Timeline for Cohomology algebra of a fiber bundle
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5 events
| when toggle format | what | by | license | comment | |
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| Dec 29, 2016 at 6:32 | comment | added | user43326 | Yes, just use the multiplicativity of the Serre SS. | |
| Dec 28, 2016 at 18:57 | comment | added | Rami | @ user43326 Thank you for the comments. I am wondering whether the following statement can be true. Suppose the spectral sequence collapses at $E_n$, for some natural $n>1$, $H^*(F;\mathbb{Z}_2)=\mathbb{Z}_2[x_1,...,x_a]/I_F$ and $H^*(B;\mathbb{Z}_2)=\mathbb{Z}_2[y_1,...,y_b]/I_B$ the corresponding algebras. Then there exists an isomorphism of graded algebras $H^*(E;\mathbb{Z}_2)\cong\mathbb{Z}_2[x_1,...,x_a,y_1,...,y_b]/I_E$ for some ideal $I_E\lhd\mathbb{Z}_2[x_1,...,x_a,y_1,...,y_b]$ . | |
| Dec 28, 2016 at 9:26 | comment | added | user43326 | When the spectral sequence collapses at $E_2$, the number of the generators of the total space is bounded by the sum of generators of the two other spaces. In general, I don't believe that there is such a simple answer, but since you are supposing finiteness, if you give, for example, the dimensions of F and B, then using the numbers of generators, you can bound the dimension as a vector space of cohomology of the base and fiber, which in turn leads to a bound on the dimension of $H^*(E)$, which in turn gives a bound on the number of algebra generators. | |
| Dec 28, 2016 at 8:30 | review | First posts | |||
| Dec 28, 2016 at 8:37 | |||||
| Dec 28, 2016 at 8:26 | history | asked | Rami | CC BY-SA 3.0 |