• • • • • •

E7

•

• • • • • • •

E8

•

Figure 6.5: Dynkin diagrams.

6.6. CLASSIFICATION OF THE IRREDUCIBLE ROOT SYSTEMS. 107

The right hand side is a non-negative integer between 0 and 4. So assuming

that ± = ±β and β ≥ ± The possibilities are listed in the following table:

2 2

0¤θ¤π

±, β β, ± β /±

0 0 π/2 undetermined

1 1 π/3 1

’1 ’1 2π/3 1

1 2 π/4 2

’1 ’1 3π/4 2

1 3 π/6 3

’1 ’3 5π/6 3

Proposition 20 If ± = ±β and if (±, β) > 0 then ± ’ β is a root. If (±, β) < 0

then ± + β is a root.

Proof. The second assertion follows from the ¬rst by replacing β by ’β. So

we need to prove the ¬rst assertion. From the table, one or the other of β, ±

or ±, β equals one. So either s± β = β ’ ± is a root or sβ ± = ± ’ β is a root.

But roots occur along with their negatives so in either event ± ’ β is a root.

QED

Proposition 21 Suppose that ± = ±β are roots. Let r be the largest integer

such that β ’ r± is a root, and let q be the largest integer such that β + q± is a

root. Then β + i± is a root for all ’r ¤ i ¤ q. Furthermore r ’ q = β, ± so

in particular |q ’ r| ¤ 3.

Proof. Suppose not. Then we can ¬nd a p and an s such that ’r ¤ p < s ¤ q

such that β + p± is a root, but β + (p + 1)± is not a root, and β + s± is a root

but β + (s ’ 1)± is not. The preceding proposition then implies that

β + p±, ± ≥ 0 β + s±, ± ¤ 0

while

which is impossible since ±, ± = 2 > 0.

Now s± adds a multiple of ± to any root, and so preserves the string of roots

β ’ r±, β ’ (r ’ 1)±, . . . , β + q±. Furthermore

s± (β + i±) = β ’ ( β, ± + i)±

so s± reverses the order of the string. In particular

s± (β + q±) = β ’ r±.

The left hand side is β’( β, ± +q)± so r’q = β, ± as stated in the proposition.

QED

We can now apply all the preceding de¬nitions and arguments to conclude

that the Dynkin diagrams above classify all the irreducible bases ∆ of root

systems.

108 CHAPTER 6. THE SIMPLE FINITE DIMENSIONAL ALGEBRAS.

Since every root is conjugate to a simple root, we can use the Dynkin dia-

grams to conclude that in an irreducible root system, either all roots have the

same length (cases A, D, E) or there are two root lengths - the remaining cases.

Furthermore, if β denotes a long root and ± a short root, the ratios β 2 / ± 2

are 2 in the cases B, C, and F4 , and 3 for the case G2 .

Proposition 22 In an irreducible root system, the Weyl group W acts irre-

ducibly on E. In particular, the W -orbit of any root spans E.

Proof. Let E be a proper invariant subspace. Let E denote its orthogonal

complement, so

E =E •E .

For any root ±, If e ∈ E then s± e = e ’ e, ± ± ∈ E . So either (e, ±) = 0 for

all e, and so ± ∈ E or ± ∈ E . Since the roots span, they can™t all belong to

the same subspace. This contradicts the irreducibility. QED

Proposition 23 If there are two distinct root lengths in an irreducible root

system, then all roots of the same length are conjugate under the Weyl group.

Also, the maximal weight is long.

Proof. Suppose that ± and β have the same length. We can ¬nd a Weyl group

element W such that wβ is not orthogonal to ± by the preceding proposition.

So we may assume that β, ± = 0. Since ± and β have the same length, by the

table above we have β, ± = ±1. Replacing β by ’β = sβ β we may assume

that β, ± = 1. Then

= (sβ s± )(± ’ β)

(sβ s± sβ )(±)

= sβ (’± ’ β + ±)

= sβ (’β)

= β. QED

Let (E, ¦) and (E , ¦ ) be two root systems. We say that a linear map

f : E ’ E is an isomorphism from the root system (E, ¦) to the root system

(E , ¦ ) if f is a linear isomorphism of E onto E with f (¦) = ¦ and

f (β), f (±) = β, ±

for all ±, β ∈ ¦.

Theorem 14 Let ∆ = {±1 , . . . , ± } be a base of ¦. Suppose that (E , ¦ ) is a

second root system with base ∆ = {±1 , . . . , ± } and that

±i , ±j = ±i , ±j , ∀ 1 ¤ i, j ¤ .

Then the bijection

±i ’ ±i

6.7. THE CLASSIFICATION OF THE POSSIBLE SIMPLE LIE ALGEBRAS.109

extends to a unique isomorphism f : (E, ¦) ’ (E , ¦ ). In other words, the

Cartan matrix A of ∆ determines ¦ up to isomorphism. In particular, The

Dynkin diagrams characterize all possible irreducible root systems.

Proof. Since ∆ is a basis of E and ∆ is a basis of E , the map ±i ’ ±i

extends to a unique linear isomorphism of E onto E . The equality in the

theorem implies that for ±, β ∈ ∆ we have

sf (±) f (β) = f (β) ’ f (β), f (±) f (±) = f (s± β).

Since the Weyl groups are generated by these simple re¬‚ections, this implies

that the map

w ’ f —¦ w —¦ f ’1

is an isomorphism of W onto W . Every β ∈ ¦ is of the form w(±) where w ∈ W

and ± is a simple root. Thus

f (β) = f —¦ w —¦ f ’1 f (±) ∈ ¦

so f (¦) = ¦ . Since s± (β) = β ’ β, ± ±, the number β, ± is determined by the

re¬‚ection s± acting on β. But then the corresponding formula for ¦ together

with the fact that

sf (±) = f —¦ s± —¦ f ’1

implies that

f (β), f (±) = β, ± .

QED

6.7 The classi¬cation of the possible simple Lie

algebras.