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.
 A: 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.
A: 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*}
