# Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ?

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 ?

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I imagine so ,since the $E_|infty$ page might have an extension problem that is not uniquely solvable. –  Spice the Bird Jun 25 '12 at 19:36
I also assume that the spectral sequence doesn't determine the cohomology in general. But it would be nice to have an explicit example at hand. –  Ralph Jun 25 '12 at 19:49

## 1 Answer

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 $\mathbb{Z} \times \mathbb{Z}/2$).

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 $$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})= \begin{cases}\mathbb{Z} \oplus \mathbb{Z}_2^n & p= 2n \newline 0 & p \text{ odd.} \end{cases}$$

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Great example, Tyler. Thanks so much. For the convenience of the readers I've added some details of the spectral sequences. If you mind, just let me know or feel free to delete it yourself. –  Ralph Jun 25 '12 at 22:31
Notice that this shows more than you asked for. Not only is there an isomorphism from any given term in the first spectral sequence to the corresponding term in the other; these isomorphims can be chosen to commute with all the differentials (trivially in this case because all the differentials are zero). –  Steven Landsburg Jun 26 '12 at 5:29
... and the spectral sequences are not only isomorphic as groups but even as graded rings. –  Ralph Jun 26 '12 at 6:11