Clifford Algebra - axiomatic definition by Hestenes

Question

I am reading David Hestenes' book "Clifford Algebra to Geometric Calculus" and am already getting stuck on the first few pages.

My university math is rusty and I've never studied Clifford Algebra before. So I'm not sure if I just can't fill the gaps in the introductory definitions here or if there is really something missing.

At the outset the author introduces the/a (?) Clifford Algebra in an axiomatic way as follows.

The/a (?) Clifford Algebra $G$ is a set of so-called "multi-vectors" that is closed under an addition and multiplication operation which fulfill these axioms:

• A1. Addition is commutative.
• A2. Addition and multiplication are associative.
• A3. Multiplication is distributive with respect to addition.
• A4. There exist unique additive and multiplicative identity elements 0 and 1.
• A5. Every element has a unique additive inverse.
• A6. Every $A \in G$ can be written as a sum of parts of different "grades": $A = \sum_{r=0}^{r=\infty} \langle A\rangle_r$, where $\langle A\rangle_r$ is the grade-r part (or r-vector part) of $A$, such that:
• A6.1 $\langle A+B\rangle_r = \langle A\rangle_r + \langle B\rangle_r$ for $A,B \in G$
• A6.2 $\langle \lambda A\rangle_r = \lambda\langle A\rangle_r = \langle A\rangle_r\lambda$ for $\lambda=\langle\lambda\rangle_0 \in G$.

So far no reference to a simple vector space (the subspace of elements of grade 1), a quadratic form on that space, or even an underlying field for the algebra has been made.

The author now states that:

Axioms 6.1 and 6.2 imply that the space $G^r$ of all r-vectors is a linear sub-space of $G$, and indeed, that $G$ itself is a linear space. Axiom 6.2 also implies that the scalars compose a commutative subalgebra of $G$. Without further ado, we assume that the space $G^0$ of all scalars is identical with the set of real numbers. As argued elsewhere in the book, we regard any wider definition of the scalars (for example as the complex numbers) to be entirely unnecessary and, indeed, inimical to the purposes of geometric algebra.

This leaves me with questions such as:

1. Can we really conclude from these axioms that $G$ is a vector space, without presupposing the field over which this is a vector space?
2. Can we conclude that the decomposition $A = \sum_{r=0}^{r=\infty} \langle A\rangle_r$ is unique? Or is this assumed and not stated?
3. Can we conclude that 6.2 holds only for all $\lambda \in G^0$? Is it not possible that the same could hold for all $\lambda \in G^r$ for some other grade $r$, making the choice of "scalars" ambiguous?
4. Can we conclude that $1 \in G^0$?

I can't answer any of these questions.

.

Note: The axioms above are not the full set of axioms. It is just here that I stopped and questioned the already stated conclusions about $G$ being a vector space and $G^r$ being linear sub-spaces etc.

To complete the set of axioms given in thet book, we have:

7. For any 1-vector (or simply "vector") $a \in G^1$ we have $aa = a^2 = |a|^2 > 0$ where $|a| > 0$ is the magnitude of $a$ (no reference to a bilinear form here, $|.|$ could be any positive definite function, no mention of $|0|=0$, but these things should probably all be assumed here, as well as $G^0=\mathbb{R}$.)
8. For every non-zero product of $r$ anti-commuting vectors $\prod_{k=1}^r a_r \in G^r$ (called an "r-blade") there is a non-zero vector $a$ such that $a\prod_{k=1}^r a_r \in G^{r+1}$. (Not sure if the "gradedness" of $G$, i.e. $G^mG^n\subset G^{m+n}$, should already be assumed here as well or can be concluded from this. And any properties of the geometric product indicating that "anti-commuting" = "linear independent" has not been stated either yet.)

Answer 1

I will try and flesh out the helpful comments given by others into an answer.

If one assumes that the decomposition $A = \sum_{r=0}^{r=\infty} \langle A\rangle_r$ given in A6 is unique, it follows that $\langle\langle A\rangle_k\rangle_l = 0 \; \forall k\neq l \in \mathbb{N}_0$ and $\{0\} = \cap_{r=0}^\infty G^r$.

A6.2 states elements $\lambda = \langle \lambda\rangle_0$ of grade 0 commute with all other multi-vectors and preserve any pure r-vector's grade when multiplied with it. The same then cannot be true for any multi-vector $A$ of grade $r\neq 0$, because the grade of the product $\lambda A$ can't be equal to both 0 and $r\neq 0$.

It follows then that $1 =\langle1\rangle_0 \in G^0$ and the identification of $G^0$ as "scalars" is unambiguous. $G^0$ is a commutative unital ring. $G = \sum G^k$ is a module over $G^0$ and all the $G^k$ are submodules.

From the implicit axiom $G^0 \cong \mathbb{R}$ it then follows that $G$ is a vector space over $\mathbb{R}$ and the $G^k$ are linear sub-spaces of $G$.