so L1 , . . . , Ln is the basis of h— dual to the basis p1 q1 , . . . , pn qn of h.

If h = a1 p1 q1 + · · · + an pn qn then

[h, q i q j ] = (ai + aj )q i q j

[h, q i pj ] = (ai ’ aj )q i pj

[h, pi pj ] = ’(ai + aj )pi pj

54 CHAPTER 3. THE CLASSICAL SIMPLE ALGEBRAS.

so the roots are

±(Li + Lj ) all i, j and Li ’ Lj i = j.

We can divide the roots ¦ into positive and negative roots by setting

¦+ = {Li + Lj }all ∪ {Li ’ Lj }i<j .

ij

If we set

±1 := L1 ’ L2 , . . . , ±n’1 := Ln’1 ’ Ln , ±n := 2Ln

then every positive root is a sum of the ±i . Indeed, Ln’1 + Ln = ±n’1 + ±n

and 2Ln’1 = 2±n’1 + ±n and so on. In particular 2±n’1 + ±n is a root.

If we set

h1 := p1 q1 ’ p2 q2 , . . . , hn’1 := pn’1 qn’1 ’ pn qn , hn := pn qn

then

±i (hi ) = 2

while for i = j

= ’1, i = 1, . . . , n ’ 1

±i (hi±1 )

= 0, j = i ± 1, i = 1, . . . , n

±i (hj ) (3.12)

= ’2.

±n (hn’1 )

In particular, the elements hi , qi pi+1 , qi+1 pi for i = 1, . . . , n’1 form a subalgebra

isomorphic to sl(2) as do the elements hn , 1 qn , ’ 1 p2 . We call these subalgebras

2

2n

2

sl(2)i , i = 1, . . . , n.

Dn = o(2n), n ≥ 3.

3.5.3

We choose a basis u1 , . . . , un , v1 , . . . , vn of our orthogonal vector space V such

that

(ui , uj ) = (vi , vj ) = 0, ∀ i, j, (ui , vj ) = δij .

We let h be the subalgebra of o(V ) spanned by the Aui vi , i = 1, . . . , n. Here we

have written Axy instead of Ax§y in order to save space. We take

Au1 v1 , . . . , Aun vn

as a basis of h and let L1 , . . . , Ln be the dual basis. Then

±Lk ± L k =

are the roots since from (3.7) we have

[Aui vi , Auk u ] = (δik + δi )Auk u

[Aui vi , Auk v ] = (δik ’ δi )Auk v

[Aui vi , Avk v ] = ’(δik + δi )Avk v .

3.5. THE ROOT STRUCTURES. 55

We can choose as positive roots the

Lk + L , Lk ’ L , k <

and set

±i := Li ’ Li+1 , i = 1, . . . , n ’ 1, ±n := Ln’1 + Ln .

Every positive root is a sum of these simple roots. If we set

hi := Aui vi ’ Aui+1 vi+1 , i = 1, . . . n ’ 1,

and

hn = Aun’1 vn’1 + Aun vn

then

±i (hi ) = 2

and for i = j

0 j = i ± 1, i = 1, . . . n ’ 2

±i (hj ) =

’1 i = 1, . . . , n ’ 2

±i (hi±1 ) =

’1

±n’1 (hn’2 ) = (3.13)

’1

±n (hn’2 ) =

±n (hn’1 ) = 0.

For i = 1, . . . , n ’ 1 the elements hi , Aui vi+1 , Aui+1 vi form a subalgebra isomor-

phic to sl(2) as do hn , Aun’1 un , Avn’1 vn .

Bn = o(2n + 1) n ≥ 2.

3.5.4

We choose a basis u1 , . . . , un , v1 , . . . , vn , x of our orthogonal vector space V such

that

(ui , uj ) = (vi , vj ) = 0, ∀ i, j, (ui , vj ) = δij ,

and

(x, ui ) = (x, vi ) = 0 ∀ i, (x, x) = 1.

As in the even dimensional case we let h be the subalgebra of o(V ) spanned by

the Aui vi , i = 1, . . . , n and take

Au1 v1 , . . . , Aun vn

as a basis of h and let L1 , . . . , Ln be the dual basis. Then

±Li ± Lj i = j, ±Li

are roots. We take

Li ± Lj , 1 ¤ i < j ¤ n, together with Li , i = 1, . . . , n

56 CHAPTER 3. THE CLASSICAL SIMPLE ALGEBRAS.

to be the positive roots, and

±i := Li ’ Li+1 , i = 1, . . . , n ’ 1, ±n := Ln

to be the simple roots. We let

hi := Aui vi ’ Aui+1 vi+1 , i = 1, . . . n ’ 1,

as in the even case, but set

hn := 2Aun vn .

Then every positive root can be written as a sum of the simple roots,

±i (hi ) = 2, i = 1, . . . n,

and for i = j

±i (hj ) = 0 j = i ± 1, i = 1, . . . n

±i (hi±1 ) = ’1 i = 1, . . . , n ’ 2, n (3.14)

±n’1 (hn ) = ’2

Notice that in this case ±n’1 + 2±n = Ln’1 + Ln is a root. Finally we can

construct subalgebras isomorphic to sl(2), with the ¬rst n ’ 1 as in the even

orthogonal case and the last sl(2) spanned by hn , Aun x , ’Avn x .

3.5.5 Diagrammatic presentation.

The information of the last four subsections can be summarized in each of the

following four diagrams:

The way to read this diagram is as follows: each node in the diagram stands

for a simple root, reading from left to right, starting with ±1 at the left. (In the

diagram D the two rightmost nodes are ± ’1 and ± , say the top ± ’1 and the

bottom ± .) Two nodes ±i and ±j are connected by (one or more) edges if and

only if ±i (hj ) = 0.

In all cases, the di¬erence, ±i ’ ±j is never a root, and, for i = j, ±i (hj ) ¤ 0

and is an integer. If, for i = j, ±i (hj ) < 0 then ±i + ±j is a root.

In two of the cases (B and C ) it happens that ±i (hj ) = ’2. Then ±i + ±j