Let $\omega$ be a closed non-exact differential $k$-form ($k \geq 1$) on a closed orientable manifold $M$.
Question: Is there always a Riemannian metric $g$ on $M$ such that $\omega$ is $g$-harmonic, i.e., $\Delta_g \omega = 0$?
Here $\Delta_g$ is the Laplace-deRham operator, defined as usual by
$\Delta_g = d \delta + \delta d$, where $\delta$ is
the $g$-codifferential. Note that non-exactness is important, since if $\omega$ were to be exact and harmonic, then by the Hodge decomposition theorem $\omega = 0$.
For instance, if $\omega$ is a 1-form on the unit circle, then it is not hard to see that $\omega$ is harmonic with respect to some metric $g$ if and only if it is a volume form (i.e., it doesn't vanish). This observation generalizes to forms of top degree on any $M$.
What can be said in general for forms which are not of top degree?
Best Answer
A closed $k$-form is called intrinsically harmonic if there is some Riemannian metric with respect to which it is harmonic. E. Calabi (Calabi, Eugenio, An intrinsic characterization of harmonic one-forms, Global Analysis, Papers in Honor of K. Kodaira 101-117 (1969). ZBL0194.24701.) showed that a one-form having non-degenerate zeros on a compact manifold without boundary is intrinsically harmonic if and only if it satisfies a property called transitivity. The precise statement and proof can be found in chapter 9 of M. Farber's book "Topology of closed one-forms". In what comes I am following Farber. That a closed one-form $\omega$ have non-degenerate zeros means that near each zero it can be written in the form $\omega = df$ with $f$ a Morse function. For such a one-form, the additional assumption of harmonicity means that the Morse index of a zero cannot be $0$ or $n$ (write $\omega = df$ near the zero; because $\omega$ is co-closed, $f$ is harmonic, so by the maximum principle cannot have a max or min at the zero). That $\omega$ be transitive means that for any point $p$ of $M$ which is not a zero of $\omega$ there is a smooth $\omega$-positive loop $\gamma: [0, 1] \to M$; that is, $\gamma(0) = p = \gamma(1)$, and $\omega(\dot{\gamma}(t)) > 0$ for $t \in [0, 1]$. Then Calabi's theorem states that a closed one-form with non-degenerate zeros is intrinsically harmonic if and only if it is transitive. Near a non-degenerate index $0$ zero of a closed one-form the one-form can be written in the form $\delta_{ij}x^{i}dx^{j}$, for which it can be checked there are no positive loops beginning sufficiently near the origin.
(If one can handle $k$-forms then by Hodge duality one expects to be able to get somewhere with $(n-k)$-forms. The intrinsic harmonicity of $(n-1)$-forms was characterized in terms of transitivity in the thesis of Ko Honda, available on his web page). Evgeny Volkov (Volkov, Evgeny, Characterization of intrinsically harmonic forms, J. Topol. 1, No. 3, 643-650 (2008). ZBL1148.57036.) weakens the non-degeneracy condition, replacing it with the condition that the closed one-form be locally intrinsically harmonic - that is, the restriction of the form to a suitable open neighborhood of its zero set is intrinsically harmonic.
As far as I know, for higher degree forms nothing much is known at all, though for some special cases, like $2$-forms on $4$-manifolds, something more has been said. One imagines that with further assumptions on the form, perhaps more can be said - for example a symplectic form is always intrinsically harmonic (use the metric determined by a compatible almost complex structure). On the other hand, Volkov's paper exhibits a closed $2$-form of rank $2$ on a $4$-manifold which is transitive but not intrinsically harmonic.