Are there simple necessary and sufficient conditions for an (oriented) even-dimensional compact smooth manifold to fiber over an (oriented) odd-dimensional manifold (with oriented fibers)?

For example, if $M \to N$ is a fiber bundle of compact manifolds with fiber $F$, then their Euler characteristics satisfy $\chi(M) = \chi(N)\chi(F)$. But if $N$ and $F$ are odd-dimensional, $\chi(N) = \chi(F) = 0$, and so $\chi(M)$ must also vanish. Is the converse true? I.e. if $M$ has vanishing Euler class, does it fiber over an odd-dimensional manifold? Or perhaps there's a hint that $\chi(M) = 0^2$, not just $0$, and so maybe some combination of Massey products also must vanish?

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    $\begingroup$ Just take connected sum of two even-dimensional manifolds with Euler characteristics $-k, k+2$, where $k$ is positive. $\endgroup$
    – Misha
    May 5, 2013 at 21:16
  • $\begingroup$ Fibering is a highly subtle question in general as the difficulty of the virtual fibering conjecture shows. If we substitute your question by the question whether the manifold fibers up to homotopy, then slightly more might be known. You might check out arxiv.org/pdf/0901.1250.pdf and miami.uni-muenster.de/servlets/DerivateServlet/Derivate-5587/… $\endgroup$ May 6, 2013 at 0:29

2 Answers 2


Being the total space of a fiber bundle is not invariant under homotopy equivalences, so I doubt there is a criterion of the type you state (eg in terms of homological invariants).

However, one can say something a little weaker. If a manifold $M^k$ fibers over a manifold $N^{\ell}$, then the fibers form a codimension $\ell$ foliation of $M^k$. A deep theorem of Thurston says that if $\chi(M^k)=0$, then $M^k$ supports a codimension $1$ foliation (the converse is an easy exercise). By the way, to appreciate how deep this is, it implies in particular that an odd-dimensional closed oriented manifold always supports a codimension $1$ foliation. This is nontrivial to prove even in dimension $3$ (where it was first proved by Lickorish).

  • $\begingroup$ compactness of $M$ is essential in establishing the converse (left as `easy' exercise). $\endgroup$
    – JHM
    Dec 5, 2013 at 1:23
  • $\begingroup$ @J.Martel : Of course. $\endgroup$ Dec 5, 2013 at 3:44
  • $\begingroup$ I only mention it because there is another question floating around MO, on whether or not $R^3-0$ can be foliated by $2$-tori. There appears some vague (and unjustified) comments that the existence of such a foliation requires the total space (an open $3$-mfld) to have $\chi=0$. I find the question interesting and trying to work it out for myself. $\endgroup$
    – JHM
    Dec 5, 2013 at 4:07
  • $\begingroup$ It is also not unusual for the compactness assumption to be (incorrectly!) omitted in certain texts, e.g. Corollary 4.11 of Lawson's ``Quantitative theory of foliations". $\endgroup$
    – JHM
    Dec 5, 2013 at 4:11

No, I don't believe there's a simple solution. But here's an approach to the problem which indicates how it can be fractured up.

Assume $M,N$ are closed and connected. If $f\: M \to N$ is homotopic to a smooth fiber bundle with $M$ and $N$ compact, then the fibers are homotopy finite (i.e., they are homotopy equivalent to a finite complex).

Conversely, it is a result first stated by Quinn (later proved by Gottlieb, and then differently by me) that if $f\: M^m \to N^n$ is such that its homotopy fiber $F$ (at some basepoint in $N$) is homotopy finite, then $F$ is a Poincare duality space of dimension $m-n$. Thus, $f$ gives rise to a fibered surgery problem.

One can approach this problem in two steps:

Step 1: find a block bundle $E \to N$ and a fiber homotopy equivalence $E\simeq M$. This step can be attacked classical surgery techniques (here the dimension of the fiber should be $\ge 6$). What one studies here is the map $\tilde S_N(M) \to \tilde S(M)$ from the fiberwise block structure space to the block structure space.

Step 2: Study the map $S_N(M) \to \tilde S_N(M)$ from the fiberwise structure space to the fiberwise block structure space. This step involves higher algebraic $K$-theory a la Waldhausen. This step is only really understood in the "concordance stable range" which in this case requires $4n \le m$ (approximately).${}^\dagger$

The above is only meant to be an outline. I first learned about these ideas from the papers of Weiss and Williams, most notably:

Automorphisms of manifolds. Surveys on surgery theory, Vol. 2, 165–220, Ann. of Math. Stud., 149, Princeton Univ. Press, Princeton, NJ, 2001

An alternative approach which packages Step 1 and Step 2 into a single step is in the third WW paper which can be obtained from Michael Weiss' website.

More recently, see the papers of Wolfgang Steimle, especially

Obstructions to stably fibering manifolds. Geom. Topol. 16 (2012), no. 3, 1691–1724

${}^\dagger$ Added Later: According to Steimle, the "stable range" for the fibering problem is more complicated than what I wrote above. Rather than write it down, let me refer to his paper for the actual range.


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