I am trying to learn Geometric Algebra from the textbook by Doran and Lasenby.

They claim in chapter 4 that the geometric product $ab$ between two vectors $a$ and $b$ is defined according to the axioms i) associativity: $(ab)c = a(bc) = abc$ ii) distributive over addition: $a(b+c) = ab+ac$ iii) The square of any vector is a real scalar

Then they claim that the inner and outer product are defined as $ a \cdot b = \frac{1}{2} (ab+ba) $ $ a \wedge b = \frac{1}{2} (ab-ba) $ so that $ a b = a \cdot b + a \wedge b $

My problem is that if you are given two vectors, say $ a = 1e_1 + 3 e_2 - 2e_3 $ $ b = 5e_1 -2 e_2 + 1e_3 $

How to actually compute $ab$?

I mean, you then have to specify how either $ a \cdot b $ and $ a \wedge b $ works, or how (in detail) $ a b $ are to be performed.

This is in my view, circular.

From another point of view, say that you start backwards by defining the inner and outer product, and then define the geometric product as $ a b = a \cdot b + a \wedge b $

Then how to show that the geometric product is associative? The usual definition of the outer product is associative, but the usual definition of the inner (dot) product is NOT associative. So how to show that the geometric product is associative if you take the inner and outer product as starting point for the geometric product?

Thank you very much in advance for any kind of feedback

  • 1
    $\begingroup$ See whether math.stackexchange.com/questions/444988/… answers your question. $\endgroup$ – Gerry Myerson Jul 8 '16 at 9:14
  • $\begingroup$ In particular , see the A by Andrey Sololov in the link provided in the above comment by Gerry Myerson. $\endgroup$ – DanielWainfleet Jul 12 '16 at 3:09
  • $\begingroup$ I think one possible idea is to reduce everything to computations involving orthonormal basis vectors using linearity/distributivity. $\endgroup$ – Chill2Macht Aug 5 '16 at 1:00

You use axioms i, ii, and iii to compute the geometric product. The dot and wedge products can then be computed.

Consider $u = 3e_1 + 2e_2$ and $v = -e_1 + 4e_2$.

Then, by axioms (i) and (ii), we can write

$$uv = 3e_1 e_1 + 12 e_1 e_2 + 2 e_2 e_1 + 8 e_2 e_2.$$

This distributes over addition and drops parentheses, as they would be redundant or unnecessary.

Now, using axiom (iii), we can evaluate the products $e_1 e_1 = e_2 e_2 = 1$ to get

$$uv = 3 + 12 e_1 e_2 + 2 e_2 e_1 + 8.$$

Now, to simplify this further, we typically use a derived result: let $x,y$ be vectors, and consider the geometric product

$$(x+y)(x+y) = xx + xy + yx + yy$$

Let's amend axiom (iii) somewhat: it is not enough that the product of a vector with itself be a scalar. Rather, we presume the presence of some symmetric bilinear form (usually positive definite, but in pseudo-Riemannian geometry, this condition is relaxed), so there must exist some map $g: V\times V \to K$, and $xx = g(x,x)$ for any $x$.

Hence, we can consider the case in which $g(x,y) = 0$, or $x \perp y$ in other words. Then

$$(x+y)(x+y) = g(x+y, x+y) = g(x,x) + g(y,y) = xx + yy$$

Hence, we conclude that, when $x \perp y$ under $g$, $xy = -yx$. This result is often used in simplifying geometric products. Applying it to our original problem yields

$$uv = 11 + 10 e_1 e_2$$

and in turn,

$$vu = 11 - 10 e_1 e_2$$

from which the dot and wedge products can be computed.

However, unlike Doran and Lasenby, I generally prefer not to use those definitions of the products. I usually compute the dot and wedge products in terms of grade projections. The symmetry properties change with grades, which makes those rules almost meaningless in my opinion.

  • $\begingroup$ Is there a stray minus sign in your definition of $v$? $\endgroup$ – Don Hatch Nov 7 '18 at 9:15

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