Dot-product and grade-projection, is grade proj. "algebraic"?

I am not a mathematician, so forgive me if I don’t really get stuff right.
Initially I was pulled in by the seeming simplicity of the equations

ab = a \cdot b + a \wedge b \\ a \cdot b = (1/2) (ab + ba) \\ a \wedge b = (1/2)(ab - ba)

which at the beginning of my journey with GA thought meant that every other operator we use can be defined in terms of the GA op. and hence their properties will be consequences of the axioms which define it.

Now that some time passed, I know that the above in general is only true for 1-vectors. Also know that the definition of the products: \cdot, \wedge; are not as straightforward as I initially anticipated but rather they are defined in terms of the grade-projection operator:
(so that they work across PGAs with differing dimensionality)

A_j \wedge B_k = \langle A_j B_k \rangle_{j+k} \\ A_j \cdot B_k = \langle A_j B_k \rangle_{|j+k|} \\ A_j \rfloor B_k = \langle A_j B_k \rangle_{k-j} \\ A_j \lfloor B_k = \langle A_j B_k \rangle_{j-k}

where A_j is a j-vector and B_k is a k-vector, and I have also included the left and right contraction although their usefulness for 2D and 3D PGA is still elusive for me.

Given my not so proficient level with GA hence not knowing an algebraic definition for the grade-projection operator and the incompatibility of the above definitions with my initial expectation of generality and derivativeness, I must ask:

  1. is there an algebraic definition of the grade-proj. op. in terms of the geometric product?
    (Of course I understand that \langle 16 + \alpha e_1 + e_{01} + \beta e_{20} + e_{012} \rangle_2 = e_{01} + \beta e_{20}
  2. are there other definitions which fully define the above or something like the above operations which would work for me while continuing my journey towards physical applications across both 2D and 3D PGA for multivectors?
  3. am I overcomplicating something simple? :smile:

I also “came up” with the following definitions for the wedge and inner product in an attempt to force the top equations to hold for multi-vectors:

a \cdot b = \left\lbrace \begin{aligned} a &\neq b \implies 0 \\ a &= b \implies ab \end{aligned} \right. \hspace{3em} a \wedge b = \left\lbrace \begin{aligned} a &\neq b \implies ab \\ a &= b \implies 0 \end{aligned} \right.

after which I realized then later verified through the video:
An Overview of the Operations in Geometric Algebra - YouTube
(Inner product comment relevant here starts at 30:45)
that the product currently in the literature that is called the inner product, according to abstract algebra is actually not an inner product, while my above definition if I did things right seems to be…
So next question then becomes:

  1. are these two definitions any good for understanding the theory?

Dominik

The grade projection operator is nothing more than the filtering out of those blades that have a different grade than the desired grade. Its use in the definition of the wedge, dot, and contraction products indicates that the result is that portion of the geometric product.

The contraction products are equivalent to the dot product when k-j = |k-j| and j-k = |k-j|. I’ve read that the contraction products provide some kind of algebraic clarity over the dot product, but I haven’t read the argument for that or worked it out myself (the Dorst paper cited by Wikipedia is behind rentier journal paywalls), so I’m just as lost as you.

The fundamental unit of the geometric product is the product of two blades. k-vectors are linear combinations of k-blades, so the geometric product of a k-vector and j-vector is the products of their k- and j-blades using the distributive property. Whether a product blade is part of the wedge, dot or contraction product depends on whether the grade of the product blade matches the grade predicate for the wedge (k+j), dot (|k-j|), or contraction (k-j and j-k) product. When the blades overlap, as in e_1 e_{12} = e_2, the grade of the product will be included in the dot product because e_2 = {\langle e_2 \rangle}_{|1-2|}. When the blades do not overlap, as in e_1 e_{23} = e_{123} the product will be part of the wedge product because e_{123} = {\langle e_{123} \rangle}_{1+2}.

The definitions given on Wikipedia are actually sums over the grades of the multivectors C and D, which generalizes the definitions to multivectors.

\displaystyle C \wedge D = \underset{r,s} \sum \langle \langle C \rangle_r \langle D \rangle_s \rangle_{r+s}
\displaystyle C \cdot D = \underset{r,s} \sum \langle \langle C \rangle_r \langle D \rangle_s \rangle_{|r-s|}
\displaystyle C ⌋ D = \underset{r,s} \sum \langle \langle C \rangle_r \langle D \rangle_s \rangle_{s-r}
\displaystyle C ⌊ D = \underset{r,s} \sum \langle \langle C \rangle_r \langle D \rangle_s \rangle_{r-s}

Thank you, especially for the following two golden sentences which I see will be helpful while doing hand computations:

  • when the blades overlap, the grade of the product will be included in the dot product,
  • when the blades do not overlap, the product will be part of the wedge product.

Be careful that the grades follow that |r-s| and r+s relation though. Partially overlapping blades aren’t included in the dot product.

e_{12} \cdot e_{23} = {\langle e_{13} \rangle}_{|2-2|} = {\langle e_{13} \rangle}_0 = 0

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