Were these hypercomplex number sets already defined in the literature for some $n\neq 3$? Are they associative?

Question

I've been studying about hypercomplex numbers for the past few days, and then a question arised:

It looks quite natural to me to define the following set (based on the definition of quaternions):

$X = \{a_0+a_1i_1+...+a_ni_n : a_j\in \mathbb{R}$ for all $j\}$

And define a product on $X$ simply extending the following rules, for all $j$:

$$i_j^2 = -1$$ $$i_{j}i_{j+1}=i_{j+2}$$ $$i_{j+1}i_{j}=-i_{j+2}$$ (here we are considering the indices modulo $n$, so that $i_{n}i_{1} = i_2$, for example).

And define addition in the natural/obvious way.

Then, by extending this we also would have defined any product of type $i_ji_k$, and by applyig the distributive law, we would have a product on all $X$ - unless I missed something wrong here, off course!

My question: is this set with the multiplication given above already defined in the literature (for some $n\neq 3$ - since the case $n=3$ is exactly the quaternions case - I suppose that maybe $n$ should be odd or $\equiv 3 \;(mod \; 4)$, but I'm not sure)? Where could I find it? Also, would this algebra be associative? (I know it would not be a division ring for $n>3$, and it is obviously not commutative)

Any book or article recomendation would be appreciated (since the ones I've read do not answer these questions), and also any comments about the above.

(rmk.: it doesn't matter if this algebra is "nice" or not, but I would like to know if it is well defined and associative, since it looks very "natural" and similar to the quaternions)

Thanks in advance.

Answer 1

As soon as $n=4$ you will need additional rules to fill in the product table so you can determine things such as $i_1i_3=?, i_2i_4=?$, etc.

Recursive definition of basis elements can be a productive idea, however.

For example, the basis elements of the Cayley-Dickson algebras can be defined that way as follows:

1. $i_p^2=-1$
2. If $i_pi_q=i_r$ then $i_qi_r=-i_p$
3. $i_1i_{2k}=i_{2k+1}$
4. $i_{2k}i_{2k+1}=i_1$
5. $i_{2k+1}i_1=i_{2k}$

These last three compose a quaternion triple and can be abbreviated: $(1, 2k, 2k+1)$

When $k=1$ this will give the product table for the quaternions.

To get higher dimensions (which will be powers of two} you need additional doubling rules. There are several options for these. Here is one such.

If $(p,q,r)$ then

1. $(2p,2q,2r)$
2. $(2p,2r+1,2q+1)$
3. $(2q,2p+1,2r+1)$
4. $(2r,2q+1,2p+1)$

If you start with the quaternion triples from the basis elements of the quaternions and add the triples for $k=2,k=3$, namely $(1,4,5),(1,6,7)$ and apply the doubling rules you will get the basis elements of the Octonions. Repeat the process and you get the sedenions, etc.

So you should experiment with your rule and see if you can find additional rules which allow you to fill in your product table in a meaningful way.