A friend of mine is interested in examples of the following situation: an oriented smooth fiber bundle $\pi \colon M \to B$ with $M$ and $B$ compact and a nonzero class $a \in H^3(B; \mathbb{Q})$ such that $\pi^* a=0$ in $H^3(M; \mathbb{Q})$. It is easy to construct such an example if the class $a$ is a product of a degree $1$ class and degree $2$ class; are there examples not of this kind?

This is a update of my comments above. As I am unable to construct a requested example, I go in the opposite direction and describe what definitely cannot be an example. Standing assumptions are that $\pi:M\to B$ is a fibration with homotopy fiber $F$, the spaces $F$, $B$ are pathconnected finite CWcomplexes, and the fibration is homologically simple over $\mathbb Q$.



The BeckerGottlieb transfer implies that $\pi^*$ is rationally a (split) monomorphism unless the Euler characteristic of the fibre is zero. Thus any proposed example must have this property. 


All I was able to came up with is the following.



Here's my two cents although it's rather sketchy. For any CW complex $X$, $H^3(X;\mathbb{Z})=[X,K(\mathbb{Z},3)]$, where $K(\mathbb{Z},3)$ comes equipped with a fibration $\mathbb{CP}^\infty\to P\to K(\mathbb{Z},3)$. The total space $P$ is contractible. Now suppose $X$ is a compact manifold of dimension $n$ which is $2$connected and $H^3(X;\mathbb{Z})=\mathbb{Z}$. Then choosing a generator of $H^3(X;\mathbb{Z})$ corresponds to a (homotopy class of) map $f:X\to K(\mathbb{Z},3)$. The pullback bundle $f^\ast P\to X$ has the property that $H^3(f^\ast P;\mathbb{Z})=0$. Since we need a finite dimensional manifold which $f^\ast P$ isn't, let $E$ denote the $(n+5)$skeleta of $f^\ast P$. It is compact and locally looks like $X\times\mathbb{CP}^2$. I think(?) that $\pi:E\to X$ is a fibre bundle. Since $\pi_3$ is unchanged for $4$skeleta or higher, it follows that $0=\pi_3(E)=\pi_3(f^\ast P)$, whence $H^3(E;\mathbb{Z})=0$. Feel free to tweak the answer if need be. Edit As pointed out by algori and Igor, the second paragraph doesn't give you a fibre bundle. 


There's an example of a smooth but infinitedimensional fibre bundle $M \to S^3$ with $H^3(M)=0$. It involves some ideas that algori and Somnath Basu have already noted. The fibre is $\mathbb{P}(H)$, projective infinitedimensional separable complex Hilbert space. Kuiper's theorem that $U(H)$ is contractible in the operatornorm topology has the wellknown consequence that $PU(H)$ is a $K(\mathbb{Z},2)$. Take as clutching function for such a bundle $M\to S^3$ a smooth map $S^2\to PU(H)$ representing a generator of $\pi_2 PU(H) \cong \mathbb{Z}$. Then $M$ is 3connected by the homotopy exact sequence of the fibration, so has vanishing $H^3$ by Hurewicz. Since $\pi_2 G =0$ for $G$ a finitedimensional Lie group (in particular, $PU_n$), this bundle isn't the stabilisation of a finitedimensional projective vector bundle. To find an example over $S^3$ (or more generally, one that is trivial over the 2skeleton) with compact smooth fibre $F$, you'll need $\pi_2 Diff(F)\otimes \mathbb{Q} \neq 0$. One can't take $F$ to a surface; I wonder if anything is known about $\pi_2 Diff(\mathbb{CP}^n)$ for $n>1$. Added: More precisely, one needs $\pi_2 Diff_0(F)\otimes \mathbb{Q}$ to have nonzero image in $\pi_2 Aut_0(F)\otimes \mathbb{Q}$, where $Aut_0(F)$ is the identity component of the space of selfhomotopy equivalences. As shown in a paper noted by Igor Belogradek in his comments above, "Rational type of classifying spaces for fibrations" by Samuel B. Smith, this fails when $F=\mathbb{CP}^n$. 


Here's another way to look at it, essentially a variation of Tim's comment: a smooth fiber bundle $M$ over $B$ with fiber $F$ is defined by a map $B\to BDiff(F)$. A class $a\in H^3(BDiff(F);Q)$ will always pull back to zero in $H^3(M;Q)$. Since any class in $H_3(,Q)$ of any space is, after taking multiples if necessary, represented by a 3dimensional bordism class, you might as well assume that $B$ is a compact oriented 3manifold. Since the fundamental class of a 3manifold is not a product of 1 and 2 diml classes iff $B$ is a rational homology sphere, this tells you what you are looking for is a manifold $F$ and a class in $H_3(BDiff(F);Q)$ represented by a map from a Q homology 3sphere. As Tim points out, if the Hurewicz map $\pi_3(BDiff(F))\to H_3(BDiff(F))$ is rationally nonzero you can take $B=S^3$, but since many Q homology 3spheres are aspherical this may give you additional flexibility. 


Ignore this answer. 

