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I would like to prove that $X\rfloor (I_n^{-1}e_0^{-1})=X^\star e_0^{-1}$ as claimed in Geometric Algebra for Computer Science (Dorst et al). I don't see where this pattern matches to any of the established contraction identities shown so far.

The five-pointed star refers to $X^\star \equiv X\rfloor I_n^{-1}$. The asterix (six points) refers to $X^*=X \rfloor I_{n+1}^{-1}$. The context is the homogeneous model, where $\mathbb{R}^n$ is modeled within $\mathbb{R}^{n+1}$. The pseudoscalar of $\mathbb{R}^{n+1}$ is $I_{n+1}=e_0\wedge I_n=e_0I_n$ and the inverse is $I_{n+1}^{-1}=I_n^{-1}e_0^{-1}$. The authors seem to be treating $X$ as a general multivector in $\mathbb{R}^{n+1}$.

A bigger context for the equation is: $$ X^* = X\rfloor I_{n+1}^{-1} = X\rfloor (I_n^{-1}e_0^{-1}) = X^\star e_0^{-1} $$

edit: $X$ is not a general multivector. It is a flat.

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  • $\begingroup$ To indulge readers who don't know all the notation... could you also explain that symbol that is a vertical line with a left-turning "foot" on it? Thanks. :) $\endgroup$ Jan 21, 2022 at 23:59
  • $\begingroup$ The symbol is "left contraction", which is introduced in this paper: researchgate.net/publication/… . Explicit formulas on pdf pg 6. $\endgroup$
    – foghorn
    Jan 24, 2022 at 23:18

2 Answers 2

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Appendix C, equation (C.6) demonstrates the distribution of the contraction over geometric product. Basically you have:

$X \rfloor (I_n^{-1}e_0^{-1}) = (X \rfloor I_n^{-1}) e_0^{-1} + \hat I_n^{-1} (X \rfloor e_0^{-1})$

Since $X$ is a $k$-flat (a $k$-blade in $R^{n+1}$ where $k > 1$) then $\hat I_n^{-1} (X \rfloor e_0^{-1}) = 0$ and the identity follows.

The text is not explicit about $X$ being a flat, but it must be for the above identity to hold. Also the identity is introduced in a section named "Direct and dual representation of flats" so $X$ must be taken as a flat.

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  • $\begingroup$ Sorry, I did not notice your answer until after I answered. Personally, the recursive/inductive nature of the solution was important to me. (C.6) addresses one of the base cases, but not the recursive part with higher grades. $\endgroup$
    – foghorn
    Jan 24, 2022 at 23:39
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X is a flat that can have the form $A$ or $e_0\wedge B$, where $A$ and $B$ are blades that don't contain $e_0$. We will factorize the blades into vectors and move those vectors across the left contraction one at a time according to the rule $(X\wedge Y)\rfloor Z\equiv X\rfloor (Y\rfloor Z)$.

For the $A$ blade:

\begin{align*} A_k\rfloor I_{n+1}^{-1} &= (A_{k-1}\wedge a_k)\rfloor (I_n^{-1}\wedge e_0^{-1}) \\ &= A_{k-1}\rfloor (a_k\rfloor (I_n^{-1}\wedge e_0^{-1})) \\ &= A_{k-1}\rfloor ((a_k\rfloor I_n^{-1})\wedge e_0^{-1}) + \widehat{I_n^{-1}}\wedge \underbrace{(a_k\rfloor e_0^{-1})}_{=0}) \\ &= (A_{k-2}\wedge a_{k-1})\rfloor ((a_k\rfloor I_n^{-1})\wedge e_0^{-1}) \\ &= \dots \\ &= (A_k\rfloor I_n^{-1}) \wedge e_0^{-1} \\ &= A_k^\star e_0^{-1} \end{align*}

For the $e_0\wedge B$ form, it starts the same way and then:

\begin{align*} (e_0\wedge B_k)\rfloor I_{n+1}^{-1} &= e_0\rfloor ((B_k\rfloor I_n^{-1})\wedge e_0^{-1}) \\ &= \underbrace{(e_0\rfloor (B_k\rfloor I_n^{-1}))}_{=0}\wedge e_0^{-1} + \widehat{B_k\rfloor I_n^{-1}}\wedge \underbrace{(e_0\rfloor e_0^{-1})}_{=1} \\ &= e_0e_0^{-1}\widehat{B_k\rfloor I_n^{-1}} \\ &= e_0\wedge (B_k\rfloor I_n^{-1}) e_0^{-1}\\ &= ((e_0\wedge B_k)\rfloor I_n^{-1})e_0^{-1} \\ &= (e_0\wedge B)^\star e_0^{-1} \end{align*}

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  • $\begingroup$ Note that the identity doesn't extend to multivectors. For instance it doesn't hold for a point $p=e_0 + \alpha e_1 + \beta e_2 + \gamma e_3$ which is a $1$-blade. It also doesn't hold for a scalar which is a $0$-blade. It also doesn't hold for the pseudoscalar itself. So it cannot hold for general multivectors. Dorst define Flats as $k$-blades where $1 < k < n$. You can extend this to some multivectors made out of Flats, but they don't have geometric interpretation. $\endgroup$ Jan 25, 2022 at 20:35
  • $\begingroup$ Thanks, I edited this answer to apply it to flats only. $\endgroup$
    – foghorn
    Jan 25, 2022 at 21:19

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