Odd valued dimensional number impossible to build?

Question

Using numbers of the form $$\alpha_1+\alpha_2e_1+\alpha_3e_2+...+\alpha_ne_{n-1}$$ where $\alpha_n\in\Bbb R$ and for all $a≠b, \alpha_a≠\alpha_b$ with $e_n^2=-1$, can these numbers exist for an odd n? I'm asking this question because if n is odd there will be at least one product of the form $e_ae_b$ that wouldn't be defined. Of course, for $n=1$, we have the Reals but then, we have $n=2, n=4, n=8, n=16$ respectively the Complexes, the Quaternions, the Octonions and Sedenions. Note that none of these values of n is odd and all are powers of 2.

Answer 1

In the context of Clifford/geometric algebras (which we have been talking about recently), I can give you a crash course on how a special case of a real geometric algebra is built. Keep in mind that what I'm describing is a special case, and there are more possibilities.

Here is a brief introduction to why Clifford algebras for finite dimensional vector spaces always have dimension a power of $2$. I think the main prerequisite is that you need to understand what a basis is for a vector space.

Given a finite dimensional $\Bbb R$ vector space $V$, find an orthonormal basis $e_1\dots e_n$ for the usual inner product (the dot product). Create a multiplication table for these elements using the following rules:

1. $e_ie_i^2=1$

2. $e_ie_j=-e_je_i$ for all $i\neq j$.

With just these two rules, and assuming that the $e_i$'s commute with all scalars, you can compute all products of $e_i$'s. For example,

$e_1e_2e_1=e_1(-e_1e_2)=-e_1e_1e_2=-e_2$

and $(e_1e_3)(e_2e_4)=e_1e_3e_2e_4=-e_1e_2e_3e_4$

and $e_2e_5e_1=-e_2e_1e_5=e_1e_2e_5$.

It's not hard to show that if you take all products of basis vectors whose subscripts are strictly increasing, that list of new things is a basis for a Clifford algebra $C\ell_{n,0}(\Bbb R)$. (The "empty" product of $e_i$'s is 1 by convention.) Being a basis means that everything in the algebra will be uniquely expressible as a sum of these things.

So for $n=1$, you just have one $e_1$, and along with the empty product $1$, your basis for the whole Clifford algebra is $\{1,e_1\}$, where $e_1^2=1$. (These are the split-complex numbers.)

For $n=2$, you would have $e_1$ and $e_2$, but you would also need the empty product $1$, as well as $e_1e_2$ in the basis. So you see, this one has four elements.

For $n=3$, you start with $e_1,e_2,e_3$, and throw in $1,e_1e_2,e_1e_3,e_2e_3$ and also you don't forget to throw in $e_1e_2e_3$. This is a basis of 8 elements! For the 3-dimensional vector space.

If you count up how many basis elements are produced this way, you'll find that there are always $2^n$ elements, explaining part of the pattern you saw so far.

This departs from your pattern though: the octonions do not fit the Clifford algebra construction since all Clifford algebras are associative rings (and octions and sedenions are not).

Answer 2

You may be intrested in Hurwitz' theorem.

A "simpler" proof that the dimension is always a power of two goes like this:

Theorem. If there is a continuous odd map $S^{n-1}\times S^{n-1}\to S^{n-1}$, then $n$ is a power of $2$.

Applying this theorem to the map $(x,y)\mapsto \frac{xy}{\Vert xy\Vert}$ (no zero divisors!) gives the desired result baout "dimensional numbers" (or rather division algebras).

The proof of the theorem itself can be done using homology and cohomology with coefficients in $\mathbb F_2$. I won't go into details, see for example Ebbinghaus et. al., Zahlen, Springer.