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- *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.

**Visit for more related articles at** Journal of Generalized Lie Theory and Applications

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

Lie triple derivation; Triangular matrix; Lie algebra

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

(1)

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

(2)

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

1. For any , the linear map ad X defined by for all 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 to the center of and 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 . Let M_{m,n} be the set of all m × n matrices over , and put . Let denotes the set of all strictly block upper triangular matrices (resp. block upper triangular matrices) in M_{n} relative to a given partition. Then and are Lie subalgebras of , i. e. M_{n} with the standard Lie bracket. In this paper, we explicitly determine the Lie triple derivations of which are as follows:

• When , Theorem 3.1 shows that every Lie triple derivation of is a sum of the adjoint action of a block upper triangular matrix in , 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 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 , we factor out the effects of adjoint action of block upper triangular matrices and those of central Lie triple derivations of , 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 in section 3. Section 3 describes the Lie triple derivations of over when char()≠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 denote the matrix with the only nonzero entry 1 on the (p,q) position for

**Lemma 2.1**

Suppose is an arbitrary field. If X∈M_{m} and Y∈M_{n} satisfy that:

XA=AY (3)

For all A∈M_{mn}, then X=λI_{m} and Y=λI_{n} for certain λ∈.

Proof: Suppose X(x_{ip})∈M_{m} and Y=(y_{qj})∈M_{n}. For any (i,j)∈[m] ×[n], by eqn (3),

(4)

Comparing the (i,j) entry of the matrices in eqn (4), we get x_{ii}=y_{jj}. Similarly, comparing the (p,j) entry for p≠i, we get x_{pi}=0 and comparing the (i,q) entry for q≠j, we get 0=y_{jq}. Therefore, X=λ_{Im} and Y=λ_{In} for some

**Lemma 2.2**

If linear maps : M_{m,p} M_{m,q} and φ :M_{n,p}→M_{n,q} satisfy that:

φ(AB)=Aφ(B) (5)

For all A∈M_{m,n}, B∈M_{n,p}, then there is X∈M_{p,q} such that φ(C)=CX for C∈M_{m,p} and φ(D)=DX for D∈M_{n,p}.

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

(6)

All such span the first row space of M_{m,p}. So φ sends the first row of M_{m,p} to the first row of M_{m,q}. There exists a unique X∈M_{p,q} such that:

(7)

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

**Lemma 2.3**

If linear maps : M_{m,p}→ M_{n,p} and φ : M_{m,q}→ M_{n,q} satisfy that:

φ(BA)φ(B)A (8)

For all A∈M_{q,p}, B∈M_{m,q}, then there is X∈M_{n,m} such that φ(C)=XC for C∈M_{m,p} and φ(D)=XD for D∈M_{m,q}.

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

**Lemma 2.4**

If linear maps : M_{m,p}→ M_{m,q} and φ : M_{q,n}→ M_{p,n} satisfy that:

φ(A)B=Aφ(B) (9)

For all A∈M_{m,p}, B∈M_{q,n}, then there is X∈M_{p,q} such that φ(C)=CX for C∈M_{m,p} and φ(D)=XD for D∈M_{q,n}.

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

(10)

which shows that the only possibly nonzero row of is the first row. So maps the first row of M_{m,p} to the first row of M_{m,q}. There exists a unique X∈M_{p,q} such that:

, forall (11)

where the first equality in eqn (11) is by eqn (10). Therefore, . So φ(B)=XB for B∈M_{q,n}. Then φ(A)B=A_{XB} for any A∈M_{m,p} and B∈M_{q,n}. Hence φ(A)=A_{X} for all A∈M_{m,p}.

**Lemma 2.5**

If linear maps satisfy that:

(12)

then there exist X∈M_{p}, Y∈M_{r}, and Z∈M_{q} such that:

(13)

(14)

(15)

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

(16)

We further discuss eqn (16) in two cases:

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

(17)

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

(18)

(19)

(20)

Next we define a linear map such that property eqn (12) still holds. For C∈M_{p,r}, let:

(21)

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

which implies that the only possibly nonzero entries of are:

(22)

(23)

(24)

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

(25)

We get f(C)=f′(C)=XC+CY for C∈M_{p,r}. So eqn (13) is done. Now for A∈M_{p,q} and B∈M_{q,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 : M_{p,q}→ M_{p,q} defined by φ(A)=g(A)−XA and φ : M_{q,r}→ M_{q,r} defined by φ(B)=BY−h(B) in eqn (26), we find Z∈M_{q} such that:

which imply eqns (14) and (15).

**Lemma 2.6**

If linear maps : M_{p,s}→ M_{p,s}, α : M_{p,q}→ M_{p,q}, : M_{q,r}→ M_{q,r}, and γ : M_{r,s}→M_{r,s} satisfy that:

(27)

then there exists X∈M_{p}, Y∈M_{q}, Z∈M_{r}, W∈M_{s} such that:

(28)

(29)

(30)

(31)

Proof: We first define a linear map by:

(32)

By eqns (27) and (32), for C∈M_{r,s}

(33)

Applying Lemma 2.5 to defined by f(D)=φ(D) for defined by g(F)=f′(F) for F∈M_{p,r}, and h : M_{r,s}→M_{r,s} defined by h(G)=γ(G) for G∈M_{r,s} in eqn (33), we find X∈M_{p}, Z∈M_{r} W∈M

(34)

(35)

(36)

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

(37)

(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 ,and Ir∈Mr such that:

(40)

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

Define Y:=−(λI_{q}+Y′), we get eqns (29) and (30).

We make some notations that will be frequently used later. A sequence (n_{1},n_{2}…,n_{t}) is called an ordered partition of n if t, n_{1},…,n_{t}∈+ and n_{1}+n_{2}+…+n_{t}=n. The t × t block matrix form associate with an order partition(n_{1},n_{2},…,n_{t}) is an expression of n × n matrices A=[A_{ij}]_{t × t} where the (i,j) block for i,j∈[t].

From now on, let us fix an order partition (n_{1},n_{2},…,n_{t} ) of n and the corresponding block matrix form. Given A∈M_{n}, let A_{ij} denote the matrix in Mn that has the same (i,j) block as A and 0’s elsewhere. Given a subset denote the set of A^{ij} (resp. A^{ij}) for all ; for examples, we will use and in this manner for defined in Definition 2.7. Let 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)∈[n_{i}] × [n_{j}]. Any notation of double index, say A_{ij} (resp. A_{ij}), may be written as A_{i,j} (resp. A_{i,j}) for clarity purpose. If A∈M_{n} is not given in advance, A_{ij} (resp. A_{ij}) may refer to an arbitrary matrix in .

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

**Definition 2.7:** Consider the matrices in M_{n} corresponding to the t × t block matrix form associate with an ordered partition (n_{1},n_{2},…,n_{t}) of n.

1. Let (abbr. ) denote the Lie Theory algebra of all block upper triangular matrices in M_{n}.

2. Let denote the Lie algebra of all strictly block upper triangular matrices in M_{n}.

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

(41)

(42)

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

is a derivation (resp. Lie triple derivation) of

The goal of this section is to describe the Lie triple derivations of the Lie algebra of strictly block upper triangular n × n matrices over a field with char()≠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()≠2. When t ≥ 5, every Lie triple derivation f of the Lie algebra can be written (notuniquely) as:

(43)

where the summand components are given below:

1. X∈.

2. φ_{1t}∈End() is a central Lie triple derivation in:

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

**Lemma 3.2**

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

(44)

(45)

**Proof:** To get eqn (44), first we prove that f(A^{12})_{ij}=0 for any , 1<i<j, and . Either i>3 or j<t.

1. Suppose j<t. Then for any

Therefore, 0=(A^{1i})_{1i}f(A^{12})_{ij}(A^{jt})_{jt} for any . So:

(46)

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

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

(48)

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

2. Suppose i>3. Then for any

Therefore, for any . So f(A^{12})_{ij}=0.

Next we show that

(48)

Since char by eqn (48), 0=(A^{12})_{12}(A^{23})_{23} f (A^{12})_{3t}. Given A^{12}, the matrix (A^{12})_{12}(A^{23})_{23} for any could be any matrix in with the rank no more than rank (A^{12})_{12}. Therefore, f (A^{12})_{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 and

**Lemma 3.3**

Let f be a Lie triple derivation of . Then:

(49)

(50)

Proof: To get eqn (49), first we prove that f(A^{23})_{ij}=0 for any , i<j, i≠2, j≠3, and (i,j)∉{(1, t−1), (1,t)}. Either i>2 or j<t−1.

(51)

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

When j=2, 0=f (A^{23})_{i2}(A^{23})_{i2}(A^{23})_{23}(A^{3t})_{3t}. Given A^{23}, the matrix (A^{23})_{23}(A^{3t})_{3t} for any could be any matrix in with rank no more then rank (A^{23})_{23}. Therefore, f (A^{23})_{i2}=0.

2. Suppose i>2. Then for any

(52)

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

- When i=0, 0=(A^{12})_{12}(A^{23})_{23}f(A^{23})_{3j}. Given A^{23}, the matrix (A^{12})_{12}(A^{23})_{23} for any could be any matrix in with rank no more then rank (A^{23})_{23}. Therefore, f(A^{23})_{3j}=0.

- When i>3, 0=(A^{12})_{12}(A^{21})_{2i} f(A^{23})_{ij} for any . Therefore, f(A^{23})_{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 34,…, t−3,t−2.

**Lemma 3.4**

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

(53)

Proof: Given any it suffices to prove that 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

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

- When j=k, 0=f(^{Ak,k+1})_{ik}(A^{k,k+1})_{k,k+1}(A^{k+1,t})_{k+1,t}. Given A^{k,k+1}, the matrix (A^{k,k+1})_{k,k+1}(A^{k+1,t})_{k+1,t} for any could be any matrix in with rank no more then rank (A^{k,k+1})_{k,k+1}. Therefore, f(A^{k,k+1})_{ij}=0.

- When j≠k, 0=f(A^{k,k+1})_{ij}(A^{j.j+1})_{i,j+1}(A^{j+1,t})_{j+1,t} for any and . Therefore, f(A^{k,k+1})_{ij}=0.

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

Therefore . So

It remains to show that

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

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

**Lemma 3.5**

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

(55)

**Proof:** Given any , it suffices to prove that f(A^{k,k+2})_{k,k+1=0}, f(A^{k,k+2})_{k+1,k+2}=0, and f (A^{k,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 (A^{k,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

Therefore, for any and . So f(A^{k,k+2})_{ij}=0.

2. Suppose i>2. Then for any

Therefore, for any and . So

Next we show that For any

Therefore, f(A^{k,k+2})_{2,t−1}=0. It remains to show that f(A^{k,k+2})_{k,k+1}=0, f(A_{k,k+2})_{k+1,k+2}=0. Now we show that f(A^{k,k+1})_{k,k+1}=0 (similarlyfor f(A^{k,k+2})_{k+1,k+2}=0). For any ,

Therefore, f(A^{k,k+2})_{k,k+1}=0.

Overall, we have proved eqn (55).

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

**Lemma 3.6**

Suppose satisfies that:

(56)

Then f is a Lie triple derivation of .

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

Similarly, we have the following lemma.

**Lemma 3.7**

Suppose satisfies that:

(57)

Then f is a Lie triple derivation of

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

**Lemma 3.8**

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

(58)

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

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

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

(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 .

So far all possible nonzero blocks of for have been located. The next lemma explicitly describes most of those nonzero blocks of . 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:

**Lemma 3.9**

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

(61)

(62)

Proof: Given 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 , ,

Therefore,

(63)

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

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 such that

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

(64)

Applying lemma 2.4 to defined by and

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

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 such that f(A^{23})_{2q}=−(A^{23})_{23}Y_{3q} for .

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

(65)

Applying lemma 2.4 to defined by and defined by φ(D)=f(D^{t2,t})_{p,t} in eqn (65), we find such that

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 such that

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

(66)

Applying lemma 2.4 to defined by and defined by φ(D)=f(D^{t3,t1})_{p,t−1} in eqn (66), we find such that f(A^{t−3,t−1})_{p,t−1}=X_{p,t−3}(A^{t−3,t−1})_{t−3,t−1} for all

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 such that f(A^{24})_{2q}=−(A^{24})_{24}Y_{4q}

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

(67)

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

10. Similarly, we can prove eqn (67) for via Lemma 2.2. In other words, there exists such that f(A^{ip})_{iq}=−(A^{ip})_{ip}Y_{pq} for all .

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

(a) For any we have

Therefore, X_{pq}=Y_{pq}.

(b) Similarly, for any and we have

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 that satisfies eqns (61) and (62). Let be the matrix such that the (p,q) block is Xpq and 0’s elsewhere, and let:

(68)

Then f_{0} is a Lie triple derivation. The equalities eqns (61) and (62) imply that:

(69)

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

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

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

Define a linear map

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

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

Satisfies that 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:

(70)

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

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

(71)

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

(72)

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

Let . It is easy to show that is done.

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

(73)

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

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

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

Let . It is easy to show that is proved.

Overall, the claim is completely proved; in particular,

Let , then we get (3.1).

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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