Timeline for Even, non liftable Stiefel-Whitney class
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2021 at 11:25 | comment | added | Georges Elencwajg | Thank you, Bertram. | |
| Apr 8, 2021 at 11:07 | comment | added | Bertram Arnold | The universal coefficient theorem gives $H^2(M,\mathbb Z) \cong \operatorname{Hom}(\mathbb Z/4,\mathbb Z) = 0$ (the torsion shows up in $H^3(M;\mathbb Z)\cong \mathbb Z/4$, generated by the Bockstein of the lift of the Stiefel-Whitney class to $\mathbb Z/4$-cohomology). | |
| Apr 8, 2021 at 10:56 | comment | added | Bertram Arnold | The natural example is a homotopy type/CW complex. One can use standard techniques to reduce it first to a finite CW complex and then find a (open) manifold with the same homotopy type. I think the minimal example is $4$-dimensional and has integral homology $\mathbb Z,0,\mathbb Z/4,0,\dots$. The bundle $E$ is $3$-dimensional (to get an even-dimensional bundle, just add a trivial line bundle), its odd Stiefel-Whitney classes vanish as requested, and $w_2(E)$ is a generator of the second cohomology with $\mathbb Z/2$-coefficients. | |
| Apr 8, 2021 at 10:40 | comment | added | Georges Elencwajg | Dear Bertram, thank you very much for your answer. Since I am not familiar with the topological concepts you use [$ K(\mathbb Z/4,2), K(\mathbb Z/2,2), BSO(3)$, $Sq^i$], could you please sum up in a conclusion the properties of the resulting manifold $M$ : dimension, cohomology groups with coefficients in $\mathbb Z$ and $\mathbb Z/2$, as well as the rank of $E$ and its Stiefel- Whitney classes? I'll try to educate myself afterwards in the powerful techniques of algebraic topology you use in order to understand your arguments in detail. | |
| Apr 8, 2021 at 10:00 | history | answered | Bertram Arnold | CC BY-SA 4.0 |