Is there an exotic $\mathbb{R}^4$ admitting an integrable almost complex structure?
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8$\begingroup$ What's more: Each small exotic $R^4$ is an open subset of ${\mathbb C}^2$, hence, has the induced complex structure from this embedding. $\endgroup$– Moishe KohanCommented Dec 30, 2020 at 18:45
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$\begingroup$ @MoisheKohan But not large exotic $\mathbb{R}^4$, that can't be embedded in $\mathbb{R}^4$ ? $\endgroup$– bonifCommented Apr 1 at 17:28
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$\begingroup$ @bonif: No, but there is an immersion in $C^2$. $\endgroup$– Moishe KohanCommented Apr 1 at 18:03
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$\begingroup$ @MoisheKohan Thanks. And can they be complexified to $\mathbb{C^4}$? $\endgroup$– bonifCommented Apr 1 at 18:19
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$\begingroup$ @bonif, no, you just pull back the complex structure from $\mathbb C^2$. $\endgroup$– Moishe KohanCommented Apr 1 at 18:55
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1 Answer
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It is a result of Gromov that an open manifold of dimension six or less admits a complex structure if and only if it admits an almost complex structure; see the corollary on page 103 of his book Partial Differential Relations. As $\mathbb{R}^4$ is contractible, every exotic $\mathbb{R}^4$ is parallelisable. Therefore, they all admit almost complex structures, and hence complex structures by Gromov's result.
answered Dec 30, 2020 at 18:27
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10$\begingroup$ An alternative argument is to use the earlier result (Hirsch-Smale) that each open parallelizable $n$-dimensional manifold admits an immersion in $R^n$. $\endgroup$ Commented Dec 30, 2020 at 18:47