I was reading this question and it got me thinking more about harmonic maps.
A smooth circle-valued map $\varphi : M \to S^1$ from a Riemannian manifold is harmonic if $\varphi^*(d \theta)$ is in the kernel of Hodge Laplace operator $\Delta^1 : \Omega^1(M) \to \Omega^1(M)$.
It seems to me that, given a nice enough harmonic $\omega \in \Omega^1(M)$, one might be able to induce a harmonic map. Suppose that $\omega$ is harmonic, and in addition, that $$\int_\gamma \omega \in \mathbb{Z}$$ for all $[\gamma] \in \pi_1(M)$. Since $\omega$ is closed, this doesn't depend on the homotopy class of $\gamma$. One should be able to define a map $\varphi_\omega : M \to S^1 = \mathbb{R}/\mathbb{Z}$ by mapping a base-point $x$ of $M$ to $1$ and defining the rest of the map via $$\varphi_\omega(x') \mapsto exp(2 \pi i \int_\gamma \omega)$$ for any path $\gamma$ from $x$ to $x'$.
My question: is it possible that $$ \varphi_\omega^*(d \theta) = \omega$$ thus recovering the original harmonic and generating a harmonic map? The example in my head is a harmonic on $S^1 \times S^1$ which 'wraps' around the first coordinate.