# Strongly Ad-Nilpotent Elements of the Lie Algebra of Upper Triangular Matrices

Authors: Zhiguang Hu
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, the strongly ad-nilpotent elements of the Lie algebratn,ℂ of upper triangular complex matrices are studied. We prove that all the nilpotent matrices in tn,ℂ are strongly ad-nilpotent if and only if n≤6. Additionally, we prove that all the elements expadx, x strongly ad-nilpotent generate the inner automorphism group Inttn,ℂ.


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## Body

## 1. Introduction

In this paper, all the algebras are assumed to be finite dimensional. LetL be a complex Lie algebra. Since ady is an endomorphism of L for any y∈L, L is the direct sum of all the generalized eigenspaces Lλady:=kerady−λidm, where m is the multiplicity of λ as a root of the characteristic polynomial of ady. When Lλady=kerady−λid, it is the ordinary eigenspace, denoted by Eλady instead.Definition 1.
Call an elementx∈L strongly ad-nilpotent if there exists y∈L and some non-zero eigenvalue λ of ady such that x∈Lλady.By the definition, every element of the generalized eigenspacesLλady associated with the non-zero eigenvalue λ of ady is strongly ad-nilpotent. From the fact that Lλady,Lμady⊂Lλ+μady, we know that a strongly ad-nilpotent element must be ad-nilpotent.We first introduce some notations used in this paper. LetAutL denote the group of all the automorphisms of L and IntL denote the subgroup of AutL generated by all expadx, adx nilpotent. Denote by NL the set of all strongly ad-nilpotent elements of L and by EL the subgroup of IntL generated by all expadx, x∈NL.The strongly ad-nilpotent elements and the groupEL are important tools for the proof of the conjugacy of Cartan subalgebras; see [1]. In the case that L is semisimple, for any ad-nilpotent y∈L, there exist h,x∈L such that h,x,y span a subalgebra isomorphic to sl2,ℂ. Especially, we have adhy=−2y, which shows y is strongly ad-nilpotent. So, it is an equivalence between ad-nilpotency and strong ad-nilpotency in a semisimple Lie algebra. Furthermore, εL=IntL. If L is nilpotent, it has no non-zero strongly ad-nilpotent element, and EL has order one.A Lie algebraL is said to be solvable if the derived series(1)L=L0L1⋯Li⋯,satisfies Lk=0 for some k≥1, where Li=Li−1,Li−1,i=1,2,…. In this paper, we consider the strongly ad-nilpotent elements of a special class of solvable Lie algebras, the linear Lie algebras of upper triangular complex matrices. Let t=tn,ℂ be the Lie algebra of upper triangular complex n×n matrices and n=nn,ℂ be the subalgebra of strictly upper triangular matrices.The paper is organized as follows. Section2 is devoted to introducing some results about the strongly ad-nilpotent elements in general Lie algebras. In Section 3, we present the Cartan decomposition of t and the inner automorphism group Intt. In Section 4, we give a characterization of strongly ad-nilpotent elements of t. A simple sufficient condition is also given to determine a strongly ad-nilpotent element. In Section 5, we prove the main results of this paper. Though the Lie algebras we discussed are over ℂ, all the results are also valid for algebraically closed field of characteristic 0.


## 2. Preliminaries

In this section, we introduce some results about the strongly ad-nilpotent elements for general Lie algebras.Proposition 1.
Assume thatL is a Lie algebra which is not nilpotent. Then, NL has non-zero elements.Proof.
SinceL is not nilpotent, by Engel’s Theorem, there exists x∈L such that adx is not nilpotent. So, there exists non-zero eigenvalue λ of adx such that Lλadx≠0, which deduces the desired result.Lemma 1.
LetL be a Lie algebra, and x∈L. Then, x∈NL if and only if there exists y∈L such that x∈L1ady.Proof.
We only need to prove the necessity. Suppose thatx∈Lλady for some y∈L and λ≠0. Then, ady−λidkx=0 for some positive integer k. Thus, ady/λ−idkx=0. It follows that x∈L1ady/λ, which completes the proof.A Lie algebraL is called decomposable if there exist ideals L1 and L2 of L such that L=L1⊕L2. Otherwise, L is called indecomposable. About the decomposable Lie algebra, we have the following.Proposition 2.
LetL be a Lie algebra, and x∈L. If L is the direct sum of its ideals L1 and L2, then x=x1+x2 is (strongly) ad-nilpotent if and only if x1 and x2 are (strongly) ad-nilpotent elements of L1 and L2, respectively, where xi∈Li,i=1,2.Proof.
First suppose thatxi∈NLi for i=1,2. By Lemma 1, there exists yi∈Li such that adyi−idkixi=0 for some positive integer ki. Then, we get(2)ady1+y2−idk1+k2x1+x2=ady1+y2−idk1+k2x1+ady1+y2−idk1+k2x2=ady1−idk1+k2x1+ady2−idk1+k2x2=0,from the fact L1,L2=0. Therefore, x1+x2 is a strongly ad-nilpotent element of L.
Conversely, letx=x1+x2 be a strongly ad-nilpotent element of L. There exists y∈L and some positive integer k such that x∈kerady−λidk. Write y=y1+y2, where yi∈Li,i=1,2. Then,(3)0=ady1+y2−idkx1+x2=ady1−idkx1+ady2−idkx2.
Thus, fori=1,2, we have adyi−idkxi=0, which shows xi∈NLi.
Applying(4)adx1+x2ky1+y2=adx1ky1+adx2ky2,∀y1∈L1,y2∈L2,by a similar argument as above, we get that x=x1+x2 is ad-nilpotent if and only if x1 and x2 are ad-nilpotent elements of L1 and L2, respectively.According to a theorem of Lan [2], a Lie algebra L is solvable if and only if the set of all the ad-nilpotent elements of L is a linear subspace of L. Furthermore, if L is solvable, then the set {x ∈ L|adx is nilpotent} is the nilpotent radical of L, the maximal nilpotent ideal of L; see [3].


## 3. The Cartan Decomposition oftn,ℂ

Let(5)h=diagh1,h2,…,hn|∑i=1nhi=0.It is easily known that(6)h˜=h+ℂI,where I is the identity matrix, is a Cartan subalgebra of t, i.e., a nilpotent subalgebra with self-normalizer in t.Furthermore,t has the Cartan decomposition as follows:(7)t=h˜⊕∐α∈Φtα,where Φ⊂h∗. Here, h∗ is the dual space of h. Set αi=ei−ei+1,1≤i≤n−1, where eih=hi for all h=diagh1,h2,…,hn∈h. Then,(8)Φ=∑k=ijαk|1≤i≤j≤n−1,tαi+αi+1+⋯+αj=ℂei,j+1,i≤j.Here and thereafter,eij denotes the matrix having one in the i,j position and zeros elsewhere. It is well known that h⊕∐α∈Φtα is a Borel subalgebra of sln,ℂ. In general, a Borel subalgebra of a Lie algebra L is defined by the maximal solvable subalgebra of L.Lemma 2.
Letx∈t. Then, x is ad-nilpotent if and only if x=kI+x0, where k∈ℂ and x0∈n.Proof.
Ifx=kI+x0, then adx=adx0. Since x0 is nilpotent, adx0 is also nilpotent. On the other hand, set x=xij. Assume xii≠xjj for some i<j. We have(9)adxmeij=xii−xjjmeij+terms linearly independent toeij,for m=1,2,…. So, adx cannot be nilpotent. This is a contradiction.Proposition 3.
The inner automorphism groupIntt is given by(10)Intt=expadtx|x∈n.Proof.
By the above lemma, we know thatIntt is just the subgroup of Autt generated by expadtx, x∈n. Note that Autt is a linear Lie group with the Lie algebra Dert consisting of all derivations of t. Since adtn⊂Dert, let G be the connected Lie subgroup of Autt whose Lie algebra is adtn. Since adtn is a nilpotent algebra, the exponential mapping is a surjection. Hence, G=expadtx|x∈n. From G⊂Intt, by the definition of Intt, we get the desired result.Remark 1.
It is easy to get thatIntt is isomorphic to the Lie group of the unipotent upper triangular complex matrices.


## 4. A Characterization of Strongly Ad-Nilpotent Elements

We first give some equivalent conditions to depict strongly ad-nilpotent elements oft.Theorem 1.
Letx∈t. The following statements are equivalent:(1)
x is strongly ad-nilpotent(2)
There existsy∈t such that x∈L1ady(3)
There existsy∈t such that x∈E1ady, i.e., y,x=x(4)
There exists a semisimple elementy of t such that y,x=xProof.
1⇒2 is followed from Lemma 1.2⇒4 Let y=ys+yn be the Jordan decomposition, i.e., ys is semisimple, yn is nilpotent, and ys,yn=0. Since ys can be written as a polynomial in y without constant term, we have ys∈t. It is obvious that ady=adys+adyn is also the Jordan decomposition of ady. According to the fact adys and ady have the same generalized eigenspaces, we have x∈L1adys. For the reason that adys is diagonalizable, we get L1adys=E1adys.4⇒3 and 3⇒1 are obvious.Corollary 1.
Nt is a subset of n.Proof.
For anyx∈Nt, there exists y such that y,x=x. Thus, x is nilpotent, and hence x∈n.Recall the basic result about the automorphism and the strongly ad-nilpotent elements.Lemma 3 (see [1]).
LetL be a Lie algebra. If ϕ∈AutL, then ϕx∈NL for all x∈NL and ϕL1ady=L1adϕy for all y∈L.From the above lemma, in order to determine which ones inn are strongly ad-nilpotent, we only need to consider the equivalence classes under the acting of Autt. For Autt, the readers are referred to literatures [4, 5].Denote byt∗ the set of invertible matrices of t. Then, t∗ is a Lie group with the Lie algebra t. For any a∈t∗, we can define an automorphism of t by φa:x↦axa−1, for all x∈t. There is a natural group epimorphism from t∗ to φa|a∈t∗. Thus, we can only consider the upper triangular nilpotent matrices under upper triangular similarity (shortly written as under t∗). It has been studied by many researchers (see [6] and the references therein); related research can be seen in [7–9]. Note that there are only finitely upper triangular similarity classes in the size 5 or less and infinitely classes in the size 6 or higher.Proposition 4.
Letx=xij and y=yij be two elements of tn,ℂ.(1)
Ifx and y are similar under t∗, then xii=yii for 1≤i≤n(2)
Ifx is semisimple, then x is similar to diagx11,x22,…,xnn under t∗Proof.
Ifx=zyz−1 for an invertible upper triangular matrix z=zij with z−1=zij, then the i,i element yields(11)xii=∑i≤j≤kzijyjkzki=ziiziiyii=yii.
Here, we have used the facts thatz−1 is upper triangular and zii=1/zii.
Ifx is semisimple, then there is an invertible matrix t and a diagonal matrix d such that x=t−1dt. By the QR decomposition, we have t=ur, where u is a unitary matrix and r is an upper triangular matrix. Then,(12)rxr−1=u−1du.
From the above equation, the left is upper triangular and the right is normal. By the fact that an upper triangular normal matrix must be diagonal, we get thatx is similar to a diagonal matrix under t∗. According to the first result of the proposition, we get the second.For anyh∈h, define(13)Δh=λ∈Φ|λh=1.We callh∈h a maximal element with one-eigenvalue if for any h˜∈h with E1adh⊂E1adh˜, then h=h˜. Denote by D the set of all the maximal elements with one-eigenvalue.For anyx∈n, we can write x as x=∑λ∈Φxλ. Define(14)Δx=λ∈Φ|xλ≠0.Lemma 4.
With the notations as above, the following claims hold:(1)
Leth∈h. If spanΔh≠h∗, then there exists h˜∈h such that Δh⊂Δh˜ and spanΔh˜=h∗.(2)
Leth,h˜∈h. If spanΔh=h∗ and Δh⊂Δh˜, then h=h˜.(3)
Leth∈h. Then, h∈D if and only if spanΔh=h∗.Proof.
(1)
Take a maximal linearly independent subsetλ1,…,λs of Δh and extend it to a basis λ1,…,λn−1⊂Φ for h∗. It is apparent that there exists only one element h˜∈h such that λih˜=1 for all 1≤i≤n−1. Obviously, spanΔh˜=h∗. For any λ∈Δh, there exist a1,…,as∈ℂ such that λ=∑i=1saiλi. Thus,(15)1=λh=∑i=1saiλih=∑i=1sai.Then,(16)λh˜=∑i=1saiλih˜=∑i=1sai=1.Therefore,λ∈Δh˜.(2)
Notice that bothh and h˜ are the solution of λl=1 for all λ∈Δh. Since spanΔh=h∗, by the uniqueness of solution, we know h=h˜.(3)
The “if” part is from (2) and the “only if” part is from (1).From the above lemma, we knowD is a finite set. It seems that it is an interesting question to determine the number of elements in D.Proposition 5.
Letx∈n. If the linear equations(17)λh=1,for allλ∈Δx,have at least one solution h in h, then x∈Nt. Particularly, if Δx consists of linearly independent elements, then x∈Nt.Proof.
From(18)x∈∐λ∈Δxtλ⊂∐λh=1tλ=E1adh,we see x is strongly ad-nilpotent.Example 1.
Letk≠0. Consider(19)A=01k000000100110000010∈n6,ℂ.
Then,(20)ΔA=α1,α1+α2,α2+α3+α4,α3,α3+α4,α5.
It is easy to obtain the unique solution(21)h=diag1.5,0.5,0.5,−0.5,−0.5,−1.5satisfying λh=1 for all λ∈ΔA. So, A is a strongly ad-nilpotent element of t6,ℂ.Theorem 2.
With the notations as above, we have(22)Nt=∪t∈t∗∪y∈DtE1adyt−1.Proof.
For anyx∈Nt, by Theorem 1, we have(23)y,x=x,for some semisimple elementyoft.
Without loss of generality, we can assumetry=0. By Proposition 4, there exists t∈t∗ such that t−1yt=h for some h∈h. So, t−1xt∈E1adh. By Lemma 4, there is h˜∈D such that E1adh⊂E1adh˜. Therefore, x∈tE1adh˜t−1.


## 5. The Main Results

Theorem 3.
Forn≤6, we have(24)Ntn,ℂ=nn,ℂ.Proof.
The casen=1 is trivial. By Corollary 1, it remains to prove n⊂Nt. Since every upper triangular matrix is similar to a direct sum of indecomposable matrices under t∗, we need only to consider the indecomposable elements. The canonical forms of all the indecomposable nilpotent matrices in nn,ℂ2≤n≤6 are given by [6] (list in Theorem 2.1 and Theorem 2.2 in [6]). In the lists, the last matrix has been proved to be strongly ad-nilpotent in Example 1. Apart from the last matrix, it is easy to check that Δx consists of linearly independent elements for any x in the rest matrices. By Proposition 5, we know x∈Nt, which completes the proof.Corollary 2.
Ifn≤6, then for any A∈nn,ℂ, the matrix equation(25)x,A=A,has at least one solution in tn,ℂ.Proposition 6.
Let(26)A=0110000000010010000110000010∈n7,ℂ.Then,A is not strongly ad-nilpotent in t7,ℂ.Proof.
Assume thatA is strongly ad-nilpotent. By Theorem 1, there exists x=xij∈t such that(27)x,A=A.
We can setx77=0; otherwise, replace x by x+cI for some suitable c∈ℂ. First, comparing the 1,2,3,4,6,7 elements of (27), respectively, we have(28)x11−x22=x33−x44=x66−x77=1.
Thus,x66=1. Next, from the 2,5,2,6,5,7,4,6 elements of (27), we obtain(29)x24=0,x22−x66+x24=1,x56=0,x44−x66−x56=1,which deduces x22=x44=2. By (28), we get x33=3. Then, from the 2,4,1,3 elements of (27), we obtain(30)x23=0,x11−x33−x23=1,which forces x11=4. Therefore, by (28), we see x22=3, contrary to x22=2.From Theorem3 and Proposition 6, we immediately get the following main result.Theorem 4.
Lettn,ℂn≥2 be the Lie algebra of upper triangular n×n complex matrices, nn,ℂ be the subalgebra of strictly upper triangular matrices, and Ntn,ℂ be the set of all strongly ad-nilpotent elements of tn,ℂ. Then,(31)Ntn,ℂ⊂nn,ℂ,with equality holds if and only if n≤6.Although forn>6, Ntn,ℂ is a proper subset of nn,ℂ as well as the set ℂI+nn,ℂ of all ad-nilpotent elements of tn,ℂ, we have the following result.Theorem 5.
For anyn≥1, we have(32)Etn,ℂ=Inttn,ℂ.Proof.
We supposen≥2, the case n=1 being trivial. Let Zt denote the center of t. Since(33)Ker ad=Zt=ℂI,we have Ker ad∩n=0. Thus, adtn is isomorphic to n. Let U be the Lie group of the unipotent upper triangular complex matrices. Then, U=expn. Since n is nilpotent and U is simply connected, Intt is isomorphic to U.
For anyi<j, it is clear that eij∈Nt by Proposition 5. Then, we get(34)expkeij=I+keij,for allk∈ℂ,i<j.
It is well known that any element ofU can be written as a product of some matrices of the form I+keij,i<j. Hence, the subgroup of U generated by all expx,x∈Nt, is just U itself. Therefore, the subgroup of Intt generated by all expadx,x∈Nt, is Intt itself; that is, Et=Intt.



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