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).
 A: 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: 


*

*$\omega$ is intrinsically harmonic

*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: 


*

*Topology of closed one-forms by M. Farber, Chapter 9.

*MathOverflow answer by Dan Fox.

A: 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).
