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Recall that an almost complex structure $J$ on a manifold $M^{2n}$ is called tamed if there exists a symplectic form $\omega$ on $M^{2n}$ such that $\omega(v,Jv)>0$ for any non-zero tangent vector $v$.

Question. Is there an example of an almost complex structure on $\mathbb CP^2$ such that any $C^{\infty}$ small perturbation of $J$ is not tamed?

Added. It turns out that there exists as well a purely local obstruction for any small perturbations of $J$ to be tamed. The precise statement and the answer is here: Almost complex structures on a 4-ball that are not tamed

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  • $\begingroup$ Start with a $J$ which admits a null-homologous $J$-holomorphic sphere $C$. This can always be achieved by deforming a given almost complex structure in a small neighborhood of a point. Then $J$ cannot be tamed, and furthermore $C$ should persist as a null-homologous $J'$-holomorphic sphere (i.e. it should deform) for any small perturbation $J'$ of $J$. This is because the linearized operator is surjective since it satisfies the numerical inequality $c_1(\nu)=0\geq -1$ ($\nu$ normal bundle of $C$) needed to apply $2.1.C_1$ in Gromov's paper, see also Hofer-Lizan-Sikorav. $\endgroup$
    – YangMills
    Commented Feb 16, 2017 at 5:11
  • $\begingroup$ I think that $c_1(v)=-2$, so this argument does not work. Indeed, since $C$ is null-homologous, $c_1(\mathbb CP^2)$ restricts to $C$ as zero, on the other hand $c_1(TC)=2$, since $C$ is a sphere. $\endgroup$
    – aglearner
    Commented Feb 16, 2017 at 10:11
  • $\begingroup$ yes, you are right, sorry. $\endgroup$
    – YangMills
    Commented Feb 16, 2017 at 14:07
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    $\begingroup$ The strategy of YangMills is still a good one. But instead, you should take a J such that a smoothly_knotted sphere of degree one becomes J-holomorphic. Tameness would imply that the standard CP^2 has a smoothly knotted symplectic embedding of degree one, which is in contradiction with a result by Gromov. $\endgroup$
    – Nikolaki
    Commented Feb 26, 2017 at 14:43
  • $\begingroup$ Great! Would you mind to put this as an answer? $\endgroup$
    – aglearner
    Commented Feb 26, 2017 at 21:09

1 Answer 1

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Summarising the discussion above and Daniel Ruberman's helpful clarifications below.

Any symplectic structure on $\mathbb{C}P^2$ is standard by a result due to Gromov and Taubes. By Siebert-Tian every symplectic surface in $\mathbb{C}P^2$ of degree at most 17 is smoothly isotopic to an algebraic surface. In particular, there is a unique smooth isotopy class of such surfaces.

Now, take $S \subset \mathbb{C}P^2$ a surface of low degree which is not ambient diffeomorphic to an algebraic surface, but which satisfies the adjunction formula ($g=(d-1)(d-2)/2$). Any almost complex structure for which $S$ is pseudoholomorphic can then not be tamed. (To see that such an almost complex structure exists in the right homotopy class, we can find a smooth homtopy of $\mathbb{C}P^2$ from the identity to a smooth map (not a diffeomorphisms obviously!) which sends $S$ to an algebraic surface.)

Finally, by the automatic transversality of Hofer-Lizan-Sikorav, for a small perturbation of the almost complex structure we can find a pseudoholomorphic surface isotopic to $S$; these almost complex structures hence do not admit taming forms either. (In order to apply the automatic transversality, we must use the assumptions that $c_1([S]) \ge 1$ and that $S$ is immersed.)

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  • $\begingroup$ But a sphere S whose self-intersection is $+1$ has simply connected complement! If you take a pushoff $S'$ that hits S in one point (as you can) then $S'$ minus a meridional disk is another disk that provides a null-homotopy for the meridian. A well-known conjecture in 4-manifold topology (see Kirby's problem list, Problem 4.23) states that any 2-sphere in the standard homology class in $\mathbb{C}P^2$ is isotopic to $\mathbb{C}P^1$. $\endgroup$
    – Danny Ruberman
    Commented Mar 1, 2017 at 12:23
  • $\begingroup$ Oh, you're right of course. I wanted to say something more precise than just "knotted", but I obviously screwed up. $\endgroup$
    – Nikolaki
    Commented Mar 1, 2017 at 14:49
  • $\begingroup$ I should point out that having a simply connected complement implies that the sphere S is topologically isotopic to $\mathbb{C}P^1$. The conjecture I cited is really asking about the smooth situation. Also, there are knotted surfaces in $\mathbb{C}P^2$ of higher degrees, so perhaps you could repeat the argument starting with one of these. $\endgroup$
    – Danny Ruberman
    Commented Mar 1, 2017 at 14:57
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    $\begingroup$ In that situation the Chern class vanishes on the surface, so the automatic transversality theorem cannot be applied. (We need an immersed surface, and we need its normal bundle times the anticanonical bundle to be positive.) $\endgroup$
    – Nikolaki
    Commented Mar 2, 2017 at 1:11
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    $\begingroup$ I'm not aware of any references. If one could construct knotted surfaces which are diffeomorphic but not isotopic to algebraic ones, then it would be possible to use a similar method to show nonconnectedness of the subset. Does someone know if such examples exist? (I would guess that we don't know the answer.) $\endgroup$
    – Nikolaki
    Commented Mar 2, 2017 at 18:33

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