Why is $u\times(v\times w)=-u\mathbin{\lrcorner} (v\wedge w)$?

Question

Consider the proof of the Triple product expansion I found on Wikipedia: $\newcommand{\imult}{\mathbin{\lrcorner}}$ $$u\times(v\times w)=-u\imult (v\wedge w)=v\wedge(u\imult w)-(u\imult v)\wedge w=(u\cdot w)v-(u\cdot v)w$$ Can someone explain the first equality? I know that the LHS equals$$*(u\wedge*(v\wedge w))$$but I don't know how to proceed further...

Answer 1

$ \newcommand\lcontr{\mathbin\lrcorner} \newcommand\grade[1]{\langle#1\rangle} \newcommand\Grade[1]{\Bigl\langle#1\Bigr\rangle} $

Refer to The Inner Products of Geometric Algebra (2002) by Leo Dorst for the definition and properties of the contraction $\lcontr$. Note that there are different conventions in defining the contractions that give slight variations; see for example the Appendix of Compendium of Multivector Contractions (2022) by André Mandolesi.

On vectors, the contraction is just $a\lcontr b = a\cdot b$. We will make use of the identity $$ a\lcontr(b\wedge c) = (a\lcontr b)c - b(a\lcontr c). $$ The contractions and exterior product are dual in the sense that $$ X\wedge Y\:I = X\lcontr(YI),\quad X\lcontr Y\:I = X\wedge(YI), $$ for arbitrary multivectors $X, Y$ and pseudoscalar $I$. The geometric product takes the least precedence, i.e. you should compute other products like $\wedge$ and $\lcontr$ first.

By $*X$ I would assume you mean $XI^{-1}$ for $I$ the right-handed unit pseudoscalar, but if you're referring to the Hodge star then the equivalence between these (up to some sign differences) can be confirmed just be seeing how they act on a basis of the exterior algebra; I will not do so here. In three dimensional Euclidean space, $I^{-1} = -I$ and $I^2 = -1$. It is true that $$ u\times(u\times w) = u\wedge[v\wedge w\:I]\:I. $$ Continuing with the identities from above, we see $$ u\wedge[v\wedge w\:I]\:I = u\lcontr(v\wedge w\:II) = -u\lcontr(v\wedge w) = (u\lcontr w)v - (u\lcontr v)w = (u\cdot w)v - (u\cdot v)w. $$ Hence we have shown that $$ u\times(u\times w) = (u\cdot w)v - (u\cdot v)w. $$