# Two embedded symplectic spheres with zero square in a symplectic $4$-manifold

I am aware that the following result is a classical one (by now). But I am not able to understand who proved it. What should be a proper reference to this statement?

Theorem. Let $M^4$ be a compact symplectic manifold with $\pi_1\ne 0$ and let $S_1$ and $S_2$ be two symplectic spheres embedded in it with $S_1^2=S_2^2=0$. Then $S_1$ and $S_2$ are symplectically isotopic in $M^4$.

In other words, is there an article/book (say pre 2000) claiming that any two symplectic spheres with zero self-intersection in an irrational ruled (non-minimal) surface are symplectically isotopic?

• By a result of McDuff (The structure of rational and ruled symplectic 4-manifolds), $M$ is a blow-up of either $\mathbb{C}P^2$ or of a ruled manifold (i.e. the total space of an $S^2$-fibration). You might find an answer to your question in her (and Lalonde's) related works, which focus on spheres and isotopies of symplectic forms on these manifolds. Oct 21, 2017 at 21:15
• Thanks Chris. I went through all the theorems and lemmas in this article but was not able to find the statement... I suspect that the statement can be in the book of McDuff and Salamon on J-holomorphic curves but was not able to get hold of it. Otherwise I know one place from 2010 where this fact is stated, but of course it should be something much earlier... Oct 21, 2017 at 21:35
• The reason you can't find the statement is that it's false. $S^2 \times S^2$, with the standard symplectic structure, has two symplectic spheres of square 0 that are not (smoothly) isotopic. Oct 21, 2017 at 22:00
• Thanks Marco, I forgot to say that $\pi_1(M)\ne 0$, it is corrected. And this is still not in the article of McDuff. Do you think you know the reference now? Oct 21, 2017 at 22:18
• Paolo Lisca and I proved something very similar on Page 29 of On Stein ﬁllings of contact torus bundles (Bull. LMS 48, 2016), within the proof of Theorem 3.5. I think that there is an argument using adjunction alone as well. (Both arguments use McDuff's theorem, as mentioned by Chris above.) Oct 21, 2017 at 22:29

It looks indeed that this question is not as classical is it sounds, so let me provide a 2010 reference to a more general statement, at least to show that there is a reference. This is Proposition 3.2 in the following paper:

https://arxiv.org/abs/1012.4146

Partial attempt of answer if $M$ is minimal. Since $S_1$ is an embedded sphere, there exists a diffeomorphism $e_i:M(J,S_i)\times_GS^2\rightarrow M, i=1,2$ (see 1) induced by the evaluation map.

Since $\pi_1(M)$ is not trivial, the base of the fibration is a surface of genus $>0$. The Serre exact sequence of the fibration shows that $\pi_2(M)$ is generated by $\pi_2(S_1)=\pi_1(S_2)$ and the Hurewicz morphism implies that the classes $[S_1]$ of $[S_2]$ are equal in $H_2(M)$. We deduce that $[S_1].[S_2]=0$ and $S_1\cap S_2$ is empty. This implies that $S_2$ coincide with the fibre of $e_1$ which contains one of its point. We can also say that since $[S_1]=[S_2]$, $S_2$ is realized by $e_1$. We deduce that there is a symplectic isotopy between $S_1$ and $S_2$.

Introduction to symplectic topology p. 185 vi) McDuff and Solomon.

• The existence of the diffeomorphism $e_i$ requires that $M$ be minimal (which is not assumed, here). Moreover, it's not true that $H_2$ of an irrational ruled surface has one generator; e.g. $H_2(\Sigma\times S^2) = \mathbb{Z}^2$, generated by a section and a fibre. (This is true for all bundles, actually.) Oct 22, 2017 at 12:11
• Minimality is not mentioned in the reference above. Oct 22, 2017 at 12:55
• I can't find the reference you are giving in McDuff and Salamon. In any case, if you blow-up $\Sigma\times S^2$, then you still have plenty of 0-spheres, but the 4-manifold is not a fibre bundle (not even topologically). (Relative) minimality is a key assumption in McDuff's theorem. Oct 22, 2017 at 14:27