On the Initial Subalgebra of a Graded Lie Algebra

(1)

http://dx.doi.org/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: [email protected]

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

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

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} trans-formation 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 th

p 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

1 0 1 , 1, 1

q r

G=G₋ ⊕ ⊕ G₋ ⊕G ⊕G ⊕ ⊕ G r> q>

(2)

M(G) = 0 [8]. Then G contains an irreducible, transitive depth-q graded subalgebra

(

)

( 1)

0 0

ad ,i

j

p p i i I

R v − G G

≤ ≤

=

∏

+

where v_i∈G₋_q, and where I is a non-negative whole number. We have G₋_q ⊆R, G−₁⊆R, and

(

ad 0

)

p

v R=

for all v∈G₋_q.

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

r>q.

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), G 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 G₀-submodule of G, then for any m∈M,

(

ad

)

j

p

m is a G₀-endomorphism of G for all j>0.

Proof. For any g∈G₀, m∈M, and x∈G, we have

[

] [

]

, , , 0

m m g m M

  ⊆ =

 

so that modulo p

(

)

[ ]

(

)

(

)

[

]

(

)

(

)

(

)

0

1

ad , ad , ad

, , ad , ad , ad .

j j

j

j j j

j

p p k k

k p

p p p

j

p

m g x m g m x

k

p m g m x g m x g m x

− ≤ ≤ −  _{ } _ = _ _{ } _           = _ _{ }+ _{ }≡ _      

∑



Lemma 2.If m∈G is such that

(

ad 0

)

₀

j

p

m G = for some j>0, then

(

ad

)

j

p

m is a G₀-endomorphism of G.

Proof. As in the proof of Lemma 1 above, we have, for any g∈G₀ and any x∈G, that modulo p,

(

)

[ ]

(

)

(

)

(

)

(

)

[ ]

(

)

(

)

0

ad , ad , ad ad , , ad

0, , ad , ad .

j j j j

j

j j

j

p p k k p p

k p

p p

p

m g x m g m x m g x g m x

k

x g m x g m x

− ≤ ≤  _{ } _{ } _{ } _ = _ _{ } _{ }≡ _{ }+ _             = +_ _{ }= _    

∑



Lemma 3.If v∈G₋_q, and j>0 is maximal such that

(

ad 0

)

j

p

v ≠ , then

(

ad

)

j

i

p p q

v G is a Lie subalgebra. Proof. Let x₁ and x₂ be any elements of G_{p q}j , Then for any v∈G−q,

(

ad , ad

)

1

(

)

2

(

ad ad ,

) (

)

1 2

j j j j

p p p p

v x v x v v x x

 ₌  

   

   

since, as we have seen in the proofs of the previous lemmas,

(

ad

)

j

p

v is a derivation, and

(

)

(

)

2

1 ad 0 j j p p q

v x ∈G₋ = . In addition, since

(

ad

)

₁ ₀

j

p

v x ∈G we have

(

ad ,

)

j ₁ ₂ j

p

p q

v x x G

 _{ ∈}

 

  . Hence,

(

ad ad ,

) (

)

1 2

(

ad

)

j j j

i

p p p

p q

v _ v x x  ∈_ v G

  , 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 v∈G₋_q is such that

(

ad 0

)

j

p k

v I = for some

integers j and k, with j≥0 and

(

j 1

)

p − q≤ ≤k r, then

(

ad 0

)

j

p m

v I = for all m, − ≤ ≤q m r, i.e.,

(

ad 0

)

j

p

v I= .

Proof. Suppose

(

ad 0

)

j

p k

(3)

[

1, 1

]

m m

I = G− I + )

(

ad 1

) (

ad

)

(

ad ad

) (

1

)

(

ad

)

0

j j j

m k p p m k p

m m k

G− − v I = v G− − I = v I =

so

(

ad 0

)

j

p m

v I = by transitivity. If m<k, then

(

ad

)

(

ad ad

) (

1

)

(

ad 1

) (

ad

)

0

j j j

p p k m k m p

m k k

v I = v G− − I = G− − v I = 

Lemma 5. If

(

ad 0v

)

pjGk = for some k such that j

p q− ≤ ≤q k r and for some j>0 , then

(

ad 0v

)

pjG= .

Proof. We will show that

(

ad 0v

)

pjGm = for all m, j

p q− ≤ ≤q m r. (If j

m< p q−q, then

(

ad 0

)

pj _m _n

n q

v G ⊆

∑

_<−G = .) First of all, suppose that p qj − < ≤q m r. Then, since

(

ad ad 0v

)

pj Ik ⊆

(

v

)

pjGk = , we have, by Lemma 4 that

(

ad 0

)

j

p

v I = . Consequently, we have

(

)

1

(

)

[

1

]

1

(

)

0 ad ad , , ad

j j j

p p p

m m m

v I ₋ v G₋ G G₋ v G 

= = _{= } _

 

so

(

ad 0v

)

pjGm = by the transitivity of G, if j

m≥ p q, or [9] (Lemma 6) otherwise. Finally, if m= p qj −q

and

(

ad 0v

)

pjGm ≠ , then by Lemma 1 (or Lemma 2),

(

ad

)

j

p m

v G is a non-zero G₀-submodule of G−_q. But

by, for example, [9] (Lemma 9), G−_q is irreducible as a G0module; therefore,

(

ad

)

j

p

m q

v G =G− , and we have

(

)

(

)

[

]

(

)

, ad ,pj ad ,pj ad 0pj

q m m

G₋ I  v G I v G I v I

  = = ⊆ =

 _{ }_ __

(by Lemma 4, as we noted earlier in the proof). But then, since G−₁⊂I, we would have

[

]

1 1 2 1 2

0=G−q,I ⊇G−q, G− ,G =G−,G−q,G 

so G−q,G₂ = 0 by, for example, [9] (Lemma 6), to contradict, for example, [9] (Lemma 8). Thus, we must

have

(

ad 0v

)

pjGm = in this case, also, so

(

ad 0

)

j

p

v G= as required. 

Lemma 6. If

(

ad 0

)

j

p

v G≠ for some v∈G₋_q and j>0, then both G_{p q}j and

(

ad

)

j

p p q

v G are non-

zero, and

(

ad

)

j

p q

G− ⊆ v G.

Proof. If

(

ad 0v

)

pjG≠ , then r≥ p qj −q, since otherwise we would have

(

ad 0

)

j

p

n n q

v G⊆

∑

_<−G = , con-

trary to hypothesis. By Lemma 5,

(

ad

)

j j

p p q q

v G ₋ is not zero, and by Lemma 1 (or Lemma 2),

(

ad

)

j j

p p q q

v G ₋

is a G₀-submodule of G₋_q; hence, by, for example, [9] (Lemma 9),

(

ad

)

j j

p

q p q q

v G ₋ =G₋

Since

(

ad v

)

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

(

)

(

)

(

)

0 , ad ,j j ad j j , ad .j j

p p p

q q _{p q q} q _{p q q} q _{p q}

G₋ G  v G ₋ G  v G ₋ G  v G

 

≠_ _{ }= _= _ _⊆

 

Thus, both _j

p q

G and

(

ad

)

j

p p q

v G are non-zero, and Lemma 6 is proved. 

Lemma 7. Let v be a non-zero element of G₋_q. If j>0 is maximal such that

(

ad 0

)

j

p

v G≠ , then

(

)

( 1) ad 0

j

p p

v − G≠ .

Proof. Suppose

(

ad 0

)

( 1)

j

p p

v − G= . Then for any x∈G_{p q}j , which is non-zero by Lemma 6, we have that

(

)

(

)

1

(

)

1

(

)

(

)

1

1 1

0 ad ad 1 ! ad ad .

j p j p

p p p

v x G p v x G

− −

−

− −

= = −

Thus

(

)

1

ad ad 0

j p

p

v x G

−

− = , so ad 1

(

ad

)

j

p

(4)

this nil set of endomorphisms is weakly closed, so by Jacobson’s theorem on nil weakly closed sets [10],

(

)

1 ad ad j j p

G− v Gp q acts nilpotently on G−1 and therefore annihilates some non-zero element of G−1 By Lemma 1 (or Lemma 2),

(

ad

)

j j

p p q

v G is a G₀-submodule of G₀ (i.e., an ideal of G₀). Hence, the annihi-

lator of

(

ad

)

j

p p q

v G in G₋₁ must be a G₀-submodule of G₋₁. By the assumed irreducibility of G, G₋₁

is irreducible as a G₀-module. Consequently,

( )

0 Ann ₁

(

ad

)

1

j j

p

G− v Gp q G−

≠ =

i.e., ₁, ad 0

(

)

j j

p p q

G₋ v G

 _{ =}

 

  , But then, we have by transitivity that

(

ad 0

)

j

p p q

v G = , so that, by Lemma 6

again

(

ad 0

)

j

p

v G= contrary to the choice of j. Thus,

(

ad

)

( 1)

j

p p

v − G must be non-zero, as asserted. 

Lemma 8.Let v be a non-zero element of G₋_q, and let j>0 be maximal such that

(

ad 0

)

j

p

v G≠ . Then

(

)

( 1) ad

j

p p

v − G is a Lie algebra, and we have that both

(

ad

)

( 1)

j

p p

q

G− ⊆ v − G and

(

)

( 1)

1 ad

j

p p

G− ⊆ v − G. Con-

sequently,

(

ad

)

( 1) ₀

j

p p

v − G+G is an irreducible, transitive, depth- q graded Lie algebra which is annihilated by

(

ad

)

j

p

v .

Proof. For j≥1 (since

(

) (

)

( )

(

)

1

ad ad ad 0

j j j

p p p p

v v G v G

+

−

= = , by the definition of j), we have

(

)

( )

₍

₎

( )

₍

₎

( )

(

₍

₎

)

( )

₍

₎

( )

(

)

( )

₍

₎

( )

1

1 1 1 1

1 1

ad ad ad , ad ad ,

ad , ad .

j j j j j

j j

p

p p p p p p p p p

p p p p

v G v v G G v v G G

v G v G

− − − − − − −     ⊇ _ _= _ _       = _ _

so _

(

ad , ad v

)

(p−1)pjG

(

v

)

(p−1)pjG ⊆_

(

ad v

)

(p−1)pjG

  ; i.e.

(

)

( 1) ad

j

p p

v − G is a Lie algebra whenever j≥1, its

closure under addition being obvious. Note that we must have _r≥_{q p}

(

−1

)

_pj−_q_{, since otherwise we would}

have

(

ad 0

)

( 1)

j

p p

n n q

v − G⊆

∑

_<−G = , to contradict Lemma 7.

By Lemma 6,

(

ad 0

)

j j

p p q

v G ≠ . By Lemma 1 (or Lemma 2),

(

ad

)

j j

p p q

v G is a non-zero ideal of G₀.

Thus, by transitivity and irreducibility, ₁, ad

(

)

₁ j

j

p p q

G₋ v G G₋

 _{ =}

 

  . Thus, we have

(

)

(

)

(

)

1 1, ad ad ,1 ad 1 1

j j j

p p p

p q p q p q

G₋ =_G₋ v G _= v _G₋ G _⊆ v G ₋ ⊆G₋

 

Consequently, we conclude that ₁

(

)

1 ad j j p p q

G₋ = v G ₋ , so ₁

(

ad

)

j

p

G− ⊆ v G. By Lemma 6,

(

ad

)

j

p q

G− ⊆ v G,

also. Thus,

(

ad

)

₀

j

p

v G+G is an irreducible, transitive depth-q graded Lie algebra. Since by Lemma 7,

(

)

( 1)

ad 0

j

p p

v − G≠ , it follows that

(

ad ad

) (

(

)

0

j j

p p

v v G ≠ , so we may repeat the argument just made to con-

clude that

(

ad

)

2

)

j

p

v G is an irreducible, transitive depth-q graded Lie algebra. Repeating the argument p−3

more times, we conclude the proof of Lemma 8. 

3. Proof of Main Theorem

Let j₁ be the maximum whole number such that

(

ad 0

)

1

j

p

v G≠ for some v∈G₋_q. Such a maximal j₁

must exist, since the height r of the finite-dimensional graded Lie algebra G is finite. If j₁=0, then we are done. Suppose then that j₁≥1. Let v₁ be an element of G₋_q such that

(

)

1

ad 0v j G≠ . Then by Lemma 8,

{ }₁ def

₍

₎

( ₁) 1

1 0

ad j

p p

G = v − G+G

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{ }2 def

₍

₎

( 1) 2

₍

₎

( 1) 1

2 1 0

ad ad

j j

p p p p

G = v − v − G+G

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

(

)

1 { }₁

₍

₎

1 { }₁

₍

₎

1 { }₁

₍

₎

1 { }₁

1 2 1 2 2

ad ad ad ad ad

j j j j

p p p p

v + v G = v G + v G = v G

Consequently, if j₁ = j₂, then

(

)

1 {1} 2

ad 0

j

p

v G ≠ , so v₂ is linearly independent of v₁. Since G₋_q, like

G is finite-dimensional, we can, by repeating this process, arrive at an integer t₁≤dimG₋_q such that

1 1

t

j = j ,

but

1 1

t m t

j + < j for any m>0 for which jt1+m is ultimately defined through the repetitive process we just

de-scribed. Then

(

)

1

(

)

( )

)

1

1 1

ad ad 0, 1

jk t ji

p p p

k i

i

v v − G k t

=

= ≤ ≤



Since the sequence

1

1, 2, , t

j j  j is non-increasing, the aforementioned commutativity of G₋_q entails that

(

)

₁ 1

(

)

( )

)

1 1

1

ad ad 0 j j t _p _pi

p

i i

G₋ v − G

=



.

If, in the above argument, we replace v₁ and j₁ with

11

t

v + and jt1+1, we eventually, by the finite

dimen-sionality of G₋_q, obtain a t₂ such that

2 11

t t

j = j + , but jt2+1< jt2. Continuing in this way, using vti+1 and

1, >1

i

t

j + i , in the above argument, we see that the series jti, 1, 2,i=  must eventually decrease to zero; i.e., we obtain a Lie algebra G{ }n such that

(

ad 0v

)

pG{ }n = for all v∈G₋_q, as required. 

Remark. Note that if we define def

0 0

t = , and

1

1, ,

i i i

j t t

V = v +  v₊ for i>0, and def

0 q

V =G₋ , then we get, for

depth q, something analogous to a flag in the sense of [1].

References

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[2] Premet, A. and Strade, H. (2006) Classification of Finite-Dimensional Simple Lie Algebras in Prime Characteristics. arXiv: math/0601380v2 [math. RA].

[3] Skryabin, S.M. (1992) New Series of Simple Lie Algebras of Characteristic 3. Russian Academy of Sciences Sbornik Mathematics, 70, 389-406. (English)

[4] Benkart, G.M., Kostrikin, A.I. and Kuznetsov, M.I. (1996) The Simple Graded Lie Algebras of Characteristic Three with Classical Reductive Component L0. Communications in Algebra, 24, 223-234.

http://dx.doi.org/10.1080/00927879608825563

[5] Benkart, G.M., Gregory, T.B. and Kuznetsov, M.I. (1998) On Graded Lie Algebras of Characteristic Three with Clas-sical Reductive Null Component. In: Ferrar, J.C. and Harada, K., Eds., The Monsteer and Lie Algebras, Vol. 7, Ohio State University Mathematical Research Publications, 149-164.

[6] Gregory, T.B. and Kuznetsov, M.I. (2004) On Depth-Three Graded Lie Algebras of Characteristic Three with Classical Reductive Null Component. Communications in Algebra, 32, 3339-3371. http://dx.doi.org/10.1081/AGB-120039401

[7] Gregory, T.B. and Kuznetsov, M.I. On Graded Lie Algebras of Characteristic Three with Classical Reductive Null Component. (In Preparation)

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Journal of Algebra, 53, 344-361. http://dx.doi.org/10.1016/0021-8693(78)90280-6

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