Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T21:59:47Zhttp://mathoverflow.net/feeds/question/100617http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/100617/fibrations-with-isomorphic-leray-serre-spectral-sequences-and-non-isomorphic-cohoFibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? Ralph2012-06-25T18:14:59Z2012-06-25T22:33:40Z
<p>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 ? </p>
http://mathoverflow.net/questions/100617/fibrations-with-isomorphic-leray-serre-spectral-sequences-and-non-isomorphic-coho/100628#100628Answer by Tyler Lawson for Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? Tyler Lawson2012-06-25T20:02:35Z2012-06-25T22:33:40Z<p>Such an example is given by the pair of fibrations
<code>$$
K(\mathbb{Z}/2,1) \to K(\mathbb{Z},2) \to K(\mathbb{Z},2)
$$</code>
(coming from the Bockstein exact sequence) and by
<code>$$
K(\mathbb{Z}/2,1) \to K(\mathbb{Z}/2,1) \times K(\mathbb{Z},2) \to K(\mathbb{Z},2).
$$</code>
(a trivial fibration).</p>
<p>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 $\mathbb{Z} \times \mathbb{Z}/2$).</p>
<hr>
<p><strong>Added (by Ralph):</strong> 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 $E_2^{p,q}(1)=H^P(\mathbb{C}P^\infty;H^q(\mathbb{R}P^\infty;\mathbb{Z}))$:
$$E_2(1)=\;
\begin{array}{cccccccc}
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \newline<br>
\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots \newline
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline
\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots\newline
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline
\mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \cdots\newline
\end{array}$$
Now, for positional reasons, $E_2(1)=E_\infty(1)$. </p>
<p>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
$$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})=
\begin{cases}\mathbb{Z} \oplus \mathbb{Z}_2^n & p= 2n \newline 0 & p \text{ odd.}<br>
\end{cases}$$</p>