Commutative Post-Lie algebra structures on nilpotent Lie algebras and Poisson algebras

TL;DR

This paper classifies commutative post-Lie algebra structures on certain nilpotent Lie algebras, showing they are associative and relate to Poisson-admissible algebras, thus advancing understanding of algebraic structures in Lie theory.

Contribution

It provides explicit descriptions of CPA-structures on specific nilpotent Lie algebras, revealing their associative nature and connection to Poisson algebras.

Findings

All CPA-structures on non-metabelian filiform nilpotent Lie algebras are associative.

CPA-structures on Lie algebras of strictly upper-triangular matrices are associative.

These structures induce Poisson-admissible algebras.

Abstract

We give an explicit description of commutative post-Lie algebra structures on some classes of nilpotent Lie algebras. For non-metabelian filiform nilpotent Lie algebras and Lie algebras of strictly upper-triangular matrices we show that all CPA-structures are associative and induce an associated Poisson-admissible algebra.

Equations

$$x\cdot y-y\cdot x$$

$$[x,y]\cdot z$$

$$x\cdot \{y,z\}$$

$$L:\mathfrak{g}\to \mathrm{Der}(\mathfrak{n})\subseteq \mathrm{End}(V),x\mapsto L(x)$$

$$x\cdot y-y\cdot x$$

$$[x,y]\cdot z$$

$$x\cdot y-y\cdot x$$

$$x\cdot (y\cdot z)$$

$$x\cdot \{y,z\}$$

$$x\cdot y$$

$$[x,y]\cdot z$$

$$x\cdot [y,z]$$

$$[x,y\circ z]$$

$$[x,y]$$

$$x\circ y$$

$$(x\cdot y)\cdot z-x\cdot (y\cdot z)$$

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

$$=y\cdot [z,x]$$

$$x\circ y$$

$$[x,y]$$

$$x\cdot [y,z]=[x\cdot y,z]+[y,x\cdot z]=0$$

$$[x\cdot y,z]=x\cdot [y,z]-[y,x\cdot z]=0.$$

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

$$x\cdot (y\cdot z)-(x\cdot y)\cdot z$$

$$=x\cdot (y\cdot z)-x\cdot (z\cdot y)-\{z,x\cdot y\}$$

$$=x\cdot \{z,y\}-\{z,x\cdot y\}$$

$$=\{x\cdot z,y\},$$

$$\mathfrak{n}\cdot (\mathfrak{n}\cdot \mathfrak{n})=(\mathfrak{n}\cdot \mathfrak{n})\cdot \mathfrak{n}=0.$$

$$\mathfrak{n}\cdot (\mathfrak{n}\cdot \mathfrak{n})$$

$$\subseteq \{\mathfrak{n}\cdot \mathfrak{n},\mathfrak{n}\}+\{\mathfrak{n},\mathfrak{n}\cdot \mathfrak{n}\}$$

$$\subseteq \{Z(\mathfrak{n}),\mathfrak{n}\}=0.$$

$$e_2\cdot e_1$$

$$e_2\cdot e_2$$

$${\mathfrak{g}}^0=\mathfrak{g}\supseteq {\mathfrak{g}}^1\supseteq {\mathfrak{g}}^2\supseteq {\mathfrak{g}}^3\supseteq \cdots$$

$${\mathfrak{g}}^{(0)}=\mathfrak{g}\supseteq {\mathfrak{g}}^{(1)}\supseteq {\mathfrak{g}}^{(2)}\supseteq {\mathfrak{g}}^{(3)}\supseteq \cdots$$

$$I_j=\mathrm{span}\{e_j,\dots ,e_n\},$$

$$\mathfrak{g}\cdot \mathfrak{g}\subseteq I_3$$

$$\mathfrak{g}\cdot I_2\subseteq I_4$$

$$e_1\cdot e_j$$

Full text

1. Introduction

Post-Lie algebras and post-Lie algebra structures arise in many different contexts. We have introduced these structures in [10] to characterize simply transitive nil-affine actions of a nilpotent Lie group $G$ on another nilpotent Lie group $N$. This plays an important role in the theory of nil-affine crystallographic groups and complete nil-affinely flat manifolds. Post-Lie algebras arise there as a natural common generalization of pre-Lie algebras [20, 21, 24, 4, 5, 6] and LR-algebras [8, 9], and the geometric questions can be formulated on the level of post-Lie algebras.

On the other hand, post Lie algebras have been introduced independently by Vallette [25] in connection with the homology of partition posets and the study of Koszul operads. Since then several authors have studied these algebras in various other contexts, e.g., for algebraic operad triples [22], in connection with modified Yang-Baxter equations, Rota-Baxter operators, universal enveloping algebras, double Lie algebras, classical $R$-matrices, isospectral flows, Lie-Butcher series and many other topics [1, 14, 15, 17].

An important question arising from geometry here is the existence question of post-Lie algebra structures for a given pair of Lie algebras. This is in general a very hard question and the answer depends very much on the algebraic properties of the two given Lie algebras. For a survey on the results and open problems see [7, 10, 11]. An important class of post-Lie algebra structures is given by commutative structures, so-called CPA-structures. These structures are much more accessible than general post-Lie algebra structures and we can answer several questions concerning existence and classification, thereby finding directions to understand the general case of post-Lie algebra structures. For CPA-structures on semisimple, perfect and complete Lie algebras, see [12, 13]. For CPA-structures on nilpotent Lie algebras, see [14].

In this paper we also show that CPA-structures are related to other algebraic structures coming from geometry, namely to Poisson structures [2] and Poisson algebras [18]. Indeed, for many classes of Lie algebras, CPA-structures are associative and induce a Poisson admissible algebra structure.

The paper is structured as follows. In section $2$ we show that for a CPA structure $(V,\cdot)$ on $\mathfrak{g}$ we have $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ if and only if $(V,\cdot)$ is an associative algebra, if and only if it is a Poisson-admissible algebra. We prove that $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g})$ implies $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ and that $(V,\cdot,[,\,])$ is a Poisson algebra if and only if the CPA-structure is central, i.e., satisfies $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g})$.

In section $3$ we show that every CPA-structure $(V,\cdot)$ on a non-metabelian complex filiform Lie algebra is associative, i.e., satisfies $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ so that $(V,\cdot)$ is Poisson-admissible. The result does not hold for metabelian filiform Lie algebras, where we only obtain $[\mathfrak{g},\mathfrak{g}]\cdot[\mathfrak{g},\mathfrak{g}]=0$. For special families of filiform Lie algebras we can give an explicit description of all CPA-structures.

In section $4$ we show that every CPA-structure $(V,\cdot)$ on the nilpotent Lie algebra $\mathfrak{n}_{n}(K)$, $n\geq 5$ of $n\times n$ strictly upper-triangular matrices over $K$ is associative and $(V,\cdot)$ is Poisson-admissible.

This paper is based on results of the PhD thesis [16] of the second author, where further details can be found.

2. Preliminaries

Let $K$ denote a field of characteristic zero. We have defined a post-Lie algebra structure on a pair of Lie algebras $(\mathfrak{g},\mathfrak{n})$ over $K$ in [10] as follows:

Definition 2.1**.**

Let $\mathfrak{g}=(V,[\,,])$ and $\mathfrak{n}=(V,\{\,,\})$ be two Lie brackets on a vector space $V$ over $K$. A post-Lie algebra structure, or PA-structure on the pair $(\mathfrak{g},\mathfrak{n})$ is a $K$-bilinear product $x\cdot y$ satisfying the identities

[TABLE]

for all $x,y,z\in V$.

Define by $L(x)(y)=x\cdot y$ and $R(x)(y)=y\cdot x$ the left respectively right multiplication operators of the algebra $A=(V,\cdot)$. By (3), all $L(x)$ are derivations of the Lie algebra $(V,\{,\})$. Moreover, by (2), the left multiplication

[TABLE]

is a linear representation of $\mathfrak{g}$. The right multiplication $R\colon V\rightarrow V,\;x\mapsto R(x)$ is a linear map, but in general not a Lie algebra representation.

If $\mathfrak{n}$ is abelian, then a post-Lie algebra structure on $(\mathfrak{g},\mathfrak{n})$ corresponds to a pre-Lie algebra structure on $\mathfrak{g}$. In other words, if $\{x,y\}=0$ for all $x,y\in V$, then the conditions reduce to

[TABLE]

i.e., $x\cdot y$ is a pre-Lie algebra structure on the Lie algebra $\mathfrak{g}$. If $\mathfrak{g}$ is abelian, then the conditions reduce to

[TABLE]

i.e., $-x\cdot y$ is an LR-structure on the Lie algebra $\mathfrak{n}$. For details see [10].

An important case of a post-Lie algebra structure arises if the algebra $A=(V,\cdot)$ is commutative, i.e., if $x\cdot y=y\cdot x$ is satisfied for all $x,y\in V$, so that we have $L(x)=R(x)$ for all $x\in V$. Then the two Lie brackets $[x,y]=\{x,y\}$ coincide, and we obtain a commutative algebra structure on $V$ associated with only one Lie algebra [12]:

Definition 2.2**.**

A commutative post-Lie algebra structure, or CPA-structure on a Lie algebra $\mathfrak{g}$ is a $K$-bilinear product $x\cdot y$ satisfying the identities:

[TABLE]

for all $x,y,z\in V$.

We often write $(V,\cdot)$ for a CPA-structure on $\mathfrak{g}$.

It turns out that certain CPA-structures are related to Poisson algebras and Poisson-admissible algebras. Such algebras also have been studied in a geometric and algebraic context, see for example [2, 18]. The definition is as follows.

Definition 2.3**.**

A Poisson algebra is a triple $(V,\circ,[\,,\,])$, where $(V,\circ)$ is a commutative, associative algebra, and $\mathfrak{g}=(V,[\,,\,])$ is a Lie algebra such that the identity

[TABLE]

holds for all $x,y,z\in V$.

Recall that a non-associative algebra is a vector space $V$ together with a bilinear product $V\times V\rightarrow V$, $(x,y)\mapsto x\cdot y$. The product need not be associative in general.

Definition 2.4**.**

A non-associative algebra $(V,\cdot)$ is called Poisson-admissible if $(V,\circ,[\,,\,])$, given by

[TABLE]

is a Poisson algebra.

A CPA-structure $(V,\cdot)$ on $\mathfrak{g}$ often satisfies $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$. This means that $(V,\cdot)$ is a Poisson-admissible algebra.

Lemma 2.5**.**

Let $(V,\cdot)$ be a CPA-structure on a Lie algebra $\mathfrak{g}$. Then the following properties are equivalent.

• $(1)$

$\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$.

• $(2)$

$(V,\cdot)$* is an associative algebra.*

• $(3)$

$(V,\cdot)$* is a Poisson-admissible algebra.*

Proof.

Using (4) and (5) we have

[TABLE]

for all $x,y,z\in V$. This shows that $(1)\Leftrightarrow(2)$. Since $(V,\cdot)$ is commutative we have

[TABLE]

and (7) is trivially satisfied. Hence $(V,\circ,[\,,\,])$ is a Poisson algebra if and only if $(V,\cdot)$ is associative. This shows that $(2)\Leftrightarrow(3)$. ∎

The lemma motivates the following definition.

Definition 2.6**.**

A CPA-structure $(V,\cdot)$ on $\mathfrak{g}$ is called associative if $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$. It is called central if $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g})$.

We have the following implications.

Lemma 2.7**.**

Every central CPA-structure on $\mathfrak{g}$ is associative and every associative CPA-structure on $\mathfrak{g}$ satisfies $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z([\mathfrak{g},\mathfrak{g}])$.

Proof.

Assume that $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g})$. Then by (6) we have

[TABLE]

for all $x,y,z\in V$. Hence we obtain that $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$. Conversely suppose that $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ and let $x,y\in\mathfrak{g}$ and $z\in[\mathfrak{g},\mathfrak{g}]$. Then $x\cdot z=0$ by assumption, so that by (6) we have

[TABLE]

Hence we have $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z([\mathfrak{g},\mathfrak{g}])$. ∎

We have studied central CPA-structures on nilpotent Lie algebras in [14] and shown the following result, see Theorem $4.3$.

Theorem 2.8**.**

All CPA-structures on a free-nilpotent Lie algebra $F_{3,c}$ with $3$ generators and nilpotency class $c\geq 3$ are central.

This result should also hold for every number of generators $g\geq 3$ and nilpotency class $c\geq 3$.

Lemma 2.9**.**

Let $(V,\cdot)$ be a CPA-structure on $\mathfrak{g}$. Then $(V,\cdot,[\,,\,])$ is a Poisson algebra if and only if $(V,\cdot)$ is central.

Proof.

Suppose that $(V,\cdot)$ is central. Then it is also associative by Lemma 2.7. Hence $(V,\cdot)$ is commutative and associative and (7) for the product $x\cdot y$ becomes

[TABLE]

which is satisfied, because every term is zero. Hence $(V,\cdot,[\,,\,])$ is a Poisson algebra. Conversely, if $(V,\cdot,[\,,\,])$ is a Poisson algebra, then $(V,\cdot)$ is associative, hence satisfies $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ by Lemma 2.5. Hence we have $[x,y\cdot z]=[x,y]\cdot z+y\cdot[x,z]=0$ for all $x,y,z\in V$ so that $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g})$. ∎

Also certain LR-structures are related to Poisson-admissible algebras.

Lemma 2.10**.**

Let $(V,\cdot)$ be an LR-structure on a Lie algebra $(\mathfrak{n},\{,\,\})$. Then the following properties are equivalent.

• $(1)$

$\mathfrak{n}\cdot\mathfrak{n}\subseteq Z(\mathfrak{n})$.

• $(2)$

$(V,\cdot)$* is an associative algebra.*

• $(3)$

$(V,\cdot)$* is a Poisson-admissible algebra.*

Proof.

Since $u\cdot v-v\cdot u=-\{u,v\}$ we have $z\cdot(x\cdot y)-(x\cdot y)\cdot z=-\{z,x\cdot y\}$ and also $x\cdot(z\cdot y)-x\cdot(y\cdot z)=-x\cdot\{z,y\}$. Hence the associator is given by

[TABLE]

which vanishes if and only if $\mathfrak{n}\cdot\mathfrak{n}\subseteq Z(\mathfrak{n})$. This shows that $(1)\Leftrightarrow(2)$. For the equivalence $(2)\Leftrightarrow(3)$, see Proposition $3.2$ in [2]. ∎

Again we call an LR-structure on $\mathfrak{n}$ associative, if one of the above conditions is satisfied. We obtain the following interesting consequence.

Proposition 2.11**.**

Let $(V,\cdot)$ be an associative LR-structure on a Lie algebra $\mathfrak{n}$. Then $\mathfrak{n}$ is $2$-step nilpotent.

Proof.

By Lemma 2.10 $(V,\cdot)$ is associative if and only it is Poisson-admissible. In this case the Lie algebra $\mathfrak{n}$ is $2$-step nilpotent by Corollary $3.1$ in [2]. ∎

Proposition 2.12**.**

Let $(V,\cdot)$ be an associative LR-structure on a Lie algebra $\mathfrak{n}$ and suppose that $Z(\mathfrak{n})\subseteq\{\mathfrak{n},\mathfrak{n}\}$. Then we have

[TABLE]

In particular, all left multiplications $L(x)$ are nilpotent.

Proof.

Because of $\mathfrak{n}\cdot\mathfrak{n}\subseteq Z(\mathfrak{n})$ and $Z(\mathfrak{n})\subseteq\{\mathfrak{n},\mathfrak{n}\}$ we have

[TABLE]

Similarly we have $(\mathfrak{n}\cdot\mathfrak{n})\cdot\mathfrak{n}=0$. ∎

Remark 2.13*.*

Proposition 2.12 is no longer true for LR-structures which are not associative. The classification of LR-structures on the $3$-dimensional Heisenberg Lie algebra $\mathfrak{n}=\mathfrak{n}_{3}(K)$, see Proposition $3.1$ of [8], gives a counterexample. Let $e_{1},e_{2},e_{3}$ be a basis of $V$ with $\{e_{1},e_{2}\}=e_{3}$. The LR-structure $A_{4}$, given by the products

[TABLE]

is not associative, since $e_{2}\cdot(e_{1}\cdot e_{2})-(e_{2}\cdot e_{1})\cdot e_{2}=e_{3}$, and $L(e_{2})$ is not nilpotent. In particular, $\mathfrak{n}\cdot(\mathfrak{n}\cdot\mathfrak{n})\neq 0$.

3. CPA-structures on filiform Lie algebras

Let $\mathfrak{g}$ be a Lie algebra. The lower central series of $\mathfrak{g}$ is given by

[TABLE]

where the ideals $\mathfrak{g}^{i}$ are defined by $\mathfrak{g}^{i}=[\mathfrak{g},\mathfrak{g}^{i-1}]$ for all $i\geq 1$. The derived series of $\mathfrak{g}$ is given by

[TABLE]

where the ideals $\mathfrak{g}^{(i)}$ are defined by $\mathfrak{g}^{(i)}=[\mathfrak{g}^{(i-1)},\mathfrak{g}^{(i-1)}]$ for all $i\geq 1$. The nilpotency class $c(\mathfrak{g})$ is the smallest integer $k\geq 1$ with $\mathfrak{g}^{k}=0$ and the solvability class $d(\mathfrak{g})$ is the smallest integer $k\geq 1$ with $\mathfrak{g}^{(k)}=0$.

Definition 3.1**.**

A Lie algebra $\mathfrak{g}$ is called metabelian if it is solvable of class $d(\mathfrak{g})\leq 2$. It is called filiform if it is nilpotent of class $c(\mathfrak{g})=\dim(\mathfrak{g})-1$. Furthermore $\mathfrak{g}$ is called a stem Lie algebra if $Z(\mathfrak{g})\subseteq[\mathfrak{g},\mathfrak{g}]$.

Let $\mathfrak{g}$ be a complex filiform Lie algebra of dimension $n$. Then there exists an adapted basis $(e_{1},\ldots,e_{n})$ of $\mathfrak{g}$, which among other relations satisfies $[e_{1},e_{i}]=e_{i+1}$ for $2\leq i\leq n-1$. For $j\geq 1$ define the characteristic ideals

[TABLE]

which refine the lower central series of $\mathfrak{g}$ with $I_{1}=\mathfrak{g}$ and $I_{j}=\mathfrak{g}^{j-2}$ for $j\geq 3$. Note that $I_{n}=Z(\mathfrak{g})\subseteq[\mathfrak{g},\mathfrak{g}]=I_{3}$ and $I_{n+i}=0$ for all $i\geq 1$.

Lemma 3.2**.**

Let $(V,\cdot)$ be a CPA-structure on a filiform Lie algebra $\mathfrak{g}$ and $(e_{1},\ldots,e_{n})$ be an adapted basis of $\mathfrak{g}$. Then $L(e_{i})(I_{j})\subseteq I_{j+1}$ for $1\leq i,j\leq n$. In particular we have $\mathfrak{g}\cdot\mathfrak{g}\subseteq I_{2}$ and $\mathfrak{g}\cdot I_{2}\subseteq I_{3}$. Furthermore it holds

[TABLE]

Proof.

Since the ideals $I_{j}$ are characteristic and the left multiplications $L(e_{i})$ are derivations, it follows that $L(e_{i})(I_{j})\subseteq I_{j}$ for all $1\leq i,j\leq n$. Since filiform Lie algebras are nilpotent stem Lie algebras, all left multiplications are nilpotent by Theorem $3.6$ in [14]. Hence we have $L(e_{i})(I_{j})\subseteq I_{j+1}$ for $1\leq i,j\leq n$. This implies that $\mathfrak{g}\cdot I_{2}\subseteq I_{3}$, so that $\mathfrak{g}\cdot\mathfrak{g}\subseteq I_{3}$ follows from $e_{1}\cdot e_{1}\in I_{3}$. We also have $\mathfrak{g}\cdot I_{3}\subseteq I_{4}$, so that $\mathfrak{g}\cdot I_{2}\subseteq I_{4}$ follows from $e_{1}\cdot e_{2},\,e_{2}\cdot e_{2}\in I_{4}$. ∎

Lemma 3.3**.**

Let $(V,\cdot)$ be a CPA-structure on a filiform Lie algebra $\mathfrak{g}$ and $(e_{1},\ldots,e_{n})$ be an adapted basis of $\mathfrak{g}$ with $\mathfrak{g}\cdot\mathfrak{g}\subseteq I_{3}$ and $\mathfrak{g}\cdot I_{2}\subseteq I_{4}$. Suppose that for some $\ell\geq 0$ we have

[TABLE]

Then the same is true for $\ell+1$, i.e., we have

[TABLE]

Proof.

We first show by induction on $j\geq 3$ that $e_{2}\cdot e_{j}\in I_{3+j+\ell}$. For $j=3$ we have using (4) and (5)

[TABLE]

Since $e_{2}\cdot e_{2}\in I_{4}$, $e_{1}\cdot e_{2}\in I_{4}$ we have $e_{1}\cdot(e_{2}\cdot e_{2})\in e_{1}\cdot I_{4}\subseteq I_{4+\ell+2}=I_{6+\ell}$ by assumption and also $e_{2}\cdot(e_{1}\cdot e_{2})\in e_{2}\cdot I_{4}\subseteq I_{6+\ell}$. It follows that $e_{2}\cdot e_{3}\in I_{6+\ell}$. For the induction step $j\mapsto j+1$ we have

[TABLE]

By induction hypothesis we have $e_{j}\cdot e_{2}\in I_{3+j+\ell}$, so that $e_{1}\cdot(e_{j}\cdot e_{2})\in I_{3+j+\ell+(\ell+2)}\subseteq I_{4+j+\ell}$. Also $e_{1}\cdot e_{2}\in I_{4}$ implies that $e_{j}\cdot(e_{1}\cdot e_{2})\in I_{j+4+\ell}$. It follows that $e_{2}\cdot e_{j+1}\in I_{4+j+\ell}$ and we have shown that $e_{2}\cdot e_{j}\in I_{3+j+\ell}$ for all $j\geq 3$. We can replace $e_{2}$ by $e_{3},\ldots,e_{n}$ and use induction as above. Then we obtain $e_{i}\cdot e_{j}\in I_{1+i+j+\ell}$ for all pairs $(i,j)$ which are not of the form $((1,j),(i,1),(2,2)$. To complete this for the remaining pairs $(i,j)$ it is enough to show the first claim, i.e., that $e_{1}\cdot e_{j}\in I_{j+\ell+3}$ for all $j\geq 3$. For $j=3$ we see that $e_{1}\cdot e_{3}\in I_{6+\ell}$ as above and for the induction step $j\mapsto j+1$ we have

[TABLE]

By induction hypothesis, $e_{j}\cdot e_{1}\in I_{j+\ell+3}$, so that $e_{1}\cdot(e_{j}\cdot e_{1})\in I_{j+2\ell+5}\subseteq I_{4+j+\ell}$. Furthermore $e_{1}\cdot e_{1}\in I_{3}$. Because of $e_{j}\cdot e_{i}\in I_{i+j+\ell+1}$ for $i,j\geq 3$, which we have shown before, it follows that $e_{j}\cdot(e_{1}\cdot e_{1})\in I_{3+j+\ell+1}=I_{4+j+\ell}$, and we are done. ∎

We can state now our main result for filiform Lie algebras.

Theorem 3.4**.**

Let $\mathfrak{g}$ be a complex filiform Lie algebra of solvability class $d(\mathfrak{g})\geq 3$. Then every CPA-structure $(V,\cdot)$ on $\mathfrak{g}$ is associative and the algebra $(V,\cdot)$ is Poisson-admissible.

Proof.

Let $\dim(\mathfrak{g})=n$. Since every filiform Lie algebra of dimension $n\leq 5$ is metabelian, we have $n\geq 6$ because of $d(\mathfrak{g})\geq 3$. For $n=6$ we can prove the result by a direct computation. Every $6$-dimensional filiform Lie algebra $\mathfrak{g}$ has an adapted basis $(e_{1},\ldots,e_{6})$ such that the Lie brackets are given by

[TABLE]

for some $\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{C}$. Here $\mathfrak{g}$ is metabelian if and only if $\alpha_{3}=0$. So we have $\alpha_{3}\neq 0$ and $d(\mathfrak{g})=3$. Let $(V,\cdot)$ be a CPA-structure on $\mathfrak{g}$. Denote the matrices for the left multiplications by

[TABLE]

All matrices are lower-triangular with respect to the basis $(e_{1},\ldots,e_{6})$ because of Lemma 3.2. The identities (4),(5),(6) are equivalent to a system of equations in the variables $\zeta_{ij}^{k}$, which can be easily solved directly. Indeed, (4) and (6) yield linear equations, which enables one to solve the quadratic equations coming from (5). We obtain the following CPA-structures:

[TABLE]

Recall that $e_{1}\cdot e_{2}=e_{2}\cdot e_{1}$. This shows that $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$, so that $(V,\cdot)$ is associative and a Poisson-admissible algebra by Lemma 2.5.

We may now assume that $(V,\cdot)$ is a CPA-structure on $\mathfrak{g}$ with $n\geq 7$. Denote again by $L(e_{k})=(\zeta_{ij}^{k})$ the left multiplications. We distinguish two cases, namely whether or not $\mathfrak{g}\cdot\mathfrak{g}\subseteq I_{3}$. This does not depend on the choice of an adapted basis for $\mathfrak{g}$. The first case is again divided into two subcases whether or not $\mathfrak{g}\cdot I_{2}\subseteq I_{4}$.

Case 1a: It holds $\mathfrak{g}\cdot\mathfrak{g}\subseteq I_{3}$ and $\mathfrak{g}\cdot I_{2}\subseteq I_{4}$. An adapted basis has the property that $[e_{i},e_{j}]\in I_{i+j}$ holds for all $1\leq i,j\leq n$ except for the case where $n$ is even, where we have the following exceptional Lie brackets

[TABLE]

Here the scalar $\alpha\in\mathbb{C}$ is zero if and only if $\mathfrak{g}^{\lceil\frac{n-4}{2}\rceil}$ is abelian.

Case 1a1: $\mathfrak{g}^{\lceil\frac{n-4}{2}\rceil}$ is abelian. In this case we have $[I_{i},I_{j}]\subseteq I_{i+j}$ for all $1\leq i,j\leq n$ with the convention that $I_{m}=0$ for all $m>n$. Using (6) we obtain

[TABLE]

for all $2\leq i\leq n-2$. We have $e_{1}\cdot e_{i+1}=\zeta_{i+2,i+1}^{1}e_{i+2}+z$ with $z\in I_{i+3}$. On the other hand it follows from $e_{1}\cdot e_{1}\in I_{3}$ that $[e_{1}\cdot e_{1},e_{i}]\in I_{i+3}$ for all $2\leq i\leq n-2$. The case $i=n-2$ requires the assumption that $\mathfrak{g}^{\lceil\frac{n-4}{2}\rceil}$ is abelian, because we need that $[I_{3},I_{n-2}]\subseteq I_{n+1}=0$. Furthermore we have $[e_{1},e_{1}\cdot e_{i}]=\zeta_{i+1,i}^{1}e_{i+2}+w$ with $w\in I_{3}$, so that

[TABLE]

for all $2\leq i\leq n-2$. This implies that $\zeta_{j+1,j}^{1}=\zeta_{3,2}^{1}$ for all $2\leq j\leq n-2$. Since $e_{1}\cdot e_{2}\in I_{4}$ we have $\zeta_{3,2}^{1}=0$, so that

[TABLE]

Hence the first subdiagonal of $L(e_{1})$ is equal to zero. The same argument for $L(e_{2})$ gives that $\zeta_{j+1,j}^{2}=\zeta_{3,2}^{2}$ for all $2\leq j\leq n-2$. However, $\zeta_{3,2}^{2}=0$ because we have

[TABLE]

Indeed, $e_{2}\cdot e_{3}=\zeta_{3,2}^{2}e_{4}+z$ with $z\in I_{5}$, and $e_{1}\cdot(e_{2}\cdot e_{2}),e_{2}\cdot(e_{1}\cdot e_{2})\in I_{5}$. So we have shown so far that $e_{1}\cdot e_{i}\in I_{i+2}$ and $e_{2}\cdot e_{i}\in I_{i+2}$ for all $1\leq i\leq n$. It follows by induction on $i\geq 3$ as before that also $e_{i}\cdot e_{j}\in I_{i+j}$ for all $i,j\geq 3$. Now we can apply Lemma 3.3 for $\ell=0$, then for $\ell=1$ and so on. Finally we obtain that $e_{i}\cdot e_{j}\in I_{n+1}=0$ for all pairs $(i,j)\neq(1,1),(1,2),(2,1),(2,2)$. This exactly says that $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$, i.e., that $L(e_{i})=0$ for all $i\geq 3$. Hence $(V,\cdot)$ is associative and a Poisson-admissible algebra.

Case 1a2: $\mathfrak{g}^{\lceil\frac{n-4}{2}\rceil}$ is not abelian. The proof is the same as above except for a slight modification. We have the additional non-trivial Lie brackets $[e_{i},e_{n-i+1}]=(-1)^{i+1}\alpha e_{n}$ for $2\leq i\leq n-1$ with $\alpha\neq 0$. Therefore we can show $[e_{1}\cdot e_{1},e_{i}]\in I_{i+3}$ only for all $2\leq i\leq n-3$, but not for $i=n-2$. Consequently the first subdiagonal of $L(e_{1})$ is not necessarily zero, but of the form $(0,\ldots,0,\lambda)$ for some $\lambda\in\mathbb{C}$. Similarly for $L(e_{2})$ it is of the form $(0,\ldots,0,\mu)$. By induction we see that

[TABLE]

We want to show that $e_{i}\cdot e_{j}\in I_{i+j}$ for $2\leq i,j\leq n$. We already know this, except for the case where $i\geq 2,j\geq 3$ with $i+j=n+1$, i.e., with $j=n-i+1$. But then $e_{i}\cdot e_{j-1}=e_{i}\cdot e_{n-i}\in I_{n}$ and $e_{i}\cdot e_{j}=e_{i}\cdot e_{n-i+1}\in I_{n}$, so that we can write $e_{i}\cdot e_{j-1}=\zeta e_{n}$ and $e_{i}\cdot e_{j}=\eta e_{n}$. We have

[TABLE]

so that $\eta=0$ and hence $e_{i}\cdot e_{j}=0$ for these $i,j$. Hence we have $e_{i}\cdot e_{j}\in I_{i+j}$ for all $2\leq i,j\leq n$. Similarly we see that $e_{1}\cdot e_{j}\in I_{j+2}$ for all $j\geq 3$. Now we can apply repeatedly Lemma 3.3, starting with $\ell=0$. It follows again that $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ and we are done.

Case 1b: It holds $\mathfrak{g}\cdot\mathfrak{g}\subseteq I_{3}$ and $\mathfrak{g}\cdot I_{2}\nsubseteq I_{4}$. It follows that $\mathfrak{g}$ is necessarily metabelian, which gives a contradiction. Hence this case cannot occur. The method of proof is very similar to the one of case $1a$, so that we will only sketch a few steps. For a detailed proof see Proposition $A.2$ in [16]. First one can write $e_{1}\cdot e_{3}=\lambda e_{4}+z$ for some $\lambda\in\mathbb{C}$ and $z\in I_{5}$. Because of

[TABLE]

we see that $\lambda e_{4}+z=\lambda^{2}e_{4}+w$ for $z,w\in I_{5}$ so that $\lambda(\lambda-1)=0$. By assumption $\mathfrak{g}\cdot I_{2}\nsubseteq I_{4}$, so that $\lambda\neq 0$ and hence $\lambda=1$. This implies that both $e_{\frac{n}{2}}\cdot e_{\frac{n}{2}}$, $e_{\frac{n}{2}}\cdot e_{\frac{n+2}{2}}$ are multiples of $e_{n}$ and $[e_{\frac{n}{2}},e_{\frac{n}{2}}]=0$. It follows that $\mathfrak{g}^{\lceil\frac{n-4}{2}\rceil}$ is abelian. Using several involved inductions we finally obtain that $\mathfrak{g}$ is metabelian and that $[\mathfrak{g},\mathfrak{g}]\cdot[\mathfrak{g},\mathfrak{g}]=0$.

Case 2: It holds $\mathfrak{g}\cdot\mathfrak{g}\nsubseteq I_{3}$. In this case one can even show that $\mathfrak{g}$ is isomorphic to the standard graded filiform Lie algebra $L_{n}$, defined by the Lie brackets $[e_{1},e_{i}]=e_{i+1}$ for $2\leq i\leq n-1$, see Proposition $A.4$ in [16]. Since $L_{n}$ is metabelian, we obtain a contradiction. This finishes the proof. ∎

Remark 3.5*.*

By the theorem every CPA-structure on a non-metabelian filiform Lie algebra with adapted basis $(e_{1},\ldots,e_{n})$ is of a very simple form. Because of $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$ the left multiplications $L(e_{i})$ are zero for $i=3,\ldots,n$, so that the only non-zero products are given by $e_{1}\cdot e_{1},e_{1}\cdot e_{2}=e_{2}\cdot e_{1},e_{2}\cdot e_{2}$, which lie in $Z([\mathfrak{g},\mathfrak{g}])$ by Lemma 2.7. For example, in dimension $6$ we have $Z([\mathfrak{g},\mathfrak{g}])=I_{5}={\rm span}\{e_{5},e_{6}\}$, which recovers the result of our direct computation at the beginning of the proof of the theorem.

The following example shows that Theorem 3.4 cannot be extended to other nilpotent stem Lie algebras, which are not filiform.

Example 3.6**.**

There exists a nilpotent stem Lie algebra $\mathfrak{g}$ of solvability class $3$ and nilpotency class $5$ admitting a CPA-structure $(V,\cdot)$ which is not associative.

Let $(e_{1},\ldots,e_{9})$ be a basis of $V$ and define the Lie brackets of $\mathfrak{g}$ by

[TABLE]

A CPA-structure, which is not associative since $(e_{2}\cdot e_{1})\cdot e_{1}-e_{2}\cdot(e_{1}\cdot e_{1})=e_{8}$, is defined as follows:

[TABLE]

We also see that Theorem 3.4 does not hold for metabelian filiform Lie algebras, because we have the following result.

Proposition 3.7**.**

Let $\mathfrak{g}$ be a metabelian filiform Lie algebra of dimension $n\geq 4$. Then there exists a CPA-structure $(V,\cdot)$ on $\mathfrak{g}$ which is not associative.

Proof.

According to [3] there exists an adapted basis $(e_{1},\ldots,e_{n})$ for $\mathfrak{g}$ such that the Lie brackets are given by

[TABLE]

with structure constants $\{\alpha_{2,k}\mid 5\leq k\leq n\}$. By convention we set these constants equal to zero for $n=4$. We define an algebra $(V,\cdot)$ as follows:

[TABLE]

It is easy to verify that $(V,\cdot)$ defines a CPA-structure on $\mathfrak{g}$. It satisfies $[\mathfrak{g},\mathfrak{g}]\cdot[\mathfrak{g},\mathfrak{g}]=0$, but not $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$. ∎

Note that the proposition does not hold for $n=3$, because all CPA-structures on the Heisenberg Lie algebra $\mathfrak{n}_{3}(\mathbb{C})$ are associative, see Proposition $6.3$ in [12].

Remark 3.8*.*

One can show in general that all CPA-structures on a filiform Lie algebra $\mathfrak{g}$ satisfy $[\mathfrak{g},\mathfrak{g}]\cdot[\mathfrak{g},\mathfrak{g}]=0$. If $\mathfrak{g}$ is not metabelian, this follows from Theorem 3.4. For metabelian filiform Lie algebras it is proved in [16].

We can apply the results now for special classes of filiform Lie algebras and determine all CPA-structures explicitly. The explicit form is less complicated and is often better suited for applications than a classification up to CPA-isomorphism. Therefore we do not give such a classification.

We consider the classes $L_{n},Q_{n},R_{n},W_{n}$ of filiform Lie algebras discussed in [19], Chapter $4$. We always assume that $(e_{1},\ldots,e_{n})$ is an adapted basis.

Definition 3.9**.**

The Lie algebra $L_{n}$ for $n\geq 3$ is defined by the Lie brackets

[TABLE]

The Lie algebra $Q_{n}$ for $n\geq 6$ even is defined by the Lie brackets

[TABLE]

The Lie algebra $R_{n}$ for $n\geq 5$ is defined by the Lie brackets

[TABLE]

The Witt Lie algebra $W_{n}$ for $n\geq 5$ is defined by the Lie brackets

[TABLE]

To give a CPA-structure $(V,\cdot)$ on $\mathfrak{g}$ explicitly it is enough to list the non-zero products $e_{i}\cdot e_{j}$ for all $1\leq i\leq j\leq n$.

Proposition 3.10**.**

Every CPA-structure on $L_{n}$, $n\geq 5$ with an adapted basis $(e_{1},\ldots,e_{n})$ is either of type $1$ with products

[TABLE]

with arbitrary parameters $\alpha_{i},\beta,\gamma,\delta$ satisfying $\alpha_{2}\delta+\beta=0$, or of type $2$ with products

[TABLE]

with arbitrary parameters $\alpha_{i},\beta,\gamma,\delta$ satisfying $\alpha_{2}\delta-\beta=0$. In both cases we have $[L_{n},L_{n}]\cdot[L_{n},L_{n}]=0$, but not $L_{n}\cdot[L_{n},L_{n}]=0$ in general.

Proof.

A direct verification shows that the above products indeed define a CPA-structure on $L_{n}$ for all given parameters. Conversely let $(V,\cdot)$ be a CPA-structure on $\mathfrak{g}$. We will show by induction on $n\geq 5$ that $(V,\cdot)$ is either of type $1$ or of type $2$. For $n=5$ this follows from a direct computation. For the induction step we use that $L_{n}/Z(L_{n})\cong L_{n-1}$. Since $\dim(Z(L_{n}))=1$ it follows $L_{n}\cdot Z(L_{n})=0$ from Corollary $3.4$ of [14]. Hence $(V,\cdot)$ induces a CPA-structure on $L_{n-1}$, which by induction hypothesis is already of type $1$ or of type $2$. We may assume that it is of type $1$, because the proof for type $2$ works exactly the same way. Hence we know that the products $e_{i}\cdot e_{j}$ are of the required form up to a certain multiple of $e_{n}$. Now since all left multiplications are derivations of $L_{n}$ and since we have an explicit basis of ${\rm Der}(L_{n})$ with respect to $(e_{1},\ldots,e_{n})$, it follows that

[TABLE]

Indeed, by Proposition $1$ in Chapter $4$ of [19], a basis of ${\rm Der}(\mathfrak{g})$ with respect to $(e_{1},\ldots,e_{n})$ is given by the $2n-1$ endomorphisms

[TABLE]

defined by

[TABLE]

for all $2\leq k\leq n-2$. Hence we see that the non-zero products $e_{i}\cdot e_{j}$ are given as follows:

[TABLE]

Because $L(e_{1})$ and $L(e_{2})$ are a linear combination of the above derivations, we immediately see that $\zeta_{3}=\gamma$, $\zeta_{4}=\beta$ and $\zeta_{6}=\delta$. By (5) we have

[TABLE]

which is equal to [math] for $n\geq 5$. Hence we have $\delta=0$. Similarly we obtain

[TABLE]

which gives $\beta=0$ and $\alpha_{2}\zeta_{6}+\zeta_{3}=0$. So we obtain exactly the CPA-structure of type $1$ on $L_{n}$. ∎

Remark 3.11*.*

For $n=4$ these two types of CPA-structures in Proposition 3.10 can be merged, but with a different condition, namely $\beta(\beta-1)=\alpha_{2}\delta$ and $L(e_{4})=0$,

[TABLE]

Proposition 3.12**.**

Every CPA-structure on $Q_{n}$, $n\geq 6$ even, with an adapted basis $(e_{1},\ldots,e_{n})$ is given as follows:

[TABLE]

with arbitrary parameters $\alpha,\beta,\gamma,\delta$.

Proof.

Let $(V,\cdot)$ be a CPA-structure on $Q_{n}$. Then $Q_{n}\cdot[Q_{n},Q_{n}]=0$ by Theorem 3.4. Now using a basis of ${\rm Der}(Q_{n})$ we obtain that the products are given as above by a straightforward calculation. ∎

In the same way, by using the explicit form ot the derivation algebra, one can show the following result.

Proposition 3.13**.**

Every CPA-structure on $R_{n}$, $n\geq 6$ with an adapted basis $(e_{1},\ldots,e_{n})$ is either of type $1$ with products

[TABLE]

or of type $2$ with products

[TABLE]

where $e_{i}\cdot e_{j}$ is a CPA-structure of type $1$.

Remark 3.14*.*

There is a third type of CPA-structure on $R_{n}$ for $n=5$, given by

[TABLE]

Finally, the result for $W_{n}$ is given as follows.

Proposition 3.15**.**

Every CPA-structure on $W_{n}$, $n\geq 7$ with an adapted basis $(e_{1},\ldots,e_{n})$ is given as follows:

[TABLE]

4. CPA-structures on Lie algebras of strictly upper-triangular matrices

In this section we study CPA-structures on the Lie algebra $\mathfrak{n}_{n}(K)$ of strictly upper-triangular $n\times n$-matrices over a field $K$. This Lie algebra is nilpotent of class $c=n-1$ and dimension $\frac{n(n-1)}{2}$. It has a basis $\{E_{j,k}\mid 1\leq j<k\leq n\}$, where the matrices $E_{i,j}$ have an entry $1$ at position $(i,j)$ and [math] otherwise. The non-trivial Lie brackets are given by

[TABLE]

The following lemma is well known.

Lemma 4.1**.**

Let $\mathfrak{g}$ be the Lie algebra $\mathfrak{n}_{n}(K)$ with $n\geq 4$ and suppose that $[z,[\mathfrak{g},\mathfrak{g}]]=0$ for some $z\in\mathfrak{g}$. Then we have $z\in\mathfrak{g}^{n-4}$.

Here is the main result of this section.

Theorem 4.2**.**

Let $\mathfrak{g}$ be the Lie algebra $\mathfrak{n}_{n}(K)$ with $n\geq 5$. Then every CPA-structure $(V,\cdot)$ on $\mathfrak{g}$ is associative and the algebra $(V,\cdot)$ is Poisson-admissible. Moreover we have $\mathfrak{g}\cdot\mathfrak{g}\subseteq\mathfrak{g}^{n-3}$.

Proof.

Since $\mathfrak{g}$ is a nilpotent stem Lie algebra, all left multiplications are nilpotent by Theorem $3.6$ in [14]. Hence they are nilpotent derivations of $\mathfrak{g}$. By Theorem $3.2$ in [23] there exists for every nilpotent derivation $D\in{\rm Der}(\mathfrak{g})$ an $u\in\mathfrak{g}$ and a $\psi\in{\rm Der}(\mathfrak{g})$ such that $D={\rm ad}(u)+\psi$ and $\psi(\mathfrak{g})\subseteq\mathfrak{g}^{n-3}$, $\psi([\mathfrak{g},\mathfrak{g}])=0$. Hence for every $x\in\mathfrak{g}$ there is a $z\in\mathfrak{g}$ and a $\psi\in{\rm Der}(\mathfrak{g})$, depending on $x$, such that $L(x)={\rm ad}(z)+\psi$ with these properties. We show by induction over $n$ that $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$. For $n=5,6$ this can be verified by a direct computation. So we may assume $n\geq 7$ for the induction step. Define Lie ideals $I$ and $J$ in $\mathfrak{g}$ as follows

[TABLE]

Let $\mathfrak{a}=I+\mathfrak{g}^{n-3}$, $\mathfrak{b}=J+\mathfrak{g}^{n-3}$ and note that $\mathfrak{g}^{n-3}={\rm span}\{E_{1,n-1},E_{2,n},E_{1,n}\}$. Then we have $\mathfrak{g}/\mathfrak{a}\cong\mathfrak{h}/Z(\mathfrak{h})$ with $\mathfrak{h}=\mathfrak{n}_{n-1}(K)$. By induction hypothesis every CPA-structure on $\mathfrak{h}$ satisfies $\mathfrak{h}\cdot[\mathfrak{h},\mathfrak{h}]=0$ and $\mathfrak{h}\cdot\mathfrak{h}\subseteq\mathfrak{h}^{n-4}$. Let $x\in\mathfrak{g}$. Then we have

[TABLE]

so that $\mathfrak{g}\cdot\mathfrak{a}\subseteq\mathfrak{a}$ because of $\mathfrak{g}^{n-3}\subseteq\mathfrak{a}$. Now let $(V,\cdot)$ be a CPA-structure on $\mathfrak{g}$. It induces a CPA-structure on the quotient $\mathfrak{g}/\mathfrak{a}\cong\mathfrak{h}/Z(\mathfrak{h})$, so that we have $(\mathfrak{g}/\mathfrak{a})\cdot(\mathfrak{g}/\mathfrak{a})\subseteq Z(\mathfrak{g}/\mathfrak{a})$ and $\mathfrak{g}/\mathfrak{a}\cdot[\mathfrak{g}/\mathfrak{a},\mathfrak{g}/\mathfrak{a}]=0$. This implies that $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g}/\mathfrak{a})+\mathfrak{a}$ and $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]\subseteq\mathfrak{a}$. On the other hand we also have $\mathfrak{g}/\mathfrak{b}\cong\mathfrak{h}/Z(\mathfrak{h})$, so that $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g}/\mathfrak{b})+\mathfrak{b}$ and $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]\subseteq\mathfrak{b}$. Together this yields

[TABLE]

Using this we obtain by (6) that

[TABLE]

This implies similarly that

[TABLE]

Since $n\geq 7$ we have $\mathfrak{g}^{n-4}\supseteq\mathfrak{g}^{3}$ and hence $\mathfrak{g}\cdot\mathfrak{g}^{n-4}=0$. Then, by (5),

[TABLE]

It remains to show that $\mathfrak{g}\cdot\mathfrak{g}\subseteq\mathfrak{g}^{n-3}$ and not only $\mathfrak{g}\cdot\mathfrak{g}\subseteq\mathfrak{g}^{n-4}$. Suppose that $L(x)(\mathfrak{g})$ is not contained in $\mathfrak{g}^{n-3}$ for some $x\in\mathfrak{g}$. Then there is a $z\in\mathfrak{g}$ and a $\psi\in{\rm Der}(\mathfrak{g})$ such that $[z,\mathfrak{g}]+\psi(\mathfrak{g})\nsubseteq\mathfrak{g}^{n-3}$. Then $[z,\mathfrak{g}]\nsubseteq\mathfrak{g}^{n-3}$ because of $\psi(\mathfrak{g})\subseteq\mathfrak{g}^{n-3}$, and hence $z\not\in\mathfrak{g}^{n-4}$. By Lemma 4.1 it follows that $[z,[\mathfrak{g},\mathfrak{g}]]\neq 0$. Since $\psi([\mathfrak{g},\mathfrak{g}])=0$ this implies $L(z)([\mathfrak{g},\mathfrak{g}])\neq 0$, which contradicts $\mathfrak{g}\cdot[\mathfrak{g},\mathfrak{g}]=0$. ∎

Note that the result does not hold for $n=4$. There are CPA-structures on $\mathfrak{n}_{4}(K)$, which are not associative.

Remark 4.3*.*

One can extend this result to the solvable Lie algebra $\mathfrak{t}_{n}(K)$ of all $n\times n$ upper-triangular matrices over $K$, see Proposition $5.34$ in [16]. Every CPA-structure $(V,\cdot)$ on $\mathfrak{t}_{n}(K)$ for $n\geq 3$ is associative and satisfies $\mathfrak{g}\cdot\mathfrak{g}\subseteq Z(\mathfrak{g})+Z([\mathfrak{g},\mathfrak{g}])$.

Acknowledgments

Dietrich Burde is supported by the Austrian Science Foundation FWF, grant P28079 and grant I3248 and Christof Ender is supported by the Austrian Science Foundation FWF, grant P28079.