4 1. GRADED LIE ALGEBRAS

Proposition 1.10. Assume g =

r

i=−q

gi is an irreducible, transitive

graded Lie algebra such that g−i

= [g−1, g−i+1] for all i ≥ 2. If M(g) = 0,

then g contains no abelian ideals.

Proof. By Proposition 1.8 (iii), any nonzero ideal J of g must contain

g− and hence must contain g−1 ⊕ [g−1, g1]. By transitivity (1.2), [J, J] ⊇

[g−1, [g−1, g1]] = 0. Consequently, J cannot be abelian.

Corollary 1.11. If g =

r

i=−q

gi is a irreducible, transitive graded Lie

algebra such that g−i = [g−1, g−i+1] for all i ≥ 2, then g/M(g) is semisimple.

Proof. This is an immediate consequence of Proposition 1.10 and the fact

that M(g) = 0 for g = g/M(g) by (iii) of Proposition 1.8.

Lemma 1.12. Assume g =

r

i=−q

gi is a graded Lie algebra such that

g+ is generated by g1 and M(g) = 0. If x ∈ g− and [x, g1] = 0, then x = 0.

Proof. The hypotheses imply that [x, g+] = 0, so that x ∈

M1(g)

= 0.

1.3. The minimal ideal I

In this section we show that graded Lie algebras satisfying certain condi-

tions must contain a unique minimal ideal I which is graded, and we derive

some properties of I.

Proposition 1.13. (Compare [W], Prop. 1.61, Cor. 1.62, Cor. 1.66.)

Assume g =

r

i=−q

gi is an irreducible, transitive graded Lie algebra such

that g−i = [g−1, g−i+1] for all i ≥ 2. If M(g) = 0, then g has a unique

minimal ideal I which is graded and contains g−.

Proof. Since M(g) = 0, every nonzero ideal must contain g− by Proposition

1.8 (iii). Therefore the intersection I of all the ideals is the unique minimal

ideal of g, and it contains g−. Let Ij := I ∩ gj for each j, and observe

that I−i = g−i for all i ≥ 1. The space I =

j

Ij is a nonzero ideal of g,

since [gi, Ij] ⊆ gi+j ∩ I = Ii+j. By the minimality of I, we conclude that

I =

j

Ij.

Corollary 1.14. Assume g =

r

i=−q

gi is an irreducible, transitive

graded Lie algebra such that g−i = [g−1, g−i+1] for all i ≥ 2. Then g =

g/M(g) has a unique minimal ideal which is graded and contains g−.