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In the Clifford algebra, geometric algebra, and applications, the author introduced the following operations. Some of the relevant notation: $(Proposition)$ denote iverson bracket, i.e. $(P) = 1$ iff $P$ is true, $0$ if otherwise. $A B = \tau(A, B) A \Delta B$, $\tau$ is some scalar function defined inductively.

Definition 2.7. For $A, B \in \mathcal{P}(X)$ we define

$$ \begin{array}{rll} A \wedge B & :=(A \cap B=\varnothing) A B & \text { outer product }\\ A \llcorner B & :=(A \subseteq B) A B & \text { left inner product } \\ A \lrcorner B & :=(A \supseteq B) A B & \text { right inner product } \\ A * B & :=(A=B) A B & \text { scalar product } \\ \langle A\rangle_k & :=(|A|=k) A & \text { projection on grade } k \\ A^{\star} & :=(-1)^{|A|} A & \text { grade involution } \\ A^{\dagger} & :=(-1)^{\binom{|A|}{2}} A & \text { reversion }\\ \end{array} $$ and extend linearly to $\mathcal{C l}(X, R, r)$.

How do i prove that these operations aren't dependent on the choice of basis that construct the isomorphism between geometric and combinatorial geometric algebra? Can these operation be defined using only the language of geometric algebra?

StructSeeker
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  • Someone familiar with universal properties and unafraid to use them really needs to write an algebraically literate introduction to geometric algebra. Chevalley's spinors book and my old note from undergrad times should do a good job on defining the product, as does Bourbaki's Algèbre, chapitre 9 (still not translated into English) and a couple other books on quadratic forms, but no one seems to have explicitly done the inner products and other operations. – darij grinberg Feb 16 '25 at 22:29
  • In other news, the Lundholm/Svensson reference you cite has a newer version. But it still relies on bases, probably to avoid confusing undergrads. – darij grinberg Feb 16 '25 at 22:32
  • I think Theorem 3.6 in Lundholm/Svensson is a version of basis-independence for the inner products. It's a bit tricky to say what the right statement is, since some of the vectors seem to be playing the role of covectors. – darij grinberg Feb 16 '25 at 22:35
  • Actually, in the context of the narrative surrounding the definition of the expressions guarded by propositions, they're set-valued. So, if $P$ is a proposition then $(P)A$ is $A$ if $P$ is true and $∅$ if $P$ is false. Borrowing from C's syntax for conditional expressions: $(P)A = P? A: ∅$. – NinjaDarth Feb 16 '25 at 23:29
  • The isomorphism is basis dependent, because the combinatorial Clifford algebra is a formal construction over a basis. I don't understand what you mean by "basis independent" here. The operations, in usual geometric algebra, are basis independent and usually introduced as such. For example, if $A,B$ have grade $s,t$ then $A\wedge B = \langle AB\rangle_{s+t}$. Another way to get this particular operation is to construct the natural isomorphism with the exterior algebra, which is basis independent. – Nicholas Todoroff Feb 16 '25 at 23:48
  • For grade projection, see https://math.stackexchange.com/questions/4527624/geometric-algebra-how-to-prove-that-the-grade-projection-operator-is-well-defin . All the other operations have well-known definitions in terms of grade projection and the geometric product. And for the latter, there are some old posts such as https://math.stackexchange.com/questions/444988/looking-for-a-clear-definition-of-the-geometric-product (maybe I'm forgetting a better one), but I don't see any proof that the geometric product (as defined combinatorially using an orthogonal basis) is independent of the basis. – mr_e_man Feb 28 '25 at 01:14
  • But my proof would be to first show that the combinatorial definition of the geometric product does in fact have the properties described in the axiomatic definition (it's associative, a vector squared equals the quadratic form, etc.), and then re-derive the combinatorial definition from the axioms, noting that it works for any orthogonal basis. – mr_e_man Feb 28 '25 at 01:21
  • Also the scalar $\tau(A,B)$ can be defined directly (not inductively), as in my answer here. – mr_e_man Feb 28 '25 at 01:41

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