Thus if there are no branch points, there must be at least one double bond

and at least two vertices on either side of the double bond. The graph with

(1)

exactly two vertices on either side is strictly contained in F4 and so is excluded.

So there must be at least three vertices on one side and two on the other of the

(1) (2)

double bond. But then F4 and E6 exhaust the possibilities for one double

bond and no branch points.

If there is a double bond and a branch point then either the double bond

(2) (1)

points toward the branch, as in A2 ’1 or away from the branch as in B . But

then these exhaust the possibilities for a diagram containing both a double bond

and a branch point.

(1)

If there are two branch points, the diagram must contain D and hence

(1)

must coincide with D .

So we are left with the task of analyzing the possibilities for diagrams with

no double bonds and a single branch point. Let m denote the minimum number

of vertices on some leg of a branch (excluding the branch point itself). If m ≥ 2,

(1) (1)

then the diagram contains E6 and hence must coincide with E6 . So we may

assume that m = 1. If two branches have only one vertex emanating, then the

(1)

diagram is strictly contained in D and hence excluded. So each of the two

other legs have at least two or more vertices. If both legs have more than two

(1)

vertices on them, the graph must contain, and hence coincide with E7 . We

are left with the sole possibility that one of the legs emanating from the branch

point has one vertex and a second leg has two vertices. But then either the

(1) (1)

graph contains or is contained in E8 so E8 is the only such possibility.

We have completed the proof that the diagrams listed in A¬ 1, A¬ 2 and

A¬ 3 are the only diagrams without loops with maximum eigenvalue 2.

If we allow loops, an easy extension of the above argument shows that the

only new diagrams are the ones in the table “Loops allowed”.

6.5 Classi¬cation of the irreducible ∆.

Notice that if we remove a vertex labeled 1 (and the bonds emanating from it)

from any of the diagrams in A¬ 2 or A¬ 3 we obtain a diagram which can also

be obtained by removing a vertex labeled 1 from one of the diagrams inA¬ 1.

(In the diagram so obtained we ignore the remaining labels.) Indeed, removing

(3) (1)

the right hand vertex labeled 1 from D4 yields A2 which is obtained from A2

by removing a vertex. Removing the left vertex marked 1 gives G2 , the diagram

(1)

obtained from G2 by removing the vertex marked 1.

(2)

Removing a vertex from A2 gives A1 . Removing the vertex labeled 1 from

(2) (1)

A2 yields B2 , obtained by removing one of the vertices labeled 1 from B .

6.6. CLASSIFICATION OF THE IRREDUCIBLE ROOT SYSTEMS. 105

(2)

Removing a vertex labeled 1 from A2 yields D2 or C2 , removing a vertex

’1

(2) (2)

labeled 1 from D +1 yields B +1 and removing a vertex labeled 1 from E6

yields F4 or C4 .

Thus all irreducible ∆ correspond to graphs obtained by removing a vertex

labeled 1 from the tableA¬ 1. So we have classi¬ed all possible Dynkin diagrams

of all irreducible ∆ . They are given in the table labeled Dynkin diagrams.

6.6 Classi¬cation of the irreducible root systems.

It is useful to introduce here some notation due to Bourbaki: A subset ¦ of a

Euclidean space E is called a root system if the following axioms hold:

• ¦ is ¬nite, spans E and does not contain 0.

• If ± ∈ ¦ then the only multiples of ± which are in ¦ are ±±.

• If ± ∈ ¦ then the re¬‚ection s± in the hyperplane orthogonal to ± sends ¦

into itself.

• If ±, β ∈ ¦ then β, ± ∈ Z,

Recall that

2(β, ±)

β, ± :=

(±, ±)

so that the re¬‚ection s± is given by

s± (β) = β ’ β, ± ±.

We have shown that each semi-simple Lie algebra gives rise to a root system, and

derived properties of the root system. If we go back to the various arguments,

we will ¬nd that most of them apply to a “general” root system according to the

above de¬nition. The one place where we used Lie algebra arguments directly,

was in showing that if β = ±± is a root then the collection of j such that β + j±

is a root forms an unbroken chain going from ’r to q where r ’ q = β, ± .

For this we used the representation theory of sl(2). So we now pause to give

an alternative proof of this fact based solely on that axioms above, and in the

process derive some additional useful information about roots.

For any two non-zero vectors ± and β in E, the cosine of the angle between

them is given by

± β cos θ = (±, β).

So

β

β, ± = 2 cos θ.

±

Interchanging the role of ± and β and multiplying gives

β, ± ±, β = 4 cos2 θ.

106 CHAPTER 6. THE SIMPLE FINITE DIMENSIONAL ALGEBRAS.

.........

• • • ≥1

A,

......

• • • ≥2

>• B

......

• • • • ≥2

< C

¨•

......

• • •¨ ≥4

D

r

r•

• >• G2

• • >• • F4

• • • • •

E6

•