Algebra Carbondale 1980: Lie Algebras, Group Theory, and Partially Ordered Algebraic Structures Proceedings of the Southern Illinois Algebra Conference, Carbondale, April 18 and 19, 1980

290 Pages·1981·13.29 MB·English

by  Robert Lee Wilson (auth.)

A "NEW" PROOF OF KAPLANSKY'S THEOREM ON SIMPLE LIE ALGEBRAS OF RANK ONE
Robert Lee Wilson*
Rutgers University
New Brunswick, New Jersey 08903
In 1958 Kaplansky [14] proved the following theorem classifying
certain simple Lie algebras of rank one:
THEOREM A: Let F be an algebraically closed field of character-
istic p > 0 and L be a finite-dimensional simple Lie algebra over F.
Let H = Fh be a one-dimensional Cartan subalgebra of L. Assume that
all the characteristic roots of ad L h belong to the prime field. If
p = 2 then L has basis {h,a,b} with [ha] = a, [hb] = b, [ab] = h. If
p = 3 then L ~ sl(2) or psl(3), and if p > 3 then L ~ sl(2) or W(]:~)
(= Der(F[x]/(xP))).
As a special case he obtained the complete classification of the
restricted simple Lie algebras of rank one:
COROLLARY B! Let F be an algebraically closed field of character-
istic p > 0 and L be a finite-dimensional restricted simple Lie algebra
of rank one over F. Then p > .2 If p = 3 then ~ L sl(2) or psl(3), and
if p > 3 then L ~ sl(2) or W(1:1).
Recently, Block and the author [5,6] have done the rank two case:
THEOREM C: Let E be an algebraically closed field of characteristic
p > 7 and L be a finite-dimensional restricted simple Lie algebra of
Supported in part by NSF grants MCS77-03608 and MCS-803000
rank two over .F Then L is classical (of type A2, C2, or G )2 or
L ~ W(2:~) ( = Der(F[Xl,X2]/(xlP,x2P)).
Not surprisingly, Kaplansky's Theorem has proven to be quite im-
portant in the proofs of several recent classification results for
simple Lie algebras [4,5,6,23]. However~ the methods used in Kaplansky's
proof (which are quite computational) do not seem to generalize well to
higher rank situations. Many recent classification arguments rely in-
stead on filtration theoretic techniques (as developed in [10-13, 15-17,
19-22].
Accordingly, it seems interesting to give a proof of Kaplansky's
Theorem using filtration theoretic techniques. This should be of in-
terest both as an illustration (in the s~mplest case) of these tech-
niques and as a proof which is more conceptual (though less self-con-
tained) than Kaplansky's proof. We give such a proof in Section i.
(We emphasize that the ideas of this proof are not really new. Several
parts could be shortened by citing references to known results. In
particular, Sections 1.13-1.18 follow immediately from the known clas-
sification of simple Lie algebras with a subalgebra of codimension I
(see [1-3, 15]).) Our proof requires as background the Engel-Jacobson
Theorem (on nil weakly closed systems) [9] and some results on repre-
sentations of sl(2) (from [4,8]) and W(l:~) (from [7,18]).
In Section 2 we discuss briefly how these techniques can be ex-
tended to higher rank.
SECTION 1, PROOF OF THEOREH A
We will prove the result by induction on dim L.
I.i Let L be as in Theorem A. We will define a certain set of maximal
subalgebras of L.
Let
H = {H ~ L I H = Fh is a Cartan subalgebra of L and all
characteristic roots of ad L h lie in the prime
field},
K = {K a L I H~K for some H E H and K is a maximal subalge-
bra of L},
and
Kma x = {M E K I dim M = max{dim K ] K E K}}.
1.2 Let 0 E L Kma x. Let L_I be an (ad L0)-submodule of L containing 0 L
such that L_I/L 0 is an irreducible L0-module. Following Weisfeiler
[19,20] we define
I = L [Li+I,L_I] + Li+ I for i < -i,
and
Li+ I = {x i E L I [L_l,X] ~ L i} for i ! 0.
It is easily checked (see [20]) that [Li,L j] c Li+ j for all i and
j. The maximality of 0 L and the finite-dimensionality of L imply L =~k
for some k > i. The simplicity of L implies that L = (0) for some
-- r
r > i. Thus
L = L_k _~ ... D _~ 0 ~ L ... ~ Lr_ I D r L (0)
L 1 --
is a filtered Lie algebra. We let
i = G Li/Li+ I
SO
G = G_k + ... + G_I O + I + G + G ... + Gr_ I
is the associated graded algebra. Note that the action of O G on G_I is
faithful and irreducible.
Note that I L is an ideal of L 0. Thus if H I ~ # L (0) we have
H c I L so 0 I. = L L This implies that 0 i = L L for all i ~ 0, so 0 = L (0),
a contradiction. Thus (recalling that 0 ~ L H) we have that H injects
into Lo/L I = G 0. We identify H with its image in . O G
+~V-i
1.3 Let L = H i=l L.ia be the Cartan decomposition of L. Since each
Gj is a Go-module each Gj has a weight space decomposition with respect
to H:
Gj = ~ YEYj aj,y.
Clearly Gj,y = (Lj, T + Lj+I)/Lj+ I so
jr = {i~ 0 I i i i p-l, Lj,i~ _~ Lj+I}.
Since H c 0 L and H I N L = (0) this implies 0 E Fj if and only if j = 0.
1.4 LE[.~MA: O G contains no proper nonzero ideals.
Proof. Let I be an ideal in GO, and I = ~ XEF0 Ix be its decom-
position into weight spaces with respect to H. If 0 = I I NH # (0),
then H a I so . I O = G If 0 = I (0) and x EIy then (as
0 £ F_l)(ad x)PG i = (0). Thus by the Engel-Jacobson Theorem on nil
weakly closed systems [9, Chapter 2] we have JIG_I] # G_I. Thus by
the irreducibility of G_I , [IG_I] = (0). Since O G acts faithfully on
G_I we have I = (0).
1.5 LEMMA: One of the following occurs:
a) p ~ 2, O G has basis (h,a,b) with [ha] ~ a, [hb] = b, [ab] = h,
b) p ~ 3, O G ~ sl(2) or psl(3),
c) p > 3, O G ~ sl(2) or W(1:1),
d) dim 0 G = 1.
Proof. If dim O # G i, then O G is simple by Lemma 1.4. Since O G
satisfies the hypotheses of Theorem A and dim O < G dim L the result
follows from the induction assumption.
1.6 LEMMA: The cases p = 2, O G has basis (h,a,b) with [ha] = a,
[hb] = b, [ab] ~ h, and p = 3, O G ~ psl(3) are impossible.
Proof. Since O G acts faithfully on G_I , F_I must contain at
least two elements. As 0 ~ F_I this implies p ~ 2. Now assume p = 3.
For x E sl(3) we will let x- denote x + FIE psl(3). Then we may
assume that H = F(EII-E22)-. (For if x E sl(3) is in Jordan canonical
form and x- is not nilpotent then x- E F(EII - E22)-.) Define a E H*
by a((Ell - E22)-) = I. Since the roots of psl(3) are ~ (and the
characteristic roots of some 0 ~ h E H are all in the prime field) we
must have that F_I c {+~}. Since O G acts faithfully on G_I we must
have F_I = {+ ~}. But then (as 0 ~ F_I) GO, a = FE21 + FE32 + FEI3
annihilates G_I _ a. Since {E21 , E32 , El3} generates psl(3) this implies
that O G annihilates G_in_a, contradicting the fact that O G acts irredu-
cibly on G_I.
1.7 We will now recall some information about the restricted repre-
sentations of sl(2) and W(I:~) which we will need.
Let sl(2) have basis e,f,h where [he] ~ 2e, [hf] ~ -2f, and
[ef] ~ h. By [8] there are precisely p irreducible restricted sl(2)-
modules, one each of dimension i, ..., p. The irreducible module of
dimension m+l has basis {v0, ..., v m} where hv i = (m-2i)hn ev i spans
vi_ I for i ~ i ~ m, ev 0 = n 0 fv i spans vi+ I for 0 ! i < m, and fv m = O.
Using this one can prove [4] that if V and W are irreducible restricted
sl(2)-modules with dim V ~ dim W (mod 2) then (V ® W) 0 generates V~W
as an sl(2)-module.
Now W(I:~) = Der(F[x]/(xP)) has basis {x i ~x I 0 ! i S p-l}.
There is a unique restricted W(l:~)-module V for which 0 is not a
weight. This module is (F[x]/(xP))/FI. If H is any Cartan subalge-
bra of W(I:~) then for some root ~ we have that H + W(I:~)~ + W(l:~)
= sl(2). Consequently, (V ® V) 0 generates V ® V as a W(l:~)-module.
1.8 :AMMEL If dim O = G ,1 then G_2 = (0). If dim O G > I, then
G_2 = 2 = G (0).
Proof. Let dim O = G i. Then as G_I is an irreducible G0-module
dim G_I = i. Thus G_2 = [G_I,G_I] = (0).
Now let dim O G > i and H = Fh where ad h has all its eigenvalues
in the prime field. Since (ad h) P - ad h annihilates ~0 and hence
acts as a scalar on any irreducible G0-module , we have that (ad h) p -
- ad h acts as 0 on any irreducible constituent of any G.. Since
l
0 i ~ F for i # 0 we have that any root vector x E G0, , X Y ~ 0, has
p-th power 0 on any irreducible constituent of any G.. Thus each such
l
constituent is restricted. Now if i,J,i+j ~ 0 then [Gi,G j] is a homo-
morphic image of i ® G Gj and the kernel contains (G i ® Gj) 0. Since
0 ~ Fi, 0 ~ ?j we have that the dimension of every irreducible consti-
tuent of i G and of Gj is even if O ~ G sl(2) or is p-i if O ~ G W(I:~).
In either case we have (by the results cited in the previous section)
that (G i ® Gj) 0 generates i ® G Gj so [Gi,Gj] = (0). Thus G'_2 =
= [G_I,G_I] = (0). Also [G2,G_I] = (0) which implies 2 = G (0).
1.9 :AMMEL S~ppose O k G slY2) and I ~ G (0). Then p 3 and
=
L ~ psl(3).
Proof. As [G_I,GI] is a nonzero ideal in the simple algebra O G
there exist x E G_I, y I E G such that h = [xy]. Suppose dim G_I > 2.
Then (as dim G_I is even) dim G_I > dim O G so there exists 0 # z E G_I
such that [zy] = 0. Then 0 = [[xz]y] = [[xy]z] + [x[zy]] = [hz], a
contradiction (as 0 ~ F_I). Thus dim G_I = 2 and similarly dim I = G 2.
Now let {e,f,h} be a basis for O ~ G sl(2) as usual. Then we may choose
bases {ui, v i} for Gi, i = ~ I, such that [e,u i] = vi, [e,v i] = 0,
[f,ui] = 0, [f,vi] = ui, [h,u i] = -ui, [h,v i] i = v for i = ~ I. Since
h E [GI,G_I] we may assume without loss of generality that [Ul,V_l] = h.
Then [Vl,V_l] = [e,h] = -2e and so
0 = [[Ul,Vl]V_l] = [[Ul,V_l]V]] + [Ul[Vl,V_l]] = [h,v 1] - [Ul,2e] =
= I v + 2v I = 3v I. Thus p = 3.
Thus L = Fh + V + W where V = {x E L I (ad h - l)nx = 0 for some
n] and W = {x E L I (ad h + l)nx = 0 for some n}. Clearly V has basis
{a,b,c} where c + 0 L = V_l , b + I L = f, and a = v I. Since [f,v I] = Ul,
[f,v_l] = U_l , and [Vl,V_l] = -2e are all nonzero we see that if x E V
then Fx = {y E V I Ix,y] = 0}. Furthermore, [[vl,f]v_l] = -h implies
that [[a,b]c] = -h.
Now if r,s E V then [r[rs]] = ah and [s[rs]] = Bh. Then
[r[s[r,s]]] + [s[[r,s],r]] + [[r,s],[r,s]] = 0 gives [Br - as,h] = O.
Thus r and s are linearly dependent unless a = B = 0. Hence [r[rs]] = O.
But then (ad a) 3 annihilates V and so [a[a[a[b,c]]]] = 0. But this
gives [a[a,h]] = 0 and so [a~h] E Fa. Thus [a,h] = -a. Similarly,
[[c,a]b] = -h and (using the Jacobi identity) [[b,c]a] = -h, so
[b,h] = -b and [c,h] = -c. Thus ad h acts diagonally on V. Clearly,
a' = [b,c], b' = [c,a], and c' = [a,b] form a basis for W and so h
also acts diagonally on W. But now the multiplication in L is com-
pletely determined, for we know all products of three elements of V.
Thus there is at most one algebra L satisfying the hypotheses of the
lemma. Since psl(3) does satisfy these hypotheses, we are done.
(The above argument from the point where we proved p = 3 is es-
sentially the proof of Kaplansky's Theorem 9 of [14].)
i.i0 LEMMA: p > ,3 O ~ G W(I:~) and 1 ~ G (OJ is impossible.
Proof. Write D for. ~--~ Let O G have basis {xiD I 0 <_ i <_ p-l}
and G_I have basis {x I I I _< i ~ p-l} (where we identify an element
in F[x]/(x p) with its coset in (F[x]/(xP))/FI). Let W be an irredu-
cible submodule of G I. As 0 ~ 7 F we must have that dim W = p-i and W
contains an element w # 0 with [xD,w] = -w and [xiD,w] = 0 for i > i.
Then [xJ,w] E O G is an eigenvector for xD belonging to j-l, so
[xJ,w] = ajxJD. Now 0 = [[xJ,xk]w] = [xJ[xk,w]] - [xk[xJ~w]] =
= [xJ,akXkD] - [xk,ajxJD] = (kaj -Jak)xJ+k-l. In particular, setting
j = i gives ka I = k a for i < k _< p-l. Since [G_l,W] / (0) we must
have I a # O. Multiplying w by a nonzero scalar we may assume I a = I
and thus
[xJ,w] = jxJD for i < j < p-l.
But then, as 2 G = (0), 0 = [x[w[D,w]]] = [[x,w],[D,w]] + [w[x[D,w]]] =
= [xD[D,w]] + [w[[x,D]w]] + [w[Dfix,w]]] = -2[D,w] + [w[D,xD]] =
= -2[D,w] - [D,w] = -3[D,w]. Thus [D,w] = 0, so w spans a one-dimen-
sional submodule of W, so dim W = I, a contradiction.
I.ii LEMMA: If p > ,3 then O ~ G W(I:~).
Proof. "fI p > 3 and O ~ G W(l:~) then by Lemma i.i0 G = G_I 0. + O
Furthermore 9-i has p-i weights, each of multiplicity one. Then O G
has basis {xmD I 0 ! i ! p-l} (where D is as above) and G_I has basis
{xi I 1 ! i ! p-l] (with the same convention as above). Then L has
basis {v i I 1 < i < . . p-l] . U .{xiD I 0 < i < p-l} where i v = L satisfies
i
i 0 + v = L x . Then
[xD, Vp_ I] = -Vp_ 1 + hD.
Let E = (ad xD) p - (ad xD). Then L E : ÷ 0 L and 0 ÷ L E : (0). Then
E(Vp_ I) = -hD. Hence 0 = E([Vp_l,X2D )] = [-ID, x2D] = -21xD. Thus
= 0 and so Vp_ I is an eigenvector for xD. Now [Vp_l, x2D] E Fh and
so there exists 0 # w E (D, v__.) with [w, x2D] = 0.
Now consider 0' L = (w, x~,±x2D, ..., xP-ID). As w is an eigen-
vector for xD this is a subalgebra. AS dim 0' L = dim 0 L we have that
0' L E Kma x. Now 0' L defines a filtration ... L_I' m 0' ~ L I' ~ L ...
of L with associated graded algebra @' =~G i' As dim 0' L = dim 0 L
we have dim G_I' # 1 and so by Lem~ma 1.5 ' ~ O G sl(2) or W(I:~). In
particular, ' O G is simple. But 0' L is solvable, a contradiction.
1.12 LEMMA: If p ~ ,3 ~nd 1 ~ G (0), then O 7 G s~(2).
Proof. Assume O ~ G sl(2). Let {e,f,h) be the usual basis. As
0 ~ F , i F_I ~ 0. F Since each weight space for h contains a nonzero
eigenvector, there is an eigenvector for ad h in L and outside of L 0.
Since 0 L is spanned by eigenvectors for ad h and is maximal, L is
spanned by eigenvectors for h. Thus L has basis {e,f,h) U {v0,...,v m}
where 0 ~ m ~ p-l, [h,v i] = (m-2i)v i, [f,v m] 0 E L and [e,v 0] 0. E L
Suppose w ~ (e,h) is an eigenvector for ad h satisfylng [w,e] = 0.
Then L 0' = (w,e,h) is a solvable subalgebra of L. (Recall that = I L (0)
by hypothesis) But as dlm L 0' = dim 0 L we have that 0' L E Kma x. If
• .. ~ L_I ! ~ 0' ~ L L I' ~ ... is the corresponding filtration and
G' =~G i' the associated graded algebra then dim @-i' # I (as dim ~i = '
= dim L/L O' = dim L/L 0) and so (by Lemma 1.5), ' ~ O G sl(2) or W(I:~)
contradicting the solvability of 0' L This implies that [e,v 0] # 0.
But [h[e,v0] ] = (m+2)[e,v0] 0 E L so (m+2) = p-2 or 0 (as m # 0). If
m = p-2 then there is a nonzero linear combination w of f and 0 v satis-
fying [w,e] = 0. Thus we must have p > 3, m = p-4, [e,v0] # 0, and
(by symmetry) [f,Vp_ 4] # 0. Thus (ad e)P-lvp_4 = le for some i # 0
and (ad e) p = (ad f)P = 0.
Now (following Winter [24]) we define E = exp(ad De) =
=~ p-i
i=0 (ad ~e)I/i!. Then if [h,u] = ju we have
[Eh,Eu] = [h-2 De,~ p-I (ad ~e)iu/i '] =
i=0
=~p-I (j+2i)(ad pe)iu/i!) -
i=0
- 2~i= p--2 0 ((i+l)(ad ~e)l+lu/(i+l)!) ° = jEu.
Also [Ee,Ef] =~p-I ((ad ~e)i[ef]/i )' = Eh. Thus M=(Ee,Ef,Eh)
i=0
is a subalgebra containing the Cartan subalgebra Eh and Eh E H. Thus
M E Kma x and by our work above it is enough to show [Ef,Ev ].j = 0.
~-i p-4
But [Ef, EVp_4] = E[f, Vp_4] - [(ad De)2f, (ad ~e) ~ Vp_4]/2 +
+ [(ad De)2f, (ad De)P-2vp_4]/2 - [(ad De)f, (ad De)P-lvp_4 ] :
[f,Vp_ 4] - D(ad De)P-lvp_4 - p 2 (ad ze)P-lvp_ 4 =
[f,Vp - ] - 4 D 3 P(ad e)P-lv p-4" As (ad e )P-lvp_ 4 # 0, we may choose D
so that [Ef, EVp_4] = 0, completing the proof.
1.13 COROLLARY: G2 = (0) and we may assume dim G _ 1 < . for all i.
Proof. G_2 = (0) by Lemma 1.8 and by Lemmas 1.5 - 1.12 we may
assume that dim O = G i. As O G acts irreducibly on G_I , dim G_l = I.
If G i = Fy then ad y is an injection of i G into Gi_ I for all i ~ 0,
giving the result.
1.14 LEMMA: There exists an element h E H such that ad h I Gi = iI
for all i.
Proof. As O G acts faithfully on G_I we may assume ad hI@_l
Suppose ad h I Gj = jl, x E G j+ , I and G_ I = Fy. Then [y[h,x]] =
= [[y,h]x] + [h[y,x]] = [y,x] + j[y,x] = (j+l)[y,x]. Thus
[y, (j+l)x - [h,x]] = 0 and so [h,x] = (j+l)x, as required.
1.15 COROLLARY: G = (0).
p-I
Proof. As 0 ~ rp, Lemma 1.14 implies Gp = (0). If 0 # x E Gp_ I
and G_I = Fy, then y E F(ad y)Px. Thus 0 = (ad y)P[x,(ad y)x] =
= [(ad y)Px, (ad y)x] is a nonzero scalar multiple of (ad y)2x, a con-
tradiction.
1.16 LEMMA: There exist elements b £ L and a £ 1 L such that
[a,b] = h. Thus M = (a,b,h) ~ sZ(2) if p > .2
Proof. As L is not solvable dim L > 2 and so I # L (0). Thus we
may find a and b, eigenvectors for ad h with L = Fb 0 + L and I = L
= Fa 2 + L such that [a,b] = h.
1.17 LEMMA: If M ~ L then Lp_ 2 ~ (0) and dim L = p ~ .3
Proof. If M # L then L/M is an M-module. It has one-dimensional
lowest weight space complimentary to 3 + L M/M. Since h has eigenvalue
2 on this space, p # 2. Thus M ~ sl(2) and the sl(2)-module L/M has
highest weight p-4 and so dimension P-3-
1.18 LEMMA: If M ~ L then L ~ W(I:~).
Proof. Write 0 = v b, I = v h, and 2 = v -2a. By Lemmas 1.14 and
1.17 we may find i £ v Li_l, i i ~ v L for 3 ~ i ~ p-I such that
[v0,v i] = ivi_ I and [h,v i] = (i-l)v i for 3 ~ i ~ p-l. We may set
i = v 0 for i ~ p and st111 have [v0,vi] = ivi_ I and [h,v i] = (i-l)v i
for all i. We prove, by induction on i+j, that [vi,v j] = (j-i)vi+j_ I
for all i and j > 0. The result is clearly true if i+j < 2. Assume
it is true for i+j < t and let i+j = t. Then
[v0Evi,vj]] = [[v0,vi]vj] + [vi[v0,vj]] =
= i[vi_l,V j] + J[vi,vj_ I] =
= (i(J-i+1) + j(j-l-i))vi+j_ 2 =
= (j-i)(i+j-l)vi+j_ 2.
Thus [v0,[vi,vj] - (j-i)vi+j_l] = .0 As [vi,v j] - (j-i)vi+j_ E I Li+j_ 2
I L this implies [vi,vj] = (j-i)vi+j_l, as required. Then the linear
map of L to W(I:~) defined by i ~ v xID is an isomorphism.
This completes the proof of Theorem A.
SECTION ,2 HIGHER RANK
We will indicate here the difficulties that arise when one attempts
to extend this proof to higher rank.
First assume that L is restricted simple of rank two over an al-
gebraically closed field of characteristic p > 7. Then H contains a
10
maximal torus T and dim T = i or 2. If dim T = I then L has been de-
termined in [23] by techniques almost exactly like those above. (The
only additional difficulty is that one must choose 0 L so that
H A I # L (0). This takes a little extra work.) The result in this
case is that there are no such algebras, i.e.~ simplicity implies that
a two-dimensional Cartan subalgebra is a torus. Hence from now on we
know that H is a torus.
We take 0 L to be a maximal subalgebra containing H and define a
filtration as in 1.2. The first difficulty comes in 1.4. It is no
longer true that any proper ideal of O G is nil. (The proof of that
fact depended on the observation that a string of p weights had to con-
tain 0, which is valid only in rank one.) One can prove easily that
either O G is semisimple or that O G contains a one-dimensional center
Fz and G0/solv(G 0) is simple (hence sl(2) or W(I:~)). The semisimple
algebras which can occur can be determined by induction and use of
Block's characterization of the semisimple algebras in terms of simple
algebras.
To handle the large collection of possible O G it is necessary
to make a more careful choice of 0 L (and in fact even to assume that
H is of a certain form). This is done in [6]. The proof is quite
long.
It seems likely that the rank two result will allow one to clas-
sify the restricted simple Lie algebras of arbitrary rank which contain
f ~
toral Cartan subalgebras. The basic idea is to define L <~'B) =
= ~i,j Li~+j~" Then L(~'B)/solv(L (~B)) will be known for each pair
of roots ~ and ~ and this should determine the structure of the entire
algebra. (Note that this is exactly what is done in the characteristic
zero classification.)
In order to classify arbitrary restricted simple Lie algebras it
seems that one should first classify the ones of toral rank two, i.e.,
those which contain a Cartan subalgebra H of arbitrary dimension which
contains a maximal torus of dimension 2.
References
[i] R. K. Amayo, "Quasi-ideals of Lie algebras i~ Prec. London Math.
Soc. (3) 33 (1976), 28-36.
[2] , "Quasi-ldeals of Lie algebras II", Prec. London Math.
See. (3) 3~3 (1976), 37-64.
[3] G. M. Benkart, I.M. Isaacs, and J.M. Osborn, "Lie algebras with
self-centralizing ad-nilpotent elements f' , J. Algebra 57 (1979),
279"309.