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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.

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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:

  1. 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 $G^0=\mathbb{R}$.)
  2. 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.)
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    $\begingroup$ Your questions are reasonable and stem, I guess, from the confusing way the axiomatic definition is given. For your question 2, uniqueness should be assumed. Once this is done, axiom 6.2 implies that the $\lambda$ in question must have indeed degree $0$ and that $G^0$ is closed under the product (and is closed under addition by axiom 6.1). Also, axiom 6.2 implies that the only $G^r$ closed under addition is $G^0$ (although, there is another property not explicitly stated establishing what and where is $\langle A\rangle_r\langle B\rangle_s$). However, the closeness of $G^0$ under ... $\endgroup$
    – F Zaldivar
    Dec 12, 2021 at 23:41
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    $\begingroup$ addition and multiplication do not imply that $1\neq 0$; and thus we do not know if $G^0$ is indeed a field (note that $1\in G^0$ by axiom 6.2). My guess is that in this area, it is always implicit that the base field is the real numbers, but in the axiomatization that you are reading there are some non explicit assumptions and extreme cases are not taken care of with the necessary care. $\endgroup$
    – F Zaldivar
    Dec 12, 2021 at 23:44
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    $\begingroup$ As @FZaldivar suggests, Hestenes's approach to 'axiomatics' is not the same as in much of the rest of mathematics; it is somewhat more in the spirit of physics, where the axioms are meant to state some things that are true about the objects being discussed rather than to provide a basis from which all else can be deduced. You may be interested in the other questions that turn up when you search Hestenes. $\endgroup$
    – LSpice
    Dec 13, 2021 at 0:10
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    $\begingroup$ Some quick, not double-checked thoughts. On the face of it, the axioms as written seem to immediately give a (possibly non-commutative) graded unital $\mathbb{Z}$-algebra, but where the graded nature of 0 and 1 is not asserted. Then Zaldivar's comments probably give us that 1 and 0 are in $G^0$, as they should be. Then we get that $G$ is a algebra over the commutative unital ring $G^0$. The author then states the implicit axiom that $G^0\simeq \mathbb{R}$, so that $G$ is a graded unital algebra over the real numbers. $\endgroup$
    – David Roberts
    Dec 13, 2021 at 3:37
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    $\begingroup$ This leads ultimately to delicate questions of convergence for homogeneous multivectors with infinite grade. But we will not be concerned with such questions in this book. <--- This kind of quote illustrates perfectly the attitude to rigour in the book, IMHO $\endgroup$
    – David Roberts
    Dec 13, 2021 at 22:17

1 Answer 1

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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 $G$.

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