Timeline for Example of symplectic 4-manifolds with no Lefschetz fibration structure?
Current License: CC BY-SA 3.0
11 events
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Nov 27, 2014 at 1:19 | comment | added | Jie Min | @asdfasdfgasd I think the point here is that the fiber has self-intersection 0, so it is homologically trivial in $\mathbb{CP}^2$. | |
Nov 26, 2014 at 22:40 | comment | added | nikita | I do not understand,what do you mean by fibers being disjoint? | |
Nov 26, 2014 at 11:23 | comment | added | Alex Degtyarev | OK, then $\mathbb{C}p^2$ is an example :) | |
Nov 26, 2014 at 11:13 | comment | added | Jie Min | @AlexDegtyarev Now I get it. Then if the fiber is homologically trivial, the manifold cannot be symplectic according to Gompf's theorem. Thanks very much for explaining that to me. | |
Nov 26, 2014 at 10:47 | comment | added | Alex Degtyarev | Fibers must be disjoint, hence homologically trivial in $\mathbb{C}p^2$. On the other hand, if there is a section, it intersects fibers at $1$, preventing them from being homologous to $0$. I'm not sure that every LF has a section (in fact, their might be an obstruction), but I can hardly imagine a LF with homologically trivial fibers. Anyway, I don't know a complete proof, so I can only comment. | |
Nov 26, 2014 at 9:09 | comment | added | Jie Min | @AlexDegtyarev Sorry I don't see why $\mathbb{CP}^2$ cannot have a section. Also I am not sure if every LF has a section. Any hints or references on that? | |
Nov 26, 2014 at 6:41 | comment | added | Alex Degtyarev | I would say $\mathbb{C}p^2$. It obviously cannot have a Lefschetz fibration with a section and, on the other hand, I think any LF has one. | |
S Nov 26, 2014 at 4:24 | history | suggested | Rodrigo A. Pérez | CC BY-SA 3.0 |
grammar (number)
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Nov 26, 2014 at 3:55 | review | Suggested edits | |||
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Nov 26, 2014 at 2:30 | review | First posts | |||
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Nov 26, 2014 at 2:26 | history | asked | Jie Min | CC BY-SA 3.0 |