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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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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. –  Robert Bryant Sep 10 '11 at 1:49
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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. –  Robert Bryant Sep 12 '11 at 11:46
    
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. –  user5706 Sep 23 '11 at 19:09
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2 Answers

up vote 14 down vote accepted

A closed $k$-form is called instrinsically harmonic if there is some Riemannian metric with respect to which it is harmonic. E. Calabi ("An instrinsic characterization of harmonic 1-forms", 1969) 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 instrinsically 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).

E. Volkov's Characterization of intrinsically harmonic forms. J. Topol. 1 (2008), no. 3, 643–650, weakens the non-degeneracy condition, replacing it with the condition that the closed one-form be locally instrinsically harmonic - that is, the restriction of the form to a suitable open neighborhood of its zero set is instrinsically 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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Fourth sentence, presumably you meant "holomorphic" instead of "homomorphic"? –  Willie Wong Sep 10 '11 at 1:18
    
Thank you Willie! I corrected. –  aglearner Sep 10 '11 at 1:42
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