Timeline for Smooth projective varieties with infinite abelian fundamental group and finite $\pi_2$
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 28, 2013 at 18:09 | vote | accept | Uiterloo | ||
| Jul 27, 2013 at 17:22 | answer | added | Danny Ruberman | timeline score: 8 | |
| Jul 27, 2013 at 14:58 | comment | added | Misha | @ulrich: Sure, although quite a bit was known prior to that paper. | |
| Jul 27, 2013 at 12:49 | comment | added | naf | @Misha: Thanks for the reference. I remembered later that the Shafarevich conjecture is now known for all linear groups (by Eyssidieux, Katzarkov, Pantev and Ramachandran). | |
| Jul 27, 2013 at 12:46 | answer | added | naf | timeline score: 11 | |
| Jul 27, 2013 at 11:43 | comment | added | Jason Starr | My comments and posted answer were wrong! | |
| Jul 27, 2013 at 11:39 | comment | added | Misha | @ulrich: Shafarevich conjecture is true for $X$ with nilpotent fundamental group (Katzarkov, 1997). | |
| Jul 27, 2013 at 11:04 | comment | added | naf | It seems likely that there is no such surface. If the universal cover is Stein (as claimed by user37314), then it would have to be contractible (since $\pi_1$ and $\pi_2$ are tivial). So $\pi(X)$ determines the cohomology of $X$ which must then be that of an abelian surface. This easily implies that $X$ must itself be an abelian surface since the map to the Albanese is forced to be of degree 1. | |
| Jul 27, 2013 at 10:44 | comment | added | naf | By a theorem of Gurjar, see "Two remarks on the topology of projective surfaces." Math. Ann. 328 (2004), no. 4, 701–706, $\pi_2$ is always torson free if Shafarevich's conjecture is true; this is now known in many cases. So if $\pi_2$ is finite, it should be trivial. | |
| Jul 26, 2013 at 18:35 | answer | added | Mohan Ramachandran | timeline score: 5 | |
| Jul 26, 2013 at 14:53 | comment | added | Jason Starr | There is a Hurewicz homomorphism from $\pi_2$ to $H_2$. For every projective complex manifold of dimension $>0$, $H_2$ is infinite: intersection against the first Chern class of an ample divisor defines a nonzero homomorphism from $H_2$ to $\mathbb{Z}$. | |
| Jul 25, 2013 at 16:57 | history | asked | Uiterloo | CC BY-SA 3.0 |