Some basics on octonions and quaternions

Question

There is an $\mathbb{R}$-bilinear operation on the octonions $\mathbb{O} \otimes_{\mathbb{R}} \mathbb{O} \rightarrow \mathbb{O}$, which is not associative. My question is instead about the algebra structure $\mathbb{O} \otimes_{\mathbb{H}} \mathbb{O} \rightarrow \mathbb{O}$.

My questions are:

• Do we have an isomorphism of algebras $\mathbb{O} \otimes_{\mathbb{H}} \mathbb{O} \cong \mathbb{O} \times \mathbb{O}$?
• Does the (noncommutative, non associative) algebra structure on the octonions gain any nice properties (such as associativity) when viewed as an algebra over the quaternions instead of an algebra over the reals?

I was thinking as well about a more general idea, in which the construction producing octonions from quaternions and quaternions from complex numbers is generalized to a situation in which we have $B \otimes_{A} B \cong B \oplus B$ (in a setting in which rank is defined). Perhaps this is related to some feature of the group ring $A[\mathbb{Z}/2\mathbb{Z}]$. In that case, we would indeed get something associative (and commutative). Meanwhile I am also interested in chains of extensions of degree two, and what can be said of a degree four extension (in a setting in which rank is defined) whose intermediate extensions are commutative degree 2 extensions. And from this perspective, I am interested in the similarity between sectonions as an algebra over the quaternions by comparison with octonions as an algebra over the complex numbers.

Answer 1

In this context, an "algebra" is short for "$F$-algebra" for some field $F$. When we say $A$ is an $F$-algebra, that means the multiplication on $A$ is $F$-bilinear. In particular, this requires $F$ to be commutative (not a skew field like $\mathbb{H}$) and central (so $\mathbb{H}$ is not a $\mathbb{C}$-algebra since $Z(\mathbb{H})=\mathbb{R}$). Things go wrong if you try to loosen this definition.

Note, by the way, you can change what field you consider an algebra over. For example, $A=M_2(\mathbb{C})$ can be considered as an $\mathbb{R}$-algebra or a $\mathbb{C}$-algebra. Facts like (non)commutativity and (non)associativity are properties of $A$'s multiplication operation and do not depend on which $F$ is chosen.

Suppose $M$ and $N$ are right and left $A$-modules, respectively. Then $M\otimes_AN$ is a well defined $F$-vector space. If actually $M$ is a $(X,A)$-bimodule (meaning left $X$-module, right $A$-modules, and these module actions commute) and $N$ is an $(A,Y)$-module, then $M\otimes_AN$ will be a $(X,Y)$-bimodule. (This is only a gain in generality, not a loss, because we can take $X$ and/or $Y$ to be $F$.)

Note $\mathbb{O}$ is only an $\mathbb{R}$-algebra (unital, nonassociative), not a $\mathbb{C}$-algebra or $\mathbb{H}$-algebra. It is a left $\mathbb{C}$-vector space and a right $\mathbb{C}$-vector space (i.e. module), indeed a $(\mathbb{C},\mathbb{C})$-bimodule because it is an alternative algebra, however it is not a left or right $\mathbb{H}$-vector space (let alone a bimodule).

That said, you can still define "$\mathbb{O}\otimes_\mathbb{H}\mathbb{O}$" as a vector space to be the quotient of $\mathbb{O}\otimes_\mathbb{R}\mathbb{O}$ by the vector subspace spanned by the differences $(xh\otimes y-x\otimes hy)$ for $x,y\in\mathbb{O}$, $h\in\mathbb{H}$. Using $\ell$ for any nonzero imaginary octonion perpendicular to $\mathbb{H}$, we can say $\mathbb{O}=\mathbb{H}\oplus\mathbb{H}\ell$ is a $\mathbb{Z}_2$-graded vector space, with $\ell\mathbb{H}=\mathbb{H}\ell$, and then $\mathbb{O}\otimes_{\mathbb{R}}\mathbb{O}=(\ell\mathbb{H}\oplus\mathbb{H})\otimes_{\mathbb{R}}(\mathbb{H}\oplus\mathbb{H}\ell)$ becomes $\mathbb{Z}_2\times\mathbb{Z}_2$-graded:

$$ (\ell\mathbb{H}\otimes\mathbb{H}) \,\oplus\, (\ell\mathbb{H}\otimes\mathbb{H}\ell) \,\oplus\, (\mathbb{H}\otimes\mathbb{H}) \,\oplus\, (\mathbb{H}\otimes\mathbb{H}\ell). $$

(All unadorned $\otimes$ are over $\mathbb{R}$.) Notice the subspace we're quotienting by is also $\mathbb{Z}_2\times\mathbb{Z}_2$-graded, so we can quotient in each of the four "components" separately. Notice $\ell x\otimes y=(\ell x)y\otimes1=\ell(yx)\otimes1$ but also $\ell x\otimes y=\ell\otimes xy=\ell(xy)\otimes 1$, therefore the difference $\ell z\otimes 1=0$ vanishes, where $z=xy-yx=$ $2\,\mathrm{Im}(x)\times\mathrm{Im}(y)$ can be any pure imaginary octonion. The same cancellation occurs with $\ell$ on the right of $\otimes$, but with $\ell$ on both or neither side no cancellation occurs (but quaternion scalars can be slid to once side).

Thus the four $\mathbb{Z}_2\times\mathbb{Z}_2$-components become, in the quotient,

$$ \mathbb{O}\otimes_{\mathbb{H}}\mathbb{O} = \mathbb{R}(\ell\otimes1) \,\oplus\, (\ell\otimes\mathbb{H}\ell) \,\oplus\, (1\otimes\mathbb{H}) \,\oplus\, \mathbb{R}(1\otimes\ell). $$

So it would appear this "$\mathbb{O}\otimes_{\mathbb{H}}\mathbb{O}$" is a $10$-dimensional $\mathbb{R}$-vector space. There is a natural map $\mathbb{O}\otimes_{\mathbb{R}}\mathbb{O}\to\mathbb{O}\otimes_{\mathbb{H}}\mathbb{O}$ given by $a\otimes b\mapsto a\otimes b$, but the algebra's multiplication map $\mathbb{O}\otimes_{\mathbb{R}}\mathbb{O}\to\mathbb{O}$ does not factor through it like it would in the associative world.