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ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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Lie Triple Derivations of the Lie Algebra of Strictly Block Upper Triangular Matrices

Ghimire P*

Department of Mathematics and Physical Sciences, Louisiana State University of Alexandria, Alexandria, LA 71302, USA

*Corresponding Author:
Ghimire P
Department of Mathematics and Physical Sciences
Louisiana State University of Alexandria
Alexandria, LA 71302, USA
Tel: (318) 427-4421
E-mail: [email protected]

Received Date: February 25, 2017; Accepted Date: April 24, 2017; Published Date: April 28, 2017

Citation: Ghimire P (2017) Lie Triple Derivations of the Lie Algebra of Strictly Block Upper Triangular Matrices. J Generalized Lie Theory Appl 11: 265. doi: 10.4172/1736-4337.1000265

Copyright: © 2017 Ghimire P. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

Let Equation be the Lie algebra of all n × n strictly block upper triangular matrices over a field Equation. In this paper, we explicitly describe all Lie triple derivations of Equation when char(Equation)≠2.

Keywords

Lie triple derivation; Triangular matrix; Lie algebra

Introduction

This paper is a study of the Lie triple derivations of the Lie algebra of strictly block upper triangular matrices over a field equation when char equation. Let equation be a Lie algebra over a field equation or a ring R. Recall that an equation-linear map (resp.an R−linearmap) equation is called a derivation of equation if:

equation (1)

For all equation. A Lie triple derivation is an equation-linear map (resp.an R−linearmap) equation which satisfies:

equation (2)

For all equation. Clearly, any derivation is a Lie triple derivation. However, the converse statement is, in general, not true [1]. The following elements of equation are typical examples of Lie triple derivations:

1. For any equation, the linear map ad X defined by equation for all equation is a derivation which is called an inner derivation. Clearly, an inner derivation is a Lie triple derivation.

2. Any linear map f that maps equation to the center equation of equation and equation to zero is a Lie triple derivation, called a central Lie triple derivation.

In recent years, significant progress has been made in studying the derivations and Lie triple derivations of matrix Lie algebras over a field or a ring. Wang and Li characterized the Lie triple derivations of the Lie algebra of strictly upper triangular matrices over a commutative ring [2]. Lie, Cao, and Li described the Lie triple and generalized triple derivations of the parabolic subalgebras of the general linear Lie algebra over a computational ring [3]. Benkovic determined the Lie triple derivations on triangular matrices [4]. More recently, Benkovic described the Lie derivations and Lie triple derivations of upper triangular matrix algebras over a unital algebra [5]. Some other results on the derivations of certain matrix Lie algebras are given in refs [6-12].

Fix a field equation. Let Mm,n be the set of all m × n matrices over equation, and put equation. Let equation denotes the set of all strictly block upper triangular matrices (resp. block upper triangular matrices) in Mn relative to a given partition. Then equation and equation are Lie subalgebras of equation equation, i. e. Mn with the standard Lie bracket. In this paper, we explicitly determine the Lie triple derivations of equation which are as follows:

• When equation, Theorem 3.1 shows that every Lie triple derivation of equation is a sum of the adjoint action of a block upper triangular matrix in equation, a central Lie triple derivation, and two special linear maps.

The main motivation of this work comes from Wang and Li’s work on the Lie triple derivation of the Lie algebra of strictly upper triangular matrices over a commutative ring [2], and authors’ study of derivations of the Lie algebra of strictly block upper triangular matrices over a field [9]. Our work on the Lie triple derivations of equation not only generalizes the main result of Wang and Li over a field, but also use a new approach that is promising to find the Lie triple derivations of other matrix Lie algebras with appropriate block forms. In disclosing the Lie triple derivation action on equation, we factor out the effects of adjoint action of block upper triangular matrices and those of central Lie triple derivations of equation, to explore the remaining Lie triple derivations.

Section 2 gives a basic introduction and determines some linear maps between matrix spaces that will be useful to describe the Lie triple derivations of equationin section 3. Section 3 describes the Lie triple derivations of equation over equation when char(equation)≠2.

Preliminary

In this section, we describe some linear maps between matrix spaces, and introduce some basic definitions and notations. The result in this section will be useful to prove the main result in Section 3.

Let equation denote the matrix with the only nonzero entry 1 on the (p,q) position for equation

Lemma 2.1

Suppose equation is an arbitrary field. If X∈Mm and Y∈Mn satisfy that:

XA=AY (3)

For all A∈Mmn, then X=λIm and Y=λIn for certain λ∈equation.

Proof: Suppose X(xip)∈Mm and Y=(yqj)∈Mn. For any (i,j)∈[m] ×[n], by eqn (3),

equation (4)

Comparing the (i,j) entry of the matrices in eqn (4), we get xii=yjj. Similarly, comparing the (p,j) entry for p≠i, we get xpi=0 and comparing the (i,q) entry for q≠j, we get 0=yjq. Therefore, X=λIm and Y=λIn for some equation

Lemma 2.2

If linear maps : Mm,p Mm,q and φ :Mn,p→Mn,q satisfy that:

φ(AB)=Aφ(B) (5)

For all A∈Mm,n, B∈Mn,p, then there is X∈Mp,q such that φ(C)=CX for C∈Mm,p and φ(D)=DX for D∈Mn,p.

Proof: For any j∈[n] and B∈Mn,p, by eqn (5),

equation (6)

All such equation span the first row space of Mm,p. So φ sends the first row of Mm,p to the first row of Mm,q. There exists a unique X∈Mp,q such that:

equation (7)

where the first equality in eqn (7) is by eqn (6). Therefore, (B)=BX. Hence φ(AB)=Aφ(B)=ABX for any A∈Mm,n and B∈Mn,p. All such AB span Mm,p. So φ(C)=CX for all C∈Mm,p.

Lemma 2.3

If linear maps : Mm,p→ Mn,p and φ : Mm,q→ Mn,q satisfy that:

φ(BA)φ(B)A (8)

For all A∈Mq,p, B∈Mm,q, then there is X∈Mn,m such that φ(C)=XC for C∈Mm,p and φ(D)=XD for D∈Mm,q.

Proof: The proof (omitted) is similar to that of Lemma 2.2.

Lemma 2.4

If linear maps : Mm,p→ Mm,q and φ : Mq,n→ Mp,n satisfy that:

φ(A)B=Aφ(B) (9)

For all A∈Mm,p, B∈Mq,n, then there is X∈Mp,q such that φ(C)=CX for C∈Mm,p and φ(D)=XD for D∈Mq,n.

Proof: For any j∈[P] and any equation , by eqn (9),

equation (10)

which shows that the only possibly nonzero row of equation is the first row. So maps the first row of Mm,p to the first row of Mm,q. There exists a unique X∈Mp,q such that:

equation, forall equation (11)

where the first equality in eqn (11) is by eqn (10). Therefore, equation. So φ(B)=XB for B∈Mq,n. Then φ(A)B=AXB for any A∈Mm,p and B∈Mq,n. Hence φ(A)=AX for all A∈Mm,p.

Lemma 2.5

If linear maps equation satisfy that:

equation (12)

then there exist X∈Mp, Y∈Mr, and Z∈Mq such that:

equation (13)

equation (14)

equation (15)

Proof: For any equation and equation, by eqn (12),

equation (16)

We further discuss eqn (16) in two cases:

1. j≠k: the left side of eqn (16) is zero and

equation (17)

2. j=k: the left side of eqn (16) is equation, and according to eqn (16), the only possibly nonzero entries of equation are:

equation (18)

equation (19)

equation (20)

Next we define a linear map equation such that property eqn (12) still holds. For C∈Mp,r, let:

equation (21)

Then for any n∈[p], m∈[r]and equation, by eqn (21),

equation

which implies that the only possibly nonzero entries of equation are:

equation (22)

equation (23)

equation (24)

where the last equality in eqns (22), (23) and (24) is by eqns (18), (19) and (20) respectively. Therefore, f ′=f on each equation and thus on the whole Mp,r. Denote:

equation (25)

We get f(C)=f′(C)=XC+CY for C∈Mp,r. So eqn (13) is done. Now for A∈Mp,q and B∈Mq,r, by eqns (12) and (13),

g(A)B + Ah(B) = f (AB) = XAB + ABY ⇒(g(A) − XA)B = A(BY − h(B)). (26)

Applying Lemma 2.4 to φ : Mp,q→ Mp,q defined by φ(A)=g(A)−XA and φ : Mq,r→ Mq,r defined by φ(B)=BY−h(B) in eqn (26), we find Z∈Mq such that:

equation

which imply eqns (14) and (15).

Lemma 2.6

If linear maps : Mp,s→ Mp,s, α : Mp,q→ Mp,q, : Mq,r→ Mq,r, and γ : Mr,s→Mr,s satisfy that:

equation (27)

then there exists X∈Mp, Y∈Mq, Z∈Mr, W∈Ms such that:

equation (28)

equation (29)

equation (30)

equation (31)

Proof: We first define a linear map equation by:

equation (32)

By eqns (27) and (32), for C∈Mr,s

equation (33)

Applying Lemma 2.5 to equation defined by f(D)=φ(D) for equation defined by g(F)=f′(F) for F∈Mp,r, and h : Mr,s→Mr,s defined by h(G)=γ(G) for G∈Mr,s in eqn (33), we find X∈Mp, Z∈Mr W∈M s , such that:

equation (34)

equation (35)

equation (36)

So eqns (28) and (31) are done. Again, applying Lemma 2.5 to f : Mp,r→Mp,r defined by f(F)=f′(F) for F∈Mp,r, g: Mp,q→Mp,q defined by g(A)=α(A) for A∈Mp,q, and h : Mq,r→Mq,r defined by h(B)=β(B) for B∈Mq,r in eqn (32), we find X′∈Mp, Y′∈Mq, and Z′∈Mr such that:

equation (37)

equation (38)

By eqns (33), (36), (35), and (38),

X′AB + AY′B + ABZ′ − AY′B = XAB + ABZ ⇒(X′ − X )AB = AB(Z − Z′). (39)

Applying Lemma 2.1 in eqn (39), we find equation,and Ir∈Mr such that:

equation (40)

Therefore, by eqns (37), (38), and (40),

equation

equation

Define Y:=−(λIq+Y′), we get eqns (29) and (30).

We make some notations that will be frequently used later. A sequence (n1,n2…,nt) is called an ordered partition of n if t, n1,…,ntequation+ and n1+n2+…+nt=n. The t × t block matrix form associate with an order partition(n1,n2,…,nt) is an expression of n × n matrices A=[Aij]t × t where the (i,j) block equation for i,j∈[t].

From now on, let us fix an order partition (n1,n2,…,nt ) of n and the corresponding block matrix form. Given A∈Mn, let Aij denote the matrix in Mn that has the same (i,j) block as A and 0’s elsewhere. Given a subset equation denote the set of Aij (resp. Aij) for all equation; for examples, we will use equation and equation in this manner for equation defined in Definition 2.7. Let equation denote the matrix that has the only nonzero entry 1 on the (p,q) position of the (i,j) block for (i, j)∈[t] × [t] and (p,q)∈[ni] × [nj]. Any notation of double index, say Aij (resp. Aij), may be written as Ai,j (resp. Ai,j) for clarity purpose. If A∈Mn is not given in advance, Aij (resp. Aij) may refer to an arbitrary matrix in equation.

An n × n matrix A is called block upper triangular (resp. strictlyblockuppertriangular) if Aij=0 for all 1≤j<i≤t (resp. 1≤j≤i≤t).

Definition 2.7: Consider the matrices in Mn corresponding to the t × t block matrix form associate with an ordered partition (n1,n2,…,nt) of n.

1. Let equation (abbr. equation) denote the Lie Theory algebra of all block upper triangular matrices in Mn.

2. Let equation denote the Lie algebra of all strictly block upper triangular matrices in Mn.

3. Call equation the block index set of equation. Given a subset equation we denote:

equation (41)

equation (42)

equation

The normalizer of Lie algebra equation in Mn is equation . For any equation, the adjoint action of equation

equation is a derivation (resp. Lie triple derivation) of equation

Lie triple derivations of equation for char(equation) ≠2

The goal of this section is to describe the Lie triple derivations of the Lie algebra equation of strictly block upper triangular n × n matrices over a field equation with char(equation)≠2. The cases 1≤t≤4 are trivial. So we consider t ≥5. Here is the main result for t ≥ 5.

Theorem 3.1

Suppose char(equation)≠2. When t ≥ 5, every Lie triple derivation f of the Lie algebra equation can be written (notuniquely) as:

equation (43)

where the summand components are given below:

1. X∈equation.

2. φ1t∈End(equation) is a central Lie triple derivation in:

equation

equation

equation

Before proving Theorem 3.1, we first prove several results on the images equation for a Lie triple derivation f and equation. The next lemma describes the image of f on equation and equation.

Lemma 3.2

Suppose char(equation)≠2. Then for any Lie triple derivation f of equation

equation (44)

equation (45)

Proof: To get eqn (44), first we prove that f(A12)ij=0 for any equation, 1<i<j, and equation. Either i>3 or j<t.

1. Suppose j<t. Then for any equation

equation

Therefore, 0=(A1i)1if(A12)ij(Ajt)jt for any equation. So:

equation (46)

For equation. Now we further discuss eqn (46) in the following two cases:

equation

- If i=2, it suffices to show that equation for any equation Given equation

equation (48)

Comparing the k-th row in the equality eqn (47), we see that the equation-th row of equation is zero. Since equation is arbitrary, equation

2. Suppose i>3. Then for any equation

equation

Therefore, equation for any equation. So f(A12)ij=0.

Next we show that equation

equation (48)

Since char equation by eqn (48), 0=(A12)12(A23)23 f (A12)3t. Given A12, the matrix (A12)12(A23)23 for any equation could be any matrix in equation with the rank no more than rank (A12)12. Therefore, f (A12)3t=0. Overall, we have proved eqn (44). The proof of eqn (45) (omitted) is similar to that of eqn (44).

Now we consider the image of a Lie triple derivation f on equation and equation

Lemma 3.3

Let f be a Lie triple derivation of . Then:

equation (49)

equation (50)

Proof: To get eqn (49), first we prove that f(A23)ij=0 for any equation, i<j, i≠2, j≠3, and (i,j)∉{(1, t−1), (1,t)}. Either i>2 or j<t−1.

equation (51)

Now we further discuss eqn (51) in the following two cases:

When j=2, 0=f (A23)i2(A23)i2(A23)23(A3t)3t. Given A23, the matrix (A23)23(A3t)3t for any equation could be any matrix in equation with rank no more then rank (A23)23. Therefore, f (A23)i2=0.

equation

2. Suppose i>2. Then for any equation

equation (52)

Now we further discuss eqn (52) in the following two cases:

- When i=0, 0=(A12)12(A23)23f(A23)3j. Given A23, the matrix (A12)12(A23)23 for any equation could be any matrix in equation with rank no more then rank (A23)23. Therefore, f(A23)3j=0.

- When i>3, 0=(A12)12(A21)2i f(A23)ij for any equation. Therefore, f(A23)ij=0

Overall, we have proved eqn (49). The proof of eqn (49) is similar to that of eqn (50).

Now we consider the image of a Lie triple derivation f on equation 34,…, t−3,t−2.

Lemma 3.4

Let f be a Lie triple derivation of and 3≤k≤t−3. Then:

equation (53)

Proof: Given any equation it suffices to prove that equation for i<j, i≠k, j≠k+1, and (i,j)∉{(1,t−1),(1,t),(2,t)} to get eqn (53). Either i>2 or j<t−1.

1. Suppose j<t−1. Then for any equation

equation

Now we further discuss eqn (54) in the following two cases:

- When j=k, 0=f(Ak,k+1)ik(Ak,k+1)k,k+1(Ak+1,t)k+1,t. Given Ak,k+1, the matrix (Ak,k+1)k,k+1(Ak+1,t)k+1,t for any equation could be any matrix in equationwith rank no more then rank (Ak,k+1)k,k+1. Therefore, f(Ak,k+1)ij=0.

- When j≠k, 0=f(Ak,k+1)ij(Aj.j+1)i,j+1(Aj+1,t)j+1,t for any equation and equation. Therefore, f(Ak,k+1)ij=0.

2. Suppose i>2. Then for any equation and any equation,

equation

Therefore equation. So equation

It remains to show that equation

equation

Therefore, equation. Overall, we have proved eqn (53).

Next we consider the image of a Lie triple derivation f on equation equation

Lemma 3.5

Suppose f be a Lie triple derivation of equation and 1≤k≤t−1. Then:

equation (55)

Proof: Given any equation, it suffices to prove that f(Ak,k+2)k,k+1=0, f(Ak,k+2)k+1,k+2=0, and f (Ak,k+2)ij=0 for i<j, i≠k, j≠k+2, (i,j)∉{(1,t−1),(1,t),(2,t)} to get eqn (55).

We first prove that f (Ak,k+2)ij =0 for i<j, i≠k, j≠k+2, (i,j)∉{(1,t−1),(1,t),(2,t)}. Either i>2 or j<t−1.

1. Suppose j<t−1. Then for any equation

equation

Therefore, equation for any equation and equation. So f(Ak,k+2)ij=0.

2. Suppose i>2. Then for any equation

equation

Therefore, equation for any equation and equation. So equation

Next we show that equationFor any equation

equation

Therefore, f(Ak,k+2)2,t−1=0. It remains to show that f(Ak,k+2)k,k+1=0, f(Ak,k+2)k+1,k+2=0. Now we show that f(Ak,k+1)k,k+1=0 (similarlyfor f(Ak,k+2)k+1,k+2=0). For any equation,

equation

Therefore, f(Ak,k+2)k,k+1=0.

Overall, we have proved eqn (55).

The following lemma shows that any equation that satisfies equation is a Lie triple derivation.

Lemma 3.6

Suppose equation satisfies that:

equation (56)

Then f is a Lie triple derivation of equation.

Proof: The f satisfying eqn (56) also satisfies the Lie triple derivation property:

equation

Similarly, we have the following lemma.

Lemma 3.7

Suppose equation satisfies that:

equation (57)

Then f is a Lie triple derivation of equation

Next we consider the image of a Lie triple derivation f on other equation

Lemma 3.8

Suppose char equation. For a Lie triple derivation f of equation and j>i+2, the image equation satisfies that:

equation (58)

Proof: Let j=i+k, k ≥3. We prove eqn (58) by induction on k.

equation

where the last relation in eqn (59) is by lemmas 3.2, 3.3 and 3.4. So we done for k=3.

2. equation Suppose eqn (58) holds for all equation where equation. Now equation

equation

equation (60)

where the last relation in eqn (60) is by induction hypothesis, the case k=3, and lemmas 3.2, 3.3, 3.4, and 3.5. So eqn (58) is true for equation.

So far all possible nonzero blocks of equation for equation have been located. The next lemma explicitly describes most of those nonzero blocks of equation. It essentially implies that the f-images on these blocks are the same as the images of an adjoint action of a block upper triangular matrix. Denote the index set:

equation

Lemma 3.9

Let f be a Lie triple derivation of . Then for any (p,q)∈Ω′, there exists equation such that:

equation (61)

equation (62)

Proof: Given equation we prove eqns (61) and (62) by the following steps:

1. We prove eqn (62) for (q,j)=(t−1,t). Then 2<p<t−1. For equation, equation,

equation

Therefore,

equation (63)

Applying lemma 2.4 to equation defined by equation and equation defined by equation in eqn (63), we find equation such that equation for all equation

2. Similarly, we can prove eqn (61) for (i,p)=(1,2) via lemma 2.4. In other words, for 2<q<t−1, there is equation such that equationequation

3. We prove eqn (62) for (q,j)=(t−2,t−1). For 1<p<t−2, equation, equation. By a similar computation as eqn (63), for equationequation

equation (64)

Applying lemma 2.4 to equation defined by equation and

equation defined by equation in eqn (64), we find equation such that equation for all equation

4. Similarly, we can prove eqn (61) for (i,p)=(2,3) via lemma 2.4. In other words, for 3<q<t, there is equation such that f(A23)2q=−(A23)23Y3q for equation.

5. Now we prove eqn (61) for (q,j)=(t−2,t). equation equation. By a similar computation as eqn (62), for equation and equation

equation (65)

Applying lemma 2.4 to equation defined by equation and equation defined by φ(D)=f(Dt2,t)p,t in eqn (65), we find equation such that equation

6. Similarly, we can prove eqn (62) for (i,p)=(1,3) via lemma 2.4. In other words, for 3<q<t−1, there is equation such that equationequation

7. Next we prove eqn (62) for (q,j)=(t−3,t−1). equation, equation. Then by a similar computation as eqn (63), for equation equation

equation (66)

Applying lemma 2.4 to equation defined by equation and equation defined by φ(D)=f(Dt3,t1)p,t−1 in eqn (66), we find equation such that f(At−3,t−1)p,t−1=Xp,t−3(At−3,t−1)t−3,t−1 for all equation

8. Similarly, we can prove eqn (61) for (i,p)=(2,4) via lemma 2.4. In other words, for 4<q<t, there is equation such that f(A24)2q=−(A24)24Y4q equation

9. Now we prove eqn (62) for equation. Then q<t−3. Given any equation. Then for equation

equation

equation (67)

Applying Lemma 2.3 to equation defined by equation and: equation defined by equation in eqn (67), we will find equation such that equation equation

10. Similarly, we can prove eqn (67) for equation via Lemma 2.2. In other words, there exists equation such that f(Aip)iq=−(Aip)ipYpq for all equation.

11. Now we prove that equation. We prove it by the following two steps:

(a) For any equation we have equationequation

equation

Therefore, Xpq=Ypq.

(b) Similarly, for any equation and equation we have equation

Finally, the equalities eqns (61) and (62) are proved.

Proof of Theorem 3.1. By Lemma 3.9, for (p,q)∈Ω′ we can find a matrix equation that satisfies eqns (61) and (62). Let equation be the matrix such that the (p,q) block is Xpq and 0’s elsewhere, and let:

equation (68)

Then f0 is a Lie triple derivation. The equalities eqns (61) and (62) imply that:

equation (69)

By Lemmas 3.2, 3.3, 3.4, 3.5, and 3.8 when char equation for any equation, the only possibly nonzero blocks of equation are the (i,j) block and the following:

• The (1,t),(2,t),(1,t−1) blocks when j∈{i+1,i+2}.

equation

equation

equation

Then Lemmas 3.6, and 3.7 show that equation are Lie triple derivation of equation. Now we get a Lie triple derivation:

equation

Define a linear map equation

equation

Then equation, so that 1t is a central Lie triple derivation of equation. We get a new derivation:

equation

equation

To get eqn (44), it suffices to prove the following claim regarding f2: there exist equation such that for each k∈[t−4], the Lie triple derivation:

equation

Satisfies that equation for 1≤p<q≤k+4. The proof is done by induction on k:

1. k=1: We proceed the case k=1 by the following steps:

equation (70)

Applying Lemma 2.6 in eqn (70), we find equationequation and equation such that

equation

- For any equation and equation, by a similar computation as eqn (70),

equation (71)

Applying Lemma 2.2 in eqn (71), we find equation such that equation and by eqn (71)

equation (72)

Applying lemma 2.4 in eqn (74), we find equation such that equation. We choose Y44=0, then equation and equation

Let equation. It is easy to show that equation is done.

2. k=2 : Similar to eqn (70), for any equation

equation (73)

Applying Lemma 2.2 in eqn (73), we find equation such that equation and equation. Let equation. It is easy to show that equation for equation is done.

3. equation Suppose the claim holds for all equation where equation is given. In other words, there exist equation for all equation such that equation satisfies that equation. Similar to eqn (70), for any equation and equation

equation

Applying Lemma 2.2 in the last equation, we find equation such that:

equation

Let equation. It is easy to show that equation equation is proved.

Overall, the claim is completely proved; in particular, equation

Let equation, then we get (3.1).

Acknowledgment

The author would like to thank the referee for the valuable comments and suggestions. The author would also like to thank Dr. Casey Orndorff and Mrs. Tanya Lueder for their help.

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