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.

  • $\begingroup$ I think this question has too many parts. Maybe just ask 1-2 now, and then ask other questions later if need be? $\endgroup$
    – Kimball
    Nov 16, 2023 at 17:48
  • $\begingroup$ @Kimball thanks for the advice, made some changes. $\endgroup$
    – user900250
    Nov 17, 2023 at 0:29
  • $\begingroup$ How are you defining $\Bbb O\otimes_{\Bbb H}\Bbb O$ and what is the map $\Bbb O\otimes_{\Bbb H}\Bbb O\to\Bbb O$? If you are defining $\Bbb O\otimes_{\Bbb H}\Bbb O$ itself to be an algebra, then you need a way to multiply two elements of $\Bbb O\otimes_{\Bbb H}\Bbb O$ (and then you should explain what that has anything to do with your map $\Bbb O\otimes_{\Bbb H}\Bbb O\to\Bbb O$, since that latter map doesn't define a multiplication on $\Bbb O\otimes_{\Bbb H}\Bbb O$ itself). $\endgroup$
    – coiso
    Nov 21, 2023 at 2:42
  • $\begingroup$ @coiso the tensor product is not itself an algebra (or rather it is but not in a way which we are considering here). Instead, there is a non-commutative non-associative map of algebras from the quaternions to the octonions. The existence of this map is enough to ensure that the non-commutative non-associative multiplication $\mathbb{O} \otimes_{\mathbb{R}} \mathbb{O} \rightarrow \mathbb{O}$ factors through a multiplication $\mathbb{O} \otimes_{\mathbb{H}} \mathbb{O} \rightarrow \mathbb{O}$ $\endgroup$
    – user900250
    Nov 21, 2023 at 8:14
  • $\begingroup$ Please define $\Bbb O\otimes_{\Bbb H}\Bbb O$ and say what the map $\Bbb O\otimes_{\Bbb H}\Bbb O\to\Bbb O$ is. $\endgroup$
    – coiso
    Nov 21, 2023 at 15:41

1 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.

  • $\begingroup$ Thanks, this is fantastic. If you happen to know anything about $\mathbb{H} \otimes_{\mathbb{C}} \mathbb{H}$ that is also very interesting to me. $\endgroup$
    – user900250
    Dec 17, 2023 at 13:44
  • $\begingroup$ @Cayley-Hamilton $\Bbb H\otimes_{\Bbb C}\Bbb H$ is a $\Bbb C$-vector space with scalar multiplication $\lambda(a\otimes b)=a\lambda\otimes b=a\otimes\lambda b$ and $\Bbb C$-basis $\{a\otimes b\mid a,b\in\{1,j\}\}$. This means $a\otimes b$ with $a\in\{1,j\}$, $b\in\{1,i,j,k\}$ (or vice-versa) forms an $\Bbb R$-basis. It can also be viewed as a simple $(\Bbb H,\Bbb H)$-bimodule with $x(a\otimes b)y=(xa)\otimes(by)$, not bimodule isomorphic to $\Bbb H^2$. (To see this, view it as $\Bbb C^2\otimes(\Bbb C^2)^*$ as a $\rm SU(2)\times SU(2)$-rep.) $\endgroup$
    – coiso
    Dec 17, 2023 at 17:03

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