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As observed by Calabi a long time ago, the manifold $S^2\times S^4$ admits an almost-complex structure (obtained by embedding it in $\mathbb{R}^7$ and using the octonionic product), which however is not integrable.

Is it known whether $S^2\times S^4$ admits an integrable complex structure?

A few remarks.

  • This is stated as an open problem in Calabi's paper, but perhaps it has been solved in the meantime?
  • This is similar to the case of $S^6$, which is still open (see this question).
  • One can also ask the same question for $\Sigma\times S^4$ for $\Sigma$ any compact Riemann surface
  • It seems that some people believe that every almost-complex manifold of real dimension $6$ or more admits an integrable complex structure (see this other question).
  • Even more generally (and this is obviously still open), one can ask about an arbitrary finite product of even dimensional spheres (excluding $S^0$). It is known that this is almost complex iff the only factors that appear are $S^2, S^6$ and $S^2\times S^4$.
  • If one allows connected sums, then for example $(S^2\times S^4)\#2(S^3\times S^3)$ is a complex manifold, and in fact it has complex structures with trivial canonical bundles (see for example here and here).
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I thought this question was resolved for $S^6$ a while ago (after the question you link to was posed and answered)? Am I wrong? –  Thomas Kragh Feb 29 '12 at 19:46
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I believe the question for $S^6$ is still open. There are old published papers that claimed to prove that it is not complex (and the general consensus is that these are wrong), and there is a preprint by Etesi on the arXiv that claims that it is complex (and I don't know anybody who has read it carefully). –  YangMills Mar 1 '12 at 2:52
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As an aside, it is a very weak condition for an oriented $6$-manifold $X$ to admit an almost complex structure (compatible with the orientation. Namely, this is equivalent to the existence of a complex spin structure, i.e. the vanishing of the Bockstein of second Stiefel-Whitney class $W_3=\beta w_2=0 \in H^3(X,\mathbb{Z})$. In particular, no 2-torsion in $H^3(X,\mathbb{Z})$ is enough. Moreover, the almost complex structures up to homotopy form an affine space under $H^2(X,\mathbb{Z})$, by obstruction theory. For instance, there is a unique structure up to homotopy on $S^6$. –  BS. Jul 29 '12 at 17:12

1 Answer 1

up vote 4 down vote accepted

This is still an open problem. See this paper for some progress, which was prompted by this MO question.

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