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−.
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