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).
  • $\begingroup$ 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? $\endgroup$ – Thomas Kragh Feb 29 '12 at 19:46
  • 5
    $\begingroup$ 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). $\endgroup$ – YangMills Mar 1 '12 at 2:52
  • 4
    $\begingroup$ 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 $S^6$. $\endgroup$ – BS. Jul 29 '12 at 17:12

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.