7

1

This is the question. Is it known that a symplectic $4$-fold with $b_2>1$ should have a homology class $C$ with $C^2<0$?

flag
I don't know, but here are some (known) observations: $\chi+ \sigma$ has to be divisible by 4 since symplectic=>almost complex, so a simply connected (or finite $H_1$) example would have b+ odd. For b+=3 and π1=0, the smallest known is probably b−=4 or 5. For some fundamental groups $\pi$ (eg $Z^n$ for $n>1$), you can use the injection $H^2(\pi)\to H^2(X)$ to show that $b_->0$. – Paul Dec 30 2010 at 7:29
Paul, thanks for your comment! – Dmitri Dec 30 2010 at 13:22

1 Answer

6

Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4-manifolds in this region, besides blow-ups of ruled surfaces, though by the Bogomolov-Miyaoka-Yau inequality and the Kodaira-Enriques classification, there are no complex surfaces.

I can't answer the question but I'll point out that a symplectic 4-manifold with $b_-=0$ and $b_+>1$ necessarily lies in this unknown region - in particular, it's not Kaehler.

To see this, rewrite $c_1^2-4c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)-4\chi$. For a symplectic manifold with $b_-=0$, this quantity equals $4b_1+b_2-4$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positive-definite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not.

link|flag
Tim, thanks for the answer!! So this question is open modulo this folkloric conjecture that for symplectic 4-folds $c_1^2\le 3c_2$. By the way, do you believe this conjecture? :) – Dmitri Jan 1 2011 at 11:31
1 
 Happy New Year, Dima! It's not necessarily an open question - someone might be able to give a clear "no" as the answer. About symplectic BMY: first, the conjecture that $c_1^2\leq 4c_2$ is a variant of a conjecture of Fintushel-Stern, I think (if you hope for 3 as constant, you have to take care of ruled surfaces). I know one plausibility argument, but the world of symplectic 4-manifolds has frequently been found to be larger than expected... – Tim Perutz Jan 1 2011 at 17:14
Tim, happy New Year for you too! I am really curious to know this plausibility argument. I guess this conjecture should be formulated only for symplectic 4-folds of non-negative Kodaira dimension, because indeed for $S=\mathbb CP^1\times \Sigma_g$ we have $c_1^2=-4(2g-2)$, while $c_2=-2(2g-2)$, so $c_1^2>4c_2$ if $g>1$... – Dmitri Jan 2 2011 at 18:29
Ah yes, I was in a muddle about what happens in the ruled case. As you say, it should really question about what happens when $K\cdot \omega \geq 0$. But I'll continue by email. – Tim Perutz Jan 2 2011 at 19:11

Your Answer

Get an OpenID
or

Not the answer you're looking for? Browse other questions tagged or ask your own question.