In Eric Chisolm's essay on geometric algebra, the author asserts in page 12 that a 2-blade, by definition the product of two orthogonal vectors (vectors $u$ and $v$ for which $uv+vu=0$), can't have a scalar part, because "it anticommutes with both u and v, which means that it can’t have a scalar part, because that part would have commuted with all vectors."
I'm not following. What does it mean for something to anticommute with "both" something and something? I understand that the geometric product of orthogonal vectors is an anticommutative operation, but I don't understand what it means for it to anticommute with both terms.
This should follow from the five preceding axioms he set forth:
Axiom 1. $G$ is a ring with unit. The additive identity is called $0$ and the multiplicative identity is called $1$.
Axiom 2. $G$ contains a field $G_0$ of characteristic zero which includes $0$ and $1$.
Axiom 3. $G$ contains a subset $G_1$ closed under addition, and $λ ∈ G_0, v ∈ G_1$ implies $λv = vλ ∈ G_1$.
Axiom 4. The square of every vector is a scalar.
Axiom 5. The inner product is nondegenerate.
This seems like that should be a simple consequence of these axioms and the definitions, but I can't see why $uv$ couldn't just have a scalar part $\mu$, and $vu$ have a scalar part $-\mu$.
He's saying that $uv$ anticommutes with $u$ and also anticommutes with $v$. His argument is then something like: suppose $uv = \lambda + x$ where $\lambda \in \mathcal G_0$ and $x \not\in\mathcal G_0$. Then $$ -\lambda u + xu = -(\lambda + x)u = u(\lambda + x) = \lambda u + ux $$ and he is then somehow concluding that $\lambda = 0$. Whether or not we conclude anything is highly non-obvious to me, and certainly shouldn't be left as a passing remark.
Axiom 6 is then introduced and renders this consideration moot, since it just declares the notion of scalar part as the projection onto $\mathcal G_0$ and declares this as linearly independent from $\mathcal G_2$.
Axiom 6 is "morally" wrong. He says, "If $\mathcal G_0 = \mathcal G_1$ then, $\mathcal G = \mathcal G_1$." In the case of a one-dimensional vector space, i.e. $\mathcal G_1 = \mathbb R$, then a geometric algebra should have $\mathcal G = \mathbb R\oplus\mathbb R$: we still have a scalar part and a vector part, it just turns out that all vectors are multiples of each other. In any case, equating both copies of $\mathbb R$ here would be a type error since the first is supposed to be a set of scalars and the second a set of vectors.
If we interpret the author as not making such a type error, then it would seem that he is trying to refer to the zero-dimensional case. But in this case we should have $\mathcal G = \mathcal G_0 \ne \{0\}$ with the last inequality from Axiom 2, while it ought to be that $\mathcal G_1 = \{0\}$ if this set is to be interpreted as the set of vectors. But it is true that if $\mathcal G = \mathbb R$ then $\mathcal G_1 = \mathcal G_0 = \mathbb R$ models these axioms, hence why Axiom 6 is only "morally" wrong.
The above "argument" for $uv$ not having a "scalar part" as well as the above confusion that Axiom 6 brings does not instill confidence in me. My personal recommendation for an introduction to geometric algebra is either Geometric Algebra for Physicists by Chris Doran and Anthony Lasenby or Clifford Algebras and Spinors by Pertti Lounesto. I want to explicitly point out Section 14 of Lounesto which discusses different ways to define Clifford algebras, in particular what is needed if the algebra is to be declared as generated by vectors.