Question

Are there fibrations $F_i \to X_i \to B_i$ $(i=1,2)$ with path-connected bases $B_i$ and connected fibres $F_i$ such that their corresponding Leray-Serre spectral sequences (integral coefficients) are isomorphic as abelian groups, i.e. $$E_r^{p,q}(1) \cong E_r^{p,q}(2)\;\;\;(2 \le r \le \infty,\;i,j \ge 0)$$ and such that their integral cohomology isn't isomorphic, i.e. there is $p \ge 0$ with $$H^p(X_1;\mathbb{Z}) \not\cong H^p(X_2;\mathbb{Z})$$ as abelian groups ?

Answer 1

Such an example is given by the pair of fibrations $$ K(\mathbb{Z}/2,1) \to K(\mathbb{Z},2) \to K(\mathbb{Z},2) $$ (coming from the Bockstein exact sequence) and by $$ K(\mathbb{Z}/2,1) \to K(\mathbb{Z}/2,1) \times K(\mathbb{Z},2) \to K(\mathbb{Z},2). $$ (a trivial fibration).

In both cases, the (cohomological!) Leray-Serre spectral sequence is concentrated in even total degree. However, the two spaces have differing $H^2$ with integral coefficients ($\mathbb{Z}$ versus $\mathbb{Z} \times \mathbb{Z}/2$).

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Added (by Ralph): Here are some more details for the spectral sequences. We know $K(\mathbb{Z}/2,1)=\mathbb{R}P^\infty$ and $K(\mathbb{Z},2)=\mathbb{C}P^\infty$ and $$H^p(\mathbb{R}P^\infty;\mathbb{Z})= \begin{cases} \mathbb{Z} & p=0 \newline \mathbb{Z}_2 & p > 0 \text{ even }\;\;, \newline 0 & p > 0 \text{ odd } \end{cases} \hspace{10pt} H^p(\mathbb{C}P^\infty;M)= \begin{cases} M & p> 0 \text{ even} \newline 0 & p > 0 \text{ odd } \end{cases}$$ where $\mathbb{Z}_2 := \mathbb{Z}/2$ and $M$ are trivial coefficients. Since $\mathbb{Z}_2$ has only two elements, the coefficient system in the LS spectral sequence of the first fibration is trivial and we obtain for $E_2^{p,q}(1)=H^P(\mathbb{C}P^\infty;H^q(\mathbb{R}P^\infty;\mathbb{Z}))$: $$E_2(1)=\; \begin{array}{cccccccc} \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \newline
\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots \newline 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots\newline 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \cdots\newline \end{array}$$ Now, for positional reasons, $E_2(1)=E_\infty(1)$.

As the 2nd fibration is trivial, the coefficient system in its LS spectral sequence is also trivial. Hence both spectral sequences agree (in all terms), while the cohomologies differ: $H^p(\mathbb{C}P^\infty;\mathbb{Z})$ is described above and $$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})= \begin{cases}\mathbb{Z} \oplus \mathbb{Z}_2^n & p= 2n \newline 0 & p \text{ odd.}
\end{cases}$$