Timeline for Closed orientable surfaces have even Euler characteristic
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2018 at 19:02 | comment | added | Gjergji Zaimi | I am curious whether one can cook up a combinatorial argument for $4k+2$-dimensional simplicial complexes. | |
| Sep 11, 2018 at 20:58 | comment | added | Qiaochu Yuan | Here is the opposite of what you asked for, the least elementary proofs I know: qchu.wordpress.com/2014/10/14/… | |
| Sep 11, 2018 at 19:37 | vote | accept | Neil Strickland | ||
| Sep 11, 2018 at 18:35 | answer | added | Gjergji Zaimi | timeline score: 15 | |
| Sep 11, 2018 at 14:12 | comment | added | mme | @FrancescoPolizzi Yes, but you need some form of Poincare duality. I suppose it would become very explicit in terms of dual triangulations in this case, which could be useful. | |
| Sep 11, 2018 at 13:33 | comment | added | Francesco Polizzi | Is $(2)$ independent on the classification (I mean, the part about the non-degenericity of the intersection form)? | |
| Sep 11, 2018 at 13:32 | comment | added | Francesco Polizzi | I do not think this is so easy. You must use orientability in some essential way, and at the and of the story this could be more or less equivalent to the classification. A possible strategy of proof is noticing that (1) since $S$ is orientable than $H^2(S, \, \mathbb{R})$ is one-dimensional, generated by the orientation class and (2) $H^1(S, \, \mathbb{R})$ carries a symplectic form given by the intersection of $1$-cycles, hence it is even-dimensional. | |
| Sep 11, 2018 at 13:08 | comment | added | Neil Strickland | @FrancescoPolizzi Yes. | |
| Sep 11, 2018 at 13:06 | comment | added | Francesco Polizzi | You want a direct proof avoiding the use of the classification, right? | |
| Sep 11, 2018 at 12:54 | history | asked | Neil Strickland | CC BY-SA 4.0 |