We can define the algebra of quaternions $\mathbb H$ over any field $k$, and depending on the arithmetic of $k$ it is either a division algebra or a matrix algebra.

We can also define the algebra of octonions $\mathbb O$ over any field $k$, and if over $k$ the $8$-ary quadratic form $Q=x_0^2+x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2$ is anisotropic, then $\mathbb O$ is again a division algebra —a non-associative one, but oh well.

What happens to $\mathbb O$ if $k$ is such that $Q$ is isotropic?

The classical structure theory of non-commutative Jordan algebras tells us that $\mathbb O$ is, over any field, a direct product of simple flexible power-associative algebras coming from a rather restricted list: simple commutative Jordan algebras, quasiassociative algebras, and flexible quadratic algebras with nondegenerate norm forms (Shafer's book on non-associative algebra explains all this, which is —I'd say— mostly forgotten nowadays) but I am pretty sure one can be very specific about what comes out in the case of octonions. In other words, one can probably find something playing the role of «matrix algebra» in the statement about quaternions.

N.B. All this is over fields of characteristic zero.

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    $\begingroup$ It's times like this that I wish I knew what a Jordan algebra is. $\endgroup$
    – Will Jagy
    Mar 31, 2013 at 4:24
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    $\begingroup$ Google "octonion algebra" and "split octonion algebra" (analogy: special orthogonal groups for isotropic quadratic forms, or even for a split quadratic form). $\endgroup$
    – user30379
    Mar 31, 2013 at 4:27
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    $\begingroup$ Yeah, it is: that's what I get from counting from zero. I blame the French. $\endgroup$ Mar 31, 2013 at 4:27
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    $\begingroup$ @pranavk, of course I already googled that :-) But the definitions of split octonions I found are essentially undistinguishable from those of octonions: there is a difference of complexity in going from quaternions to matrix algebras, which is not found in going to the split-octonions (at least, in the references to these that I found) $\endgroup$ Mar 31, 2013 at 4:29
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    $\begingroup$ Lam is fun. Granted, it's not Game of Thrones. $\endgroup$
    – Will Jagy
    Mar 31, 2013 at 5:09

1 Answer 1


$\DeclareMathOperator\Tr{Tr} $Suppose that $k$ is a field with $\operatorname{char}(k) \neq 2$. Let's agree that an "octonion algebra" over $k$ is an 8-dimensional unital $k$-algebra $A$, endowed with a quadratic form $N: A \rightarrow k$, whose associated bilinear form $T(x,y) = N(x+y) - N(x) - N(y)$ is nondegenerate, and which satisfies $N(xy) = N(x) N(y)$ for all $x,y \in A$.

When $x \in A \setminus k$, the trace $\Tr(x) = T(x,1)$ and norm $N(x)$ are determined by the algebra structure: $x$ is the root of a quadratic polynomial with coefficients $-\Tr(x)$ and $N(x)$. Hence the algebra structure determines the quadratic form $N$ in a convenient way. One can talk about the "isomorphism class of an octonion algebra" while carrying around the quadratic form or not; it doesn't really matter.

It is an old theorem (Jacobson? Albert? I can't recall … check "The Book of Involutions" too) that the isomorphism class of the octonion algebra $k$ is determined by the isomorphism class of the quadratic form $N$. Now, essentially by the Cayley–Dickson doubling process, the norm form $N$ is a Pfister form, i.e., $N$ is isomorphic to $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ for some $a,b,c \in k$.

It is a fact about Pfister forms that when they are isotropic, they are split. So if the norm form represents zero, then $N$ is isomorphic to $\langle1,-1\rangle \otimes \langle1,-1\rangle \otimes \langle1,-1\rangle$ and the octonion algebra is isomorphic to the split octonion algebra over $k$.

In this level of generality, the isomorphism classes of octonion algebras over $k$ are classified up to isomorphism by the Galois cohomology $H^3(k, \mu_2)$; you can see such a cohomology class too from the Pfister form perspective. The Pfister form $\langle1,-a\rangle \otimes \langle1,-b\rangle \otimes \langle1,-c\rangle$ depends only on the square-classes of $a, b, c$, giving three classes in $H^1(k, \mu_2)$ whose cup product is the element of $H^3(k, \mu_2)$ classifying the octonion algebra.

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    $\begingroup$ Hi, Marty. You get around. I found some old email from Pete L. Clark saying you were good at this, for some value of `this.' $\endgroup$
    – Will Jagy
    Mar 31, 2013 at 5:12
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    $\begingroup$ The theorem is Jacobson's; it is theorem 3.23 in [Richard D. Schafer, An introduction to nonassociative algebras, Pure and Applied Mathematics, Vol. 22, Academic Press, New York, 1966] Thanks! $\endgroup$ Mar 31, 2013 at 5:27

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