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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?

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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.

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  • $\begingroup$ 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 '16 at 22:06
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    $\begingroup$ 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 '16 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 '16 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 '16 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 '16 at 22:51

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