# Does S^6 have the structure of an algebraic variety?

More accurately, the question should be: Is it known that $S^6,$ the 6-dimensional sphere, is $not$ a (proper) complex algebraic variety, or algebraic space? And is there a reference? It's easy to see that it cannot be projective (as $H^2=0$), but I don't see how it can violate the usual general properties of algebraic varieties [e.g. by Chow's lemma one can get a projective (and smooth, by resolution of singularities) variety lying over and birational to it, but the cohomology groups can get larger when we go up...]. Maybe one needs some theory of classification of 3-folds (if there is one such theory).

Knowing that it cannot be algebraic will somehow make the analogous question of complex structure much sharper (by the way, it does have an almost complex structure).

Thank you.

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See mathoverflow.net/questions/1973/… for the discussion of the existence of a complex structure on $S^6$, as well as remarks on its non-algebraicity. –  ACL Nov 20 '11 at 10:55

Suppose that $X$ is a smooth complete positive-dimensional algebraic space over $\mathbb C$. Then $\mathrm H^2(X, \mathbb Q)$ can not be 0. In fact, every algebraic space contains an open dense subscheme. Let $U \subseteq X$ be an open affine subscheme; by a well known-result, the complement $C$ of $U$ has pure codimension 1; consider $C$ as a divisor, with its reduced scheme structure. I claim that the cohomology class of $C$ can not be 0. Let $Y \to X$ a birational morphism, where $Y$ is a smooth projective variety; this exists, by Chow's lemma for algebraic spaces, and resolution of singularities. The pullback of $C$ is a non-zero effective divisor on $Y$, so its cohomology class is not 0. Since formation of cohomology classes of divisors is compatible with pullbacks, this implies the result.

This says very little about the problem of existence of a complex structure. Complete algebraic spaces, or, if you prefer, Moishezon manifolds, are not so distant from projective varieties; it seems clear to me that if $S^6$ has a complex structure, this will be a very exotic animal, very far away from the world of algebraic geometry.

: why does $C$ have codimension 1? This is Corollaire 21.12.7 of EGAIV, when $X$ is a scheme. In general you can reduce to the case of a scheme by considering an étale morphism $Y \to X$, where $Y$ is an affine scheme. This map is affine, because $X$ is separated, so the inverse image of $U$ in $Y$ is affine.

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Thanks, Angelo. What is this "well known-result" that $C\subset X$ has pure codim. 1 (and any reference)? And I suppose that the cohom. class of an effective divisor in a projective variety is non-zero because one can find another (smooth) effective alg. cycle with non-empty intersection (a finite set of points) with $C,$ right? –  shenghao Nov 27 '11 at 4:48
For whatever it's worth: $S^6$ is a ($6$-dimensional) homogenous nearly Kähler manifold. These were classified by Butruille. In particular, they have a canonical almost complex structure. I assume one should be able to prove that this almost complex structure is not a complex structure.
[EDIT, thanks Spiro] Apparently this does not imply that one may conclude that there is no complex structure on $S^6$, so this is not much help.
There are also some interesting uniqueness results for connections on $S^6$ in Fukami and Ishihara's paper that might be helpful for this.
[EDIT, thanks David] For a short time there was a preprint on arXiv claiming that $S^6$ is a complex manifold. See discussion here.
The nearly Kahler structure on $S^6$ is definitely not integrable, this is well known. But it certainly does not follow from that fact that there could be no other integrable almost complex structure on $S^6$. What is known (due to LeBrun) is that there can be no integrable complex structure which is orthogonal with respect to the standard round metric on $S^6$. As far as I know, this question of existence of some integrable complex structure on $S^6$ is still wide-open. –  Spiro Karigiannis Nov 20 '11 at 12:56