Cup product on flat fiber bundles vs cup product on the corresponding Serre spectral sequence

Question

Let $F \rightarrow E \rightarrow B$ be a flat fiber bundle, $E, F, B$ closed manifolds. Consider $H^*(E, \mathbb{Q})$ and the corresponding Serre spectral sequence with isomorphism $$(*) \ \ \ H^n(E;\mathbb{Q}) \cong \bigoplus_{p+q=n}{E_\infty^{p,q}}.$$

Can there be $a \in E_\infty^{p,q}$, $b \in E_\infty^{p',q'}$ with $p+q+p'+q' = dim(E)$ and $p+p'< dim(B)$, $q + q'> dim(F)$ such that $a \cup b \neq 0$, where we take the cup product on the left hand side of $(*)$? An example would be great! (if one exists)

I know of the example of the mapping torus of a genus 3 surface elaborated here https://mathoverflow.net/a/11826/114528. This example has a cup product $$E_\infty^{0,1} \otimes E_\infty^{0,1} \rightarrow E_\infty^{1,1}.$$ Again, the cup product taken on the left hand side of $(*)$. This is kind of what I am looking for, only in the wrong degrees. (We want to end up in top degree.)

There is also a classic example with $F=B=S^2$, see for instance Example 1.17 in https://www.math.cornell.edu/~hatcher/SSAT/SSch1.pdf. Here we have $$E_\infty^{0,2} \otimes E_\infty^{0,2} \rightarrow E_\infty^{2,2}.$$These are suitable degrees. However, this is not a flat bundle.

Answer 1

Here's an example of this type of phenomenon.

Let $E$ be the trivial bundle $S^2 \times S^2 \xrightarrow{p_1} S^2$, given by projection onto the first factor, with fiber $S^2$. This is trivial, and hence flat. If $z$ is the generator of $H^2(S^2; \Bbb Q)$, then the cohomology of $E$ is generated by classes $x = p_1^*(z)$ and $y = p_2^*(z)$. We have $x^2 = y^2 = 0$ and $xy$ is a generator of $H^4(S^2 \times S^2; \Bbb Q)$.

In the Serre spectral sequence, the element $x+y$ is a perfectly good representative of the generator $z \in E_\infty^{0,2} = H^2(F; \Bbb Q)$. Since $z^2 = 0$ in $H^2(F; \Bbb Q)$, this element squares to zero in the Serre spectral sequence. However, in the cohomology of $E$ we have $(x+y)^2 = 2xy$ which is a generator of $H^4(S^2 \times S^2; \Bbb Q)$ (since we are using rational coefficients).

A common objection is that I picked the "wrong" generator of $E_\infty^{0,2}$ here, but the issue of having additive and multiplicative extension problems is precisely that there is no preferred isomorphism $$H^n(E;\mathbb{Q}) \cong \bigoplus_{p+q=n}{E_\infty^{p,q}}$$ without making extra assumptions.