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For a topological space M, It is known from homotopy theory that the elements of the first cohomology $H^1(M;\mathbb{Z})$ are in 1-1 correspondence with homotopy classes of maps $[M,S^1]$

In my case of interest M is a smooth manifold. Take $\alpha$ and take smooth $f\colon M \to S^1$ representing $\alpha$ under the above identification.

My question is, under what (sufficient / necessary / equiv) conditions $f$ gives a locally trivial fibration (Fiber bundle) of M over $S^1$?

I would prefer conditions in terms of homology / cohomology of M.

What comes to my mind is that, according to Ehrshmann theorem, since $S^1$ is compact,this map is a fibration iff surjective + submersion. Surjectivity can be assured by picking nontrivial cohomology class. As for submersion. I have no idea how to characterize.

Thanks

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Let me assume throughout this answer that $M$ is closed, oriented, and connected. Here are some necessary conditions.

If you ask for a smooth fiber bundle, then a necessary condition is that the tangent bundle of $M$ has a trivial quotient of rank $1$, or equivalently a trivial subbundle of rank $1$. This is possible iff the Euler class $e(M)$ vanishes. This gives

Condition #1: $\chi(M) = 0$ (automatic when $\dim M$ is odd).

Next, if $F$ denotes the fiber of $f$, then the long exact sequence in homotopy for the fibration $F \to M \to S^1$ takes the form

$$1 \to \pi_1(F) \to \pi_1(M) \to \mathbb{Z} \to \pi_0(F) \to 1$$

(and for $n \ge 2$ the maps $\pi_n(F) \to \pi_n(M)$ are isomorphisms). This gives $\pi_1(F) = \text{ker} \left( \pi_1(M) \to \mathbb{Z} \right)$. Since $M$ is compact, so is $F$, so $\pi_0(F)$ is finite. It follows that $\text{ker}(\mathbb{Z} \to \pi_0(F))$ is nonzero, so we get

Condition #2: $\pi_1(M) \to \mathbb{Z}$ is nonzero.

This is equivalent to the condition that the corresponding cohomology class in $H^1(M)$ is nonzero. If we furthermore assume that $F$ is connected, then $\pi_1(M) \to \mathbb{Z}$ must be surjective, which is equivalent to the condition that the corresponding cohomology class is indivisible.

Next, if $M$ is a closed smooth manifold, then so is $F$. This gives

Condition #3: $\text{ker} \left( \pi_1(M) \to \mathbb{Z} \right)$ is finitely presented.

Of course this kernel must in fact be the fundamental group of a closed manifold of dimension $\dim M$, so if $\dim M \le 4$ (so that $\dim F \le 3$) then that puts some extra restrictions on it. Beyond this I don't know if there's anything easy to say.

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    $\begingroup$ Are you assuming that the fiber is connected? What about the double cover map $S^1 \to S^1$? $\endgroup$
    – Chris Schommer-Pries
    Commented May 18, 2015 at 11:15
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    $\begingroup$ In dimension 4, it might suffice that $ker(\pi_1(M)\to \mathbb{Z})$ is a PD(3) group; if it were, conjecturally this group would be an aspherical 3-manifold group, hence there ought to be a homotopy equivalent 4-manifold with the same fundamental group. However, the Borel conjecture is still open for 4-manifolds. I'm not sure if there is a natural conjecture in the case that $\pi_2(M)\neq 0$. $\endgroup$
    – Ian Agol
    Commented May 18, 2015 at 17:53
  • $\begingroup$ @Chris: oops, yes, you're right. Let me fix that. $\endgroup$
    – Qiaochu Yuan
    Commented May 18, 2015 at 18:59
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If $M$ is a compact and irreducible 3-manifold, one answer is provided by a theorem of Stallings, in his 1962 paper "On fibering certain 3-manifolds": $\alpha$ is represented by a fibration $f : M \to S^1$ if and only if the kernel of the associated homomorphism $\pi_1(M) \to \mathbb{Z}$ is finitely generated and that homomorphism is nontrivial.

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    $\begingroup$ Yes, and W. Thurston has a beautiful theory of which classes will satisfy this condition. There is a dual Thurston norm polytope inside $H^1(M)$. Certain faces of this polytope are fibered faces, and everything inside the cone over the (open) face are fibered. $\endgroup$
    – Dylan Thurston
    Commented May 17, 2015 at 23:41
  • $\begingroup$ Ah, just kidding. As Chris Schommer-Pries says I neglected the possibility that the fiber is disconnected. $\endgroup$
    – Qiaochu Yuan
    Commented May 18, 2015 at 18:59
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Here is another way to approach this problem. Say $M$ is closed and orientable. The correspondance $H^1(M,\mathbb{Z})$ with $[M,S^1]$ works by pulling back the generator of $H^1(S^1,\mathbb{Z})$ along $f$ to get $\alpha$.

Using Poincare duality and universal coefficient theorem (and forgetting about torsion) we can view $H^1(M,\mathbb{Z})$ as a lattice inside $H^1(M,\mathbb{R})$. By de Rham's theorem, $\alpha$ is represented by a closed $1$-form so that if we pick $f$ to be a smooth representative in the homotopy class, then we have $\alpha = f^*dx$ where $dx$ is the $1$-form on $S^1$ generating $H^1_{dR}(S^1)$. Then $f$ is a submersion if and only if $f^*dx$ is a nonvanishing $1$-form on $M$. So $\alpha$ corresponds to a fiber bundle if and only if, after tensoring with $\mathbb{R}$, it can be represented by a nonvanishing closed $1$-form.

In fact one can prove something slightly stronger as is done in this paper. A closed orientable manifold $M$ is fibered over $S^1$ if and only if it has a nonvanishing closed $1$-form and in this case the fibers are the leaves of the foliation generated by this $1$-form. Furthermore, if we have a nonvanishing closed $1$-form $\omega$ that is a nonero real multiple of $f^*dx$, then the fiber bundle map induced by $\omega$ is isotopic to $f$.

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