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complex structure on S^n

The two sphere $S^2$ is a real manifold of dimension $2$, while the three sphere $S^3$ is a real manifold of dimension $3$. Now $S^2$ is a complex manifold, while $S^3$ being odd dimensional is not. Is it true that all spheres of the form $S^{2N}$ are complex manifolds?

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marked as duplicate by Kevin H. Lin, Qiaochu Yuan, S. Carnahan Jul 1 '10 at 15:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

That's VERY much false. This question is looking for a reference to the fact that $S^2$ and $S^6$ are the only ones with even almost complex structures, and it's open if $S^6$ admits a complex structure (the almost complex structure known is not one)

Edit: Other questions on complex structures on spheres are here and here.

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It's a nice exercise in characteristic classes to show that S^4k for all k are NOT complex manifolds.

EDIT: I will answer Charlie's comment here and provide a sketch of the proof.

Let $\omega=TS^{4k}$ be the tangent space to the $4k$-sphere. If $S^{4k}$ was actually a complex manifold then $\omega$ would be a complex vector bundle. In this case the complexification of the underlying real vector bundle $\omega_{\mathbb{R}}$ would be canonically isomorphic to the Whitney sum $\omega\oplus \bar{\omega}$ (Milnor&Stasheff page 176). Now by corollary 15.5 in Milnor&Stasheff $$p_k(\omega_{\mathbb{R}})=c_k^2(\omega)-2c_{k-1}c_{k+1}(\omega)+\cdots\mp 2c_{2k}(\omega)$$

This then shows that the top Pontrjagin number $$< p_k,[S^{4k}]>=<\mp 2c_{2k},[S^{4k}]>=\mp 4$$ but we also know that spheres are boundries of an oriented manifold and thus have higher Pontrjagin number 0. Contradiction.

On another note, according to C.C. Hsiung's book Almost Complex and Complex Structures on page 233 he says "In fact, the absence of an almost complex structure on $S^{4k}$ for $k\geq 1$ and $S^{2n}$ for $n\geq 4$ was proved by Wu and jointly Borel and Serre respectively."

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Don't characteristic classes in fact prove that $S^{2k}$ for $k\geq 3$ aren't? I thought it boiled down to $(k-1)!$ needs to divide 2, or else the characteristic classes obstruct, leaving $S^2$, $S^4$ and $S^6$, and $S^4$ needs to be ruled out by hand. – Charles Siegel Jun 30 '10 at 6:52
@Charlie, I think you mean $k>3$. – Justin Curry Jun 30 '10 at 16:02
Ah, of course, $>$ not $\geq$. – Charles Siegel Jul 1 '10 at 14:39
As an almost complex structure on $X$ endows $TX$ with the structure of a complex vector bundle, doesn't your argument show that these spheres don't even admit almost complex structures, let alone integrable ones? – Michael Albanese Dec 25 '12 at 7:01

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