Advances in Pure Mathematics
Vol.04 No.09(2014), Article ID:49912,4 pages
10.4236/apm.2014.49058
On the Initial Subalgebra of a Graded Lie Algebra
Thomas B. Gregory
Department of Mathematics, The Ohio State University at Mansfield, Mansfield, Ohio, USA
Email: gregory.6@osu.edu
Copyright © 2014 by author and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/
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 pth 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] .
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