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Well, actually the perspective should be reversed. Starting from the geometric product $a b$ having an inverse as mentioned $$ a a^{-1}=1 $$ one can define 2 identities $$ a \cdot b= \frac{1}{2} \left(a b + \alpha(a) \, \alpha (b) \right) $$ where $\alpha$ is the reflection automorphism $$ a \wedge b= \frac{1}{2} \left(a b - \alpha(a) \, \alpha (b) ...


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What's the motivation for the question? If you pick $R = M = k$ a field and $f = 0$, then $C := C(f) \cong k[\varepsilon]/(\varepsilon^2)$ with the ${\mathbb Z}/2{\mathbb Z}$-grading coming from the natural $\mathbb Z$-grading. Then the residue field of $C$ is an irreducible graded $C$-module for which $\psi$ is not an isomorphism.


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In linear algebra and geometric algebra both, you can talk about "transformations," "bilinear forms" and "vector spaces" all without referring to coordinates. They are all abstract ideas. It is only when you begin to identify $V=F^n$ for some $n$ and field $F$, transformations with matrices, bilinear forms with Gram matrices, etc., that you start to get ...


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A product $\mathbf{uv}$ of vectors can be written $r(\cos\theta + \mathbf{i}\sin\theta) = r\exp{(\mathbf{i}\theta)}$, where $\mathbf{i}$ is the unit pseudoscalar of the plane containing $\mathbf{u}$ and $\mathbf{v}$ and $r = |\textbf{u}||\textbf{v}|$ and $\theta$ is the angle between the vectors. This is interpreted geometrically as an arc of a circle of ...


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If you drop the $v^2 = \|v\|^2$ (or $v\cdot w = \tfrac12(vw+wv)$) identity of Clifford algebra, the result is a free tensor algebra. You can't derive any commutation relations for vectors in this algebra, so terms like $\hat x\hat y$ and $\hat y\hat x$ are independent. A general term in canonical form looks like $a + a_x \hat x + a_y \hat y + a_{xx} \hat x ...


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It isn't so much that the Clifford product is useful as that it's natural. Instead of defining the Clifford product as $v\cdot w + v\wedge w$, you can define the Clifford algebra as the algebra in which S+S and V+V addition, SS and SV multiplication, and V2 squared norm have their usual meanings from vector calculus, the usual algebraic rules apply (except ...


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You know that blades are directed measures. $a\wedge b$ for instance is a directed area, $a\wedge b\wedge c$ is a directed volume, so I suppose what you're asking is: what is for instance $1 + a + a\wedge b + a\wedge b\wedge c$? It's a difficult question and I'm not sure anyone has a definite answer, but I suspect it's the same kind of question people ...



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