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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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2 Answers 2

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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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A nice discussion of almost-complex structures (including this class of examples) is in chapter 4 of McDuff and Salamon's Introduction to symplectic topology. –  Martin Dec 27 '12 at 11:01
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Another nice thing to note is that the integrability of the almost complex structure induces by the multiplication is given by the vanishing of the expression $x(yz)-(xy)z$ for any three elements of the division ring; this can be done writing the Nijenhuis tensor or noticing that $J$ has to be parallel with respect to the induced levi-civita connection. Hence, the given structure is integrable in dimension $2$ but not in dimension $6$, because octonions are not associative. –  wisefool Dec 29 '12 at 14:07
    
Martin, I will definitely check out that book as I am yet to grasp all the details from your answer (through no fault of yours). –  Michael Albanese Dec 30 '12 at 13:20
    
@Michael: Take all the time you need, this stuff isn't easy! I might be able to help you or expand my answer if you tried to pin down what is causing trouble. Since most of my answer is about algebraic properties of $\mathbb{H}$ and $\mathbb{O}$ I suspect McDuff-Salamon might not be the right place to go (they are more concerned with geometric properties of almost-complex structures). I learned about Quaternions and Octonions from Numbers which is still one of the more user-friendly expositions I've seen. –  Martin Dec 30 '12 at 13:52
    
Alternatively, John Baez has a page dedicated to Octonions where you can find links to good resources after the table of contents. –  Martin Dec 30 '12 at 14:00

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 $\mathbb{CP}^1$.

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