8
$\begingroup$

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 $\mathbb{CP}^1$?

2 Answers 2

6

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 $\mathbb{CP}^1$.

  • 0
    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](http://books.google.com/books?id=DNjKAeRexE4C).2012-12-27
  • 3
    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.2012-12-29
  • 0
    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).2012-12-30
  • 1
    @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](http://books.google.com/books/about/Numbers.html?id=DOvXGeFRvdIC) which is still one of the more user-friendly expositions I've seen.2012-12-30
  • 1
    Alternatively, John Baez has [a page dedicated to Octonions](http://math.ucr.edu/home/baez/octonions/) where you can find links to good resources after the table of contents.2012-12-30
  • 0
    @Martin: How is the 'standard scalar product' $\langle\ ,\ \rangle$ defined?2013-01-04
  • 0
    @Michael: It's the usual Euclidean product $\langle x,y\rangle = \sum_{i=1}^n x_i y_i$. In terms of the algebra operations can be written as $\langle x,y\rangle = \operatorname{Re}(x\bar{y})$ and for purely imaginary quaternions or octonions there is the alternative expression $\langle x,y\rangle e = -\frac{1}{2}(xy+yx)$ where $e$ is the unit element.2013-01-04
  • 0
    @wisefool: Do you know of a reference which goes through the details of your comment?2014-06-18
  • 0
    @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
5

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 $\mathbb{CP}^1$.