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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?

user174981
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    As Qiaochu remarks, it cannot, but it does admit a nearly Kahler structure. This is again a triple $(M, g, J)$, where $g$ is a metric and $J$ is an almost complex structure, compatible in that $g(\cdot, J \cdot)$ is skew, but instead of asking that $\omega$ (equivalently, $J$) be parallel w.r.t. the Levi-Civita connection $\nabla$ of $g$, we only ask that the $3$-tensor $\nabla \omega$ be totally skew. – Travis Willse Jan 27 '15 at 09:12
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    In fact, the n.K. structure on the $6$-sphere can be described simply in terms of the algebra of the octonions; this structure is homogeneous with symmetry group the compact real form of the exceptional simple Lie group $G_2$ (!), and the stabilizer of a point under this group action is a copy of $SU(3)$. Better yet, one can make precise the notion that $G_2 / SU(3) \cong S^6$ is the model of $6$-dimensional strictly (i.e., non-Kahler) nearly Kahler geometry, in the same sense that $(\mathbb{R}^n, \bar{g})$ is the model of Euclidean geometry. – Travis Willse Jan 27 '15 at 09:14

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$S^6$ 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.

Qiaochu Yuan
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