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I don't know who first asked this question, but it's a question that I think many differential and complex geometers have tried to answer because it sounds so simple and fundamental. There are even a number of published proofs that are not taken seriously, even though nobody seems to know exactly why they are wrong.

The latest published proof to the affirmative: Even though the preprint is old it was just published in Journ. Math. Phys. 56, 043508-1-043508-21 (2015)

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A topical preprint has been posted on ArXiv (asserting that $S^6$ has a complex structure): – Ramsay Dec 7 '10 at 19:33
And there is a new version out: claiming to completely overhaul the proof. Did anyone take a look with expertise in this area? – Daniel Apr 30 '11 at 10:28
I think you'll find that very few experts are willing to study the 4th revision, if the first 3 had serious flaws. – Deane Yang Apr 30 '11 at 12:53
Curiously enough, I cannot find any serious discussions and comments by experts on this recently published paper in the online math community. I would like to see some. – Alex Fok May 20 at 15:28
Gabor Etesi posted a new paper to the arXiv on 8Sep15: "Complex structure on the six dimensional sphere explained": arXiv abstract. – Joseph O'Rourke Sep 9 at 11:53

7 Answers 7

Of course, I'm not about to answer this question one way or the other, but there are at least a couple of interesting things one might point out. Firstly, it has been shown (although I forget by whom) that there is no complex structure on S6 which is also orthogonal with respect to the round metric. The proof uses twistor theory. The twistor space of S6 is the bundle whose fibre at a point p is the space of orthogonal almost complex structures on the tangent space at p. It turns out that the total space is a smooth quadric hypersurface Q in CP7. If I remember rightly, an orthogonal complex structure would correspond to a section of this bundle which is also complex submanifold of Q. Studying the complex geometry of Q allows you to show this can't happen.

Secondly, there is a related question: does there exist a non-standard complex structure on CP3? To see the link, suppose there is a complex structure on S6 and blow up a point. This gives a complex manifold diffeomorphic to CP3, but with a non-standard complex structure, which would seem quite a weird phenomenon. On the other hand, so little is known about complex threefolds (in particular those which are not Kahler) that it's hard to decide what's weird and what isn't.

Finally, I once heard a talk by Yau which suggested the following ambitious strategy for finding complex structures on 6-manifolds. Assume we are working with a 6-manifold which has an almost complex structure (e.g. S6). Since the tangent bundle is a complex vector bundle it is pulled back from some complex Grassmanian via a classifying map. Requiring the structure to be integrable corresponds to a certain PDE for this map. One could then attempt to deform the map (via a cunning flow, continuity method etc.) to try and solve the PDE. I have no idea if anyone has actually tried to carry out part of this program.

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A little more detail to Joel's first paragraph (I can't see how to add a comment to it, sorry!).

The argument that there is no orthogonal complex structure on the 6-sphere is due to Claude Lebrun and the point is that such a thing, viewed as a section of twistor space, has as image a complex submanifold. Now, on the one hand, this submanifold is Kaehler, and so has non-trivial second cohomology, since the twistor space is Kaehler. On the other hand, the section itself provides a diffeomorphism of our submanifold with the 6-sphere which has trivial second cohomology. Neat, huh?

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This is interesting. Doesn't this bear some similarity to the argument used by Adler (who is or was a colleague of LeBrun) in his published "proof" of this conjecture? My recollection is that Adler tried to show that a Riemannian metric compatible with a complex structure on S^6 could be deformed into a Kahler metric, leading to the same contradiction. By the way, I never found anyone who was able to identify exactly why Adler's proof is wrong. – Deane Yang Oct 26 '09 at 1:13
That's right. He has a continuity argument involving a notion of "distinguished metric" for an almost complex structure that I have some difficulty making sense of: it requires embedding yr almost complex manifold in some high dimensional sphere. – Fran Burstall Oct 26 '09 at 21:10
Yes, I got completely lost when he embedded the sphere into a high dimensional space. I couldn't see why that would help at all and the calculations become a total mess. – Deane Yang Aug 8 '13 at 17:29

If such a complex structure exists, it would weird indeed! For example, as shown by Campana, Demailly and Peternell (Compositio 112, 77-91), if such a thing exists, then $S^6$ would have no non-constant meromorphic functions. In particular, $S^6$ can't be Moishezon, let alone algebraic.

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It should be possible to show that majority of complex 3-folds are not Moishezon. So, I would not say that this remark is a real argument against existsing of a complex strucutre on S^6. There is a nice phrase in the aricle of Gromov. Page 30. "How much do we gain in global understanding of a compact (V, J) by assuming that the structure J is integrable (i.e. complex)? It seems nothing at all: there is no single result concerning all compact complex manifolds" – Dmitri Jan 21 '10 at 23:25
I'm not sure I understand this remark by Gromov. In the complex analytic case we have the Dolbeault resolution -- one of the ways to state the integrability condition is precisely that Dolbeault complex is a complex. This leads to topological statements, e.g. the alternating sum of the Euler characteristics of $\Omega^i$'s (computed using the Chern classes) is the Euler characteristic of the manifold itself. This may or may not be true in the almost-complex case, but I don't see how to prove it. – algori Jan 22 '10 at 2:02
I think, the remark of Gromov is quite clear, it is quite hard to belive this remark, but the message is clear. As for Euler characteristics, David gave a correct explanation… – Dmitri Jan 22 '10 at 9:05
What I meant was precisely that: this is hard to believe. The Euler characteristic is just the first thing that comes to mind. – algori Jan 22 '10 at 17:41

Continuing Joel Fine and Fran Burstall's answer about, indeed "neat", Lebrun's result. Just want to recall that the "orthogonal" twistor space of any $2n$-dimensional pseudo-sphere $SO(2p+1,2q)/SO(2p,2q)$ can be written as $SO(2p+2,2q)/U(p+1,q)$. So the Kähler manifold in question, in case of the 6-sphere, is $SO(8)/U(4)$. One should think of each $j:T_xS^6\rightarrow T_xS^6$ as a linear map on $R^8$ with $j(x)=-1$ and $j(1)=x$. Well, proofs have been rewritten of LeBrun's result. I wish I had more opinion on this:

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Here is a shot in the dark (Disclosure: I really know nothing about this problem).

Let $G:=\mathsf{SU}(2)$ act on $G^3$ by simultaneous conjugation; namely, $$g\cdot(a,b,c)=(gag^{-1},gbg^{-1},gcg^{-1}).$$ Then the quotient space is homeomorphic to $S^6$ (see Bratholdt-Cooper).

The evaluation map shows that the character variety $\mathfrak{X}:=\mathrm{Hom}(\pi_1(\Sigma),G)/G$ is homeomorphic to $G^3/G,$ where $\Sigma$ is an elliptic curve with two punctures.

Fixing generic conjugation classes around the punctures, by results of Mehta and Seshadri (Math. Ann. 248, 1980), gives the moduli space of fixed determinant rank 2 degree 0 parabolic vector bundles over $\Sigma$ (where we now think of the punctures are marked points with parabolic structure). In particular, these subspaces are projective varieties.

Letting the boundary data vary over all possibilities gives a foliation of $\mathfrak{X}\cong G^3/G\cong S^6$. Therefore, we have a foliation of $S^6$ where generic leaves are projective varieties; in particular, complex.

Moreover, the leaves are symplectic given by Goldman's 2-form; making them Kähler (generically). The symplectic structures on the leaves globalize to a Poisson structure on all of $\mathfrak{X}$.

Is it possible that the complex structures on the generic leaves also globalize?

Here are some issues:

  1. As far as I know, the existence of complex structures on the leaves is generic. It is known to exist exactly when there is a correspondence to a moduli space of parabolic bundles. This happens for most, but perhaps not all, conjugation classes around the punctures (or marked points). So I would first want to show that all the leaves of this foliation do in fact admit a complex structure. Given how explicit this construction is, if it is true, it may be possible to establish it by brute force.
  2. Assuming item 1., then one needs to show that the structures on the leaves globalize to a complex structure on all of $\mathfrak{X}$. Given that in this setting, the foliation is given by the fibers of the map: $\mathfrak{X}\to [-2,2]\times [-2,2]$ by $[\rho]\mapsto (\mathrm{Tr}(\rho(c_1)),\mathrm{Tr}(\rho(c_2)))$ with respect to a presentation $\pi_1(\Sigma)=\langle a,b,c_1,c_2\ |\ aba^{-1}b^{-1}c_1c_2=1\rangle$, it seems conceivable that the structures on the leaves might be compatible.
  3. Moreover, $\mathfrak{X}$ is not a smooth manifold. It is singular despite being homeomorphic to $S^6$. So lastly, one would have to argue that everything in play (leaves, total space and complex structure) can by "smoothed out" in a compatible fashion. This to me seems like the hardest part, if 1. and 2. are even true.

Anyway, it is a shot in the dark, probably this is not possible...just the first thing I thought of when I read the question.

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Well, certainly the leaves can't be smoothed to define a smooth foliation of $S^6$ because the tangent bundle of $S^6$ (more generally, any even dimensional sphere) is irreducible, so it has no nontrivial foliations. – Robert Bryant Nov 2 '14 at 16:09
Nice! Thanks Robert! And certainly, since the leaves are real dimension 4, the foliation cannot be trivial upon smoothing. Great! I still wonder if 1. and 2. can be accomplished, but now it seems less interesting (but still interesting enough I think for a potential student project one day). – Sean Lawton Nov 2 '14 at 16:59
That's too bad. – Deane Yang Nov 2 '14 at 17:57

This is a famous open-problem. It is still unknown.

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Yeah, I know. But I think it's a great question. Am I supposed to post only questions for which I believe an answer is already known? – Deane Yang Oct 22 '09 at 23:25
Well, for problems that you know are open, there's already a site for collecting them: the open problem garden – Charles Siegel Oct 23 '09 at 0:06
Thanks for the link. I didn't know about it. But it seems like a less active site? I don't see any differential geometry there at all. – Deane Yang Oct 23 '09 at 1:26
I propose that we move future meta-discussion regarding open problems to this thread on meta. – Scott Morrison Oct 23 '09 at 3:05
@Charles Siegel: You mean like the famous open problem in Geometry "Heavy raw boundary Final Fantasy 14 CD Key Generator"? Doesn't seem like a very respectable page^^ – Matthias Ludewig Aug 8 '13 at 17:44

Personally, I do not think that that proof is correct. This is a simple question of a compact homogeneous spaces. Any even dimensional compact Lie group is a (homogeneous) complex torus bundle over a projective rational homogeneous space (which is also simply connected---K\"ahler-Einstein with positive Ricci curvature) and therefore is complex. The paper basically said that the complex structure J_H comes down to S^6 is integrable. His reason was that J_H is the restriction of J_{G_2} to H. However, H is not closed under the Lie bracket. That is why J_H can not simply come down to S^6.

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Which proof are you referring to in this answer? – j.c. Sep 19 '13 at 12:24
It sounds like the paper described in this talk: – Brendan Murphy Sep 21 '13 at 1:28

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