9.1. DEFINITION AND BASIC PROPERTIES 151

which is (anti)commutator in the Cli¬ord multiplication by y. We claim that

[y, w] = 2ι(y)w. (9.7)

In particular, [y, ·], which is automatically a derivation for the Cli¬ord multi-

plication, is also a derivation for the exterior multiplication. Alternatively, this

equation says that ι(y), which is a derivation for the exterior algebra multipli-

cation, is also a derivation for the Cli¬ord multiplication.

To prove (9.7) write

wy = a(ya(w)).

Then

yw = y § w + ι(y)w, wy = a(y § a(w)) + a(ι(y)aw) = w § y + (aι(y)a)w.

We may assume that w ∈ §k p. Then

y § w ’ (’1)k w § y = 0,

so we must show that

aι(y)aw = (’1)k’1 ι(y)w.

For this we may assume that y = 0 and we may write

w =u§z+z ,

where ι(y)u = 1 and ι(y)z = ι(y)z = 0. In fact, we may assume that z and z

are sums of products of linear elements all of which are orthogonal to y. Then

ι(y)az = ι(y)az = 0 so

ι(y)aw = (’1)k’1 az

since z has degree one less than w and hence

aι(y)aw = (’1)k’1 z = (’1)k’1 ι(y)w. QED

Commutator by an element of §2 p.

9.1.7

Suppose that

u ∈ §2 p.

Then for y ∈ p we have

[u, y] = ’[y, u] = ’2ι(y)u. (9.8)

In particular, if u = yi § yj where yi , yj ∈ p we have

[u, y] = 2(yj , y)yi ’ 2(yi , y)yj ∀ y ∈ p. (9.9)

If (yi , yj ) = 0 this is an “in¬nitesimal rotation” in the plane spanned by yi and

yj . Since yi § yj , i < j form a basis of §2 p if y1 , . . . , yn form an “orthonormal”

basis of p, we see that the map

u ’ [u, ·]

152CHAPTER 9. CLIFFORD ALGEBRAS AND SPIN REPRESENTATIONS.

gives an isomorphism of §2 p with the orthogonal algebra o(p). This identi¬-

cation di¬ers by a factor of two from the identi¬cation that we had been using

earlier.

Now each element of o(p) (in fact any linear transformation on p) induces

a derivation of §p. We claim that under the above identi¬cation of §2 p with

o(p), the derivation corresponding to u ∈ §2 p is Cli¬ord commutation by u. In

symbols, if θu denotes this induced derivation, we claim that

θu (w) = [u, w] = uw ’ wu ∀ w ∈ §p. (9.10)

To verify this, it is enough to check it on basis elements of the form (9.1), and

hence by the derivation property for each vj , where this reduces to (9.8).

We can now be more explicit about the degree four component of the Cli¬ord

square of an element of §2 p, i.e. the element (u2 )4 occurring on the right of

(9.2). We claim that for any three elements y, y , y ∈ p

1

ι(y )ι(y )ι(y)u2 = (y §y , u)ι(y )u+(y §y , u)ι(y)u+(y §y, u)ι(y )u. (9.11)

2

To prove this observe that

ι(y)u2 = (ι(y)u) u + u (ι(y)u)

ι(y )ι(y)u2 = (ι(y )ι(y)u) u ’ ι(y)uι(y )u + ι(y )uι(y)u + uι(y )ι(y)u

= 2 ((y § y , u)u + ι(y )u § ι(y)u)

1

ι(y )ι(y )ι(y)u2 (y § y , u)ι(y )u + ι(y )ι(y )u § ι(y)u ’ ι(y )u § ι(y )ι(y)u

=

2

(y § y , u)ι(y )u + (y § y , u)ι(y)u + (y § y, u)ι(y )u

=

as required.

We can also be explicit about the degree zero component of u2 . Indeed, it

follows from (9.9) that if u = yi § yj , i < j where y1 , . . . , yn form an “orthonor-

mal” basis of p then

tr(adp u)2 = ’8(yi , yi )(yj , yj ),

where adp u denotes the (commutator) action of u on p under our identi¬cation

of §2 p with o(p). But

(yi § yj , yi § yj ) = (yi , yi )(yj , yj )(= ±1).

So using (9.5) we see that

1

(u2 )0 = tr(adp u)2 = ’(u, u) (9.12)

8

for u ∈ §2 p.

9.2. ORTHOGONAL ACTION OF A LIE ALGEBRA. 153

9.2 Orthogonal action of a Lie algebra.

Let r be a Lie algebra. Suppose that we have a representation of r acting

as in¬nitesimal orthogonal transformations of p which means, in view of the

identi¬cation of §2 p with o(p) that we have a map

ν : r ’ §2 p

such that

x · y = ’2ι(y)ν(x) (9.13)

where x · y denotes the action of x ∈ r on y ∈ p.

9.2.1 Expression for ν in terms of dual bases.

It will be useful for us to write equation (9.13) in terms of a basis. So let

y1 , . . . , yn be a basis of p and let z1 , . . . , zn be the dual basis relative to ( , ) =

( , )p . We claim that

1

ν(x) = ’ yj § (x · zj ). (9.14)

4 j

Indeed, it su¬ces to verify (9.13) for each of the elements zi . Now

«

1 1 1

ι(zi ) ’ y j § x · zj = ’ x · zi + (zi , x · zj )yj .

4j 4 4j

But

(zi , x · zj ) = ’(x · zi , zj )

since x acts as an in¬nitesimal orthogonal transformation relative to ( , ). So

we can write the sum as

1 1 1

(zi , x · zj )yj = ’ (x · zi , zj )yj = ’ x · zi

4 4 4

j j

yielding «

1 1

ι(zi ) ’ y j § x · zj = ’ x · zi

4 2

j

which is (9.13).

9.2.2 The adjoint action of a reductive Lie algebra.