One of the most active and fertile subjects in matrix theory during the past one hundred years is the linear preserver problem (LPP). The earliest paper on such problem dates back to 1897 (see [1]), and a great deal of effort were devoted to the study of this type of question since then. One may consult the survey paper [2-3] for details. It is one of the important linear preserver problems to classify commutativity preserving linear maps on matrix spaces or algebras. A linear map $\varphi$ on an algebra or a matrix space $\mathcal{A}$ is said to be commutativity preserving in both directions when the condition $ab=ba$ holds if and only if $\varphi(a)\varphi(b)=\varphi(b)\varphi(a)$. Commutativity preserving linear maps on spaces of matrices or operators were considered by several authors, see [4-12]. There are several motivations to study this kind of maps. Problems concerning commutativity preserving maps are closely related to the study of Lie homomorphisms. Every associative algebra $\mathcal{A}$ becomes a Lie algebra if we introduce the Lie bracket $[a, b]$ by $[a, b]=ab-ba$ for $a, b\in \mathcal{A}$. A linear map $\phi: \mathcal{A}\rightarrow \mathcal{B}$ is called a Lie homomorphism if $\phi([a, b])=[\phi(a), \phi(b)]$ for every pair $a, b\in \mathcal{A}$. It is clear that every Lie homomorphism preserves commutativity. The assumption of preserving commutativity can be reformulated as the assumption of preserving zero Lie brackets. Let $L$ be a Lie algebra over a field, $\varphi$ a linear map of $L$. We say that $\varphi$ preserves zero Lie brackets in both directions if for every pair $x, y\in L$, we have $[x, y]=0$ if and only if $[\varphi(x), \varphi(y)]=0$. Choi et al.[6] mentioned that the results on linear maps preserving commutativity can be viewed in the content of Lie algebra, where one assumes that the linear map preserves zero products and the conclusion is that the map "essentially" preserves all products. Marcoux and Sourour [8] also pointed out that the linear maps that preserve zero Lie brackets in both directions differ only slightly from those that preserve all Lie brackets. But this assertion is not true here. In this paper, we obtain three types of linear maps which preserve zero Lie brackets in both directions but fail to preserve all Lie brackets.

Let $F$ be an arbitrary field and $F^{*}$ the group consisting of all non-zero elements of $F$. Let $F^{m\times n}$ denote the set of all $m\times n$ matrices over $F$, $E^{(n)}$ the $n\times n$ identity matrix ($E^{(m)}$ is abbreviated to $E$), $gl(n, F)$ the general linear Lie algebra over $F$. For $A\in F^{n\times n}$, $A^{'}$ denotes the transpose of $A$. Let $T(n, F)$ (resp., $S(n, F)$) be the subalgebra of $gl(n, F)$ consisting of all upper triangular (resp., strictly upper triangular) matrices, $T^{*}(n, F)$ the group consisting of all invertible elements in $T(n, F)$. Set $I=\left[\begin{array}{cc}0&E\\E&0\end{array}\right]$. The orthogonal algebra $o(2m, F)$ is defined to be the subalgebra of $gl(2m, F)$ consisting of all $X\in gl(2m, F)$ satisfying $X'I=-IX$. Let

$ l_{2m}(F)=\left\{\left[\begin{array}{cc}A&B\\0&-A'\end{array}\right] |\ A\in S(m, F), B\in F^{m\times m}, \ B'=-B\right\}. $

It is a maximal nilpotent subalgebra of $o(2m, F)$. In this paper, by using the main theorem of [7], we shall describe all the linear maps preserving zero Lie brackets in both directions of $l_{2m}(F)$ when $m\geq5$. The main idea of this paper is to reduce the problem on $l_{2m}(F)$ to that on $S(m, F)$.

For $1\leq i, j\leq m, $ let $E_{ij}$ denote the $2m\times 2m$ matrix whose $(i, j)$-entry is $1$ and all other entries are $0$; $E_{i, -j}$ the $2m\times 2m$ matrix whose $(i, j+m)$-entry is $1$ and all other entries are $0$; $E_{-i, j}$ the $2m\times 2m$ matrix whose $(i+m, j)$-entry is $1$ and all other entries are $0$; $E_{-j, -i}$ the $2m\times 2m$ matrix whose $(j+m, i+m)$-entry is $1$ and all other entries are $0$. For $1\leq i, j\leq m, $ let $e_{ij}$ denote the $m\times m$ matrix whose $(i, j)$-entry is $1$ and all other entries are $0$. For $a\in F, \ 1\leq i<j\leq m$, set

$ \begin{eqnarray*}&&T_{ij}(a)=a(E_{ij}-E_{-j, -i}), ~~ T_{ij}=\{T_{ij}(a)\ |\ a\in F\};\\ && T_{i, -j}(a)=a(E_{i, -j}-E_{j, -i}), ~~ T_{i, -j}=\{T_{i, -j}(a)\ | \ a\in F\}.\end{eqnarray*} $

Let $l_{2m}^{(1)}(F)=[l_{2m}(F), l_{2m}(F)]$, $l_{2m}^{(2)}(F)=[l_{2m}(F), l_{2m}^{(1)}(F)], \cdots, $ $l_{2m}^{(k)}(F)=[l_{2m}(F), l_{2m}^{(k)}(F)]$, $ \cdots, $ and denote

$ \begin{eqnarray} p(F)&=&T_{1m}+\sum\limits_{1\leq i<j\leq m, i+j\leq m+1}T_{i, -j}, \ \ \ \ \ u(F)=\sum\limits_{1\leq i<j\leq m, i+j\leq m}T_{i, -j}, \nonumber\\ x(F)&=&\sum\limits_{1\leq i<j\leq m-1}T_{i, -j}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y(F)=x(F)+T_{1m}+T_{1, -m}, \nonumber\\ q(F)&=&y(F)+T_{2m}+T_{2, -m}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z(F)=\sum\limits_{1\leq i<j\leq m}T_{i, -j}+\sum\limits_{1\leq i\leq m-1}T_{im}, \nonumber\\ s(F)&=&z(F)+T_{1, m-1}. \nonumber \end{eqnarray} $

Let $t(F)=\left\{\left[\begin{array}{cc}A&0\\0&-A^{'}\end{array}\right]\bigg|\ A\in S(m, F)\right\}, $ $w(F)=\left\{\left[\begin{array}{cc}0&B\\0&0\end{array}\right] \mid B\in F^{m\times m}, B^{'}=-B\right\}, $ $v(F)=\left\{\mbox{diag}(A, 0, -A^{'}, 0)|A\in S(m-1, F)\right\}.$ Then $l_{2m}(F)=t(F)+w(F)$.

Let $L$ be a Lie algebra. The center of $L$ is $z(L)=\{z\in L \mid [x, z]=0\ {\mbox{for all}}\ x\in L\}$, the centralizer of a subset $X$ of $L$ is $C_{L}(X)=\{x\in L \mid [x, X]=0\}.$ It is easy to know that the center of $l_{2m}(F)$ is $l_{2m}^{(2m-4)}(F)$ which is equal to $T_{1, -2}$, and the center of $q(F)$ is $x(F)$. The centralizer of $p(F)$ (resp., $u(F)$; resp., $x(F)$) in $l_{2m}(F)$ is $y(F)$ (resp., $s(F)$; resp., $z(F)$).

Denote by $\mathcal{T}$ the set of all linear maps of $l_{2m}(F)$ that preserve zero Lie brackets in both directions and by $\mathcal {T}^{'}$ the set of all bijections in $\mathcal {T}$. Denote by $1$ the identity map on $l_{2m}(F)$. It is clear that for $\varphi\in \mathcal{T}$ and a linear function $f$ from $l_{2m}(F)$ to $F$, the map $\varphi+f: X\mapsto \varphi(X)+f(X)T_{1, -2}(1)$ is in $\mathcal{T}$.

The following lemma is obvious.

Lemma 2.1 (ⅰ) If $\varphi\in \mathcal {T}$, then $Ker\varphi\subseteq T_{1, -2}$.

(ⅱ) $\varphi\in\mathcal {T}^{'}$ if and only if $\varphi(T_{1, -2}(1))\neq 0$.

(ⅲ) If $\varphi\in\mathcal {T}^{'}$, then $\varphi(T_{1, -2}(1))=T_{1, -2}(c)$ for some $c\in F^{*}$.

It is obvious that $t(F)$ is isomorphic to $S(m, F)$. Cao et al.[7] have described the linear maps preserving commutativity in both directions on $S(m, F)$. We now transfer them to $t(F)$ for later use. $t(F)$ has the following standard maps that preserve zero Lie brackets in both directions.

(a) $\psi_{t, c}: X\mapsto cX$, where $c$ is a constant in $F^{*}$.

(b) Int$_{t}P: X\mapsto P^{-1}XP$, where $P=\left[\begin{array}{cc}A&0\\0&A^{'-1}\end{array}\right]$ with $A\in T^{*}(m, F)$.

(c) $\eta_{t, f}: X\mapsto X+f(X)T_{1m}(1)$, where $f: t(F)\rightarrow F$ is a linear function.

(d) $\omega=1$ or $\omega: X=\left[\begin{array}{cc}A&0\\0&-A^{'}\end{array}\right]\mapsto \left[\begin{array}{cc}-RA^{'}R&0\\0&RAR\end{array}\right]$ with $R=e_{1m}+e_{2, m-1}+\cdots+e_{m-1, 2}+e_{m1}$.

(e) $\mu_{t, b}^{(ij)}$ for $b\in F$, $i=1, m$ and $j=1, 2$ are defined by

$ \begin{eqnarray*}&&\mu_{t, b}^{(11)}: X=\sum\limits_{1\leq i<j\leq m}T_{ij}(a_{ij})\mapsto X+T_{2m}(ba_{12});\\ &&\mu_{t, b}^{(m1)}: X=\sum\limits_{1\leq i<j\leq m}T_{ij}(a_{ij})\mapsto X+T_{1, m-1}(ba_{m-1, m});\\ &&\mu_{t, b}^{(12)}: X=\sum\limits_{1\leq i<j\leq m}T_{ij}(a_{ij})\mapsto X+T_{2m}(ba_{13})+T_{3m}(ba_{12});\\ &&\mu_{t, b}^{(m2)}: X=\sum\limits_{1\leq i<j\leq m}T_{ij}(a_{ij})\mapsto X+T_{1, m-2}(ba_{m-1, m})+T_{1, m-1}(ba_{m-2, m}).\end{eqnarray*} $

We call the linear maps of types (a)-(e) defined above standard maps of $t(F)$. By Lemma 2.2 [7], 2.3 [7] and Theorem 1.1 [7], we have the following theorem.

Theorem 3.1  Let $m\geq5$. Then a linear map $\varphi$ of $t(F)$ preserves commutativity in both directions if and only if $\varphi$ is of the form

$ \begin{eqnarray} \varphi&=&\psi_{t, c}{\mbox{Int}}_{t}T\omega\mu_{t, b_{4}}^{(m2)}\mu_{t, b_{3}}^{(12)}\mu_{t, b_{2}}^{(m1)} \mu_{t, b_{1}}^{(11)}\eta_{t, f}, \nonumber \end{eqnarray} $

where $\psi_{t, c}, \ {\mbox{Int}}_{t}T, \ \omega, \ \mu_{t, b_{4}}^{(m2)}, \ \mu_{t, b_{3}}^{(12)}, \ \mu_{t, b_{2}}^{(m1)}, \ \mu_{t, b_{1}}^{(11)}, \ \eta_{t, f}$ are the standard maps of $t(F)$.

It is obvious that $\mathcal {T}^{'}$ forms a group under multiplication of maps. We now define some standard maps of $l_{2m}(F)$ which preserve zero Lie brackets in both directions, then we use them to prove the main theorem of this paper. It is easy to check that the following linear maps of $l_{2m}(F)$ are all in $\mathcal {T}$ when $m\geq5$.

(1) Inner automorphisms

For $A\in T^{*}(m, F)$ and $B^{'}=-B\in F^{m\times m}$, set $X=\left[\begin{array}{cc}A&AB\\0&A^{'-1}\end{array}\right]$. The map Int$ X: Y\mapsto X^{-1}YX$ is an automorphism of $l_{2m}(F)$, called the inner automorphism of $l_{2m}(F)$ induced by $X$.

(2) Graph automorphisms

Let $\pi=E^{(2m)}-E_{mm}-E_{-m, -m}+E_{m, -m}+E_{-m, m}.$ The map $\mu_{\pi}: Y\mapsto\pi Y\pi$ is an automorphism of $l_{2m}(F)$, called the graph automorphism of $l_{2m}(F)$.

(3) Generalized diagonal automorphisms

For $g\in F^{*}$, let $G={\mbox{diag}}(E, gE)$. The map $\xi_{g}: l_{2m}(F)\rightarrow l_{2m}(F)$ defined by $Y\mapsto G^{-1}YG$ is an automorphism of $l_{2m}(F)$, called the generalized diagonal automorphism of $l_{2m}(F)$ induced by $G$.

(4) Central maps

Let $f$ be a linear map from $l_{2m}(F)$ to $F$, we define the map $\eta_{f}: l_{2m}(F)\rightarrow l_{2m}(F)$ by $\eta_{f}(Y)=Y+f(Y)T_{1, -2}(1).$ It is easy to check that $\eta_{f}\in \mathcal {T}$, called the central map. If $f$ satisfies the additional conditions: $f([X, Y])=0$ for any $X, Y\in l_{2m}(F)$ and $1+f(T_{1, -2}(1))\neq 0$, then $\eta_{f}$ is also a Lie automorphism of $l_{2m}(F)$.

(5) Sub-central automorphisms

For $h\in F$, $Y=\sum\limits_{1\leq i<j\leq m}T_{ij}(a_{ij})+\sum\limits_{1\leq k<l\leq m}T_{k, -l}(a_{k, -l})\in l_{2m}(F)$. The map $\lambda_{h}:l_{2m}(F)\rightarrow l_{2m}(F)$ defined by $\lambda_{h}(Y)=Y+T_{1, -3}(ha_{23})$ is an automorphism of $l_{2m}(F)$, called the sub-central automorphism of $l_{2m}(F)$ induced by $h$.

(6) Extremal maps

For $a, b, c\in F$, $Y=\sum\limits_{1\leq i<j\leq m}T_{ij}(a_{ij})+\sum\limits_{1\leq k<l\leq m}T_{k, -l}(a_{k, -l})\in l_{2m}(F)$, we define the map $\rho_{a, b, c}: l_{2m}(F)\rightarrow l_{2m}(F)$ by

$ \rho_{a, b, c}(Y)=Y+T_{1, -3}(aa_{13}+ca_{24})+T_{1, -4}(ba_{13}+ca_{23})-T_{2, -3}(aa_{23}+ba_{24}). $

It is easy to check that $\rho_{a, b, c}\in \mathcal {T}^{'}$, but it usually fails to be an automorphism of $l_{2m}(F)$. We call it the extremal map of type Ⅰ.

(7) Scalar multiplication

For $c\in F^{*}$, we define the map $\psi_{c}: l_{2m}(F)\rightarrow l_{2m}(F)$ by $\psi_{c}(X)=cX$. Clearly, $\psi_{c}\in \mathcal {T}^{'}$, called the multiple map. It is easy to see that $\psi_{c}$ is an automorphism of $l_{2m}(F)$ if and only if $c=1$.

Lemma 4.1 (ⅰ)  $\xi_{g} {\mbox{Int}} T_{1}={\mbox{Int}} T_{2}\xi_{g}$, where $T_{2}=\xi_{g}(T_{1})$.

(ⅱ) $\xi_{g}\lambda_{h_{1}}=\lambda_{h_{2}}\xi_{g}$, where $h_{2}=gh_{1}$.

(ⅲ) $\xi_{g} \rho_{a_{1}, b_{1}, c_{1}}=\rho_{a_{2}, b_{2}, c_{2}}\xi_{g}$, where $a_{2}=ga_{1}, b_{2}=gb_{1}, c_{2}=gc_{1}$.

(ⅳ) ${\mbox{Int}} T\mu_{\pi}=\mu_{\pi} {\mbox{Int}} T$ for $T={\mbox{diag}(A, 1, A^{'-1}, 1})$ with $A\in T^{*}(m-1, F)$.

(ⅴ) $\psi_{c}$ commutes with every linear map on $l_{2m}(F)$. In particular, $\psi_{c}$ commutes with every standard map.

Proof The proof is trivial, we omit it.

Let $\varphi$ be a linear map preserving zero Lie brackets in both directions. Throughout this section, without loss of generality, we assume that $\varphi$ is bijective and $m\geq5$. In fact, if $\varphi$ is not bijective, we have $\varphi(T_{1, -2}(1))=0$ by Lemma 2.1. Let $f$ be a linear function from $l_{2m}(F)$ to $F$ such that $f(T_{1, -2}(1))\neq0$, then $\varphi+f\in \mathcal {T}^{'}$ again by Lemma 2.1. Thus $\varphi$ can be replaced with $\varphi+f$. For $X\in l_{2m}(F)$, we denote by $C(X)$ the centralizer of $X$ in $l_{2m}(F)$, i.e., $C(X)=\{Y\in l_{2m}(F)\mid [X, Y]=0\}$. In order to prove the main result in this paper, we need to give some lemmas first.

Lemma 5.1  Let $\varphi\in \mathcal {T}^{'}$, then $p(F)$ and $y(F)$ are stable under $\varphi$.

Proof  If we can prove that $p(F)$ is invariant under $\varphi$, then $y(F)$, being the centralizer of $p(F)$ in $l_{2m}(F)$, is also invariant under $\varphi$. So for our goal, it suffices to prove that $p(F)$ is invariant under $\varphi$. It is clear that the set

$ \mathcal {B}=\{{T_{i, -j}(1), T_{1m}(1)| 1\leq i<j\leq m, i+j\leq m+1}\} $

is the canonical basis of $p(F)$, so we only need to show that $\varphi(X)\in p(F)$ for all $X\in\mathcal {B}$. It is not difficult to check that ${\mbox{dim}}\ C(X)\geq(m-1)m-(m-2)$ for any $X\in \mathcal {B}$. Since $\varphi$ is bijective and preserves zero Lie brackets in both directions, we have

$ \begin{eqnarray} \label{firstfor} {\mbox{dim}}\ C(\varphi(X))={\mbox{dim}}\ C(X)\geq(m-1)m-(m-2). \end{eqnarray} $

(5.1)

In the following, we prove that $\varphi(X)\in p(F)$ for all $X\in \mathcal {B}$ by two steps.

Step 1  $\varphi(X)\in w(F)+T_{1m}$ for all $X\in\mathcal {B}$.

If there exists some $X\in \mathcal {B}$ such that $\varphi(X)\notin w(F)+T_{1m}$, then we can assume that $\varphi(X)=\sum_{1\leq i<j\leq m}T_{ij}(a_{ij})+W$ with some $a_{st}\neq 0$ for $1\leq s<t\leq m, (s, t)\neq (1, m)$ and $W\in w(F)$. Let $i_{0}, j_{0}$ be such that $a_{i_{0}j_{0}}\neq 0, (i_{0}, j_{0})\neq (1, m)$ and $a_{i_{0}, k}=0$ for all $k<j_{0}$ and $a_{k, j_{0}}=0$ for all $k>i_{0}$. Set

$ \begin{eqnarray} M_{1}&=&E^{(2m)}-\sum\limits_{k=1}^{m-j_{0}}T_{j_{0}, j_{0}+k} (a_{i_{0}j_{0}}^{-1}a_{i_{0}, j_{0}+k}), \nonumber\\\ M_{2}&=&E^{(2m)}+\sum\limits_{k=1}^{i_{0}-1}T_{ki_{0}}(a_{i_{0}j_{0}}^{-1}a_{kj_{0}}), \nonumber\\ v_{0}&=&\sum\limits_{k=1}^{m-j_{0}}T_{j_{0}, j_{0}+k}+\sum\limits_{l=1}^{i_{0}-1}T_{l, -j_{0}}+ \sum\limits_{l=i_{0}+1}^{j_{0}-1}T_{l, -j_{0}}+ \sum\limits_{h=1}^{m-j_{0}}T_{j_{0}, -(j_{0}+h)}+\sum\limits_{k=1}^{i_{0}-1}T_{ki_{0}}.\nonumber \end{eqnarray} $

Then $v_{0}$ is a subspace of $l_{2m}(F)$ and $v_{0}\cap C((M_{1}M_{2})^{-1}\varphi(X)M_{1}M_{2})=\{0\}$. It is clear that dim $v_{0}\geq m-1$. So

$ \begin{eqnarray} {\mbox{dim}}\ C(\varphi(X))={\mbox{dim}}\ C((M_{1}M_{2})^{-1}\varphi(X)M_{1}M_{2})\leq m(m-1)-(m-1). \end{eqnarray} $

(5.2)

In contradiction with (5.1). So $\varphi(X)\in w(F)+T_{1m}$ for all $X\in\mathcal {B}$.

Step 2  $\varphi(X)\in p(F)$ for all $X\in\mathcal {B}$.

By Step 1, we know that $\varphi(X)\in w(F)+T_{1m}$ for all $X\in \mathcal {B}$. If there exists some $X\in \mathcal {B}$ such that $\varphi(X)\in w(F)+T_{1m}$ but $\varphi(X)\notin p(F)$, then we may write $\varphi(X)=\sum_{1\leq i<j\leq m}T_{i, -j}(b_{i, -j})+T_{1m}(b_{1m}), $ and there exist some $b_{p, -q}\neq 0$ with $p+q>m+1$ and $p< q$. Let $p_{0}, q_{0}$ be such that $b_{p_{0}, -q_{0}}\neq 0$ and $b_{p_{0}, -k}=0$ for all $k>q_{0}$ and $b_{k, -q_{0}}=0$ for all $k>p_{0}$. Set

$ \begin{eqnarray} N_{1}&=&E^{(2m)}+\sum\limits_{k=1}^{p_{0}-1}T_{k, p_{0}}(b_{p_{0}, -q_{0}}^{-1}b_{k, -q_{0}}), \nonumber\\ N_{2}&=&E^{(2m)}-\sum\limits_{k=1}^{p_{0}-1}T_{k, q_{0}}(b_{p_{0}, -q_{0}}^{-1}b_{k, -p_{0}})+ \sum\limits_{k=p_{0}+1}^{q_{0}-1}T_{k, q_{0}}(b_{p_{0}, -q_{0}}^{-1}b_{p_{0}, -k}).\nonumber \end{eqnarray} $

Then the entries of $N_{2}^{-1}N_{1}^{-1}\varphi(X)N_{1}N_{2}$ in the $p_{0}$-th row, $q_{0}$-th row, $(p_{0}+m)$-th column and $(q_{0}+m)$-th column are all zero except the $(p_{0}, q_{0}+m)$-entry and $(q_{0}, p_{0}+m)$-entry. Let

$ \begin{eqnarray} v_{1}=T_{1p_{0}}+\cdots+T_{p_{0}-1, p_{0}}+T_{1q_{0}}+\cdots+T_{p_{0}-1, q_{0}}+ T_{p_{0}+1, q_{0}}+\cdots+T_{q_{0}-1, q_{0}}\nonumber. \end{eqnarray} $

Then $v_{1}$ is a subspace of $l_{2m}(F)$. One may check that $v_{1}\cap C(N_{2}^{-1}N_{1}^{-1}\varphi(X)N_{1}N_{2})=\{0\}$ and dim $v_{1}=p_{0}-1+q_{0}-2>m-2$. So

$ \begin{eqnarray} {\mbox{dim}}\ C(\varphi(X))={\mbox{dim}}\ C(N_{2}^{-1}N_{1}^{-1}\varphi(X)N_{1}N_{2})<m(m-1)-(m-2). \end{eqnarray} $

(5.3)

In contradiction with (5.1). So $\varphi(X)\in p(F)$ for all $X\in \mathcal {B}$.

Since $\mathcal {B}$ is a basis of $p(F)$ and $\varphi$ is a bijective linear map, we have $\varphi(p(F))=p(F)$. That is to say $p(F)$ is stable under $\varphi$.

Lemma 5.2  Let $\varphi\in\mathcal {T}^{'}$, then $l_{2m}^{(k)}(F)$ is invariant under $\varphi$ for every $1\leq k\leq 2m-4$. Furthermore, $s(F)$, being the centralizer of $u(F)$ which exactly is $l_{2m}^{(m-1)}(F)$, is also invariant under $\varphi$.

Proof  The process, being similar to Lemma 5.1, is omitted.

Lemma 5.3  Let $\varphi\in\mathcal {T}^{'}$, then $q(F)$, $x(F)$ and $z(F)$ are invariant under $\varphi$.

Proof  If we can prove that $q(F)$ is invariant under $\varphi$, then $x(F)$, being the center of $q(F)$, is invariant under $\varphi$. Furthermore, $z(F)$, being the centralizer of $x(F)$ in $l_{2m}(F)$, is also invariant under $\varphi$. For our goal, we only need to prove that $q(F)$ is invariant under $\varphi$. Let $\varphi\in\mathcal {T}^{'}$. Since $\varphi(y(F))=y(F)\subseteq q(F), $ it suffices to prove that $\varphi(T_{2m}(1))$ and $\varphi(T_{2, -m}(1))$ are all contained in $q(F)$. Since $T_{2m}(1)\in l_{2m}^{(m-3)}(F)$ and $T_{m-2, -(m-1)}(1)\in y(F)$, we may assume that

$ \varphi(T_{2m}(1))\equiv T_{1, m-1}(a_{1, m-1})+T_{2m}(a_{2m})+T_{2, -m}(a_{2, -m})\ ({\mbox{mod}}\ y(F)), $

$ \varphi(T_{m-2, -(m-1)}(1))=\sum\limits_{1\leq i<j\leq m-1}T_{i, -j}(b_{i, -j})+T_{1, -m}(b_{1, -m})+T_{1m}(b_{1m}), $

where $b_{m-2, -(m-1)}\neq 0$. By considering the action of $\varphi$ on $[T_{2m}(1), T_{m-2, -(m-1)}(1)]=0$, we have $a_{1, m-1}b_{m-2, -(m-1)}=0$, so $a_{1, m-1}=0.$ That is to say $\varphi(T_{2m}(1))\in q(F)$. We can also prove that $\varphi(T_{2, -m}(1))\in q(F)$ in the similar way. So $\varphi(q(F))\subseteq q(F)$. Since $\varphi$ is invertible, we have $\varphi(q(F))= q(F).$

Now we give the main result of this paper.

Theorem 5.4  Let $m\geq5$. A linear map $\varphi$ of $l_{2m}(F)$ preserves zero Lie brackets in both directions if and only if $\varphi$ is of the form

$ \begin{eqnarray} \label{firstfour}\varphi=\psi_{c}(\mu_{\pi})^{\delta}{\mbox{Int}}X\lambda_{h}\rho_{a, b, c}\phi_{d}\xi_{g}\eta_{f}, \end{eqnarray} $

(5.4)

where $\psi_{c}, \mu_{\pi}, {\mbox{Int}}X, \lambda_{h}, \rho_{a, b, c}, \phi_{d}, \xi_{g}, \eta_{f}$ are the standard maps preserving zero Lie brackets in both directions, $\delta=0\ {\mbox{or}}\ 1$.

Proof  The "if" part of the theorem is clear. For the "only if" part, we will give the proof by steps. In the following, we assume that $m\geq5$ and $\varphi\in\mathcal {T}^{'}$.

Step 1  There exist $X_{1}={\mbox{diag}}(A, 1, A^{'-1}, 1)$ with $A\in T^{*}(m-1, F)$ and $c_{1}\in F^{*}$ such that ${\mbox{Int}}^{-1}X_{1}\psi_{c_{1}}^{-1}\varphi(T_{ij}(1))\equiv T_{ij}(1)\ ({\mbox{mod}}\ s(F))$ for $1\leq i<j\leq m-1$.

Since $z(F)$ is an ideal of $l_{2m}(F)$ and stable under $\varphi$, then $\varphi$ induces a linear map $\overline{\varphi}$ of $l_{2m}(F)/z(F)$ by $\overline{\varphi}(\overline{Y})=\overline{\varphi(Y)}$, where $\overline{Y}=Y+z(F)\in l_{2m}(F)/z(F)$, $Y\in l_{2m}(F)$. It can be proved that $\overline{\varphi}$ is invertible and preserves zero Lie brackets in both directions. Since $l_{2m}(F)/z(F)$ is isomorphic to $v(F)$, we may directly view $l_{2m}(F)/z(F)$ as $v(F)$. Thus by Theorem 3.1, $\overline{\varphi}$ can be written in the form:

$ \overline{\varphi}=\psi_{v, c_{1}}{\mbox{Int}}_{v}X_{1}\omega\mu_{v, b_{4}}^{(m2)}\mu_{v, b_{3}}^{(12)}\mu_{v, b_{2}}^{(m1)} \mu_{v, b_{1}}^{(11)}\eta_{v, f}, $

where $\psi_{v, c_{1}}, \ {\mbox{Int}}_{v}X_{1}, \ \omega, \ \mu_{v, b_{4}}^{(m2)}, \ \mu_{v, b_{3}}^{(12)}, \ \mu_{v, b_{2}}^{(m1)}, \ \mu_{v, b_{1}}^{(11)}, \ \eta_{v, f}$ are the standard maps of $v(F)$. It is easy to see that $\psi_{v, c_{1}}=\overline{\psi_{c_{1}}}$, ${\mbox{Int}}_{v}X_{1}=\overline{{\mbox{Int}}X_{1}}$. So $\overline{{\mbox{Int}}^{-1}X_{1}\psi_{c_{1}}^{-1}\varphi}= \omega\mu_{v, b_{4}}^{(m2)}\mu_{v, b_{3}}^{(12)}\mu_{v, b_{2}}^{(m1)}\mu_{v, b_{1}}^{(11)}\eta_{v, f}$. Denote ${\mbox{Int}}^{-1}X_{1} \psi_{c_{1}}^{-1}\varphi$ by $\varphi_{1}$.

We are now ready to prove that $\omega=1$, $b_{4}=b_{3}=b_{2}=b_{1}=0$. Since $x(F)$ and $p(F)$ are stable under $\varphi_{1}$, $T_{2, -(m-1)}(1)\in x(F)\cap p(F)$, we may write

$ \varphi_{1}(T_{2, -(m-1)}(1))=\sum\limits_{1\leq i<j\leq m-1, i+j\leq m+1}T_{i, -j}(a_{i, -j}), $

where $a_{2, -(m-1)}\neq 0$. If $\omega\neq1$, we have

$ \begin{eqnarray} \varphi_{1}(T_{13}(1)) &\equiv&-T_{1, m-2}(b_{3})-f(T_{13}(1))T_{1, m-1}(1)-T_{m-2, m-1}(1) \ ({\mbox{mod}}\ z(F)).\nonumber \end{eqnarray} $

From $[T_{13}(1), T_{2, -(m-1)}(1)]=0$, we have $[\varphi_{1}(T_{13}(1)), \varphi_{1}(T_{2, -(m-1)}(1))]=0$, which implies that $a_{2, -(m-1)}=0$, a contradiction. So $\omega=1$.

Since $x(F)$ is stable under $\varphi_{1}$, we may write

$ \varphi_{1}(T_{m-2, -(m-1)}(1))=\sum\limits_{1\leq i<j\leq m-1}T_{i, -j}(b_{i, -j}), $

where $b_{m-2, -(m-1)}\neq0$. By considering the action of $\varphi_{1}$ on $[T_{12}(1), T_{m-2, -(m-1)}(1)]$ $=0$, we have $b_{m-2, -(m-1)}b_{3}=b_{m-2, -(m-1)}b_{1}=0$, which implies that $b_{1}=b_{3}=0$.

Because $x(F)$ and $u(F)$ are stable under $\varphi_{1}$, $T_{2, -(m-2)}\in x(F)\cap u(F)$, we may write

$ \varphi_{1}(T_{2, -(m-2)}(1))=\sum\limits_{1\leq i<j\leq m-1, i+j\leq m}T_{i, -j}(c_{i, -j}). $

Since $[\varphi_{1}(T_{2, -(m-2)}(1)), \varphi_{1}(T_{1, m-2}(1))]\neq 0$, we have $c_{2, -(m-2)}\neq0.$ By applying $\varphi_{1}$ on $[T_{m-3, m-1}(1), T_{2, -(m-2)}(1)]=0$ and $[T_{m-2, m-1}(1), T_{2, -(m-2)}(1)]=0$, respectively, we have $b_{4}c_{2, -(m-2)}=b_{2}c_{2, -(m-2)}=0, $ which implies that $b_{4}=b_{2}=0$. So $\overline{\varphi_{1}}=\eta_{v, f}$, as desired.

Step 2  There exist a graph automorphism $\mu_{\pi}$ and a generalized diagonal automorphism $\xi_{g}$ such that $\xi_{g}^{-1}(\mu_{\pi}^{-1})^{\delta}\varphi_{1}(T_{1, -m}(1))\equiv T_{1, -m}(1)\ ({\mbox{mod}}\ x(F)), \ {\mbox{where}}\ \delta=0\ {\mbox {or}}\ \delta=1.$

Since $y(F)$ is stable under $\varphi_{1}$ and $T_{1, -m}(1)\in y(F)$, we may assume that

$ \varphi_{1}(T_{1, -m}(1))\equiv T_{1m}(a)+T_{1, -m}(b)\ ({\mbox{mod}}\ x(F)), \ {\mbox{where}}\ a, b\in F. $

It is easy to see that $a$ and $b$ can not be zero simultaneously. Since $q(F)$, $z(F)$ and $l_{2m}^{(m-4)}(F)$ are stable under $\varphi_{1}$, $T_{2, -m}(1)\in q(F)$, $T_{3, -m}(1)\in z(F)\cap l_{2m}^{(m-4)}(F)$, we may write

$ \begin{eqnarray} \varphi_{1}(T_{2, -m}(1))\equiv T_{2m}(c)+T_{2, -m}(d)\ ({\mbox{mod}}\ y(F)), \nonumber\\ \varphi_{1}(T_{3, -m}(1))\equiv T_{3m}(e)+T_{3, -m}(f)\ ({\mbox{mod}}\ q(F)), \nonumber \end{eqnarray} $

where $c, d, e, f\in F$. By applying $\varphi_{1}$ on $[T_{1, -m}(1), T_{2, -m}(1)]=0$, $[T_{2, -m}(1), T_{3, -m}(1)]$ $=0$ and $[T_{1, -m}(1), T_{3, -m}(1)]=0$, respectively, we get $ad+bc=0$, $cf+de=0$ and $af+be=0$. If $ab\neq 0$, then $cd\neq0$ and $ef\neq 0$. So we have

$ \left\{\begin{array} {r@{\quad\quad}l} \frac{a}{b}=-\frac{c}{d}, \\\frac{a}{b}=-\frac{e}{f}, \\\frac{e}{f}=-\frac{c}{d}, \end{array}\right. $

which implies that $\frac{c}{d}=-\frac{c}{d}$. That is to say $cd=0$. This contradiction shows that $ab=0$.

If $a\neq 0$ and $b=0$, let $g=a$ and $\delta=1$; if $b\neq 0$ and $a=0$, let $g=b$ and $\delta=0$. Then

$ \xi_{g}^{-1}(\mu_{\pi}^{-1})^{\delta}\varphi_{1}(T_{1, -m}(1))\equiv T_{1, -m}(1)\ ({\mbox{mod}}\ x(F)), $

where $(\mu_{\pi}^{-1})^{1}=\mu_{\pi}^{-1}$, $(\mu_{\pi}^{-1})^{0}=1$. Denote $\xi_{g}^{-1}(\mu_{\pi}^{-1})^{\delta}\varphi_{1}$ by $\varphi_{2}$.

Step 3  There exists $X_{2}=E^{(2m)}+P_{2}$, $P_{2}\in \sum_{2\leq k\leq m-1}T_{1k}$ such that $w(F)$ is stable under ${\mbox{Int}}^{-1}X_{2}\varphi_{2}$. In particular, ${\mbox{Int}}^{-1}X_{2}\varphi_{2}(T_{i, -m}(1))\equiv T_{i, -m}(a_{i, -m}^{(i)})\ ({\mbox{mod}}\ x(F)), $ where $a_{1, -m}^{(1)}=1, a_{i, -m}^{(i)}\in F^{*}$ for $2\leq i\leq m-1$.

If we can prove the latter assertion, then we shall get the former one by the fact that ${\mbox{Int}}^{-1}X_{2}\varphi_{2}(x(F))=x(F)$ and the latter assertion. Since $z(F)$ is stable under $\varphi_{2}$ and $T_{i, -m}(1)\in z(F)$, we may assume that

$ \varphi_{2}(T_{i, -m}(1))\equiv\sum\limits_{k=1}^{m-1}(T_{km}(a_{km}^{(i)})+T_{k, -m}(a_{k, -m}^{(i)}))\ ({\mbox{mod}}\ x(F)), $

where $2\leq i\leq m-1.$ For $2\leq k\neq i\leq m-1$, by applying $\varphi_{2}$ on $[T_{k-1, k}(1), T_{i, -m}(1)]=0, $ we have $a_{km}^{(i)}=a_{k, -m}^{(i)}=0$. By considering the action of $\varphi_{2}$ on $[T_{1, -m}(1), T_{i, -m}(1)]=0, $ and $[T_{i, -m}(1), T_{m-1, -m}(1)]=0$, we obtain that $a_{im}^{(i)}=0$ and $a_{1m}^{(i)}=a_{1m}^{(m-1)}=0$. Since $T_{i, -m}(1)\notin y(F)$, we have $a_{i, -m}^{(i)}\neq 0$. Set $X_{2}=E^{(2m)}+T_{12}(a_{1, -m}^{(2)}({a_{2, -m}^{(2)}})^{-1})+T_{13}(a_{1, -m}^{(3)}({a_{3, -m}^{(3)}})^{-1})+\cdots+T_{1, m-1}(a_{1, -m}^{(m-1)}({a_{m-1, -m}^{(m-1)}})^{-1})$ and denote ${\mbox{Int}}^{-1} X_{2}\varphi_{2}$ by $\varphi_{3}$. We see that $\varphi_{3}(T_{i, -m}(1))\equiv T_{i, -m}(a_{i, -m}^{(i)})\ ({\mbox{mod}}\ x(F))$, where $a_{1, -m}^{(1)}=1$ and $a_{i, -m}^{(i)}\in F^{*}$ for $2\leq i\leq m-1$.

Step 4  There exist $X_{3}={\mbox{diag}}(A, A^{'-1})$ with $A\in T^{*}(m, F)$ and $c_{2}\in F^{*}$ such that $\psi_{c_{2}}^{-1} X_{3}\varphi_{3}(T_{ij}(1))\equiv T_{ij}(1)\ ({\mbox{mod}}\ w(F)+T_{1m})\ {\mbox{for}}\ 1\leq i<j\leq m.$

By Step 3, we know that $w(F)$ is stable under $\varphi_{3}$, so $\varphi_{3}$ induces a linear map $\widetilde{\varphi_{3}}$ of $l_{2m}(F)/w(F)$ by $\widetilde{\varphi_{3}}(\widetilde{X})=\widetilde{\varphi_{3}(X)}$ for $\widetilde{X}=X+w(F)\in l_{2m}(F)/w(F)$. It is not difficult to prove that $\widetilde{\varphi_{3}}$ is bijective and preserves zero Lie brackets in both directions. Since $l_{2m}(F)/w(F)$ is isomorphic to $t(F)$, we may directly view $l_{2m}(F)/w(F)$ as $t(F)$. Thus by Theorem 3.1, $\widetilde{\varphi_{3}}$ can be written in the form:

$ \widetilde{\varphi_{3}}=\psi_{t, c_{2}}{\mbox{Int}}_{t} X_{3}\omega\mu_{t, b_{4}}^{(m2)} \mu_{t, b_{3}}^{(12)}\mu_{t, b_{2}}^{(m1)}\mu_{t, b_{1}}^{(11)} \eta_{t, f}, $

where $\psi_{t, c_{2}}, {\mbox{Int}}_{t} X_{3}, \omega, \mu_{t, b_{4}}^{(m2)}, \mu_{t, b_{3}}^{(12)}, \mu_{t, b_{2}}^{(m1)}, \mu_{t, b_{1}}^{(11)}, \eta_{t, f}$ are the standard maps of $t(F)$. It is easy to know that $\widetilde{\psi_{c_{2}}}=\psi_{t, c_{2}}$, $\widetilde{{\mbox{Int}} X_{3}}={\mbox{Int}}_{t}X_{3}$. So $\widetilde{{\mbox{Int}}^{-1}X_{3}\psi_{c_{2}}^{-1}\varphi_{3}}=\omega\mu_{t, b_{4}}^{(m2)} \mu_{t, b_{3}}^{(12)}\mu_{t, b_{2}}^{(m1)}\mu_{t, b_{1}}^{(11)} \eta_{t, f}.$ Denote ${\mbox{Int}}^{-1} X_{3}\psi_{c_{2}}^{-1}\varphi_{3}$ by $\varphi_{4}$.

If $\omega\neq1$, then $\varphi_{4}(T_{1, m-1}(1))\equiv-f(T_{1, m-1}(1))T_{1m}(1)-T_{2m}(1)\ ({\mbox{mod}}\ w(F)).$ It is easy to see that $\varphi_{4}(T_{1, -m}(1))=T_{1, -m}(a)+X_{0}\ {\mbox{for\ some}}\ X_{0}\in x(F)\ {\mbox{and}}\ a\in F^{*}.$ By applying $\varphi_{4}$ on $[T_{1, m-1}(1), T_{1, -m}(1)]=0$, we get $a=0$, a contradiction. So $\omega=1$.

Since $w(F)$ is stable under $\varphi_{4}$, we may write

$ \varphi_{4}(T_{m-1, -m}(1))=\sum\limits_{1\leq i<j\leq m}T_{i, -j}(a_{i, -j}), $

where $a_{m-1, -m}\neq 0$. By applying $\varphi_{4}$ on $[T_{12}(1), T_{m-1, -m}(1)]=0$, we have $a_{m-1, -m}b_{1}$ $=a_{m-1, -m}b_{3}=0$, which implies that $b_{1}=b_{3}=0$. Since $T_{m-2, -(m-1)}(1)\in x(F)$, we may write $\varphi_{4}(T_{m-2, -(m-1)})=\sum_{1\leq i<j\leq m-1}T_{i, -j}(b_{i, -j})$, where $b_{m-2, -(m-1)}\neq0$. By applying $\varphi_{4}$ on $[T_{m-2, m}(1), T_{m-2, -(m-1)}(1)]=0$ and $[T_{m-1, m}(1), T_{m-2, -(m-1)}(1)]=0$, respectively, we get $b_{m-2, -(m-1)}b_{4}=b_{m-2, -(m-1)}b_{2}$ $=0$, which implies that $b_{4}=b_{2}=0$. So $\widetilde{\varphi_{4}}=\eta_{t, f}$, as desired.

Step 5  There exist $X_{4}=E^{(2m)}+\sum_{2\leq i\leq m-1}T_{1i}(({b_{i, -m}^{(i)}})^{-1}b_{1, -m}^{(i)})$ and $X_{5}=E^{(2m)}+f(T_{12}(1))T_{2m}(1)$ such that ${\mbox{Int}}^{-1}X_{5}{\mbox{Int}}^{-1}X_{4}\varphi_{4}(T_{ij}(1))\equiv T_{ij}(1)\ ({\mbox{mod}}\ w(F))\ {\mbox{for}}\ 1\leq i<j\leq m.$

Since $w(F)$ is stable under $\varphi_{4}$, we may suppose that

$ \varphi_{4}(T_{i, -m}(1))\equiv\sum\limits_{k=1}^{m-1}T_{k, -m}(b_{k, -m}^{(i)})\ ({\mbox{mod}}\ x(F))\ {\mbox{for}}\ 2\leq i\leq m-1. $

By applying $\varphi_{4}$ on $[T_{im}(1), T_{i, -m}(1)]=0$, we have that $b_{k, -m}^{(i)}=0$ for $2\leq k\leq m-1$ and $k\neq i$. Since $y(F)$ is stable under $\varphi_{4}$ and $T_{i, -m}(1)\notin y(F)$, we have that $b_{i, -m}^{(i)}\neq0$. Set $X_{4}=E^{(2m)}+\sum_{2\leq i\leq m-1}T_{1i}(({b_{i, -m}^{(i)}})^{-1}b_{1, -m}^{(i)})$. We see that

$ {\mbox{Int}}^{-1}X_{4}\varphi_{4}(T_{i, -m}(1))\equiv T_{i, -m}(b_{i, -m}^{(i)})\ ({\mbox{mod}}\ x(F))\ {\mbox{for}}\ 2\leq i\leq m-1. $

Denote ${\mbox{Int}}^{-1}X_{4}\varphi_{4}$ by $\varphi_{5}$. We can write

$ \varphi_{5}(T_{2, -m}(1))=\sum\limits_{1\leq i<j\leq m-1}T_{i, -j}(b_{i, -j})+T_{2, -m}(b_{2, -m}^{(2)}). $

For $3\leq i<j\leq m-1$, by applying $\varphi_{5}$ on $[T_{ij}(1), T_{2, -m}(1)]=0$ and $[T_{2j}(1), T_{2, -m}(1)]$ $=0$, respectively, we get $f(T_{ij}(1))=f(T_{2j}(1))=0$ and $b_{k, -l}=0$ for $1\leq k\leq 3, 4\leq l\leq m-1$ and $4\leq k<l\leq m-1$. By applying $\varphi_{5}$ on $[T_{1i}(1), T_{2, -m}(1)]=0$, we have $f(T_{1i}(1))=0$ for $4\leq i\leq m-1$. By applying $\varphi_{5}$ on $[T_{im}(1), T_{i, -m}(1)]=0$ for $2\leq i\leq m-1$, we have $f(T_{im}(1))=0$. By applying $\varphi_{5}$ on $[T_{23}(1), T_{m-1, -m}(1)]=0$ and $[T_{13}(1), T_{m-1, -m}(1)]=0$, respectively, we have $f(T_{23}(1))=f(T_{13}(1))=0$. By considering the action of $\varphi_{5}$ on $[T_{3m}(1)-T_{3, m-1}(1), T_{2, -m}(1)+T_{2, -(m-1)}(1)]=0$ and $[T_{1m}(1)-T_{1, m-1}(1), T_{2, -m}(1)+T_{2, -(m-1)}(1)]=0$, we have $f(T_{1m}(1))=0$. Let $X_{5}=E^{(2m)}+f(T_{12}(1))T_{2m}(1)$. Then we have

$ {\mbox{Int}}^{-1}X_{5}\varphi_{5}(T_{ij}(1))\equiv T_{ij}(1)\ ({\mbox{mod}}\ w(F))\ {\mbox{for}}\ 1\leq i<j\leq m. $

Denote ${\mbox{Int}}^{-1} X_{5}\varphi_{5}$ by $\varphi_{6}$.

Step 6  There exist $h\in F$, $X_{6}=E^{(2m)}+W$ with $W\in w(F)$ and $a, b, c\in F$ such that $\rho_{a, b, c}^{-1}\lambda_{h}^{-1}{\mbox{Int}}^{-1}X_{6}\varphi_{6}(T_{ij}(1)) \equiv T_{ij}(1)\ ({\mbox{mod}}\ T_{1, -2})\ {\mbox{for}}\ 1\leq i<j\leq m.$

Suppose that

$ \varphi_{6}(T_{i, i+1}(1))=T_{i, i+1}(1)+\sum\limits_{1\leq k<l\leq m}T_{k, -l}(a_{k, -l}^{(i)}), 1\leq i\leq m-1, $

where $a_{k, -l}^{(i)}\in F$. For $1\leq t\leq m-1$ and $t\neq i-1, i+1$, by applying $\varphi_{6}$ on $[T_{i, i+1}(1), T_{t, t+1}(1)]=0$, we have $a_{t+1, -l}^{(i)}=0$ for $t+2\leq l\leq m$, $l\neq i$ and $a_{k, -(t+1)}^{(i)}=0$ for $1\leq k\leq t-1$, $k\neq i$. By considering the action of $\varphi_{6}$ on $[T_{12}(1), T_{13}(1)]=0$, $[T_{13}(1), T_{14}(1)]=0$ and $[T_{12}, T_{14}]=0$, respectively, we have $a_{2, -3}^{(1)}=0$ and $a_{3, -4}^{(1)}=0$. By applying $\varphi_{6}$ on $[T_{23}(1), T_{14}(1)]=0$, we have $a_{3, -4}^{(2)}=0$ and $a_{4, -5}^{(2)}=0$. For $3\leq k\leq m-1$, by considering the action of $\varphi_{6}$ on $[T_{2, k+1}(1), T_{k, k+1}(1)]=0$, we get $a_{1, -(k+1)}^{(k)}=a_{k+1, -(k+2)}^{(k)}=0$. For $3\leq k\leq m-2$, by applying $\varphi_{6}$ on $[T_{k, k+1}(1), T_{k-1, k+2}(1)]=0$, we obtain $a_{1, -(k+2)}^{(k)}=a_{k+2, -(k+3)}^{(k)}=0$.

For $j\neq i-1, i+1$, by considering the action of $\varphi_{6}$ on $[T_{i, i+1}(1), T_{j, j+1}(1)]=0$, we get $a_{i, j+1}^{(i)}=a_{i+1, j}^{(j)}$. Choose

$ X_{6}=E^{(2m)}+\sum\limits_{k=3}^{m}T_{1, -k}(a_{1, k-1}^{(k-1)})+ \sum\limits_{2\leq i<j\leq m}T_{i, -j}(a_{i-1, j}^{(i-1)}), \ h=a_{1, -3}^{(2)}. $

Then

$ \begin{eqnarray} \lambda_{h}^{-1}{\mbox{Int}}^{-1}X_{6}\varphi_{6}(T_{12}(1))&\;=&\;T_{12}(1)+T_{1, -2}(a_{1, -2}^{(1)}), \nonumber\\ \lambda_{h}^{-1}{\mbox{Int}}^{-1}X_{6}\varphi_{6}(T_{23}(1))&\;=&\;T_{23}(1)+T_{1, -2}(a_{1, -2}^{(2)}) +T_{2, -3}(a_{2, -3}^{(2)})+T_{1, -4}(a_{1, -4}^{(2)}), \nonumber\\ \lambda_{h}^{-1}{\mbox{Int}}^{-1}X_{6}\varphi_{6}(T_{i, i+1}(1))&\;=&\;T_{i, i+1}(1)+T_{1, -2}(a_{1, -2}^{(i)}) +T_{i, -(i+1)}(a_{i, -(i+1)}^{(i)}), \ 3\leq i\leq m-1.\nonumber \end{eqnarray} $

Denote $\lambda_{h}^{-1}{\mbox{Int}}^{-1}X_{6}\varphi_{6}$ by $\varphi_{7}$.

Now we may assume that

$ \varphi_{7}(T_{ij}(1))=T_{ij}(1)+\sum\limits_{1\leq k<l\leq m}T_{k, -l}(b_{k, -l}^{(ij)}), \ 1\leq i\leq m-2, \ i+2\leq j\leq m, $

where $b_{k, -l}^{(ij)}\in F$. For $1\leq t\leq m-1$, $t\neq i-1, j$, $2\leq s\leq m, \ 3\leq p\leq m, \ s, p\neq i$, by applying $\varphi_{7}$ on $[T_{t, t+1}(1), T_{ij}(1)]=0$, $[T_{1s}(1), T_{ij}(1)]=0$ and $[T_{2p}(1), T_{ij}(1)]=0$, respectively, we have

$ \begin{eqnarray} \varphi_{7}(T_{13}(1))&\;=&\;T_{13}(1)+T_{1, -2}(b_{1, -2}^{(13)})+T_{1, -3}(b_{1, -3}^{(13)})+T_{1, -4}(b_{1, -4}^{(13)}) +T_{2, -4}(b_{2, -4}^{(13)}), \nonumber\\ \varphi_{7}(T_{14}(1))&\;=&\;T_{14}(1)+T_{1, -2}(b_{1, -2}^{(14)})+T_{1, -4}(b_{1, -4}^{(14)}) +T_{2, -3}(b_{2, -3}^{(14)}), \nonumber\\ \varphi_{7}(T_{24}(1))&\;=&\;T_{24}(1)+T_{1, -2}(b_{1, -2}^{(24)})+T_{1, -3}(b_{1, -3}^{(24)}) +T_{2, -3}(b_{2, -3}^{(24)})+T_{2, -4}(b_{2, -4}^{(24)}), \nonumber\\ \varphi_{7}(T_{kl}(1))&\;=&\;T_{kl}(1)+T_{1, -2}(b_{1, -2}^{(kl)})+T_{k, -l}(b_{k, -l}^{(kl)}), k=1, 2, 5\leq l\leq m, \nonumber\\ \varphi_{7}(T_{ij}(1))&\;=&\;T_{ij}(1)+T_{1, -2}(b_{1, -2}^{(ij)})+T_{1, -i}(b_{1, -i}^{(ij)})+T_{i, -j}(b_{i, -j}^{(ij)}), 3\leq i<j\leq m, j\neq i+1.\nonumber \end{eqnarray} $

By applying $\varphi_{7}$ on $[T_{12}(1), T_{13}(1)]=0$ and $[T_{12}(1), T_{14}(1)]=0$, respectively, we have $b_{2, -4}^{(13)}=b_{2, -3}^{(14)}=0$. For $1\leq i<j\leq m, 1\leq k< j\leq m, k\neq i, 4\leq j\leq m$, by applying $\varphi_{7}$ on $[T_{ij}(1), T_{kj}(1)]=0$, we obtain that $b_{i, -j}^{(ij)}=0$ and $a_{i, -(i+1)}^{(i)}=0$. For $3\leq i<j\leq m$ and $j\neq i+1$, by applying $\varphi_{7}$ on $[T_{2i}(1)+T_{2, j-1}(1), T_{ij}(1)-T_{j-1, j}(1)]=0$, we get $b_{1, -i}^{(ij)}=0$.

By applying $\varphi_{7}$ on $[T_{13}(1), T_{23}(1)]=0$, $[T_{13}(1), T_{24}(1)]=0$ and $[T_{23}(1), T_{24}(1)]=0$, respectively, we get $b_{1, -3}^{(13)}=-a_{2, -3}^{(2)}$, $b_{1, -4}^{(13)}=-b_{2, -3}^{(24)}$ and $b_{1, -3}^{(24)}=a_{1, -4}^{(2)}$. Let $a=b_{1, -3}^{(13)}$, $b=b_{1, -4}^{(13)}$, $c=a_{1, -4}^{(2)}$. Then

$ \begin{eqnarray*}&&\rho^{-1}_{a, b, c}\varphi_{7}(T_{i, i+1}(1))=T_{i, i+1}(1)+T_{1, -2}(a_{1, -2}^{(i)})\ {\mbox{for}}\ 1\leq i\leq m-1, \\ &&\rho^{-1}_{a, b, c}\varphi_{7}(T_{ij}(1))=T_{ij}(1)+T_{1, -2}(b_{1, -2}^{(ij)})\ {\mbox{for}}\ 1\leq i<j\leq m-1, j\neq i+1.\end{eqnarray*} $

That is to say

$ \rho_{a, b, c}^{-1}\varphi_{7}(T_{ij}(1)) \equiv T_{ij}(1)\ ({\mbox{mod}}\ T_{1, -2})\ {\mbox{for}}\ 1\leq i<j\leq m. $

Denote $\rho^{-1}_{a, b, c}\varphi_{7}$ by $\varphi_{8}$.

Step 7  There exist some $d, s\in F^{*}, X_{7}=E^{(2m)}+T_{1m}(s^{-1}d)$ and a linear map $f$ from $l_{2m}(F)$ to $F$ such that $\xi^{-1}_{s}IntX_{7}\varphi_{8}=\eta_{f}.$

Now we assume that $\varphi_{8}(T_{i, -j}(1))=\sum\limits_{1\leq k<l\leq m}T_{k, -l}(c_{k, -l}^{(ij)})$. For $2\leq k\leq m$, $k\neq i, j$, $2\leq t\leq m-1$, $t+1\neq i, j$, by applying $\varphi_{8}$ on $[T_{1k}(1), T_{i, -j}(1)]=0$, $[T_{t, t+1}(1), T_{i, -j}(1)]=0$ and $[T_{ij}(1), T_{i, -j}(1)]=0$, respectively, we get

$ \begin{eqnarray} \varphi_{8}(T_{1, -2}(1))&\;=&\;T_{1, -2}(c_{1, -2}^{(12)}), \nonumber\\ \varphi_{8}(T_{i, -j}(1))&\;=&\;T_{i, -j}(c_{i, -j}^{(ij)})+T_{1, -2}(c_{1, -2}^{(ij)})\ {\mbox{for}}\ 1\leq i\leq2, i<j\leq m\ {\mbox{and}}\ (i, j)\neq (1, 2), \nonumber\\ \varphi_{8}(T_{i, -j}(1))&\;=&\;T_{i, -j}(c_{i, -j}^{(ij)})+T_{1, -2}(c_{1, -2}^{(ij)})+T_{1, -i}(c_{1, -i}^{(ij)})\ {\mbox{for}}\ 3\leq i<j\leq m.\nonumber\end{eqnarray} $

For $3\leq i<j\leq m-1$, by applying $\varphi_{8}$ on

$ [T_{i-1, i}(1)-T_{j, j+1}(1), T_{i, -j}(1)+T_{i-1, -(j+1)}(1)]=0, $

we obtain $c_{1, -i}^{(ij)}=0.$ For $3\leq i<k\leq m-1$, by considering the action of $\varphi_{8}$ on

$ [T_{2i}(1)-T_{2k}(1), T_{i, -m}(1)+T_{k, -m}(1)]=0, $

we get that $c_{1, -i}^{(im)}=c_{1, -3}^{(3m)}$ for $4\leq i\leq m-1$.

For $3\leq l\neq k\leq m$, by applying $\varphi_{8}$ on

$ [T_{2l}(1)-T_{2k}(1), T_{1, -l}(1)+T_{1, -k}(1)]=0, $

we get $c_{1, -l}^{(1l)}=c_{1, -k}^{(1k)}.$

For $2\leq i<k<l\leq m$, by applying $\varphi_{8}$ on $[T_{1k}(1)-T_{1l}(1), T_{i, -k}(1)+T_{i, -l}(1)]=0$, we have $c_{i, -k}^{(ik)}=c_{i, -l}^{(il)}$. By applying $\varphi_{8}$ on

$ [T_{13}(1)+T_{23}(1), T_{1, -3}(1)+T_{2, -3}(1)]=0, $

and

$ [T_{12}(1)-T_{13}(1), T_{2, -4}(1)+T_{3, -4}(1)]=0, $

we obtain $c_{1, -3}^{(13)}=c_{2, -3}^{(23)}$ and $c_{2, -4}^{(24)}=c_{3, -4}^{(34)}$. By applying $\varphi_{8}$ on

$ [T_{12}(1)+T_{1j}(1), T_{2, -k}(1)+T_{k, -j}(1)]=0 $

with $4\leq k<j\leq m$, we get $c_{2, -k}^{(2k)}=c_{k, -j}^{(kj)}$. So all $c_{i, -j}^{(ij)}$ are equal for $1\leq i<j\leq m$ except $(i, j)=(1, 2)$. Let $d=c_{1, -3}^{(3m)}$, $s=c_{i, -j}^{(ij)}$, $X_{7}=E^{(2m)}+T_{1m}(s^{-1}d), $ then

$ \begin{eqnarray} \xi^{-1}_{s}{\mbox{Int}}X_{7}\varphi_{8}(T_{1, -2}(1))&\;=&\;T_{1, -2}(s^{-1}c_{1, -2}^{(12)}), \nonumber\\ \xi^{-1}_{s}{\mbox{Int}}X_{7}\varphi_{8}(T_{2, -m}(1))&\;=&\;T_{2, -m}(s^{-1}c_{1, -2}^{(2m)}-d), \nonumber\\ \xi^{-1}_{s}{\mbox{Int}}X_{7}\varphi_{8}(T_{i, -j}(1))&\;=&\;T_{i, -j}(1)+T_{1, -2}(s^{-1}c_{1, -2}^{(ij)})\ {\mbox{for}}\ 1\leq i<j\leq m, (i, j)\neq (1, 2), (2, m).\nonumber \end{eqnarray} $

It is easy to see that

$ \begin{eqnarray} \xi^{-1}_{s}{\mbox{Int}}X_{7}\varphi_{8}(T_{i, i+1}(1))&\;=&\;T_{i, i+1}(1)+T_{1, -2}(s^{-1}a_{1, -2}^{(i)})\ {\mbox{for}}\ 1\leq i\leq m-1, \nonumber\\ \xi^{-1}_{s}{\mbox{Int}}X_{7}\varphi_{8}(T_{ij}(1))&\;=&\;T_{ij}(1)+T_{1, -2}(s^{-1}b_{1, -2}^{(ij)})\ {\mbox{for}}\ 1\leq i<j\leq m, j\neq i+1.\nonumber \end{eqnarray} $

Let $f(T_{i, i+1}(1))=s^{-1}a_{1, -2}^{(i)}$, $f(T_{1, -2}(1))=s^{-1}c_{1, -2}^{(12)}-1$, $f(T_{2, -m}(1))=s^{-1}c_{1, -2}^{(2m)}-d$, $f(T_{ij}(1))=s^{-1}b_{1, -2}^{(ij)}$, $j\neq i+1$, $f(T_{k, -l}(1))=s^{-1}c_{1, -2}^{(kl)}$, $k, l\neq (1, 2), (2, m)$. Then

$ \xi^{-1}_{s}{\mbox{Int}}X_{7}\varphi_{8}=\eta_{f}. $

Above discussion shows that

$ \varphi=\psi_{c_{1}}{\mbox{Int}}X_{1}(\mu_{\pi})^{\delta}\xi_{g}\psi_{c_{2}} {\mbox{Int}}(X_{2}X_{3}X_{4}X_{5}X_{6}X_{7})\lambda_{h}\rho_{a, b, c}\xi_{s}\eta_{f}. $

By Lemma 4.1, it is easy to show that $\varphi$ is of form (5.4). This completes the proof.