Transposed Poisson structures on solvable and perfect Lie algebras111 The work
is supported by FCT UIDB/00212/2020, UIDP/00212/2020, 2022.02474.PTDC; grant
FZ-202009269, Ministry of higher education, science and innovations of the
Republic of Uzbekistan.

Ivan Kaygorodov222CMA-UBI, Universidade da Beira Interior, Covilhã, Portugal;
<EMAIL_ADDRESS>& Abror Khudoyberdiyev333 V.I.Romanovskiy Institute
of Mathematics Academy of Science of Uzbekistan; National University of
Uzbekistan<EMAIL_ADDRESS>

Abstract. We described all transposed Poisson algebra structures on oscillator
Lie algebras, i.e., on one-dimensional solvable extensions of the
$(2n+1)$-dimensional Heisenberg algebra; on solvable Lie algebras with
naturally graded filiform nilpotent radical; on $(n+1)$-dimensional solvable
extensions of the $(2n+1)$-dimensional Heisenberg algebra; and on
$n$-dimensional solvable extensions of the $n$-dimensional algebra with the
trivial multiplication. We also gave an answer to one question on transposed
Poisson algebras early posted in a paper by Beites, Ferreira and Kaygorodov.
Namely, we found a finite-dimensional Lie algebra with non-trivial
$\frac{1}{2}$-derivations, but without non-trivial transposed Poisson algebra
structures.

Keywords: Lie algebra, transposed Poisson algebra, $\delta$-derivation.

MSC2020: 17A30, 17B40, 17B61, 17B63.

## Introduction

Since their origin in the 1970s in Poisson geometry, Poisson algebras have
appeared in several areas of mathematics and physics, such as algebraic
geometry, operads, quantization theory, quantum groups, and classical and
quantum mechanics. One of the natural tasks in the theory of Poisson algebras
is the description of all such algebras with fixed Lie or associative part
[28, 1, 14, 15].

Recently, Bai, Bai, Guo and Wu have introduced a dual notion of the Poisson
algebra [2], called a transposed Poisson algebra, by exchanging the roles of
the two multiplications in the Leibniz rule defining a Poisson algebra. A
transposed Poisson algebra defined this way not only shares some properties of
a Poisson algebra, such as the closedness under tensor products and the Koszul
self-duality as an operad but also admits a rich class of identities [25, 2,
11, 5]. It is important to note that a transposed Poisson algebra naturally
arises from a Novikov-Poisson algebra by taking the commutator Lie algebra of
its Novikov part [2]. Any unital transposed Poisson algebra is a particular
case of a “contact bracket” algebra and a quasi-Poisson algebra [5]. Each
transposed Poisson algebra is a commutative Gelfand-Dorfman algebra [25] and
it is also an algebra of Jordan brackets [11]. Transposed Poisson algebras are
related to weak Leibniz algebras [10]. In a recent paper by Ferreira,
Kaygorodov, Lopatkin a relation between $\frac{1}{2}$-derivations of Lie
algebras and transposed Poisson algebras has been established [12]. These
ideas were used to describe all transposed Poisson structures on Witt and
Virasoro algebras in [12]; on twisted Heisenberg-Virasoro, Schrödinger-
Virasoro and extended Schrödinger-Virasoro algebras in [29]; on Schrödinger
algebra in $(n+1)$-dimensional space-time in [27]; on Witt type Lie algebras
in [17]; on generalized Witt algebras in [18]; Block Lie algebras in [16, 18];
on the Lie algebra of upper triangular matrices in [19]; and on Lie incidence
algebras in [20]. Any complex finite-dimensional solvable Lie algebra was
proved to admit a non-trivial transposed Poisson structure [21]. The algebraic
and geometric classification of $3$-dimensional transposed Poisson algebras
was given in [4]. For the list of actual open questions on transposed Poisson
algebras, see [5].

The class of oscillator Lie algebras, which are the only noncommutative
solvable Lie algebras that carry a bi-invariant Lorentzian metric [22], is a
class of quadratic Lie algebras. The former algebras, which are the Lie
algebras of oscillator Lie groups, have aroused some interest [6, 1, 8, 9, 7,
3]. Ballesteros and Herranz obtained all possible coboundary Lie bialgebras
for the four-dimensional oscillator algebra [3]. Biggs and Remsing examined
the algebraic structure of the four-dimensional oscillator Lie algebra[6].
Firstly, the adjoint orbits are determined (and graphed); these orbits turn
out to be linearly isomorphic to the coadjoint orbits. This prompts the
identification of a family of invariant scalar products (from which the
Casimir functions can be recovered). They identified the group of
automorphisms (resp. inner automorphisms); subsequently, they classified all
linear subspaces. As a by-product, they arrived at classifications of the
full-rank subspaces, the subalgebras, and the ideals. Based on the four-
dimensional oscillator Lie algebra, Crampé, van de Vijver, and Vinet
introduced an associative algebra which they call the oscillator Racah
algebra, and explain a relation between the representation theory of and
multivariate Krawtchouk polynomials [9]. In [8], Calvaruso and Zaeim obtained
a classification of all Ricci, curvature, Weyl, and matter collineations for
the four-dimensional oscillator group equipped with a 1-parameter family of
left-invariant Lorentzian metrics. Boucetta and Medina determined the Lie
bialgebra structures and the solutions of the classical Yang–Baxter equation
on a generic class of oscillator Lie algebras [7]. Albuquerque, Barreiro,
Benayadi, Boucetta, and Sánchez described all Poisson algebra structures and
symmetric Leibniz bialgebra structures on generic oscillator Lie algebras [1].
The present paper is dedicated to studying transposed Poisson algebra
structures on oscillator Lie algebras. Namely, we have new examples of non-
trivial ${\rm Hom}$-Lie algebra structures and transposed Poisson structures
on oscillator Lie algebras.

In Section 2, we give the full description of all non-isomorphic transposed
Poisson structures on generic oscillator Lie algebras, i.e., on one-
dimensional solvable extensions of the $(2n+1)$-dimensional Heisenberg
algebra. Section 3 is dedicated to a study of transposed Poisson algebra
structures on some solvable algebras with a special type of nilpotent radical,
obtained in some papers by Winternitz with co-authors [26, 23, 24]. Namely, we
described all transposed Poisson algebra structures on solvable Lie algebras
with naturally graded filiform nilpotent radical; on $(n+1)$-dimensional
solvable extensions of the $(2n+1)$-dimensional Heisenberg algebra; and on
$n$-dimensional solvable extensions of the $n$-dimensional algebra with the
trivial multiplication. In the last section, we also gave an answer to one
question on transposed Poisson algebras early posted in a paper by Beites,
Ferreira, and Kaygorodov. Namely, we found a finite-dimensional Lie algebra
with non-trivial $\frac{1}{2}$-derivations, but without non-trivial transposed
Poisson algebra structures.

##  1\. Preliminaries

All the algebras below will be over the complex field and all the linear maps
will be $\mathbb{C}$-linear, unless otherwise stated. The notation $\langle
S\rangle$ means the $\mathbb{C}$-subspace generated by $S$.

######  Definition 1.

Let ${\mathfrak{L}}$ be a vector space equipped with two nonzero bilinear
operations $\cdot$ and $[-,-].$ The triple $({\mathfrak{L}},\cdot,[-,-])$ is
called a transposed Poisson algebra if $({\mathfrak{L}},\cdot)$ is a
commutative associative algebra and $({\mathfrak{L}},[-,-])$ is a Lie algebra
that satisfies the following compatibility condition

$2z\cdot[x,y]=[z\cdot x,y]+[x,z\cdot y].$

Transposed Poisson algebras were first introduced in a paper by Bai, Bai, Guo
and Wu [2].

######  Definition 2.

Let $({\mathfrak{L}},[-,-])$ be a Lie algebra. A transposed Poisson algebra
structure on $({\mathfrak{L}},[-,-])$ is a commutative associative
multiplication $\cdot$ on $\mathfrak{L}$ which makes
$({\mathfrak{L}},\cdot,[-,-])$ a transposed Poisson algebra.

######  Definition 3.

Let $({\mathfrak{L}},[-,-])$ be an algebra with a multiplication $[-,-],$
$\varphi$ be a linear map and $\phi$ be a bilinear map. Then $\varphi$ is a
$\frac{1}{2}$-derivation if it satisfies

$\varphi[x,y]=\frac{1}{2}\big{(}[\varphi(x),y]+[x,\varphi(y)]\big{)};$

$\phi$ is a $\frac{1}{2}$-biderivation if it satisfies

$\phi([x,y],z)$ | $=$ | $\frac{1}{2}\big{(}[\phi(x,z),y]+[x,\phi(y,z)]\big{)},$
---|---|---
$\phi(x,[y,z])$ | $=$ | $\frac{1}{2}\big{(}[\phi(x,y),z]+[y,\phi(x,z)]\big{)}.$

Observe that $\frac{1}{2}$-derivations are a particular case of
$\delta$-derivations introduced by Filippov (see, for example, [30] and
references therein). It is easy to see from Definition 3 that
$[\mathfrak{L},\mathfrak{L}]$ and $\operatorname{Ann}(\mathfrak{L})$ are
invariant under any $\frac{1}{2}$-derivation of $\mathfrak{L}$.

Definitions 1 and 3 immediately implies the following key Lemma.

######  Lemma 4.

Let $({\mathfrak{L}},[-,-])$ be a Lie algebra and $\cdot$ a new binary
(bilinear) operation on ${\mathfrak{L}}$. Then $({\mathfrak{L}},\cdot,[-,-])$
is a transposed Poisson algebra if and only if $\cdot$ is commutative and
associative and for every $z\in{\mathfrak{L}}$ the multiplication by $z$ in
$({\mathfrak{L}},\cdot)$ is a $\frac{1}{2}$-derivation of
$({\mathfrak{L}},[-,-]).$

The basic example of a $\frac{1}{2}$-derivation is the multiplication by a
field element. Such $\frac{1}{2}$-derivations will be called trivial.

######  Theorem 5.

Let ${\mathfrak{L}}$ be a Lie algebra without non-trivial
$\frac{1}{2}$-derivations. Then all transposed Poisson algebra structures on
${\mathfrak{L}}$ are trivial.

Let $\cdot$ be a transposed Poisson algebra structure on a Lie algebra
$({\mathfrak{L}},[-,-])$. Then any automorphism $\phi$ of
$({\mathfrak{L}},[-,-])$ induces the transposed Poisson algebra structure $*$
on $({\mathfrak{L}},[-,-])$ given by

$\displaystyle x*y=\phi\big{(}\phi^{-1}(x)\cdot\phi^{-1}(y)\big{)},\ \
x,y\in{\mathfrak{L}}.$

Clearly, $\phi$ is an isomorphism of transposed Poisson algebras
$({\mathfrak{L}},\cdot,[-,-])$ and $({\mathfrak{L}},*,[-,-])$.

##  2\. Transposed Poisson algebra structures on oscillator Lie algebras

######  Definition 6.

The oscillator Lie algebra $\mathfrak{L_{\lambda}}$ is a $(2n+2)$-dimensional
Lie algebra with its canonical basis
$\mathbb{B}=\\{e_{-1},e_{0},e_{j},\check{e}_{j}\\}_{j=1,\ldots,n}$ and the Lie
brackets are given by

$[e_{-1},e_{j}]=\lambda_{j}\check{e}_{j},\ \ \
[e_{-1},\check{e}_{j}]=-\lambda_{j}e_{j},\ \ \ [e_{j},\check{e}_{j}]=e_{0},$

for $j=1,\ldots,n$ and
$\lambda=(\lambda_{1},\ldots,\lambda_{n})\in\mathbb{R}^{n}$ with
$0<\lambda_{1}\leq\ldots\leq\lambda_{n}$.

######  Proposition 7.

A linear map $\varphi:\mathfrak{L}_{\lambda}\rightarrow\mathfrak{L}_{\lambda}$
is a $\frac{1}{2}$-derivation of the algebra $(\mathfrak{L}_{\lambda},[-,-])$
if and only if

$\varphi(e_{-1})$ | $=$ |  $\gamma e_{-1}+\mu e_{0}-\sum\limits_{j=1}^{n}2\lambda_{j}\alpha_{j}e_{j}-\sum\limits_{j=1}^{n}2\lambda_{j}\beta_{j}\check{e}_{j}$,
---|---|---
$\varphi(e_{0})$ | $=$ |  $\gamma e_{0}$,
$\varphi(e_{j})$ | $=$ |  $\alpha_{j}e_{0}+\gamma e_{j},\ \ \ (j=1,\ldots,n)$,
$\varphi(\check{e}_{j})$ | $=$ | $\beta_{j}e_{0}+\gamma\check{e}_{j},\ \ \ (j=1,\ldots,n).$

###### Proof.

For each $\ell\in\mathbb{B},$ we consider

$\varphi(\ell)=\varphi(\ell)_{-1}e_{-1}+\varphi(\ell)_{0}e_{0}+\sum\limits_{k=1}^{n}(\varphi(\ell)_{k}e_{k}+\varphi(\ell)_{\check{k}}\check{e}_{k})$.

Since $[\mathfrak{L},\mathfrak{L}]$ and $\operatorname{Ann}(\mathfrak{L})$ are
invariant under any $\frac{1}{2}$-derivation, we obtain that
$\varphi(e_{0})=\gamma e_{0}$ and
$\varphi(e_{j})_{-1}=\varphi(\check{e}_{j})_{-1}=0.$

Now from $e_{0}=[e_{j},\check{e}_{j}]$ for each $j\in\\{1,\ldots,n\\},$ we get

$\varphi(e_{0})=\varphi([e_{j},\check{e}_{j}])=\frac{1}{2}([\varphi(e_{j}),\check{e}_{j}]+[e_{j},\varphi(\check{e}_{j})])=\frac{1}{2}(\varphi(e_{j})_{j}+\varphi(\check{e}_{j})_{\check{j}}e_{0}),$

which implies $\varphi(e_{j})_{j}+\varphi(\check{e}_{j})_{\check{j}}=2\gamma.$

Since $[e_{-1},e_{j}]=\lambda_{j}\check{e}_{j}$ and
$[e_{-1},\check{e}_{j}]=-\lambda_{j}e_{j}$ then

$\varphi(\lambda_{j}\check{e}_{j})$ | $=$ | $\varphi([e_{-1},e_{j}])=\frac{1}{2}([\varphi(e_{-1}),e_{j}]+[e_{-1},\varphi(e_{j})])$
---|---|---
| $=$ | $\frac{1}{2}\lambda_{j}\varphi(e_{-1})_{-1}\check{e}_{j}-\frac{1}{2}\varphi(e_{-1})_{\check{j}}e_{0}+\frac{1}{2}\sum\limits_{k=1}^{n}\Big{(}\lambda_{k}\varphi(e_{j})_{k}\check{e}_{k}-\lambda_{k}\varphi(e_{j})_{\check{k}}e_{k}\Big{)},$

and

$-\varphi(\lambda_{j}e_{j})$ | $=$ | $\varphi([e_{-1},\check{e}_{j}])=\frac{1}{2}([\varphi(e_{-1}),\check{e}_{j}]+[e_{-1},\varphi(\check{e}_{j})])$
---|---|---
| $=$ | $-\frac{1}{2}\lambda_{j}\varphi(e_{-1})_{-1}e_{j}+\frac{1}{2}\varphi(e_{-1})_{j}e_{0}+\frac{1}{2}\Big{(}\sum\limits_{k=1}^{n}\lambda_{k}\varphi(\check{e}_{j})_{k}\check{e}_{k}-\lambda_{k}\varphi(\check{e}_{j})_{\check{k}}e_{k}\Big{)}.$

Hence, for any $j\in\\{1,\ldots,n\\},$ we have

$\varphi(e_{-1})_{-1}=\varphi(e_{j})_{j}=\varphi(\check{e}_{j})_{\check{j}}=\gamma,\quad\varphi(e_{-1})_{j}=-2\lambda_{j}\varphi(e_{j})_{0},\quad\varphi(e_{-1})_{\check{j}}=-2\lambda_{j}\varphi(e_{\check{j}})_{0},$

and

$2\lambda_{j}\varphi(\check{e}_{j})_{k}=-\lambda_{k}\varphi(e_{j})_{\check{k}},\quad
2\lambda_{j}\varphi(e_{j})_{\check{k}}=-\lambda_{k}\varphi(\check{e}_{j})_{k},\quad
1\leq j,k\leq n,$ (1)

$2\lambda_{j}\varphi(\check{e}_{j})_{\check{k}}=\lambda_{k}\varphi(e_{j})_{k},\quad
2\lambda_{j}\varphi(e_{j})_{k}=\lambda_{k}\varphi(\check{e}_{j})_{\check{k}},\quad
1\leq j,k(j\neq k)\leq n.$ (2)

Finally, from

$0$ | $=$ | $\varphi([e_{j},e_{k}])$ | $=$ | $\frac{1}{2}([\varphi(e_{j}),e_{k}]+[e_{j},\varphi(e_{k})]),$
---|---|---|---|---
$0$ | $=$ | $\varphi([e_{j},\check{e}_{k}])$ | $=$ | $\frac{1}{2}([\varphi(e_{j}),\check{e}_{k}]+[e_{j},\varphi(\check{e}_{k})]),$
$0$ | $=$ | $\varphi([\check{e}_{j},\check{e}_{k}])$ | $=$ | $\frac{1}{2}([\varphi(\check{e}_{j}),\check{e}_{k}]+[\check{e}_{j},\varphi(\check{e}_{k})]),$

for $1\leq j,k(j\neq k)\leq n,$ we derive that

$\varphi(e_{j})_{\check{k}}=\varphi(e_{k})_{\check{j}},\quad\varphi(e_{j})_{k}=-\varphi(\check{e}_{k})_{\check{j}},\quad\varphi(\check{e}_{j})_{k}=\varphi(\check{e}_{k})_{j},$
(3)

It is not difficult to deduce from (1), (2) and (3) that

$\varphi(e_{j})_{t}=0$ for
$t\in\\{1,\ldots,n,\check{1},\ldots,\check{n}\\}\setminus\\{j\\},$

$\varphi(e_{\check{j}})_{t}=0$ for
$t\in\\{1,\ldots,n,\check{1},\ldots,\check{n}\\}\setminus\\{\check{j}\\}.$

Hence, our map $\varphi$ has the following type:

$\varphi(e_{-1})$ | $=$ |  $\gamma e_{-1}+\mu e_{0}-\sum\limits_{j=1}^{n}2\lambda_{j}\alpha_{j}e_{j}-\sum\limits_{j=1}^{n}2\lambda_{j}\beta_{j}\check{e}_{j}$,
---|---|---
$\varphi(e_{0})$ | $=$ |  $\gamma e_{0}$,
$\varphi(e_{j})$ | $=$ |  $\alpha_{j}e_{0}+\gamma e_{j},\ \ \ (j=1,\ldots,n)$,
$\varphi(\check{e}_{j})$ | $=$ | $\beta_{j}e_{0}+\gamma\check{e}_{j},\ \ \ (j=1,\ldots,n).$

It is easy to see that all linear maps given by the present way are
$\frac{1}{2}$-derivations. It finishes the proof.

∎

Filippov proved that each $\delta$-derivation ($\delta\neq 0,1$) gives a non-
trivial ${\rm Hom}$-Lie algebra structure [30, Theorem 1]. Hence, by
Proposition 7, we have the following corollary.

######  Corollary 8.

Each oscillator Lie algebra admits a non-trivial ${\rm Hom}$-Lie algebra
structure.

######  Theorem 9.

Let $(\mathfrak{L}_{\lambda},[-,-])$ be the oscillator Lie algebra. Then every
symmetric $\frac{1}{2}$-biderivation gives a transposed Poisson structure. If
$(\mathfrak{L}_{\lambda},\cdot,[-,-])$ is a transposed Poisson algebra, then
$(\mathfrak{L}_{\lambda},\cdot)$ has the following multiplication:

$e_{-1}\cdot e_{-1}$ | $=$ | $\gamma e_{-1}+\mu e_{0}-2\sum\limits_{k=1}^{n}\lambda_{k}(\alpha_{k}e_{k}+\beta_{k}\check{e}_{k}),$
---|---|---
$e_{-1}\cdot e_{0}$ | $=$ | $\gamma e_{0},$
$e_{-1}\cdot e_{j}$ | $=$ | $\alpha_{j}e_{0}+\gamma e_{j},$
$e_{-1}\cdot\check{e}_{j}$ | $=$ | $\beta_{j}e_{0}+\gamma\check{e}_{j},$
$e_{j}\cdot e_{j}$ | $=$ | $\check{e}_{j}\cdot\check{e}_{j}=-\frac{\gamma}{2\lambda_{j}}e_{0}.$

$(\mathfrak{L}_{\lambda},\cdot,[-,-])$ is a Poisson algebra if and only if
$(\gamma,\alpha_{1},\ldots\alpha_{n},\beta_{1},\ldots,\beta_{n})=0.$

###### Proof.

We aim to describe the multiplication $\cdot.$ By Lemma 4, for each
$\ell\in\mathbb{B}$, there is a related $\frac{1}{2}$-derivation
$\varphi_{\ell}$ of $(\mathfrak{L}_{\lambda},[-,-])$ such that
$\varphi_{\ell_{1}}(\ell_{2})=\ell_{1}\cdot\ell_{2}=\varphi_{\ell_{2}}(\ell_{1}),$
with $\ell_{1},\ell_{2}\in\mathbb{B}.$ By Proposition 7, for each
$\ell\in\mathbb{B}$, we have

$\varphi_{\ell}(e_{-1})$ | $=$ |  $\gamma^{\ell}e_{-1}+\mu^{\ell}e_{0}-\sum\limits_{k=1}^{n}2\lambda_{k}\alpha^{\ell}_{k}e_{k}-\sum\limits_{k=1}^{n}2\lambda_{k}\beta^{\ell}_{k}\check{e}_{k}$,
---|---|---
$\varphi_{\ell}(e_{0})$ | $=$ |  $\gamma^{\ell}e_{0}$,
$\varphi_{\ell}(e_{j})$ | $=$ |  $\alpha^{\ell}_{j}e_{0}+\gamma^{\ell}e_{j},\ \ \ (j=1,\ldots,n)$,
$\varphi_{\ell}(\check{e}_{j})$ | $=$ |  $\beta^{\ell}_{j}e_{0}+\gamma^{\ell}\check{e}_{j},\ \ \ (j=1,\ldots,n)$.

In the first, for all $i,j\in\\{1,\ldots,n\\},$ we have

$\alpha^{e_{i}}_{j}e_{0}+\gamma^{e_{i}}e_{j}$ | $=$ | $\varphi_{e_{i}}(e_{j})$ | $=$ | $\varphi_{e_{j}}(e_{i})$ | $=$ |  $\alpha^{e_{j}}_{i}e_{0}+\gamma^{e_{j}}e_{i}$,
---|---|---|---|---|---|---
$\alpha^{\check{e}_{i}}_{j}e_{0}+\gamma^{\check{e}_{i}}e_{j}$ | $=$ | $\varphi_{\check{e}_{i}}(e_{j})$ | $=$ | $\varphi_{e_{j}}(\check{e}_{i})$ | $=$ |  $\beta^{e_{j}}_{i}e_{0}+\gamma^{e_{j}}\check{e}_{i}$,
$\beta^{\check{e}_{i}}_{j}e_{0}+\gamma^{\check{e}_{i}}\check{e}_{j}$ | $=$ | $\varphi_{\check{e}_{i}}(\check{e}_{j})$ | $=$ | $\varphi_{\check{e}_{j}}(\check{e}_{i})$ | $=$ | $\beta^{\check{e}_{j}}_{i}e_{0}+\gamma^{\check{e}_{j}}\check{e}_{i}.$

Hence $\beta^{\check{e}_{j}}_{i}=\beta^{\check{e}_{i}}_{j}$,
$\alpha^{e_{i}}_{j}=\alpha^{e_{j}}_{i}$,
$\alpha^{\check{e}_{i}}_{j}=\beta^{e_{j}}_{i}$ and
$\gamma^{e_{j}}=\gamma^{\check{e}_{j}}=0$.

Second, we have

$\gamma^{e_{-1}}e_{0}=\varphi_{e_{-1}}(e_{0})=\varphi_{e_{0}}(e_{-1})=\gamma^{e_{0}}e_{-1}+\mu^{e_{0}}e_{0}-2\sum\limits_{k=1}^{n}\lambda_{k}(\alpha_{k}^{e_{0}}e_{k}+\beta_{k}^{e_{0}}\check{e}_{k}),$

which implies $\alpha^{e_{0}}_{j}=\beta^{e_{0}}_{j}=\gamma^{e_{0}}=0$ and
$\gamma^{e_{-1}}=\mu^{e_{0}}$.

Also, for all $j\in\\{1,\ldots,n\\},$ we have

$\alpha^{e_{-1}}_{j}e_{0}+\gamma^{e_{-1}}e_{j}$ | $=$ | $\varphi_{e_{-1}}(e_{j})$ | $=$ | $\varphi_{e_{j}}(e_{-1})$ | $=$ | $\mu^{e_{j}}e_{0}-2\sum\limits_{k=1}^{n}\lambda_{k}(\alpha^{e_{j}}_{k}e_{k}+\beta^{e_{j}}_{k}\check{e}_{k}),$
---|---|---|---|---|---|---
$\beta^{e_{-1}}_{j}e_{0}+\gamma^{e_{-1}}\check{e}_{j}$ | $=$ | $\varphi_{e_{-1}}(\check{e}_{j})$ | $=$ | $\varphi_{\check{e}_{j}}(e_{-1})$ | $=$ | $\mu^{\check{e}_{j}}e_{0}-2\sum\limits_{k=1}^{n}\lambda_{k}(\alpha^{\check{e}_{j}}_{k}e_{k}+\beta^{\check{e}_{j}}_{k}\check{e}_{k}),$

which implies

$\gamma^{e_{-1}}=-2\lambda_{j}\alpha^{e_{j}}_{j},$ | $\alpha^{e_{j}}_{j}=\beta^{\check{e}_{j}}_{j},$ | $\alpha^{\check{e}_{j}}_{k}=0,$ | $\beta^{e_{j}}_{k}=0,$
---|---|---|---
$\alpha^{e_{j}}_{k}=0\ (j\neq k),$ | $\beta^{\check{e}_{j}}_{k}=0\ (j\neq k),$ | $\alpha^{e_{-1}}_{j}=\mu^{e_{j}},$ |  $\beta^{e_{-1}}_{j}=\mu^{\check{e}_{j}}$.

It follows that, for all $j\in\\{1,\ldots,n\\}$, the multiplication $\cdot$
can be written as

$e_{-1}\cdot e_{-1}$ | $=$ | $\gamma e_{-1}+\mu e_{0}-2\sum\limits_{k=1}^{n}\lambda_{k}(\alpha_{k}e_{k}+\beta_{k}\check{e}_{k}),$
---|---|---
$e_{-1}\cdot e_{0}$ | $=$ | $\gamma e_{0},$
$e_{-1}\cdot e_{j}$ | $=$ | $\alpha_{j}e_{0}+\gamma e_{j},$
$e_{-1}\cdot\check{e}_{j}$ | $=$ | $\beta_{j}e_{0}+\gamma\check{e}_{j},$
$e_{j}\cdot e_{j}$ | $=$ | $\check{e}_{j}\cdot\check{e}_{j}=-\frac{\gamma}{2\lambda_{j}}e_{0}.$

It is easy to see that $\cdot$ is a commutative associative multiplication.

∎

### Isomorphism problem for transposed Poisson algebras on generic oscillator
Lie algebras

An oscillator Lie algebra is called generic if
$0<\lambda_{1}<\ldots<\lambda_{n}$ and
$\lambda_{i}+\lambda_{j}\neq\lambda_{k}$ for all $1\leq i<j<k\leq n.$ It is
easy to see that
$(\mathfrak{L}_{\\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\\}},[-,-])$ is
isomorphic to
$(\mathfrak{L}_{\left\\{1,\frac{\lambda_{2}}{\lambda_{1}},\ldots,\frac{\lambda_{n}}{\lambda_{1}}\right\\}},[-,-]).$
Hence, we will consider only the case $\lambda_{1}=1.$

To prove the main result of this part of the paper, we should give the
following useful Lemma.

######  Lemma 10.

Let $\phi$ be an automorphism of $(\mathfrak{L}_{\lambda},[-,-])$. Then $\phi$
has the following types:

$\phi(e_{-1})$ | $=$ | $\pm e_{-1}+\nu e_{0}+\sum\limits_{i=1}^{n}\nu_{i}e_{i}+\sum\limits_{i=1}^{n}\check{\nu}_{i}\check{e}_{i},$
---|---|---
$\phi(e_{0})$ | $=$ | $\pm\xi e_{0},$
$\phi(e_{i})$ | $=$ | $\frac{\check{\nu}_{i}\check{\mu}_{i}\mp\nu_{i}\mu_{i}}{\lambda_{i}}e_{0}+\mu_{i}e_{i}\mp\check{\mu}_{i}\check{e}_{i},$
$\phi(\check{e}_{i})$ | $=$ | $\frac{-\check{\nu}_{i}\check{\mu}_{i}\mp\nu_{i}\mu_{i}}{\lambda_{i}}e_{0}+\check{\mu}_{i}e_{i}\pm\mu_{i}\check{e}_{i},$

where $\xi=\mu_{i}^{2}+\check{\mu}_{i}^{2}.$

###### Proof.

Let us say that

$\phi(e_{m})=\alpha^{m}_{-1}e_{-1}+\alpha^{m}_{0}e_{0}+\sum\limits_{i=1}^{n}(\alpha_{i}^{m}e_{i}+\beta_{i}^{m}\check{e}_{i}),$
where $m\in\\{-1,0,1,\ldots,n\\}.$

Since $\operatorname{Ann}(\mathfrak{L})$ and $[\mathfrak{L},\mathfrak{L}]$ are
invariant under any automorphism, we have that

$\alpha_{-1}^{0}=\alpha_{i}^{0}=\beta_{i}^{0}=0,\quad\alpha_{i}^{-1}=0,\quad
1\leq i\leq n.$

Then, for each $j\in\\{1,\ldots,n\\},$ we have

$\lambda_{j}^{2}\phi(e_{j})=-[\phi(e_{-1}),[\phi(e_{-1}),\phi(e_{j})]]=(\alpha^{-1}_{-1})^{2}\sum\limits_{i=1}^{n}\lambda_{i}^{2}(\alpha_{i}^{j}e_{i}+\beta_{i}^{j}\check{e}_{j})-\alpha^{-1}_{-1}\sum\limits_{i=1}^{n}(\alpha_{i}^{-1}\alpha_{i}^{j}+\beta_{i}^{-1}\beta_{i}^{j})e_{0}.$

Since $\lambda_{i}\neq\lambda_{j},$ we have that

$\alpha_{-1}^{-1}=\pm
1,\quad\alpha^{j}_{0}=-\frac{\alpha_{-1}^{-1}}{\lambda_{j}}(\alpha_{j}^{-1}\alpha_{j}^{j}+\beta_{j}^{-1}\beta_{j}^{j}),\quad\alpha_{i}^{j}=\beta^{j}_{i}=0,\quad
1\leq i(i\neq j)\leq n.$

Thus, we obtain

$\phi(e_{j})=-\frac{\alpha_{-1}^{-1}}{\lambda_{j}}(\alpha_{j}^{-1}\alpha_{j}^{j}+\beta_{j}^{-1}\beta_{j}^{j})e_{0}+\alpha_{j}^{j}e_{j}+\beta^{j}_{j}\check{e}_{j}$

In a similar way, considering
$\lambda_{j}^{2}\phi(\check{e}_{j})=-[\phi(e_{-1}),[\phi(e_{-1}),\phi(\check{e}_{j})]],$
we have

$\phi(\check{e}_{j})=-\frac{\alpha_{-1}^{-1}}{\lambda_{j}}(\alpha_{j}^{-1}\mu_{j}^{j}+\beta_{j}^{-1}\nu_{j}^{j})e_{0}+\mu_{j}^{j}e_{j}+\nu^{j}_{j}\check{e}_{j}.$

Finally, using the equalities

$\phi[e_{i},\check{e}_{i}]=[\phi(e_{i}),\phi(\check{e}_{i})],\quad\phi[e_{-1},{e}_{i}]=[\phi(e_{-1}),\phi({e}_{i})],\quad\phi[e_{i},\check{e}_{i}]=[\phi(e_{i}),\phi(\check{e}_{i})],$

we derive that $\phi$ has the form from our statement. ∎

For our main theorem from this part, we need to define the following set

$\mathbb{C}_{>0}=\\{a+b{\rm i}\\}_{a>0}\cup\\{b{\rm i}\\}_{b\geq 0}.$

######  Theorem 11.

Let $(\mathfrak{L}_{\lambda},[-,-])$ be the generic oscillator Lie algebra. If
$(\mathfrak{L}_{\lambda},\cdot,[-,-])$ is a transposed Poisson algebra
structure on $(\mathfrak{L}_{\lambda},[-,-])$, then
$(\mathfrak{L}_{\lambda},\cdot)$ has the following multiplication.

  1. (A)

If $(\mathfrak{L}_{\lambda},\cdot)$ is non-nilpotent, then $\gamma\neq 0$ and
we have:

$e_{-1}\cdot e_{-1}=\gamma e_{-1},\ e_{-1}\cdot e_{0}=\gamma e_{0},\
e_{-1}\cdot e_{j}=\gamma e_{j},\
e_{-1}\cdot\check{e}_{j}=\gamma\check{e}_{j},\ e_{j}\cdot
e_{j}=\check{e}_{j}\cdot\check{e}_{j}=-\frac{\gamma}{2\lambda_{j}}e_{0},$

corresponding transposed Poisson algebra
$(\mathfrak{L}_{\\{\lambda,\gamma\\}},\cdot,[-,-])$ is isomorphic only to
$(\mathfrak{L}_{\\{\lambda,-\gamma\\}},\cdot,[-,-])$; all these algebras are
non-Poisson.

  2. (B)

If $(\mathfrak{L}_{\lambda},\cdot)$ is nilpotent, then there is a sequence
$\\{\beta_{q}\in\mathbb{C}_{>0}\\}_{1\leq q\leq n},$ where the first nonzero
element is equal to $1,$ and $(\mathfrak{L}_{\lambda},\cdot)$ has the
following multiplication table:

    1. (a)

$e_{-1}\cdot
e_{-1}=e_{0}-2\sum\limits_{j=1}^{n}\lambda_{j}\beta_{j}\check{e}_{j},\
e_{-1}\cdot\check{e}_{j}=\beta_{j}\check{e}_{j},$

    2. (b)

$e_{-1}\cdot e_{-1}=-2\sum\limits_{j=1}^{n}\lambda_{j}\beta_{j}\check{e}_{j},\
e_{-1}\cdot\check{e}_{j}=\beta_{j}\check{e}_{j}.$

The corresponding transposed Poisson algebras
$(\mathfrak{L}_{\\{\lambda,\beta\\}},\cdot,[-,-])$ are not isomorphic; and
they are non-Poisson if there exists $\beta_{j}\neq 0$.

###### Proof.

Let us consider the transposed Poisson structure
$(\mathfrak{L}_{\lambda},\cdot,[-,-])$ on the generic oscillator Lie
$(\mathfrak{L}_{\lambda},[-,-])$ and $\cdot$ is given as in Theorem 9. Under
the action of a positive automorphism of the Lie algebra
$(\mathfrak{L}_{\lambda},[-,-]),$ given in Lemma 10, we rewrite the
multiplication table of $(\mathfrak{L}_{\lambda},\cdot)$ by the following way:

$e_{-1}\cdot e_{-1}$ | $=$ | $\gamma e_{-1}+\left(\frac{\mu+\gamma\nu}{\xi}+\sum\limits_{i=1}^{n}\frac{\gamma(\nu_{i}^{2}+\check{\nu}_{i}^{2})}{2\lambda_{i}\xi}\right)e_{0}+$
---|---|---
|  | $\sum\limits_{k=1}^{n}\left(\frac{\gamma(\mu_{k}\nu_{k}-\check{\mu}_{k}\check{\nu}_{k})-2\lambda_{k}(\alpha_{k}\mu_{k}-\beta_{k}\check{\mu}_{k})}{\xi}\right)e_{k}+\sum\limits_{k=1}^{n}\left(\frac{\gamma(\check{\mu}_{k}\nu_{k}+\mu_{k}\check{\nu}_{k})-2\lambda_{k}(\alpha_{k}\check{\mu}_{k}+\beta_{k}\mu_{k})}{\xi}\right)\check{e}_{k}$
$e_{-1}\cdot e_{0}$ | $=$ | $\gamma e_{0}$
$e_{-1}\cdot e_{k}$ | $=$ | $\left(\frac{\gamma(\mu_{k}\nu_{k}-\check{\mu}_{k}\check{\nu}_{k})-2\lambda_{k}(\alpha_{k}\mu_{k}-\beta_{k}\check{\mu}_{k})}{2\lambda_{k}\xi}\right)e_{0}+\gamma e_{k}$
$e_{-1}\cdot\check{e}_{k}$ | $=$ | $\left(\frac{\gamma(\check{\mu}_{k}\nu_{k}+\mu_{k}\check{\nu}_{k})-2\lambda_{k}(\alpha_{k}\check{\mu}_{k}+\beta_{k}\mu_{k})}{2\lambda_{k}\xi}\right)e_{0}+\gamma\check{e}_{k}$
$e_{k}\cdot e_{k}$ | $=$ | $\check{e}_{k}\cdot\check{e}_{k}=-\frac{\gamma}{2\lambda_{k}}e_{0}.$

Hence, we will consider two different situations.

  1. $\gamma\neq 0.$

By taking

$\nu_{k}=\frac{2\alpha_{k}\lambda_{k}}{\gamma},$
$\check{\nu}_{k}=\frac{2\beta_{k}\lambda_{k}}{\gamma}$ and
$\nu=-\sum\limits_{i=1}^{n}\frac{\gamma(\nu_{i}^{2}+\check{\nu}_{i}^{2})}{2\gamma\lambda_{i}}-\frac{\mu}{\gamma},$

we have a multiplication table from part $(A)$ in our statement. It is easy to
see that, under the action of a positive automorphism of the Lie algebra
$(\mathfrak{L}_{\lambda},[-,-]),$ given in Lemma 10, all multiplications
$\cdot$ give non-isomorphic transposed Poisson structures. Under the action of
a negative automorphism given in Lemma 10, we have the isomorphism between
structures corresponding to $\gamma$ and $-\gamma.$ Insofar,
$(\mathfrak{L}_{\lambda},\cdot)$ is non-nilpotent if and only if $\gamma\neq
0$.

  2. $\gamma=0.$

In this case, $(\mathfrak{L}_{\lambda},\cdot)$ is a nilpotent algebra. Note
that,

    1. $\bullet$

If $\\{\alpha_{k}\\}_{1\leq k\leq n}=0,$ then transposed Poisson structures
corresponding to $\\{\beta_{1},\ldots,\beta_{j},\ldots,\beta_{n}\\}$ and
$\\{\beta_{1},\ldots,-\beta_{j},\ldots,\beta_{n}\\}$ are isomorphic.

    2. $\bullet$

If $(\alpha_{k},\beta_{k})\neq(0,0)$ then after an action of a suitable
automorphism, we can suppose that $\alpha_{k}\neq 0$ and $\beta_{k}\neq 0.$ In
the case $(\alpha_{k},\beta_{k})=(0,0)$ after each automorphism action, we
will take the same situation $(\alpha_{k},\beta_{k})=(0,0).$ Let $t$ be the
maximal number such that
$(\alpha_{1},\beta_{1})=\ldots=(\alpha_{t-1},\beta_{t-1})=(0,0).$

We should consider two subcases.

    1. $\mu\neq 0.$

By taking

$\xi=\mu,$ $\check{\mu}_{t}=\alpha_{t}$ and $\mu_{k}=\beta_{k}$ for all $k$
such that $\alpha_{k}\beta_{k}\neq 0,$

we obtain that $(\mathfrak{L}_{\lambda},\cdot)$ has a multiplication table
from the part $(B.a)$ of our statement.

    2. $\mu=0.$

By taking

$\check{\mu}_{t}=\alpha_{t}$ and $\mu_{k}=\beta_{k}$ for all $k$ such that
$\alpha_{k}\beta_{k}\neq 0,$

we obtain that $(\mathfrak{L}_{\lambda},\cdot)$ has a multiplication table
from the part $(B.b)$ of our statement.

∎

##  3\. Transposed Poisson structures on finite-dimensional solvable Lie
algebras

It is proven that any finite-dimensional solvable Lie algebra over an
algebraically closed field of zero characteristic admits non-trivial
$\frac{1}{2}$-derivations [21]. For the solvable Lie algebra $\mathfrak{L}$
with non-zero annihilator, it is constructed non-trivial
$\frac{1}{2}$-derivation as $\varphi(x)=[x,\omega_{0}],$ by the element of
$\omega_{0}\in\operatorname{Ann}_{[\mathfrak{L},\mathfrak{L}]}([\mathfrak{L},\mathfrak{L}])$
such that $[x,\omega_{0}]=\lambda_{x}\omega_{0}$ for all $x\in\mathfrak{L},$
where $\lambda_{x}\neq 0$ for some $x.$ Thus, the dimension of the
$\frac{1}{2}$-derivations space of any solvable Lie algebra at least equal to
two. In the following example, we give a solvable Lie algebra in which the
dimension of the $\frac{1}{2}$-derivations space is equal to two.

###  3.1. Transposed Poisson structures on solvable Lie algebras with
naturally graded filiform nilpotent radical

All solvable Lie algebras with a naturally graded filiform nilpotent radical
were found in [26]. Now we consider the solvable Lie algebra
$\mathfrak{s}_{n,2}$ with the following multiplication table:

$[e_{i},e_{1}]$ | $=$ | $e_{i+1},$ | $2\leq i\leq n-1;$ | $[e_{1},x_{1}]$ | $=$ | $e_{1},$ |
---|---|---|---|---|---|---|---
$[e_{i},x_{1}]$ | $=$ | $(i-2)e_{i},$ | $3\leq i\leq n;$ | $[e_{i},x_{2}]$ | $=$ | $e_{i},$ | $2\leq i\leq n.$

Note that $\mathfrak{s}_{n,2}$ is a solvable Lie algebra with nilpotent
radical is the naturally graded filiform Lie algebra

$\mathfrak{n}_{n,1}:[e_{i},e_{1}]=e_{i+1},\quad 2\leq i\leq n-1.$

Moreover, $[\mathfrak{s}_{n,2},\mathfrak{s}_{n,2}]=\mathfrak{n}_{n,1}$ and
$\operatorname{Ann}_{[\mathfrak{s}_{n,2},\mathfrak{s}_{n,2}]}([\mathfrak{s}_{n,2},\mathfrak{s}_{n,2}])=\langle
e_{n}\rangle.$ In the following proposition we give the description of
$\frac{1}{2}$-derivations of the algebra $\mathfrak{s}_{n,2}.$

######  Proposition 12.

Any $\frac{1}{2}$-derivation $\varphi$ of the algebra $\mathfrak{s}_{n,2}$ has
the form

$\varphi(x_{1})=\alpha x_{1}+(n-2)\beta e_{n},\quad\varphi(x_{2})=\alpha
x_{2}+\beta e_{n},\quad\varphi(e_{k})=\alpha e_{k},\quad 1\leq k\leq n.$

######  Theorem 13.

Let $(\mathfrak{s}_{n,2},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{s}_{n,2}.$ Then, up to
isomorphism, there is only one non-trivial transposed Poisson algebra
structure on $\mathfrak{s}_{n,2}$. It is given by

$x_{1}\cdot x_{1}=(n-2)^{2}e_{n},\quad x_{2}\cdot x_{1}=(n-2)e_{n},\quad
x_{2}\cdot x_{2}=e_{n}.$

This structure is non-Poisson.

###### Proof.

Let $(\mathfrak{s}_{n,2},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{s}_{n,2}.$ Then for any
element of $x\in\mathfrak{s}_{n,2},$ the operator of multiplication
$\varphi_{x}(y)=x\cdot y$ is a $\frac{1}{2}$-derivation of
$(\mathfrak{s}_{n,2},[-,-])$. Hence, we have that

$\varphi_{x_{1}}(x_{1})$ | $=$ | $\alpha_{n+1}x_{1}+(n-2)\beta_{n+1}e_{n},$ | $\varphi_{x_{1}}(x_{2})$ | $=$ | $\alpha_{n+1}x_{2}+\beta_{n+1}e_{n},$ | $\varphi_{x_{1}}(e_{k})$ | $=$ | $\alpha_{n+1}e_{k},$
---|---|---|---|---|---|---|---|---
$\varphi_{x_{2}}(x_{1})$ | $=$ | $\alpha_{n+2}x_{1}+(n-2)\beta_{n+2}e_{n},$ | $\varphi_{x_{2}}(x_{2})$ | $=$ | $\alpha_{n+2}x_{2}+\beta_{n+2}e_{n},$ | $\varphi_{x_{2}}(e_{k})$ | $=$ | $\alpha_{n+2}e_{k},$
$\varphi_{e_{i}}(x_{1})$ | $=$ | $\alpha_{i}x_{1}+(n-2)\beta_{i}e_{n},$ | $\varphi_{e_{i}}(x_{2})$ | $=$ | $\alpha_{i}x_{2}+\beta_{i}e_{n},$ | $\varphi_{e_{i}}(e_{k})$ | $=$ | $\alpha_{i}e_{k}.$

Then from $\alpha_{i}e_{j}=\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i})=\alpha_{j}e_{i},$ we have that $\alpha_{i}=0$ for
$1\leq i\leq n.$ The last implies $e_{i}\cdot e_{j}=0$ for all $1\leq i,j\leq
n.$ Next from

$\alpha_{n+1}e_{1}$ | $=$ | $\varphi_{x_{1}}(e_{1})$ | $=$ | $e_{1}\cdot x_{1}$ | $=$ | $x_{1}\cdot e_{1}$ | $=$ | $\varphi_{e_{1}}(x_{1})$ | $=$ | $(n-2)\beta_{1}e_{n},$
---|---|---|---|---|---|---|---|---|---|---
$\alpha_{n+2}e_{1}$ | $=$ | $\varphi_{x_{2}}(e_{1})$ | $=$ | $e_{1}\cdot x_{2}$ | $=$ | $x_{2}\cdot e_{1}$ | $=$ | $\varphi_{e_{1}}(x_{2})$ | $=$ | $\beta_{1}e_{n},$

we have that $\alpha_{n+1}=\alpha_{n+2}=0.$ After that, from

$0$ | $=$ | $\alpha_{n+2}e_{i}$ | $=$ | $\varphi_{x_{2}}(e_{i})$ | $=$ | $e_{i}\cdot x_{2}$ | $=$ | $x_{2}\cdot e_{i}$ | $=$ | $\varphi_{e_{i}}(x_{2})$ | $=$ | $\beta_{i}e_{n},$
---|---|---|---|---|---|---|---|---|---|---|---|---

we get that $\beta_{i}=0$ for $1\leq i\leq n$ and $e_{i}\cdot x_{1}=e_{i}\cdot
x_{2}=0.$ Finally, considering

$(n-2)\beta_{n+2}e_{n}$ | $=$ | $\varphi_{x_{2}}(x_{1})$ | $=$ | $x_{2}\cdot x_{1}$ | $=$ | $x_{1}\cdot x_{2}$ | $=$ | $\varphi_{x_{1}}(x_{2})$ | $=$ | $\beta_{n+1}e_{n},$
---|---|---|---|---|---|---|---|---|---|---

we get that $\beta_{n+1}=(n-2)\beta_{n+2}.$ Thus, we obtain

$x_{1}\cdot x_{1}$ | $=$ | $(n-2)^{2}\gamma e_{n},$ | $x_{2}\cdot x_{1}$ | $=$ | $(n-2)\gamma e_{n},$ | $x_{2}\cdot x_{2}$ | $=$ | $\gamma e_{n}.$
---|---|---|---|---|---|---|---|---

Since we are interested only in non-trivial transposed Poisson algebra
structures, we can suppose $\gamma\neq 0$ and via automorphism

$\phi(x_{1})$ | $=$ | $x_{1},$ | $\phi(x_{2})=x_{2},$ | $\phi(e_{1})=e_{1},$ | $\phi(e_{i})=\gamma^{-1}{e_{i}},$ | $2\leq i\leq n,$
---|---|---|---|---|---|---

we conclude that these transposed Poisson algebra structures are isomorphic to
the case $\gamma=1.$ ∎

###  3.2. Transposed Poisson structures on solvable Lie algebras with
Heisenberg nilpotent radical

All solvable Lie algebras with Heisenberg nilpotent radical were found in
[24]. In this subsection, we consider $(3n+2)$-dimensional solvable Lie
algebra $L_{n,n+1}$ with the following multiplication table:

$[e_{n+i},e_{i}]$ | $=$ | $e_{2n+1},$ | $[e_{i},x_{i}]$ | $=$ | $e_{i},$ | $[e_{n+i},x_{i}]$ | $=$ | $-e_{n+i},$ | $1\leq i\leq n,$
---|---|---|---|---|---|---|---|---|---
|  |  | $[e_{i},x_{n+1}]$ | $=$ | $e_{i},$ | $[e_{2n+1},x_{n+1}]$ | $=$ | $e_{2n+1},$ | $1\leq i\leq n.$

Note that the nilpotent radical of this algebra is the $(2n+1)$-dimensional
Heisenberg algebra ${\rm H}_{n}$ with a basis
$\\{e_{1},e_{2},\dots,e_{2n+1}\\}.$ Here $[L_{n,n+1},L_{n,n+1}]={\rm H}_{n}$
and $\operatorname{Ann}_{{\rm H}_{n}}({\rm H}_{n})=\langle e_{2n+1}\rangle.$
In the following proposition we give the description of
$\frac{1}{2}$-derivations of the algebra $L_{n,n+1}.$

######  Proposition 14.

Any $\frac{1}{2}$-derivation $\varphi$ of the algebra $L_{n,n+1}$ has the form

$\varphi(e_{k})$ | $=$ | $\alpha e_{k},$ | $1\leq k\leq 2n+1;$ | $\varphi(x_{k})$ | $=$ | $\alpha x_{k},$ | $1\leq k\leq n;$ | $\varphi(x_{n+1})$ | $=$ | $\alpha x_{n+1}+\beta e_{2n+1}.$
---|---|---|---|---|---|---|---|---|---|---

######  Theorem 15.

Let $(L_{n,n+1},\cdot,[-,-])$ be a transposed Poisson algebra structure
defined on the Lie algebra $L_{n,n+1}.$ Then, up to isomorphism, there is only
one non-trivial transposed Poisson algebra structure on $L_{n,n+1}$. It is
given by

$x_{n+1}\cdot x_{n+1}=e_{2n+1}.$

This structure is non-Poisson.

###### Proof.

Let $(L_{n,n+1},\cdot,[-,-])$ be a transposed Poisson algebra structure
defined on the Lie algebra $L_{n,n+1}.$ Then for any element of $x\in
L_{n,n+1},$ we have the multiplication operator $\varphi_{x}(y)=x\cdot y$ is a
$\frac{1}{2}$-derivation of $(L_{n,n+1},[-,-]).$ Hence, we have that

$\varphi_{e_{i}}(e_{k})$ | $=$ | $\alpha_{i}e_{k},$ | $1\leq k\leq 2n+1;$ |  |  |
---|---|---|---|---|---|---
$\varphi_{e_{i}}(x_{k})$ | $=$ | $\alpha_{i}x_{k},$ | $1\leq k\leq n+1;$ | $\varphi_{e_{i}}(x_{n+1})$ | $=$ | $\alpha_{i}x_{n+1}+\beta_{i}e_{2n+1},$
$\varphi_{x_{j}}(e_{k})$ | $=$ | $\alpha_{2n+1+j}e_{k},$ | $1\leq k\leq 2n+1;$ |  |  |
$\varphi_{x_{j}}(x_{k})$ | $=$ | $\alpha_{2n+1+j}x_{k},$ | $1\leq k\leq n;$ | $\varphi_{x_{j}}(x_{n+1})$ | $=$ | $\alpha_{2n+1+j}x_{n+1}+\beta_{2n+1+j}e_{2n+1}.$

Then from

$\alpha_{i}e_{k}$ | $=$ | $\varphi_{e_{i}}(e_{k})$ | $=$ | $e_{i}\cdot e_{k}$ | $=$ | $e_{k}\cdot e_{i}$ | $=$ | $\varphi_{e_{k}}(e_{i})$ | $=$ | $\alpha_{k}e_{i},$
---|---|---|---|---|---|---|---|---|---|---

we have that $\alpha_{i}=0$ for $1\leq i\leq 2n+1.$ Next from

$\alpha_{2n+1+j}e_{1}$ | $=$ | $\varphi_{x_{j}}(e_{1})$ | $=$ | $e_{1}\cdot x_{j}$ | $=$ | $x_{j}\cdot e_{1}$ | $=$ | $\varphi_{e_{1}}(x_{j})$ | $=$ | $0,$
---|---|---|---|---|---|---|---|---|---|---

we have that $\alpha_{2n+1+j}=0$ for $1\leq j\leq n.$ After then considering

$\alpha_{3n+2}e_{i}$ | $=$ | $\varphi_{x_{n+1}}(e_{i})$ | $=$ | $e_{i}\cdot x_{n+1}$ | $=$ | $x_{n+1}\cdot e_{i}$ | $=$ | $\varphi_{e_{i}}(x_{n+1})$ | $=$ | $\beta_{i}e_{2n+1},$
---|---|---|---|---|---|---|---|---|---|---

we derive that $\alpha_{3n+2}=0$ and $\beta_{i}=0$ for $1\leq i\leq 2n+1.$
Finally, considering

$\beta_{2n+1+j}e_{2n+1}$ | $=$ | $\varphi_{x_{j}}(x_{n+1})$ | $=$ | $x_{j}\cdot x_{n+1}$ | $=$ | $x_{n+1}\cdot x_{j}$ | $=$ | $\varphi_{x_{n+1}}(x_{j})$ | $=$ | $0,$
---|---|---|---|---|---|---|---|---|---|---

we get that $\beta_{2n+1+j}=0$ for $1\leq j\leq n$ and $x_{i}\cdot x_{j}=0,$
$1\leq i\leq n+1,$ $1\leq j\leq n.$ Thus, we have that

$\quad x_{n+1}\cdot x_{n+1}=\gamma e_{2n+1}.$

Since we are interested only in non-trivial transposed Poisson algebra
structures, we have that $\gamma\neq 0$ and via automorphism

$\phi(e_{i})$ | $=$ | $e_{i},$ | $1\leq i\leq n;$ | $\phi(e_{j})$ | $=$ | $\gamma^{-1}{e_{j}},$ | $n+1\leq j\leq 2n+1;$ | $\phi(x_{k})=x_{k},$ | $1\leq k\leq n+1,$
---|---|---|---|---|---|---|---|---|---

we can suppose that $\gamma=1.$ ∎

###  3.3. Transposed Poisson structures on solvable Lie algebras with abelian
nilpotent radical

Now we consider solvable Lie algebras with abelian nilpotent radical and
maximal complementary vector space (these algebras were found in [23]). It is
known that the maximal dimension of complementary space for solvable Lie
algebras with $n$-dimensional abelian nilpotent radical is equal to $n.$
Moreover, up to isomorphism there exists only one such solvable Lie algebra
with the following multiplications:

$L_{n}:\left[e_{i},x_{i}\right]=e_{i},\quad 1\leq i\leq n,$

where $\\{e_{1},\dots,e_{n},x_{1},\dots,x_{n}\\}$ is a basis of $L_{n}.$ In
the following proposition we give the description of $\frac{1}{2}$-derivations
of the algebra $L_{n}.$

######  Proposition 16.

Any $\frac{1}{2}$-derivation $\varphi$ of the algebra $L_{n}$ has the form

$\varphi(e_{i})$ | $=$ | $\alpha_{i}e_{i},$ | $\varphi(x_{i})$ | $=$ |  $\alpha_{i}x_{i}+\beta_{i}e_{i},$ $1\leq i\leq n.$
---|---|---|---|---|---

######  Theorem 17.

Let $(L_{n},\cdot,[-,-])$ be a transposed Poisson algebra structure defined on
the Lie algebra $L_{n}.$ Then the multiplication of $(L_{n},\cdot)$ has the
following form:

$e_{i}\cdot x_{i}=\mu_{i}e_{i},\quad x_{i}\cdot
x_{i}=\mu_{i}x_{i}+\tau_{i}e_{i},\quad 1\leq i\leq n.$

###### Proof.

$(L_{n},\cdot,[-,-])$ be a transposed Poisson algebra structure defined on the
Lie algebra $L_{n}.$ Then for any element of $x\in L_{n},$ we have that
operator of multiplication $\varphi_{x}(y)=x\cdot y$ is a
$\frac{1}{2}$-derivation. Hence, for $1\leq i,k\leq n$ we derive

$\varphi_{e_{i}}(e_{k})$ | $=$ | $\alpha_{i,k}e_{k},$ | $\varphi_{e_{i}}(x_{k})$ | $=$ | $\alpha_{i,k}x_{k}+\beta_{i,k}e_{k},$
---|---|---|---|---|---
$\varphi_{x_{i}}(e_{k})$ | $=$ | $\gamma_{i,k}e_{k},$ | $\varphi_{x_{i}}(x_{k})$ | $=$ | $\gamma_{i,k}x_{k}+\delta_{i,k}e_{k}.$

Thus, we have that

$e_{i}\cdot e_{k}=0,\quad 1\leq i,k\leq n;\quad e_{j}\cdot x_{k}=x_{j}\cdot
x_{k}=0,\quad 1\leq j\neq k\leq n.$

Therefore, non-zero multiplications of the algebra $(L_{n},\cdot)$ have the
form:

$e_{i}\cdot x_{i}=\beta_{i,i}e_{i},\quad x_{i}\cdot
x_{i}=\beta_{i,i}x_{i}+\delta_{i,i}e_{i},\quad 1\leq i\leq n.$

∎

######  Remark 18.

From Theorem 17 we obtain that transposed Poisson algebra structure defined on
the Lie algebra $L_{n}$ is a direct sum of two-dimensional transposed Poisson
algebras ${\rm TP}_{i}^{\mu_{i},\tau_{i}},$ where

${\rm TP}_{i}^{\mu_{i},\tau_{i}}:\quad\left[e_{i},x_{i}\right]=e_{i},\quad
e_{i}\cdot x_{i}=\mu_{i}e,\quad x_{i}\cdot x_{i}=\mu_{i}x_{i}+\tau_{i}e_{i}.$

It is not difficult to see that any transposed Poisson algebra from the class
${\rm TP}_{i}^{\mu_{i},\tau_{i}}$ is isomorphic one of the following non-
isomorphic algebras

${\rm TP}_{1}^{\mu}={\rm TP}_{i}^{\mu,0}$ or ${\rm TP}_{2}={\rm
TP}_{i}^{0,1}.$

Therefore, we obtain that transposed Poisson algebra structure defined on the
Lie algebra $L_{n}$ is a direct sum of some copies of algebras ${\rm
TP}_{1}^{\mu}$ and ${\rm TP}_{2}.$

##  4\. Transposed Poisson structure on perfect Lie algebras

In this section, we consider transposed Poisson structure for some finite-
dimensional Lie algebras which is a semi-direct sum of semisimple and solvable
algebras, i.e., $\mathfrak{L}=\mathfrak{s}\ltimes\mathfrak{r}$. More
precisely, we consider Lie algebras with three-dimensional simple parts and
the solvable radical is an irreducible representation. It is known that if the
simple algebra $\mathfrak{sl}_{2}$ has an $(m+1)$-dimensional irreducible
representation $\mathfrak{r},$ where $m\geq 2$ (see, [13]), then we have the
algebra $\mathfrak{L}^{m}=\mathfrak{sl}_{2}\ltimes\mathfrak{r}$ with the
following multiplication:

$[e,f]$ | $=$ | $h,$ | $[h,e]$ | $=$ | $2e,$ | $[f,h]$ | $=$ | $2f,$
---|---|---|---|---|---|---|---|---
$[x_{k},h]$ | $=$ | $(2k-m)x_{k},$ | $0\leq k\leq m,$
$[x_{k},f]$ | $=$ | $x_{k+1},$ | $0\leq k\leq m-1,$
$[x_{k},e]$ | $=$ | $k(m+1-k)x_{k-1},$ | $1\leq k\leq m.$

######  Proposition 19.

If $m>2$, then any $\frac{1}{2}$-derivation of $\mathfrak{L}^{m}$ is trivial.

###### Proof.

Let $\varphi$ be a $\frac{1}{2}$-derivation of $\mathfrak{L}^{m}.$ Put

$\varphi(e)$ | $=$ | $\alpha^{1}_{1}e+\alpha^{1}_{2}f+\alpha^{1}_{3}h+\sum\limits_{k=0}^{m}\beta^{1}_{k}x_{k},$ | $\varphi(f)$ | $=$ | $\alpha^{2}_{1}e+\alpha^{2}_{2}f+\alpha^{2}_{3}h+\sum\limits_{k=0}^{m}\beta^{2}_{k}x_{k}.$
---|---|---|---|---|---

Then from
$\varphi(h)=\varphi([e,f])=\frac{1}{2}\big{(}[\varphi(e),f]+[e,\varphi(f)]\big{)},$
we have that

$\varphi(h)$ | $=$ | $-\alpha^{2}_{3}e-\alpha^{1}_{3}f+\frac{\alpha^{1}_{1}+\alpha^{2}_{2}}{2}h-\frac{m}{2}\beta^{2}_{1}x_{0}+\frac{1}{2}\sum\limits_{k=1}^{m-1}\Big{(}\beta^{1}_{k-1}-(k+1)(m-k)\beta^{2}_{k+1}\Big{)}x_{k}+\frac{1}{2}\beta^{1}_{m-1}x_{m}.$
---|---|---

Consider

$\varphi([h,e])$ | $=$ | $\frac{1}{2}\big{(}[\varphi(h),e]+[h,\varphi(e)]\big{)}=$
---|---|---
| $=$ | $\frac{1}{2}\Big{(}[-\alpha^{2}_{3}e-\alpha^{1}_{3}f+\frac{\alpha^{1}_{1}+\alpha^{2}_{2}}{2}h-\frac{m}{2}\beta^{2}_{1}x_{0}+\sum\limits_{k=1}^{m-1}\frac{\beta^{1}_{k-1}-(k+1)(m-k)\beta^{2}_{k+1}}{2}x_{k}+\frac{\beta^{1}_{m-1}}{2}x_{m},e]+$
$+[h,\alpha^{1}_{1}e+\alpha^{1}_{2}f+\alpha^{1}_{3}h+\sum\limits_{k=0}^{m}\beta^{1}_{k}x_{k}]\Big{)}$
| $=$ | $\frac{\alpha^{1}_{3}}{2}h+\frac{\alpha^{1}_{1}+\alpha^{2}_{2}}{2}e+\sum\limits_{k=1}^{m-1}\frac{(\beta^{1}_{k-1}-(k+1)(m-k)\beta^{2}_{k+1})k(m+1-k)}{4}x_{k-1}+$
$+\frac{m\beta^{1}_{m-1}}{4}x_{m-1}+\alpha^{1}_{1}e-\alpha^{1}_{2}f+\sum\limits_{k=0}^{m}\frac{\beta^{1}_{k}(m-2k)}{2}x_{k}$
| $=$ | $\frac{3\alpha^{1}_{1}+\alpha^{2}_{2}}{2}e-\alpha^{1}_{2}f+\frac{\alpha^{1}_{3}}{2}h+\sum\limits_{k=0}^{m-2}\frac{(\beta^{1}_{k}-(k+2)(m-k-1)\beta^{2}_{k+2})(k+1)(m-k)}{4}x_{k}+$
$+\frac{m\beta^{1}_{m-1}}{4}x_{m-1}+\sum\limits_{k=0}^{m}\frac{\beta^{1}_{k}(m-2k)}{2}x_{k}.$

On the other hand,

$\varphi([h,e])$ | $=$ | $2\varphi(e)$ | $=$ | $2\alpha^{1}_{1}e+2\alpha^{1}_{2}f+2\alpha^{1}_{3}h+2\sum\limits_{k=0}^{m}\beta^{1}_{k}x_{k}.$
---|---|---|---|---

Comparing coefficients at the basis elements we have

$\alpha^{1}_{1}=\alpha^{2}_{2},\quad\alpha^{1}_{2}=0,\quad\alpha^{1}_{3}=0,\quad\beta_{1,m}=0,\quad\beta^{1}_{m-1}=0,$

and (for $0\leq k\leq m-2$):

$\big{(}(k+1)(m-k)+2m-4k-8\big{)}\beta_{1,k}=(k+2)(k+1)(m-k)(m-k-1)\beta^{2}_{k+2}.$
(4)

Let us now consider

$\varphi([f,h])$ | $=$ | $\frac{1}{2}\big{(}[\varphi(f),h]+[f,\varphi(h)]\big{)}$
---|---|---
| $=$ | $\frac{1}{2}\big{[}\alpha^{2}_{1}e+\alpha^{1}_{1}f+\alpha^{2}_{3}h+\sum\limits_{k=0}^{m}\beta_{2,k}x_{k},h\big{]}+$
$+\frac{1}{2}\big{[}f,-\alpha^{2}_{3}e+\alpha^{1}_{1}h-\frac{m}{2}\beta^{2}_{1}x_{0}+\sum\limits_{k=1}^{m-1}\frac{\beta^{1}_{k-1}-(k+1)(m-k)\beta^{2}_{k+1}}{2}x_{k}+\frac{\beta^{1}_{m-1}}{2}x_{m}\big{]}$
| $=$ | $-\alpha^{2}_{1}e+\alpha^{1}_{1}f+\sum\limits_{k=0}^{m}\frac{(2k-m)\beta_{2,k}}{2}x_{k}+\frac{\alpha^{2}_{3}}{2}h+\alpha^{1}_{1}f+\frac{m\beta^{2}_{1}}{4}x_{1}-\sum\limits_{k=2}^{m}\frac{\beta_{1,k-2}-k(m-k+1)\beta^{2}_{k}}{4}x_{k}.$

On the other hand,

$\varphi([f,h])$ | $=$ | $2\varphi(f)$ | $=$ | $2\alpha^{2}_{1}e+2\alpha^{1}_{1}f+2\alpha^{2}_{3}h+2\sum\limits_{k=0}^{m}\beta^{2}_{k}x_{k}.$
---|---|---|---|---

Comparing coefficients at the basis elements we obtain

$\alpha^{2}_{1}=0,\quad\alpha^{2}_{3}=0,\quad\beta_{2,0}=0,\quad\beta_{2,1}=0,$

and

$\beta_{1,k-2}=\big{(}(k-2)m-k^{2}+5k-8\big{)}\beta_{2,k},\quad 2\leq k\leq
m.$ (5)

Then from (4) and (5) we have that $(m-2)(m+4)\beta_{2,k}=0,$ for $2\leq k\leq
m.$ Since $m\neq 2,$ then we have that $\beta_{2,k}=0,$ which implies
$\beta_{1,k-2}=0$ for any $2\leq k\leq m.$

Thus, we have

$\varphi(e)$ | $=$ | $\alpha^{1}_{1}e,$ | $\varphi(f)$ | $=$ | $\alpha^{1}_{1}f,$ | $\varphi(h)$ | $=$ | $\alpha^{1}_{1}h.$
---|---|---|---|---|---|---|---|---

Now put

$\varphi(x_{0})$ | $=$ | $\alpha^{4}_{1}e+\alpha^{4}_{2}f+\alpha^{4}_{3}h+\sum\limits_{k=0}^{m}\beta^{4}_{k}x_{k}.$
---|---|---

From

$0$ | $=$ | $2\varphi([x_{0},e])=[\varphi(x_{0}),e]+[x_{0},\varphi(e)]$
---|---|---
| $=$ | $[\alpha^{4}_{1}e+\alpha^{4}_{2}f+\alpha^{4}_{3}h+\sum\limits_{k=0}^{m}\beta^{4}_{k}x_{k},e]+[x_{0},\alpha^{1}_{1}e]$
| $=$ | $-\alpha^{4}_{2}h+2\alpha^{4}_{3}e+\sum\limits_{k=1}^{m}k(m+1-k)\beta^{4}_{k}x_{k-1},$

we have that $\alpha^{4}_{2}=\alpha^{4}_{3}=\beta^{4}_{k}=0,$ $1\leq k\leq m.$
Considering the following relation

$2m\varphi(x_{0})$ | $=$ | $2\varphi([h,x_{0}])=[\varphi(h),x_{0}]+[h,\varphi(x_{0})]$
---|---|---
| $=$ | $[\alpha^{1}_{1}h,x_{0}]+[h,\alpha^{4}_{1}e+\beta^{4}_{0}x_{0}]=2\alpha^{4}_{1}e+m(\alpha^{1}_{1}+\beta^{4}_{0})x_{1},$

we have that $\alpha^{4}_{1}=0$ and $\beta^{4}_{0}=\alpha^{1}_{1}.$ Thus,
$\varphi(x_{0})=\alpha^{1}_{1}x_{0}.$ Then from
$\varphi(x_{k})=\varphi([x_{k-1},f]),$ we obtain that
$\varphi(x_{k})=\alpha^{1}_{1}x_{k}$ for $1\leq k\leq m.$ Hence, any
$\frac{1}{2}$-derivation of $\mathfrak{L}^{m}$ is trivial. ∎

Surprisingly, in the case of $m=2$, we obtain a different result.

######  Proposition 20.

Any $\frac{1}{2}$-derivation $\varphi$ of $\mathfrak{L}^{2}$ has the following
form

$\varphi(e)$ | $=$ | $\alpha e-2\beta x_{0},$ | $\varphi(f)$ | $=$ | $\alpha f+\beta x_{2},$ | $\varphi(h)$ | $=$ | $\alpha h-2\beta x_{1},$
---|---|---|---|---|---|---|---|---
$\varphi(x_{0})$ | $=$ | $\alpha x_{0},$ | $\varphi(x_{1})$ | $=$ | $\alpha x_{1},$ | $\varphi(x_{2})$ | $=$ | $\alpha x_{2}.$

From the description of the $\frac{1}{2}$-derivation of the algebra
$\mathfrak{L}^{m}$ we have the following corollary.

######  Corollary 21.

There are no non-trivial transposed Poisson algebra structures defined on the
algebra $\mathfrak{L}^{m}.$

###### Proof.

Since in the case of $m\neq 2,$ the algebra $\mathfrak{L}^{m}$ has only
trivial $\frac{1}{2}$-derivation, we obtain that there are no non-trivial
transposed Poisson algebra structures for $m\neq 2.$

Thus, it is sufficient to prove the Corollary for $m=2.$ Then we have the
multiplication of $\mathfrak{L}^{2}$ has the form

$[e,f]$ | $=$ | $h,$ | $[h,e]$ | $=$ | $2e,$ | $[f,h]$ | $=$ | $2f,$
---|---|---|---|---|---|---|---|---
$[x_{0},h]$ | $=$ | $-2x_{0},$ | $[x_{0},f]$ | $=$ | $x_{1},$ | $[x_{1},e]$ | $=$ | $2x_{0},$
$[x_{2},h]$ | $=$ | $2x_{2},$ | $[x_{1},f]$ | $=$ | $x_{2},$ | $[x_{2},e]$ | $=$ | $2x_{1}.$

By Lemma 4, we have that for any element
$\ell\in\\{e,f,h,x_{0},x_{1},x_{2}\\}$, there is a related
$\frac{1}{2}$-derivation $\varphi_{\ell}$ of $\mathfrak{L}^{2}$. Thus, by
Proposition 20, we derive that

$\begin{array}[]{lll}\varphi_{\ell}(e)=\alpha_{\ell}e-2\beta_{\ell}x_{0},&\varphi_{\ell}(f)=\alpha_{\ell}f+\beta_{\ell}x_{2},&\varphi_{\ell}(h)=\alpha_{\ell}h-2\beta_{\ell}x_{1},\\\\[2.84526pt]
\varphi_{\ell}(x_{0})=\alpha_{\ell}x_{0},&\varphi_{\ell}(x_{1})=\alpha_{\ell}x_{1},&\varphi_{\ell}(x_{2})=\alpha_{\ell}x_{2}.\end{array}$

Then from

$\alpha_{x_{i}}x_{j}$ | $=$ | $\varphi_{x_{i}}(x_{j})$ | $=$ | $x_{i}\cdot x_{j}$ | $=$ | $\varphi_{x_{j}}(x_{i})=\alpha_{x_{j}}x_{i},$
---|---|---|---|---|---|---

we have that $\alpha_{x_{0}}=\alpha_{x_{1}}=\alpha_{x_{2}}=0.$

Now, considering

$\alpha_{e}x_{1}$ | $=$ | $\varphi_{e}(x_{1})$ | $=$ | $e\cdot x_{1}$ | $=$ | $\varphi_{x_{1}}(e)$ | $=$ | $-2\beta_{x_{1}}x_{0},$
---|---|---|---|---|---|---|---|---
$\alpha_{f}x_{0}$ | $=$ | $\varphi_{f}(x_{0})$ | $=$ | $f\cdot x_{0}$ | $=$ | $\varphi_{x_{0}}(f)$ | $=$ | $\beta_{x_{0}}x_{2},$
$\alpha_{h}x_{2}$ | $=$ | $\varphi_{h}(x_{2})$ | $=$ | $h\cdot x_{2}$ | $=$ | $\varphi_{x_{2}}(h)$ | $=$ | $-2\beta_{x_{2}}x_{1},$

we obtain that
$\alpha_{e}=\alpha_{f}=\alpha_{h}=\beta_{x_{0}}=\beta_{x_{1}}=\beta_{x_{2}}=0.$

Finally, from

$\beta_{e}x_{2}$ | $=$ | $\varphi_{e}(f)$ | $=$ | $e\cdot f$ | $=$ | $\varphi_{f}(e)$ | $=$ | $-2\beta_{f}x_{0},$
---|---|---|---|---|---|---|---|---
$-2\beta_{e}x_{1}$ | $=$ | $\varphi_{e}(h)$ | $=$ | $e\cdot h$ | $=$ | $\varphi_{h}(e)$ | $=$ | $-2\beta_{h}x_{0},$

we get that $\beta_{e}=\beta_{f}=\beta_{h}=0.$ Thus, we conclude that there
are no non-trivial transposed Poisson algebra structures on
$\mathfrak{L}^{2}$. ∎

######  Remark 22.

It should be noted that the algebra $\mathfrak{L}^{2}$ is an algebra that
admits non-trivial $\frac{1}{2}$-derivations, but does not admit non-trivial
transposed Poisson algebra structures. Moreover, the algebra
$\mathfrak{L}^{2}$ is a perfect algebra (i.e.,
$[\mathfrak{L}^{2},\mathfrak{L}^{2}]=\mathfrak{L}^{2}$) and admits non-trivial
$\frac{1}{2}$-derivations. The existence of algebras with this property gives
a particular answer to [4, Questions 2 and 5].

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# LXMERT Model Compression for Visual Question Answering

Maryam Hashemi Ghazaleh Mahmoudi ∗ Sara Kodeiri∗ Hadi Sheikhi∗ Sauleh Eetemadi
School of Computer Engineering, Iran University of Science and Technology,
Iran
{m_hashemi94, gh_mahmoodi, sara_kodeiri<EMAIL_ADDRESS><EMAIL_ADDRESS>These authors contributed equally.

###### Abstract

Large-scale pretrained models such as LXMERT are becoming popular for learning
cross-modal representations on text-image pairs for vision-language tasks.
According to the lottery ticket hypothesis, NLP and computer vision models
contain smaller subnetworks capable of being trained in isolation to full
performance. In this paper, we combine these observations to evaluate whether
such trainable subnetworks exist in LXMERT when fine-tuned on the VQA task. In
addition, we perform a model size cost-benefit analysis by investigating how
much pruning can be done without significant loss in accuracy. Our experiment
results demonstrate that LXMERT can be effectively pruned by 40%-60% in size
with 3% loss in accuracy.

##  1 Introduction and Related Work

Over the past few years, many single-modal pretrained models have been
proposed. Inspired by this, the vision-and-language pretraining seeks to learn
joint representations using visual and textual content to improve the
efficiency of vision-language tasks.

Both single-modality and cross-modality pretrained models often have hundreds
of millions of parameters. Unfortunately, training these overparametrized
models can be prohibitively time-consuming and costly, making them impractical
for resource-limited devices. However, cross-modality pretrained models suffer
more from the increased model size due to the higher input space dimension.
With the task of Visual Question Answering (VQA) Antol et al. (2015) in mind,
and its ultimate goal of being helpful to the visually impaired, decreasing
V+L model size makes it feasible to use them in limited-resource devices.

To address this problem, model compression techniques such as pruning have
been developed. Deep Learning recently enjoyed welcoming a new powerful
pruning method: The Lottery Ticket Hypothesis (LTH) Frankle and Carbin (2019).
LTH has been shown great success in various fields. It could be a powerful
tool to understand the parameter redundancy in the current pretrained V+L
models. Thus, we aim to apply LTH to LXMERTTan and Bansal (2020), one of the
best-performing two-stream V+L models, to fill this gap. We evaluate our work
on VQA Antol et al. (2015) and compare it with DistillVLMFang et al. (2021),
which leverages the knowledge distillation technique to compress large visual-
linguistic models.

Similar to this work, Gan et al. (2021) study LTH for UNITERChen et al.
(2020). However, UNITER is a single stream V+L model, and LXMERT is a two-
stream model; our results are consistent with theirs.

##  2 Methodology

In this section, we briefly explain the LXMERT architecture and LTH. Then, we
describe how we use LTH to compress the pretrained LXMERT model.

Method | test-dev |  | test-std
---|---|---|---
Yes/No | Number | other | Overall |  | Yes/No | Number | other | Overall
DistillVLM | - | - | - | 69.6 |  | - | - | - | 69.8
LXMERT | 88.24 | 54.45 | 63.05 | 72.45 |  | 88.29 | 54.37 | 63.18 | 72.63
LXMERT (low-magnitude) | 86.95 $\pm$ 0.95 | 52.60 $\pm$ 1.87 | 60.96 $\pm$ 1.76 | 70.72 $\pm$ 1.44 |  | 87.07 $\pm$ 1.12 | 52.28 $\pm$ 1.66 | 61.02 $\pm$ 1.83 | 70.87 $\pm$ 1.51
LXMERT (high-magnitude) | 74.11 $\pm$ 0.91 | 42.81 $\pm$ 1.36 | 50.5 $\pm$ 0.19 | 59.35 $\pm$ 0.61 |  | 74.23 $\pm$ 0.81 | 42.99 $\pm$ 0.87 | 50.71 $\pm$ 0.26 | 59.62 $\pm$ 0.55
LXMERT (random) | 69.26 $\pm$ 0.29 | 39.84 $\pm$ 0.93 | 45.96 $\pm$ 0.83 | 54.86 $\pm$ 0.52 |  | 69.27 $\pm$ 0.18 | 40.34 $\pm$ 0.66 | 46.33 $\pm$ 0.79 | 55.19 $\pm$ 0.45

Table 1:  Performance of subnetworks at 50% weights pruning on VQA v2, which
reported for both test-dev and test-std. Test-dev is used for debugging and
validation experiments. Test-standard is the default test data for the VQA
competition. We test each experiment for three different seeds and report the
mean and standard deviation of VQA accuracy across three seeds.

LXMERT is a Transformer-based model which takes two inputs: image and text.
Internally, LXMERT consists of two types of encoders: single-modality encoders
for each modality and a cross-modality encoder using bidirectional cross
attention to exchange information and align entities across the modalities.

The Lottery Ticket Hypothesis Frankle and Carbin (2019) shows that by
preserving the original weight initializations from the unpruned network, you
can train a network with the topology of the pruned network and achieve the
same or better test accuracy within the same number of training iterations.

In order to apply LTH to the LXMERT model, we use iterative magnitude pruning.
Therefore, we fine-tune LXMERT on the VQA task and iteratively prune 10% of
the lowest magnitude weights across the entire model, excluding embedding and
output layers. We keep pruning until our model loses roughly half the weights.
We use the default settings and hyperparameters of LXMERT Tan and Bansal
(2020) to finetune on the VQA v2.0 dataset.

##  3 Experimental Setups and Results

The experiments are designed to investigate the effectiveness and stability of
LTH on LXMERT in addition to cost-benefit analysis of the number of parameters
in the model. We conduct experiments on the widely-used VQA v2.0 Goyal et al.
(2017) dataset built based on the MS-COCO Lin et al. (2014) image corpus.

###  3.1 Effectiveness and Stability

The following steps are performed to compress the LXMERT model.

  1. 1.

The pretrained LXMERT model plus the VQA classifier’s randomly initialized
weights are saved.

  2. 2.

The model is fine-tuned on the 3,129 most frequent answers in the VQA v2.0
dataset.

  3. 3.

Iterative magnitude pruning is applied to find the low-magnitude subnetwork
(pruning 50% of the low-magnitude weights). The high-magnitude subnetwork is
computed as a compliment of the low-magnitude subnetwork with equal size. A
random subnetwork with an equal size is generated for comparison.

  4. 4.

The saved weights are restored for all three subnetworks.

  5. 5.

The high-magnitude, low-magnitude, and random subnetworks are fine-tuned and
evaluated on the VQA Task using three different seeds for initializing the VQA
model to ensure the stability of the results.

Results of subnetworks at 50% weights pruning on VQA v2.0 are summarized in
Table 1 where DistillVLM Fang et al. (2021) is also listed for comparison. Row
2 to row 5 reports respectively full finetuned LXMERT, low-magnitude, high-
magnitude, and random subnetworks. Low-magnitude pruning achieves 97% of full
finetuned LXMERT accuracy in overall for both test-dev and test-std and shows
marginal improvement over DistillVLM as the baseline. By comparing performance
across the subnetworks, random and high magnitude subnetworks perform far
worse than low-magnitude subnetwork. Surprisingly, the results demonstrate
high-magnitude subnetwork performing better than random subnetwork. This could
be a LXMERT specific phenomenon and required further investigation.

###  3.2 Cost-Benefit Analysis

We experiment with low-magnitude subnetwork by pruning 10% of the weights all
the way up to 90% of the weights in 10% increments. Accuracy of these pruned
models on VQA v2.0 are reported in Figure 1. Our results indicate a
significant loss of accuracy after 50% to 60% pruning.

Figure 1: Model size cost-benefit analysis.

##  4 Conclusion

We confirm that LTH pruning is an effective method for pruning V+L pretrained
models. We mainly focused on LXMERT, a two-stream V+L pretrained model, but
our findings are consistent with Gan et al. (2021)’s results while using
UNITER, a single-stream V+L pretrained model.

## References

  * Antol et al. (2015) Stanislaw Antol, Aishwarya Agrawal, Jiasen Lu, Margaret Mitchell, Dhruv Batra, C Lawrence Zitnick, and Devi Parikh. 2015.  VQA: Visual question answering.  In _Proceedings of the IEEE International Conference on Computer Vision_ , volume 2015 Inter, pages 2425–2433.
  * Chen et al. (2020) Yen-Chun Chen, Linjie Li, Licheng Yu, Ahmed El Kholy, Faisal Ahmed, Zhe Gan, Yu Cheng, and Jingjing Liu. 2020.  Uniter: Universal image-text representation learning.  In _European Conference on Computer Vision_ , pages 104–120. Springer.
  * Fang et al. (2021) Zhiyuan Fang, Jianfeng Wang, Xiaowei Hu, Lijuan Wang, Yezhou Yang, and Zicheng Liu. 2021.  Compressing visual-linguistic model via knowledge distillation.  In _Proceedings of the IEEE/CVF International Conference on Computer Vision (ICCV)_ , pages 1428–1438.
  * Frankle and Carbin (2019) Jonathan Frankle and Michael Carbin. 2019.  The lottery ticket hypothesis: Finding sparse, trainable neural networks.  In _International Conference on Learning Representations_.
  * Gan et al. (2021) Zhe Gan, Yen-Chun Chen, Linjie Li, Tianlong Chen, Yu Cheng, Shuohang Wang, and Jingjing Liu. 2021.  Playing lottery tickets with vision and language.  _arXiv preprint arXiv:2104.11832_.
  * Goyal et al. (2017) Yash Goyal, Tejas Khot, Douglas Summers-Stay, Dhruv Batra, and Devi Parikh. 2017\.  Making the v in vqa matter: Elevating the role of image understanding in visual question answering.  In _Proceedings of the IEEE Conference on Computer Vision and Pattern Recognition_ , pages 6904–6913.
  * Lin et al. (2014) Tsung-Yi Lin, Michael Maire, Serge Belongie, James Hays, Pietro Perona, Deva Ramanan, Piotr Dollár, and C Lawrence Zitnick. 2014.  Microsoft coco: Common objects in context.  In _European conference on computer vision_ , pages 740–755. Springer.
  * Tan and Bansal (2020) Hao Tan and Mohit Bansal. 2020.  LXMert: Learning cross-modality encoder representations from transformers.  _EMNLP-IJCNLP 2019 - 2019 Conference on Empirical Methods in Natural Language Processing and 9th International Joint Conference on Natural Language Processing, Proceedings of the Conference_ , pages 5100–5111.