Timeline for Are compact, complex, affinely flat manifolds geodesically complete?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2021 at 12:19 | comment | added | R. Alexandre | In complex dimension 2, the linear holonomy has only "complex" discompacity 1. Carrière's proof no longer applies because it uses the fact that a codimension 1 real subspace separates the full affine space in two (no longer true in a complex setting, where a complex hyperplane does not separate). How ever, the classification of affine complex surfaces applies and can be used to show that such surfaces are indeed complete | |
| Jan 29, 2020 at 21:47 | vote | accept | Mike Cocos | ||
| Sep 15, 2015 at 0:47 | comment | added | Mike Cocos | Thank you Misha. I will revise the question. I have to find out how first. I am very new to this but I love it already. | |
| Sep 15, 2015 at 0:33 | comment | added | Misha | I am not assuming any symplectic information. (Did I even mention a symplectic structure in my answer?) All I am assuming is that the linear holonomy is in $GL(n,C)\cap SL(2n,R)$, where $n$ is the complex dimension of your manifold. The $SL(2n,R)$ comes from your assumption on the parallel volume form; the $GL(n,C)$ assumption is your parallel almost complex structure. By the way, you should revise your question to reflect the question you are now asking. | |
| Sep 14, 2015 at 22:17 | comment | added | Mike Cocos | You are not assuming in any way that the symplectic structure is compatible with the almost complex structure of the manifold? Just parallel with respect to the flat connection. Otherwise it would be a metric connection and trivially true. | |
| Sep 13, 2015 at 18:15 | history | answered | Misha | CC BY-SA 3.0 |