CliffordAlgebraIdempotents

Last edit December 18, 2005

Idempotents are important in CliffordAlgebra. They are quantities which square to themselves i.e. for some p

	p^2 = p

The obvious ones are 0 and 1, but there are others. Consider a unit vector, which I will call e1, though I would prefer the 1 to be a suffix.

	e1^2 = 1

so that is not idempotent. Instead look at p1 = (1 + e1)/2, which is idempotent.

	p1^2 = ((1 + e1)/2)^2 = (1 + e1)(1 + e1)/4 = (1 + 2e1 +1)/4 = (1 + e1)/2 = p1

Now try p2 = (1 - e1)/2, which is also idempotent.

	p2^2 = ((1 - e1)/2)^2 = (1 - e1)(1 - e1)/4 = (1 - 2e1 +1)/4 = (1 - e1)/2 = p2

Now try

	p1 p2 = (1 + e1)(1 - e1)/4 = (1 + e1 - e1 -1)/4 = 0

so that the product of p1 and p2 is zero. This makes the BinomialTheorem, as usually understood, very simple. If a and b are scalar constants then

	(a p1 + b p2)^n = a^n p1 + b^n p2

as all the other terms are zero, containing at least one of each of p1 and p2.

What is the use of all this? Well e1 can be represented as the difference of p1 and p2.

	e1 = p1 - p2

so that the previous result can be used to show the structure of the powers of e1.

	e1^2 = (p1 - p2)^2 = 1^2 p1 + (-1)^2 p2 = p1 + p2 = 1

This can be extended in various ways. It works for negative powers, where the expression in terms of the BinomialTheorem

	e1^-1 = (p1 - p2)^-1	 -- (1)

cannot be manipulated directly because neither p1 nor p2 can be inverted. However, e1^-1 can be evaluated, see below, but not following directly from (1), giving the result

	e1^-1 = 1^-1 p1 + (-1)^-1 p2 = p1 - p2 = e1

showing that e1 is its own inverse.

I have an extended version of this called Vector Exponential on my web site: http://www.ceac.aston.ac.uk/research/staff/jpf/clifford/vecexp/index.php

Proof of why it works for negative power -1

As e1^2 = 1 the characteristic polynomial of e1 is

  m^2 = 1

which has roots m1 = +1 and m2 = -1

In fact (see Snygg) p1 and p2 are the eigenfunctions of e1 given by

  p1 = (e1 - m2) / (m1 - m2) = (1 + e1)/2
  p2 = (e1 - m1) / (m2 - m1) = (1 - e1)/2

and also

  p1 + p2 = 1				--- (1)
  m1p1 + m2p2 = e1			--- (2)

Multiplying (2) on the left by p1, right multiplying by e1^-1 and dividing by m1 (scalar and nonzero), and using p1 p2 = 0, then

  p1 e1^-1 = p1 e1 e1^-1/m1 = p1/m1	--- (3)

Similarly, multiplying (2) on the left by p2, right multiplying by e1^-1 and dividing by m2 (scalar and nonzero)

  p2 e1^-1 = p2 e1 e1^-1/m2 = p2/m2	--- (4)

Adding (3) and (4)

 (p1+p2) e1^-1 =  p1/m1 + p2/m2

and because of (1), the result is

 e1^-1 =  p1/m1 + p2/m2

Check

  e1 e1^-1 =  (m1p1 + m2p2) (p1/m1 + p2/m2) =  (m1/m1)p1 + (m2/m2)p2 = p1 + p2 = 1

Reference: John Snygg "Functions of Clifford Numbers or Square Matrices", Chapter 8 in Dorst Doran and Lasenby (eds) "Applications of Geometric Algebra in Computer Science and Engineering", Birkhauser, 2002, ISBN 0817642676

He shows how to find the expression of a Clifford Variable as a sum of multiples of idempotents.

This all depends on something which Hestenes mentions, in HestenesOerstedMedalLecture at the bottom of page 16, as a misleading piece of teaching: "it is meaningless to add scalars to vectors". This was one of the comments I had when I circulated the Vector Exponential to my colleagues as a Christmas greeting.

-- JohnFletcher