Transposed Poisson structures on solvable Lie algebras with filiform
nilradical111 The work is supported by grant F-FA-2021-423, Ministry of Higher
Education, Science and Innovations of the Republic of Uzbekistan.

Kobiljon Abdurasulov222CMA-UBI, Universidade da Beira Interior, Covilhã,
Portugal; Institute of Mathematics Academy of Sciences of Uzbekistan,
Tashkent, Uzbekistan<EMAIL_ADDRESS>Jobir Adashev 333Institute of
Mathematics Academy of Sciences of Uzbekistan, Tashkent, Uzbekistan; State
Pedagogical Institute of Tashkent Region, Tashkent, Uzbekistan;
<EMAIL_ADDRESS>& Sabohat Eshmeteva444State Pedagogical Institute of
Tashkent Region, Tashkent, Uzbekistan<EMAIL_ADDRESS>
Abstract: In this article, we described $\frac{1}{2}$-derivations of solvable
Lie algebras with a thread-like nilradical. Nontrivial transposed Poisson
algebras with solvable Lie algebras are constructed. That is, by using
$\frac{1}{2}$-derivations of Lie algebras, we have established commutative
associative multiplication to construct a transposed Poisson algebra with an
associated given Lie algebra.

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

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

## Introduction

Bai, Bai, Guo, and Wu [2] have introduced a dual notion of the Poisson
algebra, called a transposed Poisson algebra, by exchanging the roles of the
two multiplications in the Leibniz rule defining a Poisson algebra. We know
that Poisson algebra is introduced to commutative associative algebras using
its derivation. Similarly, the concept of a transposed Poisson algebra is
defined on a Lie Algebra through its a $\frac{1}{2}$-derivation. 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 [24, 2, 6, 4,
22, 23, 8].

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 [28, 9, 13]. In this
paper, we present transposed Poisson algebras that demonstrate solvable Lie
algebra with filiform nil-radical. Note that any unital transposed Poisson
algebra is a particular case of a “contact bracket” algebra and a quasi-
Poisson algebra [4]. Each transposed Poisson algebra is a commutative Gelfand-
Dorfman algebra [24] and it is also an algebra of Jordan brackets [6]. In a
paper by Ferreira, Kaygorodov, and Lopatkin a relation between
$\frac{1}{2}$-derivations of Lie algebras and transposed Poisson algebras has
been established [5]. These ideas were used to describe all transposed Poisson
structures on Witt and Virasoro algebras in [5]; on twisted Heisenberg-
Virasoro, Schrödinger-Virasoro and extended Schrödinger-Virasoro algebras in
[30]; on Schrödinger algebra in $(n+1)$-dimensional space-time in [27]; on
Witt type Lie algebras in [15]; on generalized Witt algebras in [16]; Block
Lie algebras in [14, 16] and on the Lie algebra of upper triangular matrices
in [17] and there is more a transposed Poisson structures on the Lie algebra
of upper triangular matrices; and on Lie incidence algebras in [18]. Any
complex finite-dimensional solvable Lie algebra was proved to admit a non-
trivial transposed Poisson structure [20]. The algebraic and geometric
classification of three-dimensional transposed Poisson algebras was given in
[3]. For the list of actual open questions on transposed Poisson algebras, see
[4]. Recently described [19] 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. Its paper also found an example of a finite-
dimensional Lie algebra with non-trivial $\frac{1}{2}$-derivations but without
non-trivial transposed Poisson algebra structures. Also, see [12, Section
7.3], and the references therein for similar studies. In [3] obtained the
algebraic and geometric classification of all complex $3$-dimensional
transposed Poisson algebras, and in [29] obtained the algebraic classification
of all complex $3$-dimensional transposed Poisson $3$-Lie algebras.

##  1\. PRELIMINARIES

In this section, we present the concepts and known results. All the algebras
we present in this section are given over the field $\mathbb{C}$ unless
otherwise stated.

We first recall the definition of a Poisson algebra.

######  Definition 1.

Let $\mathfrak{L}$ be a vector space equipped with two bilinear operations

$\cdot,\;[-,-]:\mathfrak{L}\otimes\mathfrak{L}\to\mathfrak{L}.$

The triple $(\mathfrak{L},\cdot,[-,-])$ is called a Poisson algebra if
$(\mathfrak{L},\cdot)$ is a commutative associative algebra and
$(\mathfrak{L},[-,-])$ is a Lie algebra which satisfies the compatibility
condition

$[x,y\cdot z]=[x,y]\cdot z+y\cdot[x,z].$ (1)

Eq. (1) is called the Leibniz rule since the adjoint operators of the Lie
algebra are derivations of the commutative associative algebra.

######  Definition 2.

Let $\mathfrak{L}$ be a vector space equipped with two bilinear operations

$\cdot,\;[-,-]:\mathfrak{L}\otimes\mathfrak{L}\to\mathfrak{L}.$

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 which satisfies the following
compatibility condition

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

Eq. (2) is called the transposed Leibniz rule because the roles played by the
two binary operations in the Leibniz rule in a Poisson algebra are switched.
Further, the resulting operation is rescaled by introducing a factor 2 on the
left-hand side.

Transposed Poisson algebras were first introduced in a paper by Bai, Bai, Guo
and Wu [2]. A transposed Poisson structure $\cdot$ on $\mathfrak{L}$ is called
trivial, if $x\cdot y=0$ for all $x,y\in\mathfrak{L}$.

We show that the compatibility relations of the Poisson algebra and those of
the Transposed Poisson algebra are independent in the following sense.

######  Proposition 3.

[2] Let $(\mathfrak{L},\cdot)$ be a commutative associative algebra and
$(\mathfrak{L},[-,-])$ be a Lie algebra. Then $(\mathfrak{L},\cdot,[-,-])$ is
both a Poisson algebra and a Transposed Poisson algebra if and only if

$x\cdot[y,z]=[x\cdot y,z]=0.$ (3)

######  Definition 4.

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

$\varphi([x,y])=\frac{1}{2}([\varphi(x),y]+[x,\varphi(y)]).$ (4)

Observe that $\frac{1}{2}$-derivations are a particular case of
$\delta$-derivations introduced by Filippov in [7] (see, [10, 11] and
references therein). It is easy to see from 11 that
$[\mathfrak{L},\mathfrak{L}]$ and $Ann(\mathfrak{L})$ are invariant under any
$\frac{1}{2}$-derivation of $\mathfrak{L}$. Definitions 2 and 4 immediately
implies the following key Lemma.

######  Lemma 5.

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
$(\mathfrak{L},[-,-])$.

######  Definition 6.

An $n$-dimensional Lie algebra $\mathfrak{L}$ is said to be filiform if
$\dim\mathfrak{L}^{i}=n-i$, for $2\leq i\leq n$.

Now let us define a natural gradation for the nilpotent Lie algebras.

######  Definition 7.

Given a nilpotent Lie algebra $\mathfrak{L}$ with nilindex $s$, put
$\mathfrak{L}_{i}=\mathfrak{L}^{i}/\mathfrak{L}^{i+1},\ 1\leq i\leq s-1$, and
$Gr(\mathfrak{L})=\mathfrak{L}_{1}\oplus\mathfrak{L}_{2}\oplus\ldots\oplus\mathfrak{L}_{s-1}$.
Define the product in the vector space $Gr(L)$ as follows:

$[x+\mathfrak{L}^{i+1},y+\mathfrak{L}^{j+1}]:=[x,y]+\mathfrak{L}^{i+j+1},$

where $x\in\mathfrak{L}^{i}\setminus\mathfrak{L}^{i+1},$
$y\in\mathfrak{L}^{j}\setminus\mathfrak{L}^{j+1}.$ Then
$[\mathfrak{L}_{i},\mathfrak{L}_{j}]\subseteq\mathfrak{L}_{i+j}$ and we obtain
the graded algebra $Gr(\mathfrak{L})$. If $Gr(\mathfrak{L})$ and
$\mathfrak{L}$ are isomorphic, then we say that the algebra $\mathfrak{L}$ is
naturally graded.

It is well known that there are two types of naturally graded filiform Lie
algebras. In fact, the second type will appear only in the case when the
dimension of the algebra is even.

######  Theorem 8.

[26] Any naturally graded filiform Lie algebra is isomorphic to one of the
following non isomorphic algebras:

$\begin{array}[]{ll}\mathfrak{n}_{n,1}:&[e_{i},e_{1}]=e_{i+1},\quad 2\leq
i\leq n-1;\\\\[2.84526pt] \mathfrak{Q}_{2n}:&[e_{i},e_{1}]=e_{i+1},\ 2\leq
i\leq 2n-2,\quad[e_{i},e_{2n+1-i}]=(-1)^{i}e_{2n},\ 2\leq i\leq n.\end{array}$

All solvable Lie algebras whose nilradical is the naturally graded filiform
Lie algebra $\mathfrak{n}_{n,1}$ are classified in [25]. Further solvable Lie
algebras whose nilradical is the naturally graded filiform Lie algebra
$\mathfrak{Q}_{2n}$ are classified in [1]. It is proved that the dimension of
a solvable Lie algebra whose nilradical is isomorphic to an $n$-dimensional
naturally graded filiform Lie algebra is not greater than $n+2$.

Here we give the list of such solvable Lie algebras. We denote by
$\mathfrak{s}^{i}_{n,1}$ solvable Lie algebras with nilradical
$\mathfrak{n}_{n,1}$ and codimension one, and by $\mathfrak{s}_{n,2}$ with
codimension two:

$\begin{array}[]{ll}\mathfrak{s}^{1}_{n,1}(\beta):&[e_{i},e_{1}]=e_{i+1},\
2\leq i\leq n-1,\quad[e_{1},x]=e_{1},\quad[e_{i},x]=(i-2+\beta)e_{i},\ 2\leq
i\leq n;\\\\[5.69054pt] \mathfrak{s}^{2}_{n,1}:&[e_{i},e_{1}]=e_{i+1},\ 2\leq
i\leq n-1,\quad[e_{i},x]=e_{i},\ 2\leq i\leq n;\\\\[2.84526pt]
\mathfrak{s}^{3}_{n,1}:&[e_{i},e_{1}]=e_{i+1},\ 2\leq i\leq
n-1,\quad[e_{1},x]=e_{1}+e_{2},\quad[e_{i},x]=(i-1)e_{i},\ 2\leq i\leq
n;\end{array}$

$\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1}):[e_{i},e_{1}]=e_{i+1},\
2\leq i\leq
n-1,\quad[e_{i},x]=e_{i}+\sum\limits_{l=i+2}^{n}\alpha_{l+1-i}e_{l},\ 2\leq
i\leq n;$

$\quad\quad\mathfrak{s}_{n,2}\ :\begin{cases}[e_{i},e_{1}]=e_{i+1},\ 2\leq
i\leq n-1,\quad[e_{1},x_{1}]=e_{1},\\\\[2.84526pt] [e_{i},x_{1}]=(i-2)e_{i},\
3\leq i\leq n,\quad[e_{i},x_{2}]=e_{i},&2\leq i\leq n.\end{cases}$

Now we give classification solvable complex Lie algebras having a nilradical
isomorphic to the graded algebra $\mathfrak{Q}_{2n}$. Any solvable Lie algebra
of dimension $2n+1$ with nilradical isomorphic to $\mathfrak{Q}_{2n}$ is
isomorphic to one of the following algebras:

$\mathfrak{r}_{2n+1}(\lambda):\begin{cases}[e_{i},e_{1}]=e_{i+1},\ 2\leq i\leq
2n-2,\quad[e_{i},e_{2n+1-i}]=(-1)^{i}e_{2n},\ 2\leq i\leq n,\\\\[2.84526pt]
[e_{1},x]=e_{1},\quad[e_{i},x]=(i-2+\lambda)e_{i},\ 2\leq i\leq
2n-1,\quad[e_{2n},x]=(2n-3+2\lambda)e_{2n};\end{cases}$

$\mathfrak{r}_{2n+1}(2-n,\varepsilon):\begin{cases}[e_{i},e_{1}]=e_{i+1},\
2\leq i\leq 2n-2,\quad[e_{i},e_{2n+1-i}]=(-1)^{i}e_{2n},&2\leq i\leq
n,\\\\[2.84526pt] [e_{1},x]=e_{1}+\varepsilon e_{2n},\
\varepsilon=-1,1,\quad[e_{i},x]=(i-n)e_{i},&2\leq i\leq 2n-1,\\\\[2.84526pt]
[e_{2n},x]=e_{2n};\end{cases}$

$\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}):\begin{cases}[e_{i},e_{1}]=e_{i+1},\
2\leq i\leq 2n-2,\quad[e_{i},e_{2n+1-i}]=(-1)^{i}e_{2n},&2\leq i\leq
n,\\\\[2.84526pt]
[e_{2+i},x]=e_{2+i}+\sum\limits_{k=2}^{\left[\frac{2n-2-i}{2}\right]}\lambda_{2k+1}e_{2k+1+i},\
0\leq i\leq 2n-6,\\\\[2.84526pt] [e_{2n-i},x]=e_{2n-i},\
i=1,2,3,\quad[e_{2n},x]=2e_{2n}.&\end{cases}$

Moreover, the first nonvanishing parameter $\lambda_{2k+1}$ can be normalized
to $1$.

Finally, for any $n\geq 3$ there is only one solvable Lie algebra
$\mathfrak{r}_{2n+2}$ of dimension $2n+2$ having a nilradical isomorphic to
$\mathfrak{Q}_{2n}:$

$\mathfrak{r}_{2n+2}:\begin{cases}[e_{i},e_{1}]=e_{i+1},\ 2\leq i\leq
2n-2,\quad[e_{i},e_{2n+1-i}]=(-1)^{i}e_{2n},\ 2\leq i\leq n,\\\\[2.84526pt]
[e_{i},x_{1}]=ie_{i},\ 1\leq i\leq
2n-1,\quad[e_{2n},x_{1}]=(2n+1)e_{2n},\\\\[2.84526pt] [e_{i},x_{2}]=e_{i},\
2\leq i\leq 2n-1,\quad[e_{2n},x_{2}]=2e_{2n}.\\\\[2.84526pt] \end{cases}$

Note that $\frac{1}{2}$-derivations of $\mathfrak{s}_{n,2}$ algebra were
studied by I. Kaygorodov and A. Khudoyberdiyev [19].

######  Theorem 9.

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

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

######  Theorem 10.

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

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

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

It’s obvious from Eq. (3) that this structure is non-Poisson.

##  2\. $\frac{1}{2}$-derivation of solvable Lie algebra with filiform
nilradical

In this section, we calculate $\frac{1}{2}$-derivation of solvable Lie
algebras with filiform nilradical algebras.

######  Theorem 11.

Any $\frac{1}{2}$-derivations of the algebra $\mathfrak{s}^{1}_{n,1}(\beta)$
has the form

$\varphi(e_{1})=\alpha_{1}e_{1}+\sum\limits_{i=2}^{n}\alpha_{i}e_{i},\
\varphi(e_{i})=\alpha_{1}e_{i},\ 2\leq i\leq n,\
\varphi(x)=\sum\limits_{i=2}^{n-1}(i-3+\beta)\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\alpha_{1}x,$

with restriction $(\beta-2)\alpha_{2}=0.$

###### Proof.

It is easy to see that $\mathfrak{s}^{1}_{n,1}(\beta)$ has $3$ generators. We
use these generators to calculate of $\frac{1}{2}$-derivation.

$\varphi(e_{1})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x,\quad\varphi(e_{2})=\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,\quad\varphi(x)=\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{1}$ and $e_{2}:$

$\varphi(e_{3})=\varphi([e_{2},e_{1}])=\frac{1}{2}([\varphi(e_{2}),e_{1}]+[e_{2},\varphi(e_{1})])$

$=\frac{1}{2}([\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,e_{1}]+[e_{2},\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x])=\frac{1}{2}(\sum\limits_{i=3}^{n}\beta_{i-1}e_{i}-\beta_{n+1}e_{1}+\alpha_{1}e_{3}+\alpha_{n+1}\beta
e_{2})$

$=\frac{1}{2}(-\beta_{n+1}e_{1}+\alpha_{n+1}\beta
e_{2}+(\alpha_{1}+\beta_{2})e_{3}+\sum\limits_{i=4}^{n}\beta_{i-1}e_{i}).$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{1}$ and $x:$

$\varphi([e_{1},x])=\frac{1}{2}([\varphi(e_{1}),x]+[e_{1},\varphi(x)])=\frac{1}{2}([\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x,x]+[e_{1},\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x])$

$=\frac{1}{2}(\alpha_{1}e_{1}+\sum\limits_{i=2}^{n}(i-2+\beta)\alpha_{i}e_{i}-\sum\limits_{i=3}^{n}\delta_{i-1}e_{i}+\delta_{n+1}e_{1})$

$=\frac{1}{2}((\alpha_{1}+\delta_{n+1})e_{1}+\beta\alpha_{2}e_{2}+\sum\limits_{i=3}^{n}((i-2+\beta)\alpha_{i}-\delta_{i-1})e_{i}).$

On the other hand

$\varphi([e_{1},x])=\varphi(e_{1})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x.$

Comparing coefficients at the basis elements we obtain that

$\delta_{n+1}=\alpha_{1},\quad(\beta-2)\alpha_{2}=0,\quad\delta_{i}=(i-3+\beta)\alpha_{i+1},\quad
2\leq i\leq n-1,\quad\alpha_{n+1}=0.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{2},x:$

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

$=\frac{1}{2}([\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,x]+[e_{2},\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x])=\frac{1}{2}(\beta_{1}e_{1}+\sum\limits_{i=2}^{n}(i-2+\beta)\beta_{i}e_{i}+\delta_{1}e_{3}+\beta\delta_{n+1}e_{2})$

$=\frac{1}{2}\Big{(}\beta_{1}e_{1}+\beta(\beta_{2}+\delta_{n+1})e_{2}+((1+\beta)\beta_{3}+\delta_{1})e_{3}+\sum\limits_{i=4}^{n}(i-2+\beta)\beta_{i}e_{i}\Big{)}.$

On the other hand

$\varphi([e_{2},x])=\beta\varphi(e_{2})=\beta(\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x).$

Comparing coefficients at the basis elements we obtain that

$(2\beta-1)\beta_{1}=0,\ \beta(\beta_{2}-\delta_{n+1})=0,\
\delta_{1}=(1-\beta)\beta_{3},\ (i-2-\beta)\beta_{i}=0,\ 4\leq i\leq n.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{3}$ and $x:$

$\varphi([e_{3},x])=\frac{1}{2}([\varphi(e_{3}),x]+[e_{3},\varphi(x)])$

$=\frac{1}{2}\Big{(}[\frac{1}{2}(-\beta_{n+1}e_{1}+(\alpha_{1}+\beta_{2})e_{3}+\sum\limits_{i=4}^{n}\beta_{i-1}e_{i}),x]+[e_{3},\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x]\Big{)}=$

$=\frac{1}{2}\Big{(}\frac{1}{2}(-\beta_{n+1}e_{1}+(1+\beta)(\alpha_{1}+\beta_{2})e_{3}+\sum\limits_{i=4}^{n}(i-2+\beta)\beta_{i-1}e_{i})+\delta_{1}e_{4}+(1+\beta)\delta_{n+1}e_{3}\Big{)}.$

On the other hand

$\varphi([e_{3},x])=(1+\beta)d(e_{3})=\frac{1}{2}(1+\beta)(-\beta_{n+1}e_{1}+(\alpha_{1}+\beta_{2})e_{3}+\sum\limits_{i=4}^{n}\beta_{i-1}e_{i}).$

Comparing coefficients at the basis elements we obtain that

$(2\beta+1)\beta_{n+1}=0,\ (1+\beta)(\alpha_{1}+\beta_{2}-2\delta_{n+1})=0,\
2\delta_{1}=\beta\beta_{3},\ (i-4+\beta)\beta_{i-1}=0,\ 5\leq i\leq n.$

Comparing the obtained equalities, we get the following relations.

$(2\beta-1)\beta_{1}=0,\ \beta_{2}=\alpha_{1},\ (3\beta-2)\beta_{3}=0,\
3\delta_{1}=\beta_{3},$ $(\beta-2)\alpha_{2}=0,\ (n-2-\beta)\beta_{n}=0,\
(2\beta+1)\beta_{n+1}=0,\ \beta_{i}=0,\ 4\leq i\leq n-1.$

We will have the following.

$\varphi(e_{1})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i},\
\varphi(e_{2})=\beta_{1}e_{1}+\alpha_{1}e_{2}+\beta_{3}e_{3}+\beta_{n+1}x,$

$\varphi(e_{3})=\frac{1}{2}(-\beta_{n+1}e_{1}+2\alpha_{1}e_{3}+\beta_{3}e_{4}),\
\varphi(x)=\frac{1}{3}\beta_{3}e_{1}+\sum\limits_{i=2}^{n-1}(i-3+\beta)\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\alpha_{1}x.$

Using property of the $\frac{1}{2}$-derivation for the product
$[e_{3},e_{1}]=e_{4}$ we have

$\varphi(e_{4})=\alpha_{1}e_{4}+\frac{1}{4}\beta_{3}e_{5}.$

If we check the situation for the elements $\\{e_{4},x\\}$ and
$\\{e_{3},e_{2}\\}$, we get the following relations

$\beta_{1}=0,\ \beta_{3}=0,\ \beta_{n+1}=0.$

By the induction and the property of $\frac{1}{2}$-derivation, we derive

$\varphi(e_{i})=\alpha_{1}e_{i},\ 5\leq i\leq n.$

Thus we have the following

$\varphi(e_{1})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i},\
\varphi(e_{i})=\alpha_{1}e_{i},\ 2\leq i\leq n,\
\varphi(x)=\sum\limits_{i=2}^{n-1}(i-3+\beta)\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\alpha_{1}x.$

This completes the proof of the theorem. ∎

Now we study the $\frac{1}{2}$-derivation of the algebra
$\mathfrak{s}^{2}_{n,1}.$

######  Theorem 12.

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

$\varphi(e_{1})=\alpha_{1}e_{1}+\sum\limits_{i=3}^{n}\alpha_{i}e_{i},\
\varphi(e_{i})=\alpha_{1}e_{i},\ 2\leq i\leq n,\
\varphi(x)=\sum\limits_{i=2}^{n-1}\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\alpha_{1}x.$

###### Proof.

From table multiplication of the algebra $\mathfrak{s}^{2}_{n,1}$ we conclude
that $e_{1},e_{2}$ and $x$ are the generator basis elements of the algebra. We
use these generators to calculate of $\frac{1}{2}$-derivation:

$\varphi(e_{1})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x,\
\varphi(e_{2})=\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,\
\varphi(x)=\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{1}$ and $e_{2}:$

$\varphi(e_{3})=\varphi([e_{2},e_{1}])=\frac{1}{2}([\varphi(e_{2}),e_{1}]+[e_{2},\varphi(e_{1})])$

$=\frac{1}{2}([\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,e_{1}]+[e_{2},\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x])=\frac{1}{2}(\sum\limits_{i=3}^{n}\beta_{i-1}e_{i}+\alpha_{1}e_{3}+\alpha_{n+1}e_{2}).$

Using property of the $\frac{1}{2}$-derivation for the products $[e_{1},x]=0$
and $[e_{2},x]=e_{2}$, we have

$\alpha_{2}=0,\ \delta_{i}=\alpha_{i+1},\ 2\leq i\leq n-1,$

$\beta_{1}=0,\ \delta_{1}=\beta_{3},\ \delta_{n+1}=\beta_{2},\ \beta_{i}=0,\
4\leq i\leq n+1.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{3},x:$

$\varphi([e_{3},x])=\frac{1}{2}([\varphi(e_{3}),x]+[e_{3},\varphi(x)])$

$=\frac{1}{2}\Big{(}[\frac{1}{2}((\beta_{2}+\alpha_{1})e_{3}+\beta_{3}e_{4}+\alpha_{n+1}e_{2}),x]+[e_{3},\beta_{3}e_{1}+\sum\limits_{i=2}^{n-1}\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\beta_{2}x]\Big{)}$

$=\frac{1}{2}\Big{(}\frac{1}{2}((\beta_{2}+\alpha_{1})e_{3}+\beta_{3}e_{4}+\alpha_{n+1}e_{2})+\beta_{4}e_{1}+\beta_{2}e_{3}\Big{)}.$

On the other hand

$\varphi([e_{3},x])=\varphi(e_{3})=\frac{1}{2}((\beta_{2}+\alpha_{1})e_{3}+\beta_{3}e_{4}+\alpha_{n+1}e_{2}).$

Comparing coefficients at the basis elements we obtain that $\alpha_{n+1}=0,\
\beta_{2}=\alpha_{1},\ \beta_{3}=0.$ Thus we have the following

$\varphi(e_{1})=\alpha_{1}e_{1}+\sum\limits_{i=3}^{n}\alpha_{i}e_{i},\
\varphi(e_{2})=\alpha_{1}e_{2},\ \varphi(e_{3})=\alpha_{1}e_{3},\
\varphi(x)=\sum\limits_{i=2}^{n-1}\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\alpha_{1}x.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{1}$ and $e_{i}$ for $3\leq i\leq n-1:$

$\varphi(e_{i+1})=\varphi([e_{i},e_{1}])=\frac{1}{2}([\varphi(e_{i}),e_{1}]+[e_{i},\varphi(e_{1})])$

$=\frac{1}{2}([\alpha_{1}e_{i},e_{1}]+[e_{3},\alpha_{1}e_{1}+\sum\limits_{i=3}^{n}\alpha_{i}e_{i}])=\frac{1}{2}(\alpha_{1}e_{i+1}+\alpha_{1}e_{i+1})=\alpha_{1}e_{i+1}.$

This completes the proof of the theorem. ∎

Now we will study the $\frac{1}{2}$-derivation of the algebra
$\mathfrak{s}^{3}_{n,1}.$

######  Theorem 13.

Any $\frac{1}{2}$-derivations of the algebra $\mathfrak{s}^{3}_{n,1}$ has the
form

$\varphi(e_{1})=\alpha_{1}e_{1}+\sum\limits_{i=3}^{n}\alpha_{i}e_{i},\
\varphi(e_{i})=\alpha_{1}e_{i},\ 2\leq i\leq n,\
\varphi(x)=\sum\limits_{i=3}^{n-1}(i-2)\alpha_{i+1}e_{i}+\delta_{n}e_{n}+\alpha_{1}x.$

###### Proof.

The algebra $\mathfrak{s}^{3}_{n,1}$ has $e_{1},e_{2}$ and $x$ generators. We
put

$\varphi(e_{1})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x,\
\varphi(e_{2})=\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,\
\varphi(x)=\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{2}$ and $x:$

$\varphi([e_{2},x])=\frac{1}{2}([\varphi(e_{2}),x]+[e_{2},\varphi(x)])=\frac{1}{2}([\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x,x]+[e_{2},\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x])$

$=\frac{1}{2}\Big{(}\beta_{1}(e_{1}+e_{2})+\sum\limits_{i=2}^{n}(i-1)\beta_{i}e_{i}+\delta_{1}e_{3}+\delta_{n+1}e_{2}\Big{)}.$

On the other hand

$\varphi([e_{2},x])=\varphi(e_{2})=\sum\limits_{i=1}^{n}\beta_{i}e_{i}+\beta_{n+1}x.$

Comparing coefficients at the basis elements we obtain that

$\beta_{1}=0,\ \delta_{1}=0,\ \delta_{n+1}=\beta_{2},\ \beta_{i}=0,\ 4\leq
i\leq n+1.$

Now consider the condition of $\frac{1}{2}$-derivation for the elements
$e_{1},x:$

$\varphi([e_{1},x])=\frac{1}{2}([\varphi(e_{1}),x]+[e_{1},\varphi(x)])=\frac{1}{2}([\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x,x]+[e_{1},\sum\limits_{i=1}^{n}\delta_{i}e_{i}+\delta_{n+1}x])$

$=\frac{1}{2}\Big{(}\alpha_{1}(e_{1}+e_{2})+\sum\limits_{i=2}^{n}(i-1)\alpha_{i}e_{i}-\sum\limits_{i=3}^{n}\delta_{i-1}e_{i}+\delta_{n+1}(e_{1}+e_{2})\Big{)}.$

On the other hand

$\varphi([e_{1},x])=\varphi(e_{1}+e_{2})=\sum\limits_{i=1}^{n}\alpha_{i}e_{i}+\alpha_{n+1}x+\beta_{2}e_{2}+\beta_{3}e_{3}.$

Comparing coefficients at the basis elements we obtain that

$\alpha_{2}=0,\ \alpha_{n+1}=0,\ \delta_{n+1}=\alpha_{1},\
\delta_{2}=-2\beta_{3},\ \delta_{i}=(i-2)\alpha_{i+1},\ 3\leq i\leq n-1.$

From the $\frac{1}{2}$-derivation property (4) we have

$\varphi(e_{3})=\varphi([e_{2},e_{1}])=\frac{1}{2}([\varphi(e_{2}),e_{1}]+[e_{2},\varphi(e_{1})])$

$=\frac{1}{2}([\alpha_{1}e_{2}+\beta_{3}e_{3},e_{1}]+[e_{2},\alpha_{1}e_{1}+\sum\limits_{i=3}^{n}\alpha_{i}e_{i}])=\alpha_{1}e_{3}+\frac{1}{2}\beta_{3}e_{4}.$

Consider
$\varphi([e_{3},x])=\frac{1}{2}([\varphi(e_{3}),x]+[e_{3},\varphi(x)]).$
Consequently, $\beta_{3}=0.$

Buy applying the induction and the $\frac{1}{2}$-derivation property (4) for
$3\leq i\leq n-1,$ we derive

$\varphi(e_{i+1})=\varphi([e_{i},e_{1}])=\frac{1}{2}([\varphi(e_{i}),e_{1}]+[e_{i},\varphi(e_{1})])$

$=\frac{1}{2}([\alpha_{1}e_{i},e_{1}]+[e_{3},\alpha_{1}e_{1}+\sum\limits_{i=3}^{n}\alpha_{i}e_{i}])=\frac{1}{2}(\alpha_{1}e_{i+1}+\alpha_{1}e_{i+1})=\alpha_{1}e_{i+1}.$

This completes the proof of the theorem. ∎

Now we will study the $\frac{1}{2}$-derivation of the algebra
$\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1}).$

######  Theorem 14.

Any $\frac{1}{2}$-derivations of the algebra
$\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1})$ has the
form

$\varphi(e_{1})=a_{1}e_{1}+\sum\limits_{i=3}^{n}a_{i}e_{i},\varphi(e_{i})=a_{1}e_{i},\
2\leq i\leq
n,\varphi(x)=\sum\limits_{i=2}^{n-1}(a_{i+1}+\sum\limits_{t=3}^{i-1}\alpha_{t}a_{i-t+2})e_{i}+c_{n}e_{n}+a_{1}x.$

###### Proof.

Similarly, the algebra
$\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1})$ has $3$
generators. We put the $\frac{1}{2}$-derivation for generators:

$\varphi(e_{1})=\sum\limits_{i=1}^{n}a_{i}e_{i}+a_{n+1}x,\
\varphi(e_{2})=\sum\limits_{i=1}^{n}b_{i}e_{i}+b_{n+1}x,\
\varphi(x)=\sum\limits_{i=1}^{n}c_{i}e_{i}+c_{n+1}x.$

From
$0=\varphi([e_{1},x])=\frac{1}{2}([\varphi(e_{1}),x]+[e_{1},\varphi(x)])$, we
obtain that

$a_{2}=0,\ c_{i}=a_{i+1}+\sum\limits_{t=3}^{i-1}\alpha_{t}a_{i-t+2},\ 2\leq
i\leq n-1.$

Now consider the $\frac{1}{2}$-derivation for the elements $e_{1}$ and
$e_{2}:$

$\varphi(e_{3})=\varphi([e_{2},e_{1}])=\frac{1}{2}([\varphi(e_{2}),e_{1}]+[e_{2},\varphi(e_{1})])$

$=\frac{1}{2}([\sum\limits_{i=1}^{n}b_{i}e_{i}+b_{n+1}x,e_{1}]+[e_{2},\sum\limits_{i=1}^{n}a_{i}e_{i}+a_{n+1}x])=\frac{1}{2}\Big{(}\sum\limits_{i=3}^{n}b_{i-1}e_{i}+a_{1}e_{3}+a_{n+1}(e_{2}+\sum\limits_{t=4}^{n}\alpha_{t-1}e_{t})\Big{)}$

$=\frac{1}{2}\Big{(}a_{n+1}e_{2}+(a_{1}+b_{2})e_{3}+\sum\limits_{i=4}^{n}(b_{i-1}+a_{n+1}\alpha_{i-1})e_{i}\Big{)}.$

Similarly, from
$0=\varphi([e_{3},e_{2}])=\frac{1}{2}([\varphi(e_{3}),e_{2}]+[e_{3},d(e_{2})])$,
we derive $b_{1}=0,\ b_{n+1}=0.$

We prove the following equality for $3\leq i\leq n$ by induction.

$\varphi(e_{i})=\frac{1}{2^{i-2}}\Big{(}(2^{i-2}-1)a_{n+1}e_{i-1}+((2^{i-2}-1)a_{1}+b_{2})e_{i}$

$+\sum\limits_{t=i+1}^{n}(b_{t-i+2}+(2^{i-2}-1)a_{n+1}\alpha_{t-i+2})e_{t}\Big{)}.$

If $i=3$, the relationship is fulfilled according to the above equality. Now
we prove that it is true for $i$ and for $i+1$. Now consider the condition of
$\frac{1}{2}$-derivation for the elements $e_{i},e_{1}:$

$\varphi(e_{i+1})=\varphi([e_{i},e_{1}])=\frac{1}{2}([\varphi(e_{i}),e_{1}]+[e_{i},\varphi(e_{1})])$

$=\frac{1}{2}\Big{(}[\frac{1}{2^{i-2}}\Big{(}(2^{i-2}-1)a_{n+1}e_{i-1}+((2^{i-2}-1)a_{1}+b_{2})e_{i}$

$+\sum\limits_{t=i+1}^{n}(b_{t-i+2}+(2^{i-2}-1)a_{n+1}\alpha_{t-i+2})e_{t}\Big{)},e_{1}]+[e_{i},\sum\limits_{i=1}^{n}a_{i}e_{i}+a_{n+1}x]\Big{)}$

$=\frac{1}{2}\Big{(}\frac{1}{2^{i-2}}\Big{(}(2^{i-2}-1)a_{n+1}e_{i}+((2^{i-2}-1)a_{1}+b_{2})e_{i+1}$

$+\sum\limits_{t=i+2}^{n}(b_{t-i+1}+(2^{i-2}-1)a_{n+1}\alpha_{t-i+1})e_{t}\Big{)}+a_{1}e_{i}+a_{n+1}(e_{i}+\sum\limits_{t=i+2}^{n}\alpha_{t-i+1}e_{t})\Big{)}$

$=\frac{1}{2^{i-1}}\left((2^{i-1}-1)a_{n+1}e_{i}+((2^{i-1}-1)a_{1}+b_{2})e_{i+1}+\sum\limits_{t=i+2}^{n}(b_{t-i+1}+(2^{i-1}-1)a_{n+1}\alpha_{t-i+1})e_{t}\right).$

Further, consistently using property of the $\frac{1}{2}$-derivation for the
products $[e_{n},x]=e_{n},[e_{n-1},x])=e_{n-1}$ and
$[e_{2},x]=e_{2}+\sum\limits_{i=4}^{n}\alpha_{i-1}e_{i}$, we have

$a_{n+1}=0,\ c_{n+1}=\frac{1}{2^{n-2}}((2^{n-2}-1)a_{1}+b_{2}),\ b_{2}=a_{1},\
b_{3}=2^{n-3}c_{1},$ $c_{1}=0,\ b_{i}=0,\ 4\leq i\leq n.$

This completes the proof of the theorem. ∎

Below, we present analogous descriptions of the $\frac{1}{2}$-derivation for
the solvable complex Lie algebras having a nilradical isomorphic to the
algebra $\mathfrak{Q}_{2n}$.

######  Theorem 15.

Any $\frac{1}{2}$-derivations of the algebras $\mathfrak{r}_{2n+1}(\lambda)$
and $\mathfrak{r}_{2n+1}(2-n,\varepsilon)$ has the form

$\varphi(e_{i})=ae_{i},\ 1\leq i_{\neq 2}\leq 2n,\
\varphi(e_{2})=ae_{2}+be_{2n},\ \varphi(x)=(2n-3)be_{2n-1}+ce_{2n}+ax.$

######  Theorem 16.

Any $\frac{1}{2}$-derivations of the algebra
$\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})$ has the form

$\varphi(e_{i})=ae_{i},\ 1\leq i_{\neq 2}\leq 2n,\
\varphi(e_{2})=ae_{2}+be_{2n},\ \varphi(x)=ce_{2n}+ax.$

######  Theorem 17.

Any $\frac{1}{2}$-derivations of the algebra $\mathfrak{r}_{2n+2}$ has the
form

$\varphi(e_{i})=ae_{i},\quad 1\leq i\leq
2n,\quad\varphi(x_{1})=(2n+1)be_{2n}+ax_{1},\quad\varphi(x_{2})=2be_{2n}+ax_{2}.$

##  3\. Transposed Poisson structure for solvable Lie algebra with filiform
nilradical

In section we give a descriptin of all transposed Poisson algebra structures
on solvable Lie algebras with naturally graded filiform nilpotent radical.

######  Theorem 18.

Let $(\mathfrak{s}^{1}_{n,1}(\beta),\cdot,[-,-])$ be a transposed Poisson
algebra structure defined on the Lie algebra $\mathfrak{s}^{1}_{n,1}(\beta)$.
Then the multiplication of $(\mathfrak{s}^{1}_{n,1}(\beta),\cdot)$ has the
following form:

$\begin{array}[]{lll}{\bf TP}_{1}(\mathfrak{s}^{1}_{n,1}(\beta)):&x\cdot
x=e_{n};\\\\[2.84526pt] {\bf
TP}_{2}(\mathfrak{s}^{1}_{n,1}(\beta)):&e_{1}\cdot x=e_{n},&x\cdot
x=(n-4+\beta)e_{n-1};\\\\[2.84526pt] {\bf
TP}_{3}(\mathfrak{s}^{1}_{n,1}(2-n)):&e_{1}\cdot x=e_{n},&x\cdot
x=e_{n}-e_{n-1};\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{s}^{1}_{n,1}(\beta),\cdot,[-,-])$ be a transposed Poisson
algebra structure defined on the Lie algebra $\mathfrak{s}^{1}_{n,1}(\beta)$.
Then for any element of $x\in\mathfrak{s}^{1}_{n,1}(\beta)$, 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

  * •

for $e_{i},\ 1\leq i\leq n:$

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=\alpha_{i,1}e_{1}+\sum\limits_{t=2}^{n}\alpha_{i,t}e_{t},\
\varphi_{e_{i}}(e_{t})=\alpha_{i,1}e_{t},\ 2\leq t\leq n,\\\\[2.84526pt]
\varphi_{e_{i}}(x)=\sum\limits_{t=2}^{n-1}(t-3+\beta)\alpha_{i,t+1}e_{t}+\gamma_{i,n}e_{n}+\alpha_{i,1}x,\\\\[2.84526pt]
\end{array}$

  * •

for $x:$

$\begin{array}[]{lll}\varphi_{x}(e_{1})=\alpha_{x,1}e_{1}+\sum\limits_{t=2}^{n}\alpha_{x,t}e_{t},\
\varphi_{x}(e_{t})=\alpha_{x,1}e_{t},\ 2\leq t\leq n,\\\\[2.84526pt]
\varphi_{x}(x)=\sum\limits_{t=2}^{n-1}(t-3+\beta)\alpha_{x,t+1}e_{t}+\gamma_{x,n}e_{n}+\alpha_{x,1}x.\\\\[2.84526pt]
\end{array}$

The following equality holds

$(\beta-2)\alpha_{x,2}=0,\ (\beta-2)\alpha_{i,2}=0,\ 1\leq i\leq n.$

It is known that $\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i}).$ For $2\leq i\leq n,\ j=1$ we have

$\alpha_{i,1}e_{1}+\sum\limits_{t=2}^{n}\alpha_{i,t}e_{t}=\alpha_{1,1}e_{i}.$

Comparing coefficients at the basis elements we obtain that

$\alpha_{i,i}=\alpha_{1,1},\ 2\leq i\leq n,\ \alpha_{i,t}=0,\ 1\leq i\neq
j\leq n.$

It is known that $\varphi_{e_{1}}(x)=e_{1}\cdot x=x\cdot
e_{1}=\varphi_{x}(e_{1}).$ We have

$\gamma_{1,n}e_{n}+\alpha_{1,1}x=\alpha_{x,1}e_{1}+\sum\limits_{t=2}^{n}\alpha_{x,t}e_{t}.$

Comparing coefficients at the basis elements we obtain that

$\alpha_{1,1}=0,\ \alpha_{x,n}=\gamma_{1,n},\ \alpha_{x,t}=0,\ 1\leq t\leq
n-1.$

It is known that $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i}).$ For $2\leq i\leq n$ we have $\gamma_{i,n}=0.$

Thus, we obtain

$e_{1}\cdot x=\alpha e_{n},\ x\cdot x=(n-4+\beta)\alpha e_{n-1}+\gamma e_{n}.$

In this [21] paper an automorphism of the $\mathfrak{s}^{1}_{n,1}(\beta)$
algebra is given. According to it, we get the following.

$\alpha^{\prime}=b_{1}^{n-3}c_{2}\alpha,\
\gamma^{\prime}=(2-n-\beta)a_{1}b_{1}^{n-3}c_{2}\alpha+b_{1}^{n-2}c_{2}\gamma.$

Since we are interested only in non-trivial transposed Poisson algebra
structures, we have the following cases:

  1. (1)

Let $\alpha=0,$ then we have that $\gamma\neq 0$ and via automorphism

$\phi(x)=x+e_{1},\ \phi(e_{1})=e_{1},\
\phi(e_{i})=\sum\limits_{k=i}^{n}\frac{(-1)^{k-i}}{(k-i)!\gamma}e_{k},\ 2\leq
i\leq n,$

we obtain the algebra ${\bf TP}_{1}(\mathfrak{s}^{1}_{n,1}(\beta))$.

  2. (2)

Let $\alpha\neq 0,$ then by choosing $c_{2}=\frac{1}{b_{1}^{n-3}\alpha}$, we
derive $\alpha^{\prime}=1.$

    1. (a)

If $\beta=2-n$. Then

       * •

$\gamma=0$ via automorphism

$\phi(x)=x+e_{1},\ \phi(e_{1})=e_{1},\
\phi(e_{i})=\sum\limits_{k=i}^{n}\frac{(-1)^{k-i}}{(k-i)!\alpha}e_{k},\ 2\leq
i\leq n,$

we obtain the algebra ${\bf TP}_{2}(\mathfrak{s}^{1}_{n,1}(2-n))$.

       * •

$\gamma\neq 0$ via automorphism

$\phi(x)=x+e_{1},\ \phi(e_{1})=\frac{\alpha}{\gamma}e_{1},\
\phi(e_{i})=\frac{\gamma^{n-i-1}}{\alpha^{n-i}}\sum\limits_{k=i}^{n}\frac{(-1)^{k-i}}{(k-i)!}e_{k},\quad
2\leq i\leq n,$

we have the algebra ${\bf TP}_{3}(\mathfrak{s}^{1}_{n,1}(2-n))$.

    2. (b)

If $\beta\neq 2-n$, then via automorphism

$\phi(x)=x+\frac{\gamma}{(n-2+\beta)\alpha}e_{1},\ \phi(e_{1})=e_{1},\
\phi(e_{i})=\sum\limits_{k=i}^{n}\frac{\alpha^{i-k-1}\gamma^{k-i}}{(k-i)!(2-n-\beta)^{k-i}}e_{k},\
2\leq i\leq n,$

we deduce the algebra ${\bf TP}_{2}(\mathfrak{s}^{1}_{n,1}(\beta\neq 2-n))$.

∎

######  Theorem 19.

Let $(\mathfrak{s}^{2}_{n,1},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{s}^{2}_{n,1}$. Then the
multiplication of $(\mathfrak{s}^{2}_{n,1},\cdot)$ has the following form:

$\begin{array}[]{lll}{\bf TP}_{1}(\mathfrak{s}^{2}_{n,1}):&x\cdot
x=e_{n},\\\\[2.84526pt] {\bf TP}_{2}(\mathfrak{s}^{2}_{n,1}):&e_{1}\cdot
x=e_{n},&x\cdot x=e_{n-1},\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{s}^{2}_{n,1},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{s}^{2}_{n,1}$. Then for any
element of $x\in\mathfrak{s}^{2}_{n,1}$, we have that operator of
multiplication $\varphi_{x}(y)=x\cdot y$ is a $\frac{1}{2}$ -derivation.
Hence, by using Theorem 12 for $1\leq i,k\leq n,$ we derive

  * •

for $e_{i},\ 1\leq i\leq n:$

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=\alpha_{i,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{i,t}e_{t},\quad\varphi_{e_{i}}(e_{t})=\alpha_{i,1}e_{t},\
2\leq t\leq n,\\\\[2.84526pt]
\varphi_{e_{i}}(x)=\sum\limits_{t=2}^{n-1}\alpha_{i,t+1}e_{t}+\gamma_{i,n}e_{n}+\alpha_{i,1}x,\\\\[2.84526pt]
\end{array}$

  * •

for $x:$

$\begin{array}[]{lll}\varphi_{x}(e_{1})=\alpha_{x,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{x,t}e_{t},\quad\varphi_{x}(e_{t})=\alpha_{x,1}e_{t},\
2\leq t\leq n,\\\\[2.84526pt]
\varphi_{x}(x)=\sum\limits_{t=2}^{n-1}\alpha_{x,t+1}e_{t}+\gamma_{x,n}e_{n}+\alpha_{x,1}x.\\\\[2.84526pt]
\end{array}$

It is known that $\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i}).$ For $2\leq i\leq n,\ j=1$ we have

$\alpha_{i,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{i,t}e_{t}=\alpha_{1,1}e_{i}.$

Comparing coefficients at the basis elements we obtain that

$\alpha_{i,i}=\alpha_{1,1},\ 3\leq i\leq n,\ \alpha_{i,t}=0,\ 1\leq i\neq
j\leq n.$

It is known that $\varphi_{e_{1}}(x)=e_{1}\cdot x=x\cdot
e_{1}=\varphi_{x}(e_{1}).$ We have

$\gamma_{1,n}e_{n}+\alpha_{1,1}x=\alpha_{x,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{x,t}e_{t}.$

Comparing coefficients at the basis elements we obtain that

$\alpha_{1,1}=0,\ \alpha_{x,n}=\gamma_{1,n},\ \alpha_{x,1}=0,\
\alpha_{x,t}=0,\ 3\leq t\leq n-1.$

It is known that $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i}).$ We obtain that $\gamma_{i,n}=0,\ 2\leq i\leq n.$

Thus, we obtain

$e_{1}\cdot x=\alpha e_{n},\ x\cdot x=\alpha e_{n-1}+\beta e_{n}.$

This [21] paper gives an automorphism of the $\mathfrak{s}^{2}_{n,1}$ algebra.
According to it, we get the following.

$\alpha^{\prime}=b_{1}^{n-3}c_{2}\alpha,\
\beta^{\prime}=b_{1}^{n-3}c_{3}\alpha+b_{1}^{n-2}c_{2}\beta.$

Since we are interested only in non-trivial transposed Poisson algebra
structures, we have the following cases:

  1. (1)

$\alpha=0,$ then we have that $\beta\neq 0$ and via automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\ \phi(e_{i})=\beta^{-1}e_{i},\ 2\leq i\leq n.$

We have ${\bf TP}_{1}(\mathfrak{s}^{2}_{n,1})$ algebras.

  2. (2)

$\alpha\neq 0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\
\phi(e_{i})=\alpha^{-1}e_{i}-\beta\alpha^{-2}e_{i+1},\ 2\leq i\leq n-1,\
\phi(e_{n})=\alpha^{-1}e_{n}.$

We have ${\bf TP}_{2}(\mathfrak{s}^{2}_{n,1})$ algebras.

∎

######  Theorem 20.

Let $(\mathfrak{s}^{3}_{n,1},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{s}^{3}_{n,1}$. Then the
multiplication of $(\mathfrak{s}^{3}_{n,1},\cdot)$ has the following form:

$\begin{array}[]{lll}{\bf TP}_{1}(\mathfrak{s}^{3}_{n,1}):&x\cdot
x=e_{n},\\\\[2.84526pt] {\bf TP}_{2}(\mathfrak{s}^{3}_{n,1}):&e_{1}\cdot
x=e_{n},&x\cdot x=(n-3)e_{n-1},\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{s}^{3}_{n,1},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{s}^{3}_{n,1}$. Then for any
element of $x\in\mathfrak{s}^{3}_{n,1}$, 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

  * •

for $e_{i},\ 1\leq i\leq n:$

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=\alpha_{i,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{i,t}e_{t},\quad\varphi_{e_{i}}(e_{t})=\alpha_{i,1}e_{t},\
2\leq t\leq n,\\\\[2.84526pt]
\varphi_{e_{i}}(x)=\sum\limits_{t=3}^{n-1}(t-2)\alpha_{i,t+1}e_{t}+\gamma_{i,n}e_{n}+\alpha_{i,1}x,\\\\[2.84526pt]
\end{array}$

  * •

for $x:$

$\begin{array}[]{lll}\varphi_{x}(e_{1})=\alpha_{x,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{x,t}e_{t},\quad\varphi_{x}(e_{t})=\alpha_{x,1}e_{t},\
2\leq t\leq n,\\\\[2.84526pt]
\varphi_{x}(x)=\sum\limits_{t=3}^{n-1}(t-2)\alpha_{x,t+1}e_{t}+\gamma_{x,n}e_{n}+\alpha_{x,1}x.\\\\[2.84526pt]
\end{array}$

It is known that $\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i}).$ For $2\leq i\leq n$ and $j=1$ we have

$\alpha_{i,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{i,t}e_{t}=\alpha_{1,1}e_{i}.$

Comparing coefficients at the basis elements we obtain that

$\alpha_{i,i}=\alpha_{1,1},\ 3\leq i\leq n,\ \alpha_{i,t}=0,\ 1\leq i\neq
j\leq n.$

Similarly, from $\varphi_{e_{1}}(x)=e_{1}\cdot x=x\cdot
e_{1}=\varphi_{x}(e_{1}),$ we reduce

$\gamma_{1,n}e_{n}+\alpha_{1,1}x=\alpha_{x,1}e_{1}+\sum\limits_{t=3}^{n}\alpha_{x,t}e_{t}.$

Comparing the coefficients for the basis elements, we obtain the following
restrictions on the coefficients:

$\alpha_{1,1}=0,\ \alpha_{x,n}=\gamma_{1,n},\ \alpha_{x,1}=0,\
\alpha_{x,t}=0,\ 3\leq t\leq n-1.$

Finally, from the equality $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i}),$ we derive $\gamma_{i,n}=0,\ 2\leq i\leq n.$

Thus, we obtain

$e_{1}\cdot x=\alpha e_{n},\ x\cdot x=(n-3)\alpha e_{n-1}+\beta e_{n}.$

This [21] paper gives an automorphism of the $\mathfrak{s}^{3}_{n,1}$ algebra.
According to it, we get the following.

$\alpha^{\prime}=b_{1}^{n-2}\alpha,\
\beta^{\prime}=(1-n)b_{1}^{n-2}a_{1}\alpha+b_{1}^{n-1}\beta.$

Since we are interested only in non-trivial transposed Poisson algebra
structures, we have the following possible cases:

  1. (1)

if $\alpha=0,$ then we conclude $\beta\neq 0$ and by using automorphism

$\phi(x)=x,\ \phi(e_{i})=\sqrt[n-1]{\beta^{-i}}e_{i},\ 1\leq i\leq n,$

we obtain the algebra ${\bf TP}_{1}(\mathfrak{s}^{3}_{n,1})$;

  2. (2)

if $\alpha\neq 0,$ then via automorphism

$\phi(x)=x+\frac{\beta\sqrt[n-2]{\alpha^{1-n}}}{(n-1)}e_{1},\
\phi(e_{1})=\sqrt[n-2]{\alpha^{-1}}e_{1},$
$\phi(e_{i})=\sqrt[n-2]{\alpha^{1-i}}e_{i}-\frac{\beta\sqrt[n-2]{\alpha^{2-i-n}}}{(n-1)}e_{i+1},\
2\leq i\leq n-1,\ \phi(e_{n})=\sqrt[n-2]{\alpha^{1-n}}e_{n},$

the algebra ${\bf TP}_{2}(\mathfrak{s}^{3}_{n,1})$ is obtained.

∎

######  Theorem 21.

Let
$(\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1}),\cdot,[-,-])$
be a transposed Poisson algebra structure defined on the Lie algebra
$\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1})$. Then the
multiplication of
$(\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1}),\cdot)$
has the following form:

$\begin{array}[]{lll}{\bf TP}_{1}(\mathfrak{s}^{4}_{n,1}):&x\cdot
x=e_{n},\\\\[2.84526pt] {\bf TP}_{2}(\mathfrak{s}^{4}_{n,1}):&e_{1}\cdot
x=e_{n},&x\cdot x=e_{n-1},\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let us consider the transposed Poisson structure
$(\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1}),\cdot,[-,-])$
on the Lie algebra
$\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1})$. Then for
any element of
$x\in\mathfrak{s}^{4}_{n,1}(\alpha_{3},\alpha_{4},\ldots,\alpha_{n-1})$, we
have that operator of multiplication $\varphi_{x}(y)=x\cdot y$ is a
$\frac{1}{2}$ -derivation. Hence, by using Theorem 14 we derive following

  * •

for $e_{i},\ 1\leq i\leq n:$

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=a_{i,1}e_{1}+\sum\limits_{t=3}^{n}a_{i,t}e_{t},\
\varphi_{e_{i}}(e_{t})=a_{i,1}e_{t},\ 2\leq t\leq n,\\\\[2.84526pt]
\varphi_{e_{i}}(x)=\sum\limits_{t=2}^{n-1}(a_{i,t+1}+\sum\limits_{r=3}^{t-1}\alpha_{r}a_{i,t-r+2})e_{t}+c_{i,n}e_{n}+a_{i,1}x,\\\\[2.84526pt]
\end{array}$

  * •

for $x:$

$\begin{array}[]{lll}\varphi_{x}(e_{1})=a_{x,1}e_{1}+\sum\limits_{t=3}^{n}a_{x,t}e_{t},\quad\varphi_{x}(e_{t})=a_{x,1}e_{t},\
2\leq t\leq n,\\\\[2.84526pt]
\varphi_{x}(x)=\sum\limits_{t=2}^{n-1}(a_{x,t+1}+\sum\limits_{r=3}^{t-1}\alpha_{r}a_{x,t-r+2})e_{t}+c_{x,n}e_{n}+a_{x,1}x.\\\\[2.84526pt]
\end{array}$

By considering equality $\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i})$ for $2\leq i\leq n$ and $j=1$, we have

$a_{i,1}e_{1}+\sum\limits_{t=3}^{n}a_{i,t}e_{t}=a_{1,1}e_{i}.$

From this we conclude

$a_{i,i}=a_{1,1},\ 3\leq i\leq n,\ a_{i,t}=0,\ 1\leq i\neq j\leq n.$

Also, from the multiplications $\varphi_{e_{1}}(x)=e_{1}\cdot x=x\cdot
e_{1}=\varphi_{x}(e_{1})$ and $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i}),$ we derive

$a_{1,1}=0,\ a_{x,n}=c_{1,n},\ a_{x,1}=0,\ a_{x,t}=0,\ 3\leq t\leq
n-1,c_{i,n}=0,\ 2\leq i\leq n.$

Thus, we obtain

$e_{1}\cdot x=\alpha e_{n},\ x\cdot x=\alpha e_{n-1}+\beta e_{n}.$

According to an automorphism of the algebra $\mathfrak{s}^{4}_{n,1}$ in [21],
we get the following

$\alpha^{\prime}=c_{2}\alpha,\ \beta^{\prime}=c_{3}\alpha+c_{2}\beta.$

Since we are interested only in non-trivial transposed Poisson algebra
structures, we will consider two different situations.

  1. (1)

If $\alpha=0,$ then we conclude $\beta\neq 0$ and by taking automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\ \phi(e_{i})=\beta^{-1}e_{i},\ 2\leq i\leq n,$

we have a multiplication table $x\cdot x=e_{n}$. Hence, we obtain the algebra
${\bf TP}_{1}(\mathfrak{s}^{4}_{n,1})$.

  2. (2)

If $\alpha\neq 0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\
\phi(e_{i})=\alpha^{-1}e_{i}-\beta\alpha^{-2}e_{i+1},\ 2\leq i\leq n-1,\
\phi(e_{n})=\alpha^{-1}e_{n},$

we can assume $\alpha^{\prime}=1$ abd $\beta^{\prime}=0.$ Therefore, the
algebra ${\bf TP}_{2}(\mathfrak{s}^{4}_{n,1})$ is obtained.

∎

Now we give descriptions of all transposed Poisson algebra structures on
solvable Lie algebras with nilradical isomorphic to the algebra
$\mathfrak{Q}_{2n}$

######  Theorem 22.

Let $(\mathfrak{r}_{2n+1}(\lambda),\cdot,[-,-])$ be a transposed Poisson
algebra structure defined on the Lie algebra $\mathfrak{r}_{2n+1}(\lambda)$.
Then the multiplication of $(\mathfrak{r}_{2n+1}(\lambda),\cdot)$ has the
following form:

$\begin{array}[]{lll}{\bf TP}_{1}(\mathfrak{r}_{2n+1}(\lambda)):&x\cdot
x=e_{2n},\\\\[2.84526pt] {\bf
TP}_{2}(\mathfrak{r}_{2n+1}(\lambda)):&e_{2}\cdot x=e_{2n},&x\cdot
x=(2n-3)e_{2n-1},\\\\[2.84526pt] {\bf
TP}_{2}(\mathfrak{r}_{2n+1}(\frac{2n-3}{2})):&e_{2}\cdot x=e_{2n},&x\cdot
x=(2n-3)e_{2n-1}+e_{2n},\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{r}_{2n+1}(\lambda),\cdot,[-,-])$ be a transposed Poisson
algebra structure defined on the Lie algebra $\mathfrak{r}_{2n+1}(\lambda)$.
Then we aim to describe the multiplication $\cdot$ by Lemma 5. So for any
element of $x\in\mathfrak{r}_{2n+1}(\lambda)$, the operator of multiplication
$\varphi_{x}(y)=x\cdot y=y\cdot x=\varphi_{y}(x)$ is a $\frac{1}{2}$
-derivation. Hence, we have

  * •

for $e_{i},\ 1\leq i\leq 2n:$

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=a_{i}e_{1},\quad\varphi_{e_{i}}(e_{2})=a_{i}e_{2}+b_{i}e_{2n},\quad\varphi_{e_{i}}(e_{j})=a_{i}e_{j},\
3\leq j\leq 2n,\\\\[2.84526pt]
\varphi_{e_{i}}(x)=(2n-3)b_{i}e_{2n-1}+c_{i}e_{2n}+a_{i}x,\\\\[2.84526pt]
\end{array}$

  * •

for $x:$

$\begin{array}[]{lll}\varphi_{x}(e_{1})=a_{x}e_{1},\quad\varphi_{x}(e_{2})=a_{x}e_{2}+b_{x}e_{2n},\quad\varphi_{x}(e_{j})=a_{x}e_{j},\
3\leq j\leq 2n,\\\\[2.84526pt]
\varphi_{x}(x)=(2n-3)b_{x}e_{2n-1}+c_{x}e_{2n}+a_{x}x.\\\\[2.84526pt]
\end{array}$

Firstly, for all $i,j$ from $\varphi_{e_{i}}(e_{j})=e_{i}\cdot
e_{j}=e_{j}\cdot e_{i}=\varphi_{e_{j}}(e_{i}),$ we obtain that

$a_{i}=0,\ 1\leq i\leq 2n,\ b_{1}=0,\ b_{t}=0,\ 3\leq t\leq 2n.$

Secondly, we have $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i}),$ which implies

$a_{x}=0,\ b_{2}=0,\ b_{x}=c_{2},\ c_{1}=0,\ c_{t}=0,\ 3\leq t\leq 2n.$

Thus, we obtain

$e_{2}\cdot x=\alpha e_{2n},\ x\cdot x=(2n-3)\alpha e_{2n-1}+\beta e_{2n}.$

We have the following possible cases:

  1. (1)

$\alpha=0,$ then we get $\beta\neq 0,$ and via automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\ \phi(e_{i})=\sqrt{\beta^{-1}}e_{i},\ 2\leq
i\leq 2n-1,\ \phi(e_{2n})=\beta^{-1}e_{2n},$

we obtain the algebra ${\bf TP}_{1}(\mathfrak{r}_{2n+1}(\lambda))$.

  2. (2)

$\alpha\neq 0,\ \lambda=\frac{2n-3}{2}$ and $\beta=0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\ \phi(e_{i})=\beta^{-1}e_{i},\ 2\leq i\leq
2n-1,\ \phi(e_{2n})=\beta^{-2}e_{2n},$

we have the algebra ${\bf TP}_{2}(\mathfrak{r}_{2n+1}(\frac{2n-3}{2}))$.

  3. (3)

$\alpha\neq 0,\ \lambda=\frac{2n-3}{2}$ and $\beta\neq 0,$ then by
transformation

$\phi(x)=x,\ \phi(e_{1})=\sqrt[2n-3]{\alpha^{-2}\beta}e_{1},$

$\phi(e_{i})=\sqrt[2n-3]{\alpha^{2n+1-i}\beta^{i+1-2n}}e_{i},\ 2\leq i\leq
2n-1,\ \phi(e_{2n})=\beta^{-1}e_{2n},$

we get the algebra ${\bf TP}_{3}(\mathfrak{r}_{2n+1}(\frac{2n-3}{2}))$.

  4. (4)

$\alpha\neq 0$ and $\lambda\neq\frac{2n-3}{2},$ then by transformation

$\phi(x)=x+\frac{\alpha^{-2}\beta\lambda}{2\lambda+3-2n}e_{2},\
\phi(e_{1})=e_{1}+\frac{\alpha^{-2}\beta}{2\lambda+3-2n}e_{3},\
\phi(e_{i})=\alpha^{-1}e_{i},\ 2\leq i\leq 2n-2,$

$\phi(e_{2n-1})=\alpha^{-1}e_{2n-1}+\frac{\alpha^{-3}\beta}{2\lambda+3-2n}e_{2n},\
\phi(e_{2n})=\alpha^{-2}e_{2n},$

we obtain the algebra ${\bf
TP}_{2}(\mathfrak{r}_{2n+1}(\lambda\neq\frac{2n-3}{2}))$.

∎

######  Theorem 23.

Let $(\mathfrak{r}_{2n+1}(2-n,\varepsilon),\cdot,[-,-])$ be a transposed
Poisson algebra structure defined on the Lie algebra
$\mathfrak{r}_{2n+1}(2-n,\varepsilon)$. Then the multiplication of
$(\mathfrak{r}_{2n+1}(2-n,\varepsilon),\cdot)$ has the following form:

$\begin{array}[]{lll}{\bf
TP}_{1}(\mathfrak{r}_{2n+1}(2-n,\varepsilon)):&x\cdot x=e_{2n},\\\\[2.84526pt]
{\bf TP}_{2}(\mathfrak{r}_{2n+1}(2-n,\varepsilon)):&e_{2}\cdot
x=e_{2n},&x\cdot x=(2n-3)e_{2n-1},\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{r}_{2n+1}(2-n,\varepsilon),\cdot,[-,-])$ be a transposed
Poisson algebra structure defined on the Lie algebra
$\mathfrak{r}_{2n+1}(2-n,\varepsilon)$. Then by Lemma 5 for any element of
$x\in\mathfrak{r}_{2n+1}(2-n,\varepsilon)$, there is a related
$\frac{1}{2}$-derivation $\varphi_{x}$ of
$(\mathfrak{r}_{2n+1}(2-n,\varepsilon),[-,-])$ such that
$\varphi_{x}(y)=x\cdot y=y\cdot x=\varphi_{y}(x)$, with
$y\in\mathfrak{r}_{2n+1}(2-n,\varepsilon)$. Therefore, by Theorem 15 for any
$i$, we have

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=a_{i}e_{1},\
\varphi_{e_{i}}(e_{2})=a_{i}e_{2}+b_{i}e_{2n},\
\varphi_{e_{i}}(e_{j})=a_{i}e_{j},\ 3\leq j\leq 2n,\\\\[2.84526pt]
\varphi_{e_{i}}(x)=(2n-3)b_{i}e_{2n-1}+c_{i}e_{2n}+a_{i}x,\\\\[2.84526pt]
\varphi_{x}(e_{1})=a_{x}e_{1},\ \varphi_{x}(e_{2})=a_{x}e_{2}+b_{x}e_{2n},\
\varphi_{x}(e_{j})=a_{x}e_{j},\ 3\leq j\leq 2n,\\\\[2.84526pt]
\varphi_{x}(x)=(2n-3)b_{x}e_{2n-1}+c_{x}e_{2n}+a_{x}x.\\\\[2.84526pt]
\end{array}$

Consediring the equalities $\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i})$ and $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i})$ for all $i,j\in\\{1,2,\ldots,2n\\}$, we have

$a_{i}=0,\ 1\leq i\leq 2n,\ b_{1}=0,\ b_{t}=0,\ a_{x}=0,\ b_{2}=0,\
b_{x}=c_{2},\ c_{1}=0,\ c_{t}=0,\ 3\leq t\leq 2n.$

Thus, we obtain

$e_{2}\cdot x=\alpha e_{2n},\quad x\cdot x=(2n-3)\alpha e_{2n-1}+\beta
e_{2n}.$

Since we are only interested in non-trivial transposed Poisson algebra
structures, therefore we have the following possible cases:

  1. (1)

$\alpha=0,$ then we have $\beta\neq 0$ and via isomorphism

$\phi(x)=x,\ \phi(e_{1})=\beta^{-1}e_{1},\ \phi(e_{i})=\beta^{n-i}e_{i},\
2\leq i\leq 2n-1,\ \phi(e_{2n})=\beta^{-1}e_{2n},$

we obtain the algebra ${\bf TP}_{1}(\mathfrak{r}_{2n+1}(2-n,\varepsilon))$.

  2. (2)

$\alpha\neq 0,$ then by choosing the isomorphism

$\phi(x)=x+(2-n)\beta\sqrt[n-1]{\alpha^{-1}}e_{2},\
\phi(e_{1})=\sqrt[n-1]{\alpha^{-1}}e_{1}+\beta\sqrt[n-1]{\alpha^{-2}}e_{2},$
$\phi(e_{i})=\sqrt[n-1]{\alpha^{n-i}}e_{i},\ 2\leq i\leq 2n-2,\
\phi(e_{2n-1})=\alpha^{-1}e_{2n-1}-\beta\sqrt[n-1]{\alpha^{-n}}e_{2n},\
\phi(e_{2n})=\sqrt[n-1]{\alpha^{-1}}e_{2n},$

we find the algebra ${\bf TP}_{2}(\mathfrak{r}_{2n+1}(2-n,\varepsilon))$.

∎

######  Theorem 24.

Let $(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}),\cdot,[-,-])$ be
a transposed Poisson algebra structure defined on the Lie algebra
$\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})$. Then the
multiplication of
$(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}),\cdot)$ has the
following form:

$\begin{array}[]{llll}{\bf
TP}_{1}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})):&x\cdot
x=e_{2n},\\\\[2.84526pt] {\bf
TP}_{2}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})):&e_{2}\cdot
x=e_{2n},\\\\[2.84526pt] {\bf
TP}_{3}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})):&e_{2}\cdot
x=e_{2n},&x\cdot x=e_{2n},\\\\[2.84526pt] {\bf
TP}_{4}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})):&e_{2}\cdot
e_{2}=e_{2n},\\\\[2.84526pt] {\bf
TP}_{5}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})):&e_{2}\cdot
e_{2}=e_{2n},&x\cdot x=e_{2n},\\\\[2.84526pt] {\bf
TP}_{6}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})):&e_{2}\cdot
e_{2}=e_{2n},&e_{2}\cdot x=e_{2n},&x\cdot x=\alpha e_{2n},\\\\[2.84526pt]
\end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}),\cdot,[-,-])$ be
a transposed Poisson algebra structure defined on the Lie algebra
$\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})$. Then for any element
of $x\in\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})$ the linear
operator $\varphi_{x}(y)=x\cdot y$ is a $\frac{1}{2}$-derivation. Therefore,
according to Theorem 16 we derive

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{1})=a_{i}e_{1},\
\varphi_{e_{i}}(e_{2})=a_{i}e_{2}+b_{i}e_{2n},\
\varphi_{e_{i}}(e_{j})=a_{i}e_{j},\ 3\leq j\leq 2n,\
\varphi_{e_{i}}(x)=c_{i}e_{2n}+a_{i}x,\\\\[2.84526pt]
\varphi_{x}(e_{1})=a_{x}e_{1},\ \varphi_{x}(e_{2})=a_{x}e_{2}+b_{x}e_{2n},\
\varphi_{x}(e_{j})=a_{x}e_{j},\ 3\leq j\leq 2n,\
\varphi_{x}(x)=c_{x}e_{2n}+a_{x}x,\\\\[2.84526pt] \end{array}$

From equalities $\varphi_{e_{i}}(e_{j})=e_{i}\cdot e_{j}=e_{j}\cdot
e_{i}=\varphi_{e_{j}}(e_{i})$ and $\varphi_{e_{i}}(x)=e_{i}\cdot x=x\cdot
e_{i}=\varphi_{x}(e_{i})$ we derive

$a_{i}=0,\ 1\leq i\leq 2n,\ b_{1}=0,\ b_{t}=0,\ a_{x}=0,\ b_{x}=c_{2},\
c_{1}=0,\ c_{t}=0,\ 3\leq t\leq 2n.$

Thus, we obtain

$e_{2}\cdot e_{2}=\alpha e_{2n},\ e_{2}\cdot x=\beta e_{2n},\ x\cdot x=\gamma
e_{2n}.$

To obtain only non-trivial transposed Poisson algebra structures, we have the
following possible cases:

  1. (1)

$\alpha=0,\ \beta=0$ and $\gamma\neq 0,$ then via automorphism

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

we get the algebra ${\bf
TP}_{1}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}))$.

  2. (2)

$\alpha=0,\ \beta\neq 0$ and $\gamma=0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=e_{1},\ \phi(e_{i})=\beta^{-1}e_{i},\ 2\leq i\leq
2n-1,\ \phi(e_{2n})=\beta^{-2}e_{2n},$

we have the algebra ${\bf
TP}_{2}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}))$.

  3. (3)

$\alpha=0,\ \beta\neq 0$ and $\gamma\neq 0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=\sqrt[2n-3]{\gamma\beta^{-2}}e_{1},\
\phi(e_{i})=\sqrt[2n-3]{(\gamma\beta^{-2})^{i+1-2n}}e_{i},\ 2\leq i\leq 2n-1,\
\phi(e_{2n})=\gamma^{-1}e_{2n},$

we obtain the algebra ${\bf
TP}_{3}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}))$.

  4. (4)

$\alpha\neq 0,\ \beta=0$ and $\gamma=0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=\sqrt[2n-3]{\alpha^{-1}}e_{1},\
\phi(e_{i})=\sqrt[2n-3]{\alpha^{2-i}}e_{i},\ 2\leq i\leq 2n-1,\
\phi(e_{2n})=\alpha^{-1}e_{2n},$

we find the algebra ${\bf
TP}_{4}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}))$.

  5. (5)

$\alpha\neq 0,\ \beta=0$ and $\gamma\neq 0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=\sqrt[2n-3]{\alpha^{-1}}e_{1},\
\phi(e_{i})=\sqrt[4n-6]{\alpha^{2n-2i+1}\gamma^{3-2n}}e_{i},\ 2\leq i\leq
2n-1,\ \phi(e_{2n})=\gamma^{-1}e_{2n},$

we derive the algebra ${\bf
TP}_{5}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}))$.

  6. (6)

$\alpha\neq 0$ and $\beta\neq 0,$ then via automorphism

$\phi(x)=x,\ \phi(e_{1})=\sqrt[2n-3]{\alpha^{-1}}e_{1},\
\phi(e_{i})=\sqrt[2n-3]{\alpha^{2n-i-1}}\beta^{-1}e_{i},\ 2\leq i\leq 2n-1,\
\phi(e_{2n})=\alpha\beta^{-2}e_{2n},$

we establish the algebra ${\bf
TP}_{6}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1}))$.

∎

Now we give a description of transposed Poisson algebra structures on solvable
Lie algebras with naturally graded filiform nilpotent radical
$\mathfrak{Q}_{2n}$ with codimension two.

######  Theorem 25.

Let $(\mathfrak{r}_{2n+2},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{r}_{2n+2}$. Then the
multiplication of $(\mathfrak{r}_{2n+2},\cdot)$ has the following form:

$\begin{array}[]{lllll}{\bf TP}(\mathfrak{r}_{2n+2}):&x_{1}\cdot
x_{1}=(2n+1)^{2}e_{2n},&x_{1}\cdot x_{2}=2(2n+1)e_{2n},&x_{2}\cdot
x_{2}=4e_{2n},\\\\[2.84526pt] \end{array}$

where it is taken into account that the transposed Poisson algebra has its
products with respect to the bracket $[-,-]$, and the remaining products are
equal to zero.

###### Proof.

Let $(\mathfrak{r}_{2n+2},\cdot,[-,-])$ be a transposed Poisson algebra
structure defined on the Lie algebra $\mathfrak{r}_{2n+2}$. Then according to
Lemma 5 for each element of $x\in\mathfrak{r}_{2n+2}$ the linear operator
$\varphi_{x}(y)=x\cdot y$ is a $\frac{1}{2}$-derivation. Hence, by using
Theorem 17 for all $1\leq i,j\leq n$ we can put

$\begin{array}[]{lll}\varphi_{e_{i}}(e_{j})=a_{i}e_{j},&\varphi_{e_{i}}(x_{1})=(2n+1)b_{i}e_{2n}+a_{i}x_{1},&\varphi_{e_{i}}(x_{2})=2b_{i}e_{2n}+a_{i}x_{2},\\\\[2.84526pt]
\varphi_{x_{1}}(e_{j})=a_{x_{1}}e_{j},&\varphi_{x_{1}}(x_{1})=(2n+1)b_{x_{1}}e_{2n}+a_{x_{1}}x_{1},&\varphi_{x_{1}}(x_{2})=2b_{x_{1}}e_{2n}+a_{x_{1}}x_{2},\\\\[2.84526pt]
\varphi_{x_{2}}(e_{j})=a_{x_{2}}e_{j},&\varphi_{x_{2}}(x_{1})=(2n+1)b_{x_{2}}e_{2n}+a_{x_{2}}x_{1},&\varphi_{x_{2}}(x_{2})=2b_{x_{2}}e_{2n}+a_{x_{2}}x_{2}.\\\\[2.84526pt]
\end{array}$

By checking the equalities of $\varphi_{e_{i}}(e_{j})=e_{i}\cdot
e_{j}=e_{j}\cdot e_{i}=\varphi_{e_{j}}(e_{i}),\
\varphi_{e_{i}}(x_{k})=e_{i}\cdot x_{k}=x_{k}\cdot
e_{i}=\varphi_{x_{k}}(e_{i})$ and $\varphi_{x_{1}}(x_{2})=x_{1}\cdot
x_{2}=x_{2}\cdot x_{1}=\varphi_{x_{2}}(x_{1})$ for all
$i,j\in\\{1,2,\ldots,2n\\}$ and $k\in\\{1,2\\},$ we have the following
restrictions:

$a_{t}=0,\ a_{x_{1}}=0,\ a_{x_{2}}=0,\ b_{t}=0,\ 1\leq t\leq 2n,\
2b_{x_{1}}=(2n+1)b_{x_{2}}.$

Thus, we obtain

$x_{1}\cdot x_{1}=(2n+1)^{2}\alpha e_{2n},\quad x_{1}\cdot x_{2}=2(2n+1)\alpha
e_{2n},\quad x_{2}\cdot x_{2}=4\alpha e_{2n}.$

We have non-trivial transposed Poisson algebra structures only in the case
when $\alpha\neq 0$. Further, by using the transformation

$\phi(x)=x,\ \phi(e_{1})=e_{1},\ \phi(e_{i})=\sqrt{\alpha^{-1}}e_{i},\ 2\leq
i\leq 2n-1,\ \phi(e_{2n})=\alpha^{-1}e_{2n},$

we get the algebra ${\bf TP}(\mathfrak{r}_{2n+2})$. ∎

According to the (3) formula, we get the following remark.

######  Remark 26.

  * •

The following algebras are Poisson algebras:

${\bf TP}_{1}(\mathfrak{s}^{1}_{n,1}(2-n)),\ {\bf
TP}_{1}(\mathfrak{r}_{2n+1}(\frac{3-2n}{2})).$

  * •

The following algebras are non-Poisson algebras:

${\bf TP}_{1}(\mathfrak{s}^{1}_{n,1}(\beta\neq 2-n)),\ {\bf
TP}_{2}(\mathfrak{s}^{1}_{n,1}(\beta)),\ {\bf
TP}_{3}(\mathfrak{s}^{1}_{n,1}(2-n)),\ {\bf TP}_{1}(\mathfrak{s}^{2}_{n,1}),$

${\bf TP}_{2}(\mathfrak{s}^{2}_{n,1}),\ {\bf TP}_{1}(\mathfrak{s}^{3}_{n,1}),\
{\bf TP}_{2}(\mathfrak{s}^{3}_{n,1}),\ {\bf TP}_{1}(\mathfrak{s}^{4}_{n,1}),\
{\bf TP}_{2}(\mathfrak{s}^{4}_{n,1}),$

${\bf TP}_{1}(\mathfrak{r}_{2n+1}(\lambda\neq\frac{3-2n}{2})),\ {\bf
TP}_{2}(\mathfrak{r}_{2n+1}(\lambda)),\ {\bf
TP}_{3}(\mathfrak{r}_{2n+1}(\frac{2n-3}{2})),\ {\bf
TP}_{1}(\mathfrak{r}_{2n+1}(2-n,\varepsilon)),$

${\bf TP}_{2}(\mathfrak{r}_{2n+1}(2-n,\varepsilon)),\ {\bf
TP}_{i}(\mathfrak{r}_{2n+1}(\lambda_{5},\ldots,\lambda_{2n-1})),\
i=\overline{1,6},\ {\bf TP}(\mathfrak{r}_{2n+2}).$

## References

  * [1] Ancochea Bermúdez J. M., Campoamor-Stursberg R., García Vergnolle L. Solvable Lie algebras with naturally graded nilradicals and their invariants, J. Phys. A, 39(6) (2006), 1339-1355.
  * [2] Bai C., Bai R., Guo L., Wu Y., Transposed Poisson algebras, Novikov-Poisson algebras, and 3-Lie algebras, Journal of Algebra, 632 (2023), 535-566.
  * [3] Beites P., Fernández Ouaridi A., Kaygorodov I., The algebraic and geometric classification of transposed Poisson algebras, Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas, 117 (2023), 2, 55.
  * [4] Beites P., Ferreira B. L. M., Kaygorodov I., Transposed Poisson structures, Results in Mathematics, to appear, arXiv:2207.00281.
  * [5] Ferreira B. L. M., Kaygorodov I., Lopatkin V., $\frac{1}{2}$-derivations of Lie algebras and transposed Poisson algebras, Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A. Matemáticas 115 (2021), 3, 142.
  * [6] Fernández Ouaridi A., On the simple transposed Poisson algebras and Jordan superalgebras, Journal of Algebra, 641 (2024), 173–198.
  * [7] Filippov V., $\delta$-Derivations of Lie algebras, Siberian Mathematical Journal, 39 (1998), 6, 1218-1230.
  * [8] Huang J., Chen X., Chen Z., Ding M., On a conjecture of transposed Poisson $n$-Lie algebras, arXiv:2312.04010.
  * [9] Jaworska-Pastuszak A., Pogorzały Z., Poisson structures for canonical algebras, Journal of Geometry and Physics, 148 (2020), 103564.
  * [10] Kaygorodov I., $\delta$-derivations of simple finite-dimensional Jordan superalgebras, Algebra and Logic, 46 (2007), 5, 318–329.
  * [11] Kaygorodov I., $\delta$-superderivations of simple finite-dimensional Jordan and Lie superalgebra, Algebra and Logic, 49 (2010), 2, 130–144.
  * [12] Kaygorodov I., Non-associative algebraic structures: classification and structure, Communications in Mathematics, 32 (3) (2024), 1–62
  * [13] Kaygorodov I., Khrypchenko M., Poisson structures on finitary incidence algebras, Journal of Algebra, 578 (2021), 402-420.
  * [14] Kaygorodov I., Khrypchenko M., Transposed Poisson structures on Block Lie algebras and superalgebras, Linear Algebra and Its Applications, 656 (2023), 167-197.
  * [15] Kaygorodov I., Khrypchenko M., Transposed Poisson structures on Witt type algebras, Linear Algebra and its Applications, 665 (2023), 196-210.
  * [16] Kaygorodov I., Khrypchenko M., Transposed Poisson structures on generalized Witt algebras and Block Lie algebras, Results in Mathematics, 78 (2023), 5, 186.
  * [17] Kaygorodov I., Khrypchenko M., Transposed Poisson structures on the Lie algebra of upper triangular matrices, Portugaliae Mathematica, to appear, arXiv:2305.00727.
  * [18] Kaygorodov I., Khrypchenko M., Transposed Poisson structures on Lie incidence algebras, arXiv:2309.00332.
  * [19] Kaygorodov I., Khudoyberdiyev A., Transposed Poisson structures on solvable and perfect Lie algebras, Journal of Physics A: Mathematical and Theoretical, 57 (2024), 035205.
  * [20] Kaygorodov I., Lopatkin V., Zhang Z., Transposed Poisson structures on Galilean and solvable Lie algebras, Journal of Geometry and Physics, 187 (2023), 104781.
  * [21] Khudoyberdiyev A., Sheraliyeva S., Extensions of solvable Lie algebras with naturally graded filiform nilradical, Journal of Algebra and Its Applications, 2024, 2450161.
  * [22] Liu G., Bai C., A bialgebra theory for transposed Poisson algebras via anti-pre-Lie bialgebras and anti-pre-Lie-Poisson bialgebras, Communications in Contemporary Mathematics, 2023, DOI: 10.1142/S0219199723500505.
  * [23] Liu G., Bai C., New splittings of operations of Poisson algebras and transposed Poisson algebras and related algebraic structures, In: Hounkonnou, M.N., Mitrović, M., Abbas, M., Khan, M. (eds) Algebra without Borders – Classical and Constructive Nonassociative Algebraic Structures. STEAM-H: Science, Technology, Engineering, Agriculture, Mathematics & Health. Springer, Cham., 2023, 49–96.
  * [24] Sartayev B., Some generalizations of the variety of transposed Poisson algebras, Communications in Mathematics, 32 (2024), 2, 55–62.
  * [25] Šnobl L. Winternitz P., A class of solvable Lie algebras and their Casimir invariants, Journal of Physics A, 38 (2005), 2687-2700.
  * [26] Vergne M. Cohomologie des algèbres de Lie nilpotentes. Application à l’étude de la variété des algèbres de Lie nilpotentes. Bull. Soc. Math. France, 98 (1970), 81-116.
  * [27] Yang Ya., Tang X., Khudoyberdiyev A., Transposed Poisson structures on Schrodinger algebra in $(n+1)$-dimensional space-time, arXiv:2303.08180.
  * [28] Yao Y., Ye Y., Zhang P., Quiver Poisson algebras, Journal of Algebra, 312 (2007), 2, 570-589.
  * [29] Yaxi J., Chuangchuang K., Jiafeng L., Classification of Transposed Poisson $3$-Lie algebras of dimension $3$, arXiv:2401.02593.
  * [30] Yuan L., Hua Q., $\frac{1}{2}$-(bi)derivations and transposed Poisson algebra structures on Lie algebras, Linear and Multilinear Algebra, 70 (2022), 22, 7672-7701.


11institutetext: JILA and Dept. Astrophysical & Planetary Sciences, Box 440,
U. Colorado Boulder, CO 80309, USA 11email<EMAIL_ADDRESS>http://jila.colorado.edu/~ajsh/

# The rules of 4-dimensional perspective: How to implement Lorentz
transformations in relativistic visualization

Andrew J. S. Hamilton  0000-0002-3816-5973

###### Abstract

This paper presents a pedagogical introduction to the issue of how to
implement Lorentz transformations in relativistic visualization. The most
efficient approach is to use the even geometric algebra in 3+1 spacetime
dimensions, or equivalently complex quaternions, which are fast, compact, and
robust, and straightforward to compose, interpolate, and spline. The approach
has been incorporated into the Black Hole Flight Simulator, an interactive
general relativistic ray-tracing program developed by the author.

###### Keywords:

Special relativistic visualization Lorentz transformations complex quaternions

##  1 Introduction

Einstein’s theory of Special Relativity revolutionized space and time, uniting
them from the separate 3-dimensional and 1-dimensional entities familiar to
everyday experience, into an inextricably entwined 4-dimensional spacetime.
The consequences of special relativity are not evident to everyday experience
because we move through our surroundings at only a tiny fraction of the speed
of light. But if you moved around at near the speed of light, then you would
very much notice [1, 2, 3].

The positions of objects in 3 dimensions are described by 3-dimensional
vectors, or 3-vectors for short. Of primary importance in 3d visualization
software is the ability to rotate 3-vectors rapidly and robustly. A rotation
in 3 dimensions decomposes into rotations about each of 3 directions. It is
well known that the best way to program rotations is using quaternions,
invented by William Rowan Hamilton in 1843 [4]. The algebra of quaternions is
precisely that of the even Clifford algebra, or geometric algebra, in 3
spatial dimensions.

In relativistic visualization, the perceived position of an object is
described by a 4-dimensional vector, a 4-vector. The analog of rotations of a
4-vector are Lorentz transformations [5]. To the usual 3 spatial rotations of
space, Lorentz transformations adjoin 3 rotations between space and time, also
known as boosts, a total of 6 spacetime rotations. The 3 boosts represent a
change of the observer’s velocity along each of the 3 spatial directions. To
someone moving through a scene at much less than the speed of light, these
boosts have a negligible effect on the appearance of a scene. But if you move
at near the speed of light, then the boosts are very evident.

The purpose of this paper is to present a pedagogical introduction to
implementing Lorentz transformations in computer visualizations. The paper
starts, §2, by presenting a conceptual picture of how a scene appears
distorted to an observer who moves through it at relativistic speed [6]. Then
§3 describes how to implement Lorentz transformations using complex
quaternions, also called biquaternions by William Rowan Hamilton. The
equivalence of the Lorentz group to complex quaternions was discovered by
Silberstein in 1912 [7], and remarked by Dirac [8]; see [9] for further
references. The algebra of complex quaternions is precisely that of the even
geometric algebra in 3+1 spacetime dimensions.

In relativity, time behaves mathematically as if it were an imaginary spatial
dimension. Lorentz transformations rotate spatial and time dimensions among
each other. In relativity it is extremely useful to use “natural” units in
which the speed of light is one, $c=1$, since then space and time are measured
in the same unit.

##  2 The rules of 4-dimensional perspective

The conceptual picture presented in this section is from the author’s website
[6], where animations may be found.

In 3-dimensional perspective, two ideas are fundamental:

  * •

A straight line in 3 dimensions remains a straight line in perspective;

  * •

Parallel lines meet at a vanishing point.

Figure 1:  The rules of 4-dimensional perspective. In special relativity, the
scene seen by an observer moving through the scene (right) is relativistically
beamed compared to the scene seen by an observer at rest relative to the scene
(left). On the left, the observer at the center of the circle is at rest
relative to the surrounding scene. On the right, the observer is moving to the
right through the same scene at $v=0.8$ times the speed of light. The scene is
distorted into a celestial ellipsoid with the observer displaced to its focus.
The arrowed lines represent energy-momenta of photons. The length of an
arrowed line is proportional to the perceived energy of the photon. The scene
ahead of the moving observer appears concentrated, blueshifted, and farther
away, while the scene behind appears expanded, redshifted, and closer.

The rules that govern the appearance of a scene when you move through it at
near the speed of light can be called the rules of 4-dimensional perspective.
These rules can be grasped, much like the rules of 3-dimensional perspective,
without needing to understand intricate mathematics. The rules of
4-dimensional perspective can be summarized as follows:

  * •

Paint the scene at rest on the surface of a celestial sphere;

  * •

Stretch the celestial sphere by Lorentz factor $\gamma$ along the direction of
motion into a celestial ellipsoid, and displace the observer to a focus of the
ellipsoid;

  * •

Adjust the brightness, color, and clock speed at any point on the ellipsoid in
proportion to the length of the radius between the point and the observer.

Figure 1 illustrates the rules of 4-dimensional perspective. On the left, you
are at rest relative to the scene. Imagine painting the scene on a celestial
sphere around you. The arrows represent the directions of lightrays (photons)
from the scene on the celestial sphere to you at the center. Technically,
these arrows represent the energy-momentum 4-vectors of the lightrays that you
see, §3.4.

On the right in Figure 1, you are moving to the right through the scene, at
velocity $v=0.8$ times the speed of light. The celestial sphere is stretched
along the direction of your motion by the Lorentz gamma-factor
$\gamma=1/\sqrt{1-0.8^{2}}=5/3$ into a celestial ellipsoid. You, the observer,
are not at the center of the ellipsoid, but rather at one of its foci (the
left one, if you are moving to the right). The focus of the celestial
ellipsoid, where you the observer are, is displaced from center by $\gamma
v=4/3$. The scene appears relativistically aberrated, which is to say
concentrated ahead of you, and expanded behind you.

The lengths of the arrows in Figure 1 are proportional to the energies, or
frequencies, of the photons that you see. When you are moving through the
scene at near light speed, the arrows ahead of you, in your direction of
motion, are longer than at rest, so you see the photons blue-shifted,
increased in energy, increased in frequency. Conversely, the arrows behind you
are shorter than at rest, so you see the photons red-shifted, decreased in
energy, decreased in frequency. Since photons are good clocks, the change in
photon frequency also tells you how fast or slow clocks attached to the scene
appear to you to run.

Figure 2:  Passing by a sphere at $0.97$ of the speed of light ($\gamma v=4$),
in the general direction of the center of our Galaxy, the Milky Way (left)
undistorted, (right) relativistically beamed. The sphere is painted with lines
of latitude and longitude. The background is an image of the Milky Way from
Gaia Data Release 3 [10]. In the undistorted (left) view, most of the sphere
is behind the observer; relativistic aberration (right) brings the sphere
behind into view. The sphere and background are colored with appropriately
blue- and red-shifted blackbody colors; the unredshifted color temperature is
$5{,}780\,\textrm{K}$, the color of the Sun. The field of view is
$105^{\circ}$ across the diagonal. The visualization was made using the Black
Hole Flight Simulator software [11].

The following table summarizes the four effects of relativistic beaming on the
appearance of a scene ahead of you and behind you as you move through it at
near the speed of light:

Effect | Ahead | Behind
---|---|---
Aberration | Concentrated | Expanded
Color | Blueshifted | Redshifted
Brightness | Brighter | Dimmer
Time | Speeded up | Slowed down

Figure 2 shows a visualization of passing by a sphere at near the speed of
light, in this case $v=0.97$ (in natural units $c=1$). The visualization was
made using the Black Hole Flight Simulator software [11]. The visualization
illustrates how relativistic aberration distorts the appearance of the sphere,
concentrating it and blueshifting it towards the direction the oberver is
moving, and expanding and redshifting it behind.

A feature of relativistic aberration is that circles transform to circles, and
angles are preserved. These facts were first pointed out by [12] and [13],
prior to which it had been widely thought that circles would appear Lorentz-
contracted and therefore squashed. Lines of latitude and longitude on the
surface of a sphere are circles, and they intersect at right angles. Figure 2
illustrates how relativistic aberration preserves these properties. The
outline of a sphere is itself a circle, and the outline of a relativistically
aberrated sphere remains circular.

##  3 How to implement Lorentz transformations on a computer

The advantages of quaternions for implementing spatial rotations are well-
known to 3d game programmers. Compared to standard rotation matrices,
quaternions offer increased speed (fewer multiplications) and require less
storage, and their algebraic properties simplify composition, interpolation,
and splining.

Complex quaternions retain similar advantages for implementing Lorentz
transformations. They are fast, compact, and straightforward to interpolate or
spline. Moreover, since complex quaternions contain real quaternions, Lorentz
transformations can be implemented as an extension of spatial rotations in 3d
programs that use quaternions to implement spatial rotations.

At a deeper level, the elegant properties of quaternions and complex
quaternions can be traced to the fact that they are the even geometric
subalgebras in respectively 3 spatial dimensions and 3+1 spacetime dimensions.

###  3.1 Real quaternions

It is useful to start by reviewing basic definitions and properties of
ordinary (real) quaternions. Many of these properties carry over to complex
quaternions.

A quaternion $q$ is a kind of souped-up complex number,

$q=w+ix+jy+kz\ ,$ (1)

where $w,x,y,z$ are real numbers, and the three imaginary numbers $i,j,k$, are
defined to satisfy

$i^{2}=j^{2}=k^{2}=-ijk=-1\ .$ (2)

The convention $ijk=1$ in the definition (2) allows $i$, $j$, $k$ to be
identified as bivectors (4) of the 3d geometric algebra, but is opposite to
the traditional definition $ijk=-1$ famously carved by William Rowan Hamilton
in the stone of Brougham Bridge while walking with his wife along the Royal
Canal to Dublin on 16 October 1843 [4]. A consequence of equations (2) is that
each pair of imaginary numbers anticommutes:

$ij=-ji=-k\ ,\quad jk=-kj=-i\ ,\quad ki=-ik=-j\ .$ (3)

Quaternions are distributive and associative, but not commutative. Similarly
to complex numbers, two quaternions are multiplied by multiplying and
combining their components.

The algebra of quaternions can be recognized as that of the even geometric
algebra in 3 spatial dimensions. The 3d geometric algebra is generated by the
$1+3+3+1=8=2^{3}$ orthornormal basis multivectors

$\begin{array}[]{c}1\,,\\\ \mbox{1 scalar}\end{array}\ \
\begin{array}[]{c}\bm{\gamma}_{x}\,,\ \bm{\gamma}_{y}\,,\
\bm{\gamma}_{z}\,,\\\ \mbox{3 vectors}\end{array}\ \
\begin{array}[]{c}I_{3}\bm{\gamma}_{x}\,,\ I_{3}\bm{\gamma}_{y}\,,\
I_{3}\bm{\gamma}_{z}\,,\\\ \mbox{3 bivectors (pseudovectors)}\end{array}\ \
\begin{array}[]{c}I_{3}\equiv\bm{\gamma}_{x}\bm{\gamma}_{y}\bm{\gamma}_{z}\,.\\\
\mbox{1 pseudoscalar}\end{array}$ (4)

The quaternionic imaginaries $i$, $j$, $k$ are just the basis bivectors
$I_{3}\bm{\gamma}_{a}$ of the 3d geometric algebra.

A quaternion $q$ can be stored as a 4-component object

$q=\bigl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \ }c}w&x&y&z\\\
\end{array}\bigr{\\}}\ .$ (5)

The convention to put the scalar component $w$ in the first position is
standard in mathematics and physics.

A basic operation on quaternions is quaternionic conjugation, also called
reversal in the geometric algebra, which is analogous to taking the complex
conjugate of a complex number. The reverse $\overline{q}$ of a quaternion $q$
is the quaternion obtained by flipping the sign of all 3 imaginary components,

$\overline{q}\equiv\bigl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}w&-x&-y&-z\\\ \end{array}\bigr{\\}}\ .$ (6)

The modulus $|q|$ of a quaternion $q$ is, similarly to the modulus of a
complex number, the square root of the product of itself with its reverse,

$|q|=\sqrt{q\overline{q}}=\sqrt{\overline{q}q}=\sqrt{w^{2}+x^{2}+y^{2}+z^{2}}\
.$ (7)

The modulus $|q|$ of a quaternion is unchanged by any rotation.

One of the advantages of quaternions is that quaternions are easy to invert:
the inverse of a quaternion $q$ is, again similarly to the inverse of a
complex number, its reverse $\overline{q}$ divided by its squared modulus,

$q^{-1}={\overline{q}\over\overline{q}q}\ .$ (8)

The positions of objects in a 3d scene are described by vectors in the 3d
geometric algebra. According to the array (4) of basis multivectors, a basis
vector $\bm{\gamma}_{a}$ in the 3d geometric algebra equals minus the
pseudoscalar times a bivector, $\bm{\gamma}_{a}=-I_{3}(I_{3}\bm{\gamma}_{a})$,
which implies that a vector in 3d rotates in the same way as a bivector. The
result is that a vector in a 3d program can be handled as though it were a
bivector, that is, a quaternion with zero scalar component. The 3-vector
position $\bm{r}$ of an object in a 3d scene can be represented by a
quaternion

$\bm{r}=ix+jy+kz=\bigl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}0&x&y&z\\\ \end{array}\bigr{\\}}\ .$ (9)

A 3d rotation is represented by a quaternion of unit modulus, a rotor in the
3d geometric algebra. For example, the rotor $R$ corresponding to a right-
handed rotation by angle $\theta$ about the $x$-direction is

$\displaystyle R$
$\displaystyle=e^{-i\theta/2}=\cos(\theta/2)-i\sin(\theta/2)$
$\displaystyle=\bigl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}\cos(\theta/2)&-\sin(\theta/2)&0&0\\\ \end{array}\bigr{\\}}\ .$ (11)

More generally, if $\bm{n}=ix+jy+kz$ is a unit vector in the direction
$\\{x,y,z\\}$, then a right-handed rotation by angle $\theta$ about the
$\bm{n}$-direction is represented by the rotor

$R=e^{-\bm{n}\theta/2}=\cos(\theta/2)-\bm{n}\sin(\theta/2)\ .$ (12)

A rotation $S$ following a rotation $R$ is just their quaternionic product
$SR$.

Rotating a 3-vector $\bm{r}$ by a rotor $R$ is accomplished by taking the
quaternionic product

$\bm{r}\rightarrow R\bm{r}\overline{R}\ ,$ (13)

a one-line calculation in a language that supports quaternionic operations.

It should be remarked that the sign conventions above are the standard ones in
physics. Software implementations of quaternions vary in their conventions.

###  3.2 Complex quaternions

Introduce yet another imaginary $I$ which commutes with the 3 quaternionic
imaginaries $i$, $j$, $k$,

$I^{2}=-1\ ,\quad Ii=iI\ ,\quad Ij=jI\ ,\quad Ik=kI\ .$ (14)

A complex quaternion $q$ is a quaternion with complex coefficients,
$w=w_{R}+Iw_{I}$, etc.,

$q=w+ix+jy+kz\ .$ (15)

A complex quaternion $q$ can be stored as the 8-component object

$q=q_{R}+Iq_{I}=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}w_{R}&x_{R}&y_{R}&z_{R}\\\ w_{I}&x_{I}&y_{I}&z_{I}\end{array}\biggr{\\}}\
.$ (16)

The top line contains the real part $q_{R}$ of the quaternion $q$, the bottom
line the imaginary part $q_{I}$. A complex quaternion can be implemented in
software as a pair of quaternions.

The algebra of complex quaternions is that of the even geometric algebra in
3+1 spacetime dimensions. The geometric algebra in 3+1 dimensions is generated
by the $1+4+6+4+1=16=2^{4}$ orthornormal basis multivectors

$\begin{array}[]{c}1\,,\\\ \mbox{1 scalar}\end{array}\ \
\begin{array}[]{c}\bm{\gamma}_{t}\,,\ \bm{\gamma}_{a}\,,\\\ \mbox{4
vectors}\end{array}\ \ \begin{array}[]{c}\bm{\gamma}_{a}\bm{\gamma}_{b}\,,\
I\bm{\gamma}_{a}\bm{\gamma}_{b}\,,\\\ \mbox{6 bivectors}\end{array}\ \
\begin{array}[]{c}I\bm{\gamma}_{t}\,,\ I\bm{\gamma}_{a}\,,\\\ \mbox{4
pseudovectors}\end{array}\ \
\begin{array}[]{c}I\equiv\bm{\gamma}_{t}\bm{\gamma}_{x}\bm{\gamma}_{y}\bm{\gamma}_{z}\,,\\\
\mbox{1 pseudoscalar}\end{array}$ (17)

where indices $a,b$ run over the 3 spatial dimensions $x$, $y$, $z$. The
imaginary $I$ in the complex quaternion (16) is just the pseudoscalar of the
geometric algebra. The 6 bivectors of the geometric algebra comprise 3 spatial
bivectors $\bm{\gamma}_{a}\bm{\gamma}_{b}$, along with 3 bivectors
$I\bm{\gamma}_{a}\bm{\gamma}_{b}$ which look like spatial bivectors multiplied
by the pseudoscalar. The geometrica algebra in 3+1 dimensions is precisely
that of Dirac $\gamma$-matrices in relativistic physics, which accounts for
the notation $\bm{\gamma}_{m}$.

The reverse $\overline{q}$ of the complex quaternion (16) is, similarly to the
reverse (6) of a real quaternion, the quaternion with all quaternionic
imaginary components flipped in sign,

$\overline{q}=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}w_{R}&-x_{R}&-y_{R}&-z_{R}\\\
w_{I}&-x_{I}&-y_{I}&-z_{I}\end{array}\biggr{\\}}\ .$ (18)

The complex conjugate $q^{\star}$ of the quaternion is the quaternion with all
imaginary $I$ components flipped,

$q^{\star}=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}w_{R}&x_{R}&y_{R}&z_{R}\\\
-w_{I}&-x_{I}&-y_{I}&-z_{I}\end{array}\biggr{\\}}\ .$ (19)

The operations of reversal and complex conjugation commute.

The modulus $|q|$ of a complex quaternion is defined, similarly to the modulus
(7) of a real quaternion, to be the square root of the product $q\overline{q}$
of the complex quaternion with its reverse, but whereas the modulus of a real
quaternion is a real number, the modulus of a complex quaternion is a complex
(with respect to $I$) number,

$|q|\equiv\sqrt{q\overline{q}}=\sqrt{\overline{q}q}=\sqrt{w^{2}+x^{2}+y^{2}+z^{2}}\
.$ (20)

The modulus $|q|$ is complex (with respect to $I$) because the components $w$,
$x$, $y$, $z$ are complex. To obtain a real number, the absolute value $\lVert
q\rVert$ of the complex quaternion $q$, it is necessary to take the absolute
value of the complex modulus,

$\lVert
q\rVert\equiv\sqrt{|q||q|^{\ast}}=\bigl{\lvert}\sqrt{w^{2}+x^{2}+y^{2}+z^{2}}\bigr{\rvert}\
.$ (21)

###  3.3 Lorentz transformations

A Lorentz transformation, a special relativistic rotation of spacetime, can be
represented as a complex quaternion of unit modulus, a Lorentz rotor $R$,
satisfying the unimodular condition

$R\overline{R}=1\ .$ (22)

The inverse of a Lorentz rotor $R$ is its reverse $\overline{R}$. The
unimodular condition (22) is a complex (with respect to $I$) condition, which
removes 2 degrees of freedom from the 8 degrees of freedom of a complex
quaternion, leaving the Lorentz group with 6 degrees of freedom, which is as
it should be.

Spatial rotations correspond to real unimodular quaternions, and account for 3
of the 6 degrees of freedom of Lorentz transformations. For example, a spatial
rotation by angle $\theta$ right-handedly about the $x$-axis is the real
Lorentz rotor

$R=e^{-i\theta/2}=\cos(\theta/2)-i\sin(\theta/2)\ ,$ (23)

or, stored as a complex quaternion,

$R=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}\cos(\theta/2)&-\sin(\theta/2)&0&0\\\ 0&0&0&0\end{array}\biggr{\\}}\ .$
(24)

The expression (24) coincides with the earlier expression (3.1) for a spatial
rotation as a real quaternion, which is as it should be.

Lorentz boosts account for the remaining 3 of the 6 degrees of freedom of
Lorentz transformations. A Lorentz boost is mathematically equivalent to a
rotation by an imaginary angle $I\theta$. Physicists call the boost angle
$\theta$ of a Lorentz boost its rapidity. The velocity $v$ of the Lorentz
boost is the hyperbolic tangent of the boost angle $\theta$,

$v=\tanh\theta\ .$ (25)

The Lorentz gamma-factor $\gamma$ and associated momentum factor $\gamma v$
that appear in Figure 1 are related to the boost angle $\theta$ by

$\gamma=\cosh\theta\ ,\quad\gamma v=\sinh\theta\ .$ (26)

A Lorentz boost by boost angle $\theta$ along, for example, the $x$-axis is
the complex Lorentz rotor

$R=e^{-Ii\theta/2}=\cosh(\theta/2)-Ii\sinh(\theta/2)\ ,$ (27)

or, stored as a complex quaternion,

$R=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}\cosh(\theta/2)&0&0&0\\\ 0&-\sinh(\theta/2)&0&0\end{array}\biggr{\\}}\ .$
(28)

More generally, a Lorentz boost by boost angle $\theta$ along a unit 3-vector
quaternion direction $\bm{n}$ is

$R=e^{-I\bm{n}\theta/2}=\cosh(\theta/2)-I\bm{n}\sinh(\theta/2)\ .$ (29)

The most general Lorentz transformation is a Lorentz rotor $R$, a unimodular
complex quaternion. Any such Lorentz rotor can be decomposed into a
combination of a spatial rotation and a Lorentz boost, but it is not necessary
to carry out such a decomposition. The arithmetic of complex quaternions takes
care of itself. The rule for composing Lorentz transformations is the same as
the rule for composing spatial rotations: a Lorentz transformation $S$
following a Lorentz transformation $R$ is just the product $SR$ of the
corresponding complex quaternions.

A general Lorentz rotor $R$ can be written as the exponential of a complex
quaternion angle $\bm{\theta}$, which is itself the product of a complex (with
respect to $I$) angle $\theta$ equal to the modulus of $\bm{\theta}$, and a
unimodular complex 3-direction $\bm{n}$,

$R=e^{-\bm{\theta}}=e^{-\theta\bm{n}}\ ,$ (30)

with

$\theta=|\bm{\theta}|=\theta_{R}+I\theta_{I}\
,\quad\bm{n}=\bm{n}_{R}+I\bm{n}_{I}\ ,\quad\overline{\bm{n}}\bm{n}=1\ .$ (31)

The complex quaternion 3-direction $\bm{n}$ is a 6-component object, but the
unimodular condition on $\bm{n}$ removes 2 of the degrees of freedom, leaving
$\bm{n}$ with 4 degrees of freedom. Those 4 degrees of freedom add to the 2
degrees of freedom of the complex angle $\theta$ to give 6 degrees of freedom,
which is the correct number of degrees of freedom of Lorentz transformations.

###  3.4 4-vectors

The positions of objects in a relativistic scene are described by 4-vectors in
the geometric algebra in 3+1 dimensions. In much the same way that 3-vectors
in 3d can be packaged as quaternions, equation (9), 4-vectors in 3+1
dimensions can be packaged as complex quaternions. In any geometric algebra,
an odd element of the geometric algebra can be written uniquely as the product
of some particular vector (an odd element) and an even element of the algebra.
It is convenient to choose the chosen particular vector to be the time basis
vector $\bm{\gamma}_{t}$. In 3+1 dimensions, an odd element of the algebra is
a sum of a vector $\bm{r}_{1}$ and a pseudovector $I\bm{r}_{2}$. Their sum,
the odd multivector $\bm{r}$, can be written as the product of the time vector
$\bm{\gamma}_{t}$ and an element of the even algebra, a complex quaternion
$q$,

$\bm{r}\equiv\bm{r}_{1}+I\bm{r}_{2}=\bm{\gamma}_{t}q\ .$ (32)

The odd multivector $\bm{r}$ Lorentz transforms under a Lorentz rotor $R$ as

$\bm{r}\rightarrow
R\bm{\gamma}_{t}q\overline{R}=\bm{\gamma}_{t}R^{\star}q\overline{R}\ .$ (33)

The conjugated rotor $R^{\star}$ appears in the transformation (33) because
commuting the time vector $\bm{\gamma}_{t}$ through the rotor $R$ converts the
latter to its complex (with respect to $I$) conjugate, which is true because
the time vector commutes with the quaternionic imaginaries $i$, $j$, $k$, but
anticommutes with the pseudoscalar $I$. Equation (33) shows that the complex
quaternion $q$ equivalent to the odd multivector $\bm{r}$ Lorentz transforms
as

$q\rightarrow R^{\star}q\overline{R}\ .$ (34)

When you are looking at a scene, it is light that carries the image of the
scene to your eyes. The perceived position of an object traces back along the
energy-momentum 4-vector of the light that the object emits and that you see.
The arrowed lightrays in Figure 1 represent the energy-momentum 4-vectors of
the lightrays. When you change your velocity through the scene, the energy-
momentum 4-vectors of the lightrays Lorentz-transform accordingly. Light
travels at light speed, so the components of the energy-momentum 4-vector
$\bm{p}$ of a lightray (a photon) take the form

$\bm{p}=E\\{1,\bm{v}\\}$ (35)

in which the magnitude of the velocity $\bm{v}$ is one, the speed of light,
$v\equiv|\bm{v}|=1$. The energy of the lightray (photon) is $E$, and its
3-momentum is $E\bm{v}$. The energy-momentum 4-vector of an object moving at
the speed of light is null; in the language of the geometric algebra, the
energy-momentum 4-vector $\bm{p}$ of a lightray satisfies the null condition

$\bm{p}\overline{\bm{p}}=0\ .$ (36)

The positions $\bm{r}$ of objects in a relativistic scene are thus not just
any old 4-vectors, but rather they are null 4-vectors, satisfying the null
condition

$\bm{r}\overline{\bm{r}}=0\ .$ (37)

The complex quaternion $q$ equivalent to the 4-vector $\bm{r}$, equation (32),
satisfies the null condition

$q\overline{q}=0\ .$ (38)

In relativistic visualization, one starts by defining a rest frame, a frame
with respect to which (most) objects are not moving. Relative to an observer
at rest in the scene, the 3-vector of an object in the scene at distance $r$
and unit 3-direction $\bm{n}$ is $r\bm{n}$. The direction $\bm{n}$ to the
object is opposite in sign to the direction $\bm{v}$ of a lightray from the
object to the observer. The null 4-vector position $\bm{r}$ of the object is
encoded in the null complex quaternion

$q=r(1+I\bm{n})\ .$ (39)

For example, the null 4-vector position of an object lying in the
$x$-direction relative to the observer is

$q=r\,\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \ }c}1&0&0&0\\\
0&1&0&0\end{array}\biggr{\\}}\ .$ (40)

More generally, if the 3-vector position of the object is
$r\bm{n}=\\{x,y,z\\}$, then the components of the corresponding null complex
quaternion position $q$ are

$q=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \ }c}r&0&0&0\\\
0&x&y&z\end{array}\biggr{\\}}\ .$ (41)

When the observer rotates spatially or accelerates, the scene appears to the
observer Lorentz-transformed by a Lorentz rotor $R$, which transforms
perceived null complex quaternion positions $q$ according to equation (34).
This is similar to the transformation (13) of a 3d position by a spatial
rotation, except that the rotor $R^{\star}$ in the Lorentz transformation (34)
is complex-conjugated. A spatial rotation corresponds to a real rotor $R$,
whose complex conjugate is itself, in which case the Lorentz transformation
(34) reduces to the spatial transformation (13). The Lorentz transformation
(34) of the position $q$ instructs to multiply three complex quaternions
$R^{\ast}$, $q$, and $\overline{R}$, a one-line expression in a language that
supports complex quaternion operations.

As an illustration of how the Lorentz transformation (34) works, consider the
(simple) example of an object unit distance from the observer in the (say)
$x$-direction, so $q=(1+Ii)$, and the Lorentz transformation is by boost angle
$\theta$ in the same $x$-direction, so the Lorentz rotor is
$R=e^{-Ii\theta/2}$, equation (28). Then the Lorentz transformation (34) is

$\displaystyle\biggl{\\{}\begin{array}[]{c@{\ \ }c}1&0\\\
0&1\end{array}\biggr{\\}}$
$\displaystyle\rightarrow\biggl{\\{}\begin{array}[]{c@{\ \
}c}\cosh\frac{\theta}{2}&0\\\
0&\sinh\frac{\theta}{2}\end{array}\biggr{\\}}\biggl{\\{}\begin{array}[]{c@{\ \
}c}1&0\\\ 0&1\end{array}\biggr{\\}}\biggl{\\{}\begin{array}[]{c@{\ \
}c}\cosh\frac{\theta}{2}&0\\\ 0&\sinh\frac{\theta}{2}\end{array}\biggr{\\}}$
(50) $\displaystyle=e^{\theta}\biggl{\\{}\begin{array}[]{c@{\ \ }c}1&0\\\
0&1\end{array}\biggr{\\}}\ ,$ (53)

which says that the Lorentz-boosted position appears farther away by the
factor $e^{\theta}$. The exponential factor $e^{\theta}$ agrees with the
conventional special relativistic Doppler shift formula

$e^{\theta}=\cosh\theta+\sinh\theta=\gamma(1+v)=\sqrt{1+v\over 1-v}\ .$ (54)

The result also agrees with Figure 1, which illustrates that the scene
directly ahead appears farther away and blueshifted by the Doppler factor
$\gamma(1+v)$.

Lorentz transforming a scene typically involves a large number of Lorentz
transformations, one for each point on the scene. A complex quaternion has 8
components, and two position vectors $\bm{r}_{1}$ and $\bm{r}_{2}$ can be
transformed for the price of one by treating one of them, $\bm{r}_{2}$, as a
pseudovector, and packing the two into a single complex quaternion,

$q=r_{1}(1+I\bm{n}_{1})-Ir_{2}(1+I\bm{n}_{2})\ .$ (55)

If the 3-vector components of the two positions are
$r_{i}\bm{n}_{i}=\\{x_{i},y_{i},z_{i}\\}$, $i=1,2$, then the components of the
packed complex quaternion $q$ are

$q=\biggl{\\{}\begin{array}[]{c@{\ \ }c@{\ \ }c@{\ \
}c}r_{1}&x_{2}&y_{2}&z_{2}\\\ -r_{2}&x_{1}&y_{1}&z_{1}\end{array}\biggr{\\}}\
.$ (56)

The same Lorentz transformation formula (34) applies.

##  4 Conclusion

In the future, children will learn the mysteries of special and general
relativity by playing with relativistic flight simulators. Relativity will
cease to be something accessible only to the cognoscenti.

The first task in coding a relativistic flight simulator is to implement
Lorentz transformations. The right way to do that is with complex quaternions.
This paper shows how.

## References

  * [1] Ute Kraus, “First-person visualizations of the special and general theory of relativity,” Eur. J. Phys. 29, 1–13 (2007) doi:10.1088/0143-0807/29/1/001 arXiv:0708.3454
  * [2] Ute Kraus & Corvin Zahn “Relativity visualized” (2001–2021) https://www.spacetimetravel.org/
  * [3] Zachary W. Sherin, Ryan Cheu, Philip Tan & Gerd Kortemeyer “Visualizing relativity: The OpenRelativity project,” Am. J. Phys. 84, 369–374 (2016) doi:10.1119/1.4938057
  * [4] Sean O’Donnell, Portrait of a Prodigy (Boole Press, Dublin, 1983) doi:10.1163/182539177X01015
  * [5] Hendrik Antoon Lorentz, “Electromagnetic phenomena in a system moving with any velocity smaller than that of light,” Proc. of the Royal Netherlands Academy of Arts and Sciences 6, 809–831 (1904)
  * [6] Andrew J. S. Hamilton (2010) “The Rules of 4-dimensional Perspective,” https://jila.colorado.edu/~ajsh/insidebh/4dperspective.html
  * [7] Ludwik Silberstein, “Quaternionic form of relativity,” Philosophical Magazine 23, 790–809 (1912) doi:10.1080/14786440508637276
  * [8] P. A. M. Dirac, “Application of Quaternions to Lorentz Transformations,” Proc. of the Royal Irish Academy. Section A: Math. and Phys. Sci. 50, 261–270 (1944)
  * [9] Thomas Berry & Matt Visser, “Lorentz boosts and Wigner rotations: self-adjoint complexified quaternions,” arXiv:2101.05971
  * [10] https://www.esa.int/ESA_Multimedia/Images/2020/12/The_colour_of_the_sky_from_Gaia_s_Early_Data_Release_32
  * [11] Black Hole Flight Simulator https://jila.colorado.edu/~ajsh/insidebh/bhfs.html
  * [12] Roger Penrose, “The apparent shape of a relativistically moving sphere,” Proc. Cambridge Phil. Soc. 55, 137–139 (1959) doi:10.1017/S0305004100033776
  * [13] James Terrell, “Invisibility of the Lorentz Contraction,” Phys. Rev. 116, 1041–1045 (1959) doi:10.1103/PhysRev.116.1041