# Do the octonions contain infinitely many copies of the quaternions?

Note: by "infinitely many", I'm confident I always mean $$\beth_1$$ many herein.

We can easily show the quaternions contain infinitely many copies of $$\Bbb C$$ because, given any unit vector $$\in\Bbb R^3$$ of components $$b,\,c,\,d$$, $$\Bbb R[h]$$ is isomorphic to $$\Bbb C$$ with $$h:=bi+cj+dk$$. Sure, these aren't "independent" copies of $$\Bbb C$$ in the same way $$\Bbb R[i],\,\Bbb R[j],\,\Bbb R[k]$$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $$\Bbb C$$ provide an easy definition of a determinant, even though general matrices of quaternions prohibit this. For example, if $$A_1,\,\cdots,\,A_n$$ are matrices and $$O$$ denotes a contextually appropriate zero matrix, the block matrix $$\left(\begin{array}{cccc} A_{1} & O & \cdots & O\\ O & A_{2} & \cdots & O\\ \vdots & \vdots & \ddots & O\\ O & O & \cdots & A_{n} \end{array}\right)$$can be said to have determinant $$\prod_l\det A_l$$. The order is important, but one in particular is natural.

I may be overlooking some details of the above benefit to arbitrarily many copies of $$\Bbb C$$ in $$\Bbb H$$, wherein a very rich commuting family of matrices is constructed, but my question isn't about that. I'm wondering how we'd prove there are infinitely many copies of $$\Bbb H$$ in $$\Bbb O$$. (Again, block matrices provide a benefit, in this case inheriting the associativity of the $$A_l$$.) I suspect a proof exists that admits the following sketch:

• In $$\Bbb O$$, create some $$h_1,\,h_2$$ each analogous to $$i\in\Bbb C$$, generating an associative algebra and satisfying $$h_1h_2=-h_2h_1$$;
• Note that any such pair of square roots of $$-1$$ can model the quaternions viz. $$i=h_1,\,j=h_2,\,k=h_1h_2$$;
• Show the above can be done in infinitely many different ways.

Of course there are infinitely many ways to choose the pair $$(h_1,\,h_2)$$, but $$\Bbb R[h_1,\,h_2]$$ won't always be a different set for such pairs. That's why I suspect the proof requires a few clever i-dotting t-crosses.

• In fact, we can identify the space of copies of $\Bbb H$ in $\Bbb O$ with the compact Riemannian symmetric space of type $G$, namely, $G_2 / SO(4)$; in particular this space has dimension $8$. – Travis Feb 2 at 5:29

Yes. One can show the following 2 things:

1. Any octonion $$x$$ not in $$\mathbb R$$ generates a subalgebra $$A$$ isomorphic to $$\mathbb C$$.

2. For any octonion $$y$$, the subalgebra generated $$B$$ by $$x$$ and $$y$$ is associative.

This should be covered in treatments of composition algebra, e.g. it is probably in Springer-Veldkamp.

Hence if we take $$y$$ outside of $$A$$ in 1, by Frobenius' classification of $$\mathbb R$$-division algebras, $$B$$ must be isomorphic to $$\mathbb H$$ (a priori it may not be obvious $$B$$ is division, but you can get this a posteriori from Frobenius' theorem as adjoining inverses to $$B$$, which necessarily lie in $$\mathbb O$$, still gives you an associative algebra), and thus the 4-dimensional space spanned by $$1, x, y, xy$$. Since no finite collection of $$\mathbb R^4$$ subspaces of $$\mathbb R^8$$ cover $$\mathbb R^8$$, you get infinitely many copies of $$\mathbb H$$.

To say that you get (at least) the cardinality of the continuum copies of $$\mathbb H$$, it suffices to show

1. For any nonreal octonion $$z$$, there is a subalgebra $$B \simeq \mathbb H$$ as above not containing $$z$$.

To see this, take $$x$$ as above not in $$\mathbb R z$$. Then $$x$$ and $$z$$ generate a space $$B' \simeq \mathbb H$$. Simply take $$y \not \in B'$$. Then the algebra $$B$$ generated by $$x$$ and $$y$$ cannot contain $$B'$$ since $$B \ne B'$$ but $$\dim B = \dim B'$$.

• For what it's worth, property (2) is called alternativity. – Travis Feb 2 at 5:30
• I upvoted first and only later noticed this messes up your perfectly round reputation of 2000. I'm sorry – Vincent Mar 19 at 22:55