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The Newlander-Nirenberg theorem states that any Integrable Almost Complex manifold is a complex manifold. I am looking for natural examples of complex structures that are not integrable.

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  • $\begingroup$ I changed the title by adding "almost", because nonintegrable complex structure is an oxymoron. Also, I added some relevant tags. $\endgroup$ – t.b. Jun 23 '11 at 3:25
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The sphere $S^6$ naturally sits inside of the imaginary octonians $\operatorname{Im}\mathbb{O}$. At the point $p\in S^6$, multiplication by $p$ on $ T_p S^6 = p^\bot \subseteq \operatorname{Im}\mathbb{O}$ defines an almost complex structure.

This almost complex structure is not integrable, due to the nonassociativity of the octonians.

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    $\begingroup$ Isn't it much more common to use "octonions" instead of "octonians" (Google gives 356K vs 28.3K hits)? Also, it might be worth pointing out that the existence of an integrable almost complex structure on $S^6$ is a famous open problem. $\endgroup$ – t.b. Jun 23 '11 at 2:58
  • $\begingroup$ @Theo: Thanks for the edits. I agree with your spelling of "octonions" - I consider my current version a typo. (I won't edit to avoid needlessly bumping this). $\endgroup$ – Jason DeVito Jun 23 '11 at 3:56
  • $\begingroup$ @JasonDeVito: Do you have a reference for your last statement? $\endgroup$ – Michael Albanese Jan 20 '15 at 21:19
  • $\begingroup$ @Michael: Actually I don't. The "almost" part is pretty easy to prove. The fact that is is not integrable follows from the fact that, as t.b. said, "the existence of an integrable almost complex structure on $S^6$ is a famous open problem." According to this MO link: mathoverflow.net/questions/1973/…, LeBrun showed that there is no orthogonal complex structure on $S^6$, but the above almost complex structure is orthogonal. $\endgroup$ – Jason DeVito Jan 21 '15 at 14:17
  • $\begingroup$ @JasonDeVito: I know that the almost complex structure you describe is not integrable, but I am interested in the statement that the nonintegrability is due to the nonassociativity of the octonions. A similar claim was made by the user wisefool here. $\endgroup$ – Michael Albanese Jan 21 '15 at 19:32

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