I posted this on math.stackexchange to no avail, so I hope it's appropriate to post here despite that it might not be research-level. I expect the answer to this is well-known to people studying non-associative algebras, but I cannot find it in my references and the more thorough literature on the topic is expensive! On a similar note, I would appreciate a recommendation for a reference covering octonion algebras over number fields and the Cayley-Dickson construction, in more generality. I don't mind expensive if it gives a good thorough treatment.

Let $K$ be a number field and let $\mathcal{B}=\Big(\frac{a,b}{K}\Big)$ be a quaternion $K$-algebra. Then its norm is the Pfister form $\langle\langle a,b\rangle\rangle$ over $K$. Apply the Cayley Dickson construction to $\mathcal{B}$, yielding an octonion $K$-algebra $\mathcal{C}$.

What is the Pfister form of the norm of $\mathcal{C}$? Is it $\langle\langle a,b,ab\rangle\rangle$?

Does the isomorphism class of $\mathcal{B}$ (i.e. different choices for $a,b$ preserving the ramification set of $\mathcal{B}$) determine the isomorphism class of $\mathcal{C}$? Equivalently, does the isometry class of the quadratic form $\langle\langle a,b\rangle\rangle$ determine the isometry class of the Pfister form of $\mathcal{C}$ as a quadratic form?

If it does, it makes me wonder about the octonion $K$-algebras with norm the Pfister form of $\langle\langle a,b,c\rangle\rangle$, non-isomorphic to $\mathcal{C}$. It would seem that these also contain $\mathcal{B}$ as a quaternion subalgebra, and that some variations of the Cayley-Dickson construction would take you from $\mathcal{B}$ to these. How does this work?


1 Answer 1


Suppose $K$ is a field, and $B$ your quaternion algebra $K$. The octonion algebra $C$ made from $B$ using the Cayley-Dickson construction depends on an auxiliary choice of an element $c \in K^{\times}$. Namely, $C$ is the set of pairs $(u,v)$ with $u,v \in B$ with addition

$(u_1,v_1) + (u_2, v_2) = (u_1 + u_2, v_1 + v_2)$

and multiplication

$(u_1,v_1)(u_2,v_2) = (u_1u_2 + c\overline{v_2}v_1, v_2u_1 +v_1\overline{u_2}).$

Here $x \mapsto \overline{x}$ is the involution on the quaternion aglebra $B$. The involution on $C$ is then $(u,v) \mapsto (\overline{u},-v)$. With this multiplication and involution, one computes that the norm on $C$ is

$(u,v)(\overline{u},-v) = u\overline{u} - cv \overline{v}$.

It follows that if the norm form on $B$ is the Pfister form $<<a,b>>$, then the norm form on $C$ is the Pfister form $<<a,b,c>>$.

The resulting octonion aglebra $C$ depends not just $B$ but also on $c$. For example, suppose $B$ is the quaternion division algebra over $\mathbf{R}$ (i.e., Hamilton's quaternions). If one chooses $c=1$, then the resulting octonion algebra $C$ contains nonzero elements with norm $0$, while if one chooses $c = -1$ then the norm form on $C$ is anisotropic.

  • $\begingroup$ The follow-up question then is whether there is a choice for $c$ that's considered canonical. Let's say I chose $c=ab$. In this event, does altering $a,b$ up to isomorphism of $\mathcal{B}$ alter the isomorphism class of $\mathcal{C}$? $\endgroup$
    – j0equ1nn
    Mar 17, 2016 at 22:06
  • 1
    $\begingroup$ The choice $c=1$ is canonical. In this case, the octonion aglebra $C$ is split no matter what $B$ is. The space $(x,x)$, for $x \in B$, is a maximal isotropic subspace. Moreover, if $c = x\overline{x}$ for $x \in B$, one gets the same octonion algebra $C$, because the norm forms are equivalent. In particular, the choice $c =ab$ yields the same octonion algebra as the choice $c=1$ $\endgroup$
    – user25514
    Mar 17, 2016 at 22:25
  • $\begingroup$ Okay, so since $ab=ij\overline{ij}$, using this for $c$ is no different from using $1$, up to isomorphism. This makes me think about what choices from $K^\times$ are available that do not lie in $N(\mathcal{B})$ and the resulting isomorphism classes on $\mathcal{C}$, which is interesting. Also this definitely answers my question. I don't suppose you have a recommended reference for this content? $\endgroup$
    – j0equ1nn
    Mar 17, 2016 at 22:44
  • $\begingroup$ There is a book called "Octonions, Jordan algebras, and exceptional groups" by Springer and Veldkamp. I have only looked at this book very briefly, but I'm betting it contains the answer to many interesting questions you could ask about octonion aglebras. $\endgroup$
    – user25514
    Mar 17, 2016 at 22:49
  • $\begingroup$ Yes, that book has been sitting in my Amazon "shopping cart" for a few days now as I deliberated about spending the cash. I agree it looks good, I'm probably going to pick it up. Thanks for your help! $\endgroup$
    – j0equ1nn
    Mar 17, 2016 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.