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It is fairly well-known that the only spheres which admit almost complex structures are $S^2$ and $S^6$. By embedding $S^6$ in the imaginary octonions, we obtain a non-integrable almost complex structure on $S^6$.

By embedding $S^2$ in the imaginary quaternions, do we obtain an almost complex structure on $S^2$? If so, is it the one induced by the complex structure corresponding to $\mathbb{CP}^1$?

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The point is that $V = \mathbb{R}^3 = \operatorname{Im}\mathbb{H}$ and $V = \mathbb{R}^7 = \operatorname{Im}\mathbb{O}$ inherit a cross-product $V \times V \to V$ from quaternion and octonion multiplication. It is given by $u \times v = \operatorname{Im}(uv) = \frac{1}{2}(uv - vu)$.

Given an oriented hypersurface $\Sigma \subset V$ with corresponding Gauß map $\nu \colon \Sigma \to S^n$ ($n \in 2,6$) sending a point $x \in \Sigma$ to the outer unit normal $\nu(x) \perp T_x\Sigma$ one obtains an almost complex structure by setting $J_x(u) = \nu(x) \times u\qquad\text{for }u \in T_x\Sigma.$

One economic way to see that this is an almost complex structure: The cross-product is bilinear and antisymmetric. It is related to the standard scalar product via $ \langle u \times v, w\rangle = \langle u, v \times w\rangle $ (which shows that $u\times v$ is orthogonal to both $u$ and $v$) and the Graßmann identity $u \times (v \times w) = \langle u,w\rangle v - \langle u,v\rangle w$ (only valid in dimension $3$) has the variant $ (u \times v) \times w + u \times (v\times w) = 2\langle u,w\rangle v-\langle v,w\rangle u - \langle v,u\rangle w $ valid in $3$ and $7$ dimensions (which shows that $u \times (u \times v) = -v$ for $u \perp v$ and $|u| = 1$).

Therefore $J_x(u) = \nu(x) \times u$ indeed defines an almost complex structure $J_x \colon T_x\Sigma \to T_x\Sigma$.

Specializing this to $\Sigma = S^2$ embedded as unit sphere in $\operatorname{Im}\mathbb{H}$ you can “see” (or calculate) that $J_x$ acts in exactly the same way as multiplication by $i$ on $\mathbb{CP}^1$.

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    @MichaelAlbanese: no, sorry, it's just a thing I heard somewhere and, when need came, I just did the computation of the Nijenhuis tensor … it's not very hard (also the covariant derivative of$J$is not hard to compute).2014-06-25
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Yes and yes.

We get an almost complex structure $J$ on the two-sphere $S = \{ ai + bj + ck \mid a^2 + b^2 + c^2 = 1\}$ embedded in the space of imaginary quaternions in exactly the same way as for the octonions.

Since $S$ is of real dimension two, every almost complex structure on $S$ is integrable. Then $X = (S,J)$ is a compact complex curve of genus 0, and any such curve is isomorphic to $\mathbb{CP}^1$.