Linear Algebra and its Applications

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1 Linear Algebra and its Applications 437 (2012) Contents lists available at SciVerse ScienceDirect Linear Algebra and its Applications journal homepage: Solvable extensions of a class of nilpotent linear Lie algebras < Dengyin Wang,HuiGe,XiaoweiLi Department of Mathematics, China University of Mining and Technology, Xuzhou , PR China ARTICLE INFO ABSTRACT Article history: Received 12 November 2010 Accepted 3 February 2012 Availableonline10March2012 Submitted by C.K. Li AMS classification: 17B05 17B20 17B30 17B40 Let n(> 2) be a positive integer, n a maximal nilpotent subalgebra of the symplectic algebra sp(2n, F) over a field F of characteristic not 2, s a solvable Lie algebra whose nilradical is isomorphic to n. The derivations of n are shown to be the sum of several types standard derivations. The dimension of s is shown at most dim(n) + n,ands is isomorphic to the standard Borel subalgebra b of sp(2n, F) if and only if dim(s) = dim(n) + n Elsevier Inc. All rights reserved. Keywords: Solvable Lie algebras Derivations Nilradicals 1. Introduction The Levi theorem [1,2] tells us that every finite dimensional Lie algebra L is a semi-direct sum of a semisimple Lie algebra S and the maximal solvable ideal R of L, i.e., the radical of L: L = S R, [S, S] =S, [S, R] R, [R, R] R. Semisimple Lie algebras over the field of complex numbers have been classified by Cartan [3], over the field of real numbers by Gantmacher [4]. So the problem of classifying finite dimensional Lie < Supported by the Fundamental Research Funds for the Central Universities (2010LKSX05) and the National Natural Science Foundation of China (No )". Corresponding author. address: wdengyin@126.com (D. Wang) /$ - see front matter 2012 Elsevier Inc. All rights reserved.

2 D. Wang et al. / Linear Algebra and its Applications 437 (2012) algebras is reduced to the problem of classifying finite dimensional solvable Lie algebras. However, the classification of solvable Lie algebras is only complete for the case when the dimension is not bigger than six [5,6]. It seems to be impossible to classify solvable Lie algebras in an arbitrary large finite dimension. Some recent papers [7 13] provided another way for classifying solvable Lie algebras: construct all solvable Lie algebras having certain given nilpotent Lie algebra as their nilradicals. In previous articles the classification has been performed on the following nilpotent Lie algebras: Heisenberg algebras H n (n 1) [8]; Abelian Lie algebras a n (n 1) [9]; triangular Lie algebras T n (n 2) [10]; filiform Lie algebras Q 2n [11]; quasifiliform algebras q n (n 4) [12]; and filiform Lie algebras Q 2n+1 [13]. For the latest progress concerning classifications of solvable Lie algebras containing certain given Lie algebras we mention the articles [14 17]. Let n be the standard maximal nilpotent subalgebra of the symplectic algebra sp(2n, F). We will characterize the solvable Lie algebras, denoted by s, whose nilradical are isomorphic to n. Without loss of generality we may identity n with the nilradical of the solvable Lie algebra s throughout the paper. We organize this article as follows. In Section 2, we introduce the symplectic algebra sp(2n, F) and some subalgebras of it. In Section 3, we determine all derivations of n. In the last section, we announce and prove the main result. 2. The symplectic algebras Some notations concerning the symplectic algebra are introduced as follows: gl(n, F): the general linear Lie algebra of all n n matrices over F with the bracket operation [x, y] =xy yx. d(n, F): the Cartan subalgebra of gl(n, F) of all diagonal matrices. u(n, F): the nilpotent subalgebra of gl(n, F) of all strict upper triangular matrices. s(n, F): thesetofalln n symmetric matrices. sp(2n, F): the symplectic algebra of rank n over F consisting of all the matrices of the form x y z x, withx gl(n, F) and y, z s(n, F). h: the Cartan subalgebra of sp(2n, F) consisting of all the matrices of the form x 0 0 x with x d(n, F). n: the standard maximal nilpotent subalgebra of sp(2n, F) consisting of all the matrices of the form x y 0 x with x u(n, F) and y s(n, F). b: the standard Borel subalgebra of sp(2n, F) spanned by h and n, i.e., b = h + n. We denote by e i,j the n n matrix unit with 1 in the (i, j)-entry and 0 elsewhere. For 1 i j n we set E i,j = e i,j 0, E i, j = 0 e i,j + e j,i in case that i = j, E i, j = 0 e i,j in case that 0 e j,i i = j. Then the set ={E i,j, E k, l 1 i < j n, 1 k l n} formsabasisofn, and E 1,2, E 2,3,...,E n 1,n, E n, n generate n. The operations of the elements in are as follows.

3 16 D. Wang et al. / Linear Algebra and its Applications 437 (2012) [E i,j, E k,l ]=δ j,k E i,l δ l,i E k,j ; [E i, j, E k, l ]=0; [E i,j, E k, j ]=E i, k when i < k; or [E i,j, E k, j ]=E k, i when k < i; [E i,j, E i, j ]=2E i, i ; [E i,j, E j, k ]=E i, k ; [E i,j, E k, l ]=0 when j = l, j = k. Each element X n can be uniquely written as the linear combination of the basis in the form X = a i,j E i,j + b k, l E k, l. 1 i<j n 1 k l n Since the coefficient a i,j of E i,j is uniquely determined by X and (i, j), wewilldenoteitby{x} i,j.also we will denote the coefficient b k, l of E k, l by {X} k, l. 3. Derivations and automorphisms of n In the following of this paper we will always suppose that F is a field of characteristic not 2 and n is a positive integer satisfying n > 2. We now introduce four types of standard derivations of n and then we prove that every derivation of n is just the sum of the standard ones. Inner derivations: AfixedelementX n induces an inner derivation of n (which we denote by ad(x)) by sending any Y n to [X, Y]. Diagonal derivations: AfixedH h induces a diagonal derivation of n (which we denote by ad n (H))by sending any Y n to [H, Y]. Extremal derivations: If char(f) = 3, we define τ 0 to be the map sending any Y n to {Y} 1,2 E 2, 3 + {Y} 1,3 E 2, 2.Otherwise,ifchar(F) = 3, we define τ 0 to be the zero map. One can verify that τ 0 (particularly when char(f) = 3) is a derivation of n. Defineτ 1 to be the map sending any Y n to {Y} 1,2 E 2, 2. Then it is easy to verify that τ 1 is also a derivation of n. Wecallτ 0 and τ 1 extremal derivations of n. Central derivations: For k = 2, 3,...,n 1, define τ k to be the map sending any Y nto {Y} k,k+1 E 1, 1, and define τ n to be the map sending any Y n to {Y} n, n E 1, 1. Then it is easy to verify that all τ k,for k = 2, 3,...,n, are derivations of n. Wecallτ k, k = 2, 3,...,n, central derivations of n, sincethe center of n is of one dimensional and is spanned by E 1, 1. We denote by Der(n) the derivation algebra and by ad(n) the inner derivation algebra of n, respectively. For the purpose of describing all derivations of n we need some lemmas. Lemma 3.1. Let φ be a derivation of n and suppose that φ(e i,i+1 ) = a (i) k,l E k,l + b (i) s, t E s, t, for i = 1, 2,...,n 1. 1 k<l n 1 s t n Then a (i) k,l = 0 whenever k = iandl = i + 1. Proof. We give the proof for two different cases. Case 1. i > 1. Let k, l (k < l) be positive integers satisfying k = i and l = i+1. If k = 1, then by [E 1,k, E i,i+1 ]=0 we have the equality [E 1,k,φ(E i,i+1 )]=[E i,i+1,φ(e 1,k )]. Express the left and the right side of the above equality as the linear composition of the basis respectively, one will see that the coefficient of E 1,l oftheleftsideisa (i) k,l.namely{[e 1,k,φ(E i,i+1 )]} 1,l

4 D. Wang et al. / Linear Algebra and its Applications 437 (2012) = a (i) k,l.however{[e i,i+1,φ(e 1,k )]} 1,l = 0. This shows that a (i) k,l = 0 for the case that k = 1. In the case that k = 1, since l = i + 1, we have that [E 1, l, E i,i+1 ]=0. Applying φ we have that [E 1, l,φ(e i,i+1 )]=[E i,i+1,φ(e 1, l )]. It is not difficult to see that So a (i) k,l {[E 1, l,φ(e i,i+1 )]} 1, 1 = 2a (i) 1,l and {[E i,i+1,φ(e 1, l )]} 1, 1 = 0. = 0 for the case that k = 1. Now we see that a(i) k,l = 0 whenever k = i and l = i + 1. Case 2. i = 1. Assume that k, l (k < l) are positive integers satisfying k = 1 and l = 2. Applying φ on [E l, n, E 1,2 ]=0, we have that We find that Thus a (1) k,l [E l, n,φ(e 1,2 )]=[E 1,2,φ(E l, n )]. {[E l, n,φ(e 1,2 )]} k, n = a (1) k,l and {[E 1,2,φ(E l, n )]} k, n = 0. = 0 whenever k = 1 and l = 2. Lemma 3.2. Let φ be a derivation of n and suppose that φ(e i,i+1 ) = a (i) k,l E k,l + b (i) s, t E s, t, for i = 1, 2,...,n 1. 1 k<l n 1 s t n (i) If i 2, thenb (i) s, t = 0 whenever s = i, t = iand(s, t) = (1, 1). (ii) If i = 1, thenb (1) s, t = 0 whenever s = 1, (s, t) = (2, 2) and (s, t) = (2, 3); ifthe condition that char(f) = 3 is further assumed then b (1) 2, 3 = 0. Proof. For (i), let s, t (s t) be two positive integers satisfying s = i, t = i and (s, t) = (1, 1).If s > 1, then it follows from [E 1,s, E i,i+1 ]=0 that We find that [E 1,s,φ(E i,i+1 )]=[E i,i+1,φ(e 1,s )]. {[E 1,s,φ(E i,i+1 )]} 1, t = b (i) s, t and {[E i,i+1,φ(e 1,s )]} 1, t = 0. Thus b (i) s, t = 0 for the case that s > 1. For the case that s = 1 (thus t = 1), then by [E 1,t, E i,i+1 ]=0 we have that [E 1,t,φ(E i,i+1 )]=[E i,i+1,φ(e 1,t )]. It is not difficult to see that {[E 1,t,φ(E i,i+1 )]} 1, 1 = 2b (i) 1, t and {[E i,i+1,φ(e 1,t )]} 1, 1 = 0. Thus b (i) 1, t = 0 in case that 1 < t = i. Sob(i) s, t = 0 whenever s = i, t = i and (s, t) = (1, 1). This completes the proof of (i). Now we consider the case that i = 1. Let s, t (s t) be two positive integers. If s > 3, then it follows from [E 3,s, E 1,2 ]=0that [E 3,s,φ(E 1,2 )]=[E 1,2,φ(E 3,s )].

5 18 D. Wang et al. / Linear Algebra and its Applications 437 (2012) We find that {[E 3,s,φ(E 1,2 )]} 3, t = b (1) s, t and {[E 1,2,φ(E 3,s )]} 3, t = 0. So b (1) s, t = 0 when s > 3. If t > 3, also we have that Then by [E 3,t,φ(E 1,2 )]=[E 1,2,φ(E 3,t )]. {[E 3,t,φ(E 1,2 )]} 3, 3 = 2b (1) 3, t, {[E 1,2,φ(E 3,t )]} 3, 3 = 0, we get b (1) 3, t = 0fort > 3. Also by {[E 3,t,φ(E 1,2 )]} 2, 3 = b (1) 2, t and {[E 1,2,φ(E 3,t )]} 2, 3 = 0, we get b (1) 2, t = 0fort > 3. Now we know that φ(e 1,2 ) = n j=1 a (1) 1,j E 1,j + n t=1 b (1) 1, t E 1, t + b (1) 2, 2 E 2, 2 + b (1) 3, 3 E 3, 3 + b (1) 2, 3 E 2, 3. We now show that b (1) 3, 3 = 0. Applying φ on [E 1,2, E 1,3 ]=0, we get We find that [E 1,2,φ(E 1,3 )]=[E 1,3,φ(E 1,2 )]. {[E 1,3,φ(E 1,2 )]} 1, 3 = b (1) 3, 3, {[E 1,2,φ(E 1,3 )]} 1, 3 ={φ(e 1,3 )} 2, 3. Thus {φ(e 1,3 )} 2, 3 = b (1) 3, 3. Applying φ on [E 1,3, E 2,3 ]=0weget [E 1,3,φ(E 2,3 )]=[E 2,3,φ(E 1,3 )]. It is easy to see that {[E 2,3,φ(E 1,3 )]} 2, 2 = 2{φ(E 1,3 )} 2, 3 = 2b (1) 3, 3, {[E 1,3,φ(E 2,3 )]} 2, 2 = 0. So b (1) 3, 3 = 0. To complete the proof we now only need to show that b(1) 2, 3 = 0 in case that char(f) = 3. By E 1,3 =[E 1,2, E 2,3 ] we get φ(e 1,3 ) =[φ(e 1,2 ), E 2,3 ]+[E 1,2,φ(E 2,3 )] from which we get {φ(e 1,3 )} 2, 2 = 2b (1) 2, 3. Applying φ on [E 1,2, E 1,3 ]=0, we get We find that [E 1,2,φ(E 1,3 )]=[E 1,3,φ(E 1,2 )]. {[E 1,2,φ(E 1,3 )]} 1, 2 ={φ(e 1,3 )} 2, 2 = 2b (1) 2, 3 and {[E 1,3,φ(E 1,2 )]} 1, 2 = b (1) 2, 3. Thus 3b (1) 2, 3 = 0. If char(f) = 3, then we have that b(1) 2, 3 = 0.

6 D. Wang et al. / Linear Algebra and its Applications 437 (2012) Lemma 3.3. Then a (n) k,l φ(e n, n ) = Let φ be a derivation of n and suppose that 1 k<l n a (n) k,l E k,l + 1 s t n b (n) s, t E s, t. = 0 for 1 k < l n, and b (n) s, t = 0 whenever 1 < t < n. Proof. Let k, l be two positive integers satisfying 1 k < l n. By[E k, l, E n, n ]=0wehavethat [E k, l,φ(e n, n )]=[E n, n,φ(e k, l )]. Since {[E k, l,φ(e n, n )]} k, k = 2a (n) k,l, and {[E n, n,φ(e k, l )} k, k = 0, we get a (n) k,l = 0. Let s, t (s t) be two positive integers satisfying 1 < t < n. By[E 1,t, E n, n ]=0wehave that [E 1,t,φ(E n, n )]=[E n, n,φ(e 1,t )]. It is easy to see that {[E 1,t,φ(E n, n )]} 1, s = b (n) s, t and {[E n, n,φ(e 1,t )]} 1, s = 0. This implies that b (n) s, t = 0. Lemma 3.4. Let φ be a derivation of n, then there exist N n, H h, c i F(i = 0, 1, 2,...,n,) such that φ = ad n (H) + ad(n) + n i=0 c iτ i,wherec 0 = 0 in case that char(f) = 3. Proof. Suppose that φ(e i,i+1 ) = 1 k<l n a (i) k,l E k,l + 1 s t n b (i) s, t E s, t, for i = 1, 2,...,n 1, φ(e n, n ) = 1 k<l n a (n) k,l E k,l + 1 s t n b (n) s, t E s, t. Choose N to be the matrix of the form N = U V 0 U, where 0 a (2) 1,3 a(3) 1,4 a(4) 1,5 a(n 1) 1,n b (n) 1, n 0 0 a (1) 1,3 a (1) 1,4 a (1) 1,n 1 a (1) 1,n a (2)... (2) 2,4 a 2,n 1 a (2) 2,n U = (3) a 3,n 1 a (3) ; 3,n a (n 2) n 2,n

7 20 D. Wang et al. / Linear Algebra and its Applications 437 (2012) V = b(1) 1, 1 b (2) 1, 2 b (3) 1, 3 b (n 2) 1, (n 2) b (n 1) 1, (n 1) 1 2 b(1) 1, 1 b (1) 1, 2 b (1) 1, 3 b (1) 1, 4 b (1) 1, (n 1) b (1) 1, n b (2) 1, 2 b (1) 1, 3 b (2) 2, 3 b (2) 2, 4 b (2) 2, (n 1) b (2) 2, n b (3) 1, 3 b (1) 1, 4 b (2) 2, 4 b (3) 3, 4 b (3) 3, (n 1) b (3) 3, n b (n 2) 1, (n 2) b(1) 1, (n 1) b(2) 2, (n 1) b(3) 3, (n 1) b(n 2) n 2, (n 1) b (n 1) 1, (n 1) b (1) 1, n b (2) 2, n b (3) 3, n b (n 2) n 2, n.. b (n 2) n 2, n b (n 1) n 1, n. Denote φ + ad(n) by φ 1. By calculation we find that the action of φ 1 on each generator of n takes the following form (thanks to Lemma ): (1) φ 1 (E 1,2 ) = a (1) 1,2 E 1,2 + b (1) 2, 2 E 2, 2 + b (1) 2, 3 E 2, 3, where b (1) 2, 3 = 0 in case that char(f) = 3. (2) φ 1 (E n, n ) = n s=2 d(n) s, n E s, n + b (n) 1, 1 E 1, 1, where d (n) n, n = b (n) n, n. (3) For i = 2, 3,...,n 1, φ 1 (E i,i+1 ) = i k=2 c (i) k,i+1 E k,i+1 + i s=2 Furthermore we find that the following assertions hold. d (n) s, n = 0fors = 2, 3,...,n 1. Actually, for 2 s n 1, by [E s 1,s, E n, n ]=0wehavethat [E s 1,s,φ 1 (E n, n )]=[E n, n,φ 1 (E s 1,s )]. d (i) s, i E s, i + b (i) 1, 1 E 1, 1, where c (i) i,i+1 = a(i) i,i+1. It is easy to see that {E s 1,s,φ 1 (E n, n )]} s 1, n = d (n) s, n.by{φ 1(E s 1,s )} s 1,n = 0wehavethat {[E n, n,φ 1 (E s 1,s )]} s 1, n = 0. This forces that d (n) s, n = 0fors = 2, 3,...,n 1. Let 2 i n 1, then c (i) = k,i+1 d(i) by [E k 1,k, E i,i+1 ]=0wehavethat k, i [E k 1,k,φ 1 (E i,i+1 )]=[E i,i+1,φ 1 (E k 1,k )]. It is easy to see that {E k 1,k,φ 1 (E i,i+1 )]} k 1,i+1 = c (i) k,i+1. Recalling that {φ 1 (E k 1,k )} k 1,i = 0, we have {[E i,i+1,φ 1 (E k 1,k )]} k 1,i+1 = 0. = 0fork = 2, 3,...,i 1. Actually, for 2 k i 1, Thus we have that c (i) k,i+1 = 0. Similarly, we can prove that d(i) = k, i 0. d (i) i, i = 0fori = 2, 3,...,n 1. For a fixed i(2 i n 1), by[e i 1,i, E i,i+1 ]=E i 1,i+1 we have that φ 1 (E i 1,i+1 ) =[φ 1 (E i 1,i ), E i,i+1 ]+[E i 1,i,φ(E i,i+1 )].

8 D. Wang et al. / Linear Algebra and its Applications 437 (2012) Recalling that {φ 1 (E i 1,i )} i 1, (i+1) = 0, we have that {φ 1 (E i 1,i+1 )} i 1, i ={[φ 1 (E i 1,i ), E i,i+1 ]} i 1, i +[E i 1,i,φ(E i,i+1 )] i 1, i = d (i) i, i. Since [E i 1,i, E i 1,i+1 ]=0, we get [E i 1,i,φ 1 (E i 1,i+1 )]=[E i 1,i+1,φ 1 (E i 1,i )]. It is not difficult to see that {[E i 1,i,φ 1 (E i 1,i+1 )]} i 1, (i 1) = 2{φ 1 (E i 1,i+1 )} i 1, i = 2d (i) i, i. Recalling that {φ 1 (E i 1,i )} i 1, (i+1) = 0weget {[E i 1,i+1,φ 1 (E i 1,i )]} i 1, (i 1) = 0. Thus d (i) i, i = 0. Nowwehavethat φ 1 (E 1,2 ) = a (1) 1,2 E 1,2 +b (1) 2, 2 E 2, 2 +b (1) 2, 3 E 2, 3, where b (1) 2, 3 = 0 in case that char(f) = 3; φ 1 (E k,k+1 ) = a (k) k,k+1 E k,k+1 + b (k) 1, 1 E 1, 1, k = 2, 3,...,n 1; φ 1 (E n, n ) = b (n) n, n E n, n + b (n) 1, 1 E 1, 1. Choose H to be the diagonal matrix n 1 i=1 ( n 1 k=i a (k) + 1 k,k+1 2 b(n) n, n)e i,i b(n) n, n E n,n, and we construct the map ϕ = ad n (H) + n i=2 b (i) 1, 1 τ i + b (1) 2, 2 τ 1 + b (1) 2, 3 τ 0. It is easy to verify that ϕ and φ 1 act in the same way on each generator of {E 1,2, E 2,3,...,E n 1,n, E n, n }. Thus φ 1 = ϕ. Finally, we get that φ = ad( N) + ad n (H) + n i=2 b (i) 1, 1 τ i + b (1) 2, 2 τ 1 + b (1) 2, 3 τ 0. Lemma 3.5. (i) τ i τ j = 0 for 0 i, j n; (ii) [ad n (E i,i ), τ j ] Fτ j for 1 i nand0 j n. Particularly, [ad n (E 1,1 ), τ 0 ]= τ 0, [ad n (E 1,1 ), τ 1 ]= τ 1,and[ad n (E 1,1 ), τ i ]=2τ i for i = 2, 3,...,n. (iii) Let T denote the subspace of Der(n) spanned by ad(n) together with all τ i.thent is a nilpotent ideal of Der(n). Proof. By direct verification one can easily obtain (i) and (ii). Now we verify (iii). By Lemma 3.4 we have Der(n) = ad n (h) + T. Since[ad n (h), ad(n)] ad(n) and [ad n (h), τ j ] Fτ j for 0 j n (by (ii)), we know that T is an ideal of Der(n). Eachad n (x) (for x n) and each τ j (for 0 j n) acts nilpotent on n. This implies that T is a linear Lie algebra consisting of nilpotent linear transformations and thus a nilpotent Lie algebra (by Engel s theorem). Now we construct one type of automorphism of n. LetI denote the identity map on n, c i F for i = 0, 1, 2,...,n. Denotebyρ the map I + n i=0 c iτ i.

9 22 D. Wang et al. / Linear Algebra and its Applications 437 (2012) Lemma 3.6. (i) ρ is an automorphism of n,andρ 1 = I n i=0 c iτ i. (ii) ρ ad n (E 1,1 ) ρ 1 = ad n (E 1,1 ) + c 0 τ 0 + c 1 τ 1 2(c 2 τ 2 + +c n τ n ). Proof. By Lemma 3.5, we see that n i=0 c iτ i is a nilpotent derivation of n with nil-index 2. This implies that I + n i=0 c iτ i is an automorphism of n.obviously,ρ 1 = I n i=0 c iτ i.itisnotdifficulttoverify that ρ ad n (E 1,1 ) ρ 1 and ad n (E 1,1 ) + c 0 τ 0 + c 1 τ 1 2(c 2 τ 2 + +c n τ n ) act in the same way on each generator of {E 1,2, E 2,3,...,E n 1,n, E n, n } of n. Thus (ii) holds. 4. The main result We now announce the main result of this article. Theorem 4.1. Let n(> 2) be a positive integer, F a field of characteristic not 2, andletn be the standard maximal nilpotent subalgebra of the symplectic algebra sp(2n, F), s a solvable Lie algebra with nilradical isomorphic to n.then (i) dim(s) dim(n) + n. (ii) s is isomorphic to the Borel subalgebra b of sp(2n, F) iff dim(s) = dim(n) + n. Proof. We will identify the nilradical of s with n. Suppose that dim(s) = dim(n) + r. Let K be a complementary subspace of n in s, and choose {X 1, X 2,...,X r } to be a basis of K.Theneveryad n (X i ) : Y [X i, Y], Y n, is an outer derivation of n. Thus we may assume that ad n (X i ) n x i,j ad n (E j,j )(modt), i = 1, 2,...,r, (1) j=1 where T denotes the subspace of Der(n) spanned by ad(n) together with all τ i.set x 1,1 x 1,2 x 1,n x 2,1 x 2,2 x 2,n A =. x r,1 x r,2 x r,n We claim that the row vectors of A are linear independent. Otherwise, we can find c 1, c 2,...,c r,at least one is nonzero, such that (c 1, c 2,...,c r )A = 0. Thus r c i ad n (X i ) 0 (mod T). i=1 In other words, ad n ( r i=1 c ix i ) is contained in T. Thusad n ( r i=1 c ix i ) is a nilpotent derivation of n. Denote r i=1 c ix i by Z. ThenZ / n. If the characteristic of F is zero, by [s, FZ] [s, s] n, we know FZ + n is an ideal of s. In positive characteristic, replacing FZ by its image under the space of diagonal derivations, by Lemma 3.5 we also find that FZ+nis an ideal of s.sincead n (Z) is nilpotent, we know that FZ + n is a nilpotent ideal of s properly containing n,absurd.sor n (completing the proof of (i)). If s is isomorphic to b, then obviously dim(s) = dim(b) = dim(n) + n. Conversely,if dim(s) = dim(n) + n (namely, r = n), we desire to prover that s is isomorphic to b. Since the row vectors of A is linear independent and r = n, we know that A is invertible. Assume that A 1 = (b i,j ).Thenweget, from (1), that

10 D. Wang et al. / Linear Algebra and its Applications 437 (2012) ad n (E i,i ) b i,1 ad n (X 1 ) + b i,2 ad n (X 2 ) + +b i,n ad n (X n )(mod T), i = 1, 2,...,n. Let Y i = n j=1 b i,jx j, i = 1, 2,...,n, then the set {Y 1, Y 2,...,Y n } also forms a basis of K, and ad n (E i,i ) ad n (Y i )(mod T), i = 1, 2,...,n. Namely, Assume that ad n (Y i ) ad n (E i,i )(mod T), i = 1, 2,...,n. ad n (Y i ) = ad n (E i,i ) + c i,0 τ 0 + c i,1 τ 1 + c i,2 τ 2 + +c i,n τ n + ad(n i ), i = 1, 2,...,n, where N i n. ReplacingY i by Y i N i, then we have that ad n (Y i ) = ad n (E i,i ) + c i,0 τ 0 + c i,1 τ 1 + c i,2 τ 2 + +c i,n τ n, i = 1, 2,...,n. We claim that [ad n (Y i ), ad n (Y j )]=0foranyi, j. In fact, [ad n (Y i ), ad n (Y j )]=ad n [Y i, Y j ]. This says that [ad n (Y i ), ad n (Y j )] is an inner derivation induced by [Y i, Y j ] n. On the other hand, Lemma 3.5 shows that n n n [ad n (Y i ), ad n (Y j )]= adn (E i,i ) + c i,k τ k, ad n (E j,j ) + c j,k τ k Fτ k. k=0 It is not difficult to see that ad(n) n Fτ k=0 k = 0. So [ad n (Y i ), ad n (Y j )]=0. Since ad n ([Y 1, Y i ]) =[ad n (Y 1 ), ad n (Y i )]= 0fori = 2, 3,...,n, thereexistd i F such that [Y 1, Y i ]=d i E 1, 1 for i = 2, 3,...,n. LetZ 1 = Y 1, Z i = Y i d i 2 E 1, 1 for i = 2, 3,...,n. Obviously {Z 1, Z 2,...,Z n } spans another complementary subspace of n, and ad n (Z i ) = ad n (Y i ), i = 1, 2,...,n. We claim that [Z 1, Z i ]=0fori = 2, 3,...,n. In fact, for 2 i n, we have that [ [Z 1, Z i ]= Y 1, Y i d ] i 2 E 1, 1 = d i E 1, 1 d i 2 [Y 1, E 1, 1 ] = d i E 1, 1 d i 2 ad n(y 1 )(E 1, 1 ) = d i E 1, 1 d i adn (E 1,1 ) + 2 = d i E 1, 1 2 di 2 E 1, 1 = 0 Furthermore, we have that [Z i, Z j ]=0foreachpairi, j. n c 1,j τ j (E 1, 1 ) j=0 k=0 k=0

11 24 D. Wang et al. / Linear Algebra and its Applications 437 (2012) Following ad n ([Z i, Z j ]) =[ad n (Z i ), ad n (Z j )]=[ad n (Y i ), ad n (Y j )]=0 we have that [Z i, Z j ] FE 1, 1. Assume that [Z i, Z j ]=p i,j E 1, 1 with p i,j F.Consideringtheequality [Z 1, [Z i, Z j ]]=[[Z 1, Z i ], Z j ]+[Z i, [Z 1, Z j ]], wefindthattheleftsideis2p i,j E 1, 1, but the right side is zero. So p i,j = 0, i.e., [Z i, Z j ]=0for each pair i, j. Let ρ = I c 1,0 τ 0 c 1,1 τ c 1,2τ c 1,nτ n. Then by Lemma 3.6 we know that ρ is an automorphism of n, and ρ ad n (Z 1 ) ρ 1 = ρ (ad n (E 1,1 ) + n c 1,i τ i ) ρ 1 = ad n (E 1,1 ). i=0 We further claim that ρ ad n (Z i ) ρ 1 = ad n (E i,i ) for i = 2, 3,...,n. Assume that Then ρ ad n (Z i ) ρ 1 = ad n (E i,i ) + n d i,j τ j, i = 2, 3,...,n j=0 [ad n (E 1,1 ), ρ ad n (Z i ) ρ 1 ]=[ρ ad n (Z 1 ) ρ 1,ρ ad n (Z i ) ρ 1 ]=ρ [ad n (Z 1 ), ad n (Z i )] ρ 1 =0. On the other hand, by (ii) of Lemma 3.5, we have that n n [ad n (E 1,1 ), ρ ad n (Z i ) ρ 1 ]=[ad n (E 1,1 ), ad n (E i,i )+ d i,j τ j ]= d i,0 τ 0 d i,1 τ 1 +2 d i,j τ j. This follows that d i,j = 0forj = 0, 1, 2,...,n. Soweobtainρ ad n (Z i ) ρ 1 = ad n (E i,i ) for i = 2, 3,...,n. Now we are ready to construct an isomorphic map from s to b. Definethemapσ from s to b by sending each N + n i=1 a iz i to ρ(n) + n i=1 a ie i,i, where N n. Themapσ is obviously well defined, linear, and invertible. To show that σ is an isomorphism, it suffices to verify the following equalities: (1) σ([n, M]) =[σ(n), σ (M)] for N, M n. (2) σ([z i, N]) =[σ(z i ), σ (N)] for N n, (3) σ([z i, Z j ]) =[σ(z i ), σ (Z j )] for all 1 i, j n. In fact, (1) immediately follows from the definition of σ and the fact that ρ is an automorphism of n. For (2), since [Z i, N] n, wehave,bythedefinitionofσ, that σ([z i, N]) = ρ([z i, N]). On the other hand, [σ(z i ), σ (N)] =[E i,i,ρ(n)] =ad n (E i,i )(ρ(n)). We have shown that ad n (E i,i ) = ρ ad n (Z i ) ρ 1.So j=0 [σ(z i ), σ (N)] =ρ ad n (Z i ) ρ 1 (ρ(n)) = ρ([z i, N]). Thus (2) holds true. For (3), by [Z i, Z j ]=0wehavethatσ([Z i, Z j ]) = 0. Also [σ(z i ), σ (Z j )]=[E i,i, E j,j ]=0. Hence (3) holds. Finally, we conclude that σ is an isomorphism from s to b. j=2

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