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Suppose we have a surface bundle $M$ over the unit circle $S^1$ and $M$ is assigned with Riemannian metric $g$. The projection map $\pi: M\to S^1$ now can be homotopic to a harmonic map $\phi: M\to S^1$, which is unique up to a constant function from $M$ to $S^1$. So, my question is: in general, is $\phi$ still a fiber bundle?

(The existence of $\phi$ can be shown as follows. We can find a smooth representative $f$ in the homotopy class of $\pi$, and locally we can lift $S^1$ to $\mathbb{R}$ and define $\Delta_g f$. We solve the equation $\Delta_g h=\Delta_g f$ and put $\phi=fe^{-2\pi ih}$)

If not, can we give some reasonable condition on the topology of $M$ or on the metric $g$ that will guarantee $\phi$ is a fiber bundle? In particular, in the case that $M$ is a hyperbolic 3-manifold (and $g$ is the hyperbolic metric), is $\phi$ a fiber bundle?

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  • $\begingroup$ By Ehresmann's Theorem a map between compact manifolds is a fiber bundle iff it is a surjective submersion. $\endgroup$
    – ThiKu
    May 1, 2017 at 10:57
  • $\begingroup$ In general, there exist harmonic maps which are neither null-homotopic nor a submersion. For example there are manifolds which do not fiber over the circle but have $H^1(M)\not=0$ and so allow nontrivial harmonic maps to the circle $\endgroup$
    – ThiKu
    May 1, 2017 at 11:00
  • $\begingroup$ Thanks for your comment! I understand this won't be true if $\pi$ is chosen arbitrarily in the very beginning. But if $\pi$ is a surjective submersion, can we expect $\phi$ to be a surjective submersion? I think this is the part I don't know how to prove or disprove. $\endgroup$
    – Donghao
    May 1, 2017 at 17:24
  • $\begingroup$ I think the short answer is that this is not known for hyperbolic 3-manifolds that fiber over the circle. In the case that the manifold is (congruence) arithmetic, a harmonic 1-form is a kind of automorphic form. Dunfield tried to investigate whether there are any known non-vanishing results for automorphic forms, but apparently there are no known such results. $\endgroup$
    – Ian Agol
    May 12, 2017 at 17:30

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