# Example of symplectic 4-manifolds with no Lefschetz fibration structure?

I just read about Donaldson's result on existence of Lefschetz pencil structure on symplectic manifolds (Donaldson 1999). However, one has to blow up the base locus to get a Lefschetz fibration structure.

So I wonder if every symplectic manifold (especially symplectic 4-manifold) admits a Lefschetz fibration instead of just a Lefschetz pencil? If not, are there any examples?

• 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. – Alex Degtyarev Nov 26 '14 at 6:41
• @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? – Jie Min Nov 26 '14 at 9:09
• 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. – Alex Degtyarev Nov 26 '14 at 10:47
• @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. – Jie Min Nov 26 '14 at 11:13
• OK, then $\mathbb{C}p^2$ is an example :) – Alex Degtyarev Nov 26 '14 at 11:23