I think the symplectic form on a COMPACT manifold is never exact, right? $\endgroup$ – Ali Taghavi Sep 19 '18 at 14:21
  • 1
    $\begingroup$ @AliTaghavi Yes. When I say a "closed manifold", I mean compact with no boundary. $\endgroup$ – Amitai Yuval Sep 19 '18 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.