1

A flat connected to the origin point $o$ is represented by $X=o \wedge \textbf{A}_k \wedge \infty$, where $\textbf{A}_k$ is a k-blade of directions in the Euclidean subspace, and $\infty$ is the point at infinity. I want to calculate $$||X||^2 = X \widetilde{X} = (o\wedge \textbf{A}_k \wedge \infty)(\infty \wedge \widetilde{\textbf{A}_k} \wedge o)$$

I am not sure how to simplify the expression. The answer is supposed to be $X\widetilde X = -||\textbf{A}_k||^2$.

It is scalar, so $X\widetilde X=\langle (p\wedge \textbf{A}_k \wedge \infty)(\infty \wedge \widetilde{\textbf{A}_k} \wedge p) \rangle$.

In this CGA, we have $\infty \cdot \infty=0$, $o\cdot o=0$, $o\cdot \infty = -1$.

If there was a left contraction between $X$ and $\widetilde X$, then the solution would follow, but how would that be justified?

foghorn
  • 209
  • 1
    To justify the left-contraction, see https://math.stackexchange.com/a/3320392/472818 , in particular $\langle A^\sim,B\rangle_0=\langle A^\sim,\lrcorner,B\rangle_0$, which is valid in complete generality, for any multivectors $A,B$. So you could ask a simpler question, trying to verify that general identity; I think CGA is an irrelevant complication. – mr_e_man May 05 '22 at 03:06

1 Answers1

1

Despite my comments, I can answer this directly.

Note that $A$ in the Euclidean subspace is orthogonal to both $o$ and $\infty$. (Some would be more comfortable writing these vectors as $e_o$ and $e_\infty$.) Therefore $A\wedge o\wedge\infty=A\,(o\wedge\infty)$.

$$X\,X^\sim=(A\wedge o\wedge\infty)(\infty\wedge o\wedge A^\sim)$$

$$=A\,(o\wedge\infty)(\infty\wedge o)\,A^\sim$$

$$=A\,(-1)\,A^\sim$$

$$=-\lVert A\rVert^2$$

mr_e_man
  • 5,364
  • Should I justify the middle step? It's a result of the general identity $(a\wedge b)(b\wedge a)=a^2b^2-(a\cdot b)^2$ (where $a$ and $b$ are vectors), which itself is a special case of the determinant formula given in the link. – mr_e_man May 05 '22 at 03:38
  • The only thing I would add to this is $o \wedge A \wedge \infty = \widehat{A} \wedge o \wedge \infty = \widehat{A} o \wedge \infty$. This grade involution happens to both of the factors, and the double grade involution cancels out of the result. – foghorn May 05 '22 at 19:45