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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?

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    $\begingroup$ 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. $\endgroup$
    – Chris Gerig
    Commented Oct 21, 2017 at 21:15
  • $\begingroup$ 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... $\endgroup$
    – aglearner
    Commented Oct 21, 2017 at 21:35
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    $\begingroup$ 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. $\endgroup$
    – Marco Golla
    Commented Oct 21, 2017 at 22:00
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    $\begingroup$ 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? $\endgroup$
    – aglearner
    Commented Oct 21, 2017 at 22:18
  • $\begingroup$ 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.) $\endgroup$
    – Marco Golla
    Commented Oct 21, 2017 at 22:29

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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

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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.

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  • $\begingroup$ 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.) $\endgroup$
    – Marco Golla
    Commented Oct 22, 2017 at 12:11
  • $\begingroup$ Minimality is not mentioned in the reference above. $\endgroup$
    – Tsemo Aristide
    Commented Oct 22, 2017 at 12:55
  • $\begingroup$ 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. $\endgroup$
    – Marco Golla
    Commented Oct 22, 2017 at 14:27

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