The ideal I de¬ning the Cli¬ord algebra is not Z homogeneous (unless the

bilinear form is identically zero) since its generators y1 y2 + y2 y1 ’ 2(y1 , y2 )1 are

“mixed”, being a sum of terms of degree two and degree zero in T (p). But these

terms are both even. So the Z/2Z gradation is preserved upon passing to the

quotient. In other words, C(p) is a Z/2Z graded algebra:

C(p) = C0 (p) • C1 (p)

where the elements of C0 (p) consist of sums of products of elements of p with

an even number of factors and C1 (p) consist of sums of terms each a product of

elements of p with an odd number of factors. The usual rules for multiplication

of a graded algebra obtain:

C0 (p) · C0 (p) ‚ C0 (p), C0 (p) · C1 (p) ‚ C1 (p), C1 (p) · C1 (p) ‚ C0 (p).

§p as a C(p) module.

9.1.3

Let p be a vector space with a non-degenerate symmetric bilinear form. The

exterior algebra, §p inherits a bilinear form which we continue to denote by

( , ). Here the spaces §k (p) and § (p) are orthogonal if k = while

(x1 § · · · § xk , y1 § · · · § yk ) = det ((xi , yj )) .

For v ∈ p let (v) ∈ End(§p) denote exterior multiplication by v and ι(v) be

the transpose of (v) relative to this biinear form on §p.

So ι(v) is interior multiplication by the element of p— corresponding to v

under the map p ’ p— induced by ( , )p . The map

p ’ End(§p), v ’ (v) + ι(v)

is a Cli¬ord map, i.e. satis¬es

( (v) + ι(v))2 = (v, v)p id

and so extends to a homomorphism of

C(p) ’ End §p

making §p into a C(p) module. We let xy denote the product of x and y in

C(p).

Chevalley™s linear identi¬cation of C(p) with §p.

9.1.4

Consider the linear map

C(p) ’ §p, x ’ x1

9.1. DEFINITION AND BASIC PROPERTIES 149

where 1 ∈ §0 p under the identi¬cation of §0 p with the ground ¬eld. The

element x1 on the extreme right means the image of 1 under the action of

x ∈ C(p).

For elements v1 , . . . , vk ∈ p this map sends

’

v1 v1

’ v1 § v2 + (v1 , v2 )1

v1 v2

’ v1 § v2 § v3 + (v1 , v2 )v3 ’ (v1 , v3 )v2 + (v2 , v3 )v1

v1 v2 v3

’ v1 § v2 § v3 § v4 + (v2 , v3 )v1 § v4 ’ (v2 , v4 )v1 § v3

v1 v2 v3 v4

+(v3 , v4 )v1 § v2 + (v1 , v2 )v3 § v4 ’ (v1 , v3 )v1 § v4

+(v1 , v4 )v2 § v3 + (v1 , v4 )(v2 , v3 ) ’ (v1 , v3 )(v2 , v4 ) + (v1 , v2 )(v3 , v4 )

. .

. .

. .

If the v™s form an “orthonormal” basis of p then the products

vi1 · · · vik , i1 < i2 · · · < ik , k = 0, 1, . . . , n

form a basis of C(p) while the

vi1 § · · · § vik , i1 < i2 · · · < ik , k = 0, 1, . . . , n

form a basis of §p, and in fact

v1 · · · vk ’ v1 § · · · § vk if (vi , vj ) = 0 ∀i = j. (9.1)

In particular, the map C(p) ’ §p given above is an isomorphism of vector

spaces, so we may identify C(p) with §p as a vector space if we choose, and

then consider that §p has two products: the Cli¬ord product which we denote

by juxtaposition and the exterior product which we denote with a §.

Notice that this identi¬cation preserves the Z/2Z gradation, an even element

of the Cli¬ord algebra is identi¬ed with an even element of the exterior algebra

and an odd element is identi¬ed with an odd element.

9.1.5 The canonical antiautomorphism.

The Cli¬ord algebra has a canonical anti-automorphism a which is the identity

map on p. In particular, for vi ∈ p we have a(v1 v2 ) = v2 v1 , a(v1 v2 v3 ) = v3 v2 v1 ,

etc. By abuse of language, we use the same letter a to denote the similar anti-

automorphism on §p and observe from the above computations (in particular

from the corresponding choice of bases) that a commutes with our identifying

map C(p) ’ §p so the notation is consistent. We have

1

a = (’1) 2 k(k’1) id on §k (p).

150CHAPTER 9. CLIFFORD ALGEBRAS AND SPIN REPRESENTATIONS.

For small values of k we have

1

(’1) 2 k(k’1)

k

0 1

1 1

’1

2

’1

3

4 1

5 1

’1.

6

We will use subscripts to denote the homogeneous components of elements

of §p. Notice that if u ∈ §2 p then au = ’u by the above table, while

a(u2 ) = (au)2 = u2 . Since u2 is even (and hence has only even homogeneous

components) and since the maximum degree of the homogeneous component of

u2 is 4, we conclude that

u2 = (u2 )0 + (u2 )4 ∀ u ∈ §2 p. (9.2)

For the same reason

v 2 = (v 2 )0 + (v 2 )4 ∀ v ∈ §3 p. (9.3)

We also claim the following:

1

(ww )0 = (aw, w ) = (’1) 2 k(k’1) (w, w ) ∀ w, w ∈ §k (p). (9.4)

Indeed, it is su¬cient to verify this for w, w belonging to a basis of §p, say the

basis given by all elements of the form (9.1), in which case both sides of (9.4)

vanish unless w = w . If w = w = v1 § · · · § vk (say) then

(ww)0 = ι(v1 ) · · · ι(vk )v1 § · · · § vk =

1 1

(’1) 2 k(k’1) (v1 , v1 ) · · · (vk , vk ) = (’1) 2 k(k’1) (w, w)

proving (9.4).

As special cases that we will use later on, observe that

(uu )0 = ’(u, u ) ∀ u, u ∈ §2 p (9.5)

and

(vv )0 = ’(v, v ) ∀ v, v ∈ §3 p. (9.6)

9.1.6 Commutator by an element of p.

For any y ∈ p consider the linear map