Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T21:59:47Z http://mathoverflow.net/feeds/question/100617 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100617/fibrations-with-isomorphic-leray-serre-spectral-sequences-and-non-isomorphic-coho Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? Ralph 2012-06-25T18:14:59Z 2012-06-25T22:33:40Z <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#100628 Answer by Tyler Lawson for Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? Tyler Lawson 2012-06-25T20:02:35Z 2012-06-25T22:33:40Z <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 $$H^p(\mathbb{R}P^\infty;\mathbb{Z})= \begin{cases} \mathbb{Z} &amp; p=0 \newline \mathbb{Z}_2 &amp; p > 0 \text{ even }\;\;, \newline 0 &amp; p > 0 \text{ odd } \end{cases} \hspace{10pt} H^p(\mathbb{C}P^\infty;M)= \begin{cases} M &amp; p> 0 \text{ even} \newline 0 &amp; p > 0 \text{ odd } \end{cases}$$ 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 $E_2^{p,q}(1)=H^P(\mathbb{C}P^\infty;H^q(\mathbb{R}P^\infty;\mathbb{Z}))$: $$E_2(1)=\; \begin{array}{cccccccc} \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \newline<br> \mathbb{Z}_2 &amp; 0 &amp; \mathbb{Z}_2 &amp; 0 &amp; \mathbb{Z}_2 &amp; 0 &amp; \mathbb{Z}_2 &amp; \cdots \newline 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots\newline \mathbb{Z}_2 &amp; 0 &amp; \mathbb{Z}_2 &amp; 0 &amp; \mathbb{Z}_2 &amp; 0 &amp; \mathbb{Z}_2 &amp; \cdots\newline 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots\newline \mathbb{Z} &amp; 0 &amp; \mathbb{Z} &amp; 0 &amp; \mathbb{Z} &amp; 0 &amp; \mathbb{Z} &amp; \cdots\newline \end{array}$$ 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 $$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})= \begin{cases}\mathbb{Z} \oplus \mathbb{Z}_2^n &amp; p= 2n \newline 0 &amp; p \text{ odd.}<br> \end{cases}$$</p>