Geometric Algebra Primer

Understanding the Denominator in Versor Equations

To understand this, we have to start by realizing that the denominator is always a scalar because:

† A₁² + A₂² + ... + A_k² = |A₁|² + |A₂|² + ... + |A_k|²

And since a scalar divided by itself equals one, this means that:

†† A^-1 A = 1

Furthermore, it also proves that the left and right inverse are the same. Division by a scalar α is multiplication by 1/α, which is, according to equation (3.3) commutative, proving that the left and right inverses of a versor are indeed equal.

This means that for a versor A, we have A^-1 = A^-1 = A^-1 and therefore the following:

A^-1 A = AA^-1 = A^-1 A = AA^-1 = 1

It is important to notice that in the case of vectors, the scalar represents the squared magnitude of the vector. As a consequence, the inverse of a unit vector is equal to itself.

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represent planes. Geometricalgebra introduces bivectors which can be used for the same purpose. By usingthe pseudoscalar we can get an understanding of the relationship between thetwo representations.∗The dual A of a multivector A is defined as follows:∗ −1A = AI (4.3)where I represents the pseudoscalar of the geometric algebra that is being used.The pseudoscalar is a blade (the blade with highest grade) and therefore its leftand right inverse are the same, and hence the above formula is not ambiguous.Let us consider a simple example in C` , calculating the dual of the basis3bivector e . The pseudoscalar is e . Pseudoscalars are blades and thus ver-12 123sors. You can check yourself that its inverse is e e e . We’ll use this to calculate3 2 1the dual of e :12 ∗e = e e e e12 3 2 112 = e e e e e1 2 3 2 1= −e e e e e1 3 2 2 1= −e e e1 3 1= e e e1 1 3= e 3 (4.4)Thus, the dual is basis vector e , which is exactly the normal vector of

basis3bivector e . In fact, this is true for all bivectors. If we have two arbitrary12vectors a and b ∈ C` :

a = α e₁ + α e₂ + α e₃

b = β e₁ + β e₂ + β e₃

According to equation (2.7) their outer product is:

a ∧ b = (α β - α β )e₁₂ + (α β - α β )e₁₃ + (α β - α β )e₂₃

And its dual (a ∧ b) becomes:

-1*(a ∧ b) =(a ∧ b)e₁₂₃ =(a ∧ b)e₁ e₂ e₃

=((α β - α β )e₁₂ + (α β - α β )e₁₃ + (α β - α β )e₂₃)e₁ e₂ e₃

=(α β - α β )e₁ e₂ e₃ +1(α β - α β )e₁₂ e₃ e₁ +1(α β - α β )e₁₃ e₃ e₁ +1(α β - α β )e₂₃ e₃ e₂

=(α β - α β )e₁ e₂ e₃ +1(α β - α β )e₁₂ e₃ e₁ +1(α β - α β )e₁₃ e₃ e₁ +1(α β - α β )e₂₃ e₃ e₂

• − α β )e e e e e +1 3 3 1 1 3 3 2 1(α β − α β )e e e e e2 3 3 2 2 3 3 2 1=(α β − α β )e − (α β − α β )e + (α β − α β )e1 2 2 1 3 1 3 3 1 2 2 3 3 2 1=(α β − α β )e + (α β − α β )e + (α β − α β )e (4.5)2 3 3 2 1 3 1 1 3 2 1 2 2 1 3
• Which is exactly the traditional cross product. We conclude that in three dimensions, the dual of a bivector is its normal. The dual can be used to convert between bivector and normal representations. But the dual is even more, because it is defined for any multivector.
• 4.5 Projection and Rejection
• If we have a vector a and bivector B we can decompose a in two parts. One part a that is collinear with B. We call this the projection of a onto B. The|| B 2 other part is a , and orthogonal to B. We call this the rejection of a from ⊥ BB.

vector a onto abivector B. By definition, the inner and outer product of respectively orthogonaland collinear blades are zero. In other words, the inner product of a vectororthogonal to a bivector is zero: a · B = 0 (4.8)⊥B. Likewise the outer product of a vector collinear with a bivector is zero: a ∧ B = 0 (4.9)|| B. Let’s see what happens if we multiply the orthogonal part of vector a withbivector B: a · B = a ∧ B⊥ ⊥ ⊥B B B = a ∧ B using equation (4.8)⊥B = a ∧ B + a ∧ B equation (4.9)⊥ ||B B = (a + a) ∧ B equation (2.5)⊥ ||B B = a ∧ B equation (4.6). Thus, the perpendicular part of vector a times bivector B is equal to theouter product of a and B. Now all we need to do is divide both sides of theequation by B to obtain the perpendicular part of a: a · B = a ∧ B⊥B−1 −1a B = (a ∧ B)B⊥ B −1a = (a ∧ B)B⊥ B 32. Notice that there is no ambiguity in using the inverse

because B is a blade orversor, and its left and right inverses are therefore the same. The conclusion is: −1a = (a ∧ B)B (4.10)⊥ B

Calculating the collinear part of vector a follows similar steps: a B = a · B + a ∧ B

|| || ||B B B= a · B|| B= a · B + a · B⊥

|| BB= (a + a ) · B⊥ ||B B= a · B

Again multiply both sides with the inverse bivector: −1 −1a BB = (a · B)B|| B

To conclude: −1a = (a · B)B (4.11)|| B

Using these definitions, we can now confirm that a + a = a⊥|| BB−1 −1 and (4.10)a + a = (a ∧ B)B + (a · B)B equation (4.11)⊥|| BB −1= (a ∧ B + a · B)B equation (3.3)−1= aBB equation (4.7)= a by definition of the inverse

334.6 Reflections

Armed with a way of decomposing blades in orthogonal and collinear parts we can take a look at reflections. We will get ahead of ourselves and take a3specific look at the geometric algebra of the Euclidian

space R denoted with∗C` . Suppose we have a bivector U . Its dual U will be the normal vector u.3What if we multiply a vector a with vector u, projecting and rejecting a ontoU at the same time: ua = u(a + a )⊥|| UU= ua + ua ⊥|| UUUsing (3.2) we write it in full:ua = (u · a + u ∧ a ) + (u · a + u ∧ a )⊥ ⊥|| || U UU UNote that (because u is the normal of U ) the vectors a and u are perpen-|| Udicular. This means that the inner product a · u equals zero. Likewise, the|| Uvectors a and u are collinear. This means that the outer product a ∧ u⊥ ||U Uequals zero. Removing these two 0 terms:ua = u ∧ a + u · a ⊥|| UU= u · a + u ∧ a⊥ ||U URecall that the inner product between two vectors is commutative, and the outerproduct is anticommutative, so we can write:ua = a · u − a ∧ u⊥ ||U UWe can now insert those 0-terms back in (putting in the form of equation (3.2)):ua = (a