Riemannian metric making a given function harmonic

I have a nice 3-manifold (closed, oriented) which fibers over the circle, i.e. we are given a fibration $f:M\to S^1$. Apparently $M$ should admit a metric such that $f$ is harmonic. I don't quite see this:

A harmonic $f$ would mean $(d^*d+dd^*)f=0$. And $d^*f=0$ (as $*f$ is top-dimensional), so this implies that we want $* df$ to be a closed form... How can we ensure such a metric? Take a random one and then deform it appropriately?

Ref: Seiberg-Witten Floer Homology and Symplectic Forms on $S^1\times M^3$ (Kutluhan, Taubes).

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when you say it fibres over the circle, do you have a geometer's fibration or a topologist's fibration in mind? i.e. is it a fibre bundle with diffeomorphic fibres (equivalently, $df$ of maximal rank)? – Sam Lisi Jun 18 '12 at 9:58
Eh... how do you interpret the hodge dual of a manifold-valued map $f$? I am almost certain that "harmonic" in this context should mean a harmonic map where $S^1$ is endowed with the usual Riemannian metric. Or are you thinking of $S^1$ using its usual embedding into $\mathbb{C}$? – Willie Wong Jun 18 '12 at 12:55
@WillieWong: I suspect that what Kutluhan and Taubes mean is that the closed 1-form $df$ is also co-closed (i.e. $\star df = 0$). My best guess is that this is the same as asking for $f$ to be a harmonic map $M \to S^1$, but I am not sure. (While $f$ is manifold valued, you can think of $df$ as being an honest $1$-form by identifying $T S^1 = \mathbb{R} \times S^1$.) – Sam Lisi Jun 18 '12 at 22:14
Indeed, harmonic maps into $S^1$ are in bijective correspondence (via $f\mapsto df$) with harmonic 1-forms on $M$ with integral periods. For example, section 2.2 here. – user31373 Jun 18 '12 at 22:31
@SamLisi: yes. I agree. I am just picking bones with the notation $\mathrm{d}^* f$ and $*f$ when $f$ takes values in $\mathbb{S}^1$. Harmonicity (in this case with one dimensional target) is equivalent to $f^*\omega$ (where $\omega$ is a fixed Riemannian volume form on $\mathbb{S}^1$) being both closed and co-closed; this is equivalent to what you wrote. – Willie Wong Jun 19 '12 at 6:27

The pullback under $f$ of the canonical 1-form $d\theta$ on $S^1$ is a closed form $\omega$ on $M$. Such a form is called intrinsically harmonic if there exists a Riemannian metric on $M$ making it harmonic. This term was introduced by Calabi in 1969, who proved the following:

Suppose $\omega$ has only Morse-type zeros (this is interpreted in terms of a locally-defined real-valued function $\phi$ such that $d\phi=\omega$). Then the following are equivalent:

1. $\omega$ is intrinsically harmonic
2. For any point $p\in M$ which is not a zero of $\omega$ there exists a smooth path $\gamma\colon [0,1]\to M$ such that $\gamma(0)=\gamma(1)=p$ and $\omega(\dot \gamma(t))>0$ for all $t$. (Such $\gamma$ is called an $\omega$-positive loop.)

If $f\colon M\to S^1$ is a submersion then $\omega = f^*(d\theta)$ has no zeroes, so we are in good shape on the Morse side. To get an $\omega$-positive loop at $p\in M$, we must lift some loop $\sigma_n(t)=f(p)e^{i t}$, $0\le t\le 2\pi n$ to a loop based at $p$. It's not clear to me how to do this for an arbitrary fibration over $S^1$, but in the paper the fibration is assumed to be locally trivial, in which case $\sigma_1$ should have such a lift.

Sources:

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Interesting, yet cumbersome -- do you know how this agrees with my explanation given in my post (which was the sort of short-enough-answer I was looking for)? I hope it ties together nicely. – Chris Gerig Jun 21 '12 at 3:05

I had conversed with Taubes and found the answer, using $df$ itself:
We can take an area form $\alpha$ on the fiber (a surface) and then define $*df=\alpha$. Now we can simply choose a metric along the fibers which corresponds to having $\alpha$ as the area form. We're done, since now $*df$ is closed and hence $f$ is harmonic (from what I wrote in my question).

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This is essentially the idea I described in my comment above. I guess you have to choose $\alpha$ preserved by the monodromy? I'd like some more details if you have figured them out. – Sam Lisi Jun 21 '12 at 16:57