Introduction to Geometric Algebra

Introduction to Geometric Algebra

A computational introduction to GeometricAlgebra

Juan M. Bello Rivas

ETSIT, Grupo de Tratamiento de ImágenesUniversidad Politécnica de Madrid


Outline

1 Geometric Algebra and geometric productsInner product

Applications of the inner productOuter product

Applications of the outer productGeometric product

Reduction formulaLaplace expansion of the inner productInner and outer products between multivectorsReversionMagnitude

2 Applications of Geometric AlgebraAlgebra of the Euclidean planeReflectionsRotations


Geometric Algebra and geometric products

About Geometric Algebra

It is a mathematical framework for expressing geometricalideas and doing computations.

It is being increasingly used in physics and in computer visionapplications.Its main advantages are:

Coordinate-free Geometrical ideas can be expressedcompactly without having to consider coordinatesand bases.

Extensible The concepts are extended effortlessly to spacesof arbitrary dimensions.

General It generalizes complex numbers, quaternions,Plücker coordinates, etc.




It is a mathematical framework for expressing geometricalideas and doing computations.It is being increasingly used in physics and in computer visionapplications.Its main advantages are:




















Hamilton, Grassmann and Clifford

Figure: William Rowan Hamilton, Hermann Grassmann, and WilliamKingdon Clifford



Vectors and oriented space segments

A vector can be understood as an oriented line segment.

Thekey idea behind Geometric Algebra is extending thisinterpretation of a vector to higher dimensions using the socalled geometric product.



Vectors and oriented space segments

A vector can be understood as an oriented line segment. Thekey idea behind Geometric Algebra is extending thisinterpretation of a vector to higher dimensions using the socalled geometric product.



Inner product

Outline








Inner product

Inner product

The “inner product” of two directed line segments a and b canbe defined as the oriented line segment obtained by dilating theprojection of a on b by the magnitude of b. Both magnitude andorientation are scalars.

Figure: Perpendicular projection (symmetry of the inner product).



Inner product

Inner product

The “inner product” of two directed line segments a and b canbe defined as the oriented line segment obtained by dilating theprojection of a on b by the magnitude of b. Both magnitude andorientation are scalars.

Figure: Perpendicular projection (symmetry of the inner product).



Inner product

Inner product

cos θ =OEOA

=a · b|a|

cos θ =ODOB

=a · b|b|



Inner product

Inner product

cos θ =OEOA

=a · b|a|

cos θ =ODOB

=a · b|b|



Inner product

Inner product

The inner product describes the relative directions of twovectors a and b.

a · b = |a||b| cos θ

where θ is the angle between a and b.

After the algebraic properties for the inner product have beenestablished, it can be seen as an abstract rule relating scalarsto vectors.



Inner product

Inner product

The inner product describes the relative directions of twovectors a and b.

a · b = |a||b| cos θ

where θ is the angle between a and b.After the algebraic properties for the inner product have beenestablished, it can be seen as an abstract rule relating scalarsto vectors.



Inner product

Applications. Law of Cosines

The inner product can be used to compute angles and lengthsof line segments.

It also illustrates, how once a suitable notation is found,theorems that were once hard to prove become apparent. Forexample, the “Law of Cosines” can be easily derived by takinginto account that the equation a + b = c completelycharacterizes a triangle.



Inner product

Applications. Law of Cosines

The inner product can be used to compute angles and lengthsof line segments.It also illustrates, how once a suitable notation is found,theorems that were once hard to prove become apparent. Forexample, the “Law of Cosines” can be easily derived by takinginto account that the equation a + b = c completelycharacterizes a triangle.



Inner product

Law of Cosines

From there,

c · c = (a + b) · (a + b)

= a · (a + b) + b · (a + b)

= a · a + b · b + a · b + b · a

which yields |c|2 = |a|2 + |b|2 + 2a ·b. Denoting a = |a|, b = |b|,c = |c|, and a · b = −ab cos C we obtain the “Law of Cosines”:

c2 = a2 + b2 − 2ab cos C



Inner product

Law of Cosines

From there,

c · c = (a + b) · (a + b)

= a · (a + b) + b · (a + b)

= a · a + b · b + a · b + b · a


c2 = a2 + b2 − 2ab cos C



Inner product

Law of Cosines

From there,

c · c = (a + b) · (a + b)

= a · (a + b) + b · (a + b)

= a · a + b · b + a · b + b · a


c2 = a2 + b2 − 2ab cos C



Inner product

Law of Cosines

From there,

c · c = (a + b) · (a + b)

= a · (a + b) + b · (a + b)

= a · a + b · b + a · b + b · a

which yields |c|2 = |a|2 + |b|2 + 2a ·b.

Denoting a = |a|, b = |b|,c = |c|, and a · b = −ab cos C we obtain the “Law of Cosines”:

c2 = a2 + b2 − 2ab cos C



Inner product

Law of Cosines

From there,

c · c = (a + b) · (a + b)

= a · (a + b) + b · (a + b)

= a · a + b · b + a · b + b · a


c2 = a2 + b2 − 2ab cos C



Outer product

Outline








Outer product

Outer product

Gives an algebraic expression to the fact that two non-collineardirected line segments determine a parallelogram.

Theparallelogram can be seen as a kind of “geometrical product” ofthe vectors determining its sides.Building on the principle that a vector characterizes a directedline segment, bivectors (also called 2-vectors) are introduced tocharacterize the notion of a directed plane segment.



Outer product

Outer product

Gives an algebraic expression to the fact that two non-collineardirected line segments determine a parallelogram. Theparallelogram can be seen as a kind of “geometrical product” ofthe vectors determining its sides.

Building on the principle that a vector characterizes a directedline segment, bivectors (also called 2-vectors) are introduced tocharacterize the notion of a directed plane segment.



Outer product

Outer product

Gives an algebraic expression to the fact that two non-collineardirected line segments determine a parallelogram. Theparallelogram can be seen as a kind of “geometrical product” ofthe vectors determining its sides.Building on the principle that a vector characterizes a directedline segment, bivectors (also called 2-vectors) are introduced tocharacterize the notion of a directed plane segment.



Outer product

Outer product

1 A point moving a direction and distance given by a vector asweeps out a directed line segment.

2 And the points on this line segment, each moving adistance and direction specified by a vector b sweep out aparallelogram.

3 The bivector B corresponding to this parallelogram isuniquely determined by this construction and it might bewritten as B = a ∧ b (a kind of “product” of two vectors).

4 The outer product of a and b is denoted by the bivectora ∧ b.

5 Now note that the parallelogram obtained by “sweeping balong a” differs only in orientation from the parallelogramobtained by “sweeping a along b”.

This is expressed bywriting b ∧ a = −a ∧ b = −B.



Outer product

Outer product









Outer product

Outer product









Outer product

Outer product









Outer product

Outer product









Outer product

Outer product









Outer product

Outer product





5 Now note that the parallelogram obtained by “sweeping balong a” differs only in orientation from the parallelogramobtained by “sweeping a along b”. This is expressed bywriting b ∧ a = −a ∧ b = −B.



Outer product

Outer product

Figure: a ∧ b and b ∧ a



Outer product

Following this construction, it is clear that

b ∧ a = a ∧ (−b)

= (−b) ∧ (−a) = (−a) ∧ b

The magnitude of the bivector a ∧ b is the area of thecorresponding parallelogram and is denoted by |a ∧ b|.



Outer product


b ∧ a = a ∧ (−b) = (−b) ∧ (−a)

= (−a) ∧ b




Outer product


b ∧ a = a ∧ (−b) = (−b) ∧ (−a) = (−a) ∧ b




Outer product

A few properties become apparent from these definitions. IfB,C are bivectors and λ is a scalar, it’s true that:

1 |B| = |a ∧ b| = |b ∧ a| = | − B|

2 C = λB implies |C| = |λ||B|3 λ (a ∧ b) = (λa) ∧ b = a ∧ (λb)

4 a ∧ b = 0 iff b = λa (collinearity).5 The distributive rule

a ∧ (b + c) = a ∧ b + a ∧ c



Outer product


1 |B| = |a ∧ b| = |b ∧ a| = | − B|2 C = λB implies |C| = |λ||B|

3 λ (a ∧ b) = (λa) ∧ b = a ∧ (λb)


a ∧ (b + c) = a ∧ b + a ∧ c



Outer product


1 |B| = |a ∧ b| = |b ∧ a| = | − B|2 C = λB implies |C| = |λ||B|3 λ (a ∧ b) = (λa) ∧ b = a ∧ (λb)


a ∧ (b + c) = a ∧ b + a ∧ c



Outer product



4 a ∧ b = 0 iff b = λa (collinearity).

5 The distributive rule

a ∧ (b + c) = a ∧ b + a ∧ c



Outer product




a ∧ (b + c) = a ∧ b + a ∧ c



Outer product

The construction of bivectors can be extended to trivectors andso on. It is then that the associativity of the outer productappears

(a ∧ b) ∧ c = a ∧ (b ∧ c)



Outer product

Also, if vectors are used to describe a 3-dimensional spacethen for any vectors a, b, c, and d

a ∧ b ∧ c ∧ d = 0

which reflects the intuitive notion that you can’t sweep a4-dimensional space segment in a 3-dimensional space.



Outer product

Applications of the outer product. Law of Sines

The outer product provides a simple expression for parallelism(a ∧ b = 0) in much the same way the inner product provides asimple expression for orthogonality (a · b = 0).

Let’s recall from elementary geometry that

|B| = |a ∧ b| = |a||b| sin θ = ab sin θ (1)

where a = |a|, b = |b| and θ is the angle between a and b.



Outer product

Applications of the outer product. Law of Sines

The outer product provides a simple expression for parallelism(a ∧ b = 0) in much the same way the inner product provides asimple expression for orthogonality (a · b = 0).Let’s recall from elementary geometry that

|B| = |a ∧ b| = |a||b| sin θ = ab sin θ (1)

where a = |a|, b = |b| and θ is the angle between a and b.



Outer product

Law of Sines

Taking successive outer products with a, b, and c to the triangleformula a + b = c, we arrive at:

0 + b ∧ a = c ∧ a

a ∧ b + 0 = c ∧ ba ∧ c + b ∧ c = 0

where the third equation redundant since it is a linearcombination of the first two.



Outer product

Law of Sines


0 + b ∧ a = c ∧ aa ∧ b + 0 = c ∧ b

a ∧ c + b ∧ c = 0




Outer product

Law of Sines


0 + b ∧ a = c ∧ aa ∧ b + 0 = c ∧ ba ∧ c + b ∧ c = 0




Outer product

Law of Sines

Now we have

a ∧ b = a ∧ c = c ∧ b

Applying 1 we get

ab sin C = ac sin B = cb sin A

Finally, dividing by abc and rearranging the terms in theequalities we have obtained the “Law of Sines”:

sin Aa

=sin B

b=

sin Cc



Geometric product

Outline








Geometric product

Geometric Product

By systematically surveying various possible sets of algebraicrules, Grassmann discovered several other kinds ofmultiplication besides inner and outer products.

However, heoverlooked the geometric product:

ab = a · b + a ∧ b (2)

The formula 2 can be justified on the grounds that it behaveslike a product in the sense that it inherits the properties of theinner and outer product and is consistent.



Geometric product

Geometric Product

By systematically surveying various possible sets of algebraicrules, Grassmann discovered several other kinds ofmultiplication besides inner and outer products. However, heoverlooked the geometric product:

ab = a · b + a ∧ b (2)

The formula 2 can be justified on the grounds that it behaveslike a product in the sense that it inherits the properties of theinner and outer product and is consistent.



Geometric product

Geometric Product

The product ab reflects the following facts:

two vectors are collinear if and only if their product iscommutative (ab = a · b = b · a = ba)

two vectors are orthogonal if and only if their product isanticommutative (ab = a ∧ b = −b ∧ a = −ba)



Geometric product

Geometric Product

The product ab reflects the following facts:

two vectors are collinear if and only if their product iscommutative (ab = a · b = b · a = ba)two vectors are orthogonal if and only if their product isanticommutative (ab = a ∧ b = −b ∧ a = −ba)



Geometric product

Geometric Product

It is very important to note that 2 implies the following identities:

a · b =12

(ab + ba) (3)

a ∧ b =12

(ab− ba) (4)



Geometric product

Geometric Product

The product of a vector a and a bivector B can be derived asfollows:

aB =12

(aB + aB) +12

(Ba− Ba)

=12

(aB− Ba) +12

(aB + Ba)



Geometric product

Geometric Product

The product of a vector a and a bivector B can be derived asfollows:

aB =12

(aB + aB) +12

(Ba− Ba)

=12

(aB− Ba) +12

(aB + Ba)



Geometric product

Geometric Product

Let’s introduce the following notations:

a · B =12

(aB− Ba) (5)

a ∧ B =12

(aB + Ba) (6)

Thus,

aB = a · B + a ∧ B



Geometric product

Geometric Product

Let’s introduce the following notations:

a · B =12

(aB− Ba) (5)

a ∧ B =12

(aB + Ba) (6)

Thus,

aB = a · B + a ∧ B



Geometric product

Geometric Product

This notation suggests that both a · B and a ∧ B aregeneralizations of the inner and outer products (respectively).

Indeed, 6 verifies the properties of the outer product (inparticular, the sign follows from the associative property) andcan be proved from the definitions 3 and 4.The term 5, however, deserves some more attention. Withoutloss of generality, we can write B = b ∧ c. Then, using thedefinitions 3 and 4



Geometric product

Geometric Product

This notation suggests that both a · B and a ∧ B aregeneralizations of the inner and outer products (respectively).Indeed, 6 verifies the properties of the outer product (inparticular, the sign follows from the associative property) andcan be proved from the definitions 3 and 4.

The term 5, however, deserves some more attention. Withoutloss of generality, we can write B = b ∧ c. Then, using thedefinitions 3 and 4



Geometric product

Geometric Product

This notation suggests that both a · B and a ∧ B aregeneralizations of the inner and outer products (respectively).Indeed, 6 verifies the properties of the outer product (inparticular, the sign follows from the associative property) andcan be proved from the definitions 3 and 4.The term 5, however, deserves some more attention.

Withoutloss of generality, we can write B = b ∧ c. Then, using thedefinitions 3 and 4



Geometric product

Geometric Product

This notation suggests that both a · B and a ∧ B aregeneralizations of the inner and outer products (respectively).Indeed, 6 verifies the properties of the outer product (inparticular, the sign follows from the associative property) andcan be proved from the definitions 3 and 4.The term 5, however, deserves some more attention. Withoutloss of generality, we can write B = b ∧ c. Then, using thedefinitions 3 and 4



Geometric product

Geometric Product

a · (b ∧ c) =12

(a

12

(bc− cb)− 12

(bc− cb) a)

(7)

Adding 0 = 14 (bac− cab− bac + cab) to 7 and collecting

terms, we end up with

a · (b ∧ c) = (a · b) c− (a · c) b = a · b c− a · c b (8)

which reveals that the “generalized inner product” in 3 results ina vector (the last equality emphasizes the convention that theinner and outer products have more precedence than thegeometric product).



Geometric product

Geometric Product

It can be proved that, as a general rule,

1 outer multiplication by a vector “raises the dimension”.

2 inner multiplication by a vector “lowers the dimension”.



Geometric product

Geometric Product

It can be proved that, as a general rule,

1 outer multiplication by a vector “raises the dimension”.2 inner multiplication by a vector “lowers the dimension”.



Geometric product

Axioms of Geometric Algebra

The symbol Gn will denote the geometric algebra ofn-dimensional space. This is a linear space over R and itselements are called multivectors and usually written in uppercase (in opposition to vectors, which will be written in lowercase).



Geometric product

Axioms of Geometric Algebra

Each multivector has a grade and every multivector can bewritten as a sum of pure grade terms.

A =n∑

r=0

〈A〉r

〈·〉r projects the term of grade r



Geometric product

Axioms of addition and multiplication of multivectors

Let A, B, and C be multivectors in Gn.



Geometric product

Commutativity of addition

A + B = B + A (9)



Geometric product

Associativity of addition and multiplication

(A + B) + C = A + (B + C) (10)(AB)C = A(BC) (11)



Geometric product

Distributivity of multiplication

A(B + C) = AB + AC (12)(B + C)A = BA + CA (13)



Geometric product

Additive and multiplicative identity

There exist multivectors 0 and 1 are unique and they satisfy thefollowing identities

A + 0 = A (14)1A = A (15)



Geometric product

Existence of additive inverse

A + (−A) = 0 (16)



Geometric product

Addition property of equality

If B = C, then

A + B = A + C (17)



Geometric product

Multiplication property of equality

If B = C, then

AB = AC (18)



Geometric product

Characterization of scalars

For every scalar λ and every multivector A,

λA = Aλ (19)



Geometric product

Characterization of vectors

For every vector a 6= 0,

a2 = |a|2 > 0 (20)



Geometric product

Characterization of k -vectors (with k > 1)

After exploring the geometric implications of the inner, outer,and geometric products, we turn this approach on its head bydefining the inner and outer products in terms of the geometricproduct. If Ar is an r -vector with (r > 0) then we define

aAr = a · Ar + a ∧ Ar (21)

where

a · Ar =12

(aAr − (−1)r Ara) = (−1)r+1Ar · a (22)

a ∧ Ar =12

(aAr + (−1)r Ara) = (−1)r Ar ∧ a (23)

Note that this is consistent with every derivation so far.



Geometric product


After exploring the geometric implications of the inner, outer,and geometric products, we turn this approach on its head bydefining the inner and outer products in terms of the geometricproduct. If Ar is an r -vector with (r > 0) then we define

aAr = a · Ar + a ∧ Ar (21)

where

a · Ar =12

(aAr − (−1)r Ara) = (−1)r+1Ar · a (22)

a ∧ Ar =12

(aAr + (−1)r Ara) = (−1)r Ar ∧ a (23)

Note that this is consistent with every derivation so far.



Geometric product


Based on these definitions we adopt the following axioms:

1 a · Ar is a (r − 1)-vector

2 a ∧ An = 03 a ∧ Ar is a (r + 1)-vector



Geometric product



1 a · Ar is a (r − 1)-vector2 a ∧ An = 0

3 a ∧ Ar is a (r + 1)-vector



Geometric product



1 a · Ar is a (r − 1)-vector2 a ∧ An = 03 a ∧ Ar is a (r + 1)-vector



Geometric product

Notes on division

Let A ∈ Gn, if a multiplicative inverse exists, it is denoted by A−1

or 1A and satisfies the equation.

A−1A = 1

Since multiplication not always commutes, one has to takespecial care with division of multivectors because “left division”is not equivalent (in general) to “right division”.Every non zero vector a has a multiplicative inverse a−1 = a

|a|2

(remember that, for vectors a2 = |a|2).



Geometric product

Notes on division



A−1A = 1

Since multiplication not always commutes, one has to takespecial care with division of multivectors because “left division”is not equivalent (in general) to “right division”.

Every non zero vector a has a multiplicative inverse a−1 = a|a|2




Geometric product

Notes on division



A−1A = 1

Since multiplication not always commutes, one has to takespecial care with division of multivectors because “left division”is not equivalent (in general) to “right division”.Every non zero vector a has a multiplicative inverse a−1 = a

|a|2




Geometric product

Developments in Geometric Algebra

A blade is defined as a multivector of the formB = b1 ∧ · · · ∧ bs. Note that by a similar process toGram-Schmidt orthonormalization we can always write a bladeas B = b1 ∧ · · · ∧ bs = b′1 · · ·b′s.A natural way to view the algebra is in terms of orthonormalbasis and use the following formula to do calculations



Geometric product

Reduction formula

This is the most useful identity relating inner and outer products

a · (b ∧ Cr) = a · bCr − b ∧ (a · Cr) (24)



Geometric product

Laplace expansion of the inner product

Applying the reduction formula repeatedly gives us

a · (a1 ∧ a2 ∧ · · · ∧ ar) =r∑

k=1

(−1)k+1a ·aka1∧· · · ak · · ·∧ar (25)

(where ak means that we remove that vector from the outerproduct).



Geometric product

Inner and outer products between multivectors

Given Ar and Bs multivectors of grades r and s respectively wecan define

Ar · Bs = 〈ArBs〉|r−s| (26)Ar ∧ Bs = 〈ArBs〉r+s (27)

So

ArBs = 〈ArBs〉|r−s| + 〈ArBs〉|r−s|+1 + · · ·+ 〈ArBs〉r+s



Geometric product

Inner and outer products between multivectors

Given Ar and Bs multivectors of grades r and s respectively wecan define

Ar · Bs = 〈ArBs〉|r−s| (26)Ar ∧ Bs = 〈ArBs〉r+s (27)

So

ArBs = 〈ArBs〉|r−s| + 〈ArBs〉|r−s|+1 + · · ·+ 〈ArBs〉r+s



Geometric product

Reversion

The reversion operation is defined as:

(AB)† = B†A†

(A + B)† = A† + B†⟨A†⟩

0= 〈A〉0

a† = a

In particular, this operation receives its name from the followingproperty:

(a1a2 · · · ar )† = ar · · · a2a1



Geometric product

Reversion

The reversion operation is defined as:

(AB)† = B†A†

(A + B)† = A† + B†⟨A†⟩

0= 〈A〉0

a† = a

In particular, this operation receives its name from the followingproperty:

(a1a2 · · · ar )† = ar · · · a2a1



Geometric product

Reversion

And for a bivector B = a ∧ b it works in this fashion:

B† = b ∧ a = −a ∧ b = −B

The general case for r -vectors is a straight-forwardconsequence of the previous equality and leads to the followingexpression for multivectors:

A† = 〈A〉0 + 〈A〉1 −r∑

k=2

〈A〉k



Geometric product

Reversion

And for a bivector B = a ∧ b it works in this fashion:

B† = b ∧ a = −a ∧ b = −B

The general case for r -vectors is a straight-forwardconsequence of the previous equality and leads to the followingexpression for multivectors:

A† = 〈A〉0 + 〈A〉1 −r∑

k=2

〈A〉k



Geometric product

Magnitude

To each multivector corresponds a scalar called magnitude.

A 7→ |A| =√〈A†A〉0

It can be proved that√〈A†A〉0 ≥ 0, which means that the

magnitude is well-defined.


Applications of Geometric Algebra

Algebra of the Euclidean plane

Outline









Parametric and non-parametric equations

Parametric equation for a line with direction a

x = αa

Taking left outer products on each side, leaves us with anon-parametric equation for the line:

x ∧ a = αa ∧ a = 0

It is possible to recover the parametric equation from theprevious expression by noting that

xa = x · a + x ∧ a = x · a






x = αa


x ∧ a = αa ∧ a = 0


xa = x · a + x ∧ a = x · a






x = αa


x ∧ a = αa ∧ a = 0


xa = x · a + x ∧ a = x · a





then,

x = xaa−1 = x · a a−1

= x · a a|a|2

= αa





Now consider a bivector B. The set of solutions of thenon-parametric equation for a plane x ∧ B = 0 is atwo-dimensional vector space. We’ll refer to this set as theB-plane.

The B-plane can be expressed as a multiple of a unit bivector iwith the same orientation as B.

B = Bi

So the non-parametric equation for the plane becomes

x ∧ i = 0 (28)

When working in G2, we call i the pseudoscalar





Now consider a bivector B. The set of solutions of thenon-parametric equation for a plane x ∧ B = 0 is atwo-dimensional vector space. We’ll refer to this set as theB-plane.The B-plane can be expressed as a multiple of a unit bivector iwith the same orientation as B.

B = Bi


x ∧ i = 0 (28)






Now consider a bivector B. The set of solutions of thenon-parametric equation for a plane x ∧ B = 0 is atwo-dimensional vector space. We’ll refer to this set as theB-plane.The B-plane can be expressed as a multiple of a unit bivector iwith the same orientation as B.

B = Bi


x ∧ i = 0 (28)






Now we’ll find a parametric equation for the i-plane. First, let’swrite i in terms of the orthonormal vectors σ1 and σ2,

i = σ1σ2 = σ1 ∧ σ2 = −σ2σ1

For x verifying 28, we can use the reduction formula to obtain

xi = x · i = x · (σ1 ∧ σ2)

= x · σ1 σ2 − x · σ2 σ1

Finally, multiplying on the right by i† = σ2σ1 and using again thereduction formula,

x = x · σ1 σ1 + x · σ2 σ2

= x1σ1 + x2σ2






i = σ1σ2 = σ1 ∧ σ2 = −σ2σ1


xi = x · i = x · (σ1 ∧ σ2)

= x · σ1 σ2 − x · σ2 σ1


x = x · σ1 σ1 + x · σ2 σ2

= x1σ1 + x2σ2






i = σ1σ2 = σ1 ∧ σ2 = −σ2σ1


xi = x · i = x · (σ1 ∧ σ2)

= x · σ1 σ2 − x · σ2 σ1


x = x · σ1 σ1 + x · σ2 σ2

= x1σ1 + x2σ2




Geometric interpretation of the bivector

It is clear that a bivector can be understood as an orientedplane “segment” (that is, a directed area).

Multiplication on the right by i rotates vectors counter-clockwiseby 90 degrees as exemplified by these two equalities:

σ1i = σ1σ1σ2 = σ1 · (σ1 ∧ σ2) = σ2

σ2i = σ2σ1σ1 = −σ1

Thus, σ1i2 = −σ1, which implies

i2 = −1

So this is a geometric interpretation for this equation.





It is clear that a bivector can be understood as an orientedplane “segment” (that is, a directed area).Multiplication on the right by i rotates vectors counter-clockwiseby 90 degrees as exemplified by these two equalities:

σ1i = σ1σ1σ2 = σ1 · (σ1 ∧ σ2) = σ2

σ2i = σ2σ1σ1 = −σ1


i2 = −1







σ1i = σ1σ1σ2 = σ1 · (σ1 ∧ σ2) = σ2

σ2i = σ2σ1σ1 = −σ1


i2 = −1







σ1i = σ1σ1σ2 = σ1 · (σ1 ∧ σ2) = σ2

σ2i = σ2σ1σ1 = −σ1


i2 = −1







σ1i = σ1σ1σ2 = σ1 · (σ1 ∧ σ2) = σ2

σ2i = σ2σ1σ1 = −σ1


i2 = −1





Plane spinors

A spinor is the product of two vectors in the i-plane.

For example, z = σ1x = x1σ21 + x2σ1σ2 = x1 + ix2.




Plane spinors

A spinor is the product of two vectors in the i-plane.For example, z = σ1x

= x1σ21 + x2σ1σ2 = x1 + ix2.




Plane spinors

A spinor is the product of two vectors in the i-plane.For example, z = σ1x = x1σ

21 + x2σ1σ2

= x1 + ix2.




Plane spinors

A spinor is the product of two vectors in the i-plane.For example, z = σ1x = x1σ

21 + x2σ1σ2 = x1 + ix2.




Plane spinors

Since

z† = x1 + i†x2

= x1 − i†x2

then,

x1 = Re(z) = 〈z〉0 =z + z†

2

x2 = Im(z) =〈z〉2

i=

z− z†

2iand

|z|2 =⟨

z†z⟩

0= x2

1 + x22




Plane spinors

Since

z† = x1 + i†x2 = x1 − i†x2

then,

x1 = Re(z) = 〈z〉0 =z + z†

2

x2 = Im(z) =〈z〉2

i=

z− z†

2iand

|z|2 =⟨

z†z⟩

0= x2

1 + x22




Plane spinors

Since

z† = x1 + i†x2 = x1 − i†x2

then,

x1 = Re(z) = 〈z〉0 =z + z†

2

x2 = Im(z) =〈z〉2

i=

z− z†

2iand

|z|2 =⟨

z†z⟩

0= x2

1 + x22




Plane spinors

Since

z† = x1 + i†x2 = x1 − i†x2

then,

x1 = Re(z) = 〈z〉0 =z + z†

2

x2 = Im(z) =〈z〉2

i=

z− z†

2i

and

|z|2 =⟨

z†z⟩

0= x2

1 + x22




Plane spinors

Since

z† = x1 + i†x2 = x1 − i†x2

then,

x1 = Re(z) = 〈z〉0 =z + z†

2

x2 = Im(z) =〈z〉2

i=

z− z†

2iand

|z|2 =⟨

z†z⟩

0= x2

1 + x22




Plane spinors

There’s a correspondence between vectors and spinors in thei-plane:

x = x1σ1 + x2σ2 7→ σ1x = z

z = x1 + ix2 7→ x = σ1z

The spinor z rotates and dilates σ1 by right multiplication.




Plane spinors

There’s a correspondence between vectors and spinors in thei-plane:

x = x1σ1 + x2σ2 7→ σ1x = zz = x1 + ix2 7→ x = σ1z

The spinor z rotates and dilates σ1 by right multiplication.



Reflections

Outline








Reflections

Reflections

Let a be a vector and m an unitary vector orthogonal to somehyperplane.

a = a⊥ + a||a′ = a⊥ − a||



Reflections

Reflections

Wherea|| = a ·m m (already known)

a⊥ = a− a ·m m = (am− a ·m) m = a ∧m mThus,

a′ = a⊥ − a||= a ∧m m− a ·m m= − (m · a−m ∧ a) m = −mam

This formula works for any multivector, not just vectors!It illustrates how the geometric product is suitable toproduce compact expressions of operations on vectors.



Reflections

Reflections

Wherea|| = a ·m m (already known)a⊥ = a− a ·m m

= (am− a ·m) m = a ∧m mThus,





Reflections

Reflections

Wherea|| = a ·m m (already known)a⊥ = a− a ·m m = (am− a ·m) m

= a ∧m mThus,





Reflections

Reflections

Wherea|| = a ·m m (already known)a⊥ = a− a ·m m = (am− a ·m) m = a ∧m m

Thus,





Reflections

Reflections

Wherea|| = a ·m m (already known)a⊥ = a− a ·m m = (am− a ·m) m = a ∧m mThus,

a′ = a⊥ − a||

= a ∧m m− a ·m m= − (m · a−m ∧ a) m = −mam




Reflections

Reflections


a′ = a⊥ − a||= a ∧m m− a ·m m

= − (m · a−m ∧ a) m = −mam




Reflections

Reflections






Reflections

Reflections



This formula works for any multivector, not just vectors!

It illustrates how the geometric product is suitable toproduce compact expressions of operations on vectors.



Reflections

Reflections






Rotations

Outline








Rotations

Rotation by double refection

Same construction as for single reflection but now using twounit vectors m and n.



Rotations


Let’s denote by θ the angle between m and n

First reflection,a′ = −mam

and second reflection,

a′′ = −na′n = −n (−mam) n= nmamn = RaR†

So a rotation can be succinctly expressed as a 7→ RaR†.



Rotations


Let’s denote by θ the angle between m and nFirst reflection,

a′ = −mam






Rotations



a′ = −mam


a′′ = −na′n

= −n (−mam) n= nmamn = RaR†




Rotations



a′ = −mam


a′′ = −na′n = −n (−mam) n

= nmamn = RaR†




Rotations



a′ = −mam






Rotations



a′ = −mam






Rotations

Rotors

The multivector R is called a rotor. Let’s take a closer look at it.

R = nm = n ·m + n ∧m = cos θ + n ∧m.

Taking into account that we’re only interested in the zerothprojection, the magnitude of n ∧m is⟨

(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =

〈−mn (nm− n ·m)〉0 = 〈−mnnm + mn n ·m〉0 =〈−1 + n ·m mn〉0 = 〈−1 + n ·m (m · n + m ∧ n)〉0 =

−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm

= n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m

= cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0

= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0

= 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0

=


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =

〈−mn (nm− n ·m)〉0

= 〈−mnnm + mn n ·m〉0 =〈−1 + n ·m mn〉0 = 〈−1 + n ·m (m · n + m ∧ n)〉0 =

−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =

〈−mn (nm− n ·m)〉0 = 〈−mnnm + mn n ·m〉0

=〈−1 + n ·m mn〉0 = 〈−1 + n ·m (m · n + m ∧ n)〉0 =

−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =

〈−mn (nm− n ·m)〉0 = 〈−mnnm + mn n ·m〉0 =〈−1 + n ·m mn〉0

= 〈−1 + n ·m (m · n + m ∧ n)〉0 =

−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =

〈−mn (nm− n ·m)〉0 = 〈−mnnm + mn n ·m〉0 =〈−1 + n ·m mn〉0 = 〈−1 + n ·m (m · n + m ∧ n)〉0

=

−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2

= −1 + cos2 θ = − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ

= − sin2 θ



Rotations

Rotors


R = nm = n ·m + n ∧m = cos θ + n ∧m.


(n ∧m)† (n ∧m)⟩

0= 〈−m ∧ n n ∧m〉0 = 〈−mn n ∧m〉0 =


−1 + (m · n)2 = −1 + cos2 θ = − sin2 θ



Rotations

Rotors

We can define

B =m ∧ nsin θ

Now

R = cos θ − B sin θ = e−Bθ

is a rotor involved in a rotation of angle 2θ in the plane given bym ∧ n.



Rotations

Rotors

We can define

B =m ∧ nsin θ

Now

R = cos θ − B sin θ = e−Bθ

is a rotor involved in a rotation of angle 2θ in the plane given bym ∧ n.


Bibliography

Bibliography

David Hestenes. New Foundations for ClassicalMechanics (2nd ed.)

John Vince. Geometric Algebra for Computer GraphicsAnthony Lasenby and Chris Doran. Physical Applicationsof Geometric AlgebraChristian B. U. Perwass. Applications of GeometricAlgebra in Computer Vision: The geometry of multiple viewtensors and 3D-reconstruction


Bibliography

Bibliography

David Hestenes. New Foundations for ClassicalMechanics (2nd ed.)John Vince. Geometric Algebra for Computer Graphics

Anthony Lasenby and Chris Doran. Physical Applicationsof Geometric AlgebraChristian B. U. Perwass. Applications of GeometricAlgebra in Computer Vision: The geometry of multiple viewtensors and 3D-reconstruction


Bibliography

Bibliography

David Hestenes. New Foundations for ClassicalMechanics (2nd ed.)John Vince. Geometric Algebra for Computer GraphicsAnthony Lasenby and Chris Doran. Physical Applicationsof Geometric Algebra

Christian B. U. Perwass. Applications of GeometricAlgebra in Computer Vision: The geometry of multiple viewtensors and 3D-reconstruction


Bibliography

Bibliography

David Hestenes. New Foundations for ClassicalMechanics (2nd ed.)John Vince. Geometric Algebra for Computer GraphicsAnthony Lasenby and Chris Doran. Physical Applicationsof Geometric AlgebraChristian B. U. Perwass. Applications of GeometricAlgebra in Computer Vision: The geometry of multiple viewtensors and 3D-reconstruction

Documents

Introduction to Geometric Algebra