TITLE: How to tell whether a compact manifold can be realized as a nontrivial fiber bundle?
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: 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.