Timeline for Even, non liftable Stiefel-Whitney class
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2021 at 16:23 | vote | accept | Georges Elencwajg | ||
| Apr 9, 2021 at 16:22 | comment | added | Georges Elencwajg | Thanks once more, Mark. Actually I knew that there even exist 2-dimensional complex projective smooth algebraic surfaces (thus compact 4-dimensional orientable real manifolds) with fundamental group any prescribed finite group. This was proved by Serre: here is an interesting survey by our friend Arapura. However I wanted to initiate myself into the mysteries of the algebraic topology techniques... Anyway, you have been very helpful and it is with pleasure that I "accept" your answer. | |
| Apr 9, 2021 at 14:55 | comment | added | Mark Grant | That doesn't preceisely answer your question. I think that Teichner describes both the approaches of HJRW and of Somnath Basu at the linked question for the case $\mathbb{Z}/4$, claiming that they give equivalent manifolds. Thus the resulting manifold should be orientable. | |
| Apr 9, 2021 at 14:29 | comment | added | Mark Grant | Any finitely presented group can be realized as the fundamental group of an orientable closed $4$-manifold, see HW's answer here; mathoverflow.net/questions/15411/…. (Since $\mathbb{Z}/4$ has an index $2$ subgroup, it can also be realized as the fundamental group of a non-orientable $4$-manifold.) | |
| Apr 9, 2021 at 14:07 | comment | added | Georges Elencwajg | One last question, Mark. Teichner gives an example of a 4-dimensional manifold satisfying the hypothesis of his Lemma 2. Is that example orientable? Else, how might one find an example that is orientable? | |
| Apr 9, 2021 at 13:43 | comment | added | Georges Elencwajg | Thanks for your quick confirmation, Mark. | |
| Apr 9, 2021 at 13:37 | comment | added | Mark Grant | @GeorgesElencwajg: Yes, that's entirely correct. | |
| Apr 9, 2021 at 13:33 | comment | added | Georges Elencwajg | Dear Mark, thank you very much for your reference and explanation. If $M$ is orientable we have $w_1(E)=w_1(M)=0$ . I suppose that $H^2(M;\mathbb{Z}^{w_1(E)})$ then means $H^2(M;\mathbb{Z})$. Could you please confirm that it is indeed so? | |
| Apr 9, 2021 at 12:21 | history | answered | Mark Grant | CC BY-SA 4.0 |