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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 non-zero 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?

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You probably know this but anyway such examples cannot be found if the fiber is a sphere (of any dimension). Indeed, the Gysin sequence implies that kernel of the map induced by the bundle projection on 3rd cohomology is a multiple of the Euler class. By dimension reasons the only nontrivial cases are those of S^2 and S^1 bundles. In the former case the Euler class is trivial. In the latter case any element in the kernel is the product of a 1-dimensional class and the 2-dimensional Euler class. – Igor Belegradek Mar 8 '10 at 1:05
Igor -- first, this is valid only for spherizations of vector bundles, not for general bundles with fiber sphere; second, the Euler class of an odd dimensional bundle is 2-torsion but it needn't be trivial. – algori Mar 8 '10 at 1:21
@algori: you are wrong on both counts. First of all, Petya asked the question over the rationals so 2-torsion dies, second Gysin sequence over the rationals holds for any fiber bundle whose fibers are rational homology spheres. Thus the fiber cannot be rational homology sphere. I do have one clarification: for Gysin sequence to work the fiber bundle must be orientable; this holds if e.g if the base is simply-connected or if the structure group of the bundle is path-connected. – Igor Belegradek Mar 8 '10 at 1:56
Igor -- obviously, when talking about the torsion I was referring to the integral case. Re Euler classes: if you define them via the Gysin sequence, you have to have an orientation-reversing automorphism over the identity of the base to prove that the rational Euler class of an odd-dimensional sphere bundle vanishes. This is immediate if we have a spherization of an oriented vector bundle, but how do you prove this in general? – algori Mar 8 '10 at 2:30
Igor -- by the way, why don't you post this as an answer? The fact that for any fiber bundle with fiber a product of even-dimensional spheres and complex projective spaces the Leray spectral sequence collapses (your last reference) is quite impressive. – algori Mar 8 '10 at 15:53

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 path-connected finite CW-complexes, and the fibration is homologically simple over $\mathbb Q$.

  1. A basic fact is that if the cohomological (rational) Serre spectral sequence of the bundle collapses at $E_2$, then $\pi^*: H^k(B;\mathbb Q)\to H^k(M;\mathbb Q)$ is injective. A well-known conjecture due to Halperin states that the Serre spectral sequence collapses if $F$ is rationally elliptic with positive Euler characteristic. Recall that rationally elliptic means that $F$ is simply-connected and all but finitely many homotopy groups $\pi_i(F)$ are finite. For example any homogeneous space $G/H$ where $G$, $H$ are compact Lie group is rationally elliptic, and if $G, H$ have equal rank, then $G/H$ has positive Euler characteristic. Examples include products of even-dimensional spheres or complex/quaternion projective spaces. Halperin's conjecture has been verified for $G/H$ of equal rank. Thus in the example asked by Petya the fiber cannot be $G/H$ of equal rank.

  2. A much easier argument shows that the fiber cannot be an odd-dimensional sphere. Indeed, it follows from Gysin sequence that the kernel of $\pi^*: H^3(B;\mathbb Q)\to H^3(M;\mathbb Q)$ is injective except possible when $F=S^1$ in which case the kernel is a multiple of the Euler class, i.e. it factors as the product of $1$-dimensional and $2$-dimensional classes.

  3. Suppose $H^i(B ;\mathbb Q) = 0=H^j (F;\mathbb Q)=0$ for $i < n$, $\ j< m$, and $m+n=4$, then the kernel of $\pi^*: H^3(B;\mathbb Q)\to H^3(M;\mathbb Q)$ is the image of the transgression $H^2(F;\mathbb Q)\to H^3(B;\mathbb Q)$, which can also be interpreted as the first obstruction the existence of a section of the fibration on the $3$-skeleton. Thus $F$ cannot be rationally $2$-connected (in which case $n=1$ and $m=3$).

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Thank you, very interesting. – Petya Mar 9 '10 at 11:48

The Becker--Gottlieb 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.

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This is neat. Makes part 1 of my post obsolete. – Igor Belegradek Mar 9 '10 at 16:39

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

  1. If the fiber $F$ and $M$ are not required to be manifolds and we allow $\pi:M\to B$ to be a Serre fibration, then such bundles obviously exist: take $B=S^3$, pick a base point in $B$ and take the path fibration that associates the end point to a path in $B$ that starts at the base point.

  2. If the base of a locally trivial bundle is $S^3$ and the structure group can be reduced to a Lie group, then the bundle is trivial, essentially because the sphere is made of two 3-balls with intersection $=S^2$ and $\pi_2$ of any Lie group vanishes. So if $B=S^3$ and the structure group is a Lie group, then the map of $H^3$'s with rational coefficients induced by the bundle projection is injective. I think I know how to prove this for any base $B$ which is a manifold.

  3. If one works modulo 2 rather than over the rationals, then there are plenty of examples: take e.g. the principal tautological bundle over the Grassmannian of 3-planes in $\mathbf{R}^n$ for $n$ sufficiently large.

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For $(2)$ for any manifold $B$ don't you need either $H^2(B;\mathbb{Z})$ to be zero or $G$ simply connected to get a trivial bundle. For example, take $B=S^1\times S^2$ with total space $M=S^1\times S^3$ and $\pi=\textrm{id}\times\textrm{Hopf}$. Then the required map is not injective on $H^3$'s. – Somnath Basu Mar 7 '10 at 21:34
Somnath -- yes, absolutely, there should be some assumptions; either $\pi_1(B)=e$ or $\pi_1(G)=e$ would do, I think. – algori Mar 7 '10 at 22:01

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.

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Somnath -- I think there is a potential problem when you want to take the $(n+5)$ skeleton of the pullback of $P$. $P$ does not come equipped with a preferred CW-structure. Ok, let us assume that there is a way to pick one so that $P\to K(\mathbf{Z},3)$ is a locally trivial fibration (this seems plausible but has to be proven), but then, depending on which one we pick, the skeleta of the pullback of $P$ may or may not be fibered locally trivially over $M$. – algori Mar 8 '10 at 0:35
Somnath -- unfortunately, this doesn't seem to work, see comments to the posting and in particular page 2 of the paper mentioned by Igor. – algori Mar 8 '10 at 18:19

There's an example of a smooth but infinite-dimensional 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 infinite-dimensional separable complex Hilbert space. Kuiper's theorem that $U(H)$ is contractible in the operator-norm topology has the well-known 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 3-connected by the homotopy exact sequence of the fibration, so has vanishing $H^3$ by Hurewicz.

Since $\pi_2 G =0$ for $G$ a finite-dimensional Lie group (in particular, $PU_n$), this bundle isn't the stabilisation of a finite-dimensional projective vector bundle. To find an example over $S^3$ (or more generally, one that is trivial over the 2-skeleton) 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 non-zero image in $\pi_2 Aut_0(F)\otimes \mathbb{Q}$, where $Aut_0(F)$ is the identity component of the space of self-homotopy 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 $F=\mathbb{CP}^n$.

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Tim - Gromov proved that $\textrm{Symp}(\mathbb{CP}^2,\omega_{FS})$ is homotopy equivalent to $PU(3)$. May be one can use this instead of $\textrm{Diff}(\mathbb{CP}^2)$? – Somnath Basu Mar 8 '10 at 0:38
@Somnath. Gromov's theorem implies that every symplectic CP^2-bundle over the 3-sphere is trivial - not just smoothly, but even as a symplectic fibration... – Tim Perutz Mar 8 '10 at 0:57

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 3-dimensional bordism class, you might as well assume that $B$ is a compact oriented 3-manifold. Since the fundamental class of a 3-manifold 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 3-sphere. As Tim points out, if the Hurewicz map $\pi_3(BDiff(F))\to H_3(BDiff(F))$ is rationally non-zero you can take $B=S^3$, but since many Q homology 3-spheres are aspherical this may give you additional flexibility.

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Ignore this answer.

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Ben -- this is true, but all non-zero cohomology classes of the projective space are even-dimensional. – algori Mar 7 '10 at 21:44

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