Each vector bundle is an example of a fibre bundle with some extra structure. This extra structure provides the algebraic object consisting of all sections (continuos or smooth) of given bundle. When our bundle is not necessary vector bundle but only (locally trivial) bundle the situation is somehow purely geometric. In particular each fibre bundle is a topological space (or a smooth manifold when we are in smooth situation) and the natural question arises:

Question Given a manifold $M$ how to recognize whether $M$ is homeomorphic (or diffeomorphic) to some fiber bundle?


1 Answer 1


I assume that you mean a bundle where both fiber and the base have positive dimension and the manifold is closed. Then:

  1. In dimension 2 a manifold $M$ fibers iff $\chi(M)=0$.

  2. In dimension 3 life is much more complicated. For aspherical manifolds fibering with circle or surface fiber (allowing for Seifert fibrations which are generalizations of the ordinary fibrations), one criterion is that $\pi_1(M)$ admits an infinite index normal finitely-generated nontrivial subgroup $N$. To get true fibrations you have to further insist on $\pi_1(M)/N$ torsion-free. For non-aspherical manifolds, I think, the condition is that the manifold has genus 0 or 1.

  3. In higher dimensions there is hardly anything reasonably general, but, Farrell in his thesis (see his paper with Hsiang below) proved have some criteria for manifolds to fiber over the circle. There is also a complete classification of Seifert-fibered simply-connected 5-dimensional manifolds. See

T. Farrell, W. C. Hsiang, Manifolds with $\pi_1=G\times_\alpha T$. Amer. J. Math. 95 (1973), 813–848.

  • $\begingroup$ Do you have a reference for the result of Farrell? $\endgroup$ Commented Oct 26, 2016 at 13:48
  • $\begingroup$ @MichaelAlbanese: I added a reference. $\endgroup$ Commented Oct 27, 2016 at 1:10

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