MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 '10 at 7:29
Paul, thanks for your comment! – Dmitri Dec 30 '10 at 13:22

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.

share|cite|improve this answer
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 '11 at 11:31
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 '11 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 '11 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 '11 at 19:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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