Complex Structure on S^6 using Stereographic projection. Why does not work? Consider $S^6$ and do a Stereographic projection over $\mathbb{R}^6$ (https://en.wikipedia.org/wiki/Stereographic_projection). Give to $\mathbb{R}^6$ the natural complex structure, so it is $\mathbb{C}^3$, and push back the complex structure of $\mathbb{C}^3$ on the sphere.
Actually, repeat unless you cover the whole sphere with this structure.
As far as I know, there are no known integrable complex structure on $S^6$, so my question is: why the push back of $\mathbb{C}^3$ does not work?
Thank you.
 A: $\newcommand{\Cpx}{\mathbf{C}}$While your idea does define a holomorphic structure on the complement of a point, that structure does not extend over the point at infinity.
To define a holomorphic structure on $S^{6}$, we need a collection $\{U_{\alpha}\}$ of open sets covering $S^{6}$ and for each open set a homeomorphism $\phi_{\alpha}:U_{\alpha} \to \Cpx^{3}$ such that on each overlap $\phi_{\beta}(U_{\alpha} \cap U_{\beta})$, the mapping $\phi_{\alpha} \circ \phi_{\beta}^{-1}$ is a biholomorphism of open subsets of $\Cpx^{3}$. This compatibility is the "hard part".
On the two-sphere $S^{2}$, incidentally, stereographic projections from the north and south poles don't have this property, either; the overlap map is inversion in the unit circle, i.e., $z \mapsto 1/\bar{z}$. But if we replace stereographic projection from the south pole by its complex conjugate, then we've covered $S^{2}$ by two open sets, and the overlap map is $z \mapsto 1/z$, which is a biholomorphism of the set of non-zero compex numbers.
