# How can the Cayley table for the elements of basis of a Cayley-Dickson algebra be summarized in an algebraic expression?

One would be able to construct a Cayley table that has all $e_i$ elements of the basis of algebra $A$ where $0<i<\dim A$ such that $e_0=1$, $e_1=i$, $e_2=j$ and so on. I'm looking for an algebraic expression of the table for an algebra of dimension $N$, which enables me to find the product $e_ie_j$ without looking at the table. Does such an expression exist?

P.S. The Cayley tables for quaternions, octonions and sedonions can be found in Wikipedia.

I am not aware of a single formula, only recursive formulas involving the "shuffle basis" for Cayley-Dickson spaces. The shuffle basis is not the one commonly used by most researchers. For the shuffle basis, $$e_0=1$$ and $$e_{2n}=(e_n,0)$$ and $$e_{2n+1}=(0,e_n)$$. Furthermore, the product $$e_pe_q=\pm e_{pq}$$ where $$pq$$ is defined as the 'exclusive or' of the binary representations of $$p$$ and $$q$$. There are actually eight[edit: four] different Cayley-Dickson doubling products which satisfy the quaternion properties but the one most commonly used is $$(a,b)(c,d)=(ac-db^*,a^*d+cb)$$. For each of the eight[edit: four] Cayley-Dickson doubling products there is a distinct and well-defined 'sign function' or 'twist' $$\omega(p,q)$$ such that $$e_pe_q=\omega(p,q)e_{pq}$$. It is always true that $$\omega(0,0)=\omega(p,0)=\omega(0,q)=1$$ and that $$\omega(p,p)=-1$$ for $$p>0.$$ This much is true for all eight[edit: four] Cayley-Dickson doubling products. For the product mentioned above, for all positive $$n$$, $$\omega(1,2n)=1$$ and $$\omega(2n,1)=-1$$. Furthermore, for $$0\ne p\ne q\ne 0$$ $$\omega(p,q)=-\omega(q,p)$$ is true for all eight[edit: four] doubling products. For the given doubling product it is also true that the following are all equal for $$0\ne p\ne q\ne 0$$: $$\omega(p,q),\omega(2p,2q),\omega(2q,2p+1),\omega(2q+1,2p),\omega(2q+1,2p+1).$$ From these properties, the 'sign function' or 'twist' $$\omega(p,q)$$ for all higher dimensional Cayley-Dickson algebras can be recovered inductively when using the shuffle basis.

Update: "The eight Cayley-Dickson doubling products", Adv. Appl. Clifford Algebras 26 (2016) pp 529–551, doi:10.1007/s00006-015-0638-6, arXiv:1707.07318 I now find that 4 of those 8 should also be discarded: each allows zero divisors at the eight dimensional stage. The 4 remaining products are denoted in the paper as $$P_0$$, $$P_3$$ (the standard doubling product), $$P_4$$, $$P_7$$. Those four produce algebras isomorphic to the standard Cayley-Dickson algebras.

I can now provide you with an easier answer. As in the first answer we use the convention that $$e_pe_q=\pm e_{p\oplus q}$$ where for non-negative integers $$p$$ and $$q$$, $$p\oplus q$$ is the bitwise exclusive 'or' of the binary representations of $$p$$ and $$q$$. Of the eight equivalent Cayley-Dickson doubling products there is one which, to my knowledge, has not been used previously by researchers.

[edit: unfortunately, this particular doubling product cannot be classified as a Cayley-Dickson doubling product, since it produces zero divisors at the third doubling (the octonions) rather than at the fourth (the sedenions).]

$$(a,b)(c,d)=(ac-b^*d,da^*+bc)$$

As with the other seven doubling products we have $$e_0e_p=e_pe_0=e_p$$ for all $$p$$, $$e_p^2=-1$$ for $$p>0$$ and $$e_pe_q=-e_qe_p$$ for $$0\ne p\ne q\ne 0$$. But in addition we have the wonderful formulas

$$\text{If } 2^N\le p and $$\text{If } 2^N\le p<2^{N+1}\le q \text{ then } e_pe_q=(-1)^{⌊q/2^N⌋}e_{p\oplus q}$$

where $$⌊\bullet ⌋$$ is the floor function.

For example, compute $$e_{21}e_{5}$$. First, $$e_{21}e_{5}=-e_{5}e_{21}$$. And since $$4\le5<8<21$$ we have $$e_{21}e_{5}=-e_{5}e_{21}=-(-1)^{⌊21/4⌋}e_{5\oplus 21}=-(-1)^5e_{16}=e_{16}$$.

Here is a a picture of the twist (or sign values for the basis vector multiplication table for the 1024-ions with a black pixel denoting a $$+1$$ and a white pixel denoting a $$-1$$. The table is indistinguishable to the naked eye for higher dimensional n-ions. For more information see my preprint http://arxiv.org/abs/1602.02317 which is currently being refereed for publication.

• Maybe it's worth updating this with the comment from the recent arXiv update? Feb 4 '20 at 20:20
• @DavidRoberts Good suggestion, I will have to retrieve it. Feb 4 '20 at 20:23
• Btw, I and a student independently used this style of visualisations in a related way, to keep track of sign shifts in a Moufang loop multiplication (not related to octonions, but not too far away). Our result was not so beautifully symmetric! Feb 4 '20 at 22:04
• @DavidRoberts Yes, the twist map, correspondingly, the Moufang loop signs, appear to be chaotic, but not actually. They are periodic, in a sense, when using the $\mathbb{Z}_2^n$ numbering of the basis elements. Feb 4 '20 at 23:01