Higher Jordan triple derivations on $*$-type trivial extension algebras

1.   Introduction

Let $\mathcal{R}$ be a commutative ring with identity, and $\mathcal{A}$ a unital algebra over $\mathcal{R}$. For any $X, Y\in \mathcal{A}$, denote the Jordan product of $X, Y$ by $X\circ Y = XY+YX$. An additive mapping $\Delta$ from $\mathcal{A}$ into itself is called a derivation (resp., anti-derivation) if $\Delta(XY) = \Delta(X)Y+X\Delta(Y)$ (resp., $\Delta(XY) = \Delta(Y)X+Y\Delta(X)$) for all $X, Y\in\mathcal{A}$. It is called a Jordan derivation if $\Delta(X\circ Y) = \Delta(X)\circ Y+X\circ\Delta(Y)$ for all $X, Y\in\mathcal{A}$. It is called a Jordan triple derivation if $\Delta(X\circ Y\circ Z) = \Delta(X)\circ Y\circ Z+X\circ\Delta(Y)\circ Z+X\circ Y\circ \Delta(Z)$ for all $X, Y, Z\in\mathcal{A}$. Obviously, every derivation or anti-derivation is a Jordan derivation. However, the inverse statement is not true in general (see ^[1]). If a Jordan derivation or Jordan triple derivation is not a derivation, then it is said to be proper. Otherwise, it is said to be improper.

In the past few decades, the problem of characterizing the structure of Jordan derivations and Jordan triple derivations has attracted the attention of many mathematical workers and has achieved some important research results. For example, Herstein in ^[2] proved that every Jordan derivation on a prime ring not of characteristic $2$ is a derivation. This result was extended by Cusack in ^[3] and Brešar in ^[4] to the case of semiprime. Zhang in ^[5,6] showed that every Jordan derivation on a nest algebra or a $2$-torsion free triangular algebra is a inner derivation or a derivation, respectively. Later, Hoger in ^[7] extended the result of Zhang in ^[6] and proved that, under certain conditions, each Jordan derivation on trivial extension algebras is a sum of a derivation and an anti-derivation. In addition, there have been many research results on Jordan triple derivations, as shown in references ^[8,9,10,11].

Definition 1.1. Let $\mathcal{R}$ be a commutative ring with identity, $\mathcal{A}$ a unital algebra over $\mathcal{R}$, $\mathbb{N}_{0}$ be the set of all nonnegative integers, and $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ be a family of additive maps on $\mathcal{A}$ such that $d_{0} = id_{\mathcal{A}}$ (the identity map on $\mathcal{A}$). $D$ is said to be:

$(i)$ a higher derivation if for each $n\in\mathbb{N}_{0}$,

for all $X, Y\in\mathcal{A}$;

$(ii)$ a higher anti-derivation if for each $n\in\mathbb{N}_{0}$,

for all $X, Y\in\mathcal{A}$;

$(iii)$ a higher Jordan derivation if for each $n\in\mathbb{N}_{0}$,

for all $X, Y\in\mathcal{A}$;

$(iv)$ a higher Jordan triple derivation if for each $n\in\mathbb{N}_{0}$,

for all $X, Y, Z\in\mathcal{A}$.

If a higher Jordan derivation or a higher Jordan triple derivation is not a higher derivation, then it is said to be proper. Otherwise, it is said to be improper. With the deepening of research on this topic, many research achievements have been obtained about higher Jordan derivations and higher Jordan triple derivations. For example, Xiao and Wei in ^[12] proved that every higher Jordan derivation on triangular algebras is a higher derivation; Fu, Xiao, and Du in ^[13] extended this conclusion, and proved that every nonlinear higher Jordan derivation on triangular algebras is a higher derivation. Later, Vishki, Mirzavaziri, and Moafian in ^[14] proved that, under certain conditions, every higher Jordan derivation on trivial extension algebras is a higher derivation, and this conclusion further extended the works of the authors of references ^[12,13]. Salih and Haetinger in ^[15] proved that, under certain conditions, every higher Jordan triple derivation on prime rings is a higher derivation. Ashraf and Jabeenin ^[16] proved that every nonlinear higher Jordan triple derivable mapping on triangular algebras is a higher derivation.

In this paper, we are interested in describing the form of higher Jordan triple derivation on trivial extension algebras. As a main result, we give conditions under which each higher Jordan triple derivation on trivial extension algebras is a sum of a higher derivation and a higher anti-derivation. This result extends the study of Jordan derivation on trivial extension algebras ^[7], Jordan triple derivations on $*$-type trivial extension algebras ^[17], and Jordan higher derivations on trivial extension algebras ^[14].

Let $\mathcal{R}$ be a commutative ring with identity, $\mathcal{A}$ a unital algebra over $\mathcal{R}$ and $\mathcal{M}$ be an $\mathcal{A}$-bimodule. Then the direct product $\mathcal{A} \oplus \mathcal{M}$ together with the pairwise addition, scalar product, and algebra multiplication defined by

is an $\mathcal{R}$-algebra with a unity $(1, 0)$ denoted by

and $\mathcal{T}$ is called a trivial extension algebra.

An important example of trivial extension algebra is the triangular algebra which was introduced by Cheung in ^[18]. Let ${\mathcal A}$ and ${\mathcal B}$ be unital algebras over a commutative ring $\mathcal{R}$, and ${\mathcal M}$ be a unital $({\mathcal A}, {\mathcal B})$-bimodule, which is faithful as both a left ${\mathcal A}$-module and a right ${\mathcal B}$-module. Then, the $\mathcal{R}$-algebra

under the usual matrix operations is called a triangular algebra. Basic examples of triangular algebras are upper triangular matrix algebras and nest algebras.

It is well-known that every triangular algebra can be viewed as a trivial extension algebra. Indeed, denote by $\mathcal{A} \oplus \mathcal{B}$ the direct product as an $\mathcal{R}$-algebra, and then ${\mathcal M}$ is viewed as an $\mathcal{A} \oplus \mathcal{B}$-bimodule with the module action given by $(a, b)m = am$ and $m(a, b) = mb$ for all $(a, b)\in\mathcal{A}\oplus \mathcal{B}$ and $m\in\mathcal{M}$. Then triangular algebra ${\mathcal U}$ is isomorphic to trivial extensions algebra $\mathcal{T} = (\mathcal{A} \oplus \mathcal{B})\oplus \mathcal{M}$. However, a trivial extension algebra is not necessarily a triangular algebra. For more details about trivial extension algebras, we refer the readers to ^[19,20,21].

The following notations will be used in our paper: Let $\mathcal{R}$ be a commutative ring with identity, $\mathcal{A}$ a unital algebra over $\mathcal{R}$, $\mathcal{M}$ an $\mathcal{A}$-bimodule, $\mathcal{T} = \mathcal{A} \oplus \mathcal{M}$ be a 2-torsion free trivial extension algebra (i.e., for any $X\in\mathcal{T}$, $2X = \{0\}$ implies $X = 0$), and denote by $1$ and $0$ are the unity and zero of $\mathcal{T} = \mathcal{A} \oplus \mathcal{M}$, respectively.

We say $\mathcal{T} = \mathcal{A} \oplus \mathcal{M}$ is a $*$-type trivial extension algebra if $\mathcal{A}$ has a non-trivial idempotent element $e$ and $f = 1-e$ such that

$(i)$ $e\mathcal{M}f = \mathcal{M}$;

$(ii)$ $exe\mathcal{M} = \{0\}$ implies $exe = 0, \forall x\in\mathcal{A}$;

$(iii)$ $\mathcal{M}fxf = \{0\}$ implies $fxf = 0, \forall x\in\mathcal{A}$;

$(iv)$ $exfye = 0 = fxeyf = 0, \forall x, y\in\mathcal{A}$.

For convenience, in the following we let $P_{1} = (e, 0)$, $P_{2} = (f, 0)$, and

It is not hard to see that the trivial extension algebra $\mathcal{T}$ may be represented as

Then every element $A\in\mathcal{T}$ may be represented as $A = A_{11}+A_{12}+A_{21}+A_{22}$, where $A_{ij}\in\mathcal{T}_{ij} (1\leq i\leq j\leq2)$. In the following, we give a property of $*$-type trivial extension algebras (see Lemma 1.1).

Lemma 1.1. ^[17] Let $\mathcal{T}$ be a $*$-type trivial extension algebra and $1\leq i\neq j\leq 2$. Then,

$(i)$ for any $A_{11}\in\mathcal{T}_{11}$, if $A_{11}\mathcal{T}_{12} = 0$, then $A_{11} = 0$;

$(ii)$ for any $A_{22}\in\mathcal{T}_{22}$, if $\mathcal{T}_{12}A_{22} = 0$, then $A_{22} = 0$;

$(iii)$ $A_{ij}B_{ji} = A_{ii}B_{ji} = A_{ij}B_{ii} = 0$, $\forall A_{ii}, B_{ii}\in\mathcal{T}_{ii}, \forall A_{ij}\in\mathcal{T}_{ij}, \forall B_{ji}\in\mathcal{T}_{ji}$.

For ease of reading, we provide the main conclusions of reference ^[17] as follows:

Theorem 1.1. ^[17] Let $\mathcal{T} = \mathcal{A} \oplus \mathcal{M}$ be a $2$-torsion free $*$-type trivial extension algebra and $\Delta$ be a Jordan triple derivation on $\mathcal{T}$. Then, there exists a derivation $D$ and an anti-derivation $\varphi$ on $\mathcal{T}$, respectively, such that

for all $A\in\mathcal T$.

2.   Main results

The main result of this paper is the following theorem:

Theorem 2.1. Let $\mathcal{T} = \mathcal{A} \oplus \mathcal{M}$ be a $2$-torsion free $*$-type trivial extension algebra, and $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ be a higher Jordan triple derivation on $\mathcal{T}$. Then, there exists a higher derivation $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ and a higher anti-derivation $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ on $\mathcal{T}$, respectively, such that

for any $n\geq 1$ and $X\in\mathcal{T}$.

In order to prove Theorem 2.1, we shall establish Theorems 2.2 and 2.3 in the following. We assume that $\mathcal{T}$ is a $*$-type trivial extension algebra, $\mathbb{N}_{0}$ is the set of all nonnegative integers, and $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher Jordan triple derivation on $\mathcal{T}$.

In ^[17], it is proved that if $d_{1}$ is a Jordan triple derivation on $\mathcal{T}$, then for all $A_{ij}\in \mathcal{T}_{ij}$ ($1\leq i, j\leq 2$), $d_{1}$ satisfies the following properties ($\mathcal L$):

$(i)$ $d_{1}(P_{1}) = -d_{1}(P_{2})$;

$(ii)$ $d_{1}(P_{1}) = P_{1}d_{1}(P_{1})P_{2}+P_{2}d_{1}(P_{1})P_{1}$ and $d_{1}(P_{2}) = P_{1}d_{1}(P_{2})P_{2}+P_{2}d_{1}(P_{2})P_{1}$;

$(iii)$ $P_{2}d_{1}(A_{11})P_{2} = 0$, $P_{1}d_{1}(A_{11})P_{2} = A_{11}d_{1}(P_{1})$ and $P_{2}d_{1}(A_{11})P_{1} = d_{1}(P_{1})A_{11}$;

$(iv)$ $P_{1}d_{1}(A_{22})P_{1} = 0, P_{1}d_{1}(A_{22})P_{2} = d_{1}(P_{2})A_{22}$ and $P_{2}d_{1}(A_{22})P_{1} = A_{22}d_{1}(P_{2})$;

$(v)$ $d_{1}(A_{12}) = P_{1}d_{1}(A_{12})P_{2}+P_{2}d_{1}(A_{12})P_{1}$ and $d_{1}(A_{21}) = P_{1}d_{1}(A_{21})P_{2}+P_{2}d_{1}(A_{21})P_{1}$;

$(vi)$ $d_{1}(P_{1})\circ d_{1}(P_{2}) = d_{1}(P_{1})\circ d_{1}(A_{12}) = d_{1}(P_{1})\circ d_{1}(A_{21}) = d_{1}(P_{2})\circ d_{1}(A_{12}) = d_{1}(P_{2})\circ d_{1}(A_{21}) = 0$;

$(vii)$ $d_{1}(A_{12})\circ d_{1}(A_{12}) = d_{1}(A_{21})\circ d_{1}(A_{21}) = d_{1}(A_{12})\circ d_{1}(A_{21}) = 0$.

Now, for all $A_{ij}\in \mathcal{T}_{ij}$ $(1\leq i, j\leq 2)$, we assume that $d_{k}$ ($1\leqslant k < n$) satisfy the properties $\mathcal{L}$. In the following, we show that $d_{n}$ satisfies the properties $\mathcal{L}$.

Lemma 2.1. Let $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ be a higher Jordan triple derivation on $\mathcal{T}$. Then, for each $n\geq 1$, and for any $A_{11}\in \mathcal{T}_{11}, A_{12}\in \mathcal{T}_{12}, A_{21}\in \mathcal{T}_{21}$, $A_{22}\in \mathcal{T}_{22}$,

$(i)$ $d_{n}(P_{1}) = P_{1}d_{n}(P_{1})P_{2}+P_{2}d_{n}(P_{1})P_{1}$ and $d_{n}(P_{2}) = P_{1}d_{n}(P_{2})P_{2}+P_{2}d_{n}(P_{2})P_{1}$;

$(ii)$ $d_{n}(P_{1}) = -d_{n}(P_{2})$;

$(iii)$ $P_{2}d_{n}(A_{11})P_{2} = 0$, $P_{1}d_{n}(A_{11})P_{2} = A_{11}d_{n}(P_{1})$ and $P_{2}d_{n}(A_{11})P_{1} = d_{n}(P_{1})A_{11}$;

$(iv)$ $P_{1}d_{n}(A_{22})P_{1} = 0, P_{1}d_{n}(A_{22})P_{2} = d_{n}(P_{2})A_{22}$ and $P_{2}d_{n}(A_{22})P_{1} = A_{22}d_{n}(P_{2})$;

$(v)$ $d_{n}(A_{12}) = P_{1}d_{n}(A_{12})P_{2}+P_{2}d_{n}(A_{12})P_{1}$ and $d_{n}(A_{21}) = P_{1}d_{n}(A_{21})P_{2}+P_{2}d_{n}(A_{21})P_{1}$;

$(vi)$ $d_{n}(P_{1})\circ d_{n}(P_{2}) = d_{n}(P_{1})\circ d_{n}(A_{12}) = d_{n}(P_{1})\circ d_{n}(A_{21}) = d_{n}(P_{2})\circ d_{n}(A_{12}) = d_{n}(P_{2})\circ d_{n}(A_{21}) = 0$;

$(vii)$ $d_{n}(A_{12})\circ d_{n}(A_{12}) = d_{n}(A_{21})\circ d_{n}(A_{21}) = d_{n}(A_{12})\circ d_{n}(A_{21}) = 0$.

Proof. $(i)$ For each $n\geq 1$ and for any $X, Y, Z\in \mathcal{T}$, by the definition of $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$, we get

Taking $X = Y = Z = P_{1}$ in Eq (2.1), we assume that $d_{k}$ ($1\leqslant k < n$) satisfy the properties $\mathcal{L}$, and then it follows from Lemma 1.1 $(iii)$ that

This yields from the $2$-torsion freeness of $\mathcal{T}$ that

Similarly, we get that

Therefore, $d_{n}(P_{1}) = P_{1}d_{n}(P_{1})P_{2}+P_{2}d_{n}(P_{1})P_{1}$ and $d_{n}(P_{2}) = P_{1}d_{n}(P_{2})P_{2}+P_{2}d_{n}(P_{2})P_{1}$.

$(ii)$ For each $n\geq 1$, taking $X = P_{1}, Y = P_{2}, Z = P_{1}$ in Eq (2.1), we assume that $d_{k}$ ($1\leqslant k < n$) satisfy the properties $\mathcal{L}$, then by Lemma 1.1 $(iii)$ and Lemma 2.1 $(i)$, we get that

$(iii)$–$(iv)$ For each $n\geq 1$ and for any $A_{11}\in \mathcal{T}_{11}$, taking $X = A_{11}, Y = Z = P_{2}$ in Eq (2.1), we assume that $d_{k}$ ($1\leqslant k < n$) satisfy the properties $\mathcal{L}$, and then by Lemma 1.1 $(iii)$ and Lemma 2.1 $(i, ii)$, we get that

This implies that $P_{2}d_{n}(A_{11})P_{2} = 0$. and

Similarly, for each $n\geq 1$ and for any $A_{22}\in \mathcal{T}_{22}$, we get that $P_{1}d_{n}(A_{22})P_{1} = 0$, $P_{1}d_{n}(A_{22})P_{2} = d_{n}(P_{2})A_{22}$ and $P_{2}d_{n}(A_{22})P_{1} = A_{22}d_{n}(P_{2})$.

$(v)$ For each $n\geq 1$ and for any $A_{12}\in \mathcal{T}_{12}$, taking $X = P_{1}, Y = A_{12}, Z = P_{2}$ in Eq (2.1), we assume that $d_{k}$ ($1\leqslant k < n$) satisfy the properties $\mathcal{L}$, and then by Lemma 1.1 $(iii)$ and Lemma 2.1 $(i, ii)$, we get that

Similarly, for each $n\geq 1$ and for any $A_{21}\in \mathcal{T}_{21}$, we get that $d_{n}(A_{21}) = P_{1} d_{n}(A_{21})P_{2}+P_{2}d_{n}(A_{21})P_{1}$.

$(vi)$ For each $n\geq 1$ and for any $A_{12}\in \mathcal{T}_{12}$, $A_{21}\in \mathcal{T}_{21}$, by Lemma 1.1 $(iii)$ and Lemma 2.1 $(i, ii, v)$, we can easily check that $(vi)$ holds. Similarly, we show $(vii)$ holds. The proof is complete. □

Theorem 2.2. Let $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ be a sequence of mappings on $\mathcal{T}$ (with $f_{0} = if_{\mathcal{T}}$). For each $n\geq 1$ and $X\in\mathcal{T}$, define

Then, $F$ is a higher anti-derivation on $\mathcal{T}$.

It is clear that $f_{n}(A_{ii}) = 0$ and $f_{n}(A_{ij}) = P_{j}f_{n}(A_{ij})P_{i}$ for each $n\geq 1$, and for any $A_{ii}\in \mathcal{T}_{ii}, A_{ij}\in \mathcal{T}_{ij}$ ($1\leq i\neq j \leq 2$).

In the following, we show that $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher anti-derivation, i.e., for each $n\geq 1$ and for any $X, Y\in \mathcal{T}$, $f_{n}$ satisfies $f_{n}(XY) = \sum_{p+q = n}f_{p}(Y)f_{q}(X)$. For this, we introduce Lemmas 2.2 and 2.3, and prove Lemmas 2.2 and 2.3.

Lemma 2.2. Let $f_{n}: \mathcal{T}\to\mathcal{T}$ be defined as in Theorem 2.2. Then, for each $n\geq 1$ and for any $A_{ii}, B_{ii}\in\mathcal{T}_{ii}, A_{ij}, B_{ij}\in\mathcal{T}_{ij}, B_{ji}\in\mathcal{T}_{ji}, B_{jj}\in\mathcal{T}_{jj}$ ($1\leq i\neq j \leq 2$),

$(i)$ $f_{n}(A_{ii}B_{ii}) = \sum_{p+q = n}f_{p}(B_{ii})f_{q}(A_{ii})$;

$(ii)$ $f_{n}(A_{ii}B_{jj}) = \sum_{p+q = n}f_{p}(B_{jj})f_{q}(A_{ii})$;

$(iii)$ $f_{n}(A_{ii}B_{ji}) = \sum_{p+q = n}f_{p}(B_{ji})f_{q}(A_{ii})$;

$(iv)$ $f_{n}(A_{ij}B_{ii}) = \sum_{p+q = n}f_{p}(B_{ii})f_{q}(A_{ij})$;

$(v)$ $f_{n}(A_{ij}B_{ij}) = \sum_{p+q = n}f_{p}(B_{ij})f_{q}(A_{ij})$;

$(vi)$ $f_{n}(A_{ij}B_{ji}) = \sum_{p+q = n}f_{p}(B_{ji})f_{q}(A_{ij})$.

Proof. $(i)$ For any $n\geq 1$ and $A_{ii}, B_{ii}\in\mathcal{T}_{ii}$ ($1\leq i\leq 2$), we get from $f_{n}(A_{ii}B_{ii}) = f_{n}(A_{ii}) = f_{n}(B_{ii}) = 0$ that

Similarly, we can show $(ii)$ holds.

$(iii)$ For each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{ji}\in\mathcal{T}_{ji}$ ($1\leq i\neq j \leq 2$), on the one hand, we have $f_{n}(A_{ii}B_{ji}) = f_{n}(0) = 0$. On the other hand, it follows from $f_{n}(A_{ii}) = 0$ and $f_{n}(B_{ji}) = P_{i}f_{n}(B_{ji})P_{j}$ that

Therefore, $f_{n}(A_{ii}B_{ji}) = \sum_{p+q = n}f_{p}(B_{ji})f_{q}(A_{ii})$. Similarly, we get $(iv)$.

$(v)$ For each $n\geq 1$ and for any $A_{ij}, B_{ij}\in\mathcal{T}_{ij}$ ($1\leq i\neq j \leq 2$), on the one hand, we have $f_{n}(A_{ij}B_{ij}) = f_{n}(0) = 0$. On the other hand, we get from $f_{n}(B_{ij})f_{n}(A_{ij}) = \{P_{j}f_{n}(B_{ij})P_{i}\}\{P_{j}f_{n}(A_{ij})P_{i}\} = 0$ and Lemma 1.1 $(iii)$ that

Therefore, $f_{n}(A_{ij}B_{ij}) = \sum_{p+q = n}f_{p}(B_{ij})f_{q}(A_{ij})$. Similarly, we get $(vi)$. The proof is complete. □

Lemma 2.3. Let $f_{n}: \mathcal{T}\to\mathcal{T}$ be defined as in Theorem 2.2. Then, for each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{ij}\in\mathcal{T}_{ij}, B_{jj}\in\mathcal{T}_{jj}$ ($1\leq i\neq j \leq 2$),

$(i)$ $f_{n}(A_{ii}B_{ij}) = \sum_{p+q = n}f_{p}(B_{ij})f_{q}(A_{ii})$;

$(ii)$ $f_{n}(A_{ij}B_{jj}) = \sum_{p+q = n}f_{p}(B_{jj})f_{q}(A_{ij})$.

Proof. $(i)$ For each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{ij}\in\mathcal{T}_{ij}$ ($1\leq i\neq j \leq 2$), it follows from $A_{ii}B_{ij} = A_{ii}\circ B_{ij}\circ P_{j}$ and Lemma 2.1 that

On the other hand, it is follows from $f_{n}(A_{ii}) = 0$ ($n\geq 1$) that $f_{n-k}(B_{ij})d_{k}(A_{ii}) = 0$, so we get

Similarly, we get $(ii)$. The proof is complete. □

In the following, we give the completed proof of Theorem 2.2:

Proof of Theorem 2.2. For each $n\geq 1$, let $A = A_{11}+A_{12}+A_{21}+A_{22}$ and $B = B_{11}+B_{12}+B_{21}+B_{22}$ be arbitrary elements of $\mathcal T$, where $A_{ij}, B_{ij}\in\mathcal{T}_{ij}$ ($1\leq i, j\leq 2$). Then, it follows from Lemmas 2.2 and 2.3 that

Therefore, $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher anti-derivation on $\mathcal{T}$. The proof is complete. □

Theorem 2.3. Let $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ be a sequence of mappings on $\mathcal{T}$ (with $g_{0} = ig_{\mathcal{T}}$). For each $n\geq 1$ and for any $X\in\mathcal{T}$, define

Then, $G$ is a higher derivation on $\mathcal{T}$.

Next, we show that $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation on $\mathcal{T}$. In order to prove $G$ is a higher derivation, we introduce Lemmas 2.4–2.6, and then, using the mathematical induction, we prove Lemmas 2.4–2.6.

In ^[12] Theorem 1.3, we have proved that if $g_{1} = d_{1}-f_{1}$, then $g_{1}$ is a derivation on $\mathcal{T}$, i.e., for any $X, Y\in \mathcal{T}$, $g_{1}$ satisfies

Therefore, in the following, we assume that

for each $1\leqslant k < n$ and $X, Y\in \mathcal{T}$. Next, we prove that Lemmas 2.4–2.6 hold.

By the definitions of $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ and $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$, and by Lemma 2.1, we can easily check that the following Lemma holds:

Lemma 2.4. Let $g_{n}:\mathcal{T}\to\mathcal{T}$ be defined as in Theorem 2.3. Then, for each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, A_{ij}\in\mathcal{T}_{ij}$ $(1\leq i\neq j \leq 2)$,

$(i)$ $g_{n}(P_{i}) = -g_{n}(P_{j})$ and $g_{n}(P_{i}) = P_{i}g_{n}(P_{i})P_{j}+P_{j}g_{n}(P_{i})P_{i}$;

$(ii)$ $P_{j}g_{n}(A_{ii})P_{j} = 0, P_{i}g_{n}(A_{ii})P_{j} = A_{ii}g_{n}(P_{i})$ and $P_{j}g_{n}(A_{ii})P_{i} = g_{n}(P_{i})A_{ii}$;

$(iii)$ $g_{n}(A_{ij}) = P_{i}g_{n}(A_{ij})P_{j}$.

Lemma 2.5. Let $g_{n}:\mathcal{T}\to\mathcal{T}$ be defined as in Theorem 2.3. Then, for each $n\geq 1$, and for any $A_{ii}, B_{ii}\in\mathcal{T}_{ii}$, $B_{jj}\in\mathcal{T}_{jj}, A_{ij}, B_{ij}\in\mathcal{T}_{ij}$ ($1\leq i\neq j \leq 2$),

$(i)$ $g_{n}(A_{ii}B_{ij}) = \sum_{p+q = n}g_{p}(A_{ii})g_{q}(B_{ij})$;

$(ii)$ $g_{n}(A_{ij}B_{jj}) = \sum_{p+q = n}g_{p}(A_{ij})g_{q}(B_{jj})$;

$(iii)$ $g_{n}(A_{ii}B_{ii}) = \sum_{p+q = n}g_{p}(A_{ii})g_{q}(B_{ii})$;

$(iv)$ $g_{n}(A_{ii}B_{jj}) = \sum_{p+q = n}g_{p}(A_{ii})g_{q}B_{jj})$.

Proof. $(i)$ For each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{ij}\in\mathcal{T}_{ij}$ ($1\leq i\neq j\leq 2$), taking $X = A_{ii}, Y = B_{ij}, Z = P_{j}$ in Eq (2.1), and by Lemma 1.1 $(iii)$ and Lemma 2.1, we get

Therefore, it follows from Eq (2.2), with $f_{n}(A_{ii}) = 0$ and $f_{n}(A_{ij}) = P_{j}f_{n}(A_{ij})P_{i}$ ($n\geq 1$), that

Similarly, we get that $(ii)$ holds.

$(iii)$ For each $n\geq 1$ and for any $A_{ii}, B_{ii}\in\mathcal{T}_{ii}, X_{ij}\in\mathcal{T}_{ij}$ ($1\leq i\neq j\leq 2$), by Lemma 2.5 $(i)$ and Eq (2.3), on the one hand, we get

On the other hand, we have

Comparing the above two equations, we get

This yields from Lemma 1.1 $(i)$ that

Next, we show that

Indeed, for each $n\geq 1$ and for any $A_{ii}, B_{ii}\in\mathcal{T}_{ii}$ ($1\leq i\neq j\leq 2$), taking $X = A_{ii}, Y = Z = P_{j}$ in Eq (2.1), by Lemma 2.1, we get

Therefore, we get from $\sum_{p+q = n}d_{p}(A_{ii})d_{q}(P_{j})\in\mathcal{T}_{ij}$ and $\sum_{p+q = n}d_{q}(P_{j})d_{p}(A_{ii})\in\mathcal{T}_{ji}$ that

So we get from $f_{k}(A_{ii}) = 0$ and $f_{k}(P_{j}) = 0$ ($k\geq 1$) that

Therefore,

For each $n\geq 1$ and for any $A_{ii}, B_{ii}\in\mathcal{T}_{ii}$, by Eq (2.5), we get

This implies that

Similarly, we get

Therefore, by Eqs (2.4), (2.6), (2.7) and Lemma 2.4 $(ii)$, we get that

$(iv)$ For each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{jj}\in\mathcal{T}_{jj}$ ($1\leq i\neq j\leq 2$), taking $X = A_{ii}, Y = B_{jj}, Z = P_{j}$ ($1\leq i\neq j\leq 2$) in Eq (2.1), we get from Lemma 2.1 that

Hence, we get from $\sum_{p+q = n}d_{p}(A_{ii})d_{q}(B_{jj})\in\mathcal{T}_{ij}$ and $\sum_{p+q = n}d_{p}(B_{jj})d_{q}(A_{ii})\in\mathcal{T}_{ji}$ ($1\leq i\neq j\leq 2$) that

Therefore, it follows from $g_{k}(A_{ii}) = d_{k}(A_{ii})$ and $g_{k}(B_{jj}) = d_{k}(B_{jj})$ ($k\geq 1$) that

The proof is complete. □

Lemma 2.6. Let $g_{n}:\mathcal{T}\to\mathcal{T}$ be defined as in Theorem 2.3. Then for each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{jj}\in\mathcal{T}_{jj}, A_{ij}, B_{ij}\in\mathcal{T}_{ij}, A_{ji}, B_{ji}\in\mathcal{T}_{ji}$ ($1\leq i\neq j \leq 2$),

$(i)$ $g_{n}(A_{ij}B_{ji}) = \sum_{p+q}g_{p}(A_{ij})g_{q}(B_{ji})$;

$(ii)$ $g_{n}(A_{ij}B_{ij}) = \sum_{p+q = n}g_{p}(A_{ij})g_{q}(B_{ij})$;

$(iii)$ $g_{n}(A_{ii}B_{ji}) = \sum_{p+q = n}g_{p}(A_{ii})g_{q}(B_{ji})$;

$(iv)$ $g_{n}(A_{ji}B_{jj}) = \sum_{p+q = n}g_{p}(A_{ji})g_{q}(B_{jj})$.

Proof. $(i)$ For each $n\geq 1$ and for any $A_{ij}\in\mathcal{T}_{ij}, B_{ji}\in\mathcal{T}_{ji}$ ($1\leq i\neq j \leq 2$), it follows from Lemma 1 $(iii)$ that $A_{ij}B_{ji} = 0$, and therefore we get

On the other hand, by Lemma 1 $(iii)$ and Lemma 2.4 $(iii)$, we have $g_{p}(A_{ij})g_{q}(B_{ji}) = 0$, therefore we get

Similarly, we get that $(ii)$ holds.

$(iii)$ For each $n\geq 1$ and for any $A_{ii}\in\mathcal{T}_{ii}, B_{ji}\in\mathcal{T}_{ji}$ ($1\leq i\neq j \leq 2$), by Lemma 1 $(iii)$ and Lemma 2.4 $(ii, iii)$, we get $g_{p}(A_{ii})g_{q}(B_{ji}) = 0$, and therefore we get

Similarly, we get $(iv)$ holds. The proof is complete. □

In the following, we complete the proof of Theorem 2.3.

Proof of Theorem 2.3. For any $n\geq 1$, let $A = A_{11}+A_{12}+A_{21}+A_{22}$ and $B = B_{11}+B_{12}+B_{21}+B_{22}$ be arbitrary elements of $\mathcal T$, where $A_{ij}, B_{ij}\in\mathcal{T}_{ij}$ ($1\leq i, j\leq 2$). It follows from Lemmas 2.4–2.6 that

Therefore, $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation on $\mathcal{T}$. The proof is complete. □

Next, we show that Theorem 2.1 holds.

Proof of Theorem 2.1. For each $n\geq 1$ and for any $A, B\in\mathcal{T}$, by Theorems 2.2 and 2.3, we obtain that

where $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation and $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher anti-derivation from $\mathcal{T}$ into itself such that $f_{n}(A_{ii}) = 0$ for all $A_{ii}\in\mathcal{T}_{ii}$ ($1\leq i\leq 2$). The proof is complete. □

Remark 2.1. Let $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ be a higher Jordan triple derivation from $\mathcal{T}$ into itself. Then, by Theorems 2.1 and 2.2, we obtain that the following statements are equivalent.

$(i)$ $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation;

$(ii)$ $P_{j}d_{n}(A_{ij})P_{i} = 0$ for each $n\geq 1$ and for any $A_{ij}\in\mathcal{T}_{ij}$ $(1\leq i\neq j\leq 2)$;

$(iii)$ $d_{n}(A_{ij})\in\mathcal{T}_{ij}$ for each $n\geq 1$ and for any $A_{ij}\in\mathcal{T}_{ij}$ $(1\leq i\neq j\leq 2)$.

In the following, we show that every higher Jordan triple derivation on triangular algebras is a higher derivation.

Corollary 2.1. Let ${\mathcal A}$ and ${\mathcal B}$ be unital algebras over a commutative ring $\mathcal{R}$ and ${\mathcal M}$ be a unital $({\mathcal A}, {\mathcal B})$-bimodule, which is faithful as both a left ${\mathcal A}$-module and a right ${\mathcal B}$-module, and $\mathcal{U}$ be the $2$-torsion free triangular algebra, and $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ be a higher Jordan triple derivation on $\mathcal{U}$. Then $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation.

Proof of Corollary 2.1. Let $1_{\mathcal A}$ and $1_{\mathcal B}$ be the identities of the algebras ${\mathcal A}$ and ${\mathcal B}$, respectively, and let $1$ be the identity of the triangular algebra $\mathcal{U}$. We denote

and

It is clear that the triangular algebra $\mathcal U$ may be represented as

Here $P_{1}{\mathcal U}P_{1}$ and $P_{2}{\mathcal U}P_{2}$ are subalgebras of ${\mathcal U}$ isomorphic to ${\mathcal A}$ and ${\mathcal B}$, respectively, and $P_{1}{\mathcal U}P_{2}$ is a $(P_{1}{\mathcal U}P_{1}, P_{2}{\mathcal U}P_{2})$-bimodule isomorphic to the $({\mathcal A}, {\mathcal B})$-bimodule $\mathcal{M}$.

By the definition of triangular algebra $\mathcal{U}$, we can easily check that $\mathcal{U}$ is a $*$-type trivial extension algebra, and so if $\mathcal{U}$ is a $2$-torsion free triangular algebra, then for any $n\geq 1$, $A = A_{11}+A_{12}+A_{22}\in\mathcal{U}$, where $A_{ij}\in\mathcal{U}_{ij}$ ($1\leq i, j\leq 2$), we get from Theorem 2.1 that

Where $G = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation and $F = \{f_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher anti-derivation from $\mathcal{U}$ into itself such that $f_{n}(A_{ii}) = 0$ for all $A_{ii}\in\mathcal{U}_{ii}$ $(1\leq i\leq 2)$. Next, we show that $f_{n}(A_{12}) = 0$ for each $n\geq 1$ and for any $A_{12}\in\mathcal{U}_{12}$.

Indeed, for any $A_{12}\in\mathcal{U}_{12}$, it follows from Lemma 2.1 $(v)$ and $\mathcal{U}_{21} = \{0\}$ that

And then we obtain from the definition of $f_{n}$ in Theorem 2.2 that $f_{n}(A_{12}) = P_{2}d_{n}(A_{12})P_{1} = 0$. Therefore, for any $A\in\mathcal{U}$, $f_{n}(A) = 0$, so $D = \{g_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation. The proof is complete. □

Next, we give an application of Corollary 2.1 to certain special classes of triangular algebras, such as block upper triangular matrix algebras and nest algebras.

Let $\mathcal{R}$ be a commutative ring with identity and let $M_{n\times k}(\mathcal{R})$ be the set of all $n\times k$ matrices over $\mathcal{R}$. For $n\geq 2$ and $m\leq n$, the block upper triangular matrix algebra $T_n^{\bar{k}}(\mathcal{R})$ is a subalgebra of $M_n(\mathcal{R})$ with the form

where $\bar{k} = (k_1, k_2, \cdots, k_m)$ is an ordered $m$-vector of positive integers such that $k_1+k_2+\cdots+k_m = n$.

A nest of a complex Hilbert space $\mathcal{H}$ is a chain $\mathcal{N}$ of closed subspaces of $\mathcal{H}$ containing $\{0\}$ and $\mathcal{H}$, which is closed under arbitrary intersections and closed linear span, and $\mathcal{B(H)}$ is the algebra of all bounded linear operators on $\mathcal{H}$. The nest algebra associated with $\mathcal{N}$ is the algebra

A nest $\mathcal{N}$ is called trivial if $\mathcal{N} = \{0, \mathcal{H}\}$. It is clear that every nontrivial nest algebra is a triangular algebra and every finite dimensional nest algebra is isomorphic to a complex block upper triangular matrix algebra.

Corollary 2.2. Let $T_n^{\bar{k}}(\mathcal{R})$ be a $2$-torsion free block upper triangular matrix algebra, and $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ be a higher Jordan triple derivation on $T_n^{\bar{k}}(\mathcal{R})$. Then, $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation.

Corollary 2.3. Let $\mathcal{N}$ be a nontrivial nest of a complex Hilbert space $\mathcal{H}$, $\mathrm{Alg}\mathcal{N}$ a nest algebra, and $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ a higher Jordan triple derivation on $\mathrm{Alg}\mathcal{N}$. Then, $D = \{d_{n}\}_{n\in \mathbb{N}_{0}}$ is a higher derivation.

Use of AI tools declaration

The authors declare that they have not used artificial intelligence tools in the creation of this article.

Acknowledgments

This research is supported by the National Natural Science Foundation of China (No.11901451), Talent Project Foundation of Yunnan Provincial Science and Technology Department (No.202105AC160089), Natural Science Foundation of Yunnan Province (No.202101BA070001198), and Basic Research Foundation of Yunnan Education Department (No.2021J0915).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.