mat-web.upc.edu · Abstract 1/2 Abstract. First developed as amathematical formalism for geometry...

Geometric Algebra

Mathematical Structures andApplications

S. Xambo

UPC/BSC

3-4/10/2019

S. Xambo (UPC/BSC) GA: Structures and Applications 3-4/10/2019 1 / 130

X = ”https://mat-web.upc.edu/people/sebastia.xambo”

These slides: X/GA/2019-uned-s.pdf

Main reference:

[1], X/GA/2019-uned-p.pdf (aún no accesible)

Coplementary references:

[2], X/GA/brief-spins.html

[3], X/GA/LXZ.html

[4], X/K2/K2-Xambo.pdf

[5], X/GA/2015-Xambo–EscondidasSendas-JBSG-in-memoriam.pdf

[6], X/HistoricalEssays/2018-Xambo–From-Leibniz-CG-to-GA.pdf


Abstract 1/2

Abstract. First developed as a mathematical formalism for geometryand physics, in the last three decades there has been an explosion ofnew applications of geometric algebra and geometric calculus to agreat variety of areas, like classical and relativistic mechanics, generalrelativity, cosmology, robotics, computer graphics, computer vision,molecular geometry, quantum computing, or algorithmic learning .Moreover, its very nature endows it with a rich potential for theteaching and learning of geometry and its wide range ofapplications.

In this talk we seek to present a synopsis of the conceptualframework of geometric algebra (GA), together with a sample ofapplications to geometry, particularly 2d and 3d Euclidean geometry.


Abstract 2/2

To some extent, the concepts that appear in the applications toEuclidean geometry will be known, but the perspective of geometricalgebra not only revitalizes them in form and substance, but it favorsentering, by walking a familiar path, into the spirit of the formalism,which then should enable the undertaking of more ambitious studies.


Introduction GA layers

The arena where the gamestarts is a real vector space, E , of finitedimension, n. It gives rise to ∧ = ∧E ,Grassmann’s exterior algebra of E .

When E is endowed with a metric q, itcan be extended in a natural way to ∧.The resulting structure is ∧q = ∧qE ,the metric exterior algebra.

As discovered by Clifford, ∧q can be further enriched with anassociative product (the geometric product) and the resultingpolyfacetic structure, G, is the geometric algebra of (E , q).

Geometric algebra is a powerful formalism for the study of theorthogonal groups Or ,s and their subgroups SOr ,s and SO0

r ,s

(definitions recalled on page 14)


Introduction Unfolding

GA GC

q(x, y) x · y

Geometric Product

Metricproduct

Innerproduct

Ausdehnungs-

lehre

Grassmann Algebra


Index Sections

Background, 8

Grassmann algebra, 15

Metric Grassmann algebra, 21

Geometric algebra, 27

Examples in low dimensions, 36

Spinorial groups, 57

Examples of rotors in low dimensions, 77

Topics for further study, 120

References, 125


BackgroundVectors and linear operators

MetricsBasics. Orthogonal bases and signatures. Special

signatures. Orthogonal groups


Background Vectors and linear operators

R will denote the field of real numbers. Its elements will be calledscalars and will be denoted by Greek letters (α, λ, ρ, ω, . . . ).

By a vector space we mean an R-vector space E of finite dimension n(unless said otherwise explicitely). Its elements will be called vectorsand will be denoted by latin italic letters like (x , y , u, v , . . . ).

A typical basis of E will be denoted e = e1, . . . , en.

If f is an endomorphism of E (also called a linear operator), itsmatrix with respect to e is the matrix M defined by the relation

f (e) = eM . (1)

The group of linear automorphisms of E will be denoted GL(E ). If fis a linear automorphism, then its matrix M is invertible(⇔ det(M) 6= 0), so that M ∈ GLn. The map GL(E )→ GLn,f 7→ M , is a group isomorphism.


Background Metrics

Our space E will be endowed with a bilinear symmetric regular(non-degenerate) map q : E × E → R (the metric).

Instead of q(x , x) we will simply write q(x). The function q(x) is thequadratic form associated to q. It determines q by the polarizationrelation

2q(x , x ′) = q(x + x ′)− q(x)− q(x ′). (2)

A vector x is positive (negative, null or isotropic) if q(x) > 0(q(x) < 0, q(x) = 0). A vector x is said to be a unit vector ifq(x) = ±1.

Notation: E×

will denote the set of non-isotropic vectors of E .


Background Metrics

The magnitude of x , |x |, is defined by the relation

|x | =√|q(x)|. (3)

Since |q(x)| = εxq(x), where εx ∈ {1,−1, 0} (the signature of x) isthe sign of q(x), we have q(x) = εx |x |2. If x is non-nul, thenq(x/|x |) = εx , so that x/|x | is a unit vector (the normalization of x).

Two vectors x and x ′ are said to be q-orthogonal when q(x , x ′) = 0.

Two sets of vectors F and F ′ are said to be q-orthogonal ifq(x , x ′) = 0 for all x ∈ F and x ′ ∈ F ′.

If F is a set of vectors, F⊥ denotes the set of vectors that areq-orthogonal to F . The bilinearity of q implies that F⊥ is vectorsubspace of E . We have dimF⊥ = n − dimF for all F .


Background Metrics

A basis e = e1, . . . , en is said to be q-orthogonal if q(ej , ek) = 0 forall j 6= k .

3 There exist q-orthogonal bases.

If e is orthogonal, and r (s) denotes the number of j such thatq(e j) > 0 (q(e j) < 0), we say that (r , s) is the signature of q.Notice that r + s = n, as there is no j such that q(ej) = 0.

3 The signature does not depend on the orthogonal basis used tocompute it (Sylvester’s law of inertia).

If e is an orthogonal basis, we say that it is orthonormal if q(ej) = 1for j = 1, . . . , r and q(ej) = −1 for j = r + 1, . . . , r + s = n.

Normalizing an orthogonal basis, we get an orthogonal basis of unitvectors, out of which an ortonormal basis can be obtained byreordering.


Background Metrics

We will write Er ,s to indicate that the metric q has signature r , s.

Special signatures. Instead of En,0 we will simply write En (Euclideanspace). In this case the metric is positive definite (q(x) > 0 for allnon-zero x ∈ E ) and we will use the standard notions:

• |x | = +√

q(x) (and called, in this context, norm or length of x).

• θ = θ(x , x ′) = acosq(x ,x ′)|x ||x ′| , x , x ′ 6= 0. This is the (scalar) angle

formed by x and x ′ (θ ∈ [0, π]). We have θ = 0 (θ = π) if andonly if x ′ = λx with λ > 0 (λ < 0).

The cases n = 2 and n = 3 are used in plane and space Euclideangeometry.

The space En = En = E0,n is the anti-Euclidean space of dimension n.

The space E1,3 is the ground structure for special relativity(Minkowski space) and E c

3 = E4,1 for the conformal space of the 3dspace.


Background Metrics

Orthogonal groups

An opertor f of E is said to be a q-isometry if q(fx , fx ′) = q(x , x ′).

Under composition, the set of q-isometries of E forms a group that isdenoted Oq(E ) and is called the orthogonal group of q, or of (E , q).It is also denoted Or ,s , or just On in the case of the Euclidean space.

3 It is a basic fact that if f ∈ Oq(E ), then det(f ) = ±1.

SOq(E ): subgroup of the isometries f such that det(f ) = 1 (properisometries), and it is called special orthogonal group. It is denotedSOr ,s , and SOn in the case of the Euclidean space.

The isometries of E1,3 are also called Lorentz transformations (LT),and hence O1,3 is the Lorentz group (LG) and SO1,3 is the proper orspecial LG. The restricted LG , SO0

1,3 ⊂ SO1,3 (proper orthochronous

LT) is a special case of SO0r ,s , the restricted special group to be

introduced later.


Grassmann exterior algebraThe grading

InvolutionsParity and reversal involutions. Clifford conjugation

Geometric interpretation of the blades


Exterior algebra The grading

3 (∧E ,∧): Grassmann’s exterior algebra of E . It is unital ,associative and skew-commutative: x ∧ x ′ = −x ′ ∧ x for all x , x ′ ∈ E(cf. the general rule on next page). In particular x ∧ x = x∧2 = 0 forall x ∈ E .

3 ∧kE ⊂ ∧E (k-th exterior power of E ):subspace of ∧E generated by all k-blades, which are the non-zeroexterior products x1 ∧ · · · ∧ xk (x1, . . . , xk ∈ E ).

By convention, ∧0E = R and clearly ∧1E = E .

3 ∧E = ⊕nk=0∧kE = R⊕ E ⊕ ∧2E ⊕ · · · ⊕ ∧nE (multivectors).

Grade decomposition of a multivector: a = a0 + a1 + · · ·+ an.

The elements of ∧kE are called k-vectors. Special names: k = 0,scalars; k = 1, vectors; k = 2, bivectors; k = n − 1, pseudovectors(dim n); k = n, pseudoscalars (dim 1).


Exterior algebra The grading

3 ∧E is a graded algebra, which means that a ∧ a′ ∈ ∧k+k ′E whena ∈ ∧kE and a′ ∈ ∧k ′E . The exterior product is skewcommutative(or supercommutative): for a ∈ ∧kE and a′ ∈ ∧k ′E ,

a ∧ a′ = (−1)kk′a′ ∧ a.

If e1, · · · , en is a basis of E , the(nk

)products

eJ = ej1,...,jk = ej1 ∧ · · · ∧ ejk (1 6 j1 < . . . < jk 6 n)

form a basis of ∧kE . In particular, dim∧kE =(nk

). Hence

dim∧E = 2n.

n = 1: 1, e1 = e; n = 2: 1, e1, e2, e12;

n = 3: 1, e1, e2, e3, e23, e31, e12, e123.

3 If f : E → E ′ is a linear map, ∃! algebra homomorphismf∧ : ∧E → ∧E ′, often still denoted f , such that f∧(x) = f (x) for allx ∈ E , and cosequently f (x1 ∧ · · · ∧ xk) = (fx1) ∧ · · · ∧ (fxk).


Exterior algebra Involutions

Parity involution. The linear automorphism E → E , e 7→ −e,induces a linear automorphism of ∧E that is denoted a 7→ a.

3 For a ∈ ∧E , we have a =∑

k(−1)kak .

3 a ∧ b = a ∧ b for all a, b ∈ ∧E (algebra automorphism).

Reverse involution. Reversing the order of exterior products yieldsan algebra antiautomorphism of ∧E , a 7→ a.

Since reversing a k-blade amounts to(k2

)sign changes, and since this

number has the same parity as k//2 (the integer quotient of k by 2),

3 a =∑

k(−1)k//2ak .

3 a ∧ b = b ∧ a for all a, b ∈ ∧E (algebra antiautomorphism).

Clifford conjugation. a = ˜a = ˆa =∑

k(−1)(k+1)//2ak .

3 a ∧ b = b ∧ a (algebra antiautomorphism).


Exterior algebra Geometric interpretation of the blades

Let PE the projective space of E , i.e., the set of 1-dimensional linearsubspaces of E . Given x ∈ E − {0}, we denote by [x ] ∈ PE thepoint corresponding to 〈x〉, the line spanned by x . From thedefinition it is clear that [x ] = [x ′] (x , x ′ ∈ E − {0}) if and only ifx ′ = λx for some λ ∈ R (necessarily non-zero), a relation thathenceforth will be written x ′ ∼ x .

Let b = x1 ∧ · · · ∧ xk be a k-blade. Then the relation

x ∈ 〈x1, . . . , xk〉 ⇔ x ∧ b = 0 (4)

shows that b determines the subspace 〈x1, . . . , xk〉. Moreover, ifb′ = x ′1 ∧ · · · ∧ x ′k is another k-blade, then 〈x1, . . . , xk〉 = 〈x ′1, . . . , x ′k〉if and only if b′ ∼ b. The if part follows immediately from equation(4). The only if part is a consequence of the fact that if

F = 〈x1, . . . , xk〉 = 〈x ′1, · · · , x ′k〉, then b, b′ ∈ ∧kF and hence b′ ∼ b

because ∧kF is 1-dimensional.


Exterior algebra Geometric interpretation of the blades

Thus we see that the equality [b] = [b′] in P(∧kE ) is equivalent tothe equality of the corresponding subspaces. Because of this, we willdenote by [b] the linear subspace 〈x1, . . . , xk〉 determined by b.

Notation remark . If in the expression eJ = ej1,...,jk introduced on page17 we need to indicate that the product is ∧, we will writee∧J = e∧j1,...,jk . This will be necessary when eJ will be used to indicatethe geometric product ej1 · · · ejk of the vectors instead of the exteriorproduct ej1 ∧ · · · ∧ ejk (see page 31).


The metric Grassmannalgebra

Gram determinants

Orthogonal and orthonormal bases

Inner productLaplace formula. Adjuntion formula


The metric Grassmann algebra Gram determinants

3 There is a unique metric on ∧E , still denoted q, such that thespaces ∧kE are pairwise q-orthogonal and (Gram determinants)

q(x1 ∧ · · · ∧ xk , x′1 ∧ · · · ∧ x ′k) = det((q(xi , x

′j )))

= Gq(x1, . . . , xk ; x ′1, . . . , x′k),

q(x1 ∧ · · · ∧ xk) = det((q(xi , xj))) = Gq(x1, . . . , xk) (i , j = 1, . . . , k).

In particular we observe that q(x1 ∧ · · · ∧ xk , x′1 ∧ · · · ∧ x ′k) vanishes if

one of the vectors xi is orthogonal to all vectors x ′j , or the other wayaround. Moreover, if x1, . . . , xk are pairwise orthogonal, then theGram determinant is diagonal and therefore

q(x1 ∧ · · · ∧ xk) = q(x1) · · · q(xk).

3 For a non-zero k-blade b = x1 ∧ . . . ∧ xk , q(b) = 0 if and only ifthe space F = [b] is singular.

3 If x1, . . . , xk ∈ En, q(x1 ∧ · · · ∧ xk) = V (x1, . . . , xk)2, whereV (x1, . . . , xk) is the k-volume of the parallelepiped /x1, . . . , xk/.


The metric Grassmann algebra Orthogonal and orthonormal bases

Let N = {1, . . . , n}, and denote by J the set of subsets of N(multiindices) and by Jk the subset of multiindices of rank k . Thenfrom the preceding observation it follows that the basis {e∧I }I∈Jk of

∧kE is orthogonal (orthonormal) if the basis e is orthogonal(orthonormal), and that q(e∧I ) = q(ei1) · · · q(eik ), which will bedenoted by qI . And, of course, {e∧I }I∈J is an orthogonal(orthonormal) basis of ∧E .

Exercise. The signature of this metric is (2n, 0) if s = 0 (so r = n),and (2n−1, 2n−1) otherwise. In particular (∧E , q) is:

Euclidean when (E , q) is Euclidean;

has signature (8, 8) for the Minkowski space;

has signature (16, 16) for the conformal space.


The metric Grassmann algebra Inner product

It is derived from the (left) contraction operator ix (x ∈ E ):

ix(x1 ∧ · · · ∧ xk) =∑

j(−1)j−1q(x , xj)x1 ∧ · · · ∧ xj−1 ∧ xj+1 ∧ · · · ∧ xk .3 The result is a bilinear product a · b (a, b ∈ ∧E ) uniquelydetermined by the following properties:

a · b = 0 if a or b is a scalar;

x · b = ix(b) if x ∈ E ; so x · y = q(x , y) for x , y ∈ E .

a · b = (−1)jk+mb · a (m = min(j , k)) if a ∈ ∧jE , b ∈ ∧kE . Inparticular, a · x = (−1)j+1ix(a) if x is a vector and j > 1.

(x1 ∧ · · · ∧ xj−1 ∧ xj) · b = (x1 ∧ · · · ∧ xj−1) · (xj · b) if b ∈ ∧kE and2 6 j 6 k (recursive fomula).

If a = x1 ∧ · · · ∧ xk and b = x ′1 ∧ · · · ∧ x ′k , a · b = (−1)k//2q(a, b). In

general, a · b = q(a, b) if a, b ∈ ∧kE , k > 1.



Laplace formula for the inner product a · b when a = x1 ∧ · · · ∧ xjand b = x ′1 ∧ · · · ∧ x ′k . Its general expression can be easily guessedfrom the following example: (x1 ∧ x2) · (x ′1 ∧ x ′2 ∧ x ′3) =

((x1∧x2) ·(x ′1∧x ′2))x ′3−((x1∧x2) ·(x ′1∧x ′3))x ′2 +((x1∧x2) ·(x ′2∧x ′3))x ′1.

3 For all a, b ∈ ∧E ,

a · b = a · b and a · b = b · a

Adjuntion formula. Suppose that 0 < j < k 6 n. Givenc ∈ ∧k−jE , we can consider the maps ec : ∧jE → ∧kE , a 7→ c ∧ a,and ic : LkE → ∧jE , b 7→ c · b. These maps are adjoint with respectto q, in the sense that

q(eca, b) = q(a, icb) or q(c ∧ a, b) = q(a, c · b) (5)



We may assume that c is a blade, say c = x1 ∧ · · · ∧ xk−j , for whichec = ex1 · · · exk−j

and ic = ixk−j· · · ix1 . This shows that it is enough to

consider the case k − j = 1, i.e., j = k − 1. In other words, we haveto show that q(x ∧ a, b) = q(a, x · b) for a ∈ ∧k−1E , b ∈ ∧kE andx ∈ E . Furthermore, since both sides are linear in a, b, c , it will besufficient to prove the claim when a and b are blades, saya = x2 ∧ · · · ∧ xk and b = y1 ∧ · · · ∧ yk (x2, . . . , xk , y1, . . . , yk ∈ E ).With these assumptions, q(x ∧ a, y) = Gq(x , x2, . . . , xk ; y1, . . . , yk).The development of this determinant along the first row is∑j=k

j=1(−1)j−1q(x , yj)Gq(x2, . . . , xk ; y1, . . . , yj−1, yj+1, . . . , yk) =∑j=kj=1(−1)j−1q(x , yj)q(a, b(j)), with

b(j) = y1 ∧ . . . ∧ yj−1 ∧ yj+1 ∧ . . . ∧ yk , which is equal to

q(a,∑j=k

j=1(−1)j−1q(x , yj)b(j)) = q(a, ix · b).

Remark . Equation (5) can be used, in reverse direction, to define theinner product.


Geometric algebraGeometric product

Clifford relations. Relations to ∧, ·, q, , .The even algebra. Functorialities

Working with a basis e

Artin’s formula. Commutation formula. Pseudoscalars


Geometric algebra Geometric product

3 The GA of (E , q) = Er ,s , denoted G = Gq = Gr ,s , can beconstructed by enriching ∧E with the geometric product ab(Clifford). It is unital, bilinear and associative. Moreover,

For any vectors x , x ′ ∈ E , xx ′ = x · x ′ + x ∧ x ′ (Clifford relations).

Thus xx ′ = −x ′x iff x · x ′ = 0 (anticommuting property) andx2 = q(x) (Clifford reduction).

If q(x) 6= 0 (non-isotropic , or non-null vector), x−1 = x/q(x).

For x ∈ E and a ∈ ∧E ,

xa = x · a + x ∧ a,

ax = a · x + a ∧ x .

If a ∈ G j and b ∈ Gk (j , k > 0), then (ab)i is 0 unless i is in therange |j − k |, |j − k |+ 2, . . . , j + k − 2, j + k , and

(ab)|j−k| = a · b and (ab)j+k = a ∧ b.



For any a, b ∈ G, ab = ab and ab = b a.

Riesz formulas 2x ∧ a = xa + ax , 2x · a = xa − ax . For example,2x ∧ x ′ = xx ′ − x ′x and 2x · x ′ = xx ′ + x ′x if x , x ′ ∈ E , while2x ∧ b = xb + bx and 2x · b = xb − bx if b ∈ G2. 1 (this symbolindicates that there is a note in Pointers (page 116).

The metric in terms of the geometric product: For all a, b ∈ G,

q(a, b) = (ab)0 = (ab)0.

In particular we have, for all a ∈ G,

q(a) = (aa)0 = (aa)0

If a is a k-blade, then aa is already a scalar and

q(a) = aa = aa = (−1)k//2a2

In particular we see that a is invertible if and only if a2 6= 0, or if andonly if q(a) 6= 0, and if this is the case, then we have

a−1 = a/a2 = a/q(a).S. Xambo (UPC/BSC) GA: Structures and Applications 3-4/10/2019 29 / 130


3 For any x ∈ E and a, b ∈ G, we have x · (ab) = (x · a)b + a(x · b).In other words, ix is a skew-derivation of the geometric product.

Even algebra

A multivector a is said to be even if a = a. The set G+of even

multivectors forms a subalgebra for each of the three products and itis called the even algebra. From the point of view of the lineargrading, it consists of multivectors that only have components ofeven grade. The odd multivectors form a linear subspace G− andG = G+ ⊕ G−. Moreover, dimG+

= dimG− = 2n−1.

3 Let f be an isometry of E and f∧ the corresponding exterior linearoperator of G. For x , x ′ ∈ E and a, a′ ∈ G:

f∧(a · a′) = f∧(a) · f∧(a′)

f∧(aa′) = f∧(a) f∧(a′)

f∧ is q-orthogonal.S. Xambo (UPC/BSC) GA: Structures and Applications 3-4/10/2019 30 / 130

Geometric algebra Working with a basis e

Let e = e1, . . . , en be a basis of E and N = {1, . . . , n} the set ofindices.

If K = k1, . . . , km is a sequence of indices, set

eK = ek1 · · · ekm .

3 {eI}I∈I is a basis of G = ∧E .

Remark that if I ∈ Jk , then in general eI = e∧I + lower grade terms,like e12 = e1e2 = e1 ∧ e2 + e1 · e2 = e∧12 + e1 · e2.

3 If e is orthogonal, then eI = e∧I , as

x1 · · · xk = x1 ∧ · · · ∧ xk

when x1, . . . , xk are pair-wise orthogonal vectors.



3 Artin’s formula: If I , J are multiindices, then

eIeJ = (−1)t(I ,J)qI∩JeI∆J

where t(I , J) is the number of inversions in the squence I , J , I∆J isthe symmetric difference of I and J , and qK = q (ek1) · · · q (ekm).

3 In particular, e2J = (−1)|J|//2qJ

Examples

In E2, e212 = −1 (as 2//2 = 1 and q12 = 1).

In E3, e2123 = −1 (as 3//2 = 1 and q123 = 1).

In E1,3, e20123 = −1 (as 4//2 = 2 and q0123 = (−1)3 = −1).

In E4, e21234 = 1 (as 4//2 = 2 and q1234 = (−1)4 = 1).



3 Commutation formula: Artin’s formula, in the case of anorthonormal basis, tells us that eIeJ = ±eI∆J , where the sign is(−1)t(I ,J)qI∩J . It follows that eJeI = ±eIeJ , and it turns out that thesign produced by this swapping is (−1)c(−1)|I ||J|, with c = |I ∩ J |:

eIeJ = (−1)c(−1)|I ||J|eJeI .



Pseudoscalars

Let e = e1, . . . , en be an orthonormal basis of E = Er ,s and define

ie = e1 ∧ · · · ∧ en ∈ Gn.We will say that ie is the pseudoescalar associated to e. Note that

q(ie) = q(e1) · · · q(en) = (−1)s .

If e ′ = e ′1, . . . , e′n is another orthonormal basis of E , then

ie′ = δie , δ ∈ R.

Now the equalities

q(ie) = q(ie′) = q(δie) = δ2q(ie)

allow us to conclude that δ = ±1. This means that, up to sign, thereis a unique pseudoscalar . The distinction of one of them amounts tochoose an orientation for E .



Let i ∈ Gn be a pseudoescalar and G× the group of invertiblemultivectors with respect to the geometric product. Then

(1) i ∈ G× , i−1 = (−1)s i = (−1)s(−1)n//2i, i2 = (−1)n//2(−1)s .

(2) Hodge duality. For any a ∈ Gk , ia, ai ∈ Gn−k (by Artin’s formula)and the maps a 7→ ia and a 7→ ai are linear isomorphisms Gk → Gn−k .The inverse maps are given by a 7→ i−1a and a 7→ ai−1, respectively.

(3) If n is odd, i commutes with all the elements of G (this isexpressed by saying that i is a central element of G). If n is even, icommmutes (anticommutes) with even (odd) multivectors.

(4) If q(i) = 1 (q(i) = −1), the Hodge duality maps are isometries(antiisometries): q(ia) = (aiia)0 = q(i)(aa)0 = q(i)q(a).


Examples in low dimensionsG1 ' R⊕ R = 2R. G1 ' C. G2 ' H

W = G2 (Wessel algebra)The law of cosines. The law of sines. Orthocenter and

circumcenter. Euler’s line. W ' R(2)

P = G3 (Pauli algebra)Vector algebra revisited. Cross product. P ' C(2)

G2,1, G1,2, G3

D = G1,3 (Dirac algebra)D+ ' P . γ-notation. Dirac representation. D ' H(2)


Examples in low dimensions G1 ' 2R

Let e = e1 be a unit vector of E1. The elements of G1 have the formα + βe, α, β ∈ R.

Suggested by the relation (1 + e)(1− e) = 1− e2 = 0, consider theelements u = 1

2(1 + e) and u′ = 1

2(1− e), which clearly satisfy

u + u′ = 1, uu′ = u′u = 0, u2 = u,

eu = u, eu′ = −u′.

Then we have

(α + βe) = (α + βe)u + (α + βe)u′ = (α + β)u + (α− β)u′.

and the linear isomorphism G1 → 2R, α + βe 7→ (α + β, α− β), is infact an algebra isomorphism, as

(au + a′u′)(bu + b′u′) = abu + a′b′u′.


Examples in low dimensions G1 ' C, G2 ' H

G1 ' C

If e = e1 is an orthonormal basis of E1 (i.e., e2 = −1) then theelements of G1 have the form a = α + βe and a = α− βe. Sinceaa = α2 + β2, we see that G1 is a field. Actually the linearisomorphism G1 → C, α + βe 7→ α + βi , is an algebra isomorphism.

G2 ' H

If e1, e2 is an orthonormal basis of E2 (i.e., e21 = e2

2 = −1,e1 · e2 = 0), then the linear isomorphism

G2 → H, α + β1e1 + β2e2 + γe1e2 7→ α + β1i + β2j + γk

is an algebra homomorphism. The reason is that e1, e2 and e1e2

satisfy Hamilton’s relations e21 = e2

2 = (e1e2)2 = e1e2(e1e2) = −1.


Examples in low dimensions W = G2

Wessel algebra

W = G2 = 〈1, e1, e2, e12 = i〉, i 2 = −1

W+= G+

2 = 〈1, i〉 = C ' C

e1i = −ie1 = e2; e2i = −ie2 = −e1 (right angle turn).

Let us point out how to approach plane trigonometry by means ofthe geometric algebra formalism (Wessel trigonometry). We willwork in the Euclidean plane E2 oriented by the unit area i .

Special notation. For reasons that will become clear in theexposition, we have to adapt the notations in order that they candesignate the multiple object types that enter in the relations we willbe studying. For example, here vectors will be denoted by lower-caseletters in boldface, like a or b. The norm of such vectors will bedenoted a or b.



The law of cosines

Taking the figure as a clue, we have

c2 = (a + b)2 = a2 + b2 + ab + ba = a2 + b2 + 2a · b.Since a · b = ab cos γ′ = −ab cos γ, we get c2 = a2 + b2 − 2ab cos γ.

a

bc

α

β γ

a

bc

γ γ′β

α



The law of sines

We have a + b + c = 0. Wedging with a, and then with b, we getthe relations

a ∧ b = b ∧ c = c ∧ a = A,where A is the double of the oriented area of the triangle. Using theangles, and the fact that suplementary angles have the same sine,

abi sin γ = bc i sinα = cai sin β,

and this clearly implies the law of sines (divide by abc i).

It is interesting to note that

−ba = bae iγ, −cb = cbe iα, −ac = ace iβ,

where the minus signs appear because, say, γ = ∠(b,−a).Multiplying the three expressions, and simplifying the factor a2b2c2

on both sides, we gete iαe iβe iγ = −1.

Since −1 = e iπ, the identity is equivalent to α + β + γ = π.S. Xambo (UPC/BSC) GA: Structures and Applications 3-4/10/2019 41 / 130


The orthocenter and Euler’s line

A

b

a

c b

a

cG

HG

O

−a/2

−b/2−c/2

x

yz

B C

A

B C

Orthocenter H (left) and circumcenter O (right) of the triangleABC . The point G denotes the barycenter , or centroid , of ABC ,which by definition is given by (A + B + C )/3; a, b and c are theposition vectors of the vertices A, B and C with respect to itsbaricenter G , so that a + b + c = 0. We also have x + y + z = 0.



Wessel’s algebra expression for H . The points of the heights ofA,B ,C have the form, respectively,

a + λxi , b + µyi , c + ρzi (λ, µ, ρ ∈ R).

The intersection of the first two heights, say HAB , can be obtained bysolving for λ and µ the equation a + λxi = b + µyi , or

λxi = z + µyi . (6)

Multiplying by x on the left, we get λx2i = xz + µxyi . The scalarpart of this equation is 0 = x · z + µ(x ∧ y)i = x · z + µAi . Thusµi = −(x · z)A−1, which yields HAB = b− (x · z)yA−1 (seen as apoint on the height of B). To get the expression of HAB as a point ofthe height of A, multiply equation (6) by y on the left and get, aftera similar calculation, HAB = a− (z · y)xA−1 (we have usedA = y ∧ z, as found in the discussion of the law of sines).



So we have HAB = a− (z · y)xA−1 = b− (x · z)yA−1.

Permuting cyclically once, we also haveHBC = b− (x · z)yA−1 = c− (y · x)zA−1, which show thatHAB = HBC and that the three heights are concurring there. Thisshows the existence of the orthocencter, H , and that it can beexpressed as

H = a− (z · y)xA−1 = b− (x · z)yA−1 = c− (y · x)zA−1.

Finally, summing the three expressions we get

H = −13((x · y)z + (y · z)x + (z · x)y)A−1.

Reasoning with the right figure on page 42, we see that thehomothety with center G and modulus −1/2 maps the vertices ABCto the midpoints of the respective opposite sides, so it maps theheights to the medians and therefore maps H to O: O = −1/2H(Euler), which provides an expression for O in terms of the vertices.



W ' R(2)*

Let e1, e2 be an orthonomal basis of E2. The elements of W = G2

have the form α + β1e1 + β2e2 + γi and it turns out that the linearisomorphism G2 → R(2),

α + β1e1 + β2e2 + γi 7→ αI2 + β1

[1 00 −1

]+ β2

[0 11 0

]+ γ

[0 1−1 0

]=

(α + β1 γ + β2

−γ + β2 α− β1

)is an algebra isomorphism (the basic fact is that the matrices e1 ande2 corresponding to e1 and e2 satisfy Clifford’s relations, namelye2

1 = e22 = I2 and e1e2 = −e2e1, the latter being the image of i).

* Given an algebra A we let A(m) denote the algebra of m ×mmatrices with entries in A.


Examples in low dimensions P = G3

Pauli algebra

P = G3 = 〈1, e1, e2, e3, e23, e31, e12, e123 = i〉, i 2 = −1.

e23 = ie1 = e1i , e31 = ie2 = e2i , e12 = ie3 = e3i .

General element: (α + βi) + (x + y i) (α, β ∈ R, x , y ∈ E3).

G+

2 = {α + x i} = H ' H (quaternion field).

3 q(α + x i) = (α + x i)(α + x i ) = α2 + x2

Hamilton units: i 1 = −e1i , i 2 = −e2i , i 3 = −e3i . They satisfyHamilton’s relations: i 2

1 = i 22 = i 3

3 = i 1i 2i 3 = −1.

We will return to this structure later (page 93) from a different pointof view.


Examples in low dimensions Pauli algebra

Clifford units of E3 and Hodge duality

e2

e3

e1

e1i

e3i

e2i

i = e1e2e3

1

e1, e2, e3

e1i, e2i, e3i

Scalars

Vectors–oriented segments–polar vectors

Bivectors–oriented areas–axial vectors

Pseudoscalars–oriented volumes



Vector algebra revisited

3 In the case of (oriented) E3, the Hodge duality

E3 = G1 → G2, x 7→ x∗ = ix = x i = i · x = x · i ,is an isometry, where in this case the pseudoscalar e1e2e3 (e1, e2, e3 isa positively oriented orthonormal basis) is denoted by i . The inverseisometry is given by

b 7→ b∗ = −ib = −bi , b ∈ G2.

3 For any vector x , x∗ ∧ x = (x∗x)3 = (ix2)3 = x2i = x ∧ x∗.



Cross product

The cross product x × x ′ of two vectors x , x ′ ∈ E3 is defined as theHodge dual of x ∧ x ′, namely

x × x ′ = (x ∧ x ′)∗ = −(x ∧ x ′)i , or x ∧ x ′ = i(x × x ′).

3 With A(x , x ′) = V (x , x ′) (the 2-volume (area) of theparallelogram /x , x ′/), we get |x × x ′| = A(x , x ′). Indeed,

|x × x ′|2 = (x × x ′)2 = q(x × x ′) = q(x ∧ x ′) = Gq(x , x ′) = A(x , x ′)2.

3 We also have x × x ′ = −(ix) · x ′ = −x · (ix ′). This follows easilyfrom the properties of the inner product and the fact thativ = i · v = v · i = v i for any vector v . For instance,x × x ′ = −i(x ∧ x ′) = −i · (x ∧ x ′) = −(i · x) · x ′ = −(ix) · x ′.3 x × x ′ is orthogonal to x and to x ′. For instance,(x × x ′) · x ′ = (−(ix) · x ′) · x ′ = −(ix) · (x ′ ∧ x ′) = 0.



3 The triple x , x ′, x × x ′ is positively oriented, because

x ∧ x ′ ∧ (x × x ′) = i(x × x ′) ∧ (x × x ′) = (x × x ′)2i .

3 x · (x ′ × x ′′) = −i(x ∧ x ′ ∧ x ′′) = Vi(x , x′, x ′′) (mixed product).

3 If x , x ′, x ′′ ∈ E3, (x × x ′)× x ′′ = (x ′ · x ′′)x − (x · x ′′)x ′ (doublevector product), for (x × x ′)× x ′′ = −i(x × x ′) · x ′′ = (x ∧ x ′) · x ′′,and this agrees with the claimed expression.

Remark . Since i changes sign when the orientation of space ischanged, the same occurs to the cross product: x ×i x

′ = −x ×−i x ′,which is not more than the tautology i(x ∧ x ′) = −(−i(x × x ′)).This fact is traditionally described by saying that the cross product isan axial vector , in contrast with the ordinary vectors, which aredubbed to be polar vectors. Altogether, an esoteric way ofrecognizing that the cross product is the exterior product in disguise.



Pauli representation: P = G3 ' C(2)

Let e1, e2, e3 be an orthonormal basis of E3. An element of G3 can beuniquely written in the form ξ0 + ξ1e1 + ξ2e2 + ξ3e3, whereξ0, ξ1, ξ2, ξ3 ∈ C. The the map G3 → C(2),

ξ0 + ξ1e1 + ξ2e2 + ξ3e3 7→ ξ0σ0 + ξ1σ1 + ξ2σ2 + ξ3σ3,

where ξi are elements of C on the left hand side and regarded aselements of C on the right hand side, and

σ0 = I2 =

[1 00 1

], σ1 =

[0 11 0

], σ2 =

[0 −ii 0

], σ3 =

[1 00 −1

]are the so-called Pauli matrices, is an algebra isomorphism. The mainfact for the proof is that the Pauli matrices satisfy Clifford’s relations:

σ2j = σ0, σjσk + σkσj = 0 if j 6= k .

1


Examples in low dimensions G2,1, G1,2

The isomorphism Gr+1,s+1 ' Gr ,s(2)

Consider a decomposition

Er+1,s+1 = Er ,s ⊥ 〈e, e〉,where e2 = −e2 = 1 and e · e = 0. Then any element of Gr+1,s+1 canbe written in a unique way in the form a + be + be + cee,a, b, b, c ∈ Gr ,s and the linear isomorphism Gr+1,s+1 → Gr ,s(2),

a + be + be + cee 7→(a + b c + bc − b a − b

)(7)

is an isomorphism of algebras. The key points are that the matrixcorresponding to ee is the product ee of the matrices e correspondingto e and e corresponding to e, and that e2 = −e2 = I2.

G2,1 ' G1(2) ' (2R)(2) = 2R(2).

G1,2 ' G1(2) ' C(2).


Examples in low dimensions G3

3 For any n > 0, we have

Gn+2 ' Gn ⊗ G2 ' Gn(2), Gn+2 ' Gn ⊗ G2 ' Gn ⊗H. (8)

Thus G3 ' G1 ⊗ G2 ' (2R)⊗H = 2H (biquaternion algebra)

Indeed, view En+2 as En ⊥ E2, and let e, e ′ be an orthonormal basisof E2. The linear map f : En+2 → Gn ⊗ G2,

x + λe + λ′e ′ 7→ x ⊗ ee ′ + λ(1⊗ e) + λ′(1⊗ e ′),

where x ∈ En is the vector x ∈ En but with x2 = −x2. Thenf (x + λe + λ′e ′)2 = (x2 + λ2 + λ′2)(1⊗ 1) = q(x + λe + λ′e ′)(1⊗ 1)and hence f extends uniquely to an algebra homomorphismf : Gn+2 → Gn ⊗ G2. Since any element of Gn+2 can be written in aunique way in the form a + be + b′e ′ + cee ′, with a, b, b′, c ∈ Gn, wehave f (a + be + b′e ′ + cee ′) = a ⊗ 1 + b ⊗ e + b′ ⊗ e ′ + c ⊗ ee ′,where a ∈ Gn, say, is the same as a ∈ Gn but obeying the metric ofEn instead of that of En. It is thus clear that f is a linearisomorphism and hence an algebra isomorphism. 1

The other isomorphism can be proved in an analogous way.S. Xambo (UPC/BSC) GA: Structures and Applications 3-4/10/2019 53 / 130

Examples in low dimensions D = G1,3

Dirac algebra, D = G1,3

E1,3 = 〈e0, e1, e2, e3〉.Let i = e0123, σk = eke0. Then i 2 = −1, i anticommutes withvectors, and σ1i = −e23, σ2i = −e31, σ3i = −e12. Thus

D = 〈1, e0, e1, e2, e3,σ1,σ2,σ3,σ1i ,σ2i ,σ3i , e0i , e1i , e2i , e3i , i〉.A general element has the form (α + βi) + (x + y i) + (E + Bi),α, β ∈ R, x , y ∈ E1,3, E ,B ∈ E = 〈σ1,σ2,σ3〉. The space E isEuclidean and σ1, σ2, σ3 is an orthonormal basis. Its pseudoscalaragrees with i : σ1σ2σ3 = e1e0e2e0e3e0 = −e1e2e3e0 = e0e1e2e3.

D+= 〈1,σ1,σ2,σ3, iσ1, iσ2, iσ3, i〉 ' P(E).

Its elements have the form (α + βi) + (E + Bi).



Dirac repesentation

Dirac introduced the famous matrices Γµ ∈ C(4) in his endeavor tobuild a relativistic theory of the electron, namely:

Γ0 =

[σ0 00 −σ0

], Γk =

[0 −σkσk 0

].

The matrices Γµ satisfy the Clifford relations for the signatureη = (+,−,−,−):

ΓµΓν + ΓνΓµ = 2ηµν ,and so we have a representation D → C(4) (i.e. an algebrahomomorphism) such that eµ 7→ Γµ. Note, however, that D hasdimension 16 while C(4) has dimension 32.

On the other hand, G1,3 ' G0,2(2) ' H(2). So we have G1,1 ' R(2),G1,2 ' C(2), G1,3 ' H(2).



Basis of D in the γ-notation

1∗

1

γ0 γ1 γ2 γ3

σ1 σ2 σ3 σ∗1 σ∗2 σ∗3

γ∗0 γ∗1 γ∗2 γ∗3

x∗ = x i

η(x) = +1

η(x) = −1

1


Spinorial groupsGeneration of isometries

Reflections and axial symmetries. Composing reflections.Versors. The Lipschitz group. Pinors and spinors. Rotors.

Primacy of the rotor group

Generating rotorsPlane rotors. Cartan-Dieudonne for rotors. Geometric

covariance


Spinorial groups Generation of isometries

Reflections and axial symmetries

If x ∈ E×

, define the linear automorphism mx = x : G → G by

x(a) = −xax−1 = xax−1.

3 For any vector x ′, x(x ′) is the reflection of x ′ along x (or acrossx⊥). For x(x) = −x , plainly; and x(x ′) = x ′ if x ′ ∈ x⊥, asxx ′ = −x ′x .

For effective computations, it is sometimes useful to observe that if xis a unit vector, then

x(x ′) = x ′ − 2εxx(x · x ′).Indeed, the right hand side is linear in x ′, leaves x ′ fixed if x ′ ∈ x⊥,and it maps x to x − 2εxx(x · x) = −x , as x · x = x2 = εx .



The map sx : E → E , x ′ 7→ xx ′x−1 = −x(x ′) = −mx(x ′) is the axialsymmetry about x , as sx(x) = x and sx(x ′) = −x ′ if x ′ ∈ x⊥.

3 The map x is not an algebra automorphism, but satisfies:

x(ab) = −x(ab)x−1 = −xax−1xbx−1 = −x(a)x(b).

It follows that x is grade-preserving, as this relation implies that

x(x1 · · · xk) = (−1)k−1x(x1) · · · x(xk).

3 Moreover, x is an isometry of G:

q(x(a)) = ((−xax−1)˜ (−xax−1))0

= (x−1ax2ax−1)0 = (xaax−1)0 = q(a).

In the third step we have used that x2 is a scalar, so that itcommutes with a, and that x−1x2 = x .



Composing reflections

Let x1, . . . , xk ∈ E×

and v = x1 · · · xk . Then

(x1 · · · xk)(a) = x1 · · · xkax−1k · · · x−1

1 = v av−1.

The expressions v form a group under the geometric product. Wedenote it by V ′r ,s . If we define v : G → G as the map a 7→ v av−1,then the map V ′r ,s → Or ,s , v 7→ v is surjective, by the celebratedCartan-Dieudonne theorem. So in principle V ′r ,s should suffice for thestudy of Or ,s by means of the geometric algebra. This is indeed thecase, but it turns out that it will be useful to have a more conceptualcharacterization of it. This will be accomplished by paying attentionto the following two properties: any v is even or odd, and v(E ) = E .



Versors

A versor of Er ,s is a multivector v ∈ G+ t G− (so v is an even or oddmultivector), which is invertible (so v ∈ G×) and satisfies vEv−1 = E(equivalently, vEv−1 ⊆ E ).

The set of versors will be denoted by V = Vr ,s . By the precedingcomments, we have that V ′r ,s ⊆ Vr ,s . It is also obvious that

R× ⊆ Vr ,s .



The Lipschitz group

3 The set V is a subgroup of G× .

3 For any versor v , v ∈ Or ,s , where v is defined as for the elementsof V ′, namely v(x) = v xv−1 for x ∈ E .

3 ρ : V → Or ,s , v 7→ v , is an onto group homomorphism (called thetwisted adjoint representation).

3 The kernel of ρ is R×

.

3 V is generated by E×

and hence V = V ′.



The even Lipschitz group

3 Consider the subgroup V+= V+

r ,s of Vr ,s formed by the even

versors. For any v ∈ V+, v is the product of an even number of

reflections and hence v ∈ SOr ,s .

3 The map ρ : V+

r ,s → SOr ,s is an onto group homomorphism and its

kernel is R×

.

Given an element v ∈ G±, in particular if v is a versor, the parity ofv is defined to be ±1, and will be denoted by π(v). In other words,π(v) = 1 if v is even and π(v) = −1 if v is odd. Notice that for suchelements we have v = π(v)v . The map π : V → {±1} is clearly ontoand it is easily checked that it is a group homomorphism. Its kernel isthe group V+

of even versors.



Pinors and spinors

3 The group Pin = Pinr ,s is defined as the subgroup of V formed bythe unit versors, that is, the versors v such that q(v) = v v = ±1.The elements of Pin are called pinors.

3 The group Pin coincides with the subgroup of V formed by theproducts of unit vectors.

3 The homomorphism ρ : Pinr ,s → Or ,s , v 7→ v , is onto and itskernel is {±1}.3 The group Spin = Spinr ,s = Pin

+

r ,s is the subgroup of evenelements of Pinr ,s , that is, pinors that are the product of an evennumber of unit vectors. The elements of Spin are called spinors.

3 ρ : Spinr ,s → SOr ,s is an onto homomorphism with kernel {±1}.



Rotors

3 Since spinors R are unit pinors, in general we have RR = ±1. IfRR = 1, we say that the spinor is a rotor . The set of rotors will bedenoted by Rr ,s .

In the Euclidean and anti-Euclidean cases (En = En,0 and En = E0,n),all spinors are rotors, but otherwise this is not the case: if r , s > 0and v = uu with u2 = 1 and u2 = −1, then v is a spinor, but not arotor, for v v = −1.

3 The set of rotors Rr ,s is a normal subgroup of Spinr ,s . Actually, itis the kernel of the group homomorphism Spinr ,s → {±1}, v 7→ v v .



3 The image of Rr ,s in SOr ,s by ρ will be denoted by SO0r ,s . Thus we

have an onto group homomorphism

ρ : Rr ,s → SO0r ,s (9)

whose kernel is {±1}. In the Euclidean and anti-Euclidean cases,SO0

n = SOn, but for r , s > 0 we have seen that SO0r ,s is a proper

subgroup of SOr ,s .

3 If b is a bivector (b ∈ G2), R = ±eb is a rotor. Indeed, it is even,

and R = ±e b = ±e−b, hence RR = 1. For the condition RxR ∈ Efor all x ∈ E , see [2, §5.4.6].

3 For n 6 5, R = {R ∈ G+ |RR = 1}. Here the important fact isthat R even and RR = 1 imply that RER = E . For n > 5, thisimplication is no longer true. See [2, §5.1.1].



Primacy of the rotor group

3 The rotor group R determines the structure of the spinor groupS = Spin and of the pinor group P = Pin:

If (r , s) = (n, 0) (Euclidean case) or (r , s) = (0, n) (anti-Euclideancase), then S = R and P = Rt uR for any given unit vector u. Asa consequence, On = SOn tmuSOn.

If r , s > 1, fix unit vectors u and u such that u2 = 1 and u2 = −1.Then we have: S = Rt uuR andP = Rt uRt uRt uuR = S t uS. As a consequence,SOr ,s = SO0

r ,s t ρ(uu)SO0r ,s and Or ,s = SOr ,s tmuSOr ,s .


Spinorial groups Generating rotors

Plane rotors

The most simple rotors are given as the product of two unit vectorsof the same signature. As we will see, such rotors generate all rotors.

Given two linearly independent unit vectors u and v , R = vu is aspinor. We will say that it is the plane spinor defined by u and v :

R = u · v − u ∧ v .

Since RR = u2v 2, we see that R is a rotor (in which case we will saythat it is a plane rotor) if and only if u2v 2 = 1, that is, ifu2 = v 2 = η = ±1. In both cases,

(u ∧ v)2 = −(v ∧ u)(u ∧ v) = −q(u ∧ v) = −Gq(u, v)

= (u · v)2 − u2v 2 = (u · v)2 − 1.

Thus plane rotors are naturally classified according to the sign of thisquantity. We recall that the plane [u ∧ v ] = 〈u, v〉 is singular if andonly if (u ∧ v)2 = 0 (page 22).



Elliptic plane rotors

This is the case (u ∧ v)2 < 0, or (u · v)2 < 1. Thus there exists aunique α ∈ (0, π) such that u · v = cosα and hence

(u ∧ v)2 = cos2 α− 1 = − sin2 α.

If we set i = (u ∧ v)/ sinα, then we have that i 2 = −1 and

R = cosα− i sinα = e−iα. (10)

In this case, the regular plane 〈u, v〉 does not have non-zero isotropicvectors (check that the discriminant of the quadratic (λu + v)2 = 0is 4(u ∧ v)2) and hence it is Euclidean if η = 1 and anti-Euclidean ifη = −1.

Note that in the Euclidean space all plane rotors are elliptic.



Hyperbolic plane rotors.

If (u ∧ v)2 > 0, then (u · v)2 > 1 and there exists a unique α > 0such that u · v = ε chα, where now ε is the sign of u · v , and

(u ∧ v)2 = ch 2α− 1 = sh 2α.

Setting ι = (u ∧ v)/ shα, we have that ι2 = 1 and hence

R = ε chα− ι shα = εe−ιεα. (11)

In this case the regular plane 〈u, v〉 has non-zero isotropic vectors, asthe discriminant of the quadratic equation (λu − v)2 = 0 is positive,so that the plane is a hyperbolic plane (this means that it hassignature (1,1)). Actually it is easy to see that λ± = ηεe±α are thesolutions of that quadratic equation.



Parabolic plane rotors.

If (u ∧ v)2 = 0, then (u · v)2 = 1, or u · v = ε = ±1, and so

R = ε(1− εu ∧ v) = εe−εu∧v .

If this is the case we know that the plane 〈u, v〉 is singular. This canbe checked directly on noticing that the vector εv − ηu is orthogonalto u and v .



3 Up to sign, all plane rotors are exponentials of bivectors and theyare connected to 1 or to −1 (by this we mean that there is acontinuous path on the rotor group joining the given rotor to either 1or −1).

Indeed, in the parabolic case R = εe−εu∧v is connected to ε by therotor path

R(t) = εe−εtu∧v (0 6 t 6 1),as R(1) = R and R(0) = ε.

Similarly, any elliptic plane rotor is connected to 1 (it is enough to letthe angle α in equation (10) vary in the interval [0, 1]) and anyhyperbolic rotor is connected to ε (let the parameter α in equation(11) vary in the interval [0, 1]).



Cartan-Dieudonne for rotors

3 If v is a versor and u a unit vector, then vu = u′v , with u′ a unitvector that has the same signature as u. Indeed, we can write

vu = vuv−1v = π(v) v(u)v

and u′ = π(v) v(u) satisfies that q(u′) = q(u) because v is anisometry.

3 Any rotor can be expressed as a product of plane rotors. To seethis, let R ∈ R, say R = u1 · · · u2k , with the uj unit vectors. Usingthe preceding observation, we can express R as a product of unitvectors in such a way that no positive vector appears later than anegative one. Since R is a rotor, the number of negative vectors in itmust be even, and consequently the number of positive vectors isalso even. Grouping the vectors in consecutive pairs, we get that R isthe product of k plane rotors.



3 Any rotor is connected to 1 or −1.

Indeed, since each of the k plane rotors is connected to 1 or to −1,the same is true for their product.

3 If r > 2 or s > 2, then any rotor is connected to 1. Consequently,R is connected.

Indeed, we can choose two unit vectors of the same signature, say uand v . Then (uv)2 = −1 and the rotor path

t 7→ etuv = cos t + uv sin t (0 6 t 6 π)

connects 1 (t = 0) to −1 (t = π).

1



Geometric covariance

3 Let v ∈ V and a, a′ ∈ G. Then the map v : G → G is a lineargraded automorphism that has the following additional properties:

v(a) = v(a) and v(a) = v(a).

v(aa′) = π(v) v(a) v(a′).

v(a ∧ a′) = π(v) v(x) ∧ v(a′) and v(a · a′) = π(v) v(a) · v(a′).

For any versor v , the map v : G → G is an isometry. It follws thatif u is a pinor (spinor), then v(u) is a pinor (spinor) of the samesignature as u. In particular, v(u) is a rotor when u is a rotor.

If v is even, then v is an automorphism of the geometric algebra,by which we mean that it is a linear automorphism that preserves thegrading, that is an automorphism of the geometric, outer and innerproducts, and that it commutes with the parity and reverseinvolutions.



3 For any versor v , the map v : G → G is an isometry. It follws thatif u is a pinor (spinor), then v(u) is a pinor (spinor) of the samesignature as u. In particular, v(u) is a rotor when u is a rotor.

3 If v is even, then v is an automorphism of the geometric algebra,by which we mean that it is a linear automorphism that preserves thegrading, that is an automorphism of the geometric, outer and innerproducts, and that it commutes with the parity and reverseinvolutions.


Examples of rotors in lowdimensions

Notations

Rotations in the Euclidean space

Rotors of E1 and E1

Rotors of E2

Rotors of E1,1

Rotors of E3Structure. Exponential form. Reflections of E3


AnglesAngles and rotations. Angle vector-bivector. Angle

bivector-bivector.

Spherical trigonometryArea. Fundamental relations. Cosine laws. Law of sines.

Altitudes. Polar triangle


Examples of rotors in low dimensions Notations

In the examples that follow, we set

P = Pr ,s = Pinr ,s ,

S = Sr ,s = Spinr ,s , so that Rr ,s = S+

r ,s .

In particular, Pn = Pn,0, Sn = Sn,0 and Rn = Rn,0.

For the anti-Euclidean signature, we will write

Pn = Pn = P0,n,

Sn = Sn = S0,n,

Rn = Rn = S0,n.

The same conventions are made to denote, when required, thesignature for the space E and of the groups O, SO and SO0.


Examples of rotors in low dimensions Rotations in the Euclidean space

One nice illustration of geometric algebra is how neatly it providesfresh approaches to Euclidean geometry En. We have seen someinstances before (see Wessel algebra, pages 39 and following, andPauli algrebra, pages 46 and following). Here the focus will be onwhat has geometric algebra to say about simple rotations in En.

3 Let u and u′ be unit linearly independent vectors of En andθ = ∠(u, u′). Then the rotation R produced by the rotor R = u′u isthe rotation in the plane P = 〈u, u′〉 of amplitude α = 2θ.

Indeed, since R is the identity on P⊥, it amounts to a rotation in P .Let i = iP be the unit area of P . Then u and u⊥ = ui form anorthonormal basis of P and u′ = u cos θ + u⊥ sin θ. HenceR = u′u = cos θ − i sin θ = e−iθ. We have proved what we will callspinorial formula for the rotation R :

R(x) = RxR = e−iθxe iθ. (12)



That the rotation angle is α = 2θ can be checked by its effect on u:

R(u) = e−iθue iθ = ue2iθ = u cosα + u⊥ sinα.

u

u′

θ

ui

cos θ

sin θ

i

2θ

R(u)



3 In the spinorial formula (12), what matters is the area elementa = iα = 2iθ, a bivector, not the separate factors i and α. Indeed,we can define Ra = e−a/2 and then we have

Ra(x) = e−a/2xea/2, (13)

which delivers the identity if a = 0 and otherwise the rotation in theplane P = [a] by the amplitude α = −ia.

Rotations of E3

Suppose n = 3. If u is a non-zero normal vector to the oriented planeP , its Hodge dual is u∗ = iu, where here we let i denote thepseudoscalar of E3, and we can consider the rotor Ru∗ = e−iu/2.

3 The rotation ru = Ru∗ has axis 〈u〉 and amplitude α = |u|.Indeed, the relation i = iu/|u| tells us that u∗ = iu = i |u|, so that[u∗] = [i ] = P and the rotation angle is −i(i |u|) = |u|. In sum,

ru(x) = e−iu/2xe iu/2. (14)



Olinde Rodrigues formulas

Let u, u′ ∈ E3 be unit vectors and α, α′ ∈ R. Let r = ru,α andr ′ = ru′,α′ be the rotations about u and u′ by the angles α and α′,respectively. We seek an expression for the axis an amplitude of thecomposition ru′,α′ru,α.

The key is that if R and R ′ are the right rotors of r and r ′, then RR ′

is the right rotor of r ′′ = r ′r . We have, setting C = cosα/2,S = sinα/2, with similar shorthands for α′ and α′′,

R = e iuα/2 = C + iuS , R ′ = e iu′α′/2 = C + iu′S ′. (15)

On expanding RR ′ and equating scalar and bivector parts with thoseof R ′′ = C ′′ + iu′′S ′′ we get the expressions

C ′′ = CS ′ − (u · u′)SS ′, u′′S ′′ = uSC ′ + u′CS ′ − (u × u′)SS ′ (16)


Examples of rotors in low dimensions Rotors of E1 and E1

1

u

G1 = E1 = uR

G0 = R

−u

−1

G = G0 + G1

The rotor group of E1 and E1 is {±1}, which is disconnected. Thepinor group P is {±1,±u}, where u ∈ E1 is a unit vector. In theEuclidean case, u2 = 1 and we have a group isomorphic to Z2 × Z2.In the anti-Euclidean case, u2 = −1 and we have a cyclic group oforder 4 generated by u.


Examples of rotors in low dimensions Rotors of E1 and E1

For n = 1, the possible signatures are 1 ∼ (1, 0) and 1 ∼ (0, 1), and

S1 = R1 = S1 = R1 = {±1} ' Z2,

P1 = P1 = {±1,±u},where u ∈ E is any given unit vector. If u2 = −1 (which happens inthe case E1), we have the group

{1, u,−1,−u} = {1, u, u2, u3} ' Z4,

and if u2 = 1 (which happens in the case E1), then the group isisomorphic to

Z2 × Z2 = {(0, 1), (1, 0), (0, 1), (1, 1)},as u2 = (−1)2 = (−u)2 = 1. So P1 ' Z2 × Z2 (the Klein group oforder 4) in the Euclidean case and P1 ' Z4 (the cyclic group of order4) in the anti-Euclidean case.

Finally, O1 = {±Id} and SO1 = {Id}, with ρ(±1) = Id andρ(±u) = −Id.


Examples of rotors in low dimensions Rotors of E2

The spaces E2 and E2 have a similar treatment. If i is the area unit,G+

= {α + βi |α, β ∈ R} = C. Since i 2 = −1 and i = −i ,(α + βi)(α + βi)∼ = α2 + β2 and

S2 = R2 = S2 = R2 = {α + βi ∈ G+ |α2 + β2 = 1} = U1,

where U1 = {e iθ : 0 6 θ < 2π} denotes the unit circle in C. Thisgroup is connected, but not simply connected (a turn of the unitcircle cannot be shrinked to 1, or, more precisely, the fundamentalgroup of U1 is Z, π1(U1) ' Z in symbols.

The isometry ρ(e−iθ) is the counterclockwise rotation (in relation tothe orientation i) by an angle 2θ. Indeed, in this case ianticommutes with vectors and hence ρ(e−iθ)(x) = e−iθxe iθ = xe2iθ,which is the result of rotating x by an angle 2θ.



1

iG2 =iR

G0 = R

−i

−1

G+ = G0 + G2

θ

ei θU1

cos θ

sin θ

(a)

1

i

−i

−1θ

ei θ

U1

(b)

e−i θ

−θ

e2i θ

The rotor group of E2 and E2 is the circleR2 = {e iθ | 0 6 θ < 2π} = U1 in G+

= C (a). The group SO2 is alsoisomorphic to U1, but the 2:1 group homomorphism R2 → SO2

becomes the map e−iθ 7→ e2iθ (b), so that going around the rotorgroup U1 once in a clock-wise sense produces going around therotation group U1 twice in a counterclockwise sense.



1

i

−i

−1θ

v = uei θuU1

v⊥

If u is any given unit vector, say u = e1,then

P2 = P2 = R2 t uR2 = U1 t uU1,

and ρ(ue iθ) is the reflection along thedirection v = ue iθ, for −vvv−1 = −v .Thus we see that topologicallyP2 = P2 is the disjoint unionof two circles: U1 ⊂ C = G+

anduU1 ⊂ G− = E2. The 2:1 map uU1 → O−2 is manifested in the factthat v and −v give the same reflection. In other words, topologicallyO−2 is also a circle that is circled twice when going once around thecircle uU1.


Examples of rotors in low dimensions Rotors of E1,1

Let e0, e1 be an orthonormal basis of E1,1 and i = e1e0 (in Lorentzianspaces, i.e., spaces of signature (1, n − 1), it is customary to set the“temporal axis” e0 as the “vertical direction”). We still haveG+

= {α + βi |α, β ∈ R} and (α + βi)∼ = α− βi , but

(α + βi)(α + βi)∼ = α2 − β2

inasmuch as i 2 = 1. Thus the rotor group is

R1,1 = {α + βi |α2 − β2 = 1},which shows that it has two connected (and simply connected)components: the two branches of a hyperbola in G+

. These branchesare distinguished by the sign of α and are parameterized byα = ε chλ, β = shλ (ε = ±1, λ ∈ R).



After a few calculations (in which we use that i anticommutes with e0

and e1, and basic properties of ch and sh), we find that the action of

R = Rε,λ = ε chλ + i shλ = εeελi

on a vector x is given by

R(x) = RxR = xe−ε2λi .

In particular we get, using that e0i = −e1 and e1i = −e0, thefollowing (Lorentz) transformation:

R(e0) = e0 ch 2ελ + e1 sh 2ελ, R(e1) = e0 sh 2ελ + e1 ch 2ελ.

We see that Rε,λ and R−ε,−λ produce the same isometry (as theymust, because R−ε,−λ = −Rε,λ and ±R yield the same isometry), andconsequently SO0

1,1 is isomorphic to (the additive group) R via themap (in matrix form)

λ 7→ Hλ =

(ch 2λ sh 2λsh 2λ ch 2λ

).



α2 − β2 = 1

α + βi

−(α + βi)

i

1

G+1,1 = Ca)

light cone

e1

R(e0)

e0

τ = ξτ = −ξ

R(e1)

b)

Geometric aspects of the E1,1 rotors. The “light cone” is thephysically motivated name for the isotropic cone q(x) = 0, which interms of the components of x with respect to e1 and e0, sayx = ξe1 + τe0, is given by the equation τ 2 − ξ2 = 0 and so it iscomposed of the pair of lines τ = ±ξ.



Thus we have seen that R1,1 has two connected components, whichimplies that P1,1 has eight connected components. Indeed, if we setR+ = {R1,λ} and R− = {R−1,λ}, these components are

R+, R−, e0R+, e0R−, e1R+, e1R−, iR+, iR−.From this it follows that O1,1 has four connected components:

SO01,1, me0SO0

1,1, me1SO01,1, and me1me0SO0

1,1 = −SO01,1.



Structure

Let e1, e2, e3 be an orthonormal basis of E3, i = e1e2e3 (unit volume,which satisfies i 2 = −1 and i = −i), and

H = G+

= {h = σ + x i |σ ∈ R, x ∈ E3}the even geometric algebra (see page 46). We know that H is anEuclidean space and a (skew) field: q(h) = σ2 + x2 andh−1 = h/q(h) if h 6= 0. consequently

S3 = S3 = R3 = R3 = {σ+x i ∈ H |σ2+x2 = 1} = {h ∈ H | q(h) = 1}.This shows that S3 = R3 is the unit sphere of H, which is connectedand simply connected. Then the double cover

ρ : R3 → SO3

yields that SO3 is connected. Note that for h ∈ R3 we have h = h−1

and hence h acts on E3 as the rotation

ρh : x 7→ hxh−1.



Exponential form

What is the axis and the amplitude of ρh?

3 Given h = σ + x i ∈ R3, h 6= ±1 (i.e, x 6= 0), it is clear that thereis a unique ϕ ∈ (0, π) such that σ = cosϕ and |x | = sinϕ.

3 If we set u = x/|x | = x/ sinϕ, then h = cosϕ + ui sinϕ = euiϕ.Here we use that (ui)2 = −1.

3 Therefore ρh = ru,−2ϕ (the rotation about u, or about x , throughthe angle −2ϕ).



3 If h = uv (u, v ∈ E3 linearly independent unit vectors withu · v = cos(θ), θ ∈ (0, π)), its quaternion form ish = u · v + u ∧ v = cos θ + (u × v)i . Since |u × v | = sin θ, we findϕ = θ and hence uv = ew iθ, where w = (u × v)/ sin θ, andρuv = rw ,−2θ.

For example, if we let h = ±i k = ±ek i (see page 46 for thedefinition of the Hamilton units i k), ρh is the axial symmetry aboutek (k = 1, 2, 3).



Reflections of E3

The two components of P3 are R3 and uR3, where u is any givenunit vector, and the two components of O3 are SO3 and muSO3.

If f is a rotation, muf is a reflection, say mv , except whenmuf = −Id, i.e., when f = −mu = su. To determine v , let f = rw ,αand R = e−w iα/2, so that f = R . Then

muf = ρ(u)ρ(R) = ρ(uR).

If uR were a vector, we would have muf = muR , the reflection in thedirection of uR . But in general uR is the sum of a vector and apseudoscalar, so we have to work a bit more. If we apply muf to uR ,we get −uR(uR)Ru = −uR (here we have used geometriccovariance). Therefore we also have (muf )((uR)1) = −(uR)1 andthis implies that muf = mv , where v = (uR)1. NowuR = u cos α

2− uw i sin α

2, and so



v = (uR)1 = u cos α2− (u ∧ w)i sin α

2= u cos α

2+ (u × w) sin α

2.

We have used that uw i = (u ·w)i + (u ∧w)i and that its vector partis (u ∧ w)i . Notice that v need not be a unit vector, but that it isnon-zero because w 6= ±u and α 6= π, as otherwise f = su.


Examples of rotors in low dimensions Angles

We switch, on account of the issues we wish to discuss, to thenotation introduced on page 39. Thus vectors will be denoted bylower-case letters in boldface, like a or b, their norms by a or b, the3d pseudoscalar by i, and so on.

Angles and rotations

The nature of a plane angle is a bivector (see the discussion on pages80 and following), but in 3d we can just as well, on account of Hodgeduality, take the view that a (plane) angle is determined by a vectora. If a distinction is to be made, we will write ∠a to denote the angledetermined by a, but most of the time the angle symbol will not beneeded.

This determination is mediated by the following objects: the rotorR = Ra = e−ia/2, and the rotation R , x 7→ RxR . The magnitude a ofa, which is the amplitude of the rotation R (cf. page 82), will also becalled the amplitude of ∠a.

For the null angle, determined by a = 0, we have a = 0, R = 1 andR = Id. So we may turn to non-null angles.


Examples of rotors in low dimensions Angles

Angle vector-bivector and bivector-bivector. In the remainingslides of this talk we follow the masterful Appendix A of [7] —withsome slight variations in notation and emphasis.

a

i

a·iα

a′

a′′

u

v

w=u×v

a bγ

c

iγ =icγ

iBiA

iB =bciA = ca

iAiB = ab= eicγ

cosα

sinα

cosα

Left: Angle vector-bivector. To remark the relation a ∧ i = i sinα.Right: angle bivector-bivector. To remark that iAiB = ab = e icγ.


Examples of rotors in low dimensions Spherical trigonometry

αβ

γ

∆

∆γ

∆γ

∆α

∆α

∆β

∆β

Area of a spherical triangle

The area ∆ of an spherical triangleis its angular excess:∆ = α + β + γ − π.

The double lune ∆α has area 4α.It follows that4(α + β + γ)− 2(2∆) = 4π, as∆ is counted six times in the sum4(α + β + γ) (three for the visible∆ and three for its symmetricalwith respect to the center of thesphere. In sum, ∆ = α+β+γ−π.In particular, α + β + γ > π.



Elements of a spherical triangle

An spherical triangle is determined by three unit vectors a,b, c,which are its vertexes. The oriented sides, denoted A,B ,C , are arcsof great circles. The unit bivectors on the sectors determined by thesides will be denoted iA, iB , iC , and we will set A = AiA, B = B iB ,C = C iC .The dihedral angles are denoted by α, β, γ (functionally,aα, bβ, cγ). The altitudes are denoted a,b,c.

A

C

B

O

a

b

c

α

β

γ

O

a

b

c

ab

a

cc



3 The altitudes of a spherical triangle meet at point. Indeed, thevector c · (a ∧ b) is orthogonal to the plane spanned by c and by theorthogonal projection c of c to the plane 〈a,b〉, that is, orthogonal tothe plane containing the altitude of c:c · (c · (a ∧ b)) = (c ∧ c) · (a ∧ b) = 0 andc · (c · (a ∧ b)) = c · (c · (a ∧ b)) = (c ∧ c) · (a ∧ b = 0.

Now the claim follows from the immediate identityc · (a ∧ b) + a · (b ∧ c) + b · (c ∧ a) = 0, which shows that the line〈c, c〉 ∩ 〈a, a〉 is contained in 〈b, b〉.Taking the cross product ofc · (a ∧ b) = (c · a)b− (c · b)a = b cosB − a cosA witha · (b ∧ c) = c cosC − b cosB , we easily get that the vector

a× b cosA cosB + b× c cosB cosC + c× a cosC cosA

gives the direction of the vector common to the three altitudes.



Fundamental relations

We have the following two groups of relations:

bc = eA = cosA + iA sinA, (17a)

ca = eB = cosB + iB sinB , (17b)

ab = eC = cosC + iC sinC . (17c)

iC iB = e iaα = cosα + ia sinα, (18a)

iAiC = e ibβ = cos β + ib sin β, (18b)

iB iA = e icγ = cos γ + ic sin γ, (18c)



Cosine laws

Since (ab)(bc)(ca) = 1, we get eCeAeB = 1, or e−C = eAeB . Usingthe relations (18), we can write

cosC − iC sinC = (cosA + iA sinA)(cosB + iB sinB)

= cosA cosB + iA sinA cosB + iB sinB cosA

+ (cos γ − ic sin γ) sinA sinB ,

where we have used iAiB = (iB iA)−1 = e−icγ in the last step. Thescalar and bivector parts of that relation are, respectively,

cosC = cosA cosB + sinA sinB cos γ (19a)

−iC sinC = iA sinA cosB + iB sinB cosA− ic sinA sinB (19b)

Equation (19a) relates three sides and one angle of an sphericaltriangle and is called the cosine law of sides. From any three of thequantities it involves we can determine the fourth. In particular, wecan determine C from A and B and then equation (19b) supplies iC .



Note that for a right triangle, say γ = π/2, then equation (19a)reduces to cosC = cosA cosB , which is called the Pythagoras’theorem.

Example. The angular distance C between the points a and b withspherical coordinates (ϕ1, λ1) and (ϕ2, λ2) can be obtained byapplying the cosine law of sides to the triangle abc, with c at thenorth pole. Indeed, γ = λ2 − λ1, A = π/2− ϕ1, B = π/2− ϕ2, andhence

cosC = sinϕ1 sinϕ2 + cosϕ1 cosϕ2 cos(λ2 − λ1).



From (iB iA)(iAiC )(iC iB) = −1, e icγe ibβe iaα = −1. Hencee−icγ = −e ibβe iaα. The scalar part of this equation gives the cosinelaw for angles, namely

cos γ = − cosα cos β + sinα sin β cosC . (20)

It gives a relation between the three angles and one side. Lettingc′ = −iiC (the unit vector orthogonal to iC , which satisfies iCc′ = i,or iC = ic′), the bivector part of the equations simplifies to

c sin γ = a sinα cos β + b cosα sin β + c′ sinα sin β sinC . (21)

The vector c′ = −iiC will reapear on page 112 as the third vertex ofthe polar triangle.



Remark . Equations (20) and (21) are equivalent tocos(π − γ) = cos β cosα− (b · a) sin β sinα andc sin(π − γ) = b sin β cosα + a cos β sinα− (b× a) sin β sinα, sora,2αrb,2β = rc,2π−2γ, by the Olinde Rodrigues formulas (page 83).

Remark . If we write equation (20) for α, namelycosα = − cos β cos γ + sin β sin γ cosA, and assume that γ = π/2,we get cosA = cosα/ sin β.



Law of sines

Since a ∧ b = iC sinC , we get a ∧ b ∧ c = c ∧ iC ∧ sinC . Proceedinglikewise with b ∧ c = iA sinA and c ∧ a = iB sinB , we find that

a ∧ b ∧ c = a ∧ iA sinA = b ∧ iB sinB = c ∧ iC sinC . (22)

Now consider iAiB iC = iAe−iaα = iA cosα− iiAa sinα. This gives

(iAiB iC )0 = −(iAai)0 sinα = −i(iAa)3 sinα = −i(a ∧ iA) sinα. Nowusing that (iAiB iC )0 = (iB iC iA)0 = (iAiC iB)0, and analogous reasoning,we get

i(iAiB iC )0 = (a ∧ iA) sinα = (b ∧ iB) sin β = (c ∧ iC ) sin γ. (23)

Therefore, on dividing (22) termwise (23), we get the law of sines:

a ∧ b ∧ c

i(iAiB iC )0=

sinA

sinα=

sinB

sin β=

sinC

sin γ(24)



Altitudes

The altitudes of our spherical triangle are depicted in the figure onpage 101, right. We have seen that a ∧ iA = i sin a, where a is thea-altitude, i.e., ∠(a, iA). Likewise, b ∧ iB = i sin b and c ∧ iC = i sin c.Substituting these values in the formulas (22) and (23), weimmediately get the relations

a ∧ b ∧ c

i= sin a sinA = sin b sinB = sin c sinC , (25)

which are convenient expressions for the volume of the parallelepiped/a,b, c/, and

(iAiB iC )0 = sin a sinα = sin b sin β = sin c sin γ. (26)



3 2Vi(a,b, c)ic = 2(a ∧ b ∧ c)c = eBeA − eAeB .

The first equality is obvious. The left hand side of the second relationcan be expanded (using the expression on page 116 and replacing aband ba by accb and bcca) to

13((accb− acbc) + (bcac− bcca) + (cabc− cbac))

= 13((e−Be−A − e−BeA) + (eAe−B − eAeB) + (eBeA − e−Ae−B))

= 13((eBeA − eAeB) + (eAe−B − e−BeA) + (e−Be−A − e−Ae−B))

and the claim follows because the last three expressions in parenthesiscoincide. For example,eAe−B − e−BeA = eA(e−B + eB)− eAeB − (eB + e−B)eA + eBeA,which is eBeA − eAeB because e−B + eB is a scalar and hence theother two terms cancel.



3 eBeA − eAeB = 2ic sinA sinB sin γ.

This is a straightforward calculation after replacingeA = e iAA = cosA + iA sinA and eB = e iBB = cosB + iB sinB andexpanding the products. We find that all terms cancel, except(iB iA − iAiB) sinA sinB = (e icγ − e−icγ) sinA sinB =2ic sinA sinB sin γ.

3 Vi(a,b, c) = sinA sinB sin γ = sinA sin β sinC = sinα sinB sinC .

Remarks. This can be used to get another proof of the sine law.

Moreover, with the expressions on the left hand of (25), we canobtain expressions for the altitudes:

sin a = sinB sin γ, sin b = sinC sinα, sin c = sinA sin β.

Finally Vi(a,b, c)2 = G (a,b, c) =1− 2 cosA cosB cosC + cos2 A + cos2 B + cos2 C , together with(25), can be used to obtain the altitudes from the sides.



a

b

c

c′

b′a′

α

β

γ

α′

β′

γ′

A

A′B

B′

C

C ′

Polar triangle

It is the sphericaltriangle a′b′c′

determined byiA = ia′, iB = ib′,iC = ic′, or bya′ = −iiA, b′ = −iiB ,c′ = −iiC .All other elementsare denoted as in theoriginal triangle, butwith a prime mark.



3 A′ = π − α, B ′ = π − β,C ′ = π − γ.

Indeed, cosA′ = b′ · c′ = q(ib′, ic′) = q(iB , iC ) = (iB iC )0 =−(iB iC )0 = −e−iaα = − cosα = cos(π − α).

3 The polar triagle of a′b′c′ is abc. In particular, A = π − α′.Indeed, let us calculate the vertex c′′ of the polar triangle of a′b′c′.We have c′′ = −iiC ′ , with iC ′ = a′ ∧ b′/ sinC ′ =(−iiA) ∧ (−iiB)/ sin γ = (b× c) ∧ (c× a)/ sinA sinB sin γ. Soc′′ = (b× c)× (c× a)/ sinA sinB sin γ. But(b× c)× (c× a) = ((b× c) · a)c = Vi(a,b, c)c and soc′′ = cVi(a,b, c)/ sinA sinB sin γ = c (see page 111).



Right triangles

For the right spherical triangle with γ = π/2, the following relationshold:

sinα =sinA

sinC, sin β =

sinB

sinC(law of sines, eq. (24))

cosA =cosα

sin β, cosB =

cos β

sinα(eq. (20) for A and B)

cosC = cosA cosB = cotα cot β (p. 105 and formula above)


PointersNotes labeled with the page number to which they are

attached and in which the existence of a note is indicatedby the symbol 1


Pointers

page 29: The expression x ∧ x ′ = 12(xx ′ − x ′x) generalizes to any

blade x1 ∧ · · · ∧ xk as follows:

x1 ∧ · · · ∧ xk = 1k!

∑j

(−1)t(j)xj1 · · · xjk

where j = j1, . . . , jk runs over all permutations of {1, . . . , k} and t(j)is the number of order inversions in j . This formula can be proved byinduction on k . As an example, let us consider the case k = 3. Herethe sum over the permutations of 1, 2, 3 can be arranged asx1(x2x3 − x3x2) + x2(x3x1 − x1x3) + x3(x1x2 − x2x1) =2x1(x2 ∧ x3) + 2x2(x3 ∧ x1) + 2x3(x1 ∧ x2). Now each summand splitsinto a dot product and an exterior product. The latter produces2x1 ∧ x2 ∧ x3 + 2x2 ∧ x3 ∧ x1 + 2x3 ∧ x1 ∧ x2 = 6x1 ∧ x2 ∧ x3, while theformer contribute to 0: x1 · (x2∧ x3) + x2 · (x3∧ x1) + x3 · (x1∧ x2) = 0,which is checked with a simple calculation.


Pointers P = G3

page 51: The Pauli matrices we discovered by Pauli in connectionwith his theory of the electron spin [8]. Note that σ1, σ2, σ3 have thesame characteristic polynomial, λ2 − 1, and hence the sameeigenvalues, ±1. Ultimately, this corresponds to the fact that themeasure of the electron spin along any of the three coordinate axiscan have only the two values ±1 (in suitable units). For a geometricaccount of Pauli’s theory, see [5, §2].

At this point, the advantages of working directly with G3 should beclear from all the insights we have got from it. This undelines thefact that the Pauli representation, or any other matrix representationof G3, is unnecessary, and actually extraneous to the geometricstructures involved. In other words, G3 has all the rights to beconsidered the ‘true’ Pauli algebra.


Pointers P = G3

page 53. The proof of the other isomorphism is similar and can beomitted. To note, however, that the biquaternions 2H is a very usefulalgebra, as for instance in robotics: see [9, §9.3].

page 56: The Dirac matrices were introduced by Dirac in connectionwith his relativistic theory of the electron [10]. As in the case of thePauli algebra, the Dirac representation is not only unnecessary, but itobscures the discussion of the issues involved, and hence G1,3 has tobe regarded as the ‘true’ Dirac algebra.

page 74. These statements have important applications when itcomes to study the topology of the rotor groups. See [2, §6.1.7].


Topics for further study



History: For a timeline and references, [2, §6.3], and for a study ofthe role of Leibniz and Grassmann in the early history of GA, see [6].

Conformal geometric algebra: G3+1,1 is a natural extension of thePauli algebra and it has the property that isometries of E3+1,1

represent conformal transformations of the 3d Euclidean space. For amathematical introduction, and further references, see [3, Chapter 2].In preparation: a paper on the conformal geometry for anynon-degenerate signature, with applications to conformal field theory,by the same authors.



Geometric crystallography: For the study of this beautiful subjectby means of geometric algebra, the place to start is [11]. Here is itsabstract:

We present a complete formulation of the 2D and 3Dcrystallographic space groups in the conformal geometric al-gebra of Euclidean space. This enables a simple new rep-resentation of translational and orthogonal symmetries in amultiplicative group of versors. The generators of each groupare constructed directly from a basis of lattice vectors thatdefine its crystal class. A new system of space group symbolsenables one to unambiguously write down all generators of agiven space group directly from its symbol.

In geometric crystallography there are interesting open problems, asindicated, for instance, in [12, 13] and [14].



Classification of geometric algebras and their representations:

R

R

2R

C

H

2HH

C

ν Clock

0

1

2

3

4

5

6

7 m0

m1

m1 m2

m0

m2

m1

m3

M

M

M±

M±σ

D±

D± Mσ

Mσ

This is a natural theme tocontinue the mini-course.It can be found in[2, §6.3]. In this minicourse wehave found the algebras R, 2R,C, H and 2H and one corollaryof the classification is that anygeometric algebra is isomorphicto a matrix algebra over oneof these five basic forms. Thetheory also supplies what is thebasic form and the dimension in terms of the signature for ageometric algebra and for its even algebra. In addition, it classifiesthe possible representations of these algebras and the correspondingrepresentations of the spinorial groups.



The Lie group structure of the versor groups: The theory ofspinorial groups is an excellent source of examples of Lie groups, Liealgebras, and the role of the exponential function. Again, see [2,§6.4] and the pointers to references given there. In the Euclideancase, the Lie algebra of the rotor group is the algebra of bivectorswith the commutator product, the Lie algebra of the rotation group isthe algebra of skew-symmetric endomorphism with the commutatorproduct, and the isomorphism between the two, that exists forgeneral reasons, can be described very explicitely in terms of theadjoint operator of a bivector.



Relativistic physics: For a discussion of the interplay between thegeometry of Minkoski’s space-time geometry, the correspondinggeometric algebra, the ways it accounts for space-time physics(electromagnetism and Dirac’s theory), and a guide to furtherreferences, see [3, §1.4 and Ch. 3]. Note also the important recentpaper [15].

For an account of general relativity, black holes and gravitationalwaves, see [16] and the references there.

Geometric algebra and machine learnig: See [17], and thereferences therein, for an introduction to this exciting development.


References I

[1] S. Xambo-Descamps, “On geometric calculi and their applications,”2019.

Notes for the UNED Seminar 3-4 October, 2019.

[2] S. Xambo-Descamps, Real spinorial groups—a short mathematicalintroduction.

SBMA/Springerbrief, Springer, 2018.

[3] C. Lavor, S. Xambo-Descamps, and I. Zaplana, A geometric algebrainvitation to space-time physics, robotics and molecular geometry.

Springer, 2018.


References II

[4] S. Xambo-Descamps, “A Clifford view of Klein’s Geometry,” 2009.

Slides of the invited lecture delivered at the “InternationalConference on Didactics of Mathematics as a MathematicalDiscipline” held 1-4 October 2009 in the University Madeira,Funchal, Madeira Island, Portugal: https://mat-web.upc.edu/

people/sebastia.xambo/K2/K2-Xambo.pdf.

[5] S. Xambo-Descamps, “Escondidas sendas de la geometrıa proyectivaa los formalismos cuanticos,” in El legado matematico de JuanBautista Sancho Guimera (D. Hernandez-Ruiperez and A. Campillo,eds.), pp. 233–274, Real Sociedad Matematica Espanola & EdicionesUniversidad de Salamanca, 2016.


https://mat-web.upc.edu/people/sebastia.xambo/K2/K2-Xambo.pdf

https://mat-web.upc.edu/people/sebastia.xambo/K2/K2-Xambo.pdf

References III

[6] S. Xambo-Descamps, “From Leibniz’ characteristica geometrica tocontemporary geometric algebra,” Quaderns d’Historia del’Enginyeria, vol. 16 , no. Special issue dedicated to commemorateLeibniz (1646-1716), pp. 103–134, 2018.

pdf.

[7] D. Hestenes, New foundations for classical mechanics (2nd edition).

Fundamental Theories of Physics 99, Kluwer Academic Publishers,1999. 1st edition published 1990.

[8] W. Pauli, “Zur Quantenmechanik des magnetischen Elektrons,” Z.Phys., vol. 42, pp. 601–623, 1927.


https://mat-web.upc.edu/people/sebastia.xambo/HistoricalEssays/2018-Xambo--From-Leibniz-CG-to-GA.pdf

References IV

[9] J. M. Selig, Geometric fundamentals of robotics.

Springer, 2005.

[10] P. A. M. Dirac, “The quantum theory of the electron, I, II,”Proceedings of the Royal Society of London, pp. A117:610–624 andA118:351–361, 1928.

[11] D. Hestenes and J. W. Holt, “Crystallographic space groups ingeometric algebra,” Journal of Mathematical Physics, vol. 48, no. 2,p. 023514, 2007.

[12] P. Engel, Geometric crystallography: an axiomatic introduction tocrystallography.

Reidel Publishing Company, 1986.


References V

[13] P. Engel, “Geometric crystallography,” in Handbook of convexgeometry, pp. 989–1041, Elsevier, 1993.

[14] X. Blanc and M. Lewin, “The crystallization conjecture: a review,”arXiv: 1504. 01153 , 2015.

[15] D. Hestenes, “Maxwell–Dirac electron theory,” 2019.

Preprint 07/2/2019, 20 pages.

[16] A. Lasenby, “Geometric algebra, gravity and gravitational waves,”Advances in Applied Clifford Algebras, vol. 29, no. 4, p. 41 pages,2019.


arXiv:1504.01153

References VI

[17] S. Xambo-Descamps, “Geometric calculus meets deep learning,”2019.

Slides presented at the ICIAM-2019 mini-symposium Systems,Patterns and Data Engineering eith Geometric Calculus. A volume ofProceedings has been submitted to the series SEMA-SIAM-Springer.Slides.


https://mat-web.upc.edu/people/sebastia.xambo/AL/GC&DL-08.pdf

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