Received 10 July 2014; revised 11 August 2014; accepted 21 August 2014

ABSTRACT

We show that each irreducible, transitive finite-dimensional graded Lie algebra over a field of prime characteristic p contains an initial subalgebra in which the p^th power of the adjoint transformation associated with any element in the lowest gradation space is zero.

Keywords:

Prime-Characteristic Lie Algebras

1. Introduction

In the classification of the simple finite-dimensional Lie algebras over fields of prime characteristic, irreducible transitive finite dimensional graded Lie algebras play a fundamental role [1] . The simple finite dimensional Lie algebras over algebraically closed fields of characteristic greater than three have been classified [2] . Work is being done in characteristic three [3] - [7] . It is well known that in Lie algebras of Cartan type, there is a (not necessarily proper) subalgebra, the “initial piece,” which contains the sum of the negative gradations spaces of the Lie algebra, and in which the power of the adjoint representation associated with any element of the lowest gradation space is zero. In this paper, we prove that any irreducible, transitive finite-dimensional graded Lie algebra contains such an initial subalgebra. Indeed, we prove the following theorem.

2. Main Theorem

Let

be an irreducible, transitive, finite-dimensional graded Lie algebra over a field of characteristic such that M(G) = 0 [8] . Then contains an irreducible, transitive depth- graded subalgebra

where, and where I is a non-negative whole number. We have, , and for all.

If, then the conclusion of the theorem obviously holds. In what follows, therefore, we will assume that.

2. Intermediate Results

To prove the Main Theorem, we will make use of the following series of lemmas, in which we assume the hypotheses and notation of the Main Theorem. We note that by, for example, [9] (Lemma 6), is transitive in its negative part. (Note that the lemmas we quote from [9] are valid for all prime characteristics.) As usual, we assume throughout that M(G) = 0 [8] .

Lemma 1. If M is an abelian -submodule of G, then for any, is a -endomorphism of G for all.

Proof. For any, , and, we have

so that modulo

Lemma 2. If is such that for some, then is a -endomorphism of G.

Proof. As in the proof of Lemma 1 above, we have, for any and any, that modulo,

Lemma 3. If, and is maximal such that, then is a Lie subalgebra.

Proof. Let and be any elements of, Then for any,

since, as we have seen in the proofs of the previous lemmas, is a derivation, and

. In addition, since we have. Hence,

, which, as it is obviously closed under addition, is seen to be a Lie

subalgebra, as required.

Lemma 4. Let I be the minimal (graded) ideal of G [8] . If is such that for some

integers j and k, with and, then for all m, , i.e.,

.

Proof. Suppose. Then for all m, , we have (since for all, , we have)

so by transitivity. If, then

Lemma 5. If for some k such that and for some, then.

Proof. We will show that for all,. (If, then

.) First of all, suppose that. Then, since

, we have, by Lemma 4 that. Consequently, we have

so by the transitivity of, if, or [9] (Lemma 6) otherwise. Finally, if

and, then by Lemma 1 (or Lemma 2), is a non-zero -submodule of. But

by, for example, [9] (Lemma 9), is irreducible as a module; therefore, , and we have

(by Lemma 4, as we noted earlier in the proof). But then, since, we would have

so by, for example, [9] (Lemma 6), to contradict, for example, [9] (Lemma 8). Thus, we must

have in this case, also, so as required.

Lemma 6. If for some and, then both and are non-

zero, and.

Proof. If, then, since otherwise we would have, con-

trary to hypothesis. By Lemma 5, is not zero, and by Lemma 1 (or Lemma 2),

is a -submodule of; hence, by, for example, [9] (Lemma 9),

Since is a derivation of which annihilates, we have, by, for example, [9] (Lemma 8) that

Thus, both and are non-zero, and Lemma 6 is proved.

Lemma 7. Let be a non-zero element of. If is maximal such that, then.

Proof. Suppose. Then for any, which is non-zero by Lemma 6, we have that

Thus, so is a nil set of endomorphisms of. By Lemma 3, this nil set of endomorphisms is weakly closed, so by Jacobson’s theorem on nil weakly closed sets [10] ,

acts nilpotently on and therefore annihilates some non-zero element of By Lemma 1 (or Lemma 2), is a -submodule of (i.e., an ideal of). Hence, the annihi-

lator of in must be a -submodule of. By the assumed irreducibility of,

is irreducible as a -module. Consequently,

i.e., , But then, we have by transitivity that, so that, by Lemma 6

again contrary to the choice of. Thus, must be non-zero, as asserted.

Lemma 8. Let be a non-zero element of, and let be maximal such that. Then is a Lie algebra, and we have that both and. Con- sequently, is an irreducible, transitive, depth- graded Lie algebra which is annihilated by.

Proof. For (since, by the definition of), we have

so; i.e. is a Lie algebra whenever, its

closure under addition being obvious. Note that we must have, since otherwise we would

have, to contradict Lemma 7.

By Lemma 6,. By Lemma 1 (or Lemma 2), is a non-zero ideal of.

Thus, by transitivity and irreducibility,. Thus, we have

Consequently, we conclude that, so. By Lemma 6, ,

also. Thus, is an irreducible, transitive depth-q graded Lie algebra. Since by Lemma 7,

, it follows that, so we may repeat the argument just made to con-

clude that is an irreducible, transitive depth-q graded Lie algebra. Repeating the argument

more times, we conclude the proof of Lemma 8.

3. Proof of Main Theorem

Let be the maximum whole number such that for some. Such a maximal must exist, since the height of the finite-dimensional graded Lie algebra is finite. If, then we are done. Suppose then that. Let be an element of such that. Then by Lemma 8,

is an irreducible, transitive, finite-dimensional depth- graded Lie algebra to which we may apply Lemma 8 to obtain a and such that

is an irreducible, transitive, finite-dimensional depth-q graded Lie algebra to which we may apply Lemma 8 again. Since is abelian, it follows that

Consequently, if, then, so is linearly independent of. Since, like is finite-dimensional, we can, by repeating this process, arrive at an integer such that, but for any for which is ultimately defined through the repetitive process we just described. Then

Since the sequence is non-increasing, the aforementioned commutativity of entails that

.

If, in the above argument, we replace and with and, we eventually, by the finite dimensionality of, obtain a such that, but. Continuing in this way, using and, in the above argument, we see that the series must eventually decrease to zero; i.e., we obtain a Lie algebra such that for all, as required.

Remark. Note that if we define, and for, and, then we get, for depth, something analogous to a flag in the sense of [1] .

References