# How to tell whether a compact manifold can be realized as a nontrivial fiber bundle?

Let $M$ be a compact closed path-connected manifold.

There is the Postnikov tower as in

http://en.wikipedia.org/wiki/Postnikov_system

which tells us $M$ can be realized as an inverse limit of a tower of fibrations, up to homotopy, and this tower is often infinite, if $\pi_n(M)$ does not vanish when $n$ is large enough.

So I wonder, what if we work in the homeomorphic category instead of the homotopy category, and try to realize $M$ as a fiber bundle

$F \rightarrow M \rightarrow B$

instead of a fibration (which is more general). Are there certain obstructions to this, such that one can tell whether this is possible by looking at the algebraic topology of $M$?

As an example of what I have in mind, note we have Hopf fibrations for $S^3, S^7, S^{15}$, and these are the only possibilities if the fiber $F$ and the base space $B$ are required to be spheres as well. I wonder what other spheres can be realized as non-trivial fiber bundles, if we don't put any restrictions on $F$ and $B$.

Thank you very much.

• All odd-dimensional spheres fibre over complex projective spaces. For example, see this thread: mathoverflow.net/questions/8829/… Aug 3, 2010 at 3:24
• Every sphere admits a covering map to the corresponding real projective space. Aug 3, 2010 at 6:21

I suppose you could divide this problem into suitable sub-problems.

1) Does $M$ have sub-bundles of its tangent bundle? This can be reduced to a complicated cohomology problem.

2) Are any of those sub-bundles integrable to a foliation of $M$? This is covered by the Frobenius theorem: http://en.wikipedia.org/wiki/Frobenius_theorem_%28differential_topology%29#Formulation_using_vector_fields

3) Are any of those foliations genuine bundles?

Off the top of my head I'm not sure how to approach problem (3). I believe the foliation machinery from the 70's and 80's (Haefliger and Thurston come to mind) has tools that inform on this question, though.

For example, (1) is where the story stops for $S^2$ -- if $S^2$ had a $1$-dimensional sub-bundle of its tangent bundle, its Euler class would be trivial. So $S^2$ is not a bundle over a $1$-manifold. I suppose there's more elementary proofs than that, but this is a general approach, and it works for one example. :)

• That's an old post but It got me really interested. What do you have in mind in (1)? What cohomology do you mean? Lie algebra cohomology? Dec 13, 2015 at 20:02
• No, unreduced obstruction theory, and the associated higher obstructions. This is plain old cellular cohomology with twisted coefficients. See Milnor and Stasheff's book. Dec 13, 2015 at 21:23

For manifolds fibering over $S^1$, there are certain conditions. The 2-dim case is trivial. If $dim(M)=3$, then Stallings proved that $M$ fibers over $S^1$ if and only if there is an epimorphism $\varphi:\pi_1(M)\to \mathbb{Z}$ such $ker(\varphi)$ is finitely generated. In fact, using the Thurston norm and Gabai's theory of sutured manifold hierarchies, one may algorithmically determine if a 3-manifold fibers over $S^1$.

Farrell gave algebraic conditions for fibering over $S^1$ when $dim(M)>5$.

Another low-dimensional case may be determined. If $dim(M)=3$, and one wants $M$ to fiber over a 2-dimensional manifold, then $M$ must be a Seifert-fibered space with no exceptional fibers. After possibly passing to a 2-fold cover, the group $\pi_1(M)/Z(\pi_1(M))$ should be the fundamental group of a surface (where $Z(\pi_1(M))=\mathbb{Z}$ is the fundamental group of the fiber). Again, one may check this algorithmically.

As I commented above, if you do not impose the condition that the fiber $F$ be connected, it might be difficult to answer the question. On the other hand, there is something that you can use assuming that you have a fibre bundle, but that you cannot use in the case of just a fibration: Euler characteristics are defined and the Euler characteristic of the $M$ is the product of the Euler characteristics of $F$ and $B$. Admittedly, this uses less than the full strength of a fibre bundle, but with this observation, we can resolve the case $M=S^4$! Slightly more generally, the fact that the cohomology ring has finitely many degrees, each of which is a finitely generated abelian group, might be helpful.

Suppose that $S^4$ is a non-trivial fibre bundle, with connected fiber and positive dimensional base and fiber. First of all, note that one-dimensional manifolds are circles and have vanishing Euler characteristic, so that neither the base, nor the fiber of the fibration can have dimension one. We are left with base and fiber of dimension two. Since $S^4$ is simply connected, also the base $B$ is simply connected. In particular, the Euler characteristic of the base is even (by Poincare' duality and skewness of pairing; note that the base is $S^2$, if you want to use the classification). Therefore the fiber has Euler characteristic $\pm 1$. In particular, the fiber cannot be simply connected. But then, from the long exact sequence of homotopy groups we find that either $\pi_2(S^4)$ or $\pi_1(S^4)$ are non-trivial. Since we know that this is not the case, we deduce that $S^4$ is not a fibre bundle with base and fiber of positive dimension.