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.



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.

  • 2
    $\begingroup$ Are you assuming that the fiber is connected? What about the double cover map $S^1 \to S^1$? $\endgroup$ – Chris Schommer-Pries May 18 '15 at 11:15
  • 1
    $\begingroup$ 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 May 18 '15 at 17:53
  • $\begingroup$ @Chris: oops, yes, you're right. Let me fix that. $\endgroup$ – Qiaochu Yuan May 18 '15 at 18:59

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.

  • 5
    $\begingroup$ 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 May 17 '15 at 23:41
  • $\begingroup$ Ah, just kidding. As Chris Schommer-Pries says I neglected the possibility that the fiber is disconnected. $\endgroup$ – Qiaochu Yuan May 18 '15 at 18:59

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 $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.