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> 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)?
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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?



When does an even-dimensional manifold fiber over an odd-dimensional manifold?
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).


