Rota–Baxter operators and post-Lie algebra structures on semisimple Lie algebras

Abstract

Rota–Baxter operators R of weight 1 on $\mathfrak{n}$ are in bijective correspondence to post-Lie algebra structures on pairs $(\mathfrak{g},\mathfrak{n})$, where $\mathfrak{n}$ is complete. We use such Rota–Baxter operators to study the existence and classification of post-Lie algebra structures on pairs of Lie algebras $(\mathfrak{g},\mathfrak{n})$, where $\mathfrak{n}$ is semisimple. We show that for semisimple $\mathfrak{g}$ and $\mathfrak{n}$, with $\mathfrak{g}$ or $\mathfrak{n}$ simple, the existence of a post-Lie algebra structure on such a pair $(\mathfrak{g},\mathfrak{n})$ implies that $\mathfrak{g}$ and $\mathfrak{n}$ are isomorphic, and hence both simple. If $\mathfrak{n}$ is semisimple, but $\mathfrak{g}$ is not, it becomes much harder to classify post-Lie algebra structures on $(\mathfrak{g},\mathfrak{n})$, or even to determine the Lie algebras $\mathfrak{g}$ which can arise. Here only the case $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ was studied. In this paper, we determine all Lie algebras $\mathfrak{g}$ such that there exists a post-Lie algebra structure on $(\mathfrak{g},\mathfrak{n})$ with $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$.

Keywords:

• Post-Lie algebra
• Rota–Baxter operator

2010 MATHEMATICS SUBJECT CLASSIFICATION:

• 17B20
• 17D25
• 16T25

1. Introduction

Rota–Baxter operators were introduced by Baxter [3] in 1960 as a formal generalization of integration by parts for solving an analytic formula in probability theory. Such operators $R:A\to A$ are defined on an algebra A by the identity $$R\left(x\right)R\left(y\right)=R\left(R\right(x\left)y+xR\right(y)+\lambda xy)$$ for all $x,y\in A$, where $\lambda$ is a scalar, called the weight of R. These operators were then further investigated, by Rota [30], Atkinson [1], Cartier [16] and others. In the 1980s, these operators were studied in integrable systems in the context of classical and modified Yang–Baxter equations [33,4]. Since the late 1990s, the study of Rota–Baxter operators has made great progress in many areas, both in theory and in applications [25,2,22,20,21,5,19].

Post-Lie algebras and post-Lie algebra structures also arise in many areas, e.g., in differential geometry and the study of geometric structures on Lie groups. Here, post-Lie algebras arise as a natural common generalization of pre-Lie algebras [23,26,32,6,7,8] and LR-algebras [9,10], in the context of nil-affine actions of Lie groups, see [11]. A detailed account of the differential geometric context of post-Lie algebras is also given in [18]. On the other hand, post-Lie algebras have been introduced by Vallette [34] in connection with the homology of partition posets and the study of Koszul operads. They have been studied by several authors in various contexts, e.g., for algebraic operad triples [28], in connection with modified Yang–Baxter equations, Rota–Baxter operators, universal enveloping algebras, double Lie algebras, R-matrices, isospectral flows, Lie-Butcher series and many other topics [2,18,19]. There are several results on the existence and classification of post-Lie algebra structures, in particular on commutative post-Lie algebra structures [13–15].

It is well-known [2] that Rota–Baxter operators R of weight 1 on $\mathfrak{n}$ are in bijective correspondence to post-Lie algebra structures on pairs $(\mathfrak{g},\mathfrak{n})$, where $\mathfrak{n}$ is complete. In fact, RB-operators always yield PA-structures. So it is possible (and desirable) to use results on RB-operators for the existence and classification of post-Lie algebra structures.

The paper is organized as follows. In Section 2 we give basic definitions of RB-operators and PA-structures on pairs of Lie algebras. We summarize several useful results. For a complete Lie algebra $\mathfrak{n}$ there is a bijection between PA-structures on $(\mathfrak{g},\mathfrak{n})$ and RB-operators of weight 1 on $\mathfrak{n}$. The PA-structure is given by $x·y=\{R(x),y\}$. Here we study the kernels of R and $R+\text{id}$. If $\mathfrak{g}$ and $\mathfrak{n}$ are not isomorphic, then both R and $R+\text{id}$ have a nontrivial kernel. Moreover, if one of $\mathfrak{g}$ or $\mathfrak{n}$ is not solvable, then at least one of $\ker (R)$ and $\ker (R+\text{id})$ is nontrivial.

In Section 3, we complete the classification of PA-structures on pairs of semisimple Lie algebras $(\mathfrak{g},\mathfrak{n})$, where either $\mathfrak{g}$ or $\mathfrak{n}$ is simple. We already have shown the following in [11]. If $\mathfrak{g}$ is simple, and there exists a PA-structure on $(\mathfrak{g},\mathfrak{n})$, then also $\mathfrak{n}$ is simple, and we have $\mathfrak{g}\cong \mathfrak{n}$ with $x·y=0$ or $x·y=[x,y]$. Here we deal now with the case that $\mathfrak{n}$ is simple. Again it follows that $\mathfrak{g}$ and $\mathfrak{n}$ are isomorphic. The proof via RB-operators uses results of Koszul [27] and Onishchik [29]. We also show a result concerning semisimple decompositions of Lie algebras. Suppose that $\mathfrak{g}={\mathfrak{s}}_1+{\mathfrak{s}}_2$ is the vector space sum of two semisimple subalgebras of $\mathfrak{g}$. Then $\mathfrak{g}$ is semisimple. As a corollary we show that the existence of a PA-structure on $(\mathfrak{g},\mathfrak{n})$ for $\mathfrak{g}$ semisimple and $\mathfrak{n}$ complete implies that $\mathfrak{n}$ is semisimple.

In Section 4, we determine all Lie algebras $\mathfrak{g}$ which can arise by PA-structures on $(\mathfrak{g},\mathfrak{n})$ with $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$. This turns out to be much more complicated than the case $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$, which we have done in [11]. By Theorem 3.3 of [12], $\mathfrak{g}$ cannot be solvable unimodular. On the other hand, the result we obtain shows that there are more restrictions than that.

2. Preliminaries

Let A be a nonassociative algebra over a field K in the sense of Schafer [31], with K-bilinear product $A\times A\to A,(a,b)\mapsto ab$. We will assume that K is an arbitrary field of characteristic zero, if not said otherwise.

Definition 2.1.

Let $\lambda \in K$. A linear operator $R:A\to A$ satisfying the identity (1) $$R\left(x\right)R\left(y\right)=R\left(R\right(x\left)y+xR\right(y)+\lambda xy)$$(1) for all $x,y\in A$ is called a Rota–Baxter operator on A of weight $\lambda$, or just RB-operator.

Two obvious examples are given by R = 0 and $R=\lambda \text{id}$, for an arbitrary nonassociative algebra. These are called the trivial RB-operators. The following elementary lemma was shown in [22], Proposition 1.1.12.

Lemma 2.2.

Let R be an RB-operator on A of weight λ. Then $-R-\lambda \text{id}$ is an RB-operator on A of weight λ, and ${\lambda }^{-1}R$ is an RB-operator on A of weight 1 for all $\lambda \ne 0$.

It is also easy to verify the following results.

Proposition 2.3.

[5] Let R be an RB-operator on A of weight λ and $\psi \in \text{Aut}(A)$. Then $R^{(\psi )}={\psi }^{-1}R\psi$ is an RB-operator on A of weight $\lambda$.

Proposition 2.4.

[22] Let B be a countable direct sum of an algebra A. Then the operator R defined on B by $$R\left(\right(a_1,a_2,\dots ,a_n,\dots \left)\right)=(0,a_1,a_1+a_2,a_1+a_2+a_3,\dots )$$ is an RB-operator on B of weight 1.

Proposition 2.5.

Let $B=A\oplus A$ and $\psi \in \text{Aut}(A)$. Then the operator R defined on B by (2) $$R\left(\right(a_1,a_2\left)\right)=(0,\psi (a_1\left)\right)$$(2) is an RB-operator on B of weight 1. Furthermore the operator R defined on B by (3) $$R\left(\right(a_1,a_2\left)\right)=(-a_1,-\psi (a_1\left)\right)$$(3) is an RB-operator on B of weight 1.

Proof.

Let $x=(a_1,a_2)$ and $y=(b_1,b_2)$. Then we have $$\begin{array}{l}R(R(x)y+xR(y)+\lambda xy)=R((0,\psi (a_1)b_2+(0,a_2\psi (b_1))+(a_1{b}_1,a_2{b}_2))\\ =(0,\psi (a_1{b}_1))\\ =(0,\psi (a_1)\psi (b_1))\\ =R(x)R(y).\end{array}$$

The second claim follows similarly.□

Proposition 2.6.

[25] Let $A=A_1\oplus A_2$, R₁ be an RB-operator of weight $\lambda$ on A₁, R₂ be an RB-operator of weight $\lambda$ on A₂. Then the operator $R:A\to A$ defined by $R((a_1,a_2))=(R_1(a_1),R_2(a_2))$ is an RB-operator of weight $\lambda$ on A.

Proposition 2.7.

[22] Let $A=A_1\dot{+}{A}_2$ be the direct vector space sum of two subalgebras. Then the operator R defined on A by (4) $$R\left(a_1+a_2\right)=-\lambda a_2$$(4) for $a_1\in A_1$ and $a_2\in A_2$ is an RB-operator on A of weight $\lambda$.

We call such an operator split, with subalgebras A₁ and A₂. Note that the set of all split RB-operators on A is in bijective correspondence with all decompositions $A=A_1\dot{+}{A}_2$ as a direct sum of subalgebras.

Lemma 2.8.

[5] Let R be an RB-operator of nonzero weight $\lambda$ on an algebra A. Then R is split if and only if $R(R+\lambda \text{id})=0$.

Lemma 2.9.

Let $A=A_{-}\dot{+}{A}_0\dot{+}{A}_{+}$ be a direct vector space sum of subalgebras of A. Suppose that R is an RB-operator of weight $\lambda$ on A₀, $A_{-}$ is an $(R+\text{id})(A_0)$-module and $A_{+}$ is an $R(A_0)$-module. Define an operator P on A by (5) $$P_{|A_{-}}=0,P_{|A_0}=R,P_{|A_{+}}=-\lambda \text{id}.$$(5)

Then P is an RB-operator on A of weight $\lambda$.

Definition 2.10.

Let P be an RB-operator on A defined as above such that not both $A_{-}$ and $A_{+}$ are zero. Then P is called triangular-split.

We also recall the definition of post-Lie algebra structures on a pair of Lie algebras $(\mathfrak{g},\mathfrak{n})$ over K, see [11].

Definition 2.11.

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 $x·y$ satisfying the identities: (6) $$x·y-y·x=[x,y]-\{x,y\},$$(6) (7) $$[x,y]·z=x·(y·z)-y·(x·z),$$(7) (8) $$x·\{y,z\}=\{x·y,z\}+\{y,x·z\}$$(8) for all $x,y,z\in V$.

Define by $L(x)(y)=x·y$ the left multiplication operator of the algebra $A=(V,·)$. By (8), all L(x) are derivations of the Lie algebra $(V,\{,\})$. Moreover, by (7), the left multiplication $$L:\mathfrak{g}\to \text{Der}\left(\mathfrak{n}\right)\subseteq \text{End}\left(V\right),x\mapsto L\left(x\right)$$ is a linear representation of $\mathfrak{g}$.

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 $$\begin{array}{l}x·y-y·x=[x,y],\\ [x,y]·z=x·(y·z)-y·(x·z),\end{array}$$ i.e., $x·y$ is a pre-Lie algebra structure on the Lie algebra $\mathfrak{g}$, see [11].

Definition 2.12.

Let $x·y$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$. If there exists a $\phi \in \text{End}(V)$ such that $$x·y=\{\phi (x),y\}$$ for all $x,y\in V$, then $x·y$ is called an inner PA-structure on $(\mathfrak{g},\mathfrak{n})$.

The following result is proved in [2], Corollary 5.6.

Proposition 2.13.

Let $(\mathfrak{n},\{,\},R)$ be a Lie algebra together with a Rota–Baxter operator R of weight 1, i.e., a linear operator satisfying $$\left\{R\right(x),R(y\left)\right\}=R\left(\right\{R\left(x\right),y\left\}+\right\{x,R\left(y\right)\left\}+\right\{x,y\left\}\right)$$ for all $x,y\in V$. Then $$x·y=\left\{R\right(x),y\}$$ defines an inner PA-structure on $(\mathfrak{g},\mathfrak{n})$, where the Lie bracket of $\mathfrak{g}$ is given by (9) $$[x,y]=\left\{R\right(x),y\}-\left\{R\right(y),x\}+\{x,y\}.$$(9)

Note that $\ker (R)$ is a subalgebra of $\mathfrak{n}$. For $x,y\in \ker (R)$ we have $R(\{x,y\})=0$. Recall that a Lie algebra is called complete, if it has trivial center and only inner derivations.

Proposition 2.14.

Let $\mathfrak{n}$ be a Lie algebra with trivial center. Then any inner PA-structure on $(\mathfrak{g},\mathfrak{n})$ arises by a Rota–Baxter operator of weight 1. Furthermore, if $\mathfrak{n}$ is complete, then every PA-structure on $(\mathfrak{g},\mathfrak{n})$ is inner.

Proof.

The first claim follows from Proposition 2.10 in [11]. By Lemma 2.9 in [11] every PA-structure on $(\mathfrak{g},\mathfrak{n})$ with complete Lie algebra $\mathfrak{n}$ is inner. The result can also be derived from the proof of Theorem 5.10 in [2].□

Corollary 2.15.

Let $\mathfrak{n}$ be a complete Lie algebra. Then there is bijection between PA-structures on $(\mathfrak{g},\mathfrak{n})$ and RB-operators of weight 1 on $\mathfrak{n}$.

As we have seen, any inner PA-structure on $(\mathfrak{g},\mathfrak{n})$ with $Z(\mathfrak{n})=0$ arises by a Rota–Baxter operator of weight 1. For Lie algebra $\mathfrak{n}$ with nontrivial center this need not be true.

Example 2.16.

Let $(e_1,e_2,e_3)$ be a basis of V and $\mathfrak{n}={\mathfrak{r}}_2(K)\oplus K$ with $\{e_1,e_2\}=e_2$. Then $$\phi =\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ \alpha & \beta & \gamma \end{array}\right)$$ defines an inner PA-structure on $(\mathfrak{g},\mathfrak{n})$ by $x·y=\{\phi (x),y\}$ with $\mathfrak{g}=\mathfrak{n}$, i.e., with $[e_1,e_2]=e_2$. But $\phi$ is not always a Rota–Baxter operator of weight 1 for $\mathfrak{n}$. It is easy to see that this is the case if and only if $\beta =0$.

Proposition 2.17.

Let $x·y$ be an inner PA-structure arising from an RB-operator R on $\mathfrak{n}$ of weight 1. Then R is also an RB-operator of weight 1 on $\mathfrak{g}$, i.e., it satisfies $$\left[R\right(x),R[y\left)\right]=R\left(\right[R\left(x\right),y\left]+\right[x,R\left(y\right)\left]+\right[x,y\left]\right)$$ for all $x,y\in V$.

Proof.

Because of $R([x,y])=\{R(x),R(y)\}$ and the definition of $[x,y]$ we have $$\begin{array}{l}R\left(\right[R\left(x\right),y\left]+\right[x,R\left(y\right)\left]+\right[x,y\left]\right)=\left\{R\right(R\left(x\right)),R(y\left)\right\}+\left\{R\right(x),R(R\left(y\right)\left)\right\}+\left\{R\right(x),R(y\left)\right\}\\ =\left[R\right(x),R(y\left)\right]\end{array}$$ for all $x,y\in V$.□

Corollary 2.18.

Let $x·y=\{R(x),y\}$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ defined by an RB-operator R of weight 1 on $\mathfrak{n}$. Denote by ${\mathfrak{g}}_i$ be the Lie algebra structure on V defined by $$\begin{array}{l}{[x,y]}_0=\{x,y\},\\ {[x,y]}_{i+1}={\left[R\right(x),y]}_i-{\left[R\right(y),x]}_i+{[x,y]}_i,\end{array}$$ for all $i\ge 0$. Then R defines a PA-structure on each pair $({\mathfrak{g}}_{i+1},{\mathfrak{g}}_i)$.

We have ${[x,y]}_1=[x,y]$, and both R and $R+\text{id}$ are Lie algebra homomorphisms from ${\mathfrak{g}}_{i+1}$ to ${\mathfrak{g}}_i$, see Proposition 7 in [33]. Hence, we obtain a composition of homomorphisms $${\mathfrak{g}}_i\underset{R+\text{id}}{\overset{R}{\to }}{\mathfrak{g}}_{i-1}\underset{R+\text{id}}{\overset{R}{\to }}\cdots \underset{R+\text{id}}{\overset{R}{\to }}{\mathfrak{g}}_0$$

So the kernels $\ker (R^i)$ and $\ker ({(R+\text{id})}^i)$ are ideals in ${\mathfrak{g}}_j$ for all $1\le i\le j$.

For a Lie algebra $\mathfrak{g}$, denote by ${\mathfrak{g}}^{(i)}$ the derived ideals defined by ${\mathfrak{g}}^{(1)}=\mathfrak{g}$ and ${\mathfrak{g}}^{(i+1)}=[{\mathfrak{g}}^{(i)},{\mathfrak{g}}^{(i)}]$ for $i\ge 1$. An immediate consequence of Proposition $2.13$ is the following observation.

Proposition 2.19.

Let $x·y=\{R(x),y\}$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ defined by an RB-operator R of weight 1 on $\mathfrak{n}$. Then we have $\dim {\mathfrak{g}}^{(i)}\le \dim {\mathfrak{n}}^{(i)}$ for all $i\ge 1$.

Corollary 2.20.

Let $x·y$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$, where $\mathfrak{n}$ is complete. Then we have $\dim {\mathfrak{g}}^{(i)}\le \dim {\mathfrak{n}}^{(i)}$ for all $i\ge 1$. In particular, if $\mathfrak{n}$ is solvable, so is $\mathfrak{g}$, and if $\mathfrak{g}$ is perfect, so is $\mathfrak{n}$.

Proof.

By Corollary $2.15$ this follows from the proposition.□

Proposition 2.21.

Let $x·y=\{R(x),y\}$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ defined by an RB-operator R of weight 1 on $\mathfrak{n}$. Then the following holds:

1. If $\mathfrak{g}$ and $\mathfrak{n}$ are not isomorphic, then both R and $R+\text{id}$ have a nontrivial kernel.

2. If either $\mathfrak{g}$ or $\mathfrak{n}$ is not solvable, then at least one of the operators R and $R+\text{id}$ has a nontrivial kernel.

Proof.

For (1), assume that $\ker (R)=0$. Then $R:\mathfrak{g}\to \mathfrak{n}$ is invertible, hence an isomorphism. This is a contradiction. The same is true for $R+\text{id}$. For (2) assume that $\ker (R)=\ker (R+\text{id})=0$. Then R and $R+\text{id}$ are isomorphisms from $\mathfrak{g}$ to $\mathfrak{n}$, and $\mathfrak{g}\cong \mathfrak{n}$. Then we can apply a result of Jacobson [24] to the automorphism $\psi :=(R+\text{id})°R^{-1}$ of $\mathfrak{n}$, because $\mathfrak{n}$ is not solvable. We obtain a nonzero fixed point $x\in \mathfrak{n}$, so that $$0=\psi \left(x\right)-x=\left(R+\text{id}\right)R^{-1}\left(x\right)-x=R^{-1}\left(x\right).$$

Since R is bijective, x = 0, a contradiction.□

Corollary 2.22.

Let $\mathfrak{n}$ be a simple Lie algebra and R be an invertible RB-operator of nonzero weight $\lambda$ on $\mathfrak{n}$. Then we have $R=-\lambda \text{id}$.

Proof.

By rescaling we may assume that R has weight 1. We obtain a PA-structure on $(\mathfrak{g},\mathfrak{n})$ by Proposition $2.13$, with Lie bracket (9) on $\mathfrak{g}$. Since $\mathfrak{n}$ is not solvable, either R or $R+\text{id}$ have a nontrivial kernel. But $\ker (R)=0$ by assumption, so that $\ker (R+\text{id})$ is a nontrivial ideal of $\mathfrak{n}$. Hence we have $R+\text{id}=0$.□

3. PA-structures on pairs of semisimple Lie algebras

We will assume that all algebras in this section are finite-dimensional. Let $x·y$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ over $\mathbb{C}$, where $\mathfrak{g}$ is simple and $\mathfrak{n}$ is semisimple. Then $\mathfrak{n}$ is also simple, and both $\mathfrak{g}$ and $\mathfrak{n}$ are isomorphic, see Proposition 4.9 in [11]. We have a similar result for $\mathfrak{n}$ simple and $\mathfrak{g}$ semisimple. However, its proof is more difficult than the first one.

Theorem 3.1.

Let $x·y$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ over $\mathbb{C}$, where $\mathfrak{n}$ is simple and $\mathfrak{g}$ is semisimple. Then $\mathfrak{g}$ is also simple, and both $\mathfrak{g}$ and $\mathfrak{n}$ are isomorphic.

Proof.

By Corollary $2.15$ we have $x·y=\{R(x),y\}$ for an RB-operator R of weight 1 on $\mathfrak{n}$. Assume that $\mathfrak{g}$ and $\mathfrak{n}$ are not isomorphic. By Proposition $2.21$ (2) both $\ker (R)$ and $\ker (R+\text{id})$ are proper nonzero ideals of $\mathfrak{g}$, with $\ker (R)\cap \ker (R+\text{id})=0$. So we have $$\mathfrak{g}=\ker \left(R\right)\oplus \ker \left(R+\text{id}\right)\oplus \mathfrak{s}$$ with a semisimple ideal $\mathfrak{s}$. We have $\mathfrak{n}=\text{im}(R)+\text{im}(R+\text{id})$ because of $x=R(-x)+(R+\text{id})(x)$ for all $x\in \mathfrak{n}$, and $$\begin{array}{l}\text{im}\left(R\right)\cong \mathfrak{g}/\ker \left(R\right)\cong \ker \left(R+\text{id}\right)\oplus \mathfrak{s},\\ \text{im}\left(R+\text{id}\right)\cong \mathfrak{g}/\ker \left(R+\text{id}\right)\cong \ker \left(R\right)\oplus \mathfrak{s}.\end{array}$$

This yields a semisimple decomposition $$\mathfrak{n}=\left(\ker \right(R+\text{id})\oplus \mathfrak{s})+\left(\ker \right(R)\oplus \mathfrak{s}).$$

Suppose that $\mathfrak{s}$ is nonzero. Then both summands are not simple. This is a contradiction to Theorem 4.2 in Onishchik’s paper [29], which says that at least one summand in a semisimple decomposition of a simple Lie algebra must be simple. Hence we obtain $\mathfrak{s}=0,\text{im}(R)=\ker (R+\text{id}),\text{im}(R+\text{id})=\text{im}(R)$ and $$\mathfrak{n}=\text{im}\left(R\right)\dot{+}\text{im}\left(R+\text{id}\right).$$

Then the main result of Koszul’s note [27] implies that $\mathfrak{n}=\text{im}(R)\oplus \text{im}(R+\text{id})$, which is a contradiction to the simplicity of $\mathfrak{n}$. Hence $\mathfrak{g}$ and $\mathfrak{n}$ are isomorphic.□

If $\mathfrak{g}$ is semisimple with only two simple summands, we can prove the same result for any field K of characteristic zero.

Proposition 3.2.

Let $x·y$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$, where $\mathfrak{n}$ is semisimple, and $\mathfrak{g}={\mathfrak{s}}_1\oplus {\mathfrak{s}}_2$ is the direct sum of two simple ideals of $\mathfrak{g}$. Then $\mathfrak{g}$ and $\mathfrak{n}$ are isomorphic.

The proof is the same as before. The only argument where we needed the complex numbers was the result of [29], which we do not need here.

Let $\mathfrak{n}={\mathfrak{s}}_1\oplus {\mathfrak{s}}_2$ be a direct sum of two simple isomorphic ideals ${\mathfrak{s}}_1$ and ${\mathfrak{s}}_2$. We would like to find all RB-operators of weight 1 on $\mathfrak{n}$ such that $\mathfrak{g}$ with bracket (9) is isomorphic to $\mathfrak{n}$.

Proposition 3.3.

All PA-structures on $(\mathfrak{g},\mathfrak{n})$ with $\mathfrak{g}\cong \mathfrak{n}={\mathfrak{s}}_1\oplus {\mathfrak{s}}_2$, where ${\mathfrak{s}}_1$ and ${\mathfrak{s}}_2$ simple isomorphic ideals of $\mathfrak{n}$, arise by the trivial RB-operators or by one of the following RB-operators R on $\mathfrak{n}$, and $\psi \in \text{Aut}(\mathfrak{n})$, $$\begin{array}{l}R((s_1,s_2))=(-s_1,-\psi (s_1)),\\ R((s_1,s_2))=(0,\psi (s_1)),\\ R((s_1,s_2))=(-s_1,0)),\end{array}$$ up to permuting the factors and application of $\phi (R)=-R-\text{id}$ to these operators.

Proof.

By Proposition $2.5$ and Proposition $2.7$ the given operators are RB-operators of weight 1 on $\mathfrak{n}$, because R is. By Proposition $2.21$ at least one of $\ker (R)$ and $\ker (R+\text{id})$ is nonzero. Suppose first that both $\ker (R)$ and $\ker (R+\text{id})$ are zero. Then we have $\mathfrak{g}=\ker (R)\oplus \ker (R+\text{id})$ and $\mathfrak{n}=\ker (R)\dot{+}\ker (R+\text{id})$. It is easy to see that $\ker (R)$ coincides with ${\mathfrak{s}}_1$ or ${\mathfrak{s}}_2$ by using the Theorem of Koszul [27]. Applying $\phi$ if necessary, we can assume that $\ker (R)={\mathfrak{s}}_2$. Then again by Koszul’s result we have $R((s_1,s_2))=({\psi }_1(s_1),{\psi }_2(s_1))$ or $R((s_1,s_2))=({\psi }_1(s_1),0))$ for some ${\psi }_1,{\psi }_2\in \text{Aut}(\mathfrak{n})$. Since $\text{im}(R)=\ker (R+\text{id})$ we either have $R((s_1,s_2))=(-s_1,-\psi (s_1))$ or $R((s_1,s_2))=(-s_1,0)$.

In the second case, one of the kernels is zero. Applying $\phi$ if necessary, we may assume that $\ker (R+\text{id})=0$ and $\ker (R)={\mathfrak{s}}_1$. Then $\mathfrak{g}/\ker (R)$ is a simple Lie algebra, and $-R-\text{id}$ is an invertible RB-operator of weight 1 on $\mathfrak{g}/\ker (R)$. By Corollary $2.22$ we obtain $-R-\text{id}=-\text{id}$, hence R = 0 on $\mathfrak{g}/\ker (R)$. This implies $R^2=0$ on $\mathfrak{g}$. The projections of $\text{im}(R)$ to ${\mathfrak{s}}_1$ and ${\mathfrak{s}}_2$ are either zero or an isomorphism on one factor. So we have $R((s,0))=(0,\psi (s))$ or $R((s,0))=({\psi }_1(s),{\psi }_2(s))$ for some automorphisms $\psi ,{\psi }_1,{\psi }_2$. But the second operator does not satisfy $R^2=0$, and hence is impossible. Therefore we are done.□

Proposition 3.4.

Let $x·y=\{R(x),y\}$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ defined by an RB-operator R of weight 1 on $\mathfrak{n}$. Let ${\mathfrak{n}}_1=\ker (R^n),{\mathfrak{n}}_2=\ker {(R+\text{id})}^n,{\mathfrak{n}}_3=\text{im}(R^n)\cap \text{im}({(R+\text{id})}^n)$ for $n=\dim (V)$. Then $\mathfrak{n}={\mathfrak{n}}_1\dot{+}{\mathfrak{n}}_2\dot{+}{\mathfrak{n}}_3$ with $\{{\mathfrak{n}}_1,{\mathfrak{n}}_3\}\subseteq {\mathfrak{n}}_1,\{{\mathfrak{n}}_2,{\mathfrak{n}}_3\}\subseteq {\mathfrak{n}}_2$, and ${\mathfrak{n}}_3$ is solvable.

Proof.

We first show by induction that $\ker (R^i)$ is a subalgebra of $\mathfrak{n}$, and that $$\left\{\ker \right(R^i),\text{im}({\left(R+\text{id}\right)}^i\left)\right\}\subseteq \ker \left(R^i\right)$$ for all $i\ge 1$. The case i = 1 goes as follows. We already know that $\ker (R)$ is a subalgebra of $\mathfrak{n}$. So we have to show that $\{\ker (R),\text{im}(R+\text{id})\}\subseteq \ker (R)$. Let $x\in \ker (R)$ and $y\in \mathfrak{n}$. Then by (6) we have $$\begin{array}{l}\{x,(R+\text{id}\left)\right(y\left)\right\}=\{x,R(y\left)\right\}+\{x,y\}\\ =[x,y]+\{y,R(x\left)\right\}\\ =[x,y],\end{array}$$ which is in $\ker (R)$, since this is an ideal in $\mathfrak{g}$. For the induction step $i\mapsto i+1$ consider the iteration of the Lie bracket (9) for all $i\ge 0$, given by $${[x,y]}_i={[x,y]}_{i+1}-{\left[R\right(x),y]}_i-{[x,R(y\left)\right]}_i$$ for all $i\ge 0$. Then $$\begin{array}{l}\{x,y\}={[x,y]}_1-{\left[R\right(x),y]}_0-{[x,R(y\left)\right]}_0\\ ={[x,y]}_2-{\left[R^2\right(x),y]}_0-2{\left[R\right(x),y]}_0-2{\left[R\right(x),R(y\left)\right]}_0-2{[x,R(y\left)\right]}_0-{[x,R^2(y\left)\right]}_0\end{array}$$ and so on. Define a degree of a term ${[R^l(x),R^k(y)]}_m$ by l + k + m, and let $x,y\in \ker (R^{i+1})$. We can iterate the brackets, until the degree of every summand on the right-hand side will be greater than 3i, so that all summands either have a term $R^l(x)$ with l > i, or a term $R^k(y)$ with k > i, or all summands lie in ${[\ker (R^{i+1}),\ker (R^{i+1})]}_{i+1}$. By induction hypothesis, such terms will vanish for l > i or k > i, and since $\ker (R^{i+1})$ is an ideal in ${\mathfrak{g}}_{i+1}$, we have $\{x,y\}\in \ker (R^{i+1})$, so that $\ker (R^{i+1})$ is a subalgebra of $\mathfrak{n}$. The induction step for the second claim follows similarly.

Since the image of a subalgebra under the action of an RB-operator is a subalgebra, ${\mathfrak{n}}_1,{\mathfrak{n}}_2$ and their intersection ${\mathfrak{n}}_3$ are subalgebras of $\mathfrak{n}$. We want to show that $\mathfrak{n}={\mathfrak{n}}_1\dot{+}{\mathfrak{n}}_2\dot{+}{\mathfrak{n}}_3$. Because of $\ker (R^n)\cap \text{im}(R^n)=0$ we have $\mathfrak{n}=\ker (R^n)\dot{+}\text{im}(R^n)$. In the same way we have $\mathfrak{n}=\ker ({(R+\text{id})}^n)\dot{+}\text{im}({(R+\text{id})}^n)$. We obtain $$\text{im}\left(R^n\right)\cap \ker \left({\left(R+\text{id}\right)}^n\right)\dot{+}\text{im}\left(R^n\right)\cap \text{im}\left({\left(R+\text{id}\right)}^n\right)\subseteq \text{im}\left(R^n\right).$$

We claim that $\ker ({(R+\text{id})}^n)\subseteq \text{im}(R^n)$, so that we have equality above. Indeed, for $x\in \ker ({(R+\text{id})}^n)$ we have by the binomial formula $$x+\left(\begin{array}{c}n\\ n-1\end{array}\right)R\left(x\right)+\cdots +\left(\begin{array}{c}n\\ 1\end{array}\right)R^{n-1}\left(x\right)=-R^n\left(x\right)\in \text{im}\left(R^n\right).$$

Applying $R^{n-1}$ we obtain $R^{n-1}(x)\in \text{im}(R^n)$ and $$x+nR\left(x\right)+\cdots +\left(\begin{array}{c}n\\ 2\end{array}\right)R^{n-2}\left(x\right)\in \text{im}\left(R^n\right).$$

Iterating this we obtain $x\in \text{im}(R^n)$. This yields $$\begin{array}{l}\mathfrak{n}=\ker \left(R^n\right)\dot{+}\text{im}\left(R^n\right)\\ =\ker \left(R^n\right)\dot{+}\ker \left({\left(R+\text{id}\right)}^n\right)\dot{+}\text{im}\left(R^n\right)\cap \text{im}\left({\left(R+\text{id}\right)}^n\right)\\ ={\mathfrak{n}}_1\dot{+}{\mathfrak{n}}_2\dot{+}{\mathfrak{n}}_3.\end{array}$$

On ${\mathfrak{n}}_3$ both operators R and $R+\text{id}$ are invertible. By Proposition $2.21$ part (2) it follows that ${\mathfrak{n}}_3$ is solvable.□

Corollary 3.5.

The decomposition $\mathfrak{n}={\mathfrak{n}}_1\dot{+}{\mathfrak{n}}_2\dot{+}{\mathfrak{n}}_3$ induces a decomposition ${\mathfrak{g}}_i={\mathfrak{n}}_1\dot{+}{\mathfrak{n}}_2\dot{+}{\mathfrak{n}}_3$ for each $i\ge 1$ with the same properties as in the Proposition. The Lie algebras $({\mathfrak{n}}_j,{[,]}_i)$ and $({\mathfrak{n}}_j,{[,]}_0)$ are isomorphic for j = 1, 2, 3.

Proof.

Since R and $R+\text{id}$ are RB-operators on all ${\mathfrak{g}}_i$, we obtain the same decomposition with the same subalgebras. Note that $R+\text{id}$ is invertible on ${\mathfrak{n}}_1$, R is invertible on ${\mathfrak{n}}_2$ and both are invertible on ${\mathfrak{n}}_3$. In order to show that $({\mathfrak{n}}_1,{[,]}_i$ is isomorphic to $({\mathfrak{n}}_1,{[,]}_0$, we consider a chain of isomorphisms $$({\mathfrak{n}}_1,{[,]}_n)\stackrel{R+\text{id}}{\to }({\mathfrak{n}}_1,{[,]}_{n-1})\stackrel{R+\text{id}}{\to }\cdots \stackrel{R+\text{id}}{\to }({\mathfrak{n}}_1,{[,]}_0).$$

In a similar way we can deal with ${\mathfrak{n}}_2$ and ${\mathfrak{n}}_3$.□

Proposition 3.6.

Let $\mathfrak{g}={\mathfrak{s}}_1+{\mathfrak{s}}_2$ be the vector space sum of two complex semisimple subalgebras of $\mathfrak{g}$. Then $\mathfrak{g}$ is semisimple.

Proof.

Suppose that the claim is not true and let $\mathfrak{g}$ be a counterexample of minimal dimension. Then $\mathfrak{g}$ contains a nonzero abelian ideal $\mathfrak{a}$. Then we obtain $$\mathfrak{g}/\mathfrak{a}={\mathfrak{s}}_1/({\mathfrak{s}}_1\cap \mathfrak{a})+{\mathfrak{s}}_2/({\mathfrak{s}}_2\cap \mathfrak{a}).$$

Since ${\mathfrak{s}}_1\cap \mathfrak{a}$ is an abelian ideal ${\mathfrak{s}}_1$, it must be zero, i.e., ${\mathfrak{s}}_1\cap \mathfrak{a}=0$. In the same way we have ${\mathfrak{s}}_2\cap \mathfrak{a}=0$. Hence we obtain a semisimple decomposition of $\mathfrak{g}/\mathfrak{a}$ with $\dim (\mathfrak{g}/\mathfrak{a})<\dim (\mathfrak{g})$. If $\mathfrak{g}/\mathfrak{a}$ is semisimple, this is a contradiction to the minimality of the counterexample $\mathfrak{g}$. Otherwise we may assume that $\mathfrak{g}$ has one-dimensional solvable radical. Then $\mathfrak{g}$ is reductive, and by Theorem 3.2 of [29], there are no semisimple decompositions of a complex reductive non-semisimple Lie algebra. Hence we are done.□

Proposition 3.7.

Let $x·y=\{R(x),y\}$ be a PA-structure on $(\mathfrak{g},\mathfrak{n})$ over $\mathbb{C}$, where $\mathfrak{n}$ is simple, defined by an RB-operator R of weight 1 on $\mathfrak{n}$, with associated Lie algebras ${\mathfrak{g}}_i$ for $i=1,\dots ,n=\dim (V)$. Assume that ${\mathfrak{g}}_0=\mathfrak{n}$ and ${\mathfrak{g}}_n$ are semisimple. Then all ${\mathfrak{g}}_i$ are isomorphic to $\mathfrak{n}$.

Proof.

Since ${\mathfrak{n}}_1$ and ${\mathfrak{n}}_2$ are kernels of homomorphisms, they are ideals in ${\mathfrak{g}}_n$. The quotient ${\mathfrak{g}}_n/({\mathfrak{n}}_1+{\mathfrak{n}}_2)\cong {\mathfrak{n}}_3$ is semisimple and solvable by Proposition $3.4$. Hence ${\mathfrak{n}}_3=0$, and we obtain ${\mathfrak{g}}_n=\ker (R^n)\oplus \ker ({(R+\text{id})}^n)$. Because of Corollary $3.5$ we have the decomposition ${\mathfrak{g}}_i=\ker (R^n)\dot{+}\ker ({(R+\text{id})}^n)$ for all i < n, where all Lie algebras $(\ker (R^n),{[,]}_i)$ are isomorphic, and all Lie algebras $(\ker ({(R+\text{id})}^n),{[,]}_i)$ are isomorphic. By Proposition $3.6$ all ${\mathfrak{g}}_i$ are semisimple. By Koszul’s result [27], all ${\mathfrak{g}}_i$ are isomorphic.□

Proposition 3.8.

Suppose that there is a post-Lie algebra structure on $(\mathfrak{g},\mathfrak{n})$ over $\mathbb{C}$, where $\mathfrak{g}$ is semisimple and $\mathfrak{n}$ is complete. Then $\mathfrak{n}$ must be semisimple.

Proof.

By Corollary $2.15$ the PA-structure is given by $x·y=\{R(x),y\}$, where R is an RB-operator of weight 1 on $\mathfrak{n}$. If at least one of $\ker (R)$ and $\ker (R+\text{id})$ is trivial, we obtain $\mathfrak{g}\cong \mathfrak{n}$ by Proposition $2.21$, part (1). Otherwise $\mathfrak{n}=\text{im}(R)+\text{im}(R+\text{id})$ is the sum of two nonzero semisimple subalgebras. By Proposition $3.6$ $\mathfrak{n}$ is semisimple.□

4. PA-structures on $(\mathfrak{g},\mathfrak{n})$ with $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\times \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$

In [11], Proposition 4.7 we have shown that PA-structures with $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ exist on $(\mathfrak{g},\mathfrak{n})$ if and only if $\mathfrak{g}$ is isomorphic to $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$, or to one of the solvable non-unimodular Lie algebras ${\mathfrak{r}}_{3,\lambda }(\mathbb{C})$ for $\lambda \in \mathbb{C}\setminus \{-1\}$. In this section we want to show an analogous result for $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\times \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$. Here we will use RB-operators on $\mathfrak{n}$ and an explicit classification by Douglas and Repka [17] of all subalgebras of $\mathfrak{n}$. This classification is up to inner automorphisms, but we will only need the subalgebras up to isomorphisms. Let us fix a basis $(X_1,Y_1,H_1,X_2,Y_2,H_2)$ of $\mathfrak{n}$ consisting of the following 4 × 4 matrices: $$X_1=E_{12}, Y_1=E_{21}, H_1=E_{11}-E_{22}, X_2=E_{34}, Y_2=E_{43}, H_2=E_{33}-E_{44}.$$

We use in Table 1.

Table 1. Complex 3-dimensional Lie algebras.

$\mathfrak{g}$	Lie brackets
${\mathbb{C}}^3$	–
${\mathfrak{n}}_3(\mathbb{C})$	$[e_1,e_2]=e_3$
${\mathfrak{r}}_2(\mathbb{C})\oplus \mathbb{C}$	$[e_1,e_2]=e_2$
${\mathfrak{r}}_3(\mathbb{C})$	$[e_1,e_2]=e_2, [e_1,e_3]=e_2+e_3$
${\mathfrak{r}}_{3,\lambda }(\mathbb{C}), \lambda \ne 0$	$[e_1,e_2]=e_2, [e_1,e_3]=\lambda e_3$
$\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$	$[e_1,e_2]=e_3, [e_1,e_3]=-2e_1, [e_2,e_3]=2e_2$

Table 2. Solvable subalgebras.

$\dim (\mathfrak{h})$	Representative	Isomorphism type
1	$〈X_1〉, 〈H_1〉, 〈X_1+X_2〉, 〈X_1+H_2〉, 〈H_1+a{H}_2〉, a\in {\mathbb{C}}^{*}$	$\mathbb{C}$
2	$〈X_1,X_2〉, 〈X_1,H_2〉, 〈H_1,H_2〉$	${\mathbb{C}}^2$
2	$〈X_1+X_2,H_1+H_2〉, 〈X_1,H_1+X_2〉, 〈X_1,H_1+a{H}_2〉, a\in \mathbb{C}$	${\mathfrak{r}}_2(\mathbb{C})$
3	$〈X_1,X_2,H_1+\lambda H_2〉, \lambda \ne 0$	${\mathfrak{r}}_{3,\lambda }(\mathbb{C}),\lambda \ne 0$
3	$〈X_1,H_1,H_2〉, 〈X_1,H_1,X_2〉$	${\mathfrak{r}}_2(\mathbb{C})\oplus \mathbb{C}$
4	$〈X_1,H_1,X_2,H_2〉$	${\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$

Among the family ${\mathfrak{r}}_{3,\lambda }(\mathbb{C}),\lambda \ne 0$ there are still isomorphisms. In fact, ${\mathfrak{r}}_{3,\lambda }(\mathbb{C})\cong {\mathfrak{r}}_{3,\mu }(\mathbb{C})$ if and only if $\mu ={\lambda }^{-1}$ or $\mu =\lambda$. The list of subalgebras $\mathfrak{h}$ of $\mathfrak{n}$ is given as follows. We first list the solvable subalgebras, then the semisimple ones and the subalgebras with a nontrivial Levi decomposition.

Theorem 4.1.

Suppose that there exists a post-Lie algebra structure on $(\mathfrak{g},\mathfrak{n})$, where $\mathfrak{n}=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$. Then $\mathfrak{g}$ is isomorphic to one of the following Lie algebras, and all these possibilities do occur:

1. $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$.

2. $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus {\mathfrak{r}}_{3,\lambda }(\mathbb{C}), \lambda \ne -1$.

3. ${\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus {\mathfrak{r}}_{3,\mu }(\mathbb{C}), (\lambda ,\mu )\ne (-1,-1)$.

4. ${\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$.

5. ${\mathfrak{r}}_2(\mathbb{C})\oplus ({\mathbb{C}}^3⋉\mathbb{C})=〈x_1,\dots ,x_6〉$ and Lie brackets, for $\alpha \ne 0,\beta \ne 0,-1$ $$[x_1,x_2]=x_1, [x_3,x_6]=x_3, [x_4,x_6]=\alpha x_4, [x_5,x_6]=\beta x_5.$$

6. $\mathbb{C}\oplus (({\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus \mathbb{C})⋉\mathbb{C})=〈x_1,\dots ,x_6〉$ and Lie brackets, for $\lambda \ne 0,\alpha \ne 0,-1$, $$[x_2,x_4]=x_2, [x_3,x_4]=\lambda x_3, [x_3,x_6]=x_3, [x_5,x_6]=\alpha x_5.$$

7. $({\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus {\mathbb{C}}^2)⋉\mathbb{C}=〈x_1,\dots ,x_6〉$ and Lie brackets, for $\lambda \ne 0,{\alpha }_1,{\alpha }_2\ne 0$, and $(\lambda ,{\alpha }_1,{\alpha }_2)\ne (-1,{\alpha }_1,-{\alpha }_1-1)$, $$[x_1,x_3]=x_1, [x_2,x_3]=\lambda x_2, [x_2,x_6]={\alpha }_1{x}_2, [x_4,x_6]=x_4, [x_5,x_6]={\alpha }_2{x}_5.$$

8. $({\mathbb{C}}^2\oplus {\mathbb{C}}^2)⋉{\mathbb{C}}^2=〈x_1,\dots ,x_6〉$ and Lie brackets $$\begin{array}{c}[x_1,x_5]=x_1,[x_2,x_5]={\alpha }_2{x}_2, [x_3,x_5]={\alpha }_4{x}_3, [x_4,x_5]={\alpha }_6{x}_4,\\ [x_1,x_6]={\alpha }_1{x}_1, [x_2,x_6]={\alpha }_3{x}_2, [x_3,x_6]={\alpha }_5{x}_3, [x_4,x_6]={\alpha }_7{x}_4,\end{array}$$

with one of the following conditions:

1. $(a)\hspace{1em}{\alpha }_3=1, {\alpha }_5={\alpha }_1{\alpha }_7, {\alpha }_6={\alpha }_2{\alpha }_4,{\alpha }_1{\alpha }_2\ne 1, {\alpha }_4,{\alpha }_7\ne 0,-1,$

2. $(b)\hspace{1em}{\alpha }_4={\alpha }_1-1, {\alpha }_5=-{\alpha }_1, {\alpha }_6={\alpha }_2({\alpha }_1-1), {\alpha }_7={\alpha }_1{\alpha }_3-{\alpha }_1^2{\alpha }_2-{\alpha }_3,\hspace{1em}{\alpha }_3-{\alpha }_1{\alpha }_2\ne 0, {\alpha }_1\ne 0,1.$

Proof.

By Corollary $2.15$ it is enough to consider the RB-operators R of weight 1 on $\mathfrak{n}$. Then $\ker (R)$ and $\ker (R+\text{id})$ are ideals in $\mathfrak{g}$. If R is trivial, or one of the kernels is trivial, then we have $\mathfrak{g}\cong \mathfrak{n}$, which is type (1). So we assume that R is nontrivial, both $\ker (R)$ and $\ker (R+\text{id})$ are nonzero, and $\dim (\ker (R))\ge \dim (\ker (R+\text{id}))$. Then, for $\mathfrak{n}\cancel{\cong }\mathfrak{g}$, either $\mathfrak{g}$ has a nontrivial Levi decomposition, or $\mathfrak{g}$ is solvable.

Case 1: Assume that $\mathfrak{g}$ has a nontrivial Levi decomposition, i.e., that $\mathfrak{g}\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})⋉\mathfrak{r}$. We claim that $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is a direct summand of $\mathfrak{g}$, i.e., $\mathfrak{g}\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{r}$, and that $\mathfrak{r}$ is not isomorphic to ${\mathfrak{r}}_3(\mathbb{C})$. Then we can argue as follows. Because of Remark 2.12 of [12], $\mathfrak{g}$ cannot be unimodular, except for $\mathfrak{g}\cong \mathfrak{n}$. Thus $\mathfrak{r}$ cannot be unimodular, so that $\mathfrak{g}$ is isomorphic to $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus {\mathfrak{r}}_{3,\lambda }(\mathbb{C})$ with $\lambda \ne -1$. On the other hand, all such algebras do arise by Proposition $2.6$ and Proposition 4.7 of [11].

Case 1a: Suppose that $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is not contained in $\ker (R),\ker (R+\text{id})$ as a subalgebra. Then $\dim (\ker (R+\text{id}))=1$ and $\dim (\ker (R))\in \{1,2\}$. Let us assume, both have dimension 1. The other case goes similarly. Then we have $\mathfrak{r}=〈x_1,x_2,x_3〉,\ker (R)=〈x_1〉$ and $\ker (R+\text{id})=〈x_2〉$. Furthermore $\text{im}(R)\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})⋉〈x_2,x_3〉$ and $\text{im}(R+\text{id})\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})⋉〈x_1,x_3〉$ are five-dimensional subalgebras of $\mathfrak{n}$. By Table 3, $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is a direct summand of them. This implies that $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is also a direct summand in $\mathfrak{g}$. Since both $\ker (R)$ and $\ker (R+\text{id})$ are ideals in $\mathfrak{r}$, we can exclude that $\mathfrak{r}$ is isomorphic to ${\mathfrak{r}}_3(\mathbb{C})$, and we are done.

Table 3. Semisimple subalgebras and Levi decomposable subalgebras.

$\dim (\mathfrak{h})$	Representative	Isomorphism type
3	$〈X_1,Y_1,H_1〉, 〈X_1+X_2,Y_1+Y_2,H_1+H_2〉$	$\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$
4	$〈X_1,Y_1,H_1,H_2〉, 〈X_1,Y_1,H_1,X_2〉$	$\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathbb{C}$
5	$〈X_1,Y_1,H_1,X_2,H_2〉$	$\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$

Case 1b: $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is contained in one of $\ker (R),\ker (R+\text{id})$. Without loss of generality we may assume that $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\subseteq \ker (R)$. If $\ker (R)=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$, then $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is an ideal of $\mathfrak{g}$, and we have $\mathfrak{g}\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{r}$, where $\mathfrak{r}\cong \text{im}(R)\le \mathfrak{n}$ is not isomorphic to ${\mathfrak{r}}_3(\mathbb{C})$ by Table 2, and we are done. Thus we may assume that $\dim (\ker (R))\ge 4$. If R splits with subalgebras $\ker (R)$ and $\ker (R+\text{id})$, then $\mathfrak{g}\cong \ker (R)\oplus \ker (R+\text{id})$, and $\dim (\ker (R))+\dim (\ker (R+\text{id}))=6$. By Table 3, $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is a direct summand of $\ker (R)$, and hence of $\mathfrak{g}$. So we have again $\mathfrak{g}\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{r}$, and $\mathfrak{r}$ is not isomorphic to ${\mathfrak{r}}_3(\mathbb{C})$. If R is not split, it remains to consider the case $\dim (\ker (R))=4$ and $\dim (\ker (R+\text{id}))=1$. We have $\mathfrak{r}=〈x,y,z〉$ with $\ker (R)=\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus 〈x〉,\ker (R+\text{id})=〈y〉$ and $[y,\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})]=0$. Assume that $[z,\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})]\ne 0$. Then $\mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})$ is not a direct summand of the five-dimensional subalgebra $\text{im}(R+\text{id})$ of $\mathfrak{n}$, which is a contradiction to Table 3. Thus we have $\mathfrak{g}\cong \mathfrak{s}{\mathfrak{l}}_2(\mathbb{C})\oplus \mathfrak{r}$. Since $\mathfrak{r}$ has two disjoint one-dimensional ideals $〈x〉$ and $〈y〉$, it is not isomorphic to ${\mathfrak{r}}_3(\mathbb{C})$.

Case 2: Assume that $\mathfrak{g}$ is solvable. Then $\text{im}(R)$ and $\text{im}(R+\text{id})$ are solvable subalgebras of $\mathfrak{n}$ of dimension at most 4 by Table 2. So we have $\dim (\ker (R))\ge \dim (\ker (R+\text{id}))\ge 2$. Thus we have the following four cases: $$\begin{array}{c}\left(2a\right)\hspace{1em}\dim \left(\ker \right(R\left)\right)=4,\dim \left(\ker \right(R+\text{id}\left)\right)=2,\\ \left(2b\right)\hspace{1em}\dim \left(\ker \right(R\left)\right)=3,\dim \left(\ker \right(R+\text{id}\left)\right)=3,\\ \left(2c\right)\hspace{1em}\dim \left(\ker \right(R\left)\right)=3,\dim \left(\ker \right(R+\text{id}\left)\right)=2,\\ \left(2d\right)\hspace{1em}\dim \left(\ker \right(R\left)\right)=2,\dim \left(\ker \right(R+\text{id}\left)\right)=2.\end{array}$$

For the cases $(2a)$ and $(2b)$, R is split since the dimensions add up to 6. Then $\mathfrak{g}$ is a direct sum of two solvable subalgebras, which are both isomorphic to subalgebras of $\mathfrak{n}$. So we have $\mathfrak{n}=\ker (R)\dot{+}\ker (R+\text{id})$ and $\mathfrak{g}=\ker (R)\oplus \ker (R+\text{id})$.

Case 2a: Since we have only ${\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$ as four-dimensional solvable subalgebra of $\mathfrak{n}$, we have $\mathfrak{g}\cong {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathbb{C}}^2$, which is of type (3) for $(\lambda ,\mu )=(0,0)$, or $\mathfrak{g}\cong {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$, which is of type (4). Both cases can arise. For the first one we will show this in case $(2b)$. For the second, it follows from Proposition $2.7$ with $\mathfrak{n}=〈X_1,H_1,X_2,H_2〉\dot{+}〈Y_1,Y_2+H_1〉$.

Case 2b: We have $\mathfrak{g}\cong {\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus {\mathfrak{r}}_{3,\mu }(\mathbb{C})$. The case $(\lambda ,\mu )=(-1,-1)$ cannot arise by Theorem 3.3 of [11]. The cases $(\lambda ,\mu )=(-1,\mu )$ for $\mu \ne -1$ arise by Proposition $2.7$ with $$\mathfrak{n}=〈X_1,X_2,H_1-H_2〉\dot{+}〈Y_1,Y_2,H_1+\mu H_2〉.$$

The other cases with $\lambda ,\mu \ne -1$ arise by Proposition $2.6$ and Proposition 4.7 of [11].

Case 2c: Here $\mathfrak{g}$ is isomorphic to $({\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C}))⋊\mathbb{C}$ or $({\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus {\mathbb{C}}^2)⋊\mathbb{C}$. In the first case, ${\mathfrak{r}}_2(\mathbb{C})⋊\mathbb{C}\cong \text{im}(R)$ is a solvable subalgebra of $\mathfrak{n}$, hence isomorphic to ${\mathfrak{r}}_{3,\nu }(\mathbb{C})$ by Table 2. So $\mathbb{C}$ acts trivially on ${\mathfrak{r}}_2(\mathbb{C})$, and $\text{im}(R+\text{id})\cong {\mathfrak{r}}_{3,\lambda }(\mathbb{C})⋊\mathbb{C}\cong {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$. Then $\mathfrak{g}\cong {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$, which we have already considered in Case $(2a)$. For $({\mathfrak{r}}_{3,\lambda }(\mathbb{C})\oplus {\mathbb{C}}^2)⋊\mathbb{C}$ we need to distinguish $\lambda =0$ and $\lambda \ne 0$.

Case 2c, $\lambda =0$: By Proposition $2.3$ we may assume that $\text{im}(R+\text{id})=〈X_1,H_1,X_2,H_2〉$. Since $\ker (R)$ is an ideal of $\text{im}(R+\text{id})$ isomorphic to ${\mathfrak{r}}_2(\mathbb{C})\oplus \mathbb{C}$, we have $\ker (R)=〈X_1,H_1,X_2〉$. Let us consider the characteristic polynomial χ_R of the linear operator R acting on $\mathfrak{n}$. By assumption on the kernels, ${\chi }_R(t)=t^3{(t+1)}^2(t-\rho )$.

Case 2c, $\lambda =0,\rho \ne 0,-1$: Then $R(x_6)=\rho x_6$ for $x_6=H_2+\alpha H_1+\beta X_1+\gamma X_2$. Since $\ker (R+\text{id})$ is an abelian two-dimensional subalgebra of $\mathfrak{n}$, we have $$\ker \left(R+\text{id}\right)=〈Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2〉.$$

We want to compute $[x,y]$ for x = x₆ and $y\in \ker (R+\text{id})$. By Proposition $2.13$ we have, using $R(x_6)=\rho x_6$ $$\begin{array}{l}[x,y]=\left\{R\right(x),y\}-\left\{R\right(y),x\}+\{x,y\}\\ =\left\{R\right(x),y\}\\ =\rho \{x,y\}.\end{array}$$

For $x_6=H_2+\alpha H_1+\beta X_1+\gamma X_2$ and $y\in \ker (R+\text{id})$ this yields, using the Lie brackets of $\mathfrak{n}$ in the standard basis $\{X_1,Y_1,H_1,X_2,Y_2,H_2\}$, (10) $$[x_6,Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1]=\rho \left(\right(2\alpha {\nu }_1-2\beta {\nu }_2)X_1-2\alpha Y_1+\beta H_1),$$(10) (11) $$[x_6,Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2]=\rho \left(\right(2{\nu }_3-2\gamma {\nu }_4)X_2-2Y_2+\gamma H_2).$$(11)

Since $\ker (R+\text{id})$ is an ideal in $\mathfrak{g}$ and $\rho \ne 0$, both vectors lie again in $\ker (R+\text{id})$. Comparing coefficients for the basis vectors we obtain $$\beta =-2\alpha {\nu }_2, \alpha \left({\nu }_1+{\nu }_2^2\right)=0, \gamma =-2{\nu }_4, {\nu }_3=-{\nu }_4^2.$$

Suppose that $\alpha =0$. Then $x_6=H_2-2{\nu }_4{X}_2$ and $〈X_1,H_1〉\cong {\mathfrak{r}}_2(\mathbb{C})$ is a direct summand of $\mathfrak{g}$. Therefore $\mathfrak{g}\cong {\mathfrak{r}}_2(\mathbb{C})\oplus \mathbb{C}\oplus {\mathfrak{r}}_{3,\mu }(\mathbb{C})$ with $\mathbb{C}=〈Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1〉$, ${\mathfrak{r}}_{3,\mu }(\mathbb{C})=〈X_2,H_2-2{\nu }_4{X}_2,Y_2+{\nu }_4{H}_2-{\nu }_4^2{X}_2〉,\mu =-(\rho +1)/\rho$, which we have already considered above. Hence we may assume that $\alpha \ne 0$ and ${\nu }_1=-{\nu }_2^2$. Consider a new basis for $\mathfrak{g}$ (note that we redefine x₆) given by $$\begin{array}{c}(x_1,\dots ,x_6)=(X_1,-\frac{1}{2}{H}_1+{\nu }_2{X}_1,X_2,Y_1+{\nu }_2{H}_1-{\nu }_2^2{X}_1,Y_2+{\nu }_4{H}_2-{\nu }_4^2{X}_2,\\ \frac{1}{2\rho }(H_2+\alpha H_1-2\alpha {\nu }_2{X}_1-2{\nu }_4{X}_2)),\end{array}$$ with Lie brackets $$[x_1,x_2]=x_1, [x_1,x_6]=-\frac{\rho +1}{\rho }\alpha x_1, [x_3,x_6]=-\frac{\rho +1}{\rho }{x}_3, [x_4,x_6]=\alpha x_4, [x_5,x_6]=x_6.$$

This algebra is of type (5), if we replace x₆ by $x_6+\frac{\alpha (\rho +1)}{\rho }{x}_2$. It arises for the triangular-split RB-operator R with $A_{-}=\ker (R)=〈x_1,x_2,x_3〉,\ker (R+\text{id})=〈x_4,x_5〉$ and $A_0=〈x_6〉$, where $x_6=H_2-2{\nu }_4{X}_2$, with the action $R(x_6)=\rho x_6$.

Case 2c, $\lambda =0,\rho =-1$: We may assume that there exists $x_6=Y_2+v$ such that $(R+\text{id})(x_6)=\mu (H_2+\alpha H_1+\beta X_1+\gamma X_2)$ for some nonzero μ and some $\alpha ,\beta ,\gamma \in \mathbb{C}$. Since $\ker (R+\text{id})$ is an abelian subalgebra we obtain $\alpha =\beta =0$ and $\ker (R+\text{id})=〈H_2+\gamma X_2,Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1〉$. Then we may choose $x_6=Y_2+\kappa X_2+{\nu }_3{H}_1+{\nu }_4{X}_1$. Then $$\begin{array}{l}[x_6,H_2+\gamma X_2]=\left\{R\right(x_6),H_2+\gamma X_2\}\\ =\left\{\right(R+\text{id}\left)\right(x_6)-x_6,H_2+\gamma X_2\}\\ =-\{Y_2+\kappa X_2,H_2+\gamma X_2\}\\ =-2Y_2+2\kappa X_2+\gamma H_2.\end{array}$$

This is not contained in $\ker (R+\text{id})$, which is a contradiction to the fact that $\ker (R+\text{id})$ is an ideal.

Case 2c, $\lambda =0$, ρ = 0: Then we have $R(H_2)=\alpha H_1+\beta X_1+\gamma X_2\ne 0$ and $\ker (R+\text{id})=〈Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2〉$. Since $[H_2,Y_2+{\nu }_1{X}_1+{\nu }_2{H}_1]=\{\gamma X_2,Y_2+{\nu }_1{X}_2+{\nu }_2{H}_2\}$ is in $\ker (R+\text{id})$, we obtain $\gamma =0$. Since $[H_2,Y_1+{\nu }_1{X}_1+{\nu }_2{H}_2]=\{\alpha H_1+\beta X_1,Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1\}$ is in $\ker (R+\text{id})$, we obtain $\alpha ({\nu }_1+{\nu }_2^2)=0$ and $\beta =-2\alpha {\nu }_2$. Since $R(H_2)\ne 0$ we have $\alpha \ne 0,{\nu }_1=-{\nu }_2^2$ and $R(H_2)=\alpha H_1-2\alpha {\nu }_2{X}_1$. Consider a new basis for $\mathfrak{g}$ given by $$(x_1,\dots ,x_6)=(X_1,-\frac{1}{2}{H}_1+{\nu }_2{X}_1,X_2,Y_1+{\nu }_2{H}_1-{\nu }_2^2{X}_1,Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2,-\frac{1}{2}{H}_2),$$ with Lie brackets $$[x_1,x_2]=x_1, [x_1,x_6]=\alpha x_1, [x_3,x_6]=x_3, [x_4,x_6]=-\alpha x_4.$$

This algebra is of type (3), if we replace x₆ by $x_6-\alpha x_2$.

Case 2c, $\lambda \ne 0$: Then we have $\ker (R)=〈X_1,X_2,-\frac{1}{2}(H_1+\lambda H_2)〉$. We again have ${\chi }_R(t)=t^3{(t+1)}^2(t-\rho )$, where we distinguish the cases $\rho \ne 0,-1,\rho =-1$ and ρ = 0.

Case 2c, $\lambda \ne 0,\rho \ne 0,-1$: Then we may assume that $R(x_6)=\rho x_6$ for $x_6=H_2+\alpha H_1+\beta X_1+\gamma X_2$. As $\ker (R+\text{id})$ is abelian, we have $\ker (R+\text{id})=〈Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2〉$. Since $V=\ker (R)\oplus \ker (R+\text{id})\oplus 〈x_6〉$, the two elements $H_1+\lambda H_2$ and $H_2+\alpha H_1$ need to be linearly independent, i.e., $1-\alpha \lambda \ne 0$. By (10) and (11) we obtain $\gamma =-2{\nu }_4,\beta =-2\alpha {\nu }_2,{\nu }_3=-{\nu }_4^2$, and $\alpha ({\nu }_1+{\nu }_2^2)$. Suppose that $\alpha =0$. Then $x_6=H_2-2{\nu }_4{X}_2$. Consider a new basis for $\mathfrak{g}$ given by $$(x_1,\dots ,x_6)=(Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,X_1,X_2,-\frac{1}{2}(H_1+\lambda H_2),Y_2-{\nu }_4^2{X}_2+{\nu }_4{H}_2,-\frac{1}{2(\rho +1)}{H}_2),$$ with Lie brackets $$[x_2,x_4]=x_2, [x_3,x_4]=\lambda x_3, [x_3,x_6]=x_3, [x_4,x_6]=-\lambda {\nu }_4{x}_3, [x_5,x_6]=-\frac{\rho }{1+\rho }{x}_5.$$

This is an algebra of type (6), if we replace x₄ by $x_4+\lambda {\nu }_4{x}_3$.

Now we assume that $\alpha \ne 0$. Consider a new basis for $\mathfrak{g}$ given by $$\begin{array}{c}(x_1,\dots ,x_6)=(X_1,X_2,-\frac{1}{2}(H_1+\lambda H_2),Y_2-{\nu }_4^2{X}_2+{\nu }_4{H}_2,Y_1-{\nu }_2^2{X}_1+{\nu }_2{H}_1,\\ -\frac{1}{2(\rho +1)}(H_2-2{\nu }_4{X}_2+\alpha (H_1-2{\nu }_2{X}_1\left)\right)),\end{array}$$ with Lie brackets $$\begin{array}{c}[x_1,x_3]=x_1, [x_2,x_3]=\lambda x_2, [x_1,x_6]=\alpha x_1, [x_2,x_6]=x_2,\\ [x_3,x_6]=-\alpha {\nu }_2{x}_1-\lambda {\nu }_4{x}_2, [x_4,x_6]=\delta x_4, [x_5,x_6]=\alpha \delta x_5,\end{array}$$ where $\delta =-\frac{\rho }{\rho +1}$. Replacing x₆ by $\frac{1}{\delta }(x_6-\alpha {\nu }_2{x}_1-{\nu }_4{x}_2-\alpha x_3)$ we obtain the Lie brackets $$[x_1,x_3]=x_1, [x_2,x_3]=\lambda x_2, [x_2,x_6]=\alpha \prime x_2, [x_4,x_6]=x_4, [x_5,x_6]=\alpha x_5,$$ where $$\alpha \prime =\frac{1-\alpha \lambda }{\delta }=\frac{(\rho +1)(\alpha \lambda -1)}{\rho }.$$

Note that $\alpha \prime \ne 0$ and $\alpha \prime \ne \alpha \lambda -1$ by assumption. In other words, $\alpha \ne \frac{\alpha \prime +1}{\lambda }$. Consider a new basis for $\mathfrak{g}$ given by $$(x_1^{\prime },\dots ,x_6^{\prime })=(x_2,x_1,\frac{1}{\lambda }{x}_3,x_4,x_5,x_6-\frac{\alpha \prime }{\lambda }{x}_3),$$ with Lie brackets $$[x_1^{\prime },x_3^{\prime }]=x_1^{\prime }, [x_2^{\prime },x_3^{\prime }]=\lambda \prime x_2^{\prime }, [x_2^{\prime },x_6^{\prime }]=-\alpha \prime \lambda \prime x_2^{\prime }, [x_4^{\prime },x_6^{\prime }]=x_4^{\prime }, [x_5^{\prime },x_6^{\prime }]=\alpha x_5^{\prime },$$ where $\lambda \prime =\frac{1}{\lambda }$. This is of type (7). Since ${\mathfrak{r}}_{3,\lambda }(\mathbb{C})\cong {\mathfrak{r}}_{3,\lambda \prime }(\mathbb{C})$, one may check that we do not only have $\alpha \ne \frac{\alpha \prime +1}{\lambda }$, but also $\alpha \ne \lambda -\alpha \prime$. For $\frac{\alpha \prime +1}{\lambda }\ne \lambda -\alpha \prime$ we obtain no restriction for $\alpha$. However, for $\frac{\alpha \prime +1}{\lambda }=\lambda -\alpha \prime$ we obtain $\lambda =-1$ or $\lambda =\alpha \prime +1$, which excludes both $(\lambda ,\alpha \prime ,\alpha )=(-1,\alpha \prime ,-\alpha \prime -1)$ and $(\lambda ,\alpha \prime ,\alpha )=(\lambda ,\lambda -1,1)$. Rewriting this in the parameters of the Lie brackets from type (7), we obtain all cases except for $(\lambda ,\alpha \prime ,\alpha )=(\lambda ,\lambda -1,1)$ with $\lambda \ne -1$. These PA-structures arise by a triangular-split RB-operator with $A_{-}=\ker (R),A_{+}=\ker (R+\text{id})$ and $A_0=〈x_6〉$ with the action $R(x_6)=\rho x_6,\rho \ne 0,-1$.

Case 2c, $\lambda \ne 0,\rho =-1$: This leads to a contradiction in the same way as case 2c with $\lambda =0,\rho =-1$.

Case 2c, $\lambda \ne 0$,ρ = 0: We have $R(H_2)=\alpha X_1+\beta X_2+\gamma (H_1+\lambda H_2)$ and $\ker (R+\text{id})=〈x_4,x_5〉$ with $x_4=Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,x_5=Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2$. Similarly to $(10),(11)$ we obtain $R(H_2)=\gamma (H_1-2{\nu }_2{X}_1)+\gamma \lambda (H_2-2{\nu }_4{X}_2)$. This implies that $\gamma \ne 0$ and $x_4=Y_1-{\nu }_2^2{X}_1+{\nu }_2{H}_1,x_5=Y_2-{\nu }_4^2{X}_2+{\nu }_4{H}_2$. By setting x₁ = X₁, x₂ = X₂, $x_3=-\frac{1}{2}(H_1+\lambda H_2)$ and $x_6=\frac{1}{2\gamma }{H}_2$ we obtain a new basis for $\mathfrak{g}$ with Lie brackets $$\begin{array}{c}[x_1,x_3]=x_1, [x_2,x_3]=\lambda x_2, [x_1,x_6]=-x_1, [x_2,x_6]=\delta x_2,\\ [x_3,x_6]={\nu }_2{x}_1+{\lambda }^2{\nu }_4{x}_2, [x_4,x_6]=x_4, [x_5,x_6]=\lambda x_5,\end{array}$$ where $\delta =-\frac{1+\lambda \gamma }{\gamma }$ with $\delta \ne -\lambda$. Replacing x₆ by $x_6+{\nu }_2{x}_1+\lambda {\nu }_4{x}_2+x_3$ we obtain the brackets $$[x_1,x_3]=x_1, [x_2,x_3]=\lambda x_2, [x_2,x_6]={\alpha }_1{x}_2, [x_4,x_6]=x_4, [x_5,x_6]=\lambda x_5$$ with ${\alpha }_1=\delta +\lambda =-\frac{1}{\gamma }$. This is of type (7) with ${\alpha }_2=\lambda$. It arises by the triangular-split RB-operator with $A_{-}=〈x_1,x_2〉,A_{+}=〈x_4,x_5〉$ and $A_0=〈u,v〉$, with $u=\frac{1}{\gamma }(H_2-2{\nu }_4{X}_2)$ and $v=H_1-2{\nu }_2{X}_1+\lambda (H_2-2{\nu }_4{X}_2)$, and the action R(u) = v, R(v) = 0.

Case 2d: Suppose that one of the kernels $\ker (R)$ and $\ker (R+\text{id})$ is nonabelian. Without loss of generality, let us assume that $\ker (R)\cong {\mathfrak{r}}_2(\mathbb{C})$. Write $\mathfrak{g}\cong (\ker (R)\oplus \ker (R+\text{id}))⋉〈a,b〉$. Then $\ker (R)⋉〈a〉$ is a three-dimensional solvable subalgebra of $\text{im}(R+\text{id})$. By Table 2, we see that it is isomorphic to ${\mathfrak{r}}_2(\mathbb{C})\oplus \mathbb{C}$. In this case there exist nonzero $a\prime \in \ker (R)\oplus 〈a〉$ and $b\prime \in \ker (R)\oplus 〈b〉$ such that $[a\prime ,\ker (R)]=[b\prime ,\ker (R)]=0$. Then $\mathfrak{g}\cong \ker (R)\oplus (\ker (R+\text{id})\oplus 〈a\prime ,b\prime 〉)$ with $\ker (R)\cong {\mathfrak{r}}_2(\mathbb{C})$, and $\ker (R+\text{id})\oplus 〈a\prime ,b\prime 〉\cong {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$ by Table 2. Hence we obtain $\mathfrak{g}\cong {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})\oplus {\mathfrak{r}}_2(\mathbb{C})$, which is of type (4).

So we may assume that $\ker (R)\cong \ker (R+\text{id})\cong {\mathbb{C}}^2$. Then the characteristic polynomial of R has the form ${\chi }_R(t)=t^2{(t+1)}^2(t-{\rho }_1)(t-{\rho }_2)$.

Case 2d, ${\rho }_1,{\rho }_2\ne 0,-1$: Suppose first that either ${\rho }_1\ne {\rho }_2$ or that ${\rho }_1={\rho }_2$ and the eigenspace is two-dimensional. Then by Proposition $3.4,\mathfrak{n}=\ker (R)\dot{+}\ker (R+\text{id})\dot{+}〈x_5^{\prime },x_6^{\prime }〉$ with linearly independent eigenvectors $x_5^{\prime },x_6^{\prime }$ corresponding to the eigenvalues ρ₁ and ρ₂. Since $\ker (R)$ is an abelian ideal in $\text{im}(R+\text{id})=〈X_1,H_1,X_2,H_2〉$, we may assume that $\ker (R)=〈X_1,X_2〉$ and $[x_5^{\prime },x_6^{\prime }]=0$. The decomposition $\mathfrak{n}=\ker (R+\text{id})\dot{+}\text{im}(R+\text{id})$ shows that $\ker (R+\text{id})$ has a basis $x_3=Y_1+\alpha H_1+{\nu }_3{X}_1$, $x_4=Y_2+\beta H_2+{\nu }_4{X}_2$. Since $[x_5^{\prime },x_6^{\prime }]=0$, we have $x_5^{\prime }=H_1+{\nu }_1{X}_1+{\xi }_1(H_2+{\nu }_2{X}_2)$, $x_6^{\prime }=H_2+{\nu }_2{X}_2+{\xi }_2(H_1+{\nu }_1{X}_1)$ with ${\xi }_1{\xi }_2\ne 1$. So we have by (10) and (11) $x_3=Y_1-\frac{{\nu }_1}{2}{H}_1-\frac{{\nu }_1^2}{4}{X}_1$, $x_4=Y_2-\frac{{\nu }_2}{2}{H}_2-\frac{{\nu }_2^2}{4}{X}_2$. Consider a basis for $\mathfrak{g}$ given by $$(x_1,\dots ,x_6)=(X_1,X_2,x_3,x_4,-\frac{1}{2\left(1+{\rho }_1\right)}{x}_5^{\prime },-\frac{1}{2\left(1+{\rho }_2\right)}{x}_6^{\prime }),$$ with Lie brackets $$\begin{array}{c}[x_1,x_5]=x_1, [x_1,x_6]={\xi }_2{x}_1, [x_2,x_5]={\xi }_1{x}_2, [x_2,x_6]=x_2,\\ [x_3,x_5]=\gamma x_3, [x_3,x_6]=\delta {\xi }_2{x}_3, [x_4,x_5]=\gamma {\xi }_1{x}_4, [x_4,x_6]=\delta x_4,\end{array}$$ where $\gamma =-\frac{{\rho }_1}{{\rho }_1+1}$, $\delta =-\frac{{\rho }_2}{{\rho }_2+1}$ with $\gamma ,\delta \ne 0,-1$ and ${\xi }_1{\xi }_2\ne 1$. This is type $(8a)$. It arises by the triangular-split RB-operator R with $A_{-}=〈X_1,X_2〉,A_{+}=〈x_3,x_4〉$ and $A_0=〈x_5,x_6〉$, where R acts on A₀ by $R(x_5)={\rho }_1{x}_5$ and $R(x_6)={\rho }_2{x}_6$. Note that for ${\nu }_2={\xi }_2=0$ and ${\xi }_1\ne 0$ we get type (7) without the restriction $(\lambda ,{\alpha }_1,{\alpha }_2)\ne (\lambda ,\lambda -1,1)$ for $\lambda \ne -1$, which we had in Case 2c, $\lambda \ne 0,\rho \ne 0,-1$.

Suppose now that ${\rho }_2={\rho }_1\ne 0,-1$, and the eigenspace for ρ₁ is one-dimensional. Let $R(x_5^{\prime })={\rho }_1{x}_5^{\prime }$ and $R(x_6^{\prime })=x_5^{\prime }+{\rho }_1{x}_6^{\prime }$. In the same way as before we have $x_5^{\prime }=H_1+{\nu }_1{X}_1+\xi (H_2+{\nu }_2{X}_2)$, $x_6^{\prime }=\kappa (H_2+{\nu }_2{X}_2)$ with $\kappa \ne 0$ and $x_3=Y_1-\frac{{\nu }_1}{2}{H}_1-\frac{{\nu }_1^2}{4}{X}_1$, $x_4=Y_2-\frac{{\nu }_2}{2}{H}_2-\frac{{\nu }_2^2}{4}{X}_2$. Consider a basis for $\mathfrak{g}$ given by $$(x_1,\dots ,x_6)=(X_1,X_2,x_3,x_4,-\frac{1}{2\left(1+{\rho }_1\right)}{x}_5^{\prime },-\frac{1}{2\left(1+{\rho }_1\right)}{x}_6^{\prime }),$$ with Lie brackets $$\begin{array}{c}[x_1,x_5]=x_1, [x_1,x_6]=(\gamma +1)x_1, [x_2,x_5]=\xi x_2, [x_2,x_6]=(\kappa +\xi +\gamma \xi )x_2,\\ [x_3,x_5]=\gamma x_3, [x_3,x_6]=-(\gamma +1)x_3, [x_4,x_5]=\gamma \xi x_4, [x_4,x_6]=(\kappa \gamma -\xi -\gamma \xi )x_4,\end{array}$$ where $\gamma =-\frac{{\rho }_1}{{\rho }_1+1}\ne 0,-1$ and $\kappa \ne 0$. This is type $(8b)$. It arises by the triangular-split RB-operator R with $A_{-}=〈X_1,X_2〉,A_{+}=〈x_3,x_4〉$ and $A_0=〈x_5,x_6〉$, where R acts on A₀ by $R(x_5)={\rho }_1{x}_5$ and $R(x_6)=x_5+{\rho }_1{x}_6$.

Case 2d, ${\rho }_1={\rho }_2=0$: We have $\mathfrak{g}=\ker (R+\text{id})\dot{+}\text{im}(R+\text{id})$ and we can assume that $\ker (R)=〈X_1,X_2〉$ and $\ker (R+\text{id})=〈Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,Y_2+{\nu }_3{X}_2+{\nu }_4{H}_2〉$. Suppose first that $R(v)=X_1$ and $R(w)=X_2$ for some v, w. Then $$[Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,v]=\{Y_1+{\nu }_1{X}_1+{\nu }_2{H}_1,X_1\}=-H_1+2{\nu }_2{X}_1\in \ker \left(R+\text{id}\right),$$ which is a contradiction. Otherwise we see from the possible Jordan forms of R that there exist v, w with $R(v)=\alpha X_1+\beta X_2\ne 0$ and R(w) = v. This leads to a contradiction in the same way.

Case 2d, ${\rho }_1=0,{\rho }_2\ne 0,-1$: This case is analogous to the second part of the case before.

Case 2d, ${\rho }_1=0,{\rho }_2=-1$: As above we may assume that $\text{im}(R+\text{id})=〈X_1,X_2,H_1,H_2〉$ and $\ker (R)=〈X_1,X_2〉$, and $\alpha H_1+\beta H_2+\gamma X_1+\delta X_2\in \ker (R+\text{id})\cap \text{im}(R+\text{id})$ for some $\alpha ,\beta ,\gamma ,\delta \in \mathbb{C}$. Since $\ker (R+\text{id})$ is abelian, we may assume that $\ker (R+\text{id})=〈H_1+{\nu }_1{X}_1,Y_2+{\nu }_2{X}_2+{\nu }_3{H}_2〉$ for some ${\nu }_1,{\nu }_2,{\nu }_3\in \mathbb{C}$. Let $v\in \ker (R^2)$ such that $R(v)={\nu }_4{X}_1+{\nu }_5{X}_2\ne 0$. Then $$[v,Y_2+{\nu }_2{X}_2+{\nu }_3{H}_2]=\{{\nu }_4{X}_1+{\nu }_5{X}_2,Y_2+{\nu }_2{X}_2+{\nu }_3{H}_2\}={\nu }_5(H_2-2{\nu }_3{X}_2)\in \ker \left(R+\text{id}\right)$$ implies that ${\nu }_5=0$. By $[v,H_1+{\nu }_1{X}_1]=\{{\nu }_4{X}_1,H_1+{\nu }_1{X}_1\}=-2{\nu }_4{X}_1\in \ker (R+\text{id})$ we obtain ${\nu }_4=0$, which is a contradiction to $R(v)\ne 0$.□

Remark 4.2.

The algebras from different types are nonisomorphic, except for algebras of type (8), which have intersections with types (3) and (7) for certain parameter choices.

Additional information

Funding

Dietrich Burde is supported by the Austrian Science Foundation FWF, grant P28079 and grant I3248. Vsevolod Gubarev acknowledges support by the Austrian Science Foundation FWF, grant P28079.