geometric algebra: dot products of vectors with vectors vs bivectors

Question

The book "Matrix Gateway to Geometric Algebra" by Garret Sobczyk gives the following identity:

$a\cdot(b \wedge c)=(a\cdot b)c - (a\cdot c)b$

Using the identities

$a\cdot b= (ab+ba)/2$

$a\wedge b=(ab-ba)/2$

to expand the proposed identity, including the step, e.g., $$a\cdot (b\wedge c)=(a(b\wedge c) + (b\wedge c) a)/2$$ I get $$abc + bca - acb - cba = abc + cba - acb - bca$$ which does not appear to check.

Is there a step I am missing?

On the same page 31 the author writes (equation 2.14)

$$a\cdot(b\wedge c) = (a(b\wedge c) - (b\wedge c)a)/2$$

which is the opposite sign to the basic "identity" I relied upon. Why is there sign minus in this instance? Is not $x=(b\wedge c)$, where x is an element, subject to the identity $a\cdot x = (ax+xa)/2$? There is a similar reversal of sign in equation 2.15.

Thanks,

Gary

Answer 1

The identities

$$a\cdot b= (ab+ba)/2$$

$$a\wedge b=(ab-ba)/2$$

are correct, but they are only for vectors $ a, b $. Eq 2.14 from the text:

$$a\cdot(b\wedge c) = (a(b\wedge c) - (b\wedge c)a)/2$$

(with the negative sign), is also correct. In general, if $ M $ is a k-vector, and $ a $ is a vector, then one has

$$a \cdot M=\frac{1}{{2}} \left( { a M + (-1)^{k-1} M a } \right).$$

I don't know how the Matrix Gateway book derives eq 2.14. Here's one way to show it, rewriting the dot product as a grade 1 selection (i.e. take just the vector parts of any multivector products) $$\begin{aligned}a \cdot \left( { b \wedge c } \right)&={\left\langle{{ a \left( { b \wedge c } \right) }}\right\rangle}_{1} \\ &={\left\langle{{ a \left( { b c - b \cdot c } \right) }}\right\rangle}_{1} \\ &={\left\langle{{ a b c }}\right\rangle}_{1} - a \left( { b \cdot c } \right),\end{aligned}$$ however, using $ b a = 2 a \cdot b - b a $, we have $$\begin{aligned}{\left\langle{{ a b c }}\right\rangle}_{1}&={\left\langle{{ \left( { 2 a \cdot b - b a } \right) c }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - {\left\langle{{ b a c }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - {\left\langle{{ b \left( { 2 a \cdot c - c a } \right) }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b + {\left\langle{{ b c a }}\right\rangle}_{1} \\ &=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b + \left( { b \wedge c } \right) \cdot a + \left( { b \cdot c } \right) a.\end{aligned}$$

Putting the pieces together, we have $$a \cdot \left( { b \wedge c } \right)=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b + \left( { b \wedge c } \right) \cdot a.$$ Finally, note that the reverse of vector is a vector ($\tilde{a} = a$), so if we apply the reversion operator to the last term: $$\begin{aligned}\left( { b \wedge c } \right) \cdot a&=a \cdot \left( { c \wedge b } \right) \\ &=-a \cdot \left( { b \wedge c } \right),\end{aligned}$$ and then substitute this back in, we are most of the way towards a derivation of eq 2.14 from the text: $$a \cdot \left( { b \wedge c } \right)=2 \left( { a \cdot b } \right) c - 2 \left( { a \cdot c } \right) b - a \left( { b \wedge c } \right),$$ so after some trivial rearrangement, we have: $$a \cdot \left( { b \wedge c } \right)=\left( { a \cdot b } \right) c - \left( { a \cdot c } \right) b.$$

You could also derive this from 2.14, using the reversion argument above, since the trivector terms in that antisymmetric sum cancel.

Answer 2

Assuming that we are given a multiplication that is associative and bilinear, but not commutative, then define

$$a\cdot b := \frac12(ab+ba),\qquad a\wedge b := \frac12(ab-ba),$$

which apply, at least for $1$-vectors, with perhaps some sign changes for general $k$-vectors. No equation such as $$a\cdot(b \wedge c)=(a\cdot b)c - (a\cdot c)b$$ can be an identity because, as you computed, the left side is equal to $\,\frac14(abc + cba - acb - bca)\,$ while $\, (a\cdot b)c = \frac12(abc+bac)\,$ and $\, a(b\cdot c) = \frac12(abc+acb).\,$ There is no way to get $\,\frac12(abc+cba)\,$ no matter what scalar factors are introduced. The proof is by linear algebra using as the basis $$(abc,acb,bac,bca,cab,cba).$$