TITLE: Is it known whether $S^6$ is a Kähler manifold?
QUESTION [6 upvotes]: I have just started to learn about Kähler manifolds and I now am wondering:

Is it known whether $S^6$ is a Kähler manifold?

By definition a Kähler manifold has 3 structures: a symplectic, a complex and a Riemannian structure. I know that for $n>3$ $S^{2n}$ does not have a complex structure and therefore cannot be a Kähler manifold. 
But what can be said if $n=3$? Are there other criteria that can be used to conclude that $S^6$ cannot be a Kähler manifold?

REPLY [12 votes]: $S^6$ cannot even be a symplectic manifold. The reason is that on any compact symplectic manifold $X$ of dimension $2n$, the symplectic form represents a class in $H^2(X, \mathbb{R})$ whose $n^{th}$ power is a generator of $H^{2n}(X, \mathbb{R})$, and in particular it must be nonzero. But $H^2(S^6, \mathbb{R}) = 0$. In fact $S^{2n}$ cannot be a symplectic manifold, hence cannot be a Kähler manifold, unless $n = 1$. 
It's also an open question whether $S^6$ can be given the structure of a complex manifold.