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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.

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  • $\begingroup$ 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 $8$. $\endgroup$ – Travis Willse Feb 2 at 5:29
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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 $B$ generated 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 $\dim B = \dim B'$.

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

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