# How do you prove this left contraction identity for the dual in the homogeneous model, $X\rfloor (I_n^{-1}e_0^{-1})=X^\star e_0^{-1}$?

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.

• 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. :) Jan 21, 2022 at 23:59
• The symbol is "left contraction", which is introduced in this paper: researchgate.net/publication/… . Explicit formulas on pdf pg 6. Jan 24, 2022 at 23:18

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.

• 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. Jan 24, 2022 at 23:39

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*}

• 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. Jan 25, 2022 at 20:35
• Thanks, I edited this answer to apply it to flats only. Jan 25, 2022 at 21:19