Transposed Poisson structures on Witt-type algebras 111 The first part of work
is supported by FCT UIDB/00212/2020, UIDP/00212/2020, and 2022.02474.PTDC. The
second part of this work is supported by the Russian Science Foundation under
grant 22-71-10001.

Ivan Kaygorodov222CMA-UBI, Universidade da Beira Interior, Covilhã, Portugal;
Moscow Center for Fundamental and Applied Mathematics, Russia; Saint
Petersburg University, Russia<EMAIL_ADDRESS>Abror
Khudoyberdiyev333V.I.Romanovskiy Institute of Mathematics Academy of Science
of Uzbekistan; National University of Uzbekistan<EMAIL_ADDRESS>& Zarina
Shermatova444V.I.Romanovskiy Institute of Mathematics Academy of Science of
Uzbekistan; Kimyo International University in Tashkent, Uzbekistan;
<EMAIL_ADDRESS>

Abstract. We compute $\frac{1}{2}$-derivations on the deformative Schrödinger-
Witt 555We refer the notion of “Witt” algebra to the simple Witt algebra and
the notion of “Virasoro” algebra to the central extension of the simple Witt
algebra. algebra, on not-finitely graded Witt algebras $W_{n}(G)$, and on not-
finitely graded Heisenberg-Witt algebra $HW_{n}(G)$. We classify all
transposed Poisson structures on such algebras.

Keywords: Lie algebra; transposed Poisson algebra; $\frac{1}{2}$-derivation.

MSC2020: 17A30, 17B40, 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 a fixed Lie or associative part.
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 [22, 2,
7, 18]. 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. Each
transposed Poisson algebra is a commutative Gelfand-Dorfman algebra [22] and
it is also an algebra of Jordan brackets [7]. Transposed Poisson algebras are
related to weak Leibniz algebras [5]. 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 [8]. These ideas
were used to describe all transposed Poisson structures on Witt and Virasoro
algebras in [8]; 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 solvable Lie algebra with filiform
nilradical in [1]; on oscillator Lie algebras in [17]; on Witt type Lie
algebras in [15]; on generalized Witt algebras in [16]; Block Lie algebras in
[14, 16] and on Lie incidence algebras (for all references, see the survey
[13]).

The description of transposed Poisson structures obtained on the Witt algebra
[8] opens the question of finding algebras related to the Witt algebra which
admit nontrivial transposed Poisson structures. So, some algebras related to
Witt algebra are studied in [15, 14, 16]. The present paper is a continuation
of this research. Specifically, we describe transposed Poisson structures on
the deformative Schrödinger-Witt algebra ${\mathcal{W}}(a,b,s),$ algebras
$W_{n}(G)$ and algebras $HW_{n}(G).$ The algebras ${\mathcal{W}}(a,b,s)$
contain a subalgebra isomorphic to the well-known $W$-algebra
${\mathcal{W}}(a,b).$ The algebra ${\mathcal{W}}(0,0,0)$ was introduced in
[10] and later it was generalized in [21]. Their structures have been widely
studied in [12, 21, 24, 11]. Namely, derivations and automorphism group of
${\mathcal{W}}(a,b,s)$ are described in [12, 24]; all biderivations and
commutative post-Lie algebra structures on ${\mathcal{W}}(a,b,s)$ are obtained
in [11]; graded post-Lie algebra structures and homogeneous Rota-Baxter
operators on ${\mathcal{W}}(0,0,0)$ are given in [25] and so on. The algebra
$W_{-1}(G)$ was independently introduced by Passman [20] and Xu [26]. It
naturally appeared in the theory of integrable systems in [19]. All one-
dimensional central extensions of $W_{-1}(G)$ are obtained in [23].
Derivations of $W_{-1}(G)$ with coefficients in tensor modules were found in
[9]. The algebra $W_{1}(G)$ firstly was defined in [4]. Derivations,
automorphisms, one-dimensional extensions, and the second cohomology group of
$W_{1}(G)$ were defined also in [4]. Lie bialgebras structure on $W_{1}(G)$
was described in [3]. The algebra $W_{0}(G)$ is an algebra of Witt type
defined in [28]. We introduce the most general concept generalizing of yearly
considered algebras $W_{1}(G),$ $W_{0}(G)$ and $W_{-1}(G).$ Namely, for each
$n\in\mathbb{Z}$ we define an algebra $W_{n}(G)$ and studying its properties.
The algebra $HW_{-1}(G)$ was firstly introduced in [6], and $HW_{-1}(G)$ is
constructed as the direct sum of $W_{-1}(G)$ and a module over $W_{-1}(G).$
Similarly, we define and study of algebras $HW_{n}(G)$ for an arbitrary
$n\in\mathbb{Z}.$

Summarizing, we prove that algebras ${\mathcal{W}}(a,b,s)$ admit nontrivial
transposed Poisson structures only for $b=-1$; algebras $W_{n}(G)$ admit
nontrivial transposed Poisson structures and algebras $HW_{n}(G)$ do not admit
nontrivial transposed Poisson structures.

##  1\. Preliminaries

In this section, we recall some definitions and known results for studying
transposed Poisson structures. Although all algebras and vector spaces are
considered over the complex field, many results can be proven over other
fields without modifications of proofs. 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 $[\cdot,\cdot]$. The triple
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is called a transposed Poisson algebra if
$(\mathfrak{L},\cdot)$ is a commutative associative algebra and
$(\mathfrak{L},[\cdot,\cdot])$ is a Lie algebra that satisfies the following
compatibility condition

$\begin{array}[]{c}2z\cdot[x,y]=[z\cdot x,y]+[x,z\cdot y].\end{array}$

######  Definition 2.

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

######  Definition 3.

Let $(\mathfrak{L},[\cdot,\cdot])$ be a Lie algebra,
$\varphi:\mathfrak{L}\rightarrow\mathfrak{L}$ be a linear map. Then $\varphi$
is a $\frac{1}{2}$-derivation if it satisfies

$\begin{array}[]{c}\varphi([x,y])=\frac{1}{2}\big{(}[\varphi(x),y]+[x,\varphi(y)]\big{)}.\end{array}$

Observe that $\frac{1}{2}$-derivations are a particular case of
$\delta$-derivations introduced by Filippov in 1998 (for references about the
study of $\frac{1}{2}$-derivations, see [13]) and recently the notion of
$\frac{1}{2}$-derivations of algebras was generalized to
$\frac{1}{2}$-derivations from an algebra to a module [30]. The main example
of $\frac{1}{2}$-derivations is the multiplication by an element from the
ground field. Let us call such $\frac{1}{2}$-derivations as trivial
$\frac{1}{2}$-derivations. It is easy to see that
$[\mathfrak{L},\mathfrak{L}]$ and $\operatorname{Ann}(\mathfrak{L})$ are
invariant under any $\frac{1}{2}$-derivation of $\mathfrak{L}$.

Let $G$ be an abelian group, $\mathfrak{L}=\bigoplus\limits_{g\in
G}\mathfrak{L}_{g}$ be a $G$-graded Lie algebra. We say that a
$\frac{1}{2}$-derivation $\varphi$ has degree $g$ (and denopted by
$\deg(\varphi)$) if $\varphi(\mathfrak{L}_{h})\subseteq\mathfrak{L}_{g+h}$.
Let $\triangle(\mathfrak{L})$ denote the space of $\frac{1}{2}$-derivations
and write
$\triangle_{g}(\mathfrak{L})=\\{\varphi\in\triangle(\mathfrak{L})\mid\deg(\varphi)=g\\}$.
The following trivial lemmas are useful in our work.

######  Lemma 4.

Let $\mathfrak{L}=\bigoplus\limits_{g\in G}\mathfrak{L}_{g}$ be a $G$-graded
Lie algebra. Then $\triangle(\mathfrak{L})=\bigoplus\limits_{g\in
G}\triangle_{g}(\mathfrak{L}).$

######  Lemma 5.

(see [8]) Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson
algebra and $z$ an arbitrary element from $\mathfrak{L}$. Then the left
multiplication $L_{z}$ in the commutative associative algebra
$(\mathfrak{L},\cdot)$ gives a $\frac{1}{2}$-derivation of the Lie algebra
$(\mathfrak{L},[\cdot,\cdot])$.

######  Lemma 6.

(see [8]) Let $\mathfrak{L}$ be a Lie algebra without non-trivial
$\frac{1}{2}$-derivations. Then every transposed Poisson structure defined on
$\mathfrak{L}$ is trivial.

##  2\. Transposed Poisson structures on the deformative Schrödinger-Witt
algebras

Let us give the definition of the algebra ${\mathcal{W}}(a,b,s),$ where
$a,b\in\mathbb{C}$ and $s\in\\{0,\frac{1}{2}\\}.$ ${\mathcal{W}}(a,b,s)$ is
spanned by $\\{L_{m},I_{m},Y_{m+s}\ |\ m\in\mathbb{Z}\\}$ and the
multiplication table is given by the following nontrivial relations:

$[L_{m},L_{n}]$ | $=$ | $(n-m)L_{m+n},$
---|---|---
$[L_{m},I_{n}]$ | $=$ | $(n+bm+a)I_{m+n},$
$[L_{m},Y_{n+s}]$ | $=$ | $\left(n+s+\frac{(b-1)m+a}{2}\right)Y_{m+n+s},$
$[Y_{m+s},Y_{n+s}]$ | $=$ | $(n-m)I_{m+n+2s}.$

Thanks to [11], we have
${\mathcal{W}}(a+1,b,0)\cong{\mathcal{W}}(a,b,\frac{1}{2}).$ Hence, we only
have to consider the case of $s=\frac{1}{2}$. Set $M_{n}=\langle
L_{n},I_{n}\rangle$ and $M_{n+\frac{1}{2}}=\langle Y_{n+\frac{1}{2}}\rangle.$
Then ${\mathcal{W}}(a,b,\frac{1}{2})$ is $\frac{1}{2}\mathbb{Z}$-graded, i.e.,

${\mathcal{W}}(a,b,\frac{1}{2})=\big{(}\bigoplus\limits_{n\in\mathbb{Z}}M_{n}\big{)}\bigoplus\big{(}\bigoplus\limits_{n\in\mathbb{Z}}M_{n+\frac{1}{2}}\big{)}.$

Hence $\Delta({\mathcal{W}}(a,b,\frac{1}{2}))$ has a natural
$\frac{1}{2}\mathbb{Z}$-grading, i.e.,

$\Delta({\mathcal{W}}(a,b,\frac{1}{2}))=\big{(}\bigoplus\limits_{n\in\mathbb{Z}}\Delta_{n}({\mathcal{W}}(a,b,\frac{1}{2}))\big{)}\bigoplus\big{(}\bigoplus\limits_{n\in\mathbb{Z}}\Delta_{n+\frac{1}{2}}({\mathcal{W}}(a,b,\frac{1}{2}))\big{)}.$

Note that ${\mathcal{W}}(a,b,\frac{1}{2})$ contains a subalgebra $\langle
L_{m},I_{m}\ |\ m\in\mathbb{Z}\rangle,$ which isomorphic to the well-known
algebra ${\mathcal{W}}(a,b)$ with the multiplication table:

$[L_{m},L_{n}]=(n-m)L_{m+n},\quad[L_{m},I_{n}]=(n+bm+a)I_{m+n}.$

The description of $\frac{1}{2}$-derivations of Lie algebra
${\mathcal{W}}(a,b)$ is given in the following theorem.

######  Theorem 7 (see, [8]).

If $b\neq-1,$ then there are no non-trivial $\frac{1}{2}$-derivations of
${\mathcal{W}}(a,b).$ Let $\varphi$ be a $\frac{1}{2}$-derivation of the
algebra ${\mathcal{W}}(a,-1),$ then there are two finite sets of elements from
the basic field $\\{\alpha_{t}\\}_{t\in\mathbb{Z}}$ and
$\\{\beta_{t}\\}_{t\in\mathbb{Z}},$ such that

$\varphi(L_{m})=\sum_{t\in\mathbb{Z}}\alpha_{t}L_{m+t}+\sum_{t\in\mathbb{Z}}\beta_{t}I_{m+t},\quad\varphi(I_{m})=\sum_{t\in\mathbb{Z}}\alpha_{t}I_{m+t}.$

Now we will compute $\frac{1}{2}$-derivations of the algebras
${\mathcal{W}}(a,b,\frac{1}{2}).$ Summarizing the results from lemmas 8 and 9,
we conclude that if $b\neq-1,$ then ${\mathcal{W}}(a,b,\frac{1}{2})$ does not
have nontrivial $\frac{1}{2}$-derivations. On the other side, Theorem 10 gives
the full description of nontrivial $\frac{1}{2}$-derivations of
${\mathcal{W}}(a,-1,\frac{1}{2}).$

######  Lemma 8.

If $b\neq-1,$ then $\Delta_{0}({\mathcal{W}}(a,b,\frac{1}{2}))=\langle{\rm
Id}\rangle$ and
$\Delta_{j\in\mathbb{Z}\setminus\\{0\\}}({\mathcal{W}}(a,b,\frac{1}{2}))=0.$

###### Proof.

Suppose that $b\neq-1$ and
$\varphi_{j}\in\Delta_{j}({\mathcal{W}}(a,b,\frac{1}{2}))$ be a homogeneous
$\frac{1}{2}$-derivation. In this case, we have

$\varphi_{j}(M_{n})\subseteq
M_{n+j},\quad\varphi_{j}(M_{n+\frac{1}{2}})\subseteq M_{n+j+\frac{1}{2}}.$ (1)

By Theorem 7, we get

$\varphi_{j}(L_{n})=\delta_{j,0}\lambda
L_{n},\quad\varphi_{j}(I_{n})=\delta_{j,0}\lambda I_{n}.$ (2)

By (1), we can assume that

$\varphi_{j}(Y_{n+\frac{1}{2}})=a_{j,n}Y_{n+j+\frac{1}{2}}.$ (3)

  1. (1)

If $j\neq 0,$ then applying $\varphi_{j}$ to both side of
$[Y_{m+\frac{1}{2}},Y_{n+\frac{1}{2}}]=(n-m)I_{m+n+1},$ and using (2) and (3),
we obtain

$(n-m-j)a_{j,m}+(n+j-m)a_{j,n}=0.$ (4)

Setting $n=0$ in (4), we get $(-m-j)a_{j,m}+(j-m)a_{j,0}=0.$ Then taking
$m=-j$ in this equation, we derive $2ja_{j,0}=0,$ which implies $a_{j,0}=0.$
Consequently, we have $a_{j,m}=0$ for $m\neq-j.$ Letting $m=j$ and $n=-j$ in
(4), we obtain $a_{j,-j}=0$, this shows $a_{j,m}=0$ for all $m\in\mathbb{Z}.$
This proves $\varphi_{j}=0$ for $j\in{\mathbb{Z}\setminus\\{0\\}}.$

  2. (2)

If $j=0,$ then applying $\varphi_{0}$ to both side of
$[Y_{m+s},Y_{n+s}]=(n-m)I_{m+n+2s}$ and using (2) and (3), we get

$(n-m)(a_{0,m}+a_{0,n})=2\lambda(n-m).$

Letting $m=0$ in this equation, we obtain $n(a_{0,0}+a_{0,n})=2\lambda n,$ and
$a_{0,n}=2\lambda-a_{0,0}$ for $n\neq 0.$ It is following that
$a_{0,n}=\lambda$ for all $n\in\mathbb{Z}.$ Hence $\varphi_{0}=\lambda\ {\rm
Id}.$

∎

######  Lemma 9.

If $b\neq-1,$ then $\Delta_{j+\frac{1}{2}}({\mathcal{W}}(a,b,\frac{1}{2}))=0.$

###### Proof.

Let $b\neq-1$ and
$\varphi_{j+\frac{1}{2}}\in\Delta_{j+\frac{1}{2}}({\mathcal{W}}(a,b,\frac{1}{2}))$
be a homogeneous $\frac{1}{2}$-derivation. Then we have

$\varphi_{j+\frac{1}{2}}(M_{n})\subseteq
M_{n+j+\frac{1}{2}},\quad\varphi_{j+\frac{1}{2}}(M_{n+\frac{1}{2}})\subseteq
M_{n+j+1}.$ (5)

By (5) we can assume that

$\varphi_{j+\frac{1}{2}}(L_{n})=\alpha_{j,n}Y_{n+j+\frac{1}{2}},\quad\varphi_{j+\frac{1}{2}}(I_{n})=\beta_{j,n}Y_{n+j+\frac{1}{2}},\quad\varphi_{j+\frac{1}{2}}(Y_{n+\frac{1}{2}})=\gamma_{j,n}L_{n+j+1}+\mu_{j,n}I_{n+j+1}.$

Let us say $\varphi=\varphi_{j+\frac{1}{2}}.$ The algebra
${\mathcal{W}}(a,b,\frac{1}{2})$ admits a $\mathbb{Z}_{2}$-grading. Namely,

${\mathcal{W}}(a,b,\frac{1}{2})_{\overline{0}}=\langle M_{j}\ |\
j\in\mathbb{Z}\rangle$ and
${\mathcal{W}}(a,b,\frac{1}{2})_{\overline{1}}=\langle M_{j+\frac{1}{2}}\ |\
j\in\mathbb{Z}\rangle.$

The mapping $\varphi$ changes the grading components. It is known, that the
commutator of one derivation and one $\frac{1}{2}$-derivation gives a new
$\frac{1}{2}$-derivation. Hence, $[\varphi,{\rm ad}_{Y_{n+\frac{1}{2}}}]$ is a
$\frac{1}{2}$-derivation which preserve the grading components. Namely, it is
a $\frac{1}{2}$-derivation, described in Lemma 8, i.e., $[\varphi,{\rm
ad}_{Y_{n+\frac{1}{2}}}]=\alpha_{n}{\rm Id}.$

It is easy to see, that

$\alpha_{n}I_{m}$ | $=$ | $[\varphi,{\rm ad}_{Y_{n+\frac{1}{2}}}](I_{m})=\varphi[I_{m},Y_{n+\frac{1}{2}}]-[\varphi(I_{m}),Y_{n+\frac{1}{2}}]=$
---|---|---
| $=$ | $-\beta_{j,m}[Y_{m+j+\frac{1}{2}},Y_{n+\frac{1}{2}}]=-(n-m-j)\beta_{j,m}I_{m+n+j+1}.$

For fixed elements $m$ and $j$, we can choose an element $n,$ such that $n\neq
m+j$ and $n\neq-1-j.$ The last observations give $\beta_{j,m}=0$ for each
$(j,m)\in\mathbb{Z}\times\mathbb{Z}.$ Hence, $\varphi(I_{n+\frac{1}{2}})=0$
and $\alpha_{n}=0$ for all $n\in\mathbb{Z}.$

Let us now consider the following observation

$0$ | $=$ | $[\varphi,{\rm ad}_{Y_{n+\frac{1}{2}}}](L_{m})=\varphi[L_{m},Y_{n+\frac{1}{2}}]-[\varphi(L_{m}),Y_{n+\frac{1}{2}}]=$
---|---|---
| $=$ | $(n+\frac{(b-1)m+a+1}{2})\varphi(Y_{n+m+\frac{1}{2}})-\alpha_{j,m}[Y_{m+j+\frac{1}{2}},Y_{n+\frac{1}{2}}]=$
| $=$ | $(n+\frac{(b-1)m+a+1}{2})(\gamma_{j,n+m}L_{n+m+j+\frac{1}{2}}+\mu_{j,n+m}I_{n+m+j+1})-\alpha_{j,m}(n-m-j)I_{m+j+n+1},$

which gives two relations considered below.

  1. First,

$(2n+(b-1)m+a+1)\gamma_{j,m+n}=0.$ Taking $m=0$ and $n=0,$ we obtain

$((b-1)n+a+1)\gamma_{j,n}=0$ and $(2n+a+1)\gamma_{j,n}=0.$

If $b\neq 3$ or $a\neq-1-2n,$ we have $\gamma_{j,n}=0.$ In the opposite case,
if $b=3$ and $a=-1-2n,$ we consider

$0$ | $=$ | $2(n-k)\varphi(I_{k+n+1})=2\varphi[Y_{k+\frac{1}{2}},Y_{n+\frac{1}{2}}]=$
---|---|---
| $=$ | $\mu_{j,k}[I_{k+j+\frac{1}{2}},Y_{n+\frac{1}{2}}]+\gamma_{j,n}[Y_{k+\frac{1}{2}},L_{n+j+1}]+\mu_{j,n}[Y_{k+\frac{1}{2}},I_{n+j+1}]=$
| $=$ | $-\gamma_{j,n}\Big{(}k+\frac{(b-1)(n+j+1)+a+1}{2}\Big{)}L_{n+k+j+\frac{3}{2}}.$

Taking a suitable $k$, we obtain that $\gamma_{j,n}=0$ and
$\varphi(Y_{n+\frac{1}{2}})=\mu_{j,n}I_{n+j+1}.$

  2. Second,

$\big{(}n+\frac{(b-1)m+a+1}{2}\big{)}\mu_{j,m+n}=(n-m-j)\alpha_{j,m},$ which
gives $\frac{2n+a+1}{2}\mu_{j,n}=\alpha_{j,0}(n-j).$ Let us take an element
$n,$ such that $n\neq m+j$ and $2(m+n)+a+1\neq 0,$ hence

$\alpha_{j,m}$ | $=$ | $\frac{2n+(b-1)m+a+1}{n-m-j}\mu_{j,n}$ | $=$ | $\frac{(2n+(b-1)m+a+1)(n+m-j)}{(n-m-j)(2(n+m)+a+1)}\alpha_{j,0}.$
---|---|---|---|---

The last observation, due to almost arbitrary $n$, gives that $\alpha_{j,m}=0$
for arbitrary $m\in\mathbb{Z},$ which gives $\mu_{j,m}=0$ for arbitrary
$m\in\mathbb{Z}.$

Summarizing, $\varphi=0.$

∎

######  Theorem 10.

Let $\varphi$ be a $\frac{1}{2}$-derivation of the algebra
${\mathcal{W}}(a,-1,\frac{1}{2}),$ then there are three sets of elements from
the basic field $\\{\alpha_{t}\\}_{t\in\mathbb{Z}},$
$\\{\beta_{t}\\}_{t\in\mathbb{Z}}$ and $\\{\gamma_{t}\\}_{t\in\mathbb{Z}}$
such that

$\varphi(L_{m})$ | = | $\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t}L_{m+t}+\sum\limits_{t\in{\mathbb{Z}}}\beta_{t}I_{m+t}+\sum\limits_{t\in{\mathbb{Z}}}\gamma_{t}Y_{m+t+\frac{1}{2}},$
---|---|---
$\varphi(I_{m})$ | = | $\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t}I_{m+t},$
$\varphi(Y_{m+\frac{1}{2}})$ | = | $\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t}Y_{m+t+\frac{1}{2}}+\sum\limits_{t\in{\mathbb{Z}}}\gamma_{t}I_{m+t+1}.$

###### Proof.

Suppose that $\varphi_{j}\in\Delta_{j}({\mathcal{W}}(a,-1,\frac{1}{2}))$ for
$j\in\mathbb{Z}.$ By Theorem 7, we have

$\varphi_{j}(L_{n})=\alpha_{j}L_{n+j}+\beta_{j}I_{n+j},\quad\varphi_{j}(I_{n})=\alpha_{j}I_{n+j},\quad\varphi_{j}(Y_{n+\frac{1}{2}})=a_{j,n}Y_{n+j+\frac{1}{2}}.$
(6)

Applying $\varphi_{j}$ to both side of
$[Y_{m+\frac{1}{2}},Y_{n+\frac{1}{2}}]=(n-m)I_{m+n+1}$ and using (6), we
obtain

$(n-m-j)a_{j,m}+(n-m+j)a_{j,n}=2(n-m)\alpha_{j}.$ (7)

  1. (i)

Taking $m=0$ and $n=j$ in (7), we have $ja_{j,j}=j\alpha_{j},$ so
$a_{j,j}=\alpha_{j}$ for $j\neq 0.$

  2. (ii)

Taking $m=j$ in (7), we have $na_{j,n}=n\alpha_{j},$ this implies
$a_{j,n}=\alpha_{j}$ for $n\neq 0.$

  3. (iii)

Taking $m=0$ and $n\notin\\{0,j\\}$ in (7), we have $a_{j,0}=\alpha_{j}$ for
$j\neq 0.$

  4. (iv)

Taking $j=0$ and $m=0$ in (7), we have $a_{0,n}=2\alpha_{0}-a_{0,0}$ for any
$n\neq 0$ and choosing $m\neq n,$ we get $a_{0,0}=\alpha_{0}$ and thus
$a_{0,n}=\alpha_{0}$ for all $n\in\mathbb{Z}.$

As a conclusion, we obtain that

$\varphi_{j}(L_{n})=\alpha_{j}L_{n+j}+\beta_{j}I_{n+j},\quad\varphi_{j}(I_{n})=\alpha_{j}I_{n+j},\quad\varphi_{j}(Y_{n+\frac{1}{2}})=\alpha_{j}Y_{n+j+\frac{1}{2}}.$

Let
$\varphi_{j+\frac{1}{2}}\in\Delta_{j+\frac{1}{2}}({\mathcal{W}}(a,-1,\frac{1}{2}))$
for $j\in\mathbb{Z}.$ Then, we can assume

$\varphi_{j+\frac{1}{2}}(L_{n})=\alpha_{j,n}Y_{n+j+\frac{1}{2}},\quad\varphi_{j+\frac{1}{2}}(I_{n})=\beta_{j,n}Y_{n+j+\frac{1}{2}},\quad\varphi_{j+\frac{1}{2}}(Y_{n+\frac{1}{2}})=\gamma_{j,n}L_{n+j+1}+\mu_{j,n}I_{n+j+1}.$

Applying $\varphi_{j+\frac{1}{2}}$ to both sides of the first three relations
from the multiplication table, we obtain

$\displaystyle(2(n+j)+1-2m+a)\alpha_{j,n}-(2(m+j)+1-2n+a)\alpha_{j,m}=4(n-m)\alpha_{j,m+n},$
(8) $\displaystyle(2(n+j)+1-2m+a)\beta_{j,n}=4(n-m+a)\beta_{j,m+n},$ (9)
$\displaystyle(n+j+1-m)\gamma_{j,n}=(2n+1-2m+a)\gamma_{j,m+n},$ (10)
$\displaystyle(n-m-j)\alpha_{j,m}+(n+j+1-m+a)\mu_{j,n}=(2n+1-2m+a)\mu_{j,m+n}.$
(11)

  1. Case 1.

Now we assume that $j\neq 0.$ Taking $m=0$ in (9), we have that
$(1+2j-2n-3a)\beta_{j,n}=0.$ Hence, $\beta_{j,n}=0$ if $1+2j-2n\neq 3a$. If
$1+2j-2n=3a$ and $m\neq 0,$ then $\beta_{j,m+n}=0$ and
$(2(n+j)+1-2m+a)\beta_{j,n}=0.$ The last gives $\beta_{j,n}=0$ for all
$n\in\mathbb{Z}$. Similary, from (10), we have $\gamma_{j,m}=0$ for all $m.$
Then by (8), we have $\alpha_{j,n}=\alpha_{j,0}$ for all $n\in\mathbb{Z}$.

     * A.

Let $a\notin 2\mathbb{Z}+1.$

       1. (a)

Taking $m=n=0$ in (11), we have $j\alpha_{j,0}=j\mu_{j,0},$ so this implies
$\mu_{j,0}=\alpha_{j,0}$ for $j\neq 0.$

       2. (b)

Taking $n=0$ in (11), we have $(1-2m+a)(\alpha_{j,0}-\mu_{j,m})=0.$ Notice
that $a\neq 2m-1,$ hence $\mu_{j,n}=\alpha_{j,0}$ for all $n\in\mathbb{Z}.$

     * B.

Let $a\in 2\mathbb{Z}+1\setminus\\{-1\\}$.

       1. (a)

Taking $m=0$ in (11), we have $(n-j)\alpha_{j,0}=(n-j)\mu_{j,n}.$ So this
implies $\mu_{j,n}=\alpha_{j,0}$ for $n\neq j.$

       2. (b)

Taking $m=2j,$ $n=j$ in (11), we get $(a+1)\mu_{j,j}=(a+1)\alpha_{j,0}.$ It
follows $\mu_{j,n}=\alpha_{j,0}$ for all $n\in\mathbb{Z}.$

     * C.

Let $a=-1$.

       1. (a)

Taking $m=0$ in (11), we have $(n-j)\alpha_{j,0}=(n-j)\mu_{j,n}.$ So this
implies $\mu_{j,n}=\alpha_{j,0}$ for $n\neq j.$

       2. (b)

Taking $m=2j$ and $n=-j$ in (11), we have $\mu_{j,j}=\alpha_{j,0}.$ It follows
$\mu_{j,n}=\alpha_{j,0}$ for all $n\in\mathbb{Z}.$

  2. Case 2.

Consider the case of $j=0.$ It is easy to see, that (8) gives
$\alpha_{0,n}=\alpha_{0,0}$ for all $n\in\mathbb{Z}.$

     * A.

If $a\notin 2\mathbb{Z}+1.$ Taking $m=0$ in (11), we have
$n\alpha_{0,0}=n\mu_{0,n},$ so this implies $\mu_{0,n}=\alpha_{0,0}$ for
$n\neq 0.$ Taking $m=2$ and $n=0$ in (11), we have
$(a-1)(\alpha_{0,0}-\mu_{0,0})=0.$ It follows $\mu_{0,n}=\alpha_{0,0}$ for all
$n\in\mathbb{Z}.$

     * B.

If $a\in 2\mathbb{Z}+1\setminus\\{-1\\}.$ Taking $m=0$ in (11), we have
$n\alpha_{0,0}=n\mu_{0,n},$ so this implies $\mu_{0,n}=\alpha_{0,0}$ for
$n\neq 0.$ Taking $m=1$ and $n=0$ in (11), we have $a\mu_{0,0}=a\alpha_{0,0}.$
It follows $\mu_{0,n}=\alpha_{0,0}$ for all $n\in\mathbb{Z}.$

     * C.

If $a=-1.$ Taking $m=0$ in (11), we have $n(\mu_{0,n}-\alpha_{0,0})=0$ and
this implies $\mu_{0,n}=\alpha_{0,0}$ for $n\neq 0.$ Taking $m=1$ and $n=0$ in
(11), we have $\mu_{0,0}=\alpha_{0,0}.$

Summarizing, the Theorem is proved. ∎

Now we consider the algebra ${\mathcal{W}}$ with a basis
$\\{L_{i},I_{i},Y_{i+\frac{1}{2}}\ |\ {i\in\mathbb{Z}}\\}$ given by the
following commutative multiplication table:

$L_{i}L_{j}$ | $=$ | $L_{i+j},$ | $L_{i}I_{j}$ | $=$ | $I_{i+j},$ | $L_{i}Y_{j+\frac{1}{2}}$ | $=$ | $Y_{i+j+\frac{1}{2}},$ | $Y_{i+\frac{1}{2}}Y_{j+\frac{1}{2}}$ | $=$ | $I_{i+j+1}.$
---|---|---|---|---|---|---|---|---|---|---|---

In the following, we aim to classify all transposed Poisson structures on
${\mathcal{W}}(a,-1,\frac{1}{2}).$

######  Theorem 11.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson structure
defined on the Lie algebra ${\mathcal{W}}(a,-1,\frac{1}{2}).$ Then
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is not Poisson algebra and
$(\mathfrak{L},\cdot)$ is a mutation of the algebra $\mathcal{W}$. On the
other hand, every mutation of the algebra $\mathcal{W}$ gives a transposed
Poisson structure with the Lie part isomorphic to
${\mathcal{W}}(a,-1,\frac{1}{2}).$

###### Proof.

We aim to describe the multiplication $\cdot.$ By Lemma 5, for every element
$X\in\\{L_{i},I_{i},Y_{i+\frac{1}{2}}\ |\ {i\in\mathbb{Z}}\\},$ there is a
related $\frac{1}{2}$-derivation $\varphi_{X}$ of
${\mathcal{W}}(a,-1,\frac{1}{2})$ such that $\varphi_{X}(Y)=X\cdot Y.$ Then by
Theorem 10, we have that

$\varphi_{X}(L_{m})$ | = | $\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t,X}L_{m+t}+\sum\limits_{t\in{\mathbb{Z}}}\beta_{t,X}I_{m+t}+\sum\limits_{t\in{\mathbb{Z}}}\gamma_{t,X}Y_{m+t+\frac{1}{2}},$
---|---|---
$\varphi_{X}(I_{m})$ | = | $\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t,X}I_{m+t},$
$\varphi_{X}(Y_{m+\frac{1}{2}})$ | = | $\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t,X}Y_{m+t+\frac{1}{2}}+\sum\limits_{t\in{\mathbb{Z}}}\gamma_{t,X}I_{m+t+1}.$

Now we consider $\varphi_{X}(Y)=X\cdot Y=Y\cdot X=\varphi_{Y}(X)$ for
$X,Y\in\\{L_{i},I_{i},Y_{i+\frac{1}{2}}\ |\ {i\in\mathbb{Z}}\\}.$

  1. I.

Let $X=I_{m}$ and $Y=Y_{n+\frac{1}{2}}.$ Then, from
$\varphi_{I_{m}}(Y_{n+\frac{1}{2}})=\varphi_{Y_{n+\frac{1}{2}}}(I_{m}),$ we
have

$\sum\limits_{t\in\mathbb{Z}}\alpha_{t,I_{m}}Y_{n+t+\frac{1}{2}}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,I_{m}}I_{n+t+1}=\sum\limits_{t\in\mathbb{Z}}\alpha_{t,Y_{n+\frac{1}{2}}}I_{m+t}.$

The last equality gives $\alpha_{t,I_{m}}=0$ for all $m,t\in\mathbb{Z}.$

  2. II.

Let $X=I_{m}$ and $Y=I_{n},$ then we have

$I_{m}\cdot
I_{n}=\varphi_{I_{m}}(I_{n})=\sum\limits_{t\in{\mathbb{Z}}}\alpha_{t,I_{m}}I_{n+t}=0.$

  3. III.

Let $X=I_{m}$ and $Y=L_{n}.$ Then from
$\varphi_{I_{m}}(L_{n})=\varphi_{L_{n}}(I_{m}),$ we get

$\sum\limits_{t\in\mathbb{Z}}\beta_{t,I_{m}}I_{n+t}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,I_{m}}Y_{n+t+\frac{1}{2}}=\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{n}}I_{m+t},$

which implies $\gamma_{t,I_{m}}=0.$ Hence, $I_{m}\cdot Y_{n+\frac{1}{2}}=0$
for all $m,n\in\mathbb{Z}.$

  4. IV.

Let $X=L_{m}$ and $Y=L_{0}.$ Then the equality
$\varphi_{L_{m}}(L_{0})=L_{m}\cdot L_{0}=L_{0}\cdot
L_{m}=\varphi_{L_{0}}(L_{m}),$ gives

$\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{m}}L_{t}+\sum\limits_{t\in\mathbb{Z}}\beta_{t,L_{m}}I_{t}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,L_{m}}Y_{t+\frac{1}{2}}=\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}L_{m+t}+\sum\limits_{t\in\mathbb{Z}}\beta_{t,L_{0}}I_{m+t}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,L_{0}}Y_{m+t+\frac{1}{2}}.$

Hence, we obtain $\alpha_{k,L_{m}}=\alpha_{k-m,L_{0}},$
$\beta_{k,L_{m}}=\beta_{k-m,L_{0}}$ and $\gamma_{k,L_{m}}=\gamma_{k-m,L_{0}}.$

  5. V.

Let $X=L_{0}$ and $Y=Y_{n+\frac{1}{2}},$ then from
$\varphi_{L_{0}}(Y_{n+\frac{1}{2}})=\varphi_{Y_{n+\frac{1}{2}}}(L_{0}),$ we
get

$\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}Y_{n+t+\frac{1}{2}}+\sum_{t\in\mathbb{Z}}\gamma_{t,L_{0}}I_{n+t+1}=\sum\limits_{t\in\mathbb{Z}}\beta_{t,Y_{n+\frac{1}{2}}}I_{t}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,Y_{n+\frac{1}{2}}}Y_{t+\frac{1}{2}}.$

Thus, we obtain $\gamma_{k,Y_{n+\frac{1}{2}}}=\alpha_{k-n,L_{0}},$
$\beta_{k,Y_{n+\frac{1}{2}}}=\gamma_{k-n,L_{0}}.$

Summarizing all the above parts, we have that the multiplication table of
$(\mathfrak{,}\cdot)$ is given by following non-trivial relations.

$L_{m}\cdot L_{n}$ | $=$ | $\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}L_{m+n+t}+\sum\limits_{t\in\mathbb{Z}}\beta_{t,L_{0}}I_{m+n+t}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,L_{0}}Y_{m+n+t+\frac{1}{2}},$
---|---|---
$L_{m}\cdot I_{n}$ | $=$ | $\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}I_{m+n+t},$
$L_{m}\cdot Y_{n+\frac{1}{2}}$ | $=$ | $\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}Y_{m+n+t+\frac{1}{2}}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,L_{0}}I_{m+n+t+1},$
$Y_{m+\frac{1}{2}}\cdot Y_{n+\frac{1}{2}}$ | $=$ | $\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}I_{m+n+t+1}.$

Let us define
$w:=\sum\limits_{t\in\mathbb{Z}}\alpha_{t,L_{0}}L_{t}+\sum\limits_{t\in\mathbb{Z}}\beta_{t,L_{0}}I_{t}+\sum\limits_{t\in\mathbb{Z}}\gamma_{t,L_{0}}Y_{t+\frac{1}{2}}.$

Then all nonzero multiplications on the basis elements are given below.

$L_{m}\cdot L_{n}$ | $=$ | $L_{m}wL_{n},$ | $L_{m}\cdot I_{n}$ | $=$ | $L_{m}wI_{n},$
---|---|---|---|---|---
$L_{m}\cdot Y_{n+\frac{1}{2}}$ | $=$ | $L_{m}wY_{n+\frac{1}{2}},$ | $Y_{m+\frac{1}{2}}\cdot Y_{n+\frac{1}{2}}$ | $=$ | $Y_{m+\frac{1}{2}}wY_{n+\frac{1}{2}},$

Analyzing the associative law of $\cdot,$ we can see that the multiplication
$\cdot$ is associative. The algebra $(\mathfrak{L},\cdot)$ is a mutation of
$\mathcal{W}$. On the other hand, for any element $w$ the product $\cdot$
gives a transposed Poisson structure defined on the Lie algebra
${\mathcal{W}}(a,-1,\frac{1}{2})$ for an arbitrary complex number $a.$ It
gives the complete statement of the theorem. ∎

##  3\. Transposed Poisson structures on not-finitely graded Witt algebra
$W_{n}(G)$

Let $n\in\mathbb{Z}$ and $G$ be any nontrivial additive subgroup of
$\mathbb{C}.$ The algebra $W_{n}(G)$ has a basis $\\{L_{\alpha,i}\ |\
\alpha\in G,i\in\mathbb{Z}\\}$ and the multiplication table is given below:

$[L_{\alpha,i},L_{\beta,j}]=(\beta-\alpha)L_{\alpha+\beta,i+j}+(j-i)L_{\alpha+\beta,i+j+n},$
(12)

for $\alpha,\beta\in G,$ $i,j\in\mathbb{Z}.$

######  Theorem 12.

Let $\varphi$ be a $\frac{1}{2}$-derivation of the algebra $W_{n}(G),$ then
there is a finite set $\\{a^{d,m}\\}_{d\in G,m\in\mathbb{Z}}$ of complex
elements, such that

$\varphi(L_{\alpha,i})={\sum_{d\in G,\
m\in\mathbb{Z}}}a^{d,m}L_{\alpha+d,i+m}.$

###### Proof.

The algebra $W_{0}(G)$ is a generalized Witt algebra considered in [15].
Thanks to [15, Proposition 16], the Theorem is proved for $n=0.$ Below, we
will consider only the case of $n\neq 0.$

The algebra $W_{n}(G)$ is $G$-graded

$W_{n}(G)=\bigoplus\limits_{\alpha\in G}W_{n}(G)_{\alpha},$ where
$W_{n}(G)_{\alpha}=\langle L_{\alpha,i}\ |\ i\in\mathbb{Z}\rangle$ for
$\alpha\in G.$

By Lemma 4 we get $\triangle(W_{n}(G))=\bigoplus\limits_{\alpha\in
G}\triangle_{\alpha}(W_{n}(G)).$

Let $\varphi_{d}\in\Delta_{d}(W_{n}(G)).$ We can assume that
$\varphi_{d}(L_{\alpha,i})=\sum\limits_{k\in\mathbb{Z}}a_{\alpha,i}^{d,k}L_{\alpha+d,k}.$
To determine these coefficients, we start with applying $\varphi_{d}$ to both
side of (12) and obtain

$\begin{array}[]{cl}2\sum\limits_{k\in\mathbb{Z}}&\left((\beta-\alpha)a_{\alpha+\beta,i+j}^{d,k}+(j-i)a_{\alpha+\beta,i+j+n}^{d,k}\right)L_{\alpha+\beta+d,k}=\\\\[5.69054pt]
&\lx<EMAIL_ADDRESS>&\lx@intercol\hfil+\sum\limits_{k\in\mathbb{Z}}a_{\beta,j}^{d,k}\left((\beta-\alpha+d)L_{\alpha+\beta+d,i+k}+(k-i)L_{\alpha+\beta+d,i+k+n}\right).\lx@intercol\end{array}$
(13)

Letting $\beta=i=j=0$ in (13) for $k\in\mathbb{Z},$ we get

$(d-\alpha)a_{\alpha,0}^{d,k}+(k-n)a_{\alpha,0}^{d,k-n}=(d-\alpha)a_{0,0}^{d,k}+(k-n)a_{0,0}^{d,k-n}.$
(14)

Putting $\beta=-\alpha,$ $i=j=0$ in (13) with (14) for $k\in\mathbb{Z}$, we
get

$2a_{0,0}^{d,k}=a_{\alpha,0}^{d,k}+a_{-\alpha,0}^{d,k}.$ (15)

In the equation (14) replacing $\alpha$ with $-\alpha,$ we obtain:

$(d+\alpha)a_{-\alpha,0}^{d,k}+(k-n)a_{-\alpha,0}^{d,k-n}=(d+\alpha)a_{0,0}^{d,k}+(k-n)a_{0,0}^{d,k-n}$
(16)

and adding the equations (14) and (16) with together (15) we deduce

$a_{\alpha,0}^{d,k}=a_{0,0}^{d,k}\quad\text{for all}\quad\alpha\in G.$

Putting $\beta=-\alpha,$ $j=0$ in (13), we get

$(d+2\alpha)a_{\alpha,i}^{d,k}+(k-n)a_{\alpha,i}^{d,k-n}=4\alpha
a_{0,i}^{d,k}+2ia_{0,i+n}^{d,k}+(d-2\alpha)a_{0,0}^{d,k-i}+(k-2i-n)a_{0,0}^{d,k-i-n}.$
(17)

Letting $\beta=j=0$ and in (13), we deduce

$(d-\alpha)a_{\alpha,i}^{d,k}+(k-n)a_{\alpha,i}^{d,k-n}=2ia_{\alpha,i+n}^{d,k}+(d-\alpha)a_{0,0}^{d,k-i}+(k-2i-n)a_{0,0}^{d,k-i-n}.$
(18)

Subtracting (17) from (18) it gives

$\alpha(3a_{\alpha,i}^{d,k}+a_{0,0}^{d,k-i}-4a_{0,i}^{d,k})=2i(a_{0,i+n}^{d,k}-a_{\alpha,i+n}^{d,k}).$
(19)

Now consider the case of $\beta=i=0$ in (13), we have

$(d-\alpha)a_{0,j}^{d,k}+(k-n)a_{0,j}^{d,k-n}=2ja_{\alpha,j+n}^{d,k}-2\alpha
a_{\alpha,j}^{d,k}+(d+\alpha)a_{0,0}^{d,k-j}+(k-2j-n)a_{0,0}^{d,k-j-n}.$ (20)

Setting $\alpha=0$ in (18) and substituting $j$ with $i$ in (20) then
combining them it derives

$\alpha(2a_{\alpha,i}^{d,k}-a_{0,0}^{d,k-i}-a_{0,i}^{d,k})=2i(a_{\alpha,i+n}^{d,k}-a_{0,i+n}^{d,k}).$
(21)

Finally, adding up (19) and (21), we have $a_{\alpha,i}^{d,k}=a_{0,i}^{d,k},$
and from the equation (19) it follows that
$a_{\alpha,i}^{d,k}=a_{0,i}^{d,k}=a_{0,0}^{d,k-i}.$ Thus, it proves that
$\varphi_{d}$ has the form

$\varphi_{d}(L_{\alpha,i})=\sum_{k\in\mathbb{Z}}a_{0,0}^{d,k-i}L_{\alpha+d,k}.$

∎

Now, we aim to classify all transposed Poisson structures on $W_{n}(G).$ Let
$\mathcal{W}$ be a commutative associative algebra with basis
$\\{L_{\alpha,i}\ |\ \alpha\in G,i\in\mathbb{Z}\\}$ and multiplication

$L_{\alpha,i}L_{\beta,j}=L_{\alpha+\beta,i+j}.$

######  Theorem 13.

Let $(\mathfrak{L},\cdot,[\cdot,\cdot])$ be a transposed Poisson structure
defined on the Lie algebra $W_{n}(G).$ Then
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is not Poisson algebra and
$(\mathfrak{L},\cdot)$ is a mutation of the algebra $\mathcal{W}$. On the
other hand, every mutation of the algebra $\mathcal{W}$ gives a transposed
Poisson structure with the Lie part isomorphic to the algebra $W_{n}(G).$

###### Proof.

We aim to describe the multiplication $\cdot.$ By Lemma 5, for every element
$X\in\\{L_{\alpha,i}\ |\ \alpha\in G,i\in\mathbb{Z}\\},$ there is a related
$\frac{1}{2}$-derivation $\varphi_{l}$ of $W_{n}(G)$ such that
$\varphi_{X}(Y)=X\cdot Y.$ By Theorem 12, for each
$\varphi\in\Delta(W_{n}(G)),$ we have

$\varphi(L_{\alpha,i})={\sum_{d\in G,\
k\in\mathbb{Z}}}a^{d,k}L_{\alpha+d,i+k}.$

Now we consider the multiplication $L_{0,0}\cdot L_{\alpha,i}$ for all
$i\in\mathbb{Z}\setminus\\{0\\}.$ Then we have

$\varphi_{L_{0,0}}(L_{\alpha,i})\ =\ L_{\alpha,i}\cdot L_{0,0}=L_{0,0}\cdot
L_{\alpha,i}\ =\ \varphi_{L_{\alpha,i}}(L_{0,0}),$

which implies

$\sum\limits_{d\in
G}\sum\limits_{k\in\mathbb{Z}}a_{L_{0,0}}^{d,k}L_{\alpha+d,i+k}=\sum\limits_{d\in
G}\sum\limits_{k\in\mathbb{Z}}a_{L_{\alpha,i}}^{d,k}L_{d,k}.$

Thus, we get $a_{L_{\alpha,i}}^{d,k}=a_{L_{0,0}}^{d-\alpha,k-i}$. Now, we
determine the multiplication $L_{\alpha,i}\cdot L_{\beta,j}:$

$L_{\alpha,i}\cdot L_{\beta,j}=\sum\limits_{d\in
G}\varphi_{d,L_{\alpha,i}}(L_{\beta,j})=\sum\limits_{d\in
G}\sum\limits_{k\in\mathbb{Z}}a_{L_{\alpha,i}}^{d,k}L_{\beta+d,k+j}=\sum\limits_{d\in
G}\sum\limits_{k\in\mathbb{Z}}a_{L_{0,0}}^{d-\alpha,k-i}L_{\beta+d,k+j}.$

Hence, we get

$L_{\alpha,i}\cdot L_{\beta,j}=\sum_{d\in
G}\sum_{k\in\mathbb{Z}}a_{L_{0,0}}^{d,k}L_{d+\alpha+\beta,k+i+j},$

If we denote by $w=\sum\limits_{d\in
G}\sum\limits_{k\in\mathbb{Z}}a_{L_{0,0}}^{d,k}L_{d,k},$ then we obtain
$L_{\alpha,i}\cdot L_{\beta,j}=L_{\alpha,i}wL_{\beta,j}.$ Analyzing the
associative law of $\cdot,$ we can see that the multiplication $\cdot$ is
associative. Thus, we get that the algebra $(\mathfrak{L},\cdot)$ is a
mutation of $\mathcal{W}$. On the other hand, for any element $w$ the product
$\cdot$ gives a transposed Poisson structure defined on the Lie algebra
$W_{n}(G).$ It is easy to see that
$\mathfrak{L}\cdot[\mathfrak{L},\mathfrak{L}]\neq 0$ and
$(\mathfrak{L},\cdot,[\cdot,\cdot])$ is a non-Poisson algebra. It gives the
complete statement of the theorem. ∎

######  Remark 14.

It should be mentioned that in [4] the algebra $W_{1}(G)$ was considered under
the restriction $\mathbb{Z}_{+}$ instead of $\mathbb{Z}$. Following this idea,
let $G$ be a nontrivial additive subgroups of $\mathbb{C}$, and $F$ be a
nontrivial additive subsemigroup with a neutral element of $\mathbb{C}$ and
$\mathfrak{f}\in F,$ it is possible to define a Lie algebra $W_{n}(G,F).$ The
algebra $W_{n}(G,F)$ has a basis $\\{L_{g,f}\ |\ g\in G,f\in F\\}$ and the
multiplication table is given below:

$[L_{g_{1},f_{1}},L_{g_{2},f_{2}}]=(g_{2}-g_{1})L_{g_{1}+g_{2},f_{1}+f_{2}}+(f_{2}-f_{1})L_{g_{1}+g_{2},f_{1}+f_{2}+\mathfrak{f}}.$

The description of $\frac{1}{2}$-derivations and transposed Poisson structures
on $W_{n}(G,F)$ is completely similar to Theorems 12 and 13.

##  4\. Transposed Poisson structures on not-finitely graded Heisenberg-Witt
algebra $HW_{n}(G).$

Let $n\in\mathbb{Z}$ and $G$ be any nontrivial additive subgroup of
$\mathbb{C}$. The Lie algebra $HW_{n}(G)$ has a basis $\\{L_{\alpha,i},\
H_{\alpha,i}\ |\ \alpha\in G,\ i\in\mathbb{Z}\\}$ and the following
multiplication table:

$\displaystyle[L_{\alpha,i},L_{\beta,j}]=(\beta-\alpha)L_{\alpha+\beta,i+j}+(j-i)L_{\alpha+\beta,i+j+n},$
(22) $\displaystyle[L_{\alpha,i},H_{\beta,j}]=\beta
H_{\alpha+\beta,i+j}+jH_{\alpha+\beta,i+j+n},$ (23)
$\displaystyle[H_{\alpha,i},H_{\beta,j}]=0.$ (24)

######  Theorem 15.

Let $\varphi$ be a $\frac{1}{2}$-derivation of the algebra $HW_{n}(G).$

  * •

If $n=0$, then

$\varphi(L_{\alpha,i})$ | $=$ | $\sum\limits_{d\in G\cap\mathbb{Z}}a^{d,-d}L_{\alpha+d,i-d},$
---|---|---
$\varphi(H_{\alpha,i})$ | $=$ | $\sum\limits_{d\in G\cap\mathbb{Z}}a^{d,-d}H_{\alpha+d,i-d}.$
  * •

If $n\neq 0,$ then

$\varphi(L_{\alpha,i})$ | $=$ | $\sum\limits_{d\in G\setminus\\{0\\}}\sum\limits_{k\in\mathbb{Z}}a^{d,k}L_{\alpha+d,k+i}+a^{0,0}L_{\alpha,i},$
---|---|---
$\varphi(H_{\alpha,i})$ | $=$ | $\sum\limits_{d\in G\setminus\\{0\\}}\sum\limits_{k\in\mathbb{Z}}a^{d,k}H_{\alpha+d,k+i}+a^{0,0}H_{\alpha,i},$

where $a^{d,kn}=0$ for $k\in\mathbb{Z}_{+}\setminus\\{0\\}.$ If $k=tn+m$, then

    1. (i)

for $t\in\mathbb{Z}_{+}\setminus\\{0\\},$ $m=\overline{1,|n|-1},$ then
$a^{d,k}={(-1)^{t}{d^{-t}}\prod\limits_{p=1}^{t-1}(pn+m)}a^{d,m},$

    2. (ii)

for $t\in\mathbb{Z}_{-}\setminus\\{0\\},$ $m=\overline{0,|n|-1},$ then
$a^{d,k}={d^{t}}\big{(}\prod\limits_{p=1}^{t}(pn-m)\big{)}^{-1}a^{d,m}.$

###### Proof.

The algebra $HW_{n}(G)$ is $G$-graded

$HW_{n}(G)=\bigoplus\limits_{\alpha\in
G}(HW_{n}(G))_{\alpha},\quad(HW_{n}(G))_{\alpha}=\langle
L_{\alpha,i},H_{\alpha,i}\ |\ i\in\mathbb{Z}\rangle\quad\textrm{for}\ \
\alpha\in G.$

By Lemma 4, we get $\triangle(HW_{n}(G))=\bigoplus\limits_{\alpha\in
G}\triangle_{\alpha}(HW_{n}(G)).$ Suppose that
$\varphi_{d}\in\Delta_{d}(HW_{n}(G)).$

The subalgebra $\langle L_{\alpha,i}\ |\ \alpha\in G,\ i\in\mathbb{Z}\rangle$
is isomorphic to the algebra $W_{n}(G)$ considered in the Section 3. Then by
Theorem 12, we can assume that

$\varphi_{d}(L_{\alpha,i})$ | $=$ | $\sum\limits_{k\in\mathbb{Z}}a^{d,k-i}L_{\alpha+d,k}+\sum\limits_{k\in\mathbb{Z}}b_{\alpha,i}^{d,k}H_{\alpha+d,k},$
---|---|---
$\varphi_{d}(H_{\alpha,i})$ | $=$ | $\sum\limits_{k\in\mathbb{Z}}f_{\alpha,i}^{d,k}L_{\alpha+d,k}+\sum\limits_{k\in\mathbb{Z}}c_{\alpha,i}^{d,k}H_{\alpha+d,k}.$

  1. I.

To determine the coefficients $f_{\alpha,i}^{d,k}$, we apply $\varphi_{d}$ to
(24) and obtain

$\sum\limits_{k\in\mathbb{Z}}f_{\alpha,i}^{d,k}\left(\beta
H_{\alpha+\beta+d,j+k}+jH_{\alpha+\beta+d,j+k+n}\right)=\sum\limits_{k\in\mathbb{Z}}f_{\beta,j}^{d,k}\left(\alpha
H_{\alpha+\beta+d,i+k}+iH_{\alpha+\beta+d,i+k+n}\right).$ (25)

Taking $\beta=i=j=0$ in (25), it gives $f_{0,0}^{d,k}=0.$

Taking $\beta=-\alpha,$ $i=j=0$ in (25), we have
$f_{-\alpha,0}^{d,k}=-f_{\alpha,0}^{d,k}.$

Taking $\beta=-\alpha,$ $j=0$ in (25), we have

$\alpha f_{\alpha,i}^{d,k}=-\alpha
f_{-\alpha,0}^{d,k-i}-if_{-\alpha,0}^{d,k-i-n}.$ (26)

Next, to determine the coefficients, applying $\varphi_{d}$ to both side of
(23) obtain

$\begin{array}[]{rl}2\sum\limits_{k\in\mathbb{Z}}&\left(\beta
f_{\alpha+\beta,i+j}^{d,k}+jf_{\alpha+\beta,i+j+n}^{d,k}\right)L_{\alpha+\beta+d,k}\
=\\\
\sum\limits_{k\in\mathbb{Z}}&f_{\beta,j}^{d,k}\left((\beta-\alpha+d)L_{\alpha+\beta+d,i+k}+(k-i)L_{\alpha+\beta+d,i+k+n}\right).\end{array}$
(27)

and

$\begin{array}[]{ll}2\sum\limits_{k\in\mathbb{Z}}&\left(\beta
c_{\alpha+\beta,i+j}^{d,k}+jc_{\alpha+\beta,i+j+n}^{d,k}\right)H_{\alpha+\beta+d,k}\
=\\\\[5.69054pt] &\sum\limits_{k\in\mathbb{Z}}a^{d,k-i}\left(\beta
H_{\alpha+\beta+d,j+k}+jH_{\alpha+\beta+d,j+k+n}\right)+\\\\[5.69054pt]
&\sum\limits_{k\in\mathbb{Z}}c_{\beta,j}^{d,k}\left((\beta+d)H_{\alpha+\beta+d,i+k}+kH_{\alpha+\beta+d,i+k+n}\right).\end{array}$
(28)

Putting $\beta=-\alpha,$ $i=j=0$ in (27), we obtain

$(d-2\alpha)f_{-\alpha,0}^{d,k}+(k-n)f_{-\alpha,0}^{d,k-n}=0.$ (29)

Replacing $-\alpha$ with $\alpha$ in (29), it gives

$(d+2\alpha)f_{\alpha,0}^{d,k}+(k-n)f_{\alpha,0}^{d,k-n}=0.$ (30)

Combining (29) and (30), and using $f_{-\alpha,0}^{d,k}=-f_{\alpha,0}^{d,k},$
we can get $f_{\alpha,0}^{d,k}=0.$ Moreover, from (26), we have
$f_{\alpha,i}^{d,k}=0$ for $k\in\mathbb{Z}.$

  2. II.

Now we identify the coefficients $c_{\alpha,i}^{d,k},$ so taking $\alpha=j=0$
in (28), we deduce

$2\beta c_{\beta,i}^{d,k}=\beta
a^{d,k-i}+(\beta+d)c_{\beta,0}^{d,k-i}+(k-i-n)c_{\beta,0}^{d,k-i-n}.$ (31)

Next taking $\alpha=-\beta,$ $j=0$ in (28), we have

$2\beta c_{0,i}^{d,k}=\beta
a^{d,k-i}+(\beta+d)c_{\beta,0}^{d,k-i}+(k-i-n)c_{\beta,0}^{d,k-i-n}.$ (32)

Combining (31) and (32), we obtain that

$c_{\beta,i}^{d,k}=c_{0,i}^{d,k}.$

Now putting $\beta=0,$ $i=0$ in (31), we have

$dc_{0,0}^{d,k}+(k-n)c_{0,0}^{d,k-n}=0.$ (33)

Letting $\beta\neq 0,$ $i=0$ in (31), we get

$\beta(a^{d,k}-c_{0,0}^{d,k})+dc_{0,0}^{d,k}+(k-n)c_{0,0}^{d,k-n}=0.$ (34)

Thus, we get $c_{0,0}^{d,k}=a^{d,k},$ which implies $c_{0,i}^{d,k}=a^{d,k-i},$
and

$da^{d,k}+(k-n)a^{d,k-n}=0.$ (35)

    1. A.

If $n=0,$ then we get $(d+k)a^{d,k}=0,$ which implies $a^{d,k}=0$ for
$k\neq-d.$

    2. B.

If $n\neq 0,$ then taking $d=0,$ we have $a^{0,k}=0$ for $k\neq 0.$ Then
letting $k=n,$ we get $a^{d,n}=0$ for $d\neq 0,$ which implies $a^{d,kn}=0$
for $k\in\mathbb{Z}_{+}\setminus\\{0\\}.$

Now, letting $k=tn+q$, we obtain

      1. (i)

$a^{d,k}={(-1)^{t}{d^{-t}}\prod\limits_{p=1}^{t-1}(pn+q)}a^{d,q}$ for
$t\in\mathbb{Z}_{+}\setminus\\{0\\},$ $q=\overline{1,|n|-1},$

      2. (ii)

$a^{d,k}={d^{t}}\big{(}\prod\limits_{p=1}^{t}(pn-q)\big{)}^{-1}a^{d,q}$ for
$t\in\mathbb{Z}_{-}\setminus\\{0\\},$ $q=\overline{0,|n|-1}.$

  3. III.

Next to determine the coefficients $b_{\alpha,i}^{d,k},$ we apply
$\varphi_{d}$ to both side of (22) and obtain

$\begin{array}[]{ll}2\sum\limits_{k\in\mathbb{Z}}&\left((\beta-\alpha)b_{\alpha+\beta,i+j}^{d,k}+(j-i)b_{\alpha+\beta,i+j+n}^{d,k}\right)H_{\alpha+\beta+d,k}=\\\\[5.69054pt]
&\sum\limits_{k\in\mathbb{Z}}b_{\beta,j}^{d,k}\left((\beta+d)H_{\alpha+\beta+d,i+k}+kH_{\alpha+\beta+d,i+k+n}\right)-\\\\[5.69054pt]
&\sum\limits_{k\in\mathbb{Z}}b_{\alpha,i}^{d,k}\left((\alpha+d)H_{\alpha+\beta+d,j+k}+kH_{\alpha+\beta+d,j+k+n}\right).\end{array}$
(36)

We start with taking $\beta=i=j=0$ in (36), it gives

$(d-\alpha)b_{\alpha,0}^{d,k}+(k-n)b_{\alpha,0}^{d,k-n}=db_{0,0}^{d,k}+(k-n)b_{0,0}^{d,k-n}.$
(37)

Taking $\beta=-\alpha,$ $i=j=0$ in (36) with (37) we have
$b_{\alpha,0}^{d,k}+b_{-\alpha,0}^{d,k}=2b_{0,0}^{d,k}.$ In the equation (37)
replacing $\alpha$ with $-\alpha$

$(d+\alpha)b_{-\alpha,0}^{d,k}+(k-n)b_{-\alpha,0}^{d,k-n}=db_{0,0}^{d,k}+(k-n)b_{0,0}^{d,k-n}$
(38)

and adding the equations (37) and (38) we deduce
$b_{\alpha,0}^{d,k}=b_{0,0}^{d,k},$ for all $\alpha\in G.$ Putting
$\beta=-\alpha,$ $j=0$ and $\beta=j=0$ in (36) we have

$(d+\alpha)b_{\alpha,i}^{d,k}-4\alpha
b_{0,i}^{d,k}-2ib_{0,i+n}^{d,k}=(d-\alpha)b_{0,0}^{d,k-i}-(k-n)b_{\alpha,i}^{d,k-n}+(k-i-n)b_{0,0}^{d,k-i-n}$
(39)
$(d-\alpha)b_{\alpha,i}^{d,k}-2ib_{\alpha,i+n}^{d,k}=db_{0,0}^{d,k-i}-(k-n)b_{\alpha,i}^{d,k-n}+(k-i-n)b_{0,0}^{d,k-i-n}.$
(40)

Subtracting (39) from (40), it gives this relation

$4\alpha b_{0,i}^{d,k}-2\alpha b_{\alpha,i}^{d,k}-\alpha
b_{0,0}^{d,k-i}=2ib_{\alpha,i+n}^{d,k}-2ib_{0,i+n}^{d,k}.$ (41)

Next taking $\beta=i=0$ in (36) and substituting $j$ with $i$ on it with
$\alpha=0$ in (39), then combining them we can get

$2\alpha b_{\alpha,i}^{d,k}-\alpha
b_{0,0}^{d,k-i}=2ib_{\alpha,i+n}^{d,k}-2ib_{0,i+n}^{d,k}.$ (42)

Subtracting (41) from (42) we deduce this identity
$b_{\alpha,i}^{d,k}=b_{0,i}^{d,k},$ from this with together (42 we have

$2b_{0,i}^{d,k}=b_{0,0}^{d,k-i},$ (43)

for all $k\in\mathbb{Z}.$ Finally, by (39) and (43) we have
$b_{0,0}^{d,k-i}=0$ for $k\in\mathbb{Z}.$

Thus, it proves that $\varphi_{d}$ has the form

$\varphi_{d}(L_{\alpha,i})$ | $=$ | $\sum_{k\in\mathbb{Z}}a^{d,k-i}L_{\alpha+d,k}$ | $=$ | $\sum_{k\in\mathbb{Z}}a^{d,k}L_{\alpha+d,k+i}$
---|---|---|---|---
$\varphi_{d}(H_{\alpha,i})$ | $=$ | $\sum_{k\in\mathbb{Z}}a^{d,k-i}H_{\alpha+d,k}$ | $=$ | $\sum_{k\in\mathbb{Z}}a^{d,k}H_{\alpha+d,k+i}.$

∎

######  Theorem 16.

There are no non-trivial transposed Poisson structures defined on $HW_{n}(G).$

###### Proof.

We aim to describe the multiplication $\cdot.$ By Lemma 5, for every element
$X\in\ \\{L_{\alpha,i},\ H_{\alpha,i}\ |\ \alpha\in G,\ i\in\mathbb{Z}\\},$
there is a related $\frac{1}{2}$-derivation $\varphi_{l}$ of $HW_{n}(G)$ such
that $\varphi_{X}(Y)=X\cdot Y.$ Thanks to Theorem 15, we have a description of
$\frac{1}{2}$-derivations of $HW_{n}(G).$ Now we consider the multiplication
$L_{\alpha,i}\cdot H_{\beta,j}.$

$\varphi_{H_{\beta,j}}(L_{\alpha,i})\ =\ L_{\alpha,i}\cdot H_{\beta,j}$ | $=$ | $H_{\beta,j}\cdot L_{\alpha,i}\ =\ \varphi_{L_{\alpha,i}}(H_{\beta,j}),$
---|---|---
$\sum\limits_{d\in G\setminus\\{0\\}}\sum\limits_{k\in\mathbb{Z}}a_{H_{\beta,j}}^{d,k-i}L_{\alpha+d,k}$ | $=$ | $\sum\limits_{d\in G\setminus\\{0\\}}\sum\limits_{k\in\mathbb{Z}}a_{L_{\alpha,i}}^{d,k-j}H_{\beta+d,k}.$

Thus, we get $a_{L_{\alpha,i}}^{d,k}=a_{H_{\beta,j}}^{d,k}=0$. So, it gives
$L_{\alpha,i}\cdot L_{\beta,j}=H_{\alpha,i}\cdot H_{\beta,j}=0$ and
$(HW_{n}(G),\cdot)$ is trivial.

∎

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