We know that there is an almost complex structure on $S^6$ which is not integrable. Is it always possible to find almost complex structures on $S^{2n}$? In particular does $S^4$ admit one?
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3No, it is only possible for $S^2$ and $S^6$, https://en.wikipedia.org/wiki/Almost_complex_manifold#Examples – Peter Franek May 16 '15 at 22:08
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In Sur les classes caractéristiques des structures fibrés sphériques, Wu proved that $S^{4n}$ cannot admit an almost complex structure for $n \geq 1$.
In Groupes de Lie et puissances réduites de Steenrod, Borel and Serre proved that $S^{2n}$ cannot an admit complex structure for $n \geq 4$.
Therefore, the only even-dimensional spheres which can admit an almost complex structure are $S^2$ and $S^6$. In fact, both of these do admit almost complex structures. The former is the familiar Riemann sphere, while the latter obtains an almost complex structure by viewing it as the unit length purely imaginary octonions. Note, one can obtain the almost complex structure on $S^2$ in an analogous way using quaternions; see this question.
The almost complex structure on $S^2$ is integrable; in fact, every almost complex structure on a two-dimensional manifold is integrable, see this question. However, the almost complex structure on $S^6$ described above is known to be non-integrable. It is still unknown whether $S^6$ admits an integrable almost complex structure. In fact, it is unknown whether there exists a $2n$-dimensional manifold, $n \geq 3$, which admits almost complex structures without admitting an integrable almost complex structure. In dimension $4$, there are examples of manifolds which admit almost complex structures, none of which is integrable; one example is $(S^1\times S^3)\#(S^1\times S^3)\#\mathbb{CP}^2$, see Theorem $9.2$ of Barth, Hulek, Peters, & van de Ven, Compact Complex Surfaces (second edition).
See my answer here for an update on the search for an integrable almost complex structure on $S^6$.
Michael Albanese
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There's a (recent?) proof on the existence of complex structure on S^6. On a physics journal though. (Not sure if it's correct) http://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere/1984#1984 – May 18 '15 at 09:10
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Borel-Serre explicitly say that their result applies to any differentiable structure on the topological $S^{2n\ge 8}$, but say that the proof in Steenrod for $S^4$ uses the standard differentiable structure. Is Wu's result applicable to the standard differentiable structure, or to an arbitrary differentiable structure on $S^4$? – YCor Dec 09 '18 at 00:07
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2@YCor: I don't have access to a copy of Wu's paper right now, so I don't know. However, there is a simple proof that $S^4$ does not admit an almost complex structure (regardless of which smooth structure it is endowed with): if $S^4$ did have an almost complex structure, then $p_1(TS^4) = c_1(TS^4)^2 - 2c_2(TS^4) = -2e(TS^4)$, but pairing with the fundamental class of $S^4$, we get the equation $0 = 3\sigma(S^4) = -2\chi(S^4) = -4$. In general, if two smooth four-manifolds are homeomorphic, then one admits an almost complex structure if and only if the other does. – Michael Albanese Dec 09 '18 at 02:47