for k = ’1, 0, 1.

gk» = 0

Furthermore, g±i is one dimensional, and since every root is conjugate to a

simple root, we conclude that

dim g± = 1 ∀± ∈ ¦.

We now show that

g» = {0} for » = 0, » ∈ ¦.

Indeed, suppose that g» = {0}. We know that » is not a multiple of ± for any

± ∈ ¦, since we know this to be true for simple roots, and the dimensions of the

g» are invariant under the Weyl group, each root being conjugate to a simple

root. So »⊥ does not coincide with any hyperplane orthogonal to any root. So

we can ¬nd a µ ∈ »⊥ such that (±, µ) = 0 for all roots. We may ¬nd a w ∈ W

which maps µ into the positive Weyl chamber for ∆ so that (±i , µ) ≥ 0 and

hence (±i , wµ) > 0 for i = 1, . . . , . Now

dim gw» = dim g»

144 CHAPTER 8. SERRE™S THEOREM.

and for the latter to be non-zero, we must have

w» = ki ±i

with the coe¬cients all non-negative or non-positive integers. But

0 = (», µ) = (w», wµ) = ki (±i , µ)

with (±i , µ) > 0 ∀i. Hence all the ki = 0.

So

dim g = + Card ¦.

We conclude the proof if we show that g is semi-simple, i.e. contains no abelian

ideals. So suppose that a is an abelian ideal. Since a is an ideal, it is stable

under h and hence decomposes into weight spaces. If g± © a = {0}, then g± ‚ a

and hence [g’± , g± ] ‚ a and hence the entire sl(2) generated by g± and g’±

is contained in a which is impossible since a is abelian and sl(2) is simple. So

a ‚ h. But then a must be annihilated by all the roots, which implies that

a = {0} since the roots span h— . QED

8.4 The existence of the exceptional root sys-

tems.

The idea of the construction is as follows. For each Dynkin diagram we will

chose a lattice L in a Euclidean space V , and then let ¦ consist of all vectors in

this lattice having all the same length, or having one of two prescribed lengths.

We then check that

2(±1 , ±2 )

∈Z ∀ ±1 , ±2 ∈ ¦.

(±1 , ±1 )

This implies that re¬‚ection through the hyperplane orthogonal to ±1 preserves

L, and since re¬‚ections preserve length that these re¬‚ection s preserve ¦. This

will show that ¦ is a root system and then calculation shows that it is of the

desired type. «

x

3

(G2 ). Let V be the plane in R consisting of all vectors y with

z

x + y + z = 0.

Let L be the intersection of the three dimensional standard lattice Z2 with V .

Let L1 , L2 , L3 denote the standard basis of R3 . Let ¦ consist of all vectors in

L of squared length 2 or 6. So ¦ consists of the six short vectors

±(Li ’ Lj ) i < j

8.4. THE EXISTENCE OF THE EXCEPTIONAL ROOT SYSTEMS. 145

and the six long vectors

±(2Li ’ Lj ’ Lk ) i = 1, 2, 3 j = i, j = k.

We may choose the base to consist of

±= L1 ’ L2 , ±2 = ’2L1 + L2 + L3 .

146 CHAPTER 8. SERRE™S THEOREM.

F4 . Let V = R4 and L = Z4 + Z( 1 (L1 + L2 + L3 + L4 ). Let ¦ consist of all

2

vectors of L of squared length 1 or 2. So ¦ consists of the 24 long roots

±Li ± Lj i<j

and the 24 short roots

1

±L1 , (±L1 ± L2 ± L3 ± L4 ).

2

For ∆ we may take

1

±1 = L2 ’ L3 , ±2 = L3 ’ L4 , ±3 = L4 , ±4 = (L1 ’ L2 ’ L3 ’ L4 ).

2

E8 . Let V = R8 . Let

L := { ci Li , ci ∈ Z, ci even }.

Let

1

L := L + Z( (L1 + L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8 )).

2

Let ¦ consist of all vectors in L of squared length 2. So ¦ consists of the 240

roots

8

1

±Li ± Lj (i < j), ±Li , (even number of + signs).

2 i=1

For ∆ we may take

1

(L1 ’ L1 ’ L2 ’ L3 ’ L4 ’ L5 ’ L6 ’ L7 + L8 )

±1 =

2

±2 = L1 + L2

= Li’1 ’ Li’2 (3 ¤ i ¤ 8).

±i

E7 is obtained from E8 by letting V be the span of the ¬rst 7 ±i . E6 is

obtained from E8 by letting V be the span of the ¬rst 6 ±i .

Chapter 9

Cli¬ord algebras and spin

representations.

9.1 De¬nition and basic properties

9.1.1 De¬nition.

Let p be a vector space with a symmetric bilinear form ( , ). The Cli¬ord

algebra associated to this data is the algebra

C(p) := T (p)/I

where T (p) denotes the tensor algebra

T (p) = k • p • (p — p) • · · ·

and where I denotes the ideal in T (p) generated by all elements of the form

y1 y2 + y2 y1 ’ 2(y1 , y2 )1, y1 , y2 ∈ p

and 1 is the unit element of the tensor algebra. The space p injects as a subspace

of C(p) and generates C(p) as an algebra.

A linear map f of p to an associative algebra A with unit 1A is called a

Cli¬ord map if

∀y1 , y2 ∈ p

f (y1 )f (y2 ) + f (y2 )f (y1 ) = 2(y1 , y2 )1A ,

or what amounts to the same thing (by polarization since we are not over a ¬eld

of characteristic 2) if

f (y)2 = (y, y)1A ∀ y ∈ p.

Any Cli¬ord map gives rise to a unique algebra homomorphism of C(p) to A

whose restriction to p is f . The Cli¬ord algebra is “universal” with respect to

this property.

If the bilinear form is identically zero, then C(p) = §p, the exterior algebra.

But we will be interested in the opposite extreme, the case where the bilinear

form is non-degenerate.

147

148CHAPTER 9. CLIFFORD ALGEBRAS AND SPIN REPRESENTATIONS.