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Lie algebras with almost dimensionally nilpotent inner derivations
Linear Algebra and its Applications 320 (2000) 97–114 www.elsevier.com/locate/laa
Lie algebras with almost dimensionally nilpotent inner derivations聻 Jianhua Zhou Department of Applied Mathematics, Southeast University, Nanjing 210096, People’s Republic of China Received 8 October 1999; accepted 19 June 2000 Submitted by H. Schneider
Abstract The structure of the Lie algebras with almost dimensionally nilpotent inner derivations is studied. It is proved that, if the base field is of characteristic 0, then, when d > 6 is odd, there exist just two d-dimensional Lie algebras; when d > 6 is even, there exists just one ddimensional Lie algebra such that these Lie algebras are nonsolvable and have some almost dimensionally nilpotent inner derivations. © 2000 Elsevier Science Inc. All rights reserved. AMS classification: 17B30 Keywords: Almost dimensionally nilpotent derivation; Descending basis; Self-centralizing ad-nilpotent element
1. Introduction In [3,4,6], Ding, Eberlin and Leger have studied independently the Lie algebras having nilpotent derivations of degree their dimension. Such derivations are called dimensionally nilpotent, the Lie algebras are called dimensionally nilpotent ones, although they are not necessarily nilpotent as a Lie algebra. They have studied the structure of the Lie algebras when the characteristic of the base field is 0. The results show that apart from sl2 , such Lie algebras are all solvable. Ding has proved that
聻
Project supported by the State Education Commission, National Natural Science Foundation and Southeast University. E-mait address: [email protected] (J. Zhou). 0024-3795/00/$ - see front matter 2000 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 0 ) 0 0 2 0 5 - 6
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such a Lie algebra is nilpotent if and only if its center is not trivial. Furthermore, he has classified all such non-nilpotent solvable Lie algebras. If such a derivation is inner, then the corresponding element of the Lie algebra is called self-centralizing ad-nilpotent (s.c.a.n.), the Lie algebras with s.c.a.n. elements are also called self-centralizing ad-nilpotent (s.c.a.n. in the following). In [1,2], Benkart et al. have shown that, if the characteristic of the base field is prime, then the s.c.a.n. Lie algebras are all so-called generalized Zassenhaus Lie algebras; if the characteristic is 0, then the dimension of s.c.a.n. Lie algebras 6 3. Now let L be a finite-dimensional Lie algebra over a field F . It is natural to set the question that: what would happen if L has a nilpotent derivation of degree dim L − 1? Definition 1. Let L be a finite-dimensional Lie algebra over a field F, D a derivation of L. D is called almost dimensionally nilpotent if the degree of D is dim L − 1, i.e. / 0. D dim L−1 = 0 but D dim L−2 = The aim of this paper is to study the structure of Lie algebras with almost dimensionally nilpotent inner derivations in the case char F = 0. Unless otherwise stated, in this paper, F is a field of characteristic 0, L is a Lie algebra over F, dim L = n + 1, x ∈ L is a fixed ad-nilpotent element such that ad x is almost dimensionally nilpotent in L, i.e. the degree of ad x is n. For convenience, we denote by r(u) the rank of ad u for u ∈ L. It is clear that, if ad x is almost dimensionally nilpotent, then r(x) = dim L − 2.
2. Preliminaries Lemma 2. If L can be decomposed into direct sum of proper ideals, then L is isomorphic to one of the following: (i) the two-dimensional abelian Lie algebra; (ii) the direct sum of the two-dimensional non-abelian ideal and the center of dimension 1; (iii) the four-dimensional non-abelian reducible Lie algebra. Proof. Suppose that L = I1 ⊕ I2 ⊕ · · · ⊕ Ik . P P Let x = xi , xi ∈ Ii . Then, every ad xi |Ii is nilpotent and r(x) = ki=1 r(ad xi |Ii ). Since r(x) = dim L − 2 and r(ad xi |Ii ) 6 dim Ii − 1, it follows that k = 2 and r(ad xi |Ii ) = dim Ii − 1. So, xi is s.c.a.n. in Ii . It follows from the result of [1,3] that either dim Ii 6 2 or Ii ∼ = sl2 . Since ad x is almost dimensionally nilpotent, neither I1 nor I2 is two-dimensional abelian. Thus, the lemma follows.
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Let G be a semisimple Lie algebra over an algebraically closed field of characteristic 0. If the rank, i.e. the dimension of the Cartan subalgebra of G is l, then for every g ∈ G, the dimension of the centralizer of g dim CG (g) > l. If g is ad-semisimple and dim CG (g) = l, then CG (g) must be a Cartan subalgebra of G. We have the following lemma. Lemma 3. L is not semisimple. Proof. Without loss of generality, F may be assumed to be algebraically closed. Suppose that L is semisimple. Obviously, dim CL (x) = 2. So, the rank of L 6 2. Since the ad-nilpotent degree of every ad-nilpotent element in a semisimple Lie algebra > 3, so, dim L = / 3. By Lemma 2, L can only be one of the simple Lie algebras of type A2 , B2 and G2 . Since L is simple and x is ad-nilpotent, there exist h, y ∈ L such that x, h, y is a canonical basis of a three-dimensional simple Lie subalgebra, i.e. [x y] = h, [h x] = 2x and [h y] = −2y, where h is ad-semisimple and y is ad-nilpotent in L (cf. [5]). Put T = hx, y, hi (ha, b, . . .i denotes the subspace spanned by a, b, . . .). Let R be the representation of T on L by the adjoint action. L can be decomposed into direct sum of irreducible T-submodules: L = M1 ⊕ M2 ⊕ · · · ⊕ Mm . By the representation theory of sl2 , for each i, dim Ker(x R |Mi ) = 1. Since dim CL (x) = 2, it follows that m = 2. It is clear that the degree of x R 6 max{dim Mi }. Since ad x is almost dimensionally nilpotent, dim Mi > n. So the set {dim M1 , dim M2 } = {n, 1}. Without loss of generality, one can assume that dim M1 = n and dim M2 = 1. Suppose that M2 = F u for some u ∈ L. It is clear that T R (u) = [u T ] = 0. By the representation theory of sl2 , the dimension of the kernel of the restriction of hR on M1 is 1 (It cannot be 0. For otherwise, dim CL (h) = 1, which is a contradiction.) Hence dim CL (h) = 2. So, h is regular in L, i.e. CL (h) is a Cartan subalgebra of L. It is not difficult to see that x is a root vector of L with respect to CL (h). But the length of the string of roots in the root system of a semisimple Lie algebra with respect to its Cartan subalgebra 6 4, and so, (ad x)5 = 0. But the dimensions of the simple Lie algebras of type A2 , B2 , G2 are not less than 8. That is to say, L cannot be the simple Lie algebra of type A2 , B2 or G2 . The lemma follows from this contradiction. We will need some results picked from [3]. Suppose that G is a dimensionally nilpotent Lie algebra over F , D a dimensionally nilpotent derivation of G, x1 , x2 , . . . , xm is a cyclic basis, i.e. D(x1 ) = 0 and D(xi ) = xi−1 for i > 1. By [3], if m = 3 and D is a inner derivation of L, then G ∼ = sl2 (F ). If m > 3, then G is solvable. Suppose that m > 3. Then for 1 6 i 6 m, hx1 , x2 , . . . , xi i is an ideal of G. Suppose that m X j [xi , xj ] = ai k xk . k=1
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Then j
ai k = 0
for i < m + k − j.
m m G is nilpotent if and only if am−1 m−1 = 0. If it is the case, then ai i = 0 for all j
i, 1 6 i 6 m. Moreover, if i 6 m + k − j , then ai k = 0. By the Jordan canonical form of ad x, we know that L has a basis x0 , x1 , . . . , xn such that [x0 x] = [x1 x] = 0 while [xi x] = xi−1 for 2 6 i 6 n. We call it a descending basis of L with respect to x. It is not difficult to see that we have only to study the Lie algebras in two cases: Case I: x = x0 ;
Case II: x = x1 .
If Case II happens, i.e. x = x1 , then we have the following lemma. Lemma 4. Under the above assumptions, any nontrivial ideal of L must be one of the following subspaces: (i) hx1 , x2 , . . . , xn−1 i; (ii) hx1 , x2 , . . . , xn i; (iii) hx0 i; / 0; (iv) hx0 + αx1 i for α = (v) hx0 , x1 , . . . , xn−1 i. Proof. They are all the possible subspaces invariant under ad x.
Lemma 5. If x = x1 , then dim L 6 4. Proof. If L is not solvable, denote by R the radical of L. For dim L/R > 3, by / 0. It is clear Lemmas 3 and 4, dim R = 1, i.e. R = hx0 i or hx0 + αx1 i for some α = that one may consider only the case R = hx0 i. If it is the case, then the image of x in L/R is self-centralizing ad-nilpotent. For L/R is semisimple, by [1], L/R ∼ = sl2 (F ). Now suppose that L is solvable. If hx0 i is an ideal of L, then the image of x1 in L/hx0 i is self-centralizing ad-nilpotent. So, by [1], dim L/hx0 i 6 2 and then dim L 6 / 0, then dim L 6 3. If 3. Similarly, if hx0 + αx1 i is an ideal of L for some α = J = hx1 , x2 , . . . , xn−1 i or hx1 , x2 , . . . , xn i is an ideal of L, then x1 is self-centralizing ad-nilpotent in J. But J is solvable, so, dim J 6 2. Hence, dim L 6 4. If the only ideal of L is hx0 , x1 , . . . , xn−1 i, then [L L] = J or 0. If [L L] = 0, then L is an abelian Lie algebra of two-dimension. If [L L] = J , then [J J ] = 0. So, dim J 6 2 and dim L 6 3. By Lemma 4, the lemma follows. From now on, we assume that n > 4, so x = x0 . We will need some notations and formulas similar to those in [3]. Let x0 , x1 , . . . , xn be a descending basis of L with respect to x0 . Let n X j ai k xk . [xi xj ] = k=0
J. Zhou / Linear Algebra and its Applications 320 (2000) 97–114
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j
For every j, denote by Aj = (ai k )n0 the matrices of ad xj under the basis. So, we have that j
ai k = −aji k , [Ai Aj ] =
n X
(1) j
ai k Ak .
(2)
k=0
By the Jacobi identity, we have that ( j for k = 0, n, −ai−1 k j −1 ai k = j j ai k+1 − ai−1 k for k = 1, 2, . . . , n − 1, j −1 a1 k
0 = j a1 k+1
Put
j
bi k =
j
ai k 0
for k = 0, n, for k = 1, 2, . . . , n − 1.
for 1 6 i, j, k 6 n, otherwise. j −1
j
j
Then for all i, j, k, we have that bi k = bi k+1 − bi−1 k . So j
n−j n bi−1 k+n−j −1
bi k =bin k+n−j − C1
n−j n +(−1)l Cl bi−l k+n−j −l
+ ···
n + · · · + (−1)n−j bi+j −n k .
(3)
Using (1), we have that n−j n bi−1 k+n−j −1 + · · · n −(bj k+n−i − C1n−i bjn−1 k+n−i−1
bin k+n−j − C1 =
+ · · ·)
(4)
In considering all the possibilities of the subspaces invariant under adx0 , we have the following lemma. Lemma 6. If x = x0 , then any nontrivial ideal of L must be one of the following subspaces: (i) hx1 , x2 , . . . , xi i for 1 6 i 6 n; (ii) hx0 , x1 , . . . , xn−1 i; (iii) hx1 , x2 , . . . , xs−1 , αx0 + xs i for s 6 n, α = / 0.
3. Solvable Lie algebras In this section, we suppose that L is solvable, dim L > 4 and x = x0 . The other notations are as in the preceding sections. Let K = L(1) .
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Lemma 7. If L is solvable, then K = hx1 , x2 , . . . , xn−1 i or hx0 , x1 , . . . , xn−1 i. In the last case, K 2 = hx1 , x2 , . . . , xn−2 i. In general, K l = hx1 , x2 , . . . , xn−l i for 1 < l < n. Proof. Since x1 , x2 , . . . , xn−1 ∈ K, by Lemma 6, K can only take one of the forms of: (i) hx1 , x2 , . . . , xn−1 i; (ii) hx0 , x1 , . . . , xn−1 i; (iii) hx1 , x2 , . . . , xn i; (iv) hx1 , x2 , . . . , αx0 + xn i for some α = / 0. So, it is sufficient to prove that the last two cases cannot appear. If (iii) is the case, then K is dimensionally nilpotent, ad x0 |K is a dimensionally nilpotent derivation, and x1 , x2 , . . . , xn is a descending basis with respect to it. By / K 2 . By the action of ad x0 , it can be seen that xn ∈ / K. This contradiction [3], xn ∈ shows that the case (iii) cannot happen. If we take x1 , x2 , . . . , αx0 + xn as the basis of L, then case (iv) becomes case (iii). So, case (iv) cannot occur either. If K = hx0 , x1 , . . . , xn−1 i, then x1 , x2 , . . . , xn−2 ∈ K 2 . So K 2 must be hx1 , x2 , / 0. Since . . . , xn−2 i, hx1 , x2 , . . . , xn−1 i or hx1 , x2 , . . . , αx0 + xn−1 i for some α = L is solvable, then K is nilpotent. So, dim (K/K 2 ) > 2. Hence, K 2 = hx1 , x2 , . . . , xn−2 i. By the nilpotence of K, the lemma follows. Corollary 8. Suppose that L is solvable, dim L = n + 1 and there is some element x ∈ L such that ad x is almost dimensionally nilpotent. If codim L(1) = 2, then L is the holomorph (cf. [5]) of a dimensionally nilpotent Lie algebra and a dimensionally nilpotent derivation. Proof. By Lemma 7, L2 = hx1 , x2 , . . . , xn−1 i. So, S = hx1 , x2 , . . . , xn i is an ideal of L. Clearly, S is dimensionally nilpotent and D = ad x0 |S is a dimensionally nilpotent derivation. Evidently, L ∼ = S ⊕ F D. Remark 9. It is easy to see that the converse of the corollary is also true. Now it remains to us to discuss the case codim L(1) = 1. In this case, K = L(1) = hx0 , x1 , . . . , xn−1 i. So, L cannot be nilpotent. Pick t ∈ L such that ad t is not nilpotent. If n > 4, then the restriction of ad t to K n−3 is a linear transformation of this subspace. Let ξ1 , ξ2 , ξ3 be its eigenvalues such that ξ1 and ξ1 , ξ2 are the eigenvalues of the restrictions of ad t to K n−1 , K n−2 , respectively. Then we have the following theorem. Theorem 10. Suppose that L is solvable, dim L = n + 1 and codim L(1) = n(n > 5). Let x0 ∈ L such that ad x0 is almost dimensionally nilpotent and ξi ’s be defined as above. Then:
J. Zhou / Linear Algebra and its Applications 320 (2000) 97–114
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(i) ξ2 = / ξ3 . If ξ = ξ2 /(ξ2 − ξ3 ), then ξ = / 3 − n. Moreover, ξ is independent of the choice of t. So, it is an invariant of L. (ii) If ξ (= / 3 − n) is not equal to any integer between −n + 1 and −2n + 5, then, when n is odd, ξ determines L completely; when n is even and ξ = / 1 − (n/2), ξ determines just two Lie algebras. (iii) Under the hypotheses in (ii), the elements of the set {x ∈ L | ad x is almost dimensionally nilpotent} are all conjugate under Aut (L) (up to scalar multiples). Proof. Note that S = hx1 , x2 , . . . , xn−1 i is a nilpotent dimensionally nilpotent Lie algebra and these elements form just a descending basis of S with respect to ad x0 |S . By [3], we have that j
j
ai 0 = ai k = 0
for i 6 k + n − j − 1, 1 6 i, j 6 n − 1.
By Lemma 7, we have that ain 0 = ain j = 0 for 1 6 i 6 n − 2, i < j. n (1) For L(1) = hx0 , x1 , . . . , xn−1 i, xn ∈ / L(1) , so, an−1 n = 0. But x0 ∈ L , so, we have that n / 0. an−1 0 =
(5)
For 3 < i 6 n − 2, in considering the coefficient of xi−3 in the Jacobi identity on xi , xn−3 and xn , by (3), we have that n n n n (ai−3 i−3 − 3ai−2 i−2 + 3ai−1i−1 − ai i ) n n n ×(ai−3 i−3 − ai i − an−3 n−3 ) = 0.
(6)
In the same way, in considering the coefficient of xi−1 in the Jacobi identity on xi , xn−1 and xn , we have that, for 1 < i 6 n − 2, n n n n n (ain i − ai−1 i−1 )(ai i − ai−1 i−1 + an−1 n−1 ) = −an−1 0 .
(7)
n n n n By (4), an−1 n−1 = an−2 n−2 . Put σ = an−1 0 , τ = an−1 n−1 . Let {η1 , η2 } be the set of the roots of equation X2 + τ X + σ = 0. Then
η1 + η2 = −τ.
(8)
By (7), for 1 < i 6 n − 2, we have that n ain i − ai−1 i−1 = ηs
for s = 1 or 2.
(9)
n Based on the above equalities, by a tedious computation, we can prove that ai+1 i+1 − ain i = η for 1 < i < n − 2; and a2n 2 − a1n 1 has two possibilities: (I) a2n 2 − a1n 1 = η; (II) a2n 2 − a1n 1 = −τ − η and n is even.
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n We consider firstly the case (I). Note that an−2 n−2 = τ . If 1 6 i 6 n − 3, then n−1 n−k n ai i = τ − (n − i − 2)η. So, ai+1 i = η and ai i−k = 0 for 1 < k 6 n − 2. If one takes t = xn , then one can easily obtain that
ξ1 = τ − (n − 3)η;
ξ2 = τ − (n − 4)η;
ξ3 = τ − (n − 5)η.
So, ξ=
τ − (n − 4). η
Since σ = / 0, so, τ = / −η. Therefore, ξ = / 3 − n. Under the basis x1 , x2 , . . . , xn−2 , the matrices of the actions of ad t on L(2) for all x ∈ L are all triangular, and, except that of xn , the matrices of the other elements of the chosen basis of L are all strictly triangular. Hence, ξ is independent of the choice of t. By the condition of (ii), −(τ/η) ∈ / {3, 4, . . . , n − P 1}. Then in considering the (i, i − 2) elements ci i−2 in the matrix [An−1 An ] = nk=0 Ak for 2 < i < n − 1, n by induction and a tedious computation, we can prove that ain i−t = ai−1 i−t −1 , i.e. n−1 ai i−t −1 = 0 for t − 1 < i 6 n − 1. So, we have proved that, for 1 < i < n, n n ain 1 = ai+1 2 = · · · = an−1 n−i .
Hence, An has the form 0 0 0 0 a1n 1 0 0 a2n 2 a1 0 a1 a2 . .. .. . . . . . . . .. .. .. .. . 0 an−4 0 a an−4 n−3 σ an−2 an−3 0 0 0
0
···
···
0
0
−1
0
···
···
0
0
0
0
0
0 a3n 3 .. .
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
n an−3 n−3
0
0
.
..
.
..
.
a1
τ
0
..
..
0 .. . .. .
.
an−4
···
···
a2
a1
τ
0
···
···
0
0
0
0 0 0 0 .. . , .. . 0 0 0 0
where ai ∈ F, ain i = τ − (n − 2 − i)η for i = 1, 2, . . . , n − 3. So, for 1 < k < i 6 n − 1, ain−1 i−k = 0.
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105
By (3), under the chosen basis, L has the multiplication table: [x1 xn−1 ] = 0; [xi xn−1 ] = ηxi−1
for i = 2, 3, . . . , n − 2;
[xi xn ] = (τ − (n − 2 − i)η)xi + a1 xi−1 + a2 xi−2 + · · · + ai−1 x1 for 1 6 i 6 n − 2; [xn−1 xn ] = τ xn−1 + a1 xn−2 + a2 xn−3 + · · · + an−2 x1 + σ x0 ; [xi xj ] = 0 for 1 6 i, j 6 n − 2, where ai ∈ F, η2 + τ η + σ = 0. Verification shows that this defines indeed a Lie algebra. In replacing the xi by η−1 xi for i = 1, 2, . . . , n, we may suppose that η = 1, thus σ = −(1 + τ ). Replacing xn by xn − a1 x0 , we may suppose that a1 = 0. It will be proved that one can choose a descending basis of L with respect to x0 such that a2 , a3 , . . . , an−2 are all zero. Firstly, let x10 = x1 ,
x20 = x2 ,
xi0 = xi + 12 a2 xi−2 for i = 3, 4, . . . , n;
In general, if we have had x1(k−1), x2(k−1) , . . . , xn(k−1). Let (k)
xi
(k)
xj
(k−1)
= xi
for i = 1, 2, . . . , k + 1, (k−1) a (k−1) (k−1) = xj + kk+1 + 1 xj −(k+1) for j = k + 2, . . . , n.
where as(t )’s are the structure constants of L under the basis x0 , x1(t ), . . . , xn(t ). ) By induction, one can prove that a1(t ) = · · · = at(t+1 = 0. So, under the basis (n−3)
x0 , x1
(n−3)
{xi
(n−3)
, . . . , xn
(n−3)
, the structure constants a1
(n−3)
= · · · = an−2 = 0. If we take
}i=1,2,...,n as the basis, then we have the table:
[xi xn ] = (τ − n + i + 2)xi
for 1 6 i 6 n − 2,
[xn−1 xn ] = τ xn−1 − (1 + τ )x0 , [x1 xn−1 ] = 0, [xi xn−1 ] = xi−1 [x1 x0 ] = 0,
for i = 2, 3, . . . , n − 2,
[xi x0 ] = xi−1 for 2 6 i 6 n, [xi xj ] = 0 for 1 6 i, j 6 n − 2. Here, τ = ξ + n − 4. So, for each fixed ξ , there exists just one Lie algebra of dimension n + 1 such that case (I) holds. Now we consider case (II). In this case, n is even. If we take t = xn , then we have that ξ1 = 2τ − (n − 5)η;
ξ2 = τ − (n − 4)η;
ξ3 = τ − (n − 5)η.
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So, ξ=
τ − (n − 4). η
For the same reason, we have that ξ = / 3 − n and we can suppose that η = 1. In the same way, we can prove that ξ is independent of the choice of t, so it is an invariant of L. By Lemma 6, hx1 i is an ideal of L. It is clear that L/hx1 i is a Lie algebra of case (I). By the hypothesis on ξ and the discussion on the algebras of case (I), we suppose that, when i = / j, j = / 1, ain j = 0 and a1n 1 = 2τ − n + 5; when 2 6 i 6 n n − 2, ai i = τ − (n − i − 2). Taking i = 2p, j = n, k = 1 in (4), we have that n n − 1. a2p 1 = 0 for p = 1, 2, . . . , 2 Consider the (2p 1)-entries c2p 1 in [An−1 An ] =
n X
n k an−1 kA .
k−1 k
From the left-hand side, n−1 n−1 n n−1 n n n C2p 1 = a2p 2p−1 a2p−1 1 + a2p 1 a1 1 − a2p 2p a2p 1 = (2p − τ − 2)a2p−1 1 .
From the right-hand side, n−1 n n C2p 1 = an−1 n−1 a2p−1 1 = −τ a2p−1 1 .
So, n − 1. 2 So, under the basis x0 , x1 , . . . , xn , we have that n a2p−1 1 = 0 for p = 2, 3, . . . ,
[xj xn ] = (τ − n + j + 2)xj [x1 xn ] = (2τ − n − 5)x1,
for 2 6 j 6 n − 2,
n [xn−1 xn ] = −(1 + τ )x0 + an−1 1 x1 + τ xn−1 . n Since ξ = / 1 − (n/2), 2τ − n + 6 = / 0. Put a = −an−1 1 /(2τ − n + 6). Let
xj0 = xj
for j = 0, 1, . . . , n − 2,
0 = xn−1 + ax1, xn−1 xn0 = xn + ax2 . 0 xn0 ] = Then we get a new descending basis of L with respect to x0 . We have that [xn−1 0 τ xn−1 − (1 + τ )x0 . So, the multiplication table is the following:
[x10 xn0 ] = (2τ − n + 5)x10 , [xi0 xn0 ] = (τ − n + i + 2)xi0
for i = 2, 3, . . . , n − 2,
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0 0 [xn−1 xn0 ] = τ xn−1 − (1 + τ )x00 , 0 ] = 0, [x10 xn−1
0 [x20 xn−1 ] = −(1 + τ )x10 , 0 0 ] = xi−1 for i = 3, 4, . . . , n − 2, [xi0 xn−1 0 [xi0 xn+1−i ] = (−1)i−1 (τ + 2)x10 [x10 x00 ] = 0, 0 [xi0 x00 ] = xi−1
for 2 < i 6 n − 2,
for 2 6 i 6 n.
The multiplications of the other elements of the basis vanish. Verification shows that this table defines indeed a Lie algebra. It is clear that this algebra is determined uniquely by τ , i.e. determined by ξ . By the proof, it is clear that the elements x such that ad x is almost dimensionally nilpotent are conjugate under Aut(L). Corollary 11. Suppose m > 6. Then, if m is odd, there are just two Lie algebras; if m is even, there is just one Lie algebra such that such Lie algebras satisfy the following conditions: (i) its dimension is equal to m; (ii) it is solvable but not nilpotent; (iii) its center Z is not trivial; (iv) there exists some element x ∈ L such that ad x is almost dimensionally nilpotent. If L is such a Lie algebra, then, up to scalar multiples, the elements of the set {x ∈ L | ad x is almost dimensionally nilpotent} are all conjugate under Aut(L). Proof. As L is not nilpotent and Z = / 0, dim L(1) = dim L − 1 and Z = F x1 . So, n in the proof of the theorem, a1 1 = 0. If L satisfies the condition of case (I), then ξ = 1; if L satisfies the condition of case (II), then ξ = −(n + 3)/2. So, L satisfies all conditions of the theorem and depends only on ξ . Therefore, the result follows.
4. Nonsolvable Lie algebras In this section, we discuss the nonsolvable Lie algebras. Theorem 12. Let n > 7. Then, when n is odd, there exist just two Lie algebras, when n is even, there exists just one Lie algebra such that such Lie algebras L are nonsolvable, dim L = n + 1 and there is some element x ∈ L such that ad x is almost dimensionally nilpotent. Up to scalar multiples, such ad-nilpotent elements are all conjugate under Aut(L).
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Proof. We will proceed to prove the theorem in steps. Let x0 , x1 , . . . , xn be the descending basis of L with respect to x0 . We need some lemmas. Lemma 13. There exists 0 = / β ∈ F and an integer s such that the radical R = hx1 , x2 , . . . , xs−1 , βx0 + xs i. Proof. For dim L/R > 3, by Lemma 6, it is sufficient to prove that R is not of the form of subspace hx1 , x2 , . . . , xi i for any i. Suppose to the contrary that R is of this form. Then, the image of x0 in L/R is ad-nilpotent of ad-nilpotent degree dim L/R − 1. But, L/R is semisimple. This contradicts the result of Lemma 3. The lemma follows. Lemma 14. dim R = n − 2, i.e. R = hx1 , x2 , . . . , xn−3 , βx0 + xn−2 i. Proof. Suppose that R = hx1 , x2 , . . . , xs−1 , βx0 + xs i for some β = / 0. Obviously, x¯0 is ad-nilpotent in L/R and [L/R x¯0 ] = hx¯s , x¯s+1 , . . . , x¯n−1 i, where x¯i is the image of xi in L/R. So, dim[L/R x¯0 ] = n − s. But, dim L/R = n + 1 − s. So, x¯0 is self-centralizing ad-nilpotent in L/R. By [1], L/R ∼ = sl2 (F ), i.e. R = hx1 , x2 , . . . , xn−3 , βx0 + xn−2 i. Lemma 15. R is nilpotent. Proof. Note that R is a solvable dimensionally nilpotent Lie algebra and ad x0 |R is its dimensionally nilpotent derivation, x1 , x2 , . . . , xn−3 , βx0 + xn−2 is a cyclic basis with respect to it. Suppose that [xn−3 βx0 + xn−2 ] =
n−3 X
ci xi + cn−2 (βx0 + xn−2 ).
i=1
By [3], we have that cn−2 = 0 . Since [xn−3 βx0 + xn−2 ] = βxn−4 + [xn−3 xn−2 ], n−2 / 0. It is well known cn−3 = an−3 n−3 . By [3], R is not nilpotent if and only if cn−3 = that D(R) ⊆ N the nil-radical of L. Let D = ad x0 . So we have that N = hx1 , x2 , . . . , xn−3 i. Therefore, hx1 , x2 , . . . , xn−2 i is a solvable dimensionally nilpotent Lie algen−2 / 0, it is not nilpotent. On other hand, for [xn−2 xn−1 ] = bra. But, for an−3 n−3 = [βx0 + xn−2 xn−1 ] + βxn−2 ∈ hx1 , x2 , . . . , xn−2 i, S = hx1 , x2 , . . . , xn−1 i is a subalgebra of L. For dim S/S ∩ R = 2, S is solvable. Let D = ad x0 |S . Then D(S) = hx1 , x2 , . . . , xn−2 i is contained in the nil-radical of S, and then is nilpotent. This contradiction shows that R is nilpotent.
By using above results, we give some formulas needed in the sequel. Since [xi xn−2 ] = [xi xn−2 + βx0 ] − β[xi x0 ] for 1 6 i 6 n − 3, ain−2 k = 0,
1 6 i 6 n − 2, i 6 k.
(10)
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Since for all j and 1 6 i 6 n − 3, [xi xj ] ∈ R, j
j
ai n−1 = ai n = 0,
1 6 i 6 n − 3.
(11)
Since [xn−2 xn−1 ] = [βx0 + xn−2 xn−1 ] + βxn−2 , n−1 n−1 an−2 n−1 = an−2 n = 0, n−1 an−2 0
=
n−1 β(an−2 n−2
− β).
(12) (13)
By [xn−2 xn ] = [βx0 + xn−2 xn ] + βxn−1 , we have that n an−2 n = 0,
n n an−2 n−1 = β(= an−1 n ).
(14)
By (10), (12) and (14), we obtain that ain−2 k = 0,
1 6 i 6 k.
(15)
By (11), (12) and (15), we have that ain−1 k = 0,
1 6 i < k.
(16)
By (11), (14) and (16), we have that ain k = 0, Since
n−1 an−2 n−2
1 6 i < k − 1.
(17)
n n = an−2 n−1 − an−3 n−2 , by (13) and (14), we have that
n−1 n an−2 0 = −βan−3 n−2 .
(18)
As in [3],we can prove the following lemma. n / 0. Lemma 16. an−3 n−2 =
Considering the coefficients of xi−1 in the Jacobi identity on xi , xn−3 and xn , we have that n−3 n−2 n n n n ain−3 i−2 ai−2 i−1 = ai i+1 ai+1 i−1 + an−3 n−2 ai i−1 an−3 0 , 2 < i 6 n − 4,
(19)
n−2 n n a2n 3 a3n−3 1 + an−3 n−2 a2 1 + an−3 0 = 0.
(20)
Let T = hx0 , x1 , . . . , xn−1 i. By (16)–(18), T is a subalgebra of L. Since dim (T /T ∩ R) = 2, T is solvable. Putting n−1 σ = an−2 0,
n−1 τ = an−2 n−2 ,
we have that n−1 n n n an−3 n−2 = an−2 n−1 − an−2 n−2 − an−2 n−2 = β − τ, n / 0. σ = −βan−3 n−2 =
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So, [T T ] = hx0 , x1 , . . . , xn−2 i. Therefore, T satisfies the conditions of Theorem 10. By the proof of the theorem, when 1 < i 6 n − 3, ain−1 i = τ − (n − 3 − i)η, n−2 where η is a root of the equation X2 + τ X + σ = 0. Hence, ai+1 i = β for 1 < i 6 n−3 n − 4, ai i−2 = 0 for 3 < i 6 n − 3. So, by (19), we have that n−2 n n an−3 n−2 ai i−1 + an−3 0 = 0. n−1 n−2 n n For an−3 0 = −an−2 0 = βan−3 n−2 , we have, ai i−1 = −β, i.e. η = −β. Similarly as in the proof of Theorem 10, we can prove that a2n−2 1 has two possibilities: (I) a2n−2 1 = −τ − η and n is odd; n−2 (II) a2 1 = η for any n. We now study case (I). Then a2n−2 1 = −τ − η. By (19), we have that
(2η + τ )(a2n 3 − β + τ ) = 0. So, either (i) (2η + τ ) = 0 or (ii) a2n 3 = β − τ . In fact, it can be proved that case (i) cannot happen. If case (i) holds, then τ/η = −2. By induction, we always have that, for 2 6 i 6 n − 3, n n ai−1 i =an−3 n−2 + (n − 2 − i)[−τ + (n − 3)η] η − [(n − 3)(n − 2) − i(i − 1)]. 2 n If case (ii) holds, take i = 3, we have that a2n 3 = β − τ = an−3 n−2 . By these two above formulas, we obtain that
(n − 5)[−τ + (n − 3)η] =
η [(n − 2)(n − 3) − 6] 2
i.e. τ/η = n/2 − 3. Without loss of generality, we suppose that β = −1, so, η = 1. From the above discussion, we see that, no matter which case occurs, the subalgebra T of L always satisfies the conditions of Theorem 10. By the proof of the theorem, there are x1 , x2 , . . . , xn ∈ L such that x0 , x1 , x2 , . . . , xn is a basis of L and satisfies the following conditions: (A) x0 , x1 , x2 , . . . , xn−1 is a descending basis of T, the multiplication table under this basis is given by the last form of the Lie algebra of second type in the proof of the theorem. So, we have that [x1 xn−1 ] = (2τ + 6 − n)x1 , [xi xn−1 ] = (τ − n + 3 + i)xi
for i = 2, 3, . . . , n − 3,
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[xn−2 xn−1 ] = τ xn−2 − (1 + τ )x0 , [x2 xn−2 ] = −(1 + τ )x1 , [xi xn−2 ] = xi−1 for i = 3, 4, . . . , n − 3. (B) R = hx1 , x2 , . . . , −x0 + xn−2 i. (C) There are a, αin j ∈ F such that [xn x0 ] = xn−1 + ax0 , [xi xn ] =
i+1 X
αin j xj
for 1 6 i 6 n − 1.
j =0
We will prove that, for every integer n, there exists just one Lie algebra satisfying the above conditions. That is to say, we must find out the expressions of τ, a, αin j in n. By a tedious computation, we can prove that n n n − 2, τ = − 3, a = 0, αn−3 0 = 2 2 1 n αi−1 i = 2, 3, . . . , n − 3, i = n − 1 − (n − i + 1)i, 2 n n n n αn−2 αn−3 n−2 = − + 2, n−1 = αn−1 n = −1 2 n n n n and, except αn−1 1 , αn−1 2 (= αn−2 1 ), all other αi j = 0. If we replace xn , xn−1 by xn +
n 2αn−1 2
n x2 − αn−1 1 x1 ,
2−n respectively, we can also suppose that
xn−1 +
n 2αn−1 2
2−n
x1 ,
n n n αn−1 1 = αn−2 1 = αn−1 2 = 0.
Therefore, we have found out a descending basis x0 , x1 , . . . , xn with respect to x0 such that the multiplication table of L is the following: [x1 xn−1 ] = 0, n xi , i = 1, 2, . . . , n − 3, [xi xn−1 ] = i − n2 n − 3 xn−2 − − 2 x0 , [xn−2 xn−1 ] = 2 2 [x1 xn−2 ] = 0, n [x2 xn−2 ] = 2 − x1 , 2 [xi xn−2 ] = xi−1 , i = 3, 4, . . . , n − 3, [x1 x0 ] = 0, [xi x0 ] = xi−1 ,
i = 2, 3, . . . , n,
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n [xi xn−i ] = (−1)i−1 − 1 x1 , i = 3, 4, . . . , n − 3, 2 1 [xi xn ] = n − (n − i)(i + 1) − 1 xi+1 , i = 1, 2, . . . , n − 4, 2 n n [xn−3 xn ] = − + 2 xn−2 + − 2 x0 , 2 2 [xn−2 xn ] = −xn−1 , [xn−1 xn ] = −xn , the brackets of other basis elements vanish. Verification shows that the above table defines indeed a Lie algebra. So, we have proved the following lemma. Lemma 17. For every odd integer n > 7, there is just one Lie algebra L of di/ a3n−2 mension n + 1 satisfying the condition a2n−2 1 = 2 . In such L, the elements of the set {x ∈ L|ad x is almost dimensionally nilpotent} are conjugate under Aut L (up to scalar multiples). By a similar argument, we can prove the following lemma. Lemma 18. For every integer n > 7, there is just one Lie algebra L of dimenn−2 sion n + 1 satisfying the condition a2n−2 1 = a3 2 . In such L, the elements of the set {x ∈ L |ad x is almost dimensionally nilpotent} are conjugate under Aut L (up to scalar multiples). Proof. As in the proof of the above lemma, we suppose that β = −1, so, η = 1. Our objective is to prove that, one can take x1 , x2 , . . . , xn−1 such that their products are given as in the multiplication table of the last form of the Lie algebras of first type in the proof of Theorem 10. Then, we can prove the lemma using a proof similar to that used to prove the last lemma. Note that if T satisfies the condition of Theorem 10(ii), it has been done. By induction, we can prove that, for 1 6 i 6 n − 4, ain i+1 =(n − 4)(n − 4 − i) − 12 ((n − 4)(n − 5) − i(i − 1)) −1 − τ (n − 2 − i). So, if ain i+1 = 0, then τ=
i 2 − (2n − 7)i + (n − 4)(n − 3) − 2 . 2(n − 2 − i)
Hence, it can be proved that, when 1 6 i 6 n − 3, τ (i) ∈ / {−3, −4, . . . , −n + 1}. Therefore, the condition of Theorem 10(ii) is satisfied. Hence, in the following, we / 0 for 1 6 i 6 n − 3. may suppose that, ain i+1 =
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113
n−1 n n n Note that an−2 = −1. Consider the (i i − n−2 = −an−2 0 = an−1 0 = 0, an−2 n−1 Pn n n−2 n k 1)-elements ci i−1 in the matrix equation [A A ] = k=0 an−2 kA . n−2 n n ain−2 i−2 ai−2 i−1 = ai+1 i−1 ai i+1 ,
2 < i 6 n − 3.
(21)
n In the same way, considering c2 1 , we have that a3n−1 1 a2 3 = 0, so,
a3n−2 1 = 0.
(22)
By (21) and (22), we have that ain−2 i−2 = 0,
3 6 i 6 n − 2. n−j
We now suppose that, for 2 6 j 6 k, j + 1 6 i 6 n − 2, ain−2 i−j = 0. By (3), ai i−k n−j
is the linear combination of asn−2 s−(k−j +2) for s 6 i. So, ai i−k = 0 for 2 6 j 6 k. If
n−2 n−k−1 we write ain−k−1 i−k as linear combination of aj j −1 (= 1), we see also that ai i−k = 0. When j > k + 1, by the nilpotence of hx1 , x2 , . . . , xn−2 i, we have also that n−j ai i−k = 0. Considering the (i i − k)-elements of the matrix [An−2 An ], we have that n n X X j n−2 n n n−2 n (ai j aj i−k − ai j aj i−k ) = an−2 j ai i−k . j =0
j =0
By induction, using the above results, we have that n−2 n n ain−2 i−k−1 ai−k−1 i−k = ai i+1 ai+1 i−k
k+1
n−2 n ak+1 k+2 ak+2 1 = 0. n−2 So, ak+2 1 = 0. As above, we can deduce from the above two formulas that, for k + 2 < i 6 n − 2, ain−2 i−k−1 = 0. We now can do as in the proof of Theorem 10. We can take x1 , x2 , . . . , xn−1 such that x0 , x1 , x2 , . . . , xn−1 is a descending basis of T with respect to x0 and their brackets are given as in the the multiplication table of the last form of the Lie algebras of first type in the proof of Theorem 10. Moreover, we can take xn ∈ L such that x0 , x1 , x2 , . . . , xn is a basis of L (not necessarily descending basis), R = hx1 , x2 , . . . , xn−3 , −x0 + xn−2 i, and there are a, αin j ∈ F such that [xn x0 ] = P n xn−1 + ax0 ], [xi xn ] = i+1 j =0 αi j xj , 1 6 i 6 n − 1. As in the proof of Lemma 17, we can determine the expression of a, τ, αin j on n. We have the following multiplication table:
[x1 x0 ] = 0, [xi x0 ] = xi−1 , [x1 xn−2 ] = 0,
i = 2, 3, . . . , n,
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[xi xn−2 ] = xi−1 ,
i = 2, 3, . . . , n − 3,
[xi xn−1 ] = 12 (2i + 1 − n)xi , i = 1, 2, . . . , n − 3, [xn−2 xn−1 ] = 12 (n − 5)xn−2 − 12 (n − 3)x0, [xi xn ] = 12 i(i − n + 2)xi+1 , 1 6 i 6 n − 4, [xn−3 xn ] = 12 (n − 3)(x0 − xn−2 ), [xn−2 xn ] = −xn−1 , [xn−1 xn ] = −xn , [xj xi ] = 0, 1 6 i, j 6 n − 3. Verification shows that this table defines indeed a Lie algebra. By the process of the proof, the conjugacy of the elements under Aut(L) is evident. By above two lemmas, the theorem is proved.
Acknowledgements The author would like to thank Prof. Jacques Désarménien, Prof. Gérard Duchamp and the referee for their numerous helpful suggestions.
References [1] G.M. Benkart, I.M. Isaacs, J.M. Osborn, Lie algebras with self-centralizing ad-nilpotent elements, J. Algebra 57 (1979) 279–309. [2] G.M. Benkart, I.M. Isaacs, J.M. Osborn, Albert-Zassenhaus Lie algebras and automorphisms, J. Algebra 57 (1979) 310–338. [3] S.S. Ding, Lie algebras with a cyclic nilpotent derivation, Baida Xyubao 1 (1958) 71–87. [4] V. Eberlin, Algèbres de Lie dimensionnellement nilpotenes, Comm. Algebra 28 (1) (2000) 183–191. [5] N. Jacobson, Lie Algebras, Wiley/Interscience, New York, 1962. [6] G. Leger, Dimensionally nilpotent Lie algebras, J. Algebra 117 (1988) 162–164.
Lie algebras with almost dimensionally nilpotent inner derivations聻 Jianhua Zhou Department of Applied Mathematics, Southeast University, Nanjing 210096, People’s Republic of China Received 8 October 1999; accepted 19 June 2000 Submitted by H. Schneider
Abstract The structure of the Lie algebras with almost dimensionally nilpotent inner derivations is studied. It is proved that, if the base field is of characteristic 0, then, when d > 6 is odd, there exist just two d-dimensional Lie algebras; when d > 6 is even, there exists just one ddimensional Lie algebra such that these Lie algebras are nonsolvable and have some almost dimensionally nilpotent inner derivations. © 2000 Elsevier Science Inc. All rights reserved. AMS classification: 17B30 Keywords: Almost dimensionally nilpotent derivation; Descending basis; Self-centralizing ad-nilpotent element
1. Introduction In [3,4,6], Ding, Eberlin and Leger have studied independently the Lie algebras having nilpotent derivations of degree their dimension. Such derivations are called dimensionally nilpotent, the Lie algebras are called dimensionally nilpotent ones, although they are not necessarily nilpotent as a Lie algebra. They have studied the structure of the Lie algebras when the characteristic of the base field is 0. The results show that apart from sl2 , such Lie algebras are all solvable. Ding has proved that
聻
Project supported by the State Education Commission, National Natural Science Foundation and Southeast University. E-mait address: [email protected] (J. Zhou). 0024-3795/00/$ - see front matter 2000 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 0 ) 0 0 2 0 5 - 6
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such a Lie algebra is nilpotent if and only if its center is not trivial. Furthermore, he has classified all such non-nilpotent solvable Lie algebras. If such a derivation is inner, then the corresponding element of the Lie algebra is called self-centralizing ad-nilpotent (s.c.a.n.), the Lie algebras with s.c.a.n. elements are also called self-centralizing ad-nilpotent (s.c.a.n. in the following). In [1,2], Benkart et al. have shown that, if the characteristic of the base field is prime, then the s.c.a.n. Lie algebras are all so-called generalized Zassenhaus Lie algebras; if the characteristic is 0, then the dimension of s.c.a.n. Lie algebras 6 3. Now let L be a finite-dimensional Lie algebra over a field F . It is natural to set the question that: what would happen if L has a nilpotent derivation of degree dim L − 1? Definition 1. Let L be a finite-dimensional Lie algebra over a field F, D a derivation of L. D is called almost dimensionally nilpotent if the degree of D is dim L − 1, i.e. / 0. D dim L−1 = 0 but D dim L−2 = The aim of this paper is to study the structure of Lie algebras with almost dimensionally nilpotent inner derivations in the case char F = 0. Unless otherwise stated, in this paper, F is a field of characteristic 0, L is a Lie algebra over F, dim L = n + 1, x ∈ L is a fixed ad-nilpotent element such that ad x is almost dimensionally nilpotent in L, i.e. the degree of ad x is n. For convenience, we denote by r(u) the rank of ad u for u ∈ L. It is clear that, if ad x is almost dimensionally nilpotent, then r(x) = dim L − 2.
2. Preliminaries Lemma 2. If L can be decomposed into direct sum of proper ideals, then L is isomorphic to one of the following: (i) the two-dimensional abelian Lie algebra; (ii) the direct sum of the two-dimensional non-abelian ideal and the center of dimension 1; (iii) the four-dimensional non-abelian reducible Lie algebra. Proof. Suppose that L = I1 ⊕ I2 ⊕ · · · ⊕ Ik . P P Let x = xi , xi ∈ Ii . Then, every ad xi |Ii is nilpotent and r(x) = ki=1 r(ad xi |Ii ). Since r(x) = dim L − 2 and r(ad xi |Ii ) 6 dim Ii − 1, it follows that k = 2 and r(ad xi |Ii ) = dim Ii − 1. So, xi is s.c.a.n. in Ii . It follows from the result of [1,3] that either dim Ii 6 2 or Ii ∼ = sl2 . Since ad x is almost dimensionally nilpotent, neither I1 nor I2 is two-dimensional abelian. Thus, the lemma follows.
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Let G be a semisimple Lie algebra over an algebraically closed field of characteristic 0. If the rank, i.e. the dimension of the Cartan subalgebra of G is l, then for every g ∈ G, the dimension of the centralizer of g dim CG (g) > l. If g is ad-semisimple and dim CG (g) = l, then CG (g) must be a Cartan subalgebra of G. We have the following lemma. Lemma 3. L is not semisimple. Proof. Without loss of generality, F may be assumed to be algebraically closed. Suppose that L is semisimple. Obviously, dim CL (x) = 2. So, the rank of L 6 2. Since the ad-nilpotent degree of every ad-nilpotent element in a semisimple Lie algebra > 3, so, dim L = / 3. By Lemma 2, L can only be one of the simple Lie algebras of type A2 , B2 and G2 . Since L is simple and x is ad-nilpotent, there exist h, y ∈ L such that x, h, y is a canonical basis of a three-dimensional simple Lie subalgebra, i.e. [x y] = h, [h x] = 2x and [h y] = −2y, where h is ad-semisimple and y is ad-nilpotent in L (cf. [5]). Put T = hx, y, hi (ha, b, . . .i denotes the subspace spanned by a, b, . . .). Let R be the representation of T on L by the adjoint action. L can be decomposed into direct sum of irreducible T-submodules: L = M1 ⊕ M2 ⊕ · · · ⊕ Mm . By the representation theory of sl2 , for each i, dim Ker(x R |Mi ) = 1. Since dim CL (x) = 2, it follows that m = 2. It is clear that the degree of x R 6 max{dim Mi }. Since ad x is almost dimensionally nilpotent, dim Mi > n. So the set {dim M1 , dim M2 } = {n, 1}. Without loss of generality, one can assume that dim M1 = n and dim M2 = 1. Suppose that M2 = F u for some u ∈ L. It is clear that T R (u) = [u T ] = 0. By the representation theory of sl2 , the dimension of the kernel of the restriction of hR on M1 is 1 (It cannot be 0. For otherwise, dim CL (h) = 1, which is a contradiction.) Hence dim CL (h) = 2. So, h is regular in L, i.e. CL (h) is a Cartan subalgebra of L. It is not difficult to see that x is a root vector of L with respect to CL (h). But the length of the string of roots in the root system of a semisimple Lie algebra with respect to its Cartan subalgebra 6 4, and so, (ad x)5 = 0. But the dimensions of the simple Lie algebras of type A2 , B2 , G2 are not less than 8. That is to say, L cannot be the simple Lie algebra of type A2 , B2 or G2 . The lemma follows from this contradiction. We will need some results picked from [3]. Suppose that G is a dimensionally nilpotent Lie algebra over F , D a dimensionally nilpotent derivation of G, x1 , x2 , . . . , xm is a cyclic basis, i.e. D(x1 ) = 0 and D(xi ) = xi−1 for i > 1. By [3], if m = 3 and D is a inner derivation of L, then G ∼ = sl2 (F ). If m > 3, then G is solvable. Suppose that m > 3. Then for 1 6 i 6 m, hx1 , x2 , . . . , xi i is an ideal of G. Suppose that m X j [xi , xj ] = ai k xk . k=1
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Then j
ai k = 0
for i < m + k − j.
m m G is nilpotent if and only if am−1 m−1 = 0. If it is the case, then ai i = 0 for all j
i, 1 6 i 6 m. Moreover, if i 6 m + k − j , then ai k = 0. By the Jordan canonical form of ad x, we know that L has a basis x0 , x1 , . . . , xn such that [x0 x] = [x1 x] = 0 while [xi x] = xi−1 for 2 6 i 6 n. We call it a descending basis of L with respect to x. It is not difficult to see that we have only to study the Lie algebras in two cases: Case I: x = x0 ;
Case II: x = x1 .
If Case II happens, i.e. x = x1 , then we have the following lemma. Lemma 4. Under the above assumptions, any nontrivial ideal of L must be one of the following subspaces: (i) hx1 , x2 , . . . , xn−1 i; (ii) hx1 , x2 , . . . , xn i; (iii) hx0 i; / 0; (iv) hx0 + αx1 i for α = (v) hx0 , x1 , . . . , xn−1 i. Proof. They are all the possible subspaces invariant under ad x.
Lemma 5. If x = x1 , then dim L 6 4. Proof. If L is not solvable, denote by R the radical of L. For dim L/R > 3, by / 0. It is clear Lemmas 3 and 4, dim R = 1, i.e. R = hx0 i or hx0 + αx1 i for some α = that one may consider only the case R = hx0 i. If it is the case, then the image of x in L/R is self-centralizing ad-nilpotent. For L/R is semisimple, by [1], L/R ∼ = sl2 (F ). Now suppose that L is solvable. If hx0 i is an ideal of L, then the image of x1 in L/hx0 i is self-centralizing ad-nilpotent. So, by [1], dim L/hx0 i 6 2 and then dim L 6 / 0, then dim L 6 3. If 3. Similarly, if hx0 + αx1 i is an ideal of L for some α = J = hx1 , x2 , . . . , xn−1 i or hx1 , x2 , . . . , xn i is an ideal of L, then x1 is self-centralizing ad-nilpotent in J. But J is solvable, so, dim J 6 2. Hence, dim L 6 4. If the only ideal of L is hx0 , x1 , . . . , xn−1 i, then [L L] = J or 0. If [L L] = 0, then L is an abelian Lie algebra of two-dimension. If [L L] = J , then [J J ] = 0. So, dim J 6 2 and dim L 6 3. By Lemma 4, the lemma follows. From now on, we assume that n > 4, so x = x0 . We will need some notations and formulas similar to those in [3]. Let x0 , x1 , . . . , xn be a descending basis of L with respect to x0 . Let n X j ai k xk . [xi xj ] = k=0
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101
j
For every j, denote by Aj = (ai k )n0 the matrices of ad xj under the basis. So, we have that j
ai k = −aji k , [Ai Aj ] =
n X
(1) j
ai k Ak .
(2)
k=0
By the Jacobi identity, we have that ( j for k = 0, n, −ai−1 k j −1 ai k = j j ai k+1 − ai−1 k for k = 1, 2, . . . , n − 1, j −1 a1 k
0 = j a1 k+1
Put
j
bi k =
j
ai k 0
for k = 0, n, for k = 1, 2, . . . , n − 1.
for 1 6 i, j, k 6 n, otherwise. j −1
j
j
Then for all i, j, k, we have that bi k = bi k+1 − bi−1 k . So j
n−j n bi−1 k+n−j −1
bi k =bin k+n−j − C1
n−j n +(−1)l Cl bi−l k+n−j −l
+ ···
n + · · · + (−1)n−j bi+j −n k .
(3)
Using (1), we have that n−j n bi−1 k+n−j −1 + · · · n −(bj k+n−i − C1n−i bjn−1 k+n−i−1
bin k+n−j − C1 =
+ · · ·)
(4)
In considering all the possibilities of the subspaces invariant under adx0 , we have the following lemma. Lemma 6. If x = x0 , then any nontrivial ideal of L must be one of the following subspaces: (i) hx1 , x2 , . . . , xi i for 1 6 i 6 n; (ii) hx0 , x1 , . . . , xn−1 i; (iii) hx1 , x2 , . . . , xs−1 , αx0 + xs i for s 6 n, α = / 0.
3. Solvable Lie algebras In this section, we suppose that L is solvable, dim L > 4 and x = x0 . The other notations are as in the preceding sections. Let K = L(1) .
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Lemma 7. If L is solvable, then K = hx1 , x2 , . . . , xn−1 i or hx0 , x1 , . . . , xn−1 i. In the last case, K 2 = hx1 , x2 , . . . , xn−2 i. In general, K l = hx1 , x2 , . . . , xn−l i for 1 < l < n. Proof. Since x1 , x2 , . . . , xn−1 ∈ K, by Lemma 6, K can only take one of the forms of: (i) hx1 , x2 , . . . , xn−1 i; (ii) hx0 , x1 , . . . , xn−1 i; (iii) hx1 , x2 , . . . , xn i; (iv) hx1 , x2 , . . . , αx0 + xn i for some α = / 0. So, it is sufficient to prove that the last two cases cannot appear. If (iii) is the case, then K is dimensionally nilpotent, ad x0 |K is a dimensionally nilpotent derivation, and x1 , x2 , . . . , xn is a descending basis with respect to it. By / K 2 . By the action of ad x0 , it can be seen that xn ∈ / K. This contradiction [3], xn ∈ shows that the case (iii) cannot happen. If we take x1 , x2 , . . . , αx0 + xn as the basis of L, then case (iv) becomes case (iii). So, case (iv) cannot occur either. If K = hx0 , x1 , . . . , xn−1 i, then x1 , x2 , . . . , xn−2 ∈ K 2 . So K 2 must be hx1 , x2 , / 0. Since . . . , xn−2 i, hx1 , x2 , . . . , xn−1 i or hx1 , x2 , . . . , αx0 + xn−1 i for some α = L is solvable, then K is nilpotent. So, dim (K/K 2 ) > 2. Hence, K 2 = hx1 , x2 , . . . , xn−2 i. By the nilpotence of K, the lemma follows. Corollary 8. Suppose that L is solvable, dim L = n + 1 and there is some element x ∈ L such that ad x is almost dimensionally nilpotent. If codim L(1) = 2, then L is the holomorph (cf. [5]) of a dimensionally nilpotent Lie algebra and a dimensionally nilpotent derivation. Proof. By Lemma 7, L2 = hx1 , x2 , . . . , xn−1 i. So, S = hx1 , x2 , . . . , xn i is an ideal of L. Clearly, S is dimensionally nilpotent and D = ad x0 |S is a dimensionally nilpotent derivation. Evidently, L ∼ = S ⊕ F D. Remark 9. It is easy to see that the converse of the corollary is also true. Now it remains to us to discuss the case codim L(1) = 1. In this case, K = L(1) = hx0 , x1 , . . . , xn−1 i. So, L cannot be nilpotent. Pick t ∈ L such that ad t is not nilpotent. If n > 4, then the restriction of ad t to K n−3 is a linear transformation of this subspace. Let ξ1 , ξ2 , ξ3 be its eigenvalues such that ξ1 and ξ1 , ξ2 are the eigenvalues of the restrictions of ad t to K n−1 , K n−2 , respectively. Then we have the following theorem. Theorem 10. Suppose that L is solvable, dim L = n + 1 and codim L(1) = n(n > 5). Let x0 ∈ L such that ad x0 is almost dimensionally nilpotent and ξi ’s be defined as above. Then:
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(i) ξ2 = / ξ3 . If ξ = ξ2 /(ξ2 − ξ3 ), then ξ = / 3 − n. Moreover, ξ is independent of the choice of t. So, it is an invariant of L. (ii) If ξ (= / 3 − n) is not equal to any integer between −n + 1 and −2n + 5, then, when n is odd, ξ determines L completely; when n is even and ξ = / 1 − (n/2), ξ determines just two Lie algebras. (iii) Under the hypotheses in (ii), the elements of the set {x ∈ L | ad x is almost dimensionally nilpotent} are all conjugate under Aut (L) (up to scalar multiples). Proof. Note that S = hx1 , x2 , . . . , xn−1 i is a nilpotent dimensionally nilpotent Lie algebra and these elements form just a descending basis of S with respect to ad x0 |S . By [3], we have that j
j
ai 0 = ai k = 0
for i 6 k + n − j − 1, 1 6 i, j 6 n − 1.
By Lemma 7, we have that ain 0 = ain j = 0 for 1 6 i 6 n − 2, i < j. n (1) For L(1) = hx0 , x1 , . . . , xn−1 i, xn ∈ / L(1) , so, an−1 n = 0. But x0 ∈ L , so, we have that n / 0. an−1 0 =
(5)
For 3 < i 6 n − 2, in considering the coefficient of xi−3 in the Jacobi identity on xi , xn−3 and xn , by (3), we have that n n n n (ai−3 i−3 − 3ai−2 i−2 + 3ai−1i−1 − ai i ) n n n ×(ai−3 i−3 − ai i − an−3 n−3 ) = 0.
(6)
In the same way, in considering the coefficient of xi−1 in the Jacobi identity on xi , xn−1 and xn , we have that, for 1 < i 6 n − 2, n n n n n (ain i − ai−1 i−1 )(ai i − ai−1 i−1 + an−1 n−1 ) = −an−1 0 .
(7)
n n n n By (4), an−1 n−1 = an−2 n−2 . Put σ = an−1 0 , τ = an−1 n−1 . Let {η1 , η2 } be the set of the roots of equation X2 + τ X + σ = 0. Then
η1 + η2 = −τ.
(8)
By (7), for 1 < i 6 n − 2, we have that n ain i − ai−1 i−1 = ηs
for s = 1 or 2.
(9)
n Based on the above equalities, by a tedious computation, we can prove that ai+1 i+1 − ain i = η for 1 < i < n − 2; and a2n 2 − a1n 1 has two possibilities: (I) a2n 2 − a1n 1 = η; (II) a2n 2 − a1n 1 = −τ − η and n is even.
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n We consider firstly the case (I). Note that an−2 n−2 = τ . If 1 6 i 6 n − 3, then n−1 n−k n ai i = τ − (n − i − 2)η. So, ai+1 i = η and ai i−k = 0 for 1 < k 6 n − 2. If one takes t = xn , then one can easily obtain that
ξ1 = τ − (n − 3)η;
ξ2 = τ − (n − 4)η;
ξ3 = τ − (n − 5)η.
So, ξ=
τ − (n − 4). η
Since σ = / 0, so, τ = / −η. Therefore, ξ = / 3 − n. Under the basis x1 , x2 , . . . , xn−2 , the matrices of the actions of ad t on L(2) for all x ∈ L are all triangular, and, except that of xn , the matrices of the other elements of the chosen basis of L are all strictly triangular. Hence, ξ is independent of the choice of t. By the condition of (ii), −(τ/η) ∈ / {3, 4, . . . , n − P 1}. Then in considering the (i, i − 2) elements ci i−2 in the matrix [An−1 An ] = nk=0 Ak for 2 < i < n − 1, n by induction and a tedious computation, we can prove that ain i−t = ai−1 i−t −1 , i.e. n−1 ai i−t −1 = 0 for t − 1 < i 6 n − 1. So, we have proved that, for 1 < i < n, n n ain 1 = ai+1 2 = · · · = an−1 n−i .
Hence, An has the form 0 0 0 0 a1n 1 0 0 a2n 2 a1 0 a1 a2 . .. .. . . . . . . . .. .. .. .. . 0 an−4 0 a an−4 n−3 σ an−2 an−3 0 0 0
0
···
···
0
0
−1
0
···
···
0
0
0
0
0
0 a3n 3 .. .
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
n an−3 n−3
0
0
.
..
.
..
.
a1
τ
0
..
..
0 .. . .. .
.
an−4
···
···
a2
a1
τ
0
···
···
0
0
0
0 0 0 0 .. . , .. . 0 0 0 0
where ai ∈ F, ain i = τ − (n − 2 − i)η for i = 1, 2, . . . , n − 3. So, for 1 < k < i 6 n − 1, ain−1 i−k = 0.
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By (3), under the chosen basis, L has the multiplication table: [x1 xn−1 ] = 0; [xi xn−1 ] = ηxi−1
for i = 2, 3, . . . , n − 2;
[xi xn ] = (τ − (n − 2 − i)η)xi + a1 xi−1 + a2 xi−2 + · · · + ai−1 x1 for 1 6 i 6 n − 2; [xn−1 xn ] = τ xn−1 + a1 xn−2 + a2 xn−3 + · · · + an−2 x1 + σ x0 ; [xi xj ] = 0 for 1 6 i, j 6 n − 2, where ai ∈ F, η2 + τ η + σ = 0. Verification shows that this defines indeed a Lie algebra. In replacing the xi by η−1 xi for i = 1, 2, . . . , n, we may suppose that η = 1, thus σ = −(1 + τ ). Replacing xn by xn − a1 x0 , we may suppose that a1 = 0. It will be proved that one can choose a descending basis of L with respect to x0 such that a2 , a3 , . . . , an−2 are all zero. Firstly, let x10 = x1 ,
x20 = x2 ,
xi0 = xi + 12 a2 xi−2 for i = 3, 4, . . . , n;
In general, if we have had x1(k−1), x2(k−1) , . . . , xn(k−1). Let (k)
xi
(k)
xj
(k−1)
= xi
for i = 1, 2, . . . , k + 1, (k−1) a (k−1) (k−1) = xj + kk+1 + 1 xj −(k+1) for j = k + 2, . . . , n.
where as(t )’s are the structure constants of L under the basis x0 , x1(t ), . . . , xn(t ). ) By induction, one can prove that a1(t ) = · · · = at(t+1 = 0. So, under the basis (n−3)
x0 , x1
(n−3)
{xi
(n−3)
, . . . , xn
(n−3)
, the structure constants a1
(n−3)
= · · · = an−2 = 0. If we take
}i=1,2,...,n as the basis, then we have the table:
[xi xn ] = (τ − n + i + 2)xi
for 1 6 i 6 n − 2,
[xn−1 xn ] = τ xn−1 − (1 + τ )x0 , [x1 xn−1 ] = 0, [xi xn−1 ] = xi−1 [x1 x0 ] = 0,
for i = 2, 3, . . . , n − 2,
[xi x0 ] = xi−1 for 2 6 i 6 n, [xi xj ] = 0 for 1 6 i, j 6 n − 2. Here, τ = ξ + n − 4. So, for each fixed ξ , there exists just one Lie algebra of dimension n + 1 such that case (I) holds. Now we consider case (II). In this case, n is even. If we take t = xn , then we have that ξ1 = 2τ − (n − 5)η;
ξ2 = τ − (n − 4)η;
ξ3 = τ − (n − 5)η.
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So, ξ=
τ − (n − 4). η
For the same reason, we have that ξ = / 3 − n and we can suppose that η = 1. In the same way, we can prove that ξ is independent of the choice of t, so it is an invariant of L. By Lemma 6, hx1 i is an ideal of L. It is clear that L/hx1 i is a Lie algebra of case (I). By the hypothesis on ξ and the discussion on the algebras of case (I), we suppose that, when i = / j, j = / 1, ain j = 0 and a1n 1 = 2τ − n + 5; when 2 6 i 6 n n − 2, ai i = τ − (n − i − 2). Taking i = 2p, j = n, k = 1 in (4), we have that n n − 1. a2p 1 = 0 for p = 1, 2, . . . , 2 Consider the (2p 1)-entries c2p 1 in [An−1 An ] =
n X
n k an−1 kA .
k−1 k
From the left-hand side, n−1 n−1 n n−1 n n n C2p 1 = a2p 2p−1 a2p−1 1 + a2p 1 a1 1 − a2p 2p a2p 1 = (2p − τ − 2)a2p−1 1 .
From the right-hand side, n−1 n n C2p 1 = an−1 n−1 a2p−1 1 = −τ a2p−1 1 .
So, n − 1. 2 So, under the basis x0 , x1 , . . . , xn , we have that n a2p−1 1 = 0 for p = 2, 3, . . . ,
[xj xn ] = (τ − n + j + 2)xj [x1 xn ] = (2τ − n − 5)x1,
for 2 6 j 6 n − 2,
n [xn−1 xn ] = −(1 + τ )x0 + an−1 1 x1 + τ xn−1 . n Since ξ = / 1 − (n/2), 2τ − n + 6 = / 0. Put a = −an−1 1 /(2τ − n + 6). Let
xj0 = xj
for j = 0, 1, . . . , n − 2,
0 = xn−1 + ax1, xn−1 xn0 = xn + ax2 . 0 xn0 ] = Then we get a new descending basis of L with respect to x0 . We have that [xn−1 0 τ xn−1 − (1 + τ )x0 . So, the multiplication table is the following:
[x10 xn0 ] = (2τ − n + 5)x10 , [xi0 xn0 ] = (τ − n + i + 2)xi0
for i = 2, 3, . . . , n − 2,
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0 0 [xn−1 xn0 ] = τ xn−1 − (1 + τ )x00 , 0 ] = 0, [x10 xn−1
0 [x20 xn−1 ] = −(1 + τ )x10 , 0 0 ] = xi−1 for i = 3, 4, . . . , n − 2, [xi0 xn−1 0 [xi0 xn+1−i ] = (−1)i−1 (τ + 2)x10 [x10 x00 ] = 0, 0 [xi0 x00 ] = xi−1
for 2 < i 6 n − 2,
for 2 6 i 6 n.
The multiplications of the other elements of the basis vanish. Verification shows that this table defines indeed a Lie algebra. It is clear that this algebra is determined uniquely by τ , i.e. determined by ξ . By the proof, it is clear that the elements x such that ad x is almost dimensionally nilpotent are conjugate under Aut(L). Corollary 11. Suppose m > 6. Then, if m is odd, there are just two Lie algebras; if m is even, there is just one Lie algebra such that such Lie algebras satisfy the following conditions: (i) its dimension is equal to m; (ii) it is solvable but not nilpotent; (iii) its center Z is not trivial; (iv) there exists some element x ∈ L such that ad x is almost dimensionally nilpotent. If L is such a Lie algebra, then, up to scalar multiples, the elements of the set {x ∈ L | ad x is almost dimensionally nilpotent} are all conjugate under Aut(L). Proof. As L is not nilpotent and Z = / 0, dim L(1) = dim L − 1 and Z = F x1 . So, n in the proof of the theorem, a1 1 = 0. If L satisfies the condition of case (I), then ξ = 1; if L satisfies the condition of case (II), then ξ = −(n + 3)/2. So, L satisfies all conditions of the theorem and depends only on ξ . Therefore, the result follows.
4. Nonsolvable Lie algebras In this section, we discuss the nonsolvable Lie algebras. Theorem 12. Let n > 7. Then, when n is odd, there exist just two Lie algebras, when n is even, there exists just one Lie algebra such that such Lie algebras L are nonsolvable, dim L = n + 1 and there is some element x ∈ L such that ad x is almost dimensionally nilpotent. Up to scalar multiples, such ad-nilpotent elements are all conjugate under Aut(L).
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Proof. We will proceed to prove the theorem in steps. Let x0 , x1 , . . . , xn be the descending basis of L with respect to x0 . We need some lemmas. Lemma 13. There exists 0 = / β ∈ F and an integer s such that the radical R = hx1 , x2 , . . . , xs−1 , βx0 + xs i. Proof. For dim L/R > 3, by Lemma 6, it is sufficient to prove that R is not of the form of subspace hx1 , x2 , . . . , xi i for any i. Suppose to the contrary that R is of this form. Then, the image of x0 in L/R is ad-nilpotent of ad-nilpotent degree dim L/R − 1. But, L/R is semisimple. This contradicts the result of Lemma 3. The lemma follows. Lemma 14. dim R = n − 2, i.e. R = hx1 , x2 , . . . , xn−3 , βx0 + xn−2 i. Proof. Suppose that R = hx1 , x2 , . . . , xs−1 , βx0 + xs i for some β = / 0. Obviously, x¯0 is ad-nilpotent in L/R and [L/R x¯0 ] = hx¯s , x¯s+1 , . . . , x¯n−1 i, where x¯i is the image of xi in L/R. So, dim[L/R x¯0 ] = n − s. But, dim L/R = n + 1 − s. So, x¯0 is self-centralizing ad-nilpotent in L/R. By [1], L/R ∼ = sl2 (F ), i.e. R = hx1 , x2 , . . . , xn−3 , βx0 + xn−2 i. Lemma 15. R is nilpotent. Proof. Note that R is a solvable dimensionally nilpotent Lie algebra and ad x0 |R is its dimensionally nilpotent derivation, x1 , x2 , . . . , xn−3 , βx0 + xn−2 is a cyclic basis with respect to it. Suppose that [xn−3 βx0 + xn−2 ] =
n−3 X
ci xi + cn−2 (βx0 + xn−2 ).
i=1
By [3], we have that cn−2 = 0 . Since [xn−3 βx0 + xn−2 ] = βxn−4 + [xn−3 xn−2 ], n−2 / 0. It is well known cn−3 = an−3 n−3 . By [3], R is not nilpotent if and only if cn−3 = that D(R) ⊆ N the nil-radical of L. Let D = ad x0 . So we have that N = hx1 , x2 , . . . , xn−3 i. Therefore, hx1 , x2 , . . . , xn−2 i is a solvable dimensionally nilpotent Lie algen−2 / 0, it is not nilpotent. On other hand, for [xn−2 xn−1 ] = bra. But, for an−3 n−3 = [βx0 + xn−2 xn−1 ] + βxn−2 ∈ hx1 , x2 , . . . , xn−2 i, S = hx1 , x2 , . . . , xn−1 i is a subalgebra of L. For dim S/S ∩ R = 2, S is solvable. Let D = ad x0 |S . Then D(S) = hx1 , x2 , . . . , xn−2 i is contained in the nil-radical of S, and then is nilpotent. This contradiction shows that R is nilpotent.
By using above results, we give some formulas needed in the sequel. Since [xi xn−2 ] = [xi xn−2 + βx0 ] − β[xi x0 ] for 1 6 i 6 n − 3, ain−2 k = 0,
1 6 i 6 n − 2, i 6 k.
(10)
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109
Since for all j and 1 6 i 6 n − 3, [xi xj ] ∈ R, j
j
ai n−1 = ai n = 0,
1 6 i 6 n − 3.
(11)
Since [xn−2 xn−1 ] = [βx0 + xn−2 xn−1 ] + βxn−2 , n−1 n−1 an−2 n−1 = an−2 n = 0, n−1 an−2 0
=
n−1 β(an−2 n−2
− β).
(12) (13)
By [xn−2 xn ] = [βx0 + xn−2 xn ] + βxn−1 , we have that n an−2 n = 0,
n n an−2 n−1 = β(= an−1 n ).
(14)
By (10), (12) and (14), we obtain that ain−2 k = 0,
1 6 i 6 k.
(15)
By (11), (12) and (15), we have that ain−1 k = 0,
1 6 i < k.
(16)
By (11), (14) and (16), we have that ain k = 0, Since
n−1 an−2 n−2
1 6 i < k − 1.
(17)
n n = an−2 n−1 − an−3 n−2 , by (13) and (14), we have that
n−1 n an−2 0 = −βan−3 n−2 .
(18)
As in [3],we can prove the following lemma. n / 0. Lemma 16. an−3 n−2 =
Considering the coefficients of xi−1 in the Jacobi identity on xi , xn−3 and xn , we have that n−3 n−2 n n n n ain−3 i−2 ai−2 i−1 = ai i+1 ai+1 i−1 + an−3 n−2 ai i−1 an−3 0 , 2 < i 6 n − 4,
(19)
n−2 n n a2n 3 a3n−3 1 + an−3 n−2 a2 1 + an−3 0 = 0.
(20)
Let T = hx0 , x1 , . . . , xn−1 i. By (16)–(18), T is a subalgebra of L. Since dim (T /T ∩ R) = 2, T is solvable. Putting n−1 σ = an−2 0,
n−1 τ = an−2 n−2 ,
we have that n−1 n n n an−3 n−2 = an−2 n−1 − an−2 n−2 − an−2 n−2 = β − τ, n / 0. σ = −βan−3 n−2 =
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So, [T T ] = hx0 , x1 , . . . , xn−2 i. Therefore, T satisfies the conditions of Theorem 10. By the proof of the theorem, when 1 < i 6 n − 3, ain−1 i = τ − (n − 3 − i)η, n−2 where η is a root of the equation X2 + τ X + σ = 0. Hence, ai+1 i = β for 1 < i 6 n−3 n − 4, ai i−2 = 0 for 3 < i 6 n − 3. So, by (19), we have that n−2 n n an−3 n−2 ai i−1 + an−3 0 = 0. n−1 n−2 n n For an−3 0 = −an−2 0 = βan−3 n−2 , we have, ai i−1 = −β, i.e. η = −β. Similarly as in the proof of Theorem 10, we can prove that a2n−2 1 has two possibilities: (I) a2n−2 1 = −τ − η and n is odd; n−2 (II) a2 1 = η for any n. We now study case (I). Then a2n−2 1 = −τ − η. By (19), we have that
(2η + τ )(a2n 3 − β + τ ) = 0. So, either (i) (2η + τ ) = 0 or (ii) a2n 3 = β − τ . In fact, it can be proved that case (i) cannot happen. If case (i) holds, then τ/η = −2. By induction, we always have that, for 2 6 i 6 n − 3, n n ai−1 i =an−3 n−2 + (n − 2 − i)[−τ + (n − 3)η] η − [(n − 3)(n − 2) − i(i − 1)]. 2 n If case (ii) holds, take i = 3, we have that a2n 3 = β − τ = an−3 n−2 . By these two above formulas, we obtain that
(n − 5)[−τ + (n − 3)η] =
η [(n − 2)(n − 3) − 6] 2
i.e. τ/η = n/2 − 3. Without loss of generality, we suppose that β = −1, so, η = 1. From the above discussion, we see that, no matter which case occurs, the subalgebra T of L always satisfies the conditions of Theorem 10. By the proof of the theorem, there are x1 , x2 , . . . , xn ∈ L such that x0 , x1 , x2 , . . . , xn is a basis of L and satisfies the following conditions: (A) x0 , x1 , x2 , . . . , xn−1 is a descending basis of T, the multiplication table under this basis is given by the last form of the Lie algebra of second type in the proof of the theorem. So, we have that [x1 xn−1 ] = (2τ + 6 − n)x1 , [xi xn−1 ] = (τ − n + 3 + i)xi
for i = 2, 3, . . . , n − 3,
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[xn−2 xn−1 ] = τ xn−2 − (1 + τ )x0 , [x2 xn−2 ] = −(1 + τ )x1 , [xi xn−2 ] = xi−1 for i = 3, 4, . . . , n − 3. (B) R = hx1 , x2 , . . . , −x0 + xn−2 i. (C) There are a, αin j ∈ F such that [xn x0 ] = xn−1 + ax0 , [xi xn ] =
i+1 X
αin j xj
for 1 6 i 6 n − 1.
j =0
We will prove that, for every integer n, there exists just one Lie algebra satisfying the above conditions. That is to say, we must find out the expressions of τ, a, αin j in n. By a tedious computation, we can prove that n n n − 2, τ = − 3, a = 0, αn−3 0 = 2 2 1 n αi−1 i = 2, 3, . . . , n − 3, i = n − 1 − (n − i + 1)i, 2 n n n n αn−2 αn−3 n−2 = − + 2, n−1 = αn−1 n = −1 2 n n n n and, except αn−1 1 , αn−1 2 (= αn−2 1 ), all other αi j = 0. If we replace xn , xn−1 by xn +
n 2αn−1 2
n x2 − αn−1 1 x1 ,
2−n respectively, we can also suppose that
xn−1 +
n 2αn−1 2
2−n
x1 ,
n n n αn−1 1 = αn−2 1 = αn−1 2 = 0.
Therefore, we have found out a descending basis x0 , x1 , . . . , xn with respect to x0 such that the multiplication table of L is the following: [x1 xn−1 ] = 0, n xi , i = 1, 2, . . . , n − 3, [xi xn−1 ] = i − n2 n − 3 xn−2 − − 2 x0 , [xn−2 xn−1 ] = 2 2 [x1 xn−2 ] = 0, n [x2 xn−2 ] = 2 − x1 , 2 [xi xn−2 ] = xi−1 , i = 3, 4, . . . , n − 3, [x1 x0 ] = 0, [xi x0 ] = xi−1 ,
i = 2, 3, . . . , n,
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n [xi xn−i ] = (−1)i−1 − 1 x1 , i = 3, 4, . . . , n − 3, 2 1 [xi xn ] = n − (n − i)(i + 1) − 1 xi+1 , i = 1, 2, . . . , n − 4, 2 n n [xn−3 xn ] = − + 2 xn−2 + − 2 x0 , 2 2 [xn−2 xn ] = −xn−1 , [xn−1 xn ] = −xn , the brackets of other basis elements vanish. Verification shows that the above table defines indeed a Lie algebra. So, we have proved the following lemma. Lemma 17. For every odd integer n > 7, there is just one Lie algebra L of di/ a3n−2 mension n + 1 satisfying the condition a2n−2 1 = 2 . In such L, the elements of the set {x ∈ L|ad x is almost dimensionally nilpotent} are conjugate under Aut L (up to scalar multiples). By a similar argument, we can prove the following lemma. Lemma 18. For every integer n > 7, there is just one Lie algebra L of dimenn−2 sion n + 1 satisfying the condition a2n−2 1 = a3 2 . In such L, the elements of the set {x ∈ L |ad x is almost dimensionally nilpotent} are conjugate under Aut L (up to scalar multiples). Proof. As in the proof of the above lemma, we suppose that β = −1, so, η = 1. Our objective is to prove that, one can take x1 , x2 , . . . , xn−1 such that their products are given as in the multiplication table of the last form of the Lie algebras of first type in the proof of Theorem 10. Then, we can prove the lemma using a proof similar to that used to prove the last lemma. Note that if T satisfies the condition of Theorem 10(ii), it has been done. By induction, we can prove that, for 1 6 i 6 n − 4, ain i+1 =(n − 4)(n − 4 − i) − 12 ((n − 4)(n − 5) − i(i − 1)) −1 − τ (n − 2 − i). So, if ain i+1 = 0, then τ=
i 2 − (2n − 7)i + (n − 4)(n − 3) − 2 . 2(n − 2 − i)
Hence, it can be proved that, when 1 6 i 6 n − 3, τ (i) ∈ / {−3, −4, . . . , −n + 1}. Therefore, the condition of Theorem 10(ii) is satisfied. Hence, in the following, we / 0 for 1 6 i 6 n − 3. may suppose that, ain i+1 =
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113
n−1 n n n Note that an−2 = −1. Consider the (i i − n−2 = −an−2 0 = an−1 0 = 0, an−2 n−1 Pn n n−2 n k 1)-elements ci i−1 in the matrix equation [A A ] = k=0 an−2 kA . n−2 n n ain−2 i−2 ai−2 i−1 = ai+1 i−1 ai i+1 ,
2 < i 6 n − 3.
(21)
n In the same way, considering c2 1 , we have that a3n−1 1 a2 3 = 0, so,
a3n−2 1 = 0.
(22)
By (21) and (22), we have that ain−2 i−2 = 0,
3 6 i 6 n − 2. n−j
We now suppose that, for 2 6 j 6 k, j + 1 6 i 6 n − 2, ain−2 i−j = 0. By (3), ai i−k n−j
is the linear combination of asn−2 s−(k−j +2) for s 6 i. So, ai i−k = 0 for 2 6 j 6 k. If
n−2 n−k−1 we write ain−k−1 i−k as linear combination of aj j −1 (= 1), we see also that ai i−k = 0. When j > k + 1, by the nilpotence of hx1 , x2 , . . . , xn−2 i, we have also that n−j ai i−k = 0. Considering the (i i − k)-elements of the matrix [An−2 An ], we have that n n X X j n−2 n n n−2 n (ai j aj i−k − ai j aj i−k ) = an−2 j ai i−k . j =0
j =0
By induction, using the above results, we have that n−2 n n ain−2 i−k−1 ai−k−1 i−k = ai i+1 ai+1 i−k
k+1
n−2 n ak+1 k+2 ak+2 1 = 0. n−2 So, ak+2 1 = 0. As above, we can deduce from the above two formulas that, for k + 2 < i 6 n − 2, ain−2 i−k−1 = 0. We now can do as in the proof of Theorem 10. We can take x1 , x2 , . . . , xn−1 such that x0 , x1 , x2 , . . . , xn−1 is a descending basis of T with respect to x0 and their brackets are given as in the the multiplication table of the last form of the Lie algebras of first type in the proof of Theorem 10. Moreover, we can take xn ∈ L such that x0 , x1 , x2 , . . . , xn is a basis of L (not necessarily descending basis), R = hx1 , x2 , . . . , xn−3 , −x0 + xn−2 i, and there are a, αin j ∈ F such that [xn x0 ] = P n xn−1 + ax0 ], [xi xn ] = i+1 j =0 αi j xj , 1 6 i 6 n − 1. As in the proof of Lemma 17, we can determine the expression of a, τ, αin j on n. We have the following multiplication table:
[x1 x0 ] = 0, [xi x0 ] = xi−1 , [x1 xn−2 ] = 0,
i = 2, 3, . . . , n,
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[xi xn−2 ] = xi−1 ,
i = 2, 3, . . . , n − 3,
[xi xn−1 ] = 12 (2i + 1 − n)xi , i = 1, 2, . . . , n − 3, [xn−2 xn−1 ] = 12 (n − 5)xn−2 − 12 (n − 3)x0, [xi xn ] = 12 i(i − n + 2)xi+1 , 1 6 i 6 n − 4, [xn−3 xn ] = 12 (n − 3)(x0 − xn−2 ), [xn−2 xn ] = −xn−1 , [xn−1 xn ] = −xn , [xj xi ] = 0, 1 6 i, j 6 n − 3. Verification shows that this table defines indeed a Lie algebra. By the process of the proof, the conjugacy of the elements under Aut(L) is evident. By above two lemmas, the theorem is proved.
Acknowledgements The author would like to thank Prof. Jacques Désarménien, Prof. Gérard Duchamp and the referee for their numerous helpful suggestions.
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