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.
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This is a famous open-problem. It is still unknown. |
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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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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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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: http://arxiv.org/abs/math/0509442 |
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