Is there a complex structure on the 6-sphere?

Question

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.

Answer 1

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.

Answer 2

Michael Atiyah posted a short paper "The Non-Existent Complex 6-Sphere" https://arxiv.org/abs/1610.09366 with a claimed negative solution to the problem.

Answer 3

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?

Answer 4

Here is a philosophical idea. exploit the following asymmetry in our state of knowledge about closed orientable manifolds: whereas almost complex is equivalent to almost symplectic: symplectic entails a further homological condition while being complex entails no further known homological condition.

This potential unknown further condition to be a closed complex manifold must reduce to no condition in complex dimension one as does the symplectic condition reduce to no condition in complex dimension one. Looking at known examples one can ask whether for a closed complex manifold above complex dimension one must the sum of the betti numbers necessarily be at least three. Note complex projective two space realizes three and above complex dimension two one has the circle cross odd spheres with total betti number four. So the guess is close to being sharp if true, and if true proves only the two sphere among manifolds with the betti numbers of the even sphere can be a closed complex manifold. Note: For manifolds of even complex dimension the last statement has been known since the work of rene thom on cobordism, the signature and the euler characteristic showing there is not even an almost complex structure.

Answer 5

There was a workshop on this problem at Univ. Marburg, in March 2017:

https://www.mathematik.uni-marburg.de/~agricola/Hopf2017/

It resulted in a special issue of the Journal of Differential Geometry and its Applications (April 2018):

https://www.sciencedirect.com/journal/differential-geometry-and-its-applications/vol/57/suppl/C

Most of the papers from that issue are on the ArXiv. The introductory one is a good historical overview:

Ilka Agricola, Giovanni Bazzoni, Oliver Goertsches, Panagiotis Konstantis, Sönke Rollenske, On the history of the Hopf problem, arXiv:1708.01068

The pdf of that paper has arxiv links to most of the other papers (or see below). It mentions the papers by Etesi and Atiyah, saying about them that "the community of experts does not seem to find unity" and linking these MO threads.

By all appearances though, the problem is still open ;-).

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The rest of the papers are:

• Panagiotis Konstantis, Maurizio Parton, Almost complex structures on spheres, arXiv:1707.03883

• Cristina Draper, Notes on $G_2$: The Lie algebra and the Lie group, arXiv:1704.07819

• Ilka Agricola, Aleksandra Borówka, Thomas Friedrich, $S^6$ and the geometry of nearly Kähler $6$-manifolds, arXiv:1707.08591

• Ana Cristina Ferreira, Non-existence of orthogonal complex structures on the round 6-sphere, arXiv:1906.02062

• Boris Kruglikov, Non-existence of orthogonal complex structures on 6-sphere with a metric close to the round one, arXiv:1708.07297

• Daniele Angella, Hodge numbers of a hypothetical complex structure on $S^6$, arXiv:1705.10518

• Christian Lehn, Sönke Rollenske, Caren Schinko, The complex geometry of a hypothetical complex structure on $S^6$, ResearchGate (request only)

• Aleksy Tralle, Markus Upmeier, Chern's contribution to the Hopf problem: an exposition based on Bryant's paper, arXiv:1708.02904

Answer 6

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.

Answer 7

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:

R. Albuquerque, Isabel M.C. Salavessa, On the twistor space of pseudo-spheres Differential Geometry and its Applications, 25 (2007), pp. 207-219, doi:10.1016/j.difgeo.2006.08.004, arXiv:math/0509442.

Answer 8

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.

Answer 9

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.

Answer 10

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

Answer 11

Here is another paper by Gabor Etesi claiming to contain a different proof of existence of the complex structure on 6-sphere, by describing an explicit diffeomorphism to a conjugate orbit in $G_2$: https://arxiv.org/abs/1509.02300