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Using the chern character, it can be shown that there is no complex structure on $S^n$ if $n > 6$: See May's book: if $S^{2n}$ has a complex structure, let $\tau$ be the tangent bundle. $c_n(\tau) = \chi(S^{2n}) = 2$ must be divisible by $(n-1)!$ by Husemöller, Fibre bundles, chapter 20, Theorem 9.8. So the only case left is $S^4$ and $S^6$.

Is there a complex structure on $S^4$ or $S^6$?

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    $\begingroup$ Half of your question is a duplicate: mathoverflow.net/questions/1973/… $\endgroup$ Jan 13, 2010 at 16:27
  • $\begingroup$ What about the case of $S^2$? $\endgroup$ Mar 7, 2023 at 4:19
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    $\begingroup$ @DanielAsimov Certainly: the 2-sphere is the Riemann sphere (use $z$ for a chart away from the north pole and $1/z$ for a chart away from the south pole). Or view it as $\mathbf P^1(\mathbf C)$. That's why $S^2$ wasn't mentioned in the question. $\endgroup$
    – KConrad
    Mar 7, 2023 at 13:53
  • $\begingroup$ KConrad: Of course. But it should have been mentioned, if only to point out what you did. $\endgroup$ Mar 7, 2023 at 16:51
  • $\begingroup$ @DanielAsimov, ah, okay. I misunderstood the intention behind your "What about". $\endgroup$
    – KConrad
    Mar 8, 2023 at 14:20

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It is known that $S^4$ doesn't even have an almost complex structure. (The sphere $S^6$ does have some, but whether any of them is the underlying almost complex structure of a complex manifold is open.) The lack of almost complex structure can be proved a number of ways, one way is by showing that an almost complex, compact, four manifold with $\dim_{\mathbb{Q}}H^2(X,\mathbb{Q})=0$ has $\chi(X)=0$, but the four sphere doesn't. (It follows from the index theorem, here's a quick reference, first result: Srinivasacharyulu, "Topology of complex manifolds", Canad. Math. Bull. 1966.)

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    $\begingroup$ A clarifying comment: $S^6$ DEFINITELY has an almost complex structure, but it's not integrable. It is open if there is a complex structure, and for more details, see the other question that Reid linked to. $\endgroup$ Jan 13, 2010 at 17:11
  • $\begingroup$ Charles, the link you provided is broken. Could you perhaps replace it? It would also be helpful to give the author and title of the document. $\endgroup$
    – Danu
    Jun 25, 2017 at 10:07

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