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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
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Nov 26, 2014 at 3:55 review Suggested edits
S Nov 26, 2014 at 4:24
Nov 26, 2014 at 2:30 review First posts
Nov 26, 2014 at 3:56
Nov 26, 2014 at 2:26 history asked Jie Min CC BY-SA 3.0