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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?

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    $\begingroup$ I don't have an answer, but this problem is unlikely to be solvable locally once the dimension is high enough and $k$ is far from $0$ or $n$, the dimension of $M$. For example, if $n$ and $k$ are such that $n\choose k$ is greater than $\tfrac12n(n{+}1)$, then the equation $\Delta_g\omega=0$ is an overdetermined (second order) equation for $g$ and, most likely, won't have solutions. I think that the first time this happens is $(n,k)= (7,3)$, so I would check there first. I have to admit, though, that I haven't done the calculation. $\endgroup$ Sep 10, 2011 at 1:49
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    $\begingroup$ Probably $(n,k) = (7,4)$ is a better case to investigate, since there are fewer conditions for a $4$-form on a $7$-manifold to be closed than for a $3$-form, and the condition $\delta_g\omega=0$ (which, together with $d\omega=0$ will force $\Delta_g\omega=0$) is also overdetermined (for $g$) in this case. $\endgroup$ Sep 12, 2011 at 11:46
  • $\begingroup$ Thanks to all who helped elucidate this question! I was mostly interested in the degrees $k = 1$ and $k = n-1$, but it would certainly be very interesting to see what can be said about the intermediate $k$'s. Perhaps a topic for a future Ph.D. thesis. $\endgroup$ Sep 23, 2011 at 19:09

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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.

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This question is quite subtle, I don't believe the answer is known in the general situation. But if you consider the case of $1$-forms on surfaces, one can completely characterise those that are harmonic. They form a "space" of finite dimension modulo self-diffeos of the surface. They are called minimal, and they can all be represented as real parts of some holomorphic $1$-forms. Minimal 1-forms on a surface $S$ are characterised by the property that for each point $x\in S$, where the one-form is non-vanishing there exists a circle $S^1$ on $S$ such that the one-form restricted to $S^1$ has not zeros while $x\in S^1$.

For example, in the case of a $T^2$ a one-form is harmonic for some metric iff it has no zeros at all.

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  • $\begingroup$ In general, every closed 1-form without zeros is intrinsecally harmonic (defined in closed orientable manifolds). This fact was observed by E. Calabi in the paper "An instrinsic characterization of harmonic 1-forms", 1969. $\endgroup$ Sep 19, 2019 at 17:54
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1 - Eugenio Calabi affirms that every closed non singular 1-forms in closed manifolds are intrinsecally harmonic ("An instrinsic characterization of harmonic 1-forms", 1969). This is easy proved observing first that such forms are transitive. The dual case is a open problem, but I believe that is true. A closed $(n-1)$-form non singular and non null in cohomology is intrinsecally harmonic iff the voluming-preserving flow induced by this form admit cross section (closed comdimension one submanifold cutting every orbit of the flow) or iff admit complementar foliation (foliation transversal to the flow; at leas $C^2$). Is possible proof this fact in the spirit of Calabi's work. This problem have consequences in flat characterization of circle bundles.

2 - In the case of closed $p$-forms of the rank $p$, the problem have a translate for foliation theory. Is possible to show that if a closed $p$ form $\omega$ of the rank $p$ is transitive, then exists complementar form $\eta$, that is, $\eta$ is a closed form such that $\omega\wedge\eta$ is volume form. This is proved using the theory of foliations cycles (see the Sullivan's paper "Cycles for the dynamical study of foliated manifolds and complex manifolds"). However, in a fiber bundle $\xi=(\pi,F,E,M)$ with symple connected base, compact total space, if $\Omega_M$ is any volume form in $M$, the form $\pi^*\Omega$ is intrinsecally harmonic iff $\xi$ is trivial. We can run away from trivial cases showing that $[F]\neq 0\in H_{\dim F}(E;\mathbb{R})$ and exists exemples those bundles with section. This give us the examples cited by Dan Fox. The problem is the dimension of the kernel of $\eta$. In the cases $p=1$ or $p=n-1$ (without singularities), in the condition of transitivity, the dimension of the kernel of $\eta$ is $n-1$ (case $p=1$) or 1 (case $p=n-1$), and the Calabi argument applies.

3 - Is too a open problem to show that harmonic forms are transitive, as observed by Katz ("Harmonic forms and near-minimal singular foliations"). We can to show that if $\omega$ is a harmonic $p$-form of rank $p$, then the have in $M$ two complementary $SL(*)$-foliations induced by $\ker\omega$ and $\ker *\omega$.

4 - I have studied the problem of decomposable forms. By the Tischler's argument (and others considerations; see "On fibering certain foliated manifolds over $\mathbb{S}^1$") is sufficient considering bundles $\xi=(F,\pi,E,\mathbb{T}^{p})$. If this bundle admit transversal foliation with holonomy group contained in $SL(*)$, then the form $\pi^*(\Omega_{\mathbb{T}^{p}})$ is intrinsecally harmonic. The ideia of study particular examples is know if we can rule out the hypothesis of transitivity. In any bundle $\xi=(\pi,F,E,M)$, with compact total space, $[F]\neq 0$ and $\pi_1(M)$ finite, the form $\pi^*(\Omega)$ is intrinsecally harmonic.

Ps. The above remarks is part of development of my doctoral project and are is under analysis.

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  • $\begingroup$ "Eugenio Calabi affirms that every closed non singular 1-forms in closed manifolds are intrinsecally harmonic ..." I don't think this is true, considering Calabi's theorem. That is because a closed non-singular 1-form on a closed manifold need not be transitive. Example: take a Morse function $f$ on a manifold. This has at least one minimum and maximum. Thus, $df$ is not transitive, because if you start at a point near the maximum, any $df$-increasing path has to lead away from the maximum. There are even more subtle problems with 1-forms that have non-Morse zeros, but I have no example here. $\endgroup$
    – user505117
    Oct 16, 2021 at 11:38
  • $\begingroup$ First of all we began by clarifying a definition: 1 - closed manifolds means a boundaryless and compact manifold 2 - a non singular form on $M$ means a form satisfying $\omega_x\neq 0$ for all $x\in M$. Thus, $\omega\neq df$ for any function $f$, since on compact manifolds $df$ have some zero. A proof of transitivity condition for such kind of form can be found in repositorio.unicamp.br/jspui/handle/REPOSIP/354867 chapter 2. $\endgroup$ Oct 18, 2021 at 12:43
  • $\begingroup$ I see, I had the wrong definition of "non singular" in mind. My comment does not apply, then. $\endgroup$
    – user505117
    Oct 19, 2021 at 8:48

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