and a Generalization of the Quaternions:

The Pin and Spin Groups

Jean Gallier

Department of Computer and Information Science

University of Pennsylvania

Philadelphia, PA 19104, USA

e-mail: jean@saul.cis.upenn.edu

December 19, 2002

2

Contents

1 Cli¬ord Algebras, Cli¬ord Groups, Pin and Spin 5

1.1 Introduction: Rotations As Group Actions . . . . . . . . . . . . . . . . . . . 5

1.2 Cli¬ord Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Cli¬ord Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.4 The Groups Pin(n) and Spin(n) . . . . . . . . . . . . . . . . . . . . . . . . 22

1.5 The Groups Pin(p, q) and Spin(p, q) . . . . . . . . . . . . . . . . . . . . . . 28

1.6 Periodicity of the Cli¬ord Algebras Clp,q . . . . . . . . . . . . . . . . . . . . 30

1.7 The Complex Cli¬ord Algebras Cl(n, C) . . . . . . . . . . . . . . . . . . . . 34

1.8 The Groups Pin(p, q) and Spin(p, q) as double covers . . . . . . . . . . . . . 34

1.9 More on the Topology of O(p, q) and SO(p, q) . . . . . . . . . . . . . . . . . 39

3

4 CONTENTS

Chapter 1

Cli¬ord Algebras, Cli¬ord Groups,

and the Groups Pin(n) and Spin(n)

1.1 Introduction: Rotations As Group Actions

One of the main goals of these notes is to explain how rotations in Rn are induced by the

action of a certain group, Spin(n), on Rn , in a way that generalizes the action of the unit

complex numbers, U(1), on R2 , and the action of the unit quaternions, SU(2), on R3 (i.e.,

the action is de¬ned in terms of multiplication in a larger algebra containing both the group

Spin(n) and Rn ). The group Spin(n), called a spinor group, is de¬ned as a certain subgroup

of units of an algebra, Cln , the Cli¬ord algebra associated with Rn . Furthermore, for n ≥ 3,

we are lucky, because the group Spin(n) is topologically simpler than the group SO(n).

Indeed, for n ≥ 3, the group Spin(n) is simply connected (a fact that it not so easy to prove

without some machinery), whereas SO(n) is not simply connected. Intuitively speaking,

SO(n) is more twisted than Spin(n). In fact, we will see that Spin(n) is a double cover of

SO(n).

Since the spinor groups are certain well chosen subroups of units of Cli¬ord algebras, it is

necessary to investigate Cli¬ord algebras to get a ¬rm understanding of spinor groups. These

notes provide a tutorial on Cli¬ord algebra and the groups Spin and Pin, including a study

of the structure of the Cli¬ord algebra Clp,q associated with a nondegenerate symmetric

bilinear form of signature (p, q) and culminating in the beautiful “8-periodicity theorem” of

Elie Cartan and Raoul Bott (with proofs). We alo explain when Spin(p, q) is a double-cover

of SO(n). The reader should be warned that a certain amount of algebraic (and topological)

background is expected, and that these notes are not meant for a novice. This being said,

perseverant readers will be rewarded by being exposed to some beautiful and nontrivial

concepts and results (including proofs), including Elie Cartan and Raoul Bott “8-periodicity

theorem.”

Going back to rotations as transformations induced by group actions, recall that if V is

a vector space, a linear action (on the left) of a group G on V is a map, ±: G — V ’ V ,

5

6 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

satisfying the following conditions, where, for simplicity of notation, we denote ±(g, v) by

g · v:

(1) g · (h · v) = (gh) · v, for all g, h ∈ G and v ∈ V ;

(2) 1 · v = v, for all v ∈ V , where 1 is the identity of the group G;

(3) The map v ’ g · v is a linear isomorphism of V for every g ∈ G.

For example, the (multiplicative) group, U(1), of unit complex numbers acts on R2 (by

identifying R2 and C) via complex multiplication: For every z = a + ib (with a2 + b2 = 1),

for every (x, y) ∈ R2 (viewing (x, y) as the complex number x + iy),

z · (x, y) = (ax ’ by, ay + bx).

Now, every unit complex number is of the form cos θ + i sin θ, and thus, the above action of

z = cos θ + i sin θ on R2 corresponds to the rotation of angle θ around the origin. In the case

n = 2, the groups U(1) and SO(2) are isomorphic, but this is an exception.

We can de¬ne an action of the group of unit quaternions, SU(2), on R3 . For this, we use

the fact that R3 can be identi¬ed with the pure quaternions in H, namely, the quaternions

of the form x1 i + x2 j + x3 k, where (x1 , x2 , x3 ) ∈ R3 . Then, we de¬ne the action of SU(2)

over R3 by

Z · X = ZXZ ’1 = ZXZ,

where Z ∈ SU(2) and X is any pure quaternion. Now, it turns out that the map ρZ (where

ρZ (X) = ZXZ)) is indeed a rotation, and that the map ρ: Z ’ ρZ is a surjective homomor-

phism, ρ: SU(2) ’ SO(3), whose kernel is {’1, 1}, where 1 denotes the multiplicative unit

quaternion. (For details, see Gallier [16], Chapter 8).

We can also de¬ne an action of the group SU(2) — SU(2) over R4 , by identifying R4 with

the quaternions. In this case,

(Y, Z) · X = Y XZ,

where (Y, Z) ∈ SU(2)—SU(2) and X ∈ H is any quaternion. Then, the map ρY,Z is a rotation

(where ρY,Z (X) = Y XZ), and the map ρ: (Y, Z) ’ ρY,Z is a surjective homomorphism,

ρ: SU(2) — SU(2) ’ SO(4), whose kernel is {(1, 1), (’1, ’1)}. (For details, see Gallier [16],

Chapter 8).

Thus, we observe that for n = 2, 3, 4, the rotations in SO(n) can be realized via the

linear action of some group (the case n = 1 is trivial, since SO(1) = {1, ’1}). It is also the

case that the action of each group can be somehow be described in terms of multiplication in

some larger algebra “containing” the original vector space Rn (C for n = 2, H for n = 3, 4).

However, these groups appear to have been discovered in an ad hoc fashion, and there does

not appear to be any universal way to de¬ne the action of these groups on Rn . It would

certainly be nice if the action was always of the form

Z · X = ZXZ ’1 (= ZXZ).

1.2. CLIFFORD ALGEBRAS 7

A systematic way of constructing groups realizing rotations in terms of linear action, using

a uniform notion of action, does exist. Such groups are the spinor groups, to be described

in the following sections.

1.2 Cli¬ord Algebras

We explained in Section 1.1 how the rotations in SO(3) can be realized by the linear action

of the group of unit quaternions, SU(2), on R3 , and how the rotations in SO(4) can be

realized by the linear action of the group SU(2) — SU(2) on R4 .

The main reasons why the rotations in SO(3) can be represented by unit quaternions are

the following:

(1) For every nonzero vector u ∈ R3 , the re¬‚ection su about the hyperplane perpendicular

to u is represented by the map

v ’ ’uvu’1 ,

where u and v are viewed as pure quaternions in H (i.e., if u = (u1 , u2 , u2 ), then view

u as u1 i + u2 j + u3 k, and similarly for v).

(2) The group SO(3) is generated by the re¬‚ections.

As one can imagine, a successful generalization of the quaternions, i.e., the discovery

of a group, G inducing the rotations in SO(n) via a linear action, depends on the ability

to generalize properties (1) and (2) above. Fortunately, it is true that the group SO(n) is

generated by the hyperplane re¬‚ections. In fact, this is also true for the orthogonal group,

O(n), and more generally, for the group of direct isometries, O(¦), of any nondegenerate

quadratic form, ¦, by the Cartan-Dieudonn´ theorem (for instance, see Bourbaki [6], or

e

Gallier [16], Chapter 7, Theorem 7.2.1). In order to generalize (2), we need to understand

how the group G acts on Rn . Now, the case n = 3 is special, because the underlying space,

R3 , on which the rotations act, can be embedded as the pure quaternions in H. The case

n = 4 is also special, because R4 is the underlying space of H. The generalization to n ≥ 5

requires more machinery, namely, the notions of Cli¬ord groups and Cli¬ord algebras. As we

will see, for every n ≥ 2, there is a compact, connected (and simply connected when n ≥ 3)

group, Spin(n), the “spinor group,” and a surjective homomorphism, ρ: Spin(n) ’ SO(n),

whose kernel is {’1, 1}. This time, Spin(n) acts directly on Rn , because Spin(n) is a certain

subgroup of the group of units of the Cli¬ord algebra, Cln , and Rn is naturally a subspace

of Cln .

The group of unit quaternions SU(2) turns out to be isomorphic to the spinor group

Spin(3). Because Spin(3) acts directly on R3 , the representation of rotations in SO(3)

by elements of Spin(3) may be viewed as more natural than the representation by unit

quaternions. The group SU(2) — SU(2) turns out to be isomorphic to the spinor group

Spin(4), but this isomorphism is less obvious.

8 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

In summary, we are going to de¬ne a group Spin(n) representing the rotations in SO(n),

for any n ≥ 1, in the sense that there is a linear action of Spin(n) on Rn which induces a

surjective homomorphism, ρ: Spin(n) ’ SO(n), whose kernel is {’1, 1}. Furthermore, the

action of Spin(n) on Rn is given in terms of multiplication in an algebra, Cln , containing

Spin(n), and in which Rn is also embebedded. It turns out that as a bonus, for n ≥ 3, the

group Spin(3) is topologically simpler than SO(n), since Spin(3) is simply connected, but

SO(n) is not. By being astute, we can also construct a group, Pin(n), and a linear action

of Pin(n) on Rn that induces a surjective homomorphism, ρ: Pin(n) ’ O(n), whose kernel

is {’1, 1}. The di¬cilty here is the presence of the negative sign in (2). We will see how

Atiyah, Bott and Shapiro circumvent this problem by using a “twisted adjoint action,” as

opposed to the usual adjoint action (where v ’ uvu’1 ).

These notes are heavily inspired by Br¨cker and tom Dieck [7], Chapter 1, Section 6,

o

where most details can be found. This Chapter is almost entirely taken from the ¬rst 11

pages of the beautiful and seminal paper by Atiyah, Bott and Shapiro [3], Cli¬ord Modules,

and we highly recommend it. Another excellent (but concise) exposition can be found in

Kirillov [18]. A very thorough exposition can be found in two places:

1. Lawson and Michelsohn [20], where the material on Pin(p, q) and Spin(p, q) can be

found in Chapter I.

2. Lounesto™s excellent book [21].

One may also want to consult Baker [4], Curtis [12], Porteous [23], Fulton and Harris (Lecture

20) [15], Choquet-Bruhat [11], Bourbaki [6], or Chevalley [10], a classic. The original source

is Elie Cartan™s book (1937) whose translation in English appears in [8].

We begin by recalling what is an algebra over a ¬eld. Let K denote any (commutative)

¬eld, although for our purposes, we may assume that K = R (and occasionally, K = C).

Since we will only be dealing with associative algebras with a multiplicative unit, we only

de¬ne algebras of this kind.

De¬nition 1.1 Given a ¬eld, K, a K-algebra is a K-vector space, A, together with a bilinear

operation, ·: A — A ’ A, called multiplication, which makes A into a ring with unity, 1 (or

1A , when we want to be very precise). This means that · is associative and that there is

a multiplicative identity element, 1, so that 1 · a = a · 1 = a, for all a ∈ A. Given two

K-algebras A and B, a K-algebra homomorphism, h: A ’ B, is a linear map that is also a

ring homomorphism, with h(1A ) = 1B .

For example, the ring, Mn (K), of all n — n matrices over a ¬eld, K, is a K-algebra.

There is an obvious notion of ideal of a K-algebra: An ideal, A ⊆ A, is a linear subspace

of A that is also an ideal with respect to multiplication in A. If the ¬eld K is understood,

we usually simply say an algebra instead of a K-algebra.

We also need a quick review of tensor products. The basic idea is that tensor products

allow us to view multilinear maps as linear maps. The maps become simpler, but the spaces

1.2. CLIFFORD ALGEBRAS 9

(product spaces) become more complicated (tensor products). More more details, see Atiyah

and Macdonald [2].

De¬nition 1.2 Given two K-vector spaces, E and F , a tensor product of E and F is a pair,

(E — F, —), where E — F is a K-vector space and —: E — F ’ E — F is a bilinear map, so

that for every K-vector space, G, and every bilinear map, f : E — F ’ G, there is a unique

linear map, f— : E — F ’ G, with

f (u, v) = f— (u — v) for all u ∈ E and all v ∈ V ,

as in the diagram below: —

E—F ’’ E — F

¦

¦f

f —

G

The vector space E — F is de¬ned up to isomorphism. The vectors u — v, where u ∈ E

and v ∈ F , generate E — F .

Remark: We should really denote the tensor product of E and F by E —K F , since it

depends on the ¬eld K. Since we usually deal with a ¬xed ¬eld K, we use the simpler

notation E — F .

We have natural isomorphisms

(E — F ) — G ≈ E — (F — G) and E — F ≈ F — E.

Given two linear maps f : E ’ F and g: E ’ F , we have a unique bilinear map

f — g: E — E ’ F — F so that

(f — g)(a, a ) = (f (a), g(a )) for all a ∈ A and all a ∈ A .

Thus, we have the bilinear map — —¦ (f — g): E — E ’ F — F , and so, there is a unique

linear map f — g: E — E ’ F — F , so that

(f — g)(a — a ) = f (a) — g(a ) for all a ∈ A and all a ∈ A .

Let us now assume that E and F are K-algebras. We want to make E — F into a

K-algebra. Since the multiplication operations mE : E — E ’ E and mF : F — F ’ F are

bilinear, we get linear maps mE : E — E ’ E and mF : F — F ’ F , and thus, the linear map

mE — mF : (E — E) — (F — F ) ’ E — F.

Using the isomorphism „ : (E — E) — (F — F ) ’ (E — F ) — (E — F ), we get a linear map

mE—F : (E — F ) — (E — F ) ’ E — F,

10 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

which de¬nes a multiplication m on E — F (namely, m(u, v) = mE—F (u — v)). It is easily

checked that E — F is indeed a K-algebra under the multiplication m. Using the simpler

notation · for m, we have

(a — a ) · (b — b ) = (aa ) — (bb )

for all a, b ∈ E and all a , b ∈ F .

Given any vector space, V , over a ¬eld, K, there is a special K-algebra, T (V ), together

with a linear map, i: V ’ T (V ), with the following universal mapping property: Given any

K-algebra, A, for any linear map, f : V ’ A, there is a unique K-algebra homomorphism,

f : T (V ) ’ A, so that

f = f —¦ i,

as in the diagram below:

i

’’ T (V )

V ¦

¦

f f

A

The algebra, T (V ), is the tensor algebra of V . It may be constructed as the direct sum

V —i ,

T (V ) =

i≥0

where V 0 = K, and V —i is the i-fold tensor product of V with itself. For every i ≥ 0, there

is a natural injection ιn : V —n ’ T (V ), and in particular, an injection ι0 : K ’ T (V ). The

multiplicative unit, 1, of T (V ) is the image, ι0 (1), in T (V ) of the unit, 1, of the ¬eld K.

Since every v ∈ T (V ) can be expressed as a ¬nite sum

v = v1 + · · · + vk ,

where vi ∈ V —ni and the ni are natural numbers with ni = nj if i = j, to de¬ne multiplication

in T (V ), using bilinearity, it is enough to de¬ne the multiplication V —m — V —n ’’ V —(m+n) .

Of course, this is de¬ned by

(v1 — · · · — vm ) · (w1 — · · · — wn ) = v1 — · · · — vm — w1 — · · · — wn .

(This has to be made rigorous by using isomorphisms involving the associativity of tensor

products, for details, see see Atiyah and Macdonald [2].) The algebra T (V ) is an example

of a graded algebra, where the homogeneous elements of rank n are the elements in V —n .

Remark: It is important to note that multiplication in T (V ) is not commutative. Also, in

all rigor, the unit, 1, of T (V ) is not equal to 1, the unit of the ¬eld K. However, in view

of the injection ι0 : K ’ T (V ), for the sake of notational simplicity, we will denote 1 by 1.

More generally, in view of the injections ιn : V —n ’ T (V ), we identify elements of V —n with

their images in T (V ).

1.2. CLIFFORD ALGEBRAS 11

Most algebras of interest arise as well-chosen quotients of the tensor algebra T (V ). This

is true for the exterior algebra, • V , (also called Grassmann algebra), where we take the

quotient of T (V ) modulo the ideal generated by all elements of the form v — v, where v ∈ V ,

and for the symmetric algebra, Sym V , where we take the quotient of T (V ) modulo the ideal

generated by all elements of the form v — w ’ w — v, where v, w ∈ V . A Cli¬ord algebra

may be viewed as a re¬nement of the exterior algebra, in which we take the quotient of

T (V ) modulo the ideal generated by all elements of the form v — v ’ ¦(v) · 1, where ¦ is the

quadratic form associated with a symmetric bilinear form, •: V — V ’ K. For simplicity,

let us assume that we are now dealing with real algebras.

De¬nition 1.3 Let V be a real ¬nite-dimensional vector space together with a symmetric

bilinear form, •: V — V ’ R, and associated quadratic form, ¦(v) = •(v, v). A Cli¬ord

algebra associated with V and ¦ is a real algebra, Cl(V, ¦), together with a linear map,

i¦ : V ’ Cl(V, ¦), satisfying the condition (i(v))2 = ¦(v) · 1 for all v ∈ V , and so that for

every real algebra, A, and every linear map, f : V ’ A, with

(f (v))2 = ¦(v) · 1 for all v ∈ V ,

there is a unique algebra homomorphism, f : Cl(V, ¦) ’ A, so that

f = f —¦ i¦ ,

as in the diagram below:

i

¦

’’ Cl(V, ¦)

V ¦

¦

f f

A

By a familiar argument, any two Cli¬ord algebras associated with V and ¦ are isomorphic.

We often denote i¦ by i.

To show the existence of Cl(V, ¦), observe that T (V )/A does the job, where A is the

ideal of T (V ) generated by all elements of the form v — v ’ ¦(v) · 1, where v ∈ V . The map

i¦ : V ’ Cl(V, ¦) is the composition

ι π

1

V ’’ T (V ) ’’ T (V )/A,

where π is the natural quotient map. We often denote the Cli¬ord algebra Cl(V, ¦) simply

by Cl(¦).

Remark: Observe that De¬nition 1.3 does not assert that i¦ is injective or that there is

an injection of R into Cl(V, ¦), but we will prove later that both facts are true when V is

¬nite-dimensional. Also, as in the case of the tensor algebra, the unit of the algebra Cl(V, ¦)

and the unit of the ¬eld R are not equal.

12 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

Since

¦(u + v) ’ ¦(u) ’ ¦(v) = 2•(u, v)

and

(i(u + v))2 = (i(u))2 + (i(v))2 + i(u) · i(v) + i(v) · i(u),

using the fact that

i(u)2 = ¦(u) · 1,

we get

i(u) · i(v) + i(v) · i(u) = 2•(u, v) · 1.

As a consequence, if (u1 , . . . , un ) is an orthogonal basis w.r.t. • (which means that

•(uj , uk ) = 0 for all j = k), we have

i(uj ) · i(uk ) + i(uk ) · i(uj ) = 0 for all j = k.

Remark: Certain authors drop the unit, 1, of the Cli¬ord algebra Cl(V, ¦) when writing

the identities

i(u)2 = ¦(u) · 1

and

2•(u, v) · 1 = i(u) · i(v) + i(v) · i(u),

usually written as

1

•(u, v) = (i(u) · i(v) + i(v) · i(u)).

2

This is very confusing and technically wrong, because we only have an injection of R into

Cl(V, ¦), but R is not a subset of Cl(V, ¦).

We warn the readers that Lawson and Michelsohn [20] adopt the opposite of our sign

convention in de¬ning Cli¬ord algebras, i.e., they use the condition

(f (v))2 = ’¦(v) · 1 for all v ∈ V .

The most confusing consequence of this is that their Cl(p, q) is our Cl(q, p).

Observe that when ¦ ≡ 0 is the quadratic form identically zero everywhere, then the

Cli¬ord algebra Cl(V, 0) is just the exterior algebra, • V .

Example 1.1 Let V = R, e1 = 1, and assume that ¦(x1 e1 ) = ’x2 . Then, Cl(¦) is spanned

1

by the basis (1, e1 ). We have

e2 = ’1.

1

Under the bijection

e1 ’ i,

1.2. CLIFFORD ALGEBRAS 13

the Cli¬ord algebra, Cl(¦), also denoted by Cl1 , is isomorphic to the algebra of complex

numbers, C.

Now, let V = R2 , (e1 , e2 ) be the canonical basis, and assume that ¦(x1 e1 + x2 e2 ) =

’(x2 + x2 ). Then, Cl(¦) is spanned by the basis by (1, e1 , e2 , e1 e2 ). Furthermore, we have

1 2

e2 = ’1, e2 = ’1, (e1 e2 )2 = ’1.

e2 e1 = ’e1 e2 , 1 2

Under the bijection

e1 ’ i, e2 ’ j, e1 e2 ’ k,

it is easily checked that the quaternion identities

i2 = j2 = k2 = ’1,

ij = ’ji = k,

jk = ’kj = i,

ki = ’ik = j,

hold, and thus, the Cli¬ord algebra Cl(¦), also denoted by Cl2 , is isomorphic to the algebra

of quaternions, H.

Our prime goal is to de¬ne an action of Cl(¦) on V in such a way that by restricting

this action to some suitably chosen multiplicative subgroups of Cl(¦), we get surjective

homomorphisms onto O(¦) and SO(¦), respectively. The key point is that a re¬‚ection

in V about a hyperplane H orthogonal to a vector w can be de¬ned by such an action,

but some negative sign shows up. A correct handling of signs is a bit subtle and requires

the introduction of a canonical anti-automorphism, t, and of a canonical automorphism, ±,

de¬ned as follows:

Proposition 1.1 Every Cli¬ord algebra, Cl(¦), has a unique canonical anti-automorphism,

t: Cl(¦) ’ Cl(¦), satisfying the properties

t(x · y) = t(y) · t(x), t —¦ t = id, and t(i(v)) = i(v),

for all x, y ∈ Cl(¦) and all v ∈ V .

Proof . Consider the opposite algebra Cl(¦)o , in which x ·o y = y · x. It has the universal

mapping property. Thus, we get a unique isomorphism, t, as in the diagram below:

i

’’

V Cl(¦)

¦

¦

t

i

Cl(¦)o

We also denote t(x) by xt . When V is ¬nite-dimensional, for a more palatable description

of t in terms of a basis of V , see the paragraph following Theorem 1.4.

The canonical automorphism, ±, is de¬ned using the proposition

14 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

Proposition 1.2 Every Cli¬ord algebra, Cl(¦), has a unique canonical automorphism,

±: Cl(¦) ’ Cl(¦), satisfying the properties

± —¦ ± = id, ±(i(v)) = ’i(v),

and

for all v ∈ V .

Proof . Consider the linear map ±0 : V ’ Cl(¦) de¬ned by ±0 (v) = ’i(v), for all v ∈ V . We

get a unique homomorphism, ±, as in the diagram below:

i

’’ Cl(¦)

V ¦

±0

¦±

Cl(¦)

Furthermore, every x ∈ Cl(¦) can be written as

x = x1 · · · xm ,

with xj ∈ i(V ), and since ±(xj ) = ’xj , we get ± —¦ ± = id. It is clear that ± is bijective.

Again, when V is ¬nite-dimensional, a more palatable description of ± in terms of a basis

of V can be given. If (e1 , . . . , en ) is a basis of V , then the Cli¬ord algebra Cl(¦) consists of

certain kinds of “polynomials,” linear combinations of monomials of the form J »J eJ , where

J = {i1 , i2 , . . . , ik } is any subset (possibly empty) of {1, . . . , n}, with 1 ¤ i1 < i2 · · · < ik ¤ n,

and the monomial eJ is the “product” ei1 ei2 · · · eik . The map ± is the linear map de¬ned on

monomials by

±(ei1 ei2 · · · eik ) = (’1)k ei1 ei2 · · · eik .

For a more rigorous explanation, see the paragraph following Theorem 1.4.

We now show that if V has dimension n, then i is injective and Cl(¦) has dimension 2n .

A clever way of doing this is to introduce a graded tensor product.

First, observe that

Cl(¦) = Cl0 (¦) • Cl1 (¦),

where

Cli (¦) = {x ∈ Cl(¦) | ±(x) = (’1)i x}, where i = 0, 1.

We say that we have a Z/2-grading, which means that if x ∈ Cli (¦) and y ∈ Clj (¦), then

x · y ∈ Cli+j (mod 2) (¦).

When V is ¬nite-dimensional, since every element of Cl(¦) is a linear combination of the

form J »J eJ , as explained earlier, in view of the description of ± given above, we see that

the elements of Cl0 (¦) are those for which the monomials eJ are products of an even number

of factors, and the elements of Cl1 (¦) are those for which the monomials eJ are products of

an odd number of factors.

1.2. CLIFFORD ALGEBRAS 15

Remark: Observe that Cl0 (¦) is a subalgebra of Cl(¦), whereas Cl1 (¦) is not.

Given two Z/2-graded algebras A = A0 • A1 and B = B 0 • B 1 , their graded tensor

product A — B is de¬ned by

(A — B)0 = (A0 • B 0 ) — (A1 • B 1 ),

(A — B)1 = (A0 • B 1 ) — (A1 • B 0 ),

with multiplication

(a — b) · (a — b ) = (’1)ij (a · a) — (b · b ),

for a ∈ Ai and b ∈ B j . The reader should check that A — B is indeed Z/2-graded.

Proposition 1.3 Let V and W be ¬nite dimensional vector spaces with quadratic forms ¦

and Ψ. Then, there is a quadratic form, ¦ • Ψ, on V • W de¬ned by

(¦ + Ψ)(v, w) = ¦(v) + Ψ(w).

If we write i: V ’ Cl(¦) and j: W ’ Cl(Ψ), we can de¬ne a linear map,

f : V • W ’ Cl(¦) — Cl(Ψ),

by

f (v, w) = i(v) — 1 + 1 — j(w).

Furthermore, the map f induces an isomorphism (also denoted by f )

f : Cl(V • W ) ’ Cl(¦) — Cl(Ψ).

Proof . See Br¨cker and tom Dieck [7], Chapter 1, Section 6, page 57.

o

As a corollary, we obtain the following result:

Theorem 1.4 For every vector space, V , of ¬nite dimension n, the map i: V ’ Cl(¦) is

injective. Given a basis (e1 , . . . , en ) of V , the 2n ’ 1 products

i(ei1 )i(ei2 ) · · · i(eik ), 1 ¤ i1 < i 2 · · · < i k ¤ n

and 1 form a basis of Cl(¦). Thus, Cl(¦) has dimension 2n .

Proof . The proof is by induction on n = dim(V ). For n = 1, the tensor algebra T (V ) is

just the polynomial ring R[X], where i(e1 ) = X. Thus, Cl(¦) = R[X]/(X 2 ’ ¦(e1 )), and

the result is obvious. Since

i(ej ) · i(ek ) + i(ek ) · i(ej ) = 2•(ei , ej ) · 1,

it is clear that the products

i(ei1 )i(ei2 ) · · · i(eik ), 1 ¤ i1 < i 2 · · · < i k ¤ n

16 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

and 1 generate Cl(¦). Now, there is always a basis that is orthogonal with respect to • (for

example, see Artin [1], Chapter 7, or Gallier [16], Chapter 6, Problem 6.14), and thus, we

have a splitting

n

(V, ¦) = (Vk , ¦k ),

k=1

where Vk has dimension 1. Choosing a basis so that ek ∈ Vk , the theorem follows by induction

from Proposition 1.3.

Since i is injective, for simplicity of notation, from now on, we write u for i(u). Theo-

rem 1.4 implies that if (e1 , . . . , en ) is an orthonormal basis of V , then Cl(¦) is the algebra

presented by the generators (e1 , . . . , en ) and the relations

e2 = ¦(ej ) · 1, 1 ¤ j ¤ n, and

j

ej ek = ’ek ej , 1 ¤ j, k ¤ n, j = k.

If V has ¬nite dimension n and (e1 , . . . , en ) is a basis of V , by Theorem 1.4, the maps t

and ± are completely determined by their action on the basis elements. Namely, t is de¬ned

by

t(ei ) = ei

t(ei1 ei2 · · · eik ) = eik eik’1 · · · ei1

where 1 ¤ i1 < i2 · · · < ik ¤ n, and, of course, t(1) = 1. The map ± is de¬ned by

±(ei ) = ’ei

±(ei1 ei2 · · · eik ) = (’1)k ei1 ei2 · · · eik

where 1 ¤ i1 < i2 · · · < ik ¤ n, and, of course, ±(1) = 1. Furthermore, the even-graded

elements (the elements of Cl0 (¦)) are those generated by 1 and the basis elements consisting

of an even number of factors, ei1 ei2 · · · ei2k , and the odd-graded elements (the elements of

Cl1 (¦)) are those generated by the basis elements consisting of an odd number of factors,

ei1 ei2 · · · ei2k+1 .

We are now ready to de¬ne the Cli¬ord group and investigate some of its properties.

1.3 Cli¬ord Groups

First, we de¬ne conjugation on a Cli¬ord algebra, Cl(¦), as the map

x ’ x = t(±(x)) for all x ∈ Cl(¦).

Observe that

t —¦ ± = ± —¦ t.

1.3. CLIFFORD GROUPS 17

If V has ¬nite dimension n and (e1 , . . . , en ) is a basis of V , in view of previous remarks,

conjugation is de¬ned by

ei = ’ei

ei1 ei2 · · · eik = (’1)k eik eik’1 · · · ei1

where 1 ¤ i1 < i2 · · · < ik ¤ n, and, of course, 1 = 1. Conjugation is an anti-automorphism.

The multiplicative group of invertible elements of Cl(¦) is denoted by Cl(¦)— .

De¬nition 1.4 Given a ¬nite dimensional vector space, V , and a quadratic form, ¦, on V ,

the Cli¬ord group of ¦ is the group

“(¦) = {x ∈ Cl(¦)— | ±(x) · v · x’1 ∈ V for all v ∈ V }.

The map N : Cl(Q) ’ Cl(Q) given by

N (x) = x · x

is called the norm of Cl(¦).

We see that the group “(¦) acts on V via

x · v = ±(x)vx’1 ,

where x ∈ “(¦) and v ∈ V . Actually, it is not entirely obvious why the action “(¦)—V ’’ V

is a linear action, and for that matter, why “(¦) is a group.

This is because V is ¬nite-dimensional and ± is an automorphism. As a consequence, for

any x ∈ “(¦), the map ρx from V to V de¬ned by

v ’ ±(x) · v · x’1

is linear and injective, and thus bijective, since V has ¬nite dimension. It follows that

x’1 ∈ “(¦) (the reader should ¬ll in the details).

We also de¬ne the group “+ (¦), called the special Cli¬ord group, by

“+ (¦) = “(¦) © Cl0 (¦).

Observe that N (v) = ’¦(v) · 1 for all v ∈ V . Also, if (e1 , . . . , en ) is a basis of V , we leave it

as an exercise to check that

N (ei1 ei2 · · · eik ) = (’1)k ¦(ei1 )¦(ei2 ) · · · ¦(eik ) · 1.

Remark: The map ρ: “(¦) ’ GL(V ) given by x ’ ρx is called the twisted adjoint rep-

resentation. It was introduced by Atiyah, Bott and Shapiro [3]. It has the advantage of

18 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

not introducing a spurious negative sign, i.e., when v ∈ V and ¦(v) = 0, the map ρv is the

re¬‚ection sv about the hyperplane orthogonal to v (see Proposition 1.6). Furthermore, when

¦ is nondegenerate, the kernel Ker (ρ) of the representation ρ is given by Ker (ρ) = R— · 1,

where R— = R ’ {0}. The earlier adjoint representation (used by Chevalley [10] and others)

is given by

v ’ x · v · x’1 .

Unfortunately, in this case, ρx represents ’sv , where sv is the re¬‚ection about the hyperplane

orthogonal to v. Furthermore, the kernel of the representation ρ is generally bigger than R— ·1.

This is the reason why the twisted adjoint representation is preferred (and must be used for

a proper treatment of the Pin group).

Proposition 1.5 The maps ± and t induce an automorphism and an anti-automorphism of

the Cli¬ord group, “(¦).

Proof . It is not very instructive, see Br¨cker and tom Dieck [7], Chapter 1, Section 6, page

o

58.

The following proposition shows why Cli¬ord groups generalize the quaternions.

Proposition 1.6 Let V be a ¬nite dimensional vector space and ¦ a quadratic form on V .

For every element, x, of the Cli¬ord group, “(¦), if ¦(x) = 0, then the map ρx : V ’ V

given by

v ’ ±(x)vx’1 for all v ∈ V

is the re¬‚ection about the hyperplane H orthogonal to the vector x.

Proof . Recall that the re¬‚ection s about the hyperplane H orthogonal to the vector x is

given by

•(u, x)

s(u) = u ’ 2 x.

¦(x)

However, we have

x2 = ¦(x) · 1 and u · x + x · u = 2•(u, x) · 1.

Thus, we have

1·x

•(u, x)

s(u) = u ’ 2 x = u ’ 2•(u, x)

¦(x) ¦(x)

1

= u ’ 2•(u, x) ·x

¦(x)

1

= u ’ 2•(u, x) ·x

x2

= u ’ 2•(u, x)x’1

1.3. CLIFFORD GROUPS 19

u ’ 2•(u, x)(1 · x’1 )

=

u ’ (2•(u, x)1) · x’1

=

u ’ (u · x + x · u) · x’1

=

’x · u · x’1

=

±(x) · u · x’1 ,

=

since ±(x) = ’x, for x ∈ V .

In general, we have a map

ρ: “(¦) ’ GL(V )

de¬ned by

ρ(x)(v) = ±(x) · v · x’1 ,

for all x ∈ “(¦) and all v ∈ V . We would like to show that ρ is a surjective homomorphism

from “(¦) onto O(•) and a surjective homomorphism from “+ (¦) onto SO(•). For this,

we will need to assume that • is nondegenerate, which means that for every v ∈ V , if

•(v, w) = 0 for all w ∈ V , then v = 0. For simplicity of exposition, we ¬rst assume that ¦

is the quadratic form on Rn de¬ned by

¦(x1 , . . . , xn ) = ’(x2 + · · · + x2 ).

1 n

Let Cln denote the Cli¬ord algebra Cl(¦) and “n denote the Cli¬ord group “(¦). The

following lemma plays a crucial role:

Lemma 1.7 The kernel of the map ρ: “n ’ GL(n) is R— · 1, the mutiplicative group of

nonzero scalar multiples of 1 ∈ Cln .

Proof . If ρ(x) = id, then

±(x)v = vx for all v ∈ Rn . (1)

Since Cln = Cl0 • Cl1 , we can write x = x0 + x1 , with xi ∈ Cli for i = 1, 2. Then, equation

n n n

(1) becomes

x0 v = vx0 ’ x1 v = vx1 for all v ∈ Rn .

and (2)

Using Theorem 1.4, we can express x0 as a linear combination of monomials in the canonical

basis (e1 , . . . , en ), so that

with a0 ∈ Cl0 , b1 ∈ Cl1 ,

x0 = a0 + e1 b1 , n n

where neither a0 nor b1 contains a summand with a factor e1 . Applying the ¬rst relation in

(2) to v = e1 , we get

e1 a0 + e2 b1 = a0 e1 + e1 b1 e1 . (3)

1

20 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

Now, the basis (e1 , . . . , en ) is orthogonal w.r.t. ¦, which implies that

ej ek = ’ek ej for all j = k.

Since each monomial in a0 is of even degre and contains no factor e1 , we get

a0 e1 = e1 a0 .

Similarly, since b1 is of odd degree and contains no factor e1 , we get

e1 b1 e1 = ’e2 b1 .

1

But then, from (3), we get

e1 a0 + e2 b1 = a0 e1 + e1 b1 e1 = e1 a0 ’ e2 b1 ,

1 1

and so, e2 b1 = 0. However, e2 = ’1, and so, b1 = 0. Therefore, x0 contains no monomial

1 1

with a factor e1 . We can apply the same argument to the other basis elements e2 , . . . , en ,

and thus, we just proved that x0 ∈ R · 1.

A similar argument applying to the second equation in (2), with x1 = a1 +e1 b0 and v = e1

shows that b0 = 0. We also conclude that x1 ∈ R · 1. However, R · 1 ⊆ Cl0 , and so, x1 = 0.

n

—

0

Finally, x = x ∈ (R · 1) © “n = R · 1.

Remark: If ¦ is any nondegenerate quadratic form, we know (for instance, see Artin [1],

Chapter 7, or Gallier [16], Chapter 6, Problem 6.14) that there is an orthogonal basis

(e1 , . . . , en ) with respect to • (i.e. •(ej , ek ) = 0 for all j = k). Thus, the commutation

relations

e2 = ¦(ej ) · 1, with ¦(ej ) = 0, 1 ¤ j ¤ n, and

j

ej ek = ’ek ej , 1 ¤ j, k ¤ n, j = k

hold, and since the proof only rests on these facts, Lemma 1.7 holds for any nondegenerate

quadratic form.

However, Lemma 1.7 may fail for degenerate quadratic forms. For example, if ¦ ≡ 0,

then Cl(V, 0) = • V . Consider the element x = 1 + e1 e2 . Clearly, x’1 = 1 ’ e1 e2 . But

now, for any v ∈ V , we have

±(1 + e1 e2 )v(1 + e1 e2 )’1 = (1 + e1 e2 )v(1 ’ e1 e2 ) = v.

Yet, 1 + e1 e2 is not a scalar multiple of 1.

The following proposition shows that the notion of norm is well-behaved.

Proposition 1.8 If x ∈ “n , then N (x) ∈ R— · 1.

1.3. CLIFFORD GROUPS 21

Proof . The trick is to show that N (x) is in the kernel of ρ. To say that x ∈ “n means that

±(x)vx’1 ∈ Rn for all v ∈ Rn .

Applying t, we get

t(x)’1 vt(±(x)) = ±(x)vx’1 ,

since t is the identity on Rn . Thus, we have

v = t(x)±(x)v(t(±(x))x)’1 = ±(xx)v(xx)’1 ,

so xx ∈ Ker (ρ). By Proposition 1.5, we have x ∈ “n , and so, xx = x x ∈ Ker (ρ).

Remark: Again, the proof also holds for the Cli¬ord group “(¦) associated with any non-

degenerate quadratic form ¦. When ¦(v) = ’ v 2 , where v is the standard Euclidean

norm of v, we have N (v) = v 2 · 1 for all v ∈ V . However, for other quadratic forms, it is

possible that N (x) = » · 1 where » < 0, and this is a di¬culty that needs to be overcome.

Proposition 1.9 The restriction of the norm, N , to “n is a homomorphism, N : “n ’ R— ·1,

and N (±(x)) = N (x) for all x ∈ “n .

Proof . We have

N (xy) = xyy x = xN (y)x = xxN (y) = N (x)N (y),

where the third equality holds because N (x) ∈ R— · 1. We also have

N (±(x)) = ±(x)±(x) = ±(xx) = ±(N (x)) = N (x).

Remark: The proof also holds for the Cli¬ord group “(¦) associated with any nondegen-

erate quadratic form ¦.

Proposition 1.10 We have Rn ’ {0} ⊆ “n and ρ(“n ) ⊆ O(n).

Proof . Let x ∈ “n and v ∈ Rn , with v = 0. We have

N (ρ(x)(v)) = N (±(x)vx’1 ) = N (±(x))N (v)N (x’1 ) = N (x)N (v)N (x)’1 = N (v),

since N : “n ’ R— · 1. However, for v ∈ Rn , we know that

N (v) = ’¦(v)1.

Thus, ρ(x) is norm-preserving, and so, ρ(x) ∈ O(n).

Remark: The proof that ρ(“(¦)) ⊆ O(¦) also holds for the Cli¬ord group “(¦) associated

with any nondegenerate quadratic form ¦. The ¬rst statement needs to be replaced by the

fact that every non-isotropic vector in Rn (a vector is non-isotropic if ¦(x) = 0) belongs to

“(¦). Indeed, x2 = ¦(x) · 1, which implies that x is invertible.

We are ¬nally ready for the introduction of the groups Pin(n) and Spin(n).

22 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

1.4 The Groups Pin(n) and Spin(n)

De¬nition 1.5 We de¬ne the pinor group, Pin(n), as the kernel Ker (N ) of the homomor-

phism N : “n ’ R— · 1, and the spinor group, Spin(n), as Pin(n) © “+ .

n

Observe that if N (x) = 1, then x is invertible and x’1 = x, since xx = N (x) = 1. Thus,

we can write

Pin(n) = {x ∈ Cln | xvx’1 ∈ Rn for all v ∈ Rn , N (x) = 1},

and

Spin(n) = {x ∈ Cl0 | xvx’1 ∈ Rn for all v ∈ Rn , N (x) = 1}.

n

Remark: According to Atiyah, Bott and Shapiro, the use of the name Pin(k) is a joke due

to Jean-Pierre Serre (Atiyah, Bott and Shapiro [3], page 1).

Theorem 1.11 The restriction of ρ to the pinor group, Pin(n), is a surjective homo-

morphism, ρ: Pin(n) ’ O(n), whose kernel is {’1, 1}, and the restriction of ρ to the

spinor group, Spin(n), is a surjective homomorphism, ρ: Spin(n) ’ SO(n), whose ker-

nel is {’1, 1}.

Proof . By Proposition 1.10, we have a map ρ: Pin(n) ’ O(n). The reader can easily check

that ρ is a homomorphism. By the Cartan-Dieudonn´ theorem (see Bourbaki [6], or Gallier

e

[16], Chapter 7, Theorem 7.2.1), every isometry f ∈ SO(n) is the composition f = s1 —¦· · ·—¦sk

of hyperplane re¬‚ections sj . If we assume that sj is a re¬‚ection about the hyperplane Hj

orthogonal to the nonzero vector wj , by Proposition 1.6, ρ(wj ) = sj . Since N (wj ) = wj 2 ·1,

we can replace wj by wj / wj , so that N (w1 · · · wk ) = 1, and then

f = ρ(w1 · · · wk ),

and ρ is surjective. Note that

Ker (ρ | Pin(n)) = Ker (ρ) © ker(N ) = {t ∈ R— · 1 | N (t) = 1} = {’1, 1}.

As to Spin(n), we just need to show that the restriction of ρ to Spin(n) maps “n into

SO(n). If this was not the case, there would be some improper isometry f ∈ O(n) so that

ρ(x) = f , where x ∈ “n © Cl0 . However, we can express f as the composition of an odd

n

number of re¬‚ections, say

f = ρ(w1 · · · w2k+1 ).

Since

ρ(w1 · · · w2k+1 ) = ρ(x),

we have x’1 w1 · · · w2k+1 ∈ Ker (ρ). By Lemma 1.7, we must have

x’1 w1 · · · w2k+1 = »1

1.4. THE GROUPS PIN(N ) AND SPIN(N ) 23

for some » ∈ R— , and thus,

w1 · · · w2k+1 = »x,

where x has even degree and w1 · · · w2k+1 has odd degree, which is impossible.

Let us denote the set of elements v ∈ Rn with N (v) = 1 (with norm 1) by S n’1 . We have

the following corollary of Theorem 1.11:

Corollary 1.12 The group Pin(n) is generated by S n’1 and every element of Spin(n) can

be written as the product of an even number of elements of S n’1 .

Example 1.2 The reader should verify that

Pin(1) ≈ Z/4Z, Spin(1) = {’1, 1} ≈ Z/2Z,

and also that

Pin(2) ≈ {ae1 + be2 | a2 + b2 = 1} ∪ {c1 + de1 e2 | c2 + d2 = 1}, Spin(2) = U(1).

We may also write Pin(2) = U(1) + U(1), where U(1) is the group of complex numbers

of modulus 1 (the unit circle in R2 ). It can also be shown that Spin(3) ≈ SU(2) and

Spin(4) ≈ SU(2) — SU(2). The group Spin(5) is isomorphic to the symplectic group

Sp(2), and Spin(6) is isomorphic to SU(4) (see Curtis [12] or Porteous [23]).

Let us take a closer look at Spin(2). The Cli¬ord algebra Cl2 is generated by the four

elements

1, e1 , e2 , , e1 e2 ,

and they satisfy the relations

e2 = ’1, e2 = ’1, e1 e2 = ’e2 e1 .

1 2

The group Spin(2) consists of all products

2k

(ai e1 + bi e2 )

i=1

consisting of an even number of factors and such that a2 + b2 = 1. In view of the above

i i

relations, every such element can be written as

x = a1 + be1 e2 ,

where x satis¬es the conditions that xvx’1 ∈ R2 for all v ∈ R2 , and N (x) = 1. Since

X = a1 ’ be1 e2 ,

24 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

we get

N (x) = a2 + b2 ,

and the condition N (x) = 1 is simply a2 + b2 = 1. We claim that xvx’1 ∈ R2 if x ∈ Cl0 .

2

0 1 1

’1 ’1

Indeed, since x ∈ Cl2 and v ∈ Cl2 , we have xvx ∈ Cl2 , which implies that xvx ∈ R2 ,

since the only elements of Cl1 are those in R2 . Then, Spin(2) consists of those elements

2

2 2

x = a1 + be1 e2 so that a + b = 1. If we let i = e1 e2 , we observe that

i2 = ’1,

e1 i = ’ie1 = ’e2 ,

e2 i = ’ie2 = e1 .

Thus, Spin(2) is isomorphic to U(1). Also note that

e1 (a1 + bi) = (a1 ’ bi)e1 .

Let us ¬nd out explicitly what is the action of Spin(2) on R2 . Given X = a1 + bi, with

a2 + b2 = 1, for any v = v1 e1 + v2 e2 , we have

±(X)vX ’1 = X(v1 e1 + v2 e2 )X ’1

= X(v1 e1 + v2 e2 )(’e1 e1 )X

= X(v1 e1 + v2 e2 )(’e1 )(e1 X)

= X(v1 1 + v2 i)Xe1

X 2 (v1 1 + v2 i)e1

=

(((a2 ’ b2 )v1 ’ 2abv2 )1 + (a2 ’ b2 )v2 + 2abv1 )i)e1

=

((a2 ’ b2 )v1 ’ 2abv2 )e1 + (a2 ’ b2 )v2 + 2abv1 )e2 .

=

Since a2 + b2 = 1, we can write X = a1 + bi = (cos θ)1 + (sin θ)i, and the above derivation

shows that

±(X)vX ’1 = (cos 2θv1 ’ sin 2θv2 )e1 + (cos 2θv2 + sin 2θv1 )e2 .

This means that the rotation ρX induced by X ∈ Spin(2) is the rotation of angle 2θ around

the origin. Observe that the maps

v ’ v(’e1 ), X ’ Xe1

establish bijections between R2 and Spin(2) U(1). Also, note that the action of X =

cos θ + i sin θ viewed as a complex number yields the rotation of angle θ, whereas the action

of X = (cos θ)1 + (sin θ)i viewed as a member of Spin(2) yields the rotation of angle 2θ.

There is nothing wrong. In general, Spin(n) is a two“to“one cover of SO(n).

Next, let us take a closer look at Spin(3). The Cli¬ord algebra Cl3 is generated by the

eight elements

1, e1 , e2 , , e3 , , e1 e2 , e2 e3 , e3 e1 , e1 e2 e3 ,

1.4. THE GROUPS PIN(N ) AND SPIN(N ) 25

and they satisfy the relations

e2 = ’1, ej ej = ’ej ei , 1 ¤ i, j ¤ 3, i = j.

i

The group Spin(3) consists of all products

2k

(ai e1 + bi e2 + ci e3 )

i=1

consisting of an even number of factors and such that a2 + b2 + c2 = 1. In view of the above

i i i

relations, every such element can be written as

x = a1 + be2 e3 + ce3 e1 + de1 e2 ,

where x satis¬es the conditions that xvx’1 ∈ R3 for all v ∈ R3 , and N (x) = 1. Since

X = a1 ’ be2 e3 ’ ce3 e1 ’ de1 e2 ,

we get

N (x) = a2 + b2 + c2 + d2 ,

and the condition N (x) = 1 is simply a2 + b2 + c2 + d2 = 1.

It turns out that the conditions x ∈ Cl0 and N (x) = 1 imply that xvx’1 ∈ R3 for all

3

v ∈ R . To prove this, ¬rst observe that N (x) = 1 implies that x’1 = ±x, and that v = ’v

3

for any v ∈ R3 , and so,

xvx’1 = ’xvx’1 .

Also, since x ∈ Cl0 and v ∈ Cl1 , we have xvx’1 ∈ Cl1 . Thus, we can write

3 3 3

xvx’1 = u + »e1 e2 e3 , for some u ∈ R3 and some » ∈ R.

But

e1 e2 e3 = ’e3 e2 e1 = e1 e2 e3 ,

and so,

xvx’1 = ’u + »e1 e2 e3 = ’xvx’1 = ’u ’ »e1 e2 e3 ,

which implies that » = 0. Thus, xvx’1 ∈ R3 , as claimed. Then, Spin(3) consists of those

elements x = a1 + be2 e3 + ce3 e1 + de1 e2 so that a2 + b2 + c2 + d2 = 1. Under the bijection

i ’ e2 e3 , j ’ e3 e1 , k ’ e1 e2 ,

we can check that we have an isomorphism between the group SU(2) of unit quaternions

and Spin(3). If X = a1 + be2 e3 + ce3 e1 + de1 e2 ∈ Spin(3), observe that

X ’1 = X = a1 ’ be2 e3 ’ ce3 e1 ’ de1 e2 .

26 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

Now, using the identi¬cation

i ’ e2 e3 , j ’ e3 e1 , k ’ e1 e2 ,

we can easily check that

(e1 e2 e3 )2 = 1,

i(e1 e2 e3 ) = ’e1 ,

(e1 e2 e3 )i =

j(e1 e2 e3 ) = ’e2 ,

(e1 e2 e3 )j =

k(e1 e2 e3 ) = ’e3 ,

(e1 e2 e3 )k =

’i,

(e1 e2 e3 )e1 =

’j,

(e1 e2 e3 )e2 =

’k.

(e1 e2 e3 )e3 =

Then, if X = a1 + bi + cj + dk ∈ Spin(3), for every v = v1 e1 + v2 e2 + v3 e3 , we have

±(X)vX ’1 = X(v1 e1 + v2 e2 + v3 e3 )X ’1

X(e1 e2 e3 )2 (v1 e1 + v2 e2 + v3 e3 )X ’1

=

(e1 e2 e3 )X(e1 e2 e3 )(v1 e1 + v2 e2 + v3 e3 )X ’1

=

’(e1 e2 e3 )X(v1 i + v2 j + v3 k)X ’1 .

=

This shows that the rotation ρX ∈ SO(3) induced by X ∈ Spin(3) can be viewed as the

rotation induced by the quaternion a1 + bi + cj + dk on the pure quaternions, using the maps

v ’ ’(e1 e2 e3 )v, X ’ ’(e1 e2 e3 )X

to go from a vector v = v1 e1 + v2 e2 + v3 e3 to the pure quaternion v1 i + v2 j + v3 k, and back.

We close this section by taking a closer look at Spin(4). The group Spin(4) consists of

all products

2k

(ai e1 + bi e2 + ci e3 + di e4 )

i=1

consisting of an even number of factors and such that a2 + b2 + c2 + d2 = 1. Using the

i i i i

relations

e2 = ’1, ej ej = ’ej ei , 1 ¤ i, j ¤ 4, i = j,

i

every element of Spin(4) can be written as

x = a1 1 + a2 e1 e2 + a3 e2 e3 + a4 e3 e1 + a5 e4 e3 + a6 e4 e1 + a7 e4 e2 + a8 e1 e2 e3 e4 ,

where x satis¬es the conditions that xvx’1 ∈ R4 for all v ∈ R4 , and N (x) = 1. Let

i = e1 e2 , j = e2 e3 , k = 33 e1 , i = e4 e3 , j = e4 e1 , k = e4 e2 ,

1.4. THE GROUPS PIN(N ) AND SPIN(N ) 27

and I = e1 e2 e3 e4 . The reader will easily verify that

ij = k

jk = i

ki = j

i2 ’1, j2 = ’1, k2 = ’1

=

iI = Ii = i

jI = Ij = j

kI = Ik = k

I2 = 1, I = I.

Then, every x ∈ Spin(4) can be written as

x = u + Iv, with u = a1 + bi + cj + dk and v = a 1 + b i + c j + d k,

with the extra conditions stated above. Using the above identities, we have

(u + Iv)(u + Iv ) = uu + vv + I(uv + vu ).

As a consequence,

N (u + Iv) = (u + Iv)(u + Iv) = uu + vv + I(uv + vu),

and thus, N (u + Iv) = 1 is equivalent to

uu + vv = 1 and uv + vu = 0.

As in the case n = 3, it turns out that the conditions x ∈ Cl0 and N (x) = 1 imply that

4

xvx ∈ R for all v ∈ R . The only change to the proof is that xvx’1 ∈ Cl1 can be written

’1 4 4

4

as

xvx’1 = u + »i,j,k ei ej ek , for some u ∈ R4 , with {i, j, k} ⊆ {1, 2, 3, 4}.

i,j,k

As in the previous proof, we get »i,j,k = 0. Then, Spin(4) consists of those elements u + Iv

so that

uu + vv = 1 and uv + vu = 0,

with u and v of the form a1 + bi + cj + dk. Finally, we see that Spin(4) is isomorphic to

Spin(2) — Spin(2) under the isomorphism

u + vI ’ (u + v, u ’ v).

Indeed, we have

N (u + v) = (u + v)(u + v) = 1,

28 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

and

N (u ’ v) = (u ’ v)(u ’ v) = 1,

since

uu + vv = 1 and uv + vu = 0,

and

(u + v, u ’ v)(u + v , u ’ v ) = (uu + vv + uv + vu , uu + vv ’ (uv + vu )).

Remark: It can be shown that the assertion if x ∈ Cl0 and N (x) = 1, then xvx’1 ∈ Rn for

n

n

all v ∈ R , is true up to n = 5 (see Porteous [23], Chapter 13, Proposition 13.58). However,

√

this is already false for n = 6. For example, if X = 1/ 2(1 + e1 e2 e3 e4 e5 e6 ), it is easy to see

that N (X) = 1, and yet, Xe1 X ’1 ∈ R6 .

/

1.5 The Groups Pin(p, q) and Spin(p, q)

For every nondegenerate quadratic form ¦ over R, there is an orthogonal basis with respect

to which ¦ is given by

¦(x1 , . . . , xp+q ) = x2 + · · · + x2 ’ (x2 + · · · + x2 ),

1 p p+1 p+q

where p and q only depend on ¦. The quadratic form corresponding to (p, q) is denoted ¦p,q

and we call (p, q) the signature of ¦p,q . Let n = p + q. We can de¬ne the groups O(p, q) and

SO(p, q) as the group of isometries w.r.t. ¦p,q , i.e., the group of linear maps f so that

¦p,q (f (v)) = ¦p,q (v) for all v ∈ Rn .

We denote the Cli¬ord algebra Cl(¦p,q ) where ¦p,q has signature (p, q) by Clp,q , the corre-

sponding Cli¬ord group by “p,q , and the special Cli¬ord group “p,q © Cl0 by “+ . Note that

p,q p,q

with this new notation, Cln = Cl0,n .

As we mentioned earlier, since Lawson and Michelsohn [20] adopt the opposite of our

sign convention in de¬ning Cli¬ord algebras, their Cl(p, q) is our Cl(q, p).

As we mentioned in Section 1.3, we have the problem that N (v) = ’¦(v) · 1 but ’¦(v)

is not necessarily positive (where v ∈ Rn ). The ¬x is simple: Allow elements x ∈ “p,q with

N (x) = ±1.

De¬nition 1.6 We de¬ne the pinor group, Pin(p, q), as the group

Pin(p, q) = {x ∈ “p,q | N (x) = ±1},

and the spinor group, Spin(p, q), as Pin(p, q) © “+ .

p,q

1.5. THE GROUPS PIN(P, Q) AND SPIN(P, Q) 29

Remarks:

(1) It is easily checked that the group Spin(p, q) is also given by

Spin(p, q) = {x ∈ Cl0 | xvx ∈ Rn for all v ∈ Rn , N (x) = 1}.

p,q

This is because Spin(p, q) consists of elements of even degree.

(2) One can check that if N (x) = 0, then

±(x)vx’1 = xvt(x)/N (x).

Thus, we have

Pin(p, q) = {x ∈ Clp,q | xvt(x)N (x) ∈ Rn for all v ∈ Rn , N (x) = ±1}.

When ¦(x) = ’ x 2 , we have N (x) = x 2 , and

Pin(n) = {x ∈ Cln | xvt(x) ∈ Rn for all v ∈ Rn , N (x) = 1}.

Theorem 1.11 generalizes as follows:

Theorem 1.13 The restriction of ρ to the pinor group, Pin(p, q), is a surjective homomor-

phism, ρ: Pin(p, q) ’ O(p, q), whose kernel is {’1, 1}, and the restriction of ρ to the spinor

group, Spin(p, q), is a surjective homomorphism, ρ: Spin(p, q) ’ SO(p, q), whose kernel is

{’1, 1}.

Proof . The Cartan-Dieudonn´ also holds for any nondegenerate quadratic form ¦, in the

e

sense that every isometry in O(¦) is the composition of re¬‚ections de¬ned by hyperplanes

orthogonal to non-isotropic vectors (see Dieudonn´ [13], Chevalley [10], Bourbaki [6], or

e

Gallier [16], Chapter 7, Problem 7.14). Thus, Theorem 1.11 also holds for any nondegenerate

quadratic form ¦. The only change to the proof is the following: Since N (wj ) = ’¦(wj ) · 1,

we can replace wj by wj / |¦(wj )|, so that N (w1 · · · wk ) = ±1, and then

f = ρ(w1 · · · wk ),

and ρ is surjective.

If we consider Rn equipped with the quadratic form ¦p,q (with n = p + q), we denote the

set of elements v ∈ Rn with N (v) = 1 by Sp,q . We have the following corollary of Theorem

n’1

1.13 (generalizing Corollary 1.14):

n’1

Corollary 1.14 The group Pin(p, q) is generated by Sp,q and every element of Spin(p, q)

n’1

can be written as the product of an even number of elements of Sp,q .

30 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

Example 1.3 The reader should check that

Cl0,1 ≈ C, Cl1,0 ≈ R • R.

We also have

Pin(0, 1) ≈ Z/4Z, Pin(1, 0) ≈ Z/2Z — Z/2Z,

from which we get Spin(0, 1) = Spin(1, 0) ≈ Z/2Z. Also, show that

Cl0,2 ≈ H, Cl1,1 ≈ M2 (R), Cl2,0 ≈ M2 (R),

where Mn (R) denotes the algebra of n — n matrices. One can also work out what are

Pin(2, 0), Pin(1, 1), and Pin(0, 2). Show that

Spin(0, 2) = Spin(2, 0) ≈ U(1),

and

Spin(1, 1) = {a · 1 + be1 e2 | a2 ’ b2 = 1}.

Observe that Spin(1, 1) is not connected.

More generally, it can be shown that Cl0 and Cl0 are isomorphic, from which it follows

p,q q,p

that Spin(p, q) and Spin(q, p) are isomorphic, but Pin(p, q) and Pin(q, p) are not isomorphic

in general, and in particular, Pin(p, 0) and Pin(0, p) are not isomorphic in general (see

Choquet-Bruhat [11], Chapter I). However, due to the “8-periodicity” of the Cli¬ord algebras

(to be discussed in the next section), it can be shown that Clp,q and Clq,p are isomorphic

when p ’ q = 0 mod 4.

1.6 Periodicity of the Cli¬ord Algebras Clp,q

It turns out that the real algebras Clp,q can be build up as tensor products of the basic

algebras R, C, and H. According to Lounesto (Section 23.16 [21]), the description of the real

algebras Clp,q as matrix algebras and the 8-periodicity was ¬rst discovered by Elie Cartan

in 1908. Of course, Cartan used a very di¬erent notation. These facts were rediscovered

independently by Raoul Bott in the 1960™s (see Raoul Bott™s comments in Volume 2 of his

Collected papers.).

We will use the notation R(n) (resp. C(n)) for the algebra Mn (R) of all n — n real

matrices (resp. the algebra Mn (C) of all n — n complex matrices). As mentioned in Example

1.3, it is not hard to show that

Cl1,0 = R • R

Cl0,1 = C

Cl0,2 = H Cl2,0 = R(2)

and

Cl1,1 = R(2).

The key to the classi¬cation is the following lemma:

1.6. PERIODICITY OF THE CLIFFORD ALGEBRAS CLP,Q 31

Lemma 1.15 We have the isomorphisms

Cl0,n+2 ≈ Cln,0 — Cl0,2

Cln+2,0 ≈ Cl0,n — Cl2,0

Clp+1,q+1 ≈ Clp,q — Cl1,1 ,

for all n, p, q ≥ 0.

Proof . Let ¦0,n (x) = ’ x 2 , where x is the standard Euclidean norm on Rn+2 , and let

(e1 , . . . , en+2 ) be an orthonormal basis for Rn+2 under the standard Euclidean inner product.

We also let (e1 , . . . , en ) be a set of generators for Cln,0 and (e1 , e2 ) be a set of generators

for Cl0,2 . We can de¬ne a linear map f : Rn+2 ’ Cln,0 — Cl0,2 by its action on the basis

(e1 , . . . , en+2 ) as follows:

ei — e1 e2 for 1 ¤ i ¤ n

f (ei ) =

1 — ei’n for n + 1 ¤ i ¤ n + 2.

Observe that for 1 ¤ i, j ¤ n, we have

f (ei )f (ej ) + f (ej )f (ei ) = (ei ej + ej ei ) — (e1 ee )2 = ’2δij 1 — 1,

since e1 e2 = ’e2 e1 , (e1 )2 = ’1, and (e2 )2 = ’1, and ei ej = ’ej ei , for all i = j, and

(ei )2 = 1, for all i with 1 ¤ i ¤ n. Also, for n + 1 ¤ i, j ¤ n + 2, we have

f (ei )f (ej ) + f (ej )f (ei ) = 1 — (ei’n ej’n + ej’n ei’n ) = ’2δij 1 — 1,

and

f (ei )f (ek ) + f (ek )f (ei ) = 2ei — (e1 e2 en’k + en’k e1 e2 ) = 0,

for 1 ¤ i, j ¤ n and n + 1 ¤ k ¤ n + 2 (since en’k = e1 or en’k = e2 ). Thus, we have

2

f (x)2 = ’ x · 1 — 1 for all x ∈ Rn+2 ,

and by the universal mapping property of Cl0,n+2 , we get an algebra map

f : Cl0,n+2 ’ Cln,0 — Cl0,2 .

Since f maps onto a set of generators, it is surjective. However

dim(Cl0,n+2 ) = 2n+2 = 2n · 2 = dim(Cln,0 )dim(Cl0,2 ) = dim(Cln,0 — Cl0,2 ),

and f is an isomorphism.

The proof of the second identity is analogous. For the third identity, we have

¦p,q (x1 , . . . , xp+q ) = x2 + · · · + x2 ’ (x2 + · · · + x2 ),

1 p p+1 p+q

32 CHAPTER 1. CLIFFORD ALGEBRAS, CLIFFORD GROUPS, PIN AND SPIN

and let (e1 , . . . , ep+1 , 1 , . . . , q+1 ) be an orthogonal basis for Rp+q+2 so that ¦p+1,q+1 (ei ) = +1

and ¦p+1,q+1 ( j ) = ’1 for i = 1, . . . , p+1 and j = 1, . . . , q +1. Also, let (e1 , . . . , ep , 1 , . . . , q )

be a set of generators for Clp,q and (e1 , 1 ) be a set of generators for Cl1,1 . We de¬ne a linear

map f : Rp+q+2 ’ Clp,q — Cl1,1 by its action on the basis as follows:

ei — e1 for 1 ¤ i ¤ p

1

f (ei ) =

1 — e1 for i = p + 1,

and

— e1 for 1 ¤ j ¤ q

j 1

f ( j) =

1— 1 for j = q + 1.

We can check that

f (x)2 = ¦p+1,q+1 (x) · 1 — 1 for all x ∈ Rp+q+2 ,

and we ¬nish the proof as in the ¬rst case.

To apply this lemma, we need some further isomorphisms among various matrix algebras.

Proposition 1.16 The following isomorphisms hold:

R(m) — R(n) ≈ for all m, n ≥ 0

R(mn)

R(n) —R K ≈ for K = C or K = H and all n ≥ 0

K(n)

C —R C ≈ C•C

C —R H ≈ C(2)

H —R H ≈ R(4).

Proof . Details can be found in Lawson and Michelsohn [20]. The ¬rst two isomorphisms are

quite obvious. The third isomorphism C • C ’ C — C is obtained by sending

1 1

(1, 0) ’ (1 — 1 + i — i), (0, 1) ’ (1 — 1 ’ i — i).

2 2

The ¬eld C is isomorphic to the subring of H generated by i. Thus, we can view H as a

C-vector space under left scalar multiplication. Consider the R-bilinear map

π: C — H ’ HomC (H, H) given by

πy,z (x) = yxz,

where y ∈ C and x, z ∈ H. Thus, we get an R-linear map π: C —R H ’ HomC (H, H).

However, we have HomC (H, H) ≈ C(2). Furthermore, since

πy,z —¦ πy ,z = πyy ,zz ,

the map π is an algebra homomorphism

π: C — H ’ C(2).

1.6. PERIODICITY OF THE CLIFFORD ALGEBRAS CLP,Q 33

We can check on a basis that π is injective, and since

dimR (C — H) = dimR (C(2)) = 8,

the map π is an isomorphism. The last isomorphism is proved in a similar fashion.

We now have the main periodicity theorem.

Theorem 1.17 (Cartan/Bott) For all n ≥ 0, we have the following isomorphisms:

Cl0,n+8 ≈ Cl0,n — Cl0,8

Cln+8,0 ≈ Cln,0 — Cl8,0 .

Furthermore,

Cl0,8 = Cl8,0 = R(16).

Proof . By Lemma 1.15 we have the isomorphisms

Cl0,n+2 ≈ Cln,0 — Cl0,2

Cln+2,0 ≈ Cl0,n — Cl2,0 ,

and thus,

Cl0,n+8 ≈ Cln+6,0 — Cl0,2 ≈ Cl0,n+4 — Cl2,0 — Cl0,2 ≈ · · · ≈ Cl0,n — Cl2,0 — Cl0,2 — Cl2,0 — Cl0,2 .

Since Cl0,2 = H and Cl2,0 = R(2), by Proposition 1.16, we get

Cl2,0 — Cl0,2 — Cl2,0 — Cl0,2 ≈ H — H — R(2) — R(2) ≈ R(4) — R(4) ≈ R(16).

The second isomorphism is proved in a similar fashion.

From all this, we can deduce the following table:

n 0 1 2 3 4 5 6 7 8

H • H H(2) R(8) R(8) • R(8) R(16)

Cl0,n R C H C(4)

R R • R R(2) C(2) H(2) H(2) • H(2) H(4)

Cln,0 C(8) R(16)