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Apr 13, 2017 at 12:58 history edited CommunityBot
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Mar 8, 2017 at 23:52 history edited aglearner CC BY-SA 3.0
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Mar 3, 2017 at 22:37 history edited aglearner CC BY-SA 3.0
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Mar 2, 2017 at 1:17 vote accept aglearner
Mar 1, 2017 at 7:28 answer added Nikolaki timeline score: 5
Feb 26, 2017 at 21:09 comment added aglearner Great! Would you mind to put this as an answer?
Feb 26, 2017 at 14:43 comment added Nikolaki 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.
Feb 16, 2017 at 14:07 comment added YangMills yes, you are right, sorry.
Feb 16, 2017 at 10:11 comment added aglearner 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.
Feb 16, 2017 at 5:11 comment added YangMills 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.
Feb 16, 2017 at 1:32 history edited aglearner
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Feb 15, 2017 at 17:04 history asked aglearner CC BY-SA 3.0