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With a little bit of work, I have proven to myself that the geometric product between three vectors is associative ($a$, $b$, and $c$ are 1-vectors): $$\begin{aligned}(ab)c &= a(bc) \\ &= (b \cdot c) a - (a \cdot c) b + (a \cdot b) c + a \wedge b \wedge c.\end{aligned}$$

And, by extension, that it is consistent for four vectors ($a$, $b$, $c$, $d$) as: $$\begin{aligned} (abc)d &= a(bcd) \\ &= (b \cdot c)(a \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot b)(c \cdot d) \\ &\quad+ (b \cdot c)a \wedge d - (a \cdot c)b \wedge d + (a \cdot b)c \wedge d \\ &\quad+ b \wedge c(a \cdot d) - a \wedge c(b \cdot d) + a \wedge b(c \cdot d) \\ &\quad+ a \wedge b \wedge c \wedge d\end{aligned}$$ However, the following, which I would expect to match the rules for associativity, is quite different: $$\begin{aligned} (ab)(cd) &= (a \cdot b + a \wedge b)(c \cdot d + c \wedge d) \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + (a \wedge b)(c \wedge d) \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + (a \wedge b).(c \wedge d) + a \wedge b \wedge c \wedge d \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) + a \cdot (b \cdot (c \wedge d)) + a \wedge b \wedge c \wedge d \\ &= (a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) \\ &\quad + (a \cdot d)(b \cdot c) - (a \cdot c)(b \cdot d) \\ &\quad + a \wedge b \wedge c \wedge d \\ &= (b \cdot c)(a \cdot d) - (a \cdot c)(b \cdot d) + (a \cdot b)(c \cdot d) \\ &\quad + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d) \\ &\quad + a \wedge b \wedge c \wedge d.\end{aligned}$$ Which, when compared with $a(bcd)$ and $(abc)d$ has all pure scalars and the quad-vector, but is missing four of the six scaled bi-vectors.

Is it that my understanding of associativity is incorrect in that $(ab)(cd)$ is not the same as $a(bcd)$ or $(abc)d$, did I make a mistake in my algebra, or is there something else going on?

My digging into this comes from the desire to understand $e_{12} e_{23}$ which is short-hand for $(e_1 \wedge e_2)(e_2 \wedge e_3)$ (note, geometric product for $()()$) where $e_1$, $e_2$, $e_3$ are orthonormal basis vectors and I have been unable to determine whether the result should be $0$ or $e_{13}$. Also, that $e_1 e_2 e_2 e_3$ comes up in my working out of geometric products when $e_1 e_2$ is used as shorthand for $e_1 \wedge e_2$.

Follow-up:

Many thanks to Somos for encouraging me to keep looking and to Peeter Joot for his hint about ${\langle ( \wedge )( \wedge ) \rangle}_2$ as the apparent circular definition made me dig into just what $(a \wedge b)(c \wedge d)$ looks like.

Extrapolating from appendix C.1 of Geometric Algebra for Computer Science (Dorst, Fontijne, Mann). $$ (a \wedge b)(c \wedge d) = \frac{1}{4} (ab - ba)(cd - dc) $$ This expands out into four permutations of $abcd$: $$ abcd - abdc - bacd + badc $$ I won't go into the full expansion, but it results in 12 grade-0 terms (4 of which cancel), 24 grade-2 terms (8 of which cancel), and 4 grade-4 terms. The terms that don't cancel add together, thus requiring the $\frac{1}{4}$. The terms that do cancel are exactly those terms in my $(ab)(cd)$ expansion remaining after removing $(a \wedge b)(c \wedge d)$.

Thus, while the geometric product is defined to be associative, this exercise shows that associativity in action, which is what I needed.

taniwha
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  • What is your definition of "geometric product"? According to Wikipedia Geometric algebra one of the axiomatic properties of the geometric product is associativity. This is similar to the product in groups which is axiomatically assumed to be associative. – Somos Aug 22 '22 at 11:44
  • I am using $A B = A \cdot B + A \wedge B$ ($A$ and $B$ are multi-vectors). If this is wrong, then that's probably the source of my trouble, as it would mean that (⋅+∧)(⋅+∧) is not (⋅)(⋅)+(⋅)(∧)+(∧)(⋅)+(∧)(∧). The left side is two geometric products between two vectors, the right side is my understanding of the expansion. – taniwha Aug 22 '22 at 13:51
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    That is the problem. That expression only holds for vectors. – Somos Aug 22 '22 at 14:50
  • Then I shall have to keep digging. At least I know what direction to look. – taniwha Aug 22 '22 at 15:01

1 Answers1

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It looks like you are missing a factor in your expansion of $ \left( {a \wedge b} \right) \left( {c \wedge d} \right) $. In general, a product of bivectors $ A B $, should have scalar, bivector, and quadvector components: $$A B = A \cdot B + {\left\langle{{ A B }}\right\rangle}_{2} + A \wedge B,$$ so for $ A = a \wedge b, B = c \wedge d $, we have $$ \left( { a \wedge b } \right) \left( { c \wedge d } \right) = \left( { a \wedge b } \right) \cdot \left( { c \wedge d } \right) + {\left\langle{{ \left( { a \wedge b } \right) \left( { c \wedge d } \right)}}\right\rangle}_{2} + \left( { a \wedge b } \right) \wedge \left( { c \wedge d } \right).$$

You can drop the braces in the final wedge, but that grade-two term should provide the missing bivector terms.

On your rationale.

If you only want to understand $(e_1 \wedge e_2)(e_2 \wedge e_3)$, it's helpful to revert to the GA axioms, two of which are:

  • Associativity
  • Contraction: $x^2 = x \cdot x$, for any vectors x.

With those, plus $x \wedge y = x y$, for vectors $x, y$ where $x \cdot y = 0$, you have

$$\begin{aligned}\left( {e_1 \wedge e_2} \right) \left( { e_2 \wedge e_3 } \right)&=\left( {e_1 e_2} \right) \left( { e_2 e_3 } \right) \\ &=e_1 e_2 e_2 e_3 \\ &=e_1 \left( { e_2 e_2 } \right) e_3 \\ &=e_1 e_3.\end{aligned}$$

As the dot and wedges of a pair of bivectors are the grade-0 and grade-4 terms respectively (by definition), you can immediately conclude that $$\left( {e_1 \wedge e_2} \right) \cdot \left( { e_2 \wedge e_3 } \right) = 0,$$ $$\left( {e_1 \wedge e_2} \right) \wedge \left( { e_2 \wedge e_3 } \right) = 0,$$ and $$ {\left\langle{{\left( {e_1 \wedge e_2} \right) \left( { e_2 \wedge e_3 } \right)}}\right\rangle}_{2} = e_1 e_3.$$

One expansion.

Here's one way to expand the product of four vectors, grouping the first and last pairs. $$\begin{aligned}a b c d &=\left( { a \cdot b + a \wedge b } \right) \left( { c \cdot d + c \wedge d } \right) \\ &=\left( { a \cdot b } \right) \left( { c \cdot d } \right) \\ &+\quad\left( { a \cdot b } \right) \left( { c \wedge d } \right)+\left( { c \cdot d } \right) \left( { a \wedge b } \right) \\ &+\quad \left( { a \wedge b } \right)\left( { c \wedge d } \right).\end{aligned}$$ The first term of this bivector-bivector product can be expanded with application of $ a b = a \cdot b + a \wedge b $ in reverse $$\begin{aligned}\left( { a \wedge b } \right)\left( { c \wedge d } \right)&=\left( { a b - a \cdot b } \right)\left( { c \wedge d } \right) \\ &=a b \left( { c \wedge d } \right) - \left( { a \cdot b } \right) \left( { c \wedge d } \right) \end{aligned}$$ The product of $ b $ with $ c \wedge d $ is $$\begin{aligned}b \left( { c \wedge d } \right) &=b \cdot\left( { c \wedge d } \right) +b \wedge\left( { c \wedge d } \right) \\ &=\left( { b \cdot c } \right) d- \left( { b \cdot d } \right) c+ b \wedge c \wedge d.\end{aligned}$$ Multiplying this by $ a $ on the left, we have a scalar grade $$ \left\langle{{ a b \left( { c \wedge d } \right) }}\right\rangle= \left( { b \cdot c } \right) \left( { a \cdot d } \right) - \left( { b \cdot d } \right) \left( { a \cdot c } \right),$$ a bivector grade $$\begin{aligned}{\left\langle{{ a b \left( { c \wedge d } \right) }}\right\rangle}_{2}&=\left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right)+ a \cdot \left( { b \wedge c \wedge d } \right) \\ &=\left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right) \\ &\quad + \left( { a \cdot b } \right) \left( { c \wedge d } \right) - \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right),\end{aligned}$$ and a grade-four component $${\left\langle{{ a b \left( { c \wedge d } \right) }}\right\rangle}_{4}= a \wedge b \wedge c \wedge d.$$ Putting all the pieces together, here are all the grades of the $ a b c d $ product $$\begin{aligned} \left\langle{{( a b)( c d) }}\right\rangle &=\left( { a \cdot b } \right) \left( { c \cdot d } \right) +\left( { b \cdot c } \right) \left( { a \cdot d } \right) - \left( { b \cdot d } \right) \left( { a \cdot c } \right) \\ {\left\langle{{ (a b)( c d) }}\right\rangle}_{2} &= \left( { a \cdot b } \right) \left( { c \wedge d } \right)+ \left( { c \cdot d } \right) \left( { a \wedge b } \right) - \left( { a \cdot b } \right) \left( { c \wedge d } \right) \\ &\quad+ \left( { b \cdot c } \right) \left( { a \wedge d } \right)- \left( { b \cdot d } \right) \left( { a \wedge c } \right)+ \left( { a \cdot b } \right) \left( { c \wedge d } \right) \\ &\quad- \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right) \\ &= \left( { a \cdot b } \right) \left( { c \wedge d } \right)+ \left( { c \cdot d } \right) \left( { a \wedge b } \right) + \left( { b \cdot c } \right) \left( { a \wedge d } \right) \\ &\quad - \left( { b \cdot d } \right) \left( { a \wedge c } \right)- \left( { a \cdot c } \right) \left( { b \wedge d } \right) + \left( { a \cdot d } \right) \left( { b \wedge c } \right) \\ {\left\langle{{ (a b)( c d) }}\right\rangle}_{4} &= a \wedge b \wedge c \wedge d.\end{aligned},$$

which matches your result, finishing your experimental proof of associative multiplication for a product of four vectors.

Peeter Joot
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    Thanks. This was quite helpful (along with somos's comments). I've found and, more importantly, understood the expansion for $a_1 \wedge a_2$, applied it to both halves of the product expanded it out and I'm now working through cancellations etc of my $abcd$ expansion. I can see where the missing bivectors come from, though I haven't yet verified the results, but that's just more work (probably a lot, since I tend to have to do these things three or more times to get them right). – taniwha Aug 23 '22 at 02:37
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    I've finished my cancellations and found that my $abcd$ expansion is suspect, but I at least now can see where the missing wedge products come from.Also, thanks for the edits. – taniwha Aug 23 '22 at 03:33
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    Even if my $abcd$ expansion is not correct, I still get $e_1 \wedge $e_2$ out of that mess, so progress. When I get everything sorted, I'll post my workings. – taniwha Aug 23 '22 at 03:41
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    All sorted. The reason I suspected my $abcd$ expansion was I had expanded only $(a \wedge b)(c \wedge d)$, thus I was missing three of the grades 0 and 2 terms from the $(ab)(cd)$ expansion ($(a \cdot b)(c \cdot d) + (a \cdot b)(c \wedge d) + (a \wedge b)(c \cdot d)$), but when I double checked my $abcd$ expansion, I checked my $(ab)(cd)$ expansion and found those terms were precisely those that had cancelled out in the $(a \wedge b)(c \wedge d)$ expansion. I can see why other than for understanding, geometric algebra is best not done by hand :) – taniwha Aug 23 '22 at 05:14
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    added a note about the "My digging ..." part of your question. – Peeter Joot Aug 23 '22 at 15:05
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    I've added a follow-up to my question summarizing the resolution of my confusion. – taniwha Aug 24 '22 at 03:08
  • I'd tried this expansion by hand too, and messed it up the first time. Tried again, and got it right this time, so typed it up above for completeness. – Peeter Joot Aug 24 '22 at 21:56
  • I need to go through it again, but for me, it was the $(a \circ b)(c \circ d)$ terms that cancelled ($\circ$ is either $\cdot$ or $\wedge$). – taniwha Aug 24 '22 at 23:29
  • I've gone through it again, and I have confirmed that any term with $a \cdot b$ or $c \cdot d$ gets cancelled, which makes sense as there's no path in $(a \wedge b)(c \wedge d)$ for either of those two dot products to appear. – taniwha Aug 25 '22 at 00:14