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The Geometric Algebra of 3D Euclidean Space


Let e1, e2, e3 be an orthonormal basis for 3 dimensional euclidean space. This means that e1, e2, e3 are three mutually orthogonal vectors each of them of length 1. We want to define, besides the usual operations of addition of vectors and scalar multiplication, a new operation of multiplication of vectors. We would like this new product to be associative, and distributive with respect to the sum of vectors.

We need only to define all possible products of the basis vectors and then consider all the linear combinations of them. We call the set of all the linear combinations of all the products of the vectors e1, e2, e3 the geometric algebra of the three dimensional space spanned by e1, e2, e3.

We define:

  • e1 e1 = 1
  • e2 e2 = 1
  • e3 e3 = 1
  • e1 e2 = - e2 e1
  • e1 e3 = - e3 e1
  • e2 e3 = - e2 e3
  • e1 e2 e3 = (e1 e2) e3 = e1 (e2 e3)
In other words, each basis vector is required to square to 1 and to anti-commute with the others. Notice that by taking products of the basis vectors we get out of the vector space generated by them. In particular we generate the unit scalar and we shall prove that also two new kinds of objects different from scalars and vectors are obtained. Objects of the type e1 e2 will be called bivectors and the ones of the type e1 e2 e3 will be called trivectors for obvious reasons.

Multiplication by scalar and addition are extended from their domain in the original 3D vector space to all the objects of the geometric algebra and all the usual properties for these operations are kept valid. In particular scalars are assumed to commute with ALL the elements of the geometric algebra and the unit scalar 1 to be the identity element for the geometric product.

Let us show some of the consequences of the previous assumptions. First notice, that all the scalars (generated from 1) and all the vectors (generated by e1, e2, e3) are part of the geometric algebra. Let a be a vector i.e.

a = x e1 + y e2 + z e3

then by the rules above:

a a = ( x e1 + y e2 + z e3) (x e1 + y e2 + z e3)
= x^2 + y^2 + z^2 = innerprod(a,a)

and we have that the square of every vector is the same as the positive scalar equal to the square of its length. In particular if we add two vectors a and b we obtain another vector (a + b) whose square is:

(a + b) (a + b) = a^2 + b^2 + ab + ba

and as we know, the square of every vector equals the innerprod of the vector with itself we obtain, using the linearity of innerproducts, that,

a^2 + b^2 + 2 innerprod(a,b) = a^2 + b^2 + ab + ba

From where we deduce that:

innerprod(a,b) = ( ab + ba ) / 2

So if we didn't know about the innerproduct we could have defined it as above, as the symetrized geometric product of two vectors. Clearly,

ab = ( ab + ba ) / 2 + ( ab - ba ) / 2

The first term is the innerproduct and the second, which is the anti-symmetrized product of a with b is an object of the type "bivector". We call this new product the outer or wedge product of a with b. Thus,

ab = innerprod(a,b) + wedge(a,b)

Notice that as a special case we have:

e1 e2 = wedge(e1,e2) = e12

The last equality is just the definition of a new notation that allows me to save a lot of typing. We use simmilar notations for the other outer products of basis elements: e23, e31, e32.. etc. Now that it should be clear that e1 is a vector we drop the bold-face type and we simply write e1, e2, e3.

We now state our first proposition.

The set

{ 1, e1, e2, e3, e23, e31, e12, e123 }

is a basis for the linear (vector) space generated by taking all the possible geometric products of the vectors e1,e2,e3, and all the linear combinations of these products. Thus, every element in the geometric algebra has a unique expression as a linear combination of the elements in this basis,

A = u0 + u1 e1 + u2 e2 + u3 e3 + v1 e23 + v2 e31 + v3 e12 + v0 e123

where A is an arbitrary element of the algebra (we call it a multivector) and the coefficients u0, u1,...,v3,v0 are scalars.

We only need to show that the 8 elements of the set are linearly independent for we already know that they span the algebra. So let us take a general multivector A = 0 and show that this is only possible when all the scalar coefficients u0,...,v0 are 0. To show this, first notice that by taking linear combinations of vectors we always get another vector or perhaps the scalar 0, but never a bivector or a trivector. The same is true for scalars, bivectors and trivectors. By multiplying an object of one type by a scalar and adding this to an object of the same type we can get either the scalar 0 or another object of the same type. Objects of different grades don't mix by taking linear combinations. We say that the geometric algebra is a "graded algebra". Hence, we need only to prove that the 3 vectors are l.i. and the three bivectors are l.i. If

u1 e1 + u2 e2 + u3 e3 = 0

then by multiplying through on the left by e1 we get,

u1 + u2 e12 - u3 e31 = 0

and since the bivectors don't mix with the scalars the only way a (scalar) + (bivector) = 0 is that

u1 = 0 and u2 e12 - u3 e31 = 0

had we multiplied by e2 or e3 we would have obtained u2 =0 or u3 = 0. So if the three vectors e1, e2, e3 square to 1 and anticommute they must be l.i. In a simmilar way we show that the three bivectors, e23, e31, e12 are l.i. From

v1 e23 + v2 e31 + v3 e12 = 0

we can deduce that v1 = v2 = v3 = 0. Multiplying through on the left by e32, we obtain

v1 e32 e23 + v2 e32 e31 + v3 e32 e12 = 0

but e32 e23 = e3 e2 e2 e3 = e3 (e2 e2) e3 = e3 (1) e3 = e3 e3 = 1. Also e32 e31 = - e12, and e32 e12 = e31 so the above equation becomes,

v1 - v2 e12 + v3 e31 = 0

from where we deduce that v1 = 0. If instead of multiplying by e32 we multiply by e13 we get v2 = 0, and multiplying by e21 we obtain v3 = 0. We conclude that the three bivectors are in fact l.i. and the whole geometric algebra as a linear space has dimension: 1 + 3 + 3 + 1 = choose(3,0) + choose(3,1) + choose(3,2) + choose(3,3) = 2^3 = 8.

Carlos Rodriguez <>
Last modified: Sun Aug 4 09:53:13 EDT 1996