# Almost complex structures on a 4-ball that are not tamed

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 almost complex structure $J$ on a closed ball $B^4$, such that any $C^{\infty}$-small perturbation of $J$ is not tamed by any symplectic form? (I assume that $J$ behaves nicely on the boundary of $B^4$, in particular it is smooth there).

Remark. Clearly, if such $J$ exists on $B^4$, it would exists on any $4$-manifold admitting an almost complex structure.

This question is a follow-up to the following one, where a global obstruction for "tamebility" was found for $\mathbb CP^2$ Almost complex structures on $\mathbb CP^2$ that are not tamed

Consider an almost complex structure $J$ on $B^4$ for which the contact hyperplanes of the overtwisted contact structure on $S^3=\partial B^4$ (in the same homotopy class as the standard tight contact structure) become $J$-complex. If there was a taming symplectic form, we would have constructed a so-called weak symplectic filling of an overtwisted contact manifold. However, This is not possible by a result of Eliashberg and Gromov.
Finally, observe that the contact condition is open, and by Gray's stability also a perturbation of $J$ would be tangent to $\partial B^4$ along some tangent hyperplane distribution being an overtwisted contact structure.
• What are you using to construct such a $J$? Certainly if such a $J$ exists it satisfies the conditions of the question, but I don't see where the construction is coming from. – PVAL Mar 8 '17 at 23:22
• Since I am assuming that the contact structure is homotopic to the standard contact structure on $S^3$, the almost complex structure can be taken to be homotopic to the standard integrable complex structure on the ball (we just need to deform the complex tangencies along the boundary). – Nikolaki Mar 8 '17 at 23:26
• In the definition of weak filling one asks that $\omega$ is convex at the boundary. How do you see that any $\omega$ taming $J$ has to be convex at $S^3$? – aglearner Mar 8 '17 at 23:29
• The contact condition means that the exterior differential of any contact form (which in this case is equal to the Levi form of $J$) is nondegenerate on the contact planes. – Nikolaki Mar 8 '17 at 23:31