Local and 2-Local Derivations and Automorphisms on Simple Leibniz Algebras

Abstract

This paper is devoted to study local and 2-local derivations and automorphism of complex finite-dimensional simple Leibniz algebras. We show that all local derivations and 2-local derivations on a finite-dimensional complex simple Leibniz algebra are automatically derivations. We prove that nilpotent Leibniz algebras as a rule admit local derivations and 2-local derivations which are not derivations. Further, we consider automorphisms of simple Leibniz algebras. It is established that every 2-local automorphism on a complex finite-dimensional simple Leibniz algebra is an automorphism and that nilpotent Leibniz algebras admit 2-local automorphisms which are not automorphisms. A similar problem concerning local automorphism on simple Leibniz algebras is reduced to the case of simple Lie algebras.

1 Introduction

Let $\mathcal {A}$ be an algebra. A mapping T is said to be a local automorphism (respectively, a local derivation) if for every $x\in \mathcal {A}$ there exists an automorphism (respectively, a derivation) $T_x$ of $\mathcal {A},$ depending on x,  such that $T_x(x)=T(x).$ These notions were introduced and investigated independently by Kadison [18] and Larson and Sourour [19]. Later, in 1997, P. Šemrl introduced the concepts of 2-local automorphisms and 2-local derivations [25]. A map $\varPhi :\mathcal {A} \rightarrow \mathcal {A}$ (not linear in general) is called a 2-local automorphism (respectively, a 2-local derivation) if for every $x, y\in \mathcal {A},$ there exists an automorphism (respectively, a derivation) $\varPhi _{x,y}:\mathcal {A} \rightarrow \mathcal {A}$ (depending on x, y) such that $\varPhi _{x.y}(x)=\varPhi (x),$$\varPhi _{x,y}(y)=\varPhi (y).$

The above papers gave rise to a series of works devoted to the description of mappings which are close to automorphisms and derivations of C*-algebras and operator algebras. For details, we refer to the papers [5] and [8].

Later, several papers have been devoted to similar notions and corresponding problems for derivations and automorphisms of Lie algebras. Every derivation (respectively, automorphism) of a Lie algebra $\mathcal {L}$ is a local derivation (respectively, local automorphism) and a 2-local derivation (respectively, 2-local automorphism). For a given Lie algebra $\mathcal {L},$ the main problem concerning these notions is to prove that they automatically become a derivation (respectively, an automorphism) or to give examples of local and 2-local derivations or automorphisms of $\mathcal {L},$ which are not derivations or automorphisms, respectively. In [9], it is proved that every 2-local derivation on a semisimple Lie algebra $\mathcal {L}$ is a derivation and that each finite-dimensional nilpotent Lie algebra with dimension larger than two admits 2-local derivation which is not a derivation. In [6], it was established that every local derivation on semisimple Lie algebras is a derivation and gives examples of nilpotent finite-dimensional Lie algebras with local derivations which are not derivations. Concerning 2-local automorphism, Chen and Wang in [14] prove that if $\mathcal {L}$ is a simple Lie algebra of type $\mathcal {A}_l$, $\mathcal {D}_l$ or $\mathcal {E}_k$ ($k=6,7,8$) over an algebraically closed field of characteristic zero, then every 2-local automorphism of $\mathcal {L}$ is an automorphism. Finally, in [7], Ayupov and Kudaybergenov generalized this result and proved that every 2-local automorphism of a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic zero is an automorphism. Moreover, they show also that every nilpotent Lie algebra with finite dimension larger than two admits 2-local automorphisms which is not an automorphism. Local derivations and automorphisms on various algebras were studied in [4,5,6,7,8]. It should be noted that similar problems for local automorphism of finite-dimensional Lie algebras still remain open.

Leibniz algebras present a “non-antisymmetric” extension of Lie algebras. In last decades, a series of papers have been devoted to the structure theory and classification of finite-dimensional Leibniz algebras. Several classical theorems from Lie algebras theory have been extended to the Leibniz algebras case. For some details from the theory of Leibniz algebras, we refer to the papers [1,2,3, 15, 20, 21]. In particular, for a finite-dimensional simple Leibniz algebra over an algebraically closed field of characteristic zero, derivations have recently been described in [23].

In the present paper, we study local and 2-local derivations and automorphisms of finite-dimensional simple complex Leibniz algebras.

In Sect. 2, we give some preliminaries from the Leibniz algebras theory. In Sect. 3, we prove that every 2-local derivation on a simple Leibniz algebra $\mathcal {L}$ is a derivation. We also prove that all nilpotent Leibniz algebras (except the so-called null-filiform Leibniz algebras) admit 2-local derivations which are not derivations. Similar results for local derivations on simple Leibniz algebras are obtained in Sect. 4. Namely, we show that every local derivation on a simple complex Leibniz algebra is a derivation and that each finite-dimensional filiform Leibniz algebra $\mathcal {L}$ with $\dim \mathcal {L}\ge 3$ admits a local derivation which is not a derivation.

In Sect. 5, we study automorphisms of simple Leibniz algebras.

Finally, in Sect. 6 we consider 2-local and local automorphisms of finite-dimensional Leibniz algebras. First, we show that every 2-local automorphism of a complex simple Leibniz algebra is an automorphism and prove that each n-dimensional nilpotent Leibniz algebra such that $n\ge 2$ and $\dim [\mathcal {L}, \mathcal {L}]\le n-2$ admits a 2-local automorphism which is not an automorphism. At the end of Sect. 6, we show that the problem concerning local automorphisms of simple complex Leibniz algebras is reduced to the similar problem for simple Lie algebras, which is recently solved by M. Constantini in [13]. In fact, it is proved that a local automorphism of finite-dimensional simple Lie algebras over an algebraically closed field of characteristic 0 is either an automorphism or an anti-automorphism.

2 Preliminaries

In this section, we give some necessary definitions and preliminary results.

Definition 1

An algebra $(\mathcal {L},[\cdot ,\cdot ])$ over a field $\mathbb {F}$ is called a Leibniz algebra if it is defined by the identity

$$\begin{aligned}{}[x,[y,z]]=[[x,y],z] - [[x,z],y], \ \text{ for } \text{ all }\ x,y \in \mathcal {L}, \end{aligned}$$

which is called Leibniz identity.

It is a generalization of the Jacobi identity since under the condition of antisymmetricity of the product “[$\cdot ,\cdot$],” this identity changes to the Jacobi identity. In fact, the definition above is the notion of the right Leibniz algebra, where “right” indicates that any right multiplication operator is a derivation of the algebra. In the present paper, the term “Leibniz algebra” will always mean the “right Leibniz algebra.” The left Leibniz algebra is characterized by the property that any left multiplication operator is a derivation.

Let $\mathcal {L}$ be a Leibniz algebra and $\mathcal {I}$ be the ideal generated by squares in $\mathcal {L}:$$\mathcal {I}=\text {id}\langle [x,x]\ | \ x\in \mathcal {L} \rangle .$ The quotient $\mathcal {L}/\mathcal {I}$ is called “the associated Lie algebra” of the Leibniz algebra $\mathcal {L}.$ The natural epimorphism $\varphi : \mathcal {L} \rightarrow \mathcal {L}/\mathcal {I}$ is a homomorphism of Leibniz algebras. The ideal $\mathcal {I}$ is the minimal ideal with the property that the quotient algebra is a Lie algebra. It is easy to see that the ideal $\mathcal {I}$ coincides with the subspace of $\mathcal {L}$ spanned by the squares and that $\mathcal {L}$ is the left annihilator of $\mathcal {I},$ i.e., $[\mathcal {L},\mathcal {I}]=0.$

Definition 2

A Leibniz algebra $\mathcal {L}$ is called simple if its ideals are only $\{0\}, \mathcal {I}, \mathcal {L}$ and $[\mathcal {L},\mathcal {L}]\ne \mathcal {I}.$

This definition agrees with that of simple Lie algebra, where $\mathcal {I}=\{0\}.$

Given a Leibniz algebra $\mathcal {L}$, we define the lower central sequence defined recursively as

$$\begin{aligned} \mathcal {L}^1=\mathcal {L}, \ \mathcal {L}^{k+1}=[\mathcal {L}^k,\mathcal {L}], \ \ k \ge 1. \end{aligned}$$

Definition 3

A Leibniz algebra $\mathcal {L}$ is said to be nilpotent, if there exists $t\in {\mathbb {N}}$ such that $\mathcal {L}^{t}=\{0\}$. The minimal number t with such property is said to be the index of nilpotency of the algebra $\mathcal {L}$.

Since for a nilpotent algebra we have $\dim \mathcal {L}^2\le n-1$, the index of nilpotency of an n-dimensional nilpotent Leibniz algebra is not greater than $n+1$.

A nilpotent Leibniz algebra $\mathcal {L}$ is called filiform if $\dim \mathcal {L}^k=n-k$ for $2\le k \le n.$

Recall [22] that each complex n-dimensional filiform Leibniz algebra admits a basis $\{e_1,e_2,\ldots ,e_n\}$ such that the table of multiplication of the algebra has one of the following forms:

where $\alpha \in \{0,1\}$ for odd n and $\alpha =0$ for even n. Moreover, the structure constants of an algebra from $F_3(\theta _1,\theta _2,\theta _3)$ should satisfy the Leibniz identity.

It is easy to see that algebras of the first and the second families are non-Lie algebras. Moreover, an algebra of the third family is a Lie algebra if and only if $(\theta _1,\theta _2,\theta _3)=(0,0,0).$

A Leibniz algebra $\mathcal {L}$ is called null-filiform if $\dim \mathcal {L}^k=n+1-k$ for $1\le k \le n+1.$

Clearly, a null-filiform Leibniz algebra has maximal index of nilpotency. Moreover, it is easy to show that a nilpotent Leibniz algebra is null-filiform if and only if it is a one-generated algebra (see [10]). Note that this notion has no sense in the Lie algebras case, because Lie algebras are at least two-generated.

For a given Leibniz algebra $\mathcal {L}$, we define the derived sequence as follows:

$$\begin{aligned} \mathcal {L}^{[1]}=\mathcal {L},\quad \mathcal {L}^{[s+1]}=[\mathcal {L}^{[s]},\mathcal {L}^{[s]}], \quad s \ge 1. \end{aligned}$$

Definition 4

A Leibniz algebra $\mathcal {L}$ is called solvable, if there exists $s\in {\mathbb {N}}$ such that $\mathcal {L}^{[s]}=\{0\}.$

It is known that the sum of solvable ideals of a Leibniz algebra is a solvable ideal too. Therefore, each Leibniz algebra contains a maximal solvable ideal which is called solvable radical.

The following theorem proved by A.M. Bloh [12] (see also D. Barnes [11]) presents an analogue of Levi–Malcev’s theorem for Leibniz algebras.

Theorem 5

Let $\mathcal {L}$ be a finite-dimensional Leibniz algebra over a field of characteristic zero, and let $\mathcal {R}$ be its solvable radical. Then, there exists a semisimple Lie subalgebra $\mathcal {S}$ of $\mathcal {L}$ such that $\mathcal {L}=\mathcal {S}\dot{+}\mathcal {R}.$

This theorem applied to a simple Leibniz algebra $\mathcal {L}$ gives

Corollary 6

Let $\mathcal {L}$ be a simple Leibniz algebra over a field of characteristic zero, and let $\mathcal {I}$ be the ideal generated by squares in $\mathcal {L},$ then there exists a simple Lie algebra $\mathcal {G}$ such that $\mathcal {I}$ is an irreducible module over $\mathcal {G}$ and $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}.$

Further, we shall use the following important result [16].

Theorem 7

(Schur’s Lemma) Let $\mathcal {G}$ be a complex Lie algebra, and let $\mathcal {U}$ and $\mathcal {V}$ be irreducible $\mathcal {G}$-modules. Then,

• (i) Any $\mathcal {G}$-module homomorphism $\varTheta : \mathcal {U} \rightarrow \mathcal {V}$ is either trivial or an isomorphism;

• (ii) A linear map $\varTheta : \mathcal {V} \rightarrow \mathcal {V}$ is a $\mathcal {G}$-module homomorphism if and only if $\varTheta = \lambda id_{|_\mathcal {V}}$ for some $\lambda \in \mathbb {C}.$

The notion of derivation for a Leibniz algebra is defined similar to the Lie algebras case as follows.

Definition 8

A linear transformation d of a Leibniz algebra $\mathcal {L}$ is said to be a derivation if for any $x, y\in \mathcal {L}$ one has

$$\begin{aligned} D([x,y])=[D(x),y]+[x, D(y)]. \end{aligned}$$

Let a be an element of a Leibniz algebra $\mathcal {L}.$ Consider the operator of right multiplication $R_a:\mathcal {L}\rightarrow \mathcal {L}$, defined by $R_a(x)=[x,a].$ The Leibniz identity which characterizes Leibniz algebras exactly means that every right multiplication operator $R_a$ is a derivation. Such derivations are called inner derivation on $\mathcal {L}$ . Denote by $Der (\mathcal {L})$ the space of all derivations of $\mathcal {L}$.

Now we shall present the main subjects considered in this paper, so-called local and 2-local derivation.

Definition 9

A linear operator $\varDelta : \ \mathcal {L} \ \rightarrow \mathcal {L}$ is called a local derivation if for any $x\in \mathcal {L}$ there exists a derivation $D_{x}\in Der (\mathcal {L})$ such that

$$\begin{aligned} \varDelta (x)=D_{x}(x). \end{aligned}$$

Definition 10

A map $\varDelta : \mathcal {L} \rightarrow \mathcal {L}$ (not necessary linear) is called 2-local derivation if for any $x, y\in \mathcal {L}$ there exists a derivation $D_{x,y}\in Der (\mathcal {L})$ such that

$$\begin{aligned} \varDelta (x)=D_{x,y}(x), \quad \varDelta (y)=D_{x,y}(y). \end{aligned}$$

From now on, we assume that all algebras are considered over the field of complex numbers $\mathbb {C}$ and suppose that $\mathcal {L}$ is a non-Lie Leibniz algebra, i.e., $\mathcal {I}\ne \{0\}.$

Now we give a description of derivations on simple Leibniz algebras obtained in [23].

Let $\mathcal {L}$ be a simple Leibniz algebra with $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}.$ Consider a projection operator $\text {pr}_{\mathcal {I}}$ from $\mathcal {L}$ onto $\mathcal {I},$ that is

$$\begin{aligned} \text {pr}_\mathcal {I}(x+i)=i,\, x+i\in \mathcal {L}\dot{+}\mathcal {I}. \end{aligned}$$

(1)

Suppose that $\mathcal {G}$ and $\mathcal {I}$ are not isomorphic as $\mathcal {G}$-modules. Then, any derivation D on $\mathcal {L}$ can be represented as

$$\begin{aligned} D=R_a+\lambda \text {pr}_\mathcal {I}, \end{aligned}$$

(2)

where $R_a$ is an inner derivation generated by an element $a\in \mathcal {G},$$\text {pr}_\mathcal {I}$ is a derivation of the form (1), $\lambda \in \mathbb {C}.$

Now let us assume that $\mathcal {G}$ and $\mathcal {I}$ are isomorphic as $\mathcal {G}$-modules. There exists a unique (up to multiplication by constant) isomorphism $\theta$ of linear spaces $\mathcal {G}$ and $\mathcal {I}$ such that $\theta ([x,y])=[\theta (x), y]$ for all $x, y\in \mathcal {G},$ i.e., $\theta$ is a module isomorphism of $\mathcal {G}$-modules $\mathcal {G}$ and $\mathcal {I}.$ Let us extend $\theta$ onto $\mathcal {L}$ as

$$\begin{aligned} \theta (x+i)=\theta (x),\, x+i\in \mathcal {L}\dot{+}\mathcal {I}. \end{aligned}$$

(3)

For a simple Leibniz algebra $\mathcal {L}$ with $\dim \mathcal {G}=\dim \mathcal {I}$, any derivation D on $\mathcal {L}$ can be represented as

$$\begin{aligned} D=R_a+\omega \theta +\lambda \text {pr}_\mathcal {I}, \end{aligned}$$

(4)

where $a\in \mathcal {G},$$\text {pr}_\mathcal {I}$ is a derivation of form (1) and $\theta$ is a derivation of form (3), $\lambda , \,\omega \in \mathbb {C}.$

3 2-Local Derivations on Leibniz Algebras

3.1 2-Local Derivations on Simple Leibniz Algebras

The first main result of this section is the following:

Theorem 11

Let $\mathcal {L}$ be a simple Leibniz algebra. Then, any 2-local derivation on $\mathcal {L}$ is a derivation.

For the proof of this theorem, we need several lemmas.

From theory of representation of semisimple Lie algebras [16], we have that a Cartan subalgebra $\mathcal {H}$ of simple Lie algebra $\mathcal {G}$ acts diagonalizable on $\mathcal {G}$-module $\mathcal {I}:$

$$\begin{aligned} \mathcal {I}=\bigoplus _{\alpha \in \varGamma } \mathcal {I}_\alpha , \end{aligned}$$

where

$$\begin{aligned} \mathcal {I}_\alpha= & {} \{i\in \mathcal {I}: [i, h]=\alpha (h)i, \,\, \forall \, h\in \mathcal {H}\},\\ \varGamma= & {} \{\alpha \in \mathcal {H}^*: \mathcal {I}_\alpha \ne \{0\}\} \end{aligned}$$

and $\mathcal {H}^*$ is the space of all linear functionals on $\mathcal {H}.$ Elements of $\varGamma$ are called weights of $\mathcal {I}.$

For every $\beta \in \varGamma$, take a nonzero element $i^{(0)}_\beta \in \mathcal {I}_\beta .$ Set

$$\begin{aligned} i_0=\sum \limits _{\beta \in \varGamma }i^{(0)}_\beta . \end{aligned}$$

Fix a regular element $h_0$ in $\mathcal {H},$ in particular

$$\begin{aligned} \{x\in \mathcal {G}; [x, h_0]=0\}=\mathcal {H}. \end{aligned}$$

In the following two lemmas, we assume that $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}$ is a Leibniz algebra such that $\mathcal {G}$ and $\mathcal {I}$ are non-isomorphic as $\mathcal {G}$-modules.

Lemma 12

Let $D$ be a derivation on $\mathcal {L}$ such that $D(h_0+i_0)=0.$ Then $D|_{\mathcal {I}}\equiv 0$.

Proof

By (2), there exist an element $a\in \mathcal {G}$ and a number $\lambda \in \mathbb {C}$ such that $D=R_a+\lambda \text {pr}_\mathcal {I}.$ We have

$$\begin{aligned} 0=D(h_0+i_0)=[h_0+i_0, a]+\lambda i_0=[h_0, a]+[i_0, a]+\lambda i_0. \end{aligned}$$

Since $[h_0, a]\in \mathcal {G}$ and $[i_0, a]+\lambda i_0\in \mathcal {I},$ it follows that $[h_0, a]=0$ and $[i_0, a]+\lambda i_0=0.$ Since $h_0$ is a regular element, we have that $a\in \mathcal {H}.$ Further,

$$\begin{aligned} 0=[i_0, a]+\lambda i_0=\left[ \sum \limits _{\beta \in \varGamma }i^{(0)}_\beta , a\right] +\lambda i_0=\sum \limits _{\beta \in \varGamma }\beta (a)i^{(0)}_\beta +\lambda \sum \limits _{\beta \in \varGamma }i^{(0)}_\beta . \end{aligned}$$

Thus, $\beta (a)=-\lambda$ for all $\beta \in \varGamma .$

Let i be an arbitrary element of I, then it has a decomposition $i=\sum \nolimits _{\beta \in \varGamma }i_\beta ,$ where $i_\beta \in I_\beta ,\, \beta \in \varGamma .$ From $\beta (a)=-\lambda$ for all $\beta \in \varGamma$ we get

$$\begin{aligned} D(i)= & {} [i, a]+\lambda i=\left[ \sum \limits _{\beta \in \varGamma }i_\beta , a\right] +\lambda \sum \limits _{\beta \in \varGamma }i_\beta \\= & {} \sum \limits _{\beta \in \varGamma }\beta (a) i_\beta +\lambda \sum \limits _{\beta \in \varGamma }i_\beta =0. \end{aligned}$$

$\square$

Lemma 13

Let D be a derivation on $\mathcal {L}$ such that $D(h_0+i_0)=0.$ Then, $D=0.$

Proof

Let $y\in \mathcal {L}$ be an arbitrary element. Since $[y, y]\in \mathcal {I},$ Lemma 12 implies that $D([y, y])=0.$ The derivation identity

$$\begin{aligned} D([y, y])=[D(y), y]+[y, D(y)] \end{aligned}$$

implies that

$$\begin{aligned}{}[D(y), y]+[y, D(y)]=0. \end{aligned}$$

(5)

Putting $y=x+i\in \mathcal {G}+\mathcal {I}$ in (5), we obtain that

$$\begin{aligned}{}[D(x+i), x+i]+[x+i, D(x+i)]=0. \end{aligned}$$

Taking into account $D(x+i)=D(x)$, we have that

$$\begin{aligned}{}[D(x), x]+[x, D(x)]+[i, D(x)]=0. \end{aligned}$$

Using (5) from the last equality, we have $[\mathcal {I}, D(x)]=0.$ Further, for arbitrary elements $z\in \mathcal {G}$ and $i\in \mathcal {I}$, we have

$$\begin{aligned}{}[i, [D(x), z]]= & {} [[i, D(x)], z]-[[i, z], D(x)]=0 \end{aligned}$$

and

$$\begin{aligned}{}[i, [z, D(x)]]= & {} [[i, z], D(x)]-[[i, D(x)], z]=0. \end{aligned}$$

This means that $[\mathcal {I}, \mathcal {G}_{D(x)}]=0,$ where $\mathcal {G}_{D(x)}$ is an ideal in $\mathcal {G}$ generated by the element $D(x).$ Since $\mathcal {I}$ is an irreducible module over the simple Lie algebra $\mathcal {G},$ we obtain that $\mathcal {G}_{D(x)}=\{0\},$ i.e., $D(x)=0.$ Finally, $D(y)=D(x+i)=D(x)=0.$$\square$

Consider a decomposition for $\mathcal {G},$ called the root decomposition

$$\begin{aligned} \mathcal {G}=\mathcal {H}\oplus \bigoplus \limits _{\alpha \in \varPhi }\mathcal {G}_\alpha , \end{aligned}$$

where

$$\begin{aligned} \mathcal {G}_\alpha= & {} \{x\in \mathcal {G}: [h, x] = \alpha (h)x,\, \forall \, h \in \mathcal {H}\},\\ \varPhi= & {} \{\alpha \in \mathcal {H}^*\setminus \{0\}: \mathcal {G}_\alpha \ne \{0\}\}. \end{aligned}$$

The set $\varPhi$ is called the root system of $\mathcal {G},$ and subspaces $\mathcal {G}_\alpha$ are called the root subspaces.

Further, if $\mathcal {G}$ and $\mathcal {I}$ are isomorphic $\mathcal {G}$-modules, then the decomposition for $\mathcal {I}$ can be written as

$$\begin{aligned} \mathcal {I}= \mathcal {I}_0\oplus \bigoplus \limits _{\alpha \in \varPhi }\mathcal {I}_\alpha , \end{aligned}$$

where $\mathcal {I}_0=\theta (\mathcal {H})$ and $\mathcal {I}_\beta =\theta (\mathcal {G}_\beta )$ for all $\beta \in \varPhi .$ This follows from

$$\begin{aligned}{}[\theta (x_\beta ), h]=\theta ([x_\beta , h])=\beta (h)\theta (x_\beta ) \end{aligned}$$

and

$$\begin{aligned}{}[\theta (h'), h]=\theta ([h', h])=0 \end{aligned}$$

for all $h, h'\in \mathcal {H}$ and $x_\beta \in \mathcal {G}_\beta ,$$\beta \in \varPhi .$

For every $\beta \in \varPhi$, take a nonzero element $x^{(0)}_\beta \in \mathcal {G}_\beta$ and put

$x_0=\sum \limits _{\beta \in \varGamma }x^{(0)}_\beta$ and $i_0=\theta (x_0).$

It is clear that $[x^{(0)}_\beta , h]=\beta (h)x^{(0)}_\beta$ for all $h\in \mathcal {H}$ and $\beta \in \varPhi .$

By [14, Lemma 2.2], there exists an element $h_0\in \mathcal {H}$ such that $\alpha (h_0) \ne \beta (h_0)$ for every $\alpha , \beta \in \varPhi , \alpha \ne \beta .$ In particular, $\alpha (h_0) \ne 0$ for every $\alpha \in \varPhi .$ Such elements $h_0$ are called strongly regular elements of $\mathcal {G}.$ Again by [?, Lemma 2.2], every strongly regular element $h_0$ is a regular element, i.e.,

$$\begin{aligned} \{x\in \mathcal {G}: [h_0, x]=0\}=\mathcal {H}. \end{aligned}$$

Choose a fixed strongly regular element $h_0\in \mathcal {H}.$

Lemma 14

Suppose that $a, h\in \mathcal {H},$$h\ne 0$ and $\lambda , \omega \in \mathbb {C}$ are such that

$$\begin{aligned} \omega \theta (h)+[i_0, a]+\lambda i_0=0. \end{aligned}$$

Then, $a=0$ and $\lambda =\omega =0.$

Proof

We have

$$\begin{aligned} 0= & {} [i_0, a]+\lambda i_0+\omega \theta (h)=[\theta (x_0), a]+\lambda \theta (x_0)+\omega \theta (h)\\= & {} \theta ([x_0, a])+\lambda \theta (x_0)+\omega \theta (h)=\theta ([x_0, a]+\lambda x_0+\omega h). \end{aligned}$$

Since $\theta$ is an isomorphism of linear spaces $\mathcal {G}$ and $\mathcal {I},$ it follows that $[x_0, a]+\lambda x_0+\omega h=0.$ Further

$$\begin{aligned} 0= & {} [x_0, a]+\lambda x_0+\omega h=\left[ \sum \limits _{\beta \in \varPhi }x^{(0)}_\beta , a\right] +\lambda x_0+\omega h\\= & {} \sum \limits _{\beta \in \varPhi }\beta (a)x^{(0)}_\beta +\lambda \sum \limits _{\beta \in \varPhi }x^{(0)}_\beta +\omega h. \end{aligned}$$

Now we multiply both sides of this equality by the regular element $h_0\in \mathcal {H},$ then we get

$$\begin{aligned} 0=\sum \limits _{\beta \in \varPhi }\beta (a)\beta (h_0)x^{(0)}_\beta +\lambda \sum \limits _{\beta \in \varPhi }\beta (h_0)x^{(0)}_\beta . \end{aligned}$$

From this, we obtain $\beta (a)\beta (h_0)=-\lambda \beta (h_0)$ for all $\beta \in \varPhi .$ Since $h_0$ is a strongly regular element, it follows that $\beta (h_0)\ne 0$ for all $\beta \in \varPhi ,$ and then we derive that $\beta (a)=-\lambda .$ Putting in the last equality the root $-\beta$ instead of $\beta$, we obtain $\beta (a)=\lambda .$ Thus, $\lambda =0$ and $\beta (a)=0$ for all $\beta \in \varPhi .$ Since the set $\varPhi$ contains $k=\dim \mathcal {H}$ linearly independent elements, it follows that $\varPhi$ separates points of $\mathcal {H},$ and therefore we get $a=0.$ Further, from

$$\begin{aligned} \omega \theta (h)+[i_0, a]+\lambda i_0=0 \end{aligned}$$

we obtain that $\omega \theta (h)=0.$ Since $\theta (h)$ is nonzero, it follows that $\omega =0.$$\square$

Lemma 15

Let D be a derivation on $\mathcal {L}$ such that $D(h_0+i_0)=0.$ Then, $D=0.$

Proof

By (4), there exist an element $a\in \mathcal {G}$ and numbers $\lambda , \theta \in \mathbb {C}$ such that

$$\begin{aligned} D=R_a+\omega \theta +\lambda \text {pr}_\mathcal {I}. \end{aligned}$$

We have

$$\begin{aligned} 0=D(h_0+i_0)=[h_0+i_0, a]+\omega \theta (h_0)+\lambda i_0=[h_0, a]+\omega \theta (h_0)+[i_0, a]+\lambda i_0. \end{aligned}$$

Since $[h_0, a]\in \mathcal {G}$ and $\omega \theta (h_0)+[i_0, a]+\lambda i_0\in \mathcal {I},$ it follows that $[h_0, a]=0$ and $\omega \theta (h_0)+[i_0, a]+\lambda i_0=0.$ Since $h_0$ is a strongly regular element, we have that $a\in \mathcal {H}.$ Now Lemma 14 implies that $a=0$ and $\lambda =\omega =0.$ This means that $D=0.$$\square$

Now we are in position to prove Theorem 11.

Proof of Theorem 11

Let $\varDelta$ be a 2-local derivation on $\mathcal {L}.$ Take a derivation $D$ on $\mathcal {L}$ such that

$$\begin{aligned} \varDelta (h_0+i_0)=D(h_0+i_0). \end{aligned}$$

Consider the 2-local derivation $\varDelta -D.$ Let $x\in \mathcal {L}.$ Take a derivation $\delta$ on $\mathcal {L}$ such that

$$\begin{aligned} (\varDelta -D)(h_0+i_0)=\delta (h_0+i_0),\,\, (\varDelta -D)(x)=\delta (x). \end{aligned}$$

Since $\delta (h_0+i_0)=0,$ by Lemmata 13 and 15, we obtain that $\delta (x) =0,$ i.e., $\varDelta (x)=D(x).$ This means that $\varDelta =D$ is a derivation. $\square$

Remark 16

The analogues of Lemmas 13 and 15 are not true for simple Lie algebras.

Let $\mathcal {G}$ be a simple Lie algebra, and suppose that there exists a nonzero element $x_0\in \mathcal {G}$ such that

$$\begin{aligned} D(x_0)=0,\, D\in \text {Der}(\mathcal {G})\,\Rightarrow \, D=0. \end{aligned}$$

Take the inner derivation $D=R_{x_0}$ generated by the element $x_0.$ Then, $D(x_0)=0,$ but $D$ is a nontrivial derivation.

3.2 2-Local Derivations of Nilpotent Leibniz Algebras

In this subsection, under a certain assumption, we give a general construction of 2-local derivations which are not derivations for an arbitrary variety (not necessarily associative, Lie or Leibniz) of algebras. This construction is then applied to show that nilpotent Leibniz algebras always admit 2-local derivations which are not derivations.

For an arbitrary algebra $\mathcal {L}$ with multiplication denoted as xy, let

$$\begin{aligned} \mathcal {L}^2=\text {span}\{xy: x, y \in \mathcal {L}\} \end{aligned}$$

and

$$\begin{aligned} \text {Ann}(\mathcal {L})=\{x\in \mathcal {L}: xy=yx=0\,\,\text {for all}\,\, y\in \mathcal {L}\}. \end{aligned}$$

Note that a linear operator $\delta$ on $\mathcal {L}$ such that $\delta |_{\mathcal {L}^2}\equiv 0$ and $\delta (\mathcal {L})\subseteq \text {Ann}(\mathcal {L})$ is a derivation. Indeed, for every $x,y\in \mathcal {L}$, we have

$$\begin{aligned} \delta (xy)=0=\delta (x)y+x\delta (y). \end{aligned}$$

Theorem 17

Let $\mathcal {L}$ be an n-dimensional algebra with $n\ge 2.$ Suppose that

• (i) $\dim \mathcal {L}^2\le n-2;$

• (ii) the annihilator $\text {Ann}(\mathcal {L})$ of $\mathcal {L}$ is nontrivial.

Then, $\mathcal {L}$ admits a 2-local derivation which is not a derivation.

Proof

Let us consider a decomposition of $\mathcal {L}$ in the following form:

$$\begin{aligned} \mathcal {L}=\mathcal {L}^2\oplus V. \end{aligned}$$

Due to $\dim \mathcal {L}^2\le n-2,$ we have $\dim V=k\ge 2.$ Let $\{e_1,\ldots , e_k\}$ be a basis of V.

Let us define a homogeneous nonadditive function f on $\mathbb {C}^2$ as follows

$$\begin{aligned}: f(y_1, y_2) = \left\{ \begin{array}{ll} \frac{\textstyle y_1^2}{\textstyle y_2}, &{} \text {if}\,\, y_2\ne 0\\ 0, &{} \text {if}\,\, y_2= 0, \end{array} \right. \end{aligned}$$

where $(y_1, y_2)\in \mathbb {C}^2.$

Let us fix a nonzero element $z\in \text {Ann}(\mathcal {L}).$ Define an operator $\varDelta$ on $\mathcal {L}$ by

$$\begin{aligned} \varDelta (x)=f(\lambda _1,\lambda _2)z,\, \ x=x_1+\sum \limits _{i=1}^k\lambda _i e_i, \end{aligned}$$

where $\lambda _i\in \mathbb {C},$$i=1,\ldots ,k,$$x_1\in \mathcal {L}^2.$ The operator $\varDelta$ is not a derivation since it is not linear.

Let us now show that $\varDelta$ is a 2-local derivation. Define a linear operator $\delta$ on $\mathcal {L}$ by

$$\begin{aligned} \delta (x)=(a\lambda _1+b\lambda _2)z,\, \ x=x_1+\sum \limits _{i=1}^k \lambda _i e_i, \end{aligned}$$

where $a, b\in \mathbb {C}.$ Since $\delta |_{\mathcal {L}^2}\equiv 0$ and $\delta (\mathcal {L})\subseteq \text {Ann}(\mathcal {L})$, the operator $\delta$ is a derivation.

Let $x=x_1+\sum \nolimits _{i=1}^k \lambda _i e_i$ and $y=y_1+\sum \nolimits _{i=1}^k \mu _i e_i$ be elements of $\mathcal {L}.$ We choose a and b such that

$$\begin{aligned} \varDelta (x)=\delta (x),\,\, \varDelta (y)=\delta (y). \end{aligned}$$

Let us rewrite the above equalities as system of linear equations with respect to unknowns a, b as follows

$$\begin{aligned} \left\{ \begin{array}{ll} \lambda _1 a+\lambda _2 b &{} = f(\lambda _1,\lambda _2), \\ \mu _1 a+\mu _2 b &{} = f(\mu _1,\mu _2). \end{array} \right. \end{aligned}$$

Since the function f is homogeneous, the system has nontrivial solution. Therefore, $\varDelta$ is a 2-local derivation, as required. $\square$

For every nilpotent Leibniz algebra with nilindex equal to t, we have that $\{0\}\ne \mathcal {L}^{t-1}\subseteq \text {Ann}(\mathcal {L}).$ It is known [10] that up to isomorphism, there exists a unique n-dimensional nilpotent Leibniz which satisfies the condition $\dim \mathcal {L}^2=n-1$ which is the null-filiform algebra, i.e., $\dim \mathcal {L}^2\le n-2$ for all nilpotent Leibniz algebras except the null-filiform. Therefore, Theorem 17 implies the following result:

Corollary 18

Let $\mathcal {L}$ be a finite-dimensional nonnull-filiform nilpotent Leibniz algebra with $\dim \mathcal {L}\ge 2.$ Then, $\mathcal {L}$ admits a 2-local derivation which is not a derivation.

Remark 19

Let us show that every 2-local derivation of the null-filiform algebra $NF_n$ is a derivation.

It is known that the unique null-filiform algebra, denoted by $NF_n$, admits a basis $\{e_1, e_2, \ldots , e_n\}$ in which its table of multiplications is the following (see [10]):

$$\begin{aligned}{}[e_i,e_1]=e_{i+1}, \ 1\le i \le n. \end{aligned}$$

Let $D\in Der(NF_n)$ and $D(e_1)=\sum \nolimits _{i=1}^{n}\alpha _i e_i$ for some $\alpha _i\in \mathbb {C}$, then it is easy to check that

$$\begin{aligned} D(e_j)=j \alpha _1 e_j +\sum _{i=2}^{n-j+1}\alpha _ie_{j+i-1}, \ 2\le j \le n. \end{aligned}$$

(6)

Let $\varDelta$ be a 2-local derivation. For the elements $e_1, x, y\in NF_n$, there exist derivations $D_{e_1, x}$ and $D_{e_1, y}$ such that

$$\begin{aligned} \varDelta (e_1)=D_{e_1, x}(e_1)=D_{e_1, y}(e_1)=\sum \limits _{i=1}^{n}\alpha _i e_i. \end{aligned}$$

From (6), we conclude that each derivation on $NF_n$ is uniquely defined by its value on the element $e_1.$ Therefore, $D_{e_1, x}(z)=D_{e_1, y}(z)$ for any $z\in NF_n$. Thus, we obtain that if $\varDelta (e_1)=D_{e_1, x}(e_1)$ for some $D_{e_1, x}$, then $\varDelta (z)=D_{e_1, x}(z)$ for any $z\in NF_n,$ i.e., $\varDelta$ is a derivation.

4 Local Derivations on Leibniz Algebras

4.1 Local Derivations on Simple Leibniz Algebras

Now we shall give the main result concerning local derivations on simple Leibniz algebras.

Theorem 20

Let $\mathcal {L}$ be a simple Leibniz algebra. Then, any local derivation on $\mathcal {L}$ is a derivation.

Let us first consider a simple Leibniz algebra $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}$ such that $\mathcal {G}$ and $\mathcal {I}$ are isomorphic $\mathcal {G}$-modules.

Lemma 21

Let $\varDelta$ be a local derivation on $\mathcal {L}$ such that $\varDelta$ maps $\mathcal {L}$ into $\mathcal {I}.$ Then, $\varDelta$ is a derivation.

Proof

Fix a basis $\{x_1, \ldots , x_m\}$ in $\mathcal {G}.$ In this case, the system of vectors $\{y_i: y_i=\theta (x_i), i\in \overline{1, m}\}$ is a basis in $\mathcal {I},$ where $\theta$ is a module isomorphism of $\mathcal {G}$-modules $\mathcal {G}$ and $\mathcal {I},$ in particular, $\theta ([x, y])=[\theta (x), y]$ for all $x, y\in \mathcal {G}.$

For an element $x=x_i \,(i\in \overline{1, m})$, take an element $a_i\in \mathcal {G}$ and a number $\omega _i\in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (x_i)=[x_i, a_i]+\omega _i\theta (x_i). \end{aligned}$$

Since $\varDelta (x_i)\in \mathcal {I}$ and $[x_i, a_i]\in \mathcal {G},$ it follows that $[x_i, a_i]=0.$ Thus,

$$\begin{aligned} \varDelta (x_i)=\omega _i\theta (x_i)=\omega _i y_i. \end{aligned}$$

Now for the element $x=x_i+x_j,$ where $i\ne j,$ take an element $a_{i,j}\in \mathcal {G}$ and a number $\omega _{i,j}\in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (x_i+x_j)=[x_i+x_j, a_{i,j}]+\omega _{i, j}\theta (x_i+x_j)\in \mathcal {I}. \end{aligned}$$

Then, $[x_i+x_j, a_{i,j}]=0.$ Thus,

$$\begin{aligned} \varDelta (x_i+x_j)=\omega _{i,j}\theta (x_i+x_j)=\omega _{i,j}(y_i+y_j). \end{aligned}$$

On the other hand,

$$\begin{aligned} \varDelta (x_i+x_j)=\omega _i x_i+\omega _jy_j. \end{aligned}$$

Comparing the last two equalities, we obtain $\omega _i=\omega _j$ for all $i, j.$ This means that there exists a number $\omega \in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (x_i)=\omega y_i \end{aligned}$$

(7)

for all $i=1, \ldots , m.$

Now for $x=x_i+y_i\in \mathcal {G}+\mathcal {I}$, take an element $a_x\in \mathcal {G}$ and numbers $\omega _x, \lambda _x\in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (x_i+y_i)=[x_i+y_i, a_x]+\omega _x\theta (x_i)+\lambda _x y_i\in \mathcal {I}. \end{aligned}$$

Then, $[x_i, a_x]=0,$ and

$$\begin{aligned} 0=\theta (0)=\theta ([x_i, a_x])=[\theta (x_i), a_x]=[y_i, a_x]. \end{aligned}$$

Thus,

$$\begin{aligned} \varDelta (x_i+y_i)=\omega _x\theta (x_i)+\lambda _x y_i=(\omega _x+\lambda _x)y_i. \end{aligned}$$

Taking into account (7), we obtain that

$$\begin{aligned} \varDelta (y_i)=\varDelta (x_i+y_i)-\varDelta (x_i)=(\omega _x+\lambda _x) y_i-\omega y_i=(\omega _x-\omega +\lambda _x)y_i. \end{aligned}$$

This means that for every $i\in \{1, \ldots , m\}$, there exists a number $\lambda _i \in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (y_i)=\lambda _i y_i \end{aligned}$$

for all $i=1, \ldots , m.$

Now take an element $x=x_i+x_j+y_i+y_j\in \mathcal {G}+\mathcal {I},$ where $i\ne j.$ Since

$$\begin{aligned} \varDelta (x_i+x_j+y_i+y_j)= & {} [x_i+x_j+y_i+y_j, a_x]\\&+\omega _x\theta (x_i+x_j)+\lambda _x (y_i+y_j)\in \mathcal {I}, \end{aligned}$$

we get that $[x_i+x_j, a_x]=0,$ and therefore $[y_i+y_j, a_x]=0.$ Thus,

$$\begin{aligned} \varDelta (x_i+x_j+y_i+y_j)=\omega _x\theta (x_i+x_j)+\lambda _x (y_i+y_j)=(\omega _x+\lambda _x)(y_i+y_j). \end{aligned}$$

Taking into account (7), we obtain that

$$\begin{aligned} \varDelta (y_i+y_j)= & {} \varDelta (x_i+x_j+y_i+y_j)-\varDelta (x_i+x_j)\\= & {} (\omega _x+\lambda _x)(y_i+y_j)-\omega (y_i+y_j)= (\omega _x-\omega +\lambda _x)(y_i+y_j). \end{aligned}$$

On the other hand

$$\begin{aligned} \varDelta (y_i+y_j)=\varDelta (y_i)+\varDelta (y_j)=\lambda _i y_i+\lambda _j y_j. \end{aligned}$$

Comparing the last two equalities, we obtain that $\lambda _i=\lambda _j$ for all $i$ and $j.$ This means that there exists a number $\lambda \in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (y_i)=\lambda y_i \end{aligned}$$

(8)

for all $i=1, \ldots , m.$ Combining (7) and (8) we obtain that $\varDelta =\omega \theta +\lambda \text {pr}_\mathcal {I}.$ This means that $\varDelta$ is a derivation. $\square$

Let now $\varDelta$ be an arbitrary local derivation on $\mathcal {L}.$ For an arbitrary element $x\in \mathcal {L}$, take an element $a_x\in \mathcal {G}$ and a number $\omega _x\in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (x)=[x, a_x]+\omega _x\theta (x). \end{aligned}$$

Then, the mapping

$$\begin{aligned} x\in \mathcal {G}\rightarrow [x, a_x]\in \mathcal {G} \end{aligned}$$

is a well-defined local derivation on $\mathcal {G},$ and therefore by [6, Theorem 3.1] it is an inner derivation generated by an element $a\in \mathcal {G}.$ Then, the local derivation $\varDelta -R_a$ maps $\mathcal {L}$ into $\mathcal {I}.$ By Lemma 21, we get that $\varDelta -R_a$ is a derivation and therefore $\varDelta$ is also a derivation.

In the next lemma, we consider a simple Leibniz algebra $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}$ such that $\mathcal {G}$ and $\mathcal {I}$ are not isomorphic $\mathcal {G}$-modules.

Lemma 22

Let $\varDelta$ be a local derivation on $\mathcal {L}$ such that $\varDelta$ maps $\mathcal {L}$ into $\mathcal {I}.$ Then, $\varDelta$ is a derivation.

Proof

Fix a basis $\{y_1, \ldots , y_k\}$ in $\mathcal {I}.$ We can assume that for any $y_i$ there exists a weight $\beta _i$ such that $y_i\in \mathcal {I}_{\beta _i}.$

Let $h_0$ be a strongly regular element in $\mathcal {H}.$ For $y=h_0+y_i\in \mathcal {G}+\mathcal {I}$, take an element $a_y\in \mathcal {G}$ and a number $\lambda _y\in \mathbb {C}$ such that

$$\begin{aligned} \varDelta (h_0+y_i)=[h_0, a_y]+[y_i, a_y]+\lambda _y y_i\in \mathcal {I}. \end{aligned}$$

Then, $[h_0, a_y]=0,$ and therefore $a_y\in \mathcal {H}.$ Further,

$$\begin{aligned} \varDelta (y_i)=\varDelta (h_0+y_i)=[y_i, a_y]+\lambda _y y_i=(\beta _i(a_y)+\lambda _y)y_i. \end{aligned}$$

This means that there exist numbers $\lambda _i,\, i=1, \ldots , m$ such that

$$\begin{aligned} \varDelta (y_i)=\lambda _i y_i. \end{aligned}$$

Now we will show that $\lambda _1=\ldots =\lambda _m.$ Take $y_{i_1},$$y_{i_2},$$i_1\ne i_2.$ Denote $i_{\beta _1}=y_{i_1},$$i_{\beta _2}=y_{i_2}.$ We have

$$\begin{aligned} \varDelta (i_{\beta _1}+i_{\beta _2})=\lambda _{i_1}i_{\beta _1}+\lambda _{i_2}i_{\beta _2}. \end{aligned}$$

(9)

Without lost of generality, we can assume that $\beta _1$ is a fixed highest weight of $\mathcal {I}.$ It is known [16, page 108] that difference of two weights is represented as

$$\begin{aligned} \beta _1-\beta _2=n_1\alpha _1+\ldots +n_l\alpha _l, \end{aligned}$$

where $\alpha _1, \ldots , \alpha _l$ are simple roots of $\mathcal {G},$$n_1, \ldots , n_l$ are nonnegative integers.

Case 1$\alpha _0=n_1\alpha _1+\ldots +n_l\alpha _l$ is not a root. Consider an element

$$\begin{aligned} x=n_1e_{\alpha _1}+\ldots +n_le_{\alpha _l}+i_{\beta _1}+i_{\beta _2}. \end{aligned}$$

Take an element $a_x=h+\sum \nolimits _{\alpha \in \varPhi }c_\alpha e_\alpha \in \mathcal {G}$ and number $\lambda _x$ such that

$$\begin{aligned} \varDelta (x)=[x, a_x]+\lambda _x(i_{\beta _1}+i_{\beta _2}). \end{aligned}$$

Since $\varDelta (x)\in \mathcal {I},$ we obtain that

$$\begin{aligned} \left[ \sum \limits _{s=1}^ln_s e_{\alpha _s}, h+\sum \limits _{\alpha \in \varPhi }c_\alpha e_\alpha \right] =0. \end{aligned}$$

Let us rewrite the last equality as

$$\begin{aligned} \sum \limits _{s=1}^l n_s \alpha _s(h)e_{\alpha _s}+\sum \limits _{t=1}^l\sum \limits _{\alpha \in \varPhi } (*) e_{\alpha +\alpha _t}=0, \end{aligned}$$

where the symbols $(*)$ denote appropriate coefficients. The second summand does not contain any element of the form $e_{\alpha _s}.$ Indeed, if we assume that $\alpha _s=\alpha +\alpha _t,$ we have that $\alpha =\alpha _s-\alpha _t.$ But $\alpha _s-\alpha _t$ is not a root, because $\alpha _s, \alpha _t$ are simple roots. Hence, all coefficients of the first summand are zero, i.e.,

$$\begin{aligned} n_1 \alpha _1(h)=\ldots =n_l \alpha _l(h)=0. \end{aligned}$$

Further,

$$\begin{aligned} \varDelta (i_{\beta _1}+i_{\beta _2})=\varDelta (x)=[i_{\beta _1}+i_{\beta _2}, a_x]+\lambda _x(i_{\beta _1}+i_{\beta _2}). \end{aligned}$$

Let us calculate the commutator $[i_{\beta _1}+i_{\beta _2}, a_x].$ We have

$$\begin{aligned}{}[i_{\beta _1}+i_{\beta _2}, a_x]= & {} \left[ i_{\beta _1}+i_{\beta _2}, h+\sum \limits _{\alpha \in \varPhi }c_\alpha e_\alpha \right] \\= & {} \beta _1(h)i_{\beta _1}+\beta _2(h)i_{\beta _2}+\sum \limits _{t=1}^2\sum \limits _{\alpha \in \varPhi }(*) i_{\beta _t+\alpha }. \end{aligned}$$

The last summand does not contain $i_{\beta _1}$ and $i_{\beta _2},$ because $\beta _1-\beta _2$ is not a root by the assumption. This means that

$$\begin{aligned} \varDelta (i_{\beta _1}+i_{\beta _2})=(\beta _1(h)+\lambda _x)i_{\beta _1}+(\beta _2(h)+\lambda _x)i_{\beta _2}. \end{aligned}$$

(10)

The difference of the coefficients of the right side is

$$\begin{aligned} \beta _1(h)-\beta _2(h)=\sum \limits _{s=1}^l n_s\alpha _s(h)=0, \end{aligned}$$

because $n_1\alpha _1(h)=\ldots =n_l\alpha _l(h)=0.$ Finally, comparing coefficients in (9) and (10), we get

$$\begin{aligned} \lambda _{i_1}=\beta _1(h)+\lambda _x=\beta _2(h)+\lambda _x=\lambda _{i_2}. \end{aligned}$$

Case 2$\alpha _0=n_1\alpha _1+\ldots +n_l\alpha _l$ is a root. Since $\beta _1$ is a highest weight, we get $\dim \mathcal {I}_{\beta _1}=1.$ Further, since $\beta _1-\beta _2$ is a root, [17, Lemma 3.2.9] implies that $\dim \mathcal {I}_{\beta _2}=\dim \mathcal {I}_{\beta _1},$ and therefore there exist numbers $t_{-\alpha _0}\ne 0$ and $t_{\alpha _0}$ such that $[i_{\beta _1}, e_{-\alpha _0}]=t_{-\alpha _0}i_{\beta _2},\, [i_{\beta _2}, e_{\alpha _0}]=t_{\alpha _0}i_{\beta _1}.$ Consider an element

$$\begin{aligned} x=t_{-\alpha _0}e_{\alpha _0}+t_{\alpha _0}e_{-\alpha _0}+i_{\beta _1}+i_{\beta _2}. \end{aligned}$$

Take an element $a_x=h+\sum \limits _{\alpha \in \varPhi }c_\alpha e_\alpha \in \mathcal {G}$ and number $\lambda _x$ such that

$$\begin{aligned} \varDelta (x)=[x, a_x]+\lambda _x(i_{\beta _1}+i_{\beta _2}). \end{aligned}$$

Since $\varDelta (x)\in \mathcal {I},$ we obtain that

$$\begin{aligned} \left[ t_{-\alpha _0}e_{\alpha _0}+t_{\alpha _0}e_{-\alpha _0}, h+\sum \limits _{\alpha \in \varPhi }c_\alpha e_\alpha \right] =0. \end{aligned}$$

Let us rewrite the last equality as

$$\begin{aligned} \alpha _0(h)t_{-\alpha _0}e_{\alpha _0}-\alpha _0(h)t_{\alpha _0}e_{-\alpha _0}+(t_{-\alpha _0}c_{-\alpha _0}- t_{\alpha _0}c_{\alpha _0})h_{\alpha _0}+\sum \limits _{\alpha \ne \pm \alpha _0} (*) e_{\alpha \pm \alpha _0}=0, \end{aligned}$$

where $h_{\alpha _0}=[e_{\alpha _0}, e_{-\alpha _0}]\in \mathcal {H}.$ The last summand in the sum does not contain elements $e_{\alpha _0}$ and $e_{-\alpha _0}.$ Indeed, if we assume that $\alpha _0=\alpha -\alpha _0,$ we have that $\alpha =2\alpha _0.$ But $2\alpha _0$ is not a root. Hence, the first three coefficients of this sum are zero, i.e.,

$$\begin{aligned} \alpha _0(h)=0,\, t_{\alpha _0}c_{\alpha _0}=t_{-\alpha _0}c_{-\alpha _0}. \end{aligned}$$

(11)

Further,

$$\begin{aligned} \varDelta (i_{\beta _1}+i_{\beta _2})=\varDelta (x)=[i_{\beta _1}+i_{\beta _2}, a_x]+\lambda _x(i_{\beta _1}+i_{\beta _2}). \end{aligned}$$

Let us consider the product $[i_{\beta _1}+i_{\beta _2}, a_x].$ We have

$$\begin{aligned}{}[i_{\beta _1}+i_{\beta _2}, a_x]= & {} \left[ i_{\beta _1}+i_{\beta _2}, h+\sum \limits _{\alpha \in \varPhi }c_\alpha e_\alpha \right] \\= & {} [i_{\beta _1}, h]+c_{\alpha _0}[i_{\beta _2}, e_{\alpha _0}]+[i_{\beta _2}, h]+c_{-\alpha _0}[i_{\beta _1}, e_{-\alpha _0}]\\&+\, c_{\alpha _0}[i_{\beta _1}, e_{\alpha _0}]+c_{-\alpha _0}[i_{\beta _2}, e_{-\alpha _0}]+\sum \limits _{t=1}^2\sum \limits _{\alpha \ne \pm \alpha _0}c_\alpha [i_{\beta _t}, e_\alpha ]\\= & {} (\beta _1(h)+t_{\alpha _0}c_{\alpha _0})i_{\beta _1}+(\beta _2(h)+t_{-\alpha _0}c_{-\alpha _0})i_{\beta _2}\\&+\, (*) i_{2\beta _1-\beta _2}+(*) i_{2\beta _2-\beta _1}+\sum \limits _{t=1}^2 \sum \limits _{\alpha \ne \pm \alpha _0}(*) i_{\beta _t+\alpha }. \end{aligned}$$

The last three summands do not contain $i_{\beta _1}$ and $i_{\beta _2},$ because $\beta _1-\beta _2=\alpha _0$ and $\alpha \ne \pm \alpha _0.$ This means that

$$\begin{aligned} \varDelta (i_{\beta _1}+i_{\beta _2})=(\beta _1(h)+t_{\alpha _0}c_{\alpha _0}+ \lambda _x)i_{\beta _1}+(\beta _2(h)+t_{-\alpha _0}c_{-\alpha _0}+\lambda _x)i_{\beta _2}. \end{aligned}$$

(12)

Taking into account (11), we find the difference of coefficients in the right side:

$$\begin{aligned} (\beta _1(h)+t_{\alpha _0}c_{\alpha _0})-(\beta _2(h)+t_{-\alpha _0}c_{-\alpha _0})= \alpha _0(h)+t_{\alpha _0}c_{\alpha _0}-t_{-\alpha _0}c_{-\alpha _0}=0. \end{aligned}$$

Combining (9) and (12), we obtain that

$$\begin{aligned} \lambda _{i_1}=\beta _1(h)+t_{\alpha _0}c_{\alpha _0}+\lambda _x=\beta _2(h)+t_{-\alpha _0}c_{-\alpha _0}+\lambda _x=\lambda _{i_2}. \end{aligned}$$

So, we have proved that

$$\begin{aligned} \varDelta (x_i+y_i)=\lambda y_i, \end{aligned}$$

where $\lambda \in \mathbb {C}.$ This means that $\varDelta =\lambda \text {pr}_\mathcal {I}.$$\square$

Let $\varDelta$ be an arbitrary local derivation, and let $x\in \mathcal {G}$ be an arbitrary element. Take an element $a_x\in \mathcal {G}$ such that

$$\begin{aligned} \varDelta (x)=[x, a_x]. \end{aligned}$$

As in the case $\dim \mathcal {G}=\dim \mathcal {I},$ the mapping

$$\begin{aligned} x\in \mathcal {G}\rightarrow [x, a_x]\in \mathcal {G} \end{aligned}$$

is an inner derivation generated by an element $a\in \mathcal {G},$ and $\varDelta -R_a$ maps $\mathcal {L}$ into $\mathcal {I}.$ This means that $\varDelta$ is a derivation that completes the proof of Theorem  20.

4.2 Local Derivations on Filiform Leibniz Algebras

In this subsection, we consider a special class of nilpotent Leibniz algebras, so-called filiform Leibniz algebras, and show that they admit local derivations which are not derivations.

Theorem 23

Let $\mathcal {L}$ be a finite-dimensional filiform Leibniz algebra with $\dim \mathcal {L}\ge 3.$ Then, $\mathcal {L}$ admits a local derivation which is not a derivation.

For filiform Lie algebras, this result was proved in [6, Theorem 4.1]. Hence, it is suffices to consider filiform non-Lie Leibniz algebras.

Firstly, we consider the family of algebras $\mathcal {L}=F_1(\alpha _4,\alpha _5,\dots ,\alpha _n,\theta ).$ Let us define a linear operator D on $\mathcal {L}$ by

$$\begin{aligned} D\left( \sum \limits _{k=1}^n x_ke_k\right) =\alpha (x_1 + x_2)e_{n-1} + \beta x_3 e_n, \end{aligned}$$

(13)

where $\alpha , \beta \in \mathbb {C}.$

Lemma 24

The linear operator D on $\mathcal {L}$ defined by (13) is a derivation if and only if $\alpha =\beta .$

Proof

Suppose that the linear operator D defined by (13) is a derivation. Since $[e_1, e_1]=e_3,$ we have that

$$\begin{aligned} D([e_1, e_1])= & {} D(e_3)=\beta e_n \end{aligned}$$

and

$$\begin{aligned}{}[D(e_1), e_1]+ [e_1, D(e_1)]= & {} [\alpha e_{n-1}, e_1]+[e_1, \alpha e_{n-1}]=\alpha e_n. \end{aligned}$$

Thus, $\alpha =\beta .$

Conversely, let D be a linear operator defined by (13) with $\alpha =\beta .$ We may assume that $\alpha =\beta =1.$

In order to prove that D is a derivation, it is sufficient to show that

$$\begin{aligned} D([e_i, e_j])=[D(e_i),e_j]+[e_i,D(e_j)] \end{aligned}$$

for all $1\le i, j \le n.$

Case 1$i+j=2.$ Then, $i=j=1$, and in this case, we can check as above.

Case 2$i+j=3.$ Then,

$$\begin{aligned}{}[D(e_2),e_1]+[e_2,D(e_1)]= & {} [e_{n-1}, e_1]+[e_2, e_{n-1}]\\= & {} [e_{n-1}, e_1]=e_n=D(e_3)=D([e_2, e_1]) \end{aligned}$$

and

$$\begin{aligned}{}[D(e_1),e_2]+[e_1,D(e_2)]= & {} [e_{n-1}, e_2]+[e_1, e_{n-1}]=0\\= & {} D\left( \displaystyle \sum _{k=4}^{n-1} \alpha _k e_k+\theta e_n\right) =D([e_1, e_2]). \end{aligned}$$

Case 3$i+j\ge 4.$ Then,

$$\begin{aligned} D([e_i, e_j])=0=[D(e_i), e_j]+[e_i, D(e_j)]. \end{aligned}$$

$\square$

Now we consider the linear operator $\varDelta$ defined by (13) with $\alpha =1, \beta =0.$

Lemma 25

The linear operator $\varDelta$ is a local derivation which is not a derivation.

Proof

By Lemma 24, $\varDelta$ is not a derivation.

Let us show that $\varDelta$ is a local derivation. Denote by $D_1$ the derivation defined by (13) with $\alpha =\beta =1.$ Let $D_2$ be a linear operator on $\mathcal {L}$ defined by

$$\begin{aligned} D_2\left( \sum \limits _{k=1}^n x_ke_k\right) =(x_1+x_2) e_n. \end{aligned}$$

Since $D_2|_{\mathcal {[\mathcal {L},\mathcal {L}]}}\equiv 0$ and $D_2(\mathcal {L})\subseteq Z(\mathcal {L}),$ it follows that

$$\begin{aligned} D_2([x,y])=0=[D_2(x),y]+[x,D_2(y)] \end{aligned}$$

for all $x, y\in \mathcal {L}.$ So, $D_2$ is a derivation.

Finally, for any $x=\sum \nolimits _{k=1}^n x_ke_k$, we find a derivation D such that $\varDelta (x)=D(x).$

Case 1$x_1+x_2=0.$ Then,

$$\begin{aligned} \varDelta (x) = 0 = D_2(x). \end{aligned}$$

Case 2$x_1+x_2\ne 0.$ Set

$$\begin{aligned} D=D_1+tD_2, \end{aligned}$$

where $t=-\frac{\textstyle x_3}{\textstyle x_1+x_2}.$ Then,

$$\begin{aligned} D(x)= & {} D_1(x)+t D_2(x) = (x_1+x_2)e_{n-1}+x_3e_n+t(x_1+x_2)e_n = \\= & {} (x_1+x_2)e_{n-1}+x_3e_n-x_3e_n =\varDelta (x). \end{aligned}$$

So, $\varDelta$ is a local derivation. $\square$

Now let us consider the second and third classes.

For an algebra $\mathcal {L}=F_2(\beta _3,\beta _4,\ldots ,\beta _n,\gamma )$ from the second class, define a linear operator D on $\mathcal {L}$ by

$$\begin{aligned} D\left( \sum \limits _{k=1}^n x_ke_k\right) =\alpha x_1 e_{n-1} + \beta x_3 e_n, \end{aligned}$$

(14)

where $\alpha , \beta \in \mathbb {C}.$

For an algebra $\mathcal {L}=F_3(\theta _1,\theta _2,\theta _3)$, define a linear operator D on $\mathcal {L}$ by

$$\begin{aligned} D\left( \sum \limits _{k=1}^n x_ke_k\right) =\alpha x_2 e_{n-1}+\beta x_3 e_{n}, \end{aligned}$$

(15)

where $\alpha , \beta \in \mathbb {C}.$

As in the proof of Lemma 24, we can check that the operator $D$ defined by (14) or (15) is a derivation if and only if $\alpha =\beta .$ Therefore, the operator $D$ defined by (14) or (15) gives the example of a local derivations which is not a derivation.

5 Automorphisms of Simple Leibniz Algebras

Let $\mathcal {L}=\mathcal {G}\dot{+} \mathcal {I}$ be a simple Leibniz algebra, and let $\varphi$ be an automorphism of $\mathcal {L}$. The ideal generated by squares of elements of $\mathcal {L}$ coincides with the $\text {span}\{ [x, x]: x\in \mathcal {L}\}.$ Then, for $x=\sum \nolimits _{k=1}^n\lambda _i[x_i, x_i]\in \mathcal {I}$, we have

$$\begin{aligned} \varphi (x)= & {} \varphi \left( \sum \limits _{k=1}^n\lambda _i[x_i,x_i]\right) =\sum \limits _{k=1}^n\lambda _i\varphi ([x_i,x_i])\\= & {} \sum \limits _{k=1}^n\lambda _i[\varphi (x_i),\varphi (x_i)]= \sum \limits _{k=1}^n\lambda _i[y_i,y_i], \end{aligned}$$

i.e., $\varphi (\mathcal {I})\subseteq \mathcal {I}.$

Now let $y=\sum \nolimits _{k=1}^n\lambda _i[y_i,y_i]$ be an arbitrary element of the ideal $\mathcal {I}.$ Since $\varphi$ is an automorphism, for every $y_i\in \mathcal {L}$ there exists $x_i\in \mathcal {L}$ such that $\varphi (x_i)=y_i$. Then, we have

$$\begin{aligned} y= & {} \sum \limits _{k=1}^n\lambda _i[y_i,y_i]=\sum \limits _{k=1}^n\lambda _i[\varphi (x_i),\varphi (x_i)]\\= & {} \sum \limits _{k=1}^n \lambda _i\varphi ([x_i,x_i])= \varphi \left( \sum \limits _{k=1}^n\lambda _i[x_i,x_i]\right) . \end{aligned}$$

This implies that for the element $z=\sum \nolimits _{k=1}^n\lambda _i[x_i,x_i]\in \mathcal {I}$, we have $\varphi (z)=y$. So we have proved that $\mathcal {I} \subseteq \varphi (\mathcal {I}),$ and therefore $\varphi (\mathcal {I})=\mathcal {I}$.

Now we shall show that any $\varphi \in \text {Aut}(\mathcal {L})$ can be represented as the sum $\varphi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {G},\mathcal {I}}+\varphi _{\mathcal {I},\mathcal {I}}$, where $\varphi _{\mathcal {G},\mathcal {G}} : \mathcal {G} \rightarrow \mathcal {G}$ is an automorphism on $\mathcal {G},$$\varphi _{\mathcal {G},\mathcal {I}} : \mathcal {G} \rightarrow \mathcal {I}$ is a $\mathcal {G}$-module homomorphism from $\mathcal {G}$ into $\mathcal {I},$ and $\varphi _{\mathcal {I},\mathcal {I}} :\mathcal {I} \rightarrow \mathcal {I}$ is a $\mathcal {G}$-module isomorphism of $\mathcal {I}.$ In particular,

$$\begin{aligned} \varphi (x+i)=\varphi _{\mathcal {G},\mathcal {G}}(x)+\varphi _{\mathcal {G},\mathcal {I}}(x)+\varphi _{\mathcal {I},\mathcal {I}}(i),\, x+i\in \mathcal {G}+\mathcal {I}. \end{aligned}$$

Lemma 26

Let $\mathcal {L}=\mathcal {G}\dot{+} \mathcal {I}$ be a simple Leibniz algebra, and let $\varphi \in \text {Aut}(\mathcal {L})$ be an automorphism. Then,

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}, \end{aligned}$$

if $\dim \mathcal {G}\ne \dim \mathcal {I},$ and

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}, \end{aligned}$$

if $\dim \mathcal {G}=\dim \mathcal {I}.$

Proof

Let $x,y\in \mathcal {G}$, then

$$\begin{aligned} \varphi _{\mathcal {G},\mathcal {G}}([x,y])+\varphi _{\mathcal {G},\mathcal {I}}([x,y])= & {} \varphi ([x,y])=[\varphi (x),\varphi (y)]\\= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x)+\varphi _{\mathcal {G},\mathcal {I}}(x), \varphi _{\mathcal {G},\mathcal {G}}(y)+\varphi _{\mathcal {G},\mathcal {I}}(y)]\\= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x),\varphi _{\mathcal {G},\mathcal {G}}(y)]+[\varphi _{\mathcal {G},\mathcal {I}}(x), \varphi _{\mathcal {G},\mathcal {G}}(y)]. \end{aligned}$$

This implies

$$\begin{aligned} \varphi _{\mathcal {G},\mathcal {G}}([x,y])= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x),\varphi _{\mathcal {G},\mathcal {G}}(y)], \end{aligned}$$

(16)

$$\begin{aligned} \varphi _{\mathcal {G},\mathcal {I}}([x,y])= & {} [\varphi _{\mathcal {G},\mathcal {I}}(x), \varphi _{\mathcal {G},\mathcal {G}}(y)]. \end{aligned}$$

(17)

Set

$$\begin{aligned} \psi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}. \end{aligned}$$

Let us show that $\psi$ is also an automorphism. Indeed,

$$\begin{aligned} \psi ([x+i, y+j])= & {} \psi ([x,y]+[i,y])=\varphi _{\mathcal {G},\mathcal {G}}([x,y])+\varphi _{\mathcal {I},\mathcal {I}}([i,y])\\= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x),\varphi _{\mathcal {G},\mathcal {G}}(y)]+\varphi ([i,y])\\= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x), \varphi _{\mathcal {G},\mathcal {G}}(y)]+[\varphi (i), \varphi (y)]\\= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x),\varphi _{\mathcal {G},\mathcal {G}}(y)]+[\varphi _{\mathcal {I},\mathcal {I}}(i), \varphi _{\mathcal {G},\mathcal {G}}(y)+\varphi _{\mathcal {G},\mathcal {I}}(y)]\\= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x),\varphi _{\mathcal {G},\mathcal {G}}(y)]+[\varphi _{\mathcal {I},\mathcal {I}}(i), \varphi _{\mathcal {G},\mathcal {G}}(y)]\\ {}= & {} [\varphi _{\mathcal {G},\mathcal {G}}(x)+\varphi _{\mathcal {I},\mathcal {I}}(i),\varphi _{\mathcal {G},\mathcal {G}}(y)+ \varphi _{\mathcal {I},\mathcal {I}}(j)]\\= & {} [\psi (x+i),\psi (y+j)]. \end{aligned}$$

Now consider an automorphism

$$\begin{aligned} \eta =\varphi \circ \psi ^{-1}. \end{aligned}$$

Then,

$$\begin{aligned} \eta (x+i)=x+\eta _{\mathcal {G},\mathcal {I}}(x)+i, \end{aligned}$$

where $\eta _{\mathcal {G},\mathcal {I}}=\varphi _{\mathcal {G},\mathcal {I}}\circ \varphi ^{-1}_{\mathcal {G},\mathcal {G}}.$ Applying (16) and (17) to $\eta$ we obtain that

$$\begin{aligned} \eta _{\mathcal {G},\mathcal {I}}([x,y])=[\eta _{\mathcal {G},\mathcal {I}}(x), y]. \end{aligned}$$

This means that $\eta _{\mathcal {G},\mathcal {I}}$ is a $\mathcal {G}$-module homomorphism from $\mathcal {G}$ into $\mathcal {I}.$

Case 1 Let $\dim \mathcal {G}\ne \dim \mathcal {I}.$ In this case by Schur’s Lemma, we obtain that $\eta _{\mathcal {G},\mathcal {I}}=0.$ Now the equality $\eta _{\mathcal {G},\mathcal {I}}=\varphi _{\mathcal {G},\mathcal {I}}\circ \varphi ^{-1}_{\mathcal {G},\mathcal {G}}$ implies that $\varphi _{\mathcal {G},\mathcal {I}}=0.$ Thus,

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}. \end{aligned}$$

Case 2 Let $\dim \mathcal {G}= \dim \mathcal {I}.$ In this case, again by Schur’s Lemma, we obtain that $\eta _{\mathcal {G},\mathcal {I}}=\omega \theta ,$ where $\omega \in \mathbb {C}.$ Thus, $\varphi _{\mathcal {G},\mathcal {I}}=\eta _{\mathcal {G},\mathcal {I}} \circ \varphi _{\mathcal {G},\mathcal {G}}=\omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}},$ and therefore

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}. \end{aligned}$$

$\square$

Further, we shall use the following lemma.

Lemma 27

Let $\varphi \in \text {Aut}(\mathcal {L})$ be an automorphism. Then,

• (a) if $\varphi _{\mathcal {G},\mathcal {G}}=id_\mathcal {G},$ then $\varphi _{\mathcal {I},\mathcal {I}}=\lambda id_\mathcal {I};$

• (b) if $\varphi _{\mathcal {I},\mathcal {I}}=id_\mathcal {I},$ then $\varphi _{\mathcal {G},\mathcal {G}}= id_\mathcal {G}.$

Proof

• (a) Similar to the proof of (17), we obtain that

  $$\begin{aligned} \varphi _{\mathcal {I},\mathcal {I}}([i,x])=[\varphi _{\mathcal {I},\mathcal {I}}(i), \varphi _{\mathcal {G},\mathcal {G}}(x)]=[\varphi _{\mathcal {I},\mathcal {I}}(i), x] \end{aligned}$$

  for all $i\in \mathcal {I}, x\in \mathcal {G}.$ By Schur’s lemma, we obtain that $\varphi _{\mathcal {I},\mathcal {I}}=\lambda id_\mathcal {I}.$

• (b) Since

  $$\begin{aligned}{}[i, x]=\varphi _{\mathcal {I},\mathcal {I}}([i,x])=[\varphi _{\mathcal {I},\mathcal {I}}(i), \varphi _{\mathcal {G},\mathcal {G}}(x)]=[i, \varphi _{\mathcal {G},\mathcal {G}}(x)], \end{aligned}$$

  we obtain that $[i, \varphi _{\mathcal {G},\mathcal {G}}(x)-x]=0$ for all $i\in \mathcal {I}, x\in \mathcal {G}.$ Similar as in the proof of Lemma 13, we have $[\mathcal {I}, \mathcal {G}_{\varphi _{\mathcal {G},\mathcal {G}}(x)-x}]=0,$ and $\varphi _{\mathcal {G},\mathcal {G}}(x)-x=0.$

$\square$

Lemma 28

Let $\mathcal {L}=\mathcal {G}\dot{+} \mathcal {I}$ be a simple Leibniz algebra with $\dim \mathcal {G}=\dim \mathcal {I}.$ Then, any automorphism $\varphi \in \text {Aut}(\mathcal {L})$ can be represented as

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}}+\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}, \end{aligned}$$

where $\omega \in \mathbb {C}$ and $0\ne \lambda \in \mathbb {C}.$

Proof

Let $\varphi _{\mathcal {G},\mathcal {G}}$ be an automorphism of $\mathcal {G}.$ Let us show that $\phi =\varphi _{\mathcal {G},\mathcal {G}}+\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}$ is an automorphism. It suffices to check that

$$\begin{aligned} \phi ([i, x])=[\phi (i), \phi (x)] \end{aligned}$$

for all $x\in \mathcal {G},$$i\in \mathcal {I}.$ Since $\theta$ is a $\mathcal {G}$-module isomorphism, it follows that $\theta ([\theta ^{-1}(i), x])=[i,x].$ We have

$$\begin{aligned} \phi ([i,x])= & {} \lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}([i,x])=\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}(\theta ([\theta ^{-1}(i), x])\\= & {} \lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}([\theta ^{-1}(i),x])=\lambda \theta ([\varphi _{\mathcal {G},\mathcal {G}}(\theta ^{-1}(i)),\varphi _{\mathcal {G},\mathcal {G}}(x)])\\= & {} [(\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1})(i),\varphi _{\mathcal {G},\mathcal {G}}(x)]= [\phi (i),\phi (x)]. \end{aligned}$$

Let us consider

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}. \end{aligned}$$

Set

$\psi =\varphi _{\mathcal {G},\mathcal {G}}+ \varphi _{\mathcal {I},\mathcal {I}},\,$$\phi =\varphi _{\mathcal {G},\mathcal {G}}+ \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}$ and $\eta =\psi \circ \phi ^{-1}.$

Then, $\eta =\text {id}_{\mathcal {G}}+\eta _{\mathcal {I},\mathcal {I}},$ and therefore by Lemma 27, it follows that $\eta _{\mathcal {I},\mathcal {I}}=\lambda \text {id}_\mathcal {I}.$ Thus,

$$\begin{aligned} \psi =\eta \circ \phi =(\text {id}_{\mathcal {G}}+\lambda \text {id}_\mathcal {I})\circ (\varphi _{\mathcal {G},\mathcal {G}}+ \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1})=\varphi _{\mathcal {G},\mathcal {G}}+\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}. \end{aligned}$$

Hence,

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}}+\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}. \end{aligned}$$

$\square$

6 Local and 2-Local Automorphisms on Simple Leibniz Algebras

6.1 2-Local Automorphisms of Simple Leibniz Algebras

Let $\mathcal {L}=\mathcal {G}\dot{+} \mathcal {I}$ be a simple Leibniz algebra. Then, any $\varphi \in Aut(\mathcal {L})$ decomposes into

$$\begin{aligned} \varphi= & {} \varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {G},\mathcal {I}}+\varphi _{\mathcal {I},\mathcal {I}}, \\ \varphi _{\mathcal {G},\mathcal {I}}= & {} \omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}} \end{aligned}$$

where $\omega \in \mathbb {C}$ and $\varphi _{\mathcal {G},\mathcal {I}}$ is assumed to be zero when $\dim \mathcal {G}\ne \dim \mathcal {I}.$

Lemma 29

Let $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}$ be a simple Leibniz algebra and let $\varphi \in Aut(\mathcal {L})$ be such that $\varphi (h_0)=h_0,$ where $h_0$ is a strongly regular element from $\mathcal {H}.$ Then,

• (a) $\varphi (e_\alpha )=t_\alpha e_\alpha$ and $\varphi (e_{-\alpha })=t_\alpha ^{-1} e_{-\alpha },$ where $t_\alpha \in \mathbb {C}$ for all $\alpha \in \varPhi ,$

• (b) $\varphi (h)=h$ for all $h\in \mathcal {H}.$

Proof

Let $\varphi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {G},\mathcal {I}}+\varphi _{\mathcal {I},\mathcal {I}}.$ Since

$$\begin{aligned} h_0=\varphi (h_0)= \varphi _{\mathcal {G},\mathcal {G}}(h_0)+ \varphi _{\mathcal {G},\mathcal {I}}(h_0), \end{aligned}$$

it follows that $\varphi _{\mathcal {G},\mathcal {G}}(h_0)=h_0$ and $\varphi _{\mathcal {G},\mathcal {I}}(h_0)=0$ (that is $\theta (h_0)=0$). Thus, $\varphi _{\mathcal {G},\mathcal {I}}\equiv 0.$ Now assertions (a) and (b) follow from [7, Lemma 2.2]. $\square$

Let $\mathcal {H}$ be a Cartan subalgebra of $\mathcal {G}$, and let

$$\begin{aligned} \mathcal {G}=\mathcal {H}\oplus \bigoplus _{\alpha \in \varPhi }\mathcal {G}_{\alpha }, \quad \mathcal {I}=\bigoplus _{\beta \in \varGamma }\mathcal {I}_\beta . \end{aligned}$$

It is known [16, page 108] that the $\mathcal {G}$-module $\mathcal {I}$ is generated by the elements of the form

$$\begin{aligned}{}[\ldots [[i^{+}, e_{-\alpha _1}], e_{-\alpha _2}], \ldots , e_{-\alpha _k}], \end{aligned}$$

(18)

where $i^{+}$ is a highest weight vector of $\mathcal {G}$-module $\mathcal {I},$$e_{-\alpha _i}\in \mathcal {G}_{-\alpha _i}$ and $\alpha _i \in \varPhi$ is a positive root for all $i=1,\ldots , k,$$k\in \mathbb {N}.$

Let $\dim \mathcal {I}_\beta =s_\beta , \ \beta \in \varGamma$, and let $\left\{ i^{(1)}_\beta , \ldots , i^{(s_\beta )}_\beta \right\}$ be a basis of $\mathcal {I}_\beta ,$ consisting of elements of form (18). Set

$$\begin{aligned} i_0=\sum _{\beta \in \varGamma }\sum _{k=1}^{s_\beta }i^{(k)}_{l_\beta }. \end{aligned}$$

(19)

Lemma 30

Let $\mathcal {L}=\mathcal {G}\dot{+} \mathcal {I}$ be a simple Leibniz algebra with $\dim \mathcal {G}\ne \dim \mathcal {I}$, and let $\varphi \in Aut(\mathcal {L})$ be such that $\varphi (h_0+i_0)=h_0+i_0.$ Then, $\varphi$ is an identity automorphism of $\mathcal {L}.$

Proof

Since

$$\begin{aligned} h_0+i_0=\varphi (h_0+i_0)=\varphi _{\mathcal {G},\mathcal {G}}(h_0)+\varphi _{\mathcal {I},\mathcal {I}}(i_0), \end{aligned}$$

it follows that $\varphi _{\mathcal {G},\mathcal {G}}(h_0)=h_0$ and $\varphi _{\mathcal {I},\mathcal {I}}(i_0)=i_0.$ By Lemma 29, $\varphi _{\mathcal {G},\mathcal {G}}$ acts diagonally on $\mathcal {G},$ i.e., $\varphi _{\mathcal {G},\mathcal {G}}(e_\alpha )=t_\alpha e_\alpha ,$$t_\alpha \in \mathbb {C},$$\alpha \in \varPhi .$

Let $i_+$ be the highest weight vector of $\mathcal {I}$, and let $\alpha$ be a positive root. Then,

$$\begin{aligned} {[}\varphi (i_+), e_\alpha ]= & {} t_\alpha ^{-1}[\varphi (i_+), \varphi (e_\alpha )]=t_\alpha ^{-1}\varphi ([i_+, e_\alpha ])=\varphi (0)=0 \\ {[}\varphi (i_+), h]= & {} [\varphi (i_+), \varphi (h)]=\alpha _+(h)\varphi (i_+), \end{aligned}$$

where $\alpha _+\in \varGamma$ is a highest weight. This means that $\varphi (i_+)$ is also a highest weight vector. Since the highest weight subspace is one-dimensional, it follows that $\varphi (i_{+})=\lambda _+ i_{+},$ where $\lambda _+\in \mathbb {C}.$ Now taking into account that $\varphi (i_{+})=\lambda _+ i_{+}$ from the previous lemma, we conclude that

$$\begin{aligned} \varphi \left( i^{(k)}_{\beta }\right)= & {} \lambda _+[\ldots [[i_{+}, \varphi (e_{-\alpha _1})], \varphi (e_{-\alpha _2})],\ldots , \varphi (e_{-\alpha _k})]\\= & {} \left( \prod \limits _{p=1}^k t_{-\alpha _p}\lambda _+\right) [\ldots [[i_{+}, e_{-\alpha _1}], e_{-\alpha _2}],\ldots , e_{-\alpha _k}]=c_\beta ^{(k)}\lambda _+i_\beta ^{(k)}, \end{aligned}$$

i.e.,

$$\begin{aligned} \varphi \left( i^{(k)}_{\beta }\right) =c_\beta ^{(k)}\lambda _+i_\beta ^{(k)}, \end{aligned}$$

for $1\le k\le s_\beta , \beta \in \varGamma .$ Taking into account these equalities, from $i_0=\varphi _{\mathcal {I},\mathcal {I}}(i_0),$ we obtain $\varphi (i^{(k)}_{\beta })=i^{(k)}_{\beta }$ for all $1\le k\le s_\beta , \beta \in \varGamma .$ This implies $\varphi (i)=i$ for all $i\in \mathcal {I}.$ By Lemma 27, it follows that $\varphi =id_\mathcal {L}.$$\square$

Let $\dim \mathcal {H}=k\ge 2$, and let $\alpha _1, \ldots , \alpha _k$ be simple roots. Set $i_s=\theta (h_{\alpha _s}),$$s=1,\ldots , k,$$i_\alpha =\theta (e_\alpha ),$$\alpha \in \varPhi .$ Take

$$\begin{aligned} i_0=\sum \limits _{s=1}^k i_s+\sum \limits _{\alpha \in \varPhi } i_\alpha . \end{aligned}$$

Lemma 31

Let $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}$ be a simple Leibniz algebra with $\dim \mathcal {G}=\dim \mathcal {I}$ and $\dim \mathcal {H}=k\ge 2.$ Suppose that $\varphi$ is an automorphism on $\mathcal {L}$ such that $\varphi (h_0+i_0)=h_0+i_0.$ Then, $\varphi =\text {id}_\mathcal {L}.$

Proof

Let $\varphi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {G},\mathcal {I}}+\varphi _{\mathcal {I},\mathcal {I}}.$ Since

$$\begin{aligned} \varphi _{\mathcal {G},\mathcal {G}}(h_0)=h_0,\, \varphi _{\mathcal {G},\mathcal {I}}(h_0)+\varphi _{\mathcal {I},\mathcal {I}}(i_0)=i_0, \end{aligned}$$

by Lemma 29 we have that $\varphi _{\mathcal {G},\mathcal {G}}(e_\alpha )=t_\alpha e_\alpha ,$$\alpha \in \varPhi$ and $\varphi _{\mathcal {G},\mathcal {G}}(h)=h$ for all $h\in \mathcal {H}.$ Then,

$$\begin{aligned} \varphi _{\mathcal {I},\mathcal {I}}(i_s)=\lambda \theta (\varphi _{\mathcal {G},\mathcal {G}}(\theta ^{-1}(i_s)))=\lambda \theta (\varphi _{\mathcal {G},\mathcal {G}}(h_{\alpha _s}))=\lambda \theta (h_{\alpha _s})=\lambda i_s \end{aligned}$$

and

$$\begin{aligned} \varphi _{\mathcal {I},\mathcal {I}}(i_\alpha )=\lambda \theta (\varphi _{\mathcal {G},\mathcal {G}}(\theta ^{-1}(i_\alpha )))=\lambda \theta (\varphi _{\mathcal {G},\mathcal {G}}(e_{\alpha }))=\lambda \theta (e_{\alpha })=\lambda t_\alpha i_\alpha . \end{aligned}$$

Further,

$$\begin{aligned} i_0= & {} \varphi _{\mathcal {G},\mathcal {I}}(h_0)+\varphi _{\mathcal {I},\mathcal {I}}(i_0)= \omega \theta (h_0)+\varphi _{\mathcal {I},\mathcal {I}}(i_0)\\= & {} \omega \sum \limits _{s=1}^k a_s i_s+\sum \limits _{s=1}^k \lambda i_s+\sum \limits _{\alpha \in \varPhi } \lambda t_\alpha i_\alpha , \end{aligned}$$

i.e.,

$$\begin{aligned} \sum \limits _{\alpha \in \varPhi } (1-\lambda t_\alpha ) i_\alpha = \sum \limits _{s=1}^k (\omega a_s +\lambda -1)i_s. \end{aligned}$$

Since the right side of this equality belongs to $\mathcal {I}_0$ and the left side does not belong to $\mathcal {I}_0,$ it follows that $\lambda t_\alpha =1$ for all $\alpha \in \varPhi .$ Hence, $\varphi _{\mathcal {I},\mathcal {I}}(i_\alpha )=i_\alpha$ for all $\alpha \in \varPhi .$

Since $\dim \mathcal {H}\ge 2,$ any row of the Cartan matrix of a simple Lie algebra $\mathcal {G}$ contains a negative number (see [16, page 59]). This means that for any simple root $\alpha _i\in \varPhi$, there exists a simple root $\alpha _j\in \varPhi$ such that

$$\begin{aligned} a_{i,j}=(\alpha _i, \alpha _j)<0, \end{aligned}$$

where $(\cdot , \cdot )$ is a bilinear form on $\mathcal {H}^*$ induced by the Killing form on $\mathcal {G}.$ Then, by [16, page 45, Lemma 9.4], we obtain that $\alpha _i+\alpha _j$ is also a root and $[e_{\alpha _j}, e_{\alpha _i}]=n_{\alpha _j,\alpha _i}e_{\alpha _i+\alpha _j},$ where $n_{\alpha _j,\alpha _i}$ is a non zero integer.

Further,

$$\begin{aligned}{}[i_{\alpha _j}, e_{\alpha _i}]= & {} [\theta (e_{\alpha _j}), e_{\alpha _i}]=\theta \left( [e_{\alpha _j}, e_{\alpha _i}]\right) =\theta (n_{\alpha _j,\alpha _i}e_{\alpha _i+\alpha _j})\\= & {} n_{\alpha _j,\alpha _i}i_{\alpha _i+\alpha _j}. \end{aligned}$$

Applying to this equality $\varphi ,$ we obtain that

$$\begin{aligned} n_{\alpha _j,\alpha _i}i_{\alpha _i+\alpha _j}= & {} n_{\alpha _j,\alpha _i}\varphi (i_{\alpha _i+\alpha _j})=\varphi ([i_{\alpha _j}, e_{\alpha _i}])=[\varphi (i_{\alpha _j}), \varphi (e_{\alpha _i})]\\= & {} [i_{\alpha _j},t_{\alpha _i} e_{\alpha _i}]=n_{\alpha _j,\alpha _i}t_{\alpha _i}i_{\alpha _i+\alpha _j}. \end{aligned}$$

Thus, $t_{\alpha _i}=1$ for all $i=1,\ldots , k,$ i.e., $\varphi _{\mathcal {G},\mathcal {G}}$ acts identically on the subset of all simple roots $\{h_{\alpha _i}, e_{\alpha _i}, e_{-\alpha _i}: 1\le i \le k\}.$ Since this subset generates the algebra $\mathcal {G},$ it follows that $\varphi _{\mathcal {G},\mathcal {G}}$ acts identically on $\mathcal {G},$ i.e., $\varphi _{\mathcal {G},\mathcal {G}}=\text {id}_\mathcal {G}.$ By Lemma 27, there exists a number $\lambda$ such that $\varphi _{\mathcal {I},\mathcal {I}}=\lambda \text {id}_\mathcal {I}.$ Since $\varphi _{\mathcal {I},\mathcal {I}}(i_\alpha )=i_\alpha ,$ it follows that $\lambda =1,$ i.e., $\varphi _{\mathcal {I},\mathcal {I}}=\text {id}_\mathcal {I}.$

Finally,

$$\begin{aligned} i_0= & {} \varphi _{\mathcal {G},\mathcal {I}}(h_0)+\varphi _{\mathcal {I},\mathcal {I}}(i_0)=\varphi _{\mathcal {G},\mathcal {I}}(h_0)+i_0, \end{aligned}$$

implies that $\varphi _{\mathcal {G},\mathcal {I}}(h_0)=0,$ and therefore $\varphi _{\mathcal {G},\mathcal {I}}\equiv 0.$ So,

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\varphi _{\mathcal {I},\mathcal {I}}=\text {id}_\mathcal {L}. \end{aligned}$$

$\square$

Example 32

Lemma 31 is not true for algebras with $\dim \mathcal {H}=1.$

There is a unique simple Leibniz algebra with one-dimensional Cartan subalgebra and $\dim \mathcal {G}=\dim \mathcal {I}.$ This is the six-dimensional simple Leibniz algebra

$$\begin{aligned} \mathcal {L}=\mathfrak {sl}_2\dot{+}\text {span}\{x_0, x_1, x_2\}=\text {span}\{h, e, f, x_0, x_1, x_2\}, \end{aligned}$$

and nonzero products of the basis vectors in $\mathcal {L}$ are represented as follows [24]:

$$\begin{aligned} \begin{array}{llll} \, [x_k,e]=-k(3-k)x_{k-1}, &{} k\in \{1,2\},&{}\\ \, [x_k,f]=x_{k+1}, &{} k\in \{0, 1\}, &{} \\ \, [x_k,h]=(2-2k)x_k, &{} k\in \{0, 1,2\},&{}\\ \, [e,h]=2e, &{} [h,f]=2f, &{}[e,f]=h, \\ \, [h,e]=-2e,&{} [f,h]=-2f, &{} [f,e]=-h.\\ \end{array} \end{aligned}$$

Note that the $\mathfrak {sl}_2$-module isomorphism $\theta :\mathfrak {sl}_2\rightarrow \mathcal {I}$ is defined by

$$\begin{aligned} \theta (h)=2x_1,\, \theta (e)=2x_0,\, \theta (f)=x_2. \end{aligned}$$

Let $\varphi \in Aut(\mathcal {L})$ be an automorphism such that $\varphi (h_0+i_0)=h_0+i_0,$ where $h_0=h,$$i_0=x_1+x_0+x_2.$ Then, either $\varphi =\text {id}_\mathcal {L},$ or either

$$\begin{aligned} \varphi =\varphi _{\mathcal {G},\mathcal {G}}+\theta \circ \varphi _{\mathcal {G},\mathcal {G}}- \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}, \end{aligned}$$

(20)

where

$$\begin{aligned} \varphi _{\mathcal {G},\mathcal {G}}(h)=h,\, \varphi _{\mathcal {G},\mathcal {G}}(e)=-e,\, \varphi _{\mathcal {G},\mathcal {G}}(f)=-f. \end{aligned}$$

Let $\varphi =\varphi _{\mathcal {G},\mathcal {G}}+ \omega \theta \circ \varphi _{\mathcal {G},\mathcal {G}}+\lambda \theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}.$ Then, $\varphi (h+i_0)=h+i_0$ implies that $\varphi _{\mathcal {G},\mathcal {G}}(h)=h$ and $\omega \theta (h)+\lambda \theta (\varphi _{\mathcal {G},\mathcal {G}}(\theta ^{-1}(i_0)))=i_0.$ Using Lemma 29, we obtain that

$$\begin{aligned}&\,&\varphi _{\mathcal {G},\mathcal {G}}(h)=h,\,\varphi _{\mathcal {G},\mathcal {G}}(e)=t e,\,\varphi _{\mathcal {G},\mathcal {G}}(f)=t^{-1} f. \end{aligned}$$

Since

$$\begin{aligned} x_1+x_0+x_2=i_0=\omega \theta (h)+\lambda \theta (\varphi _{\mathcal {G},\mathcal {G}}(\theta ^{-1}(i_0)))= (2\omega +\lambda )x_1+\lambda t x_0+\lambda t^{-1}x_2, \end{aligned}$$

it follows that $2\omega +\lambda =1,\, \lambda t =\lambda t^{-1}=1.$

Case 1$\lambda =t=1.$ In this case, $\omega =0.$ Thus, $\varphi _{\mathcal {G},\mathcal {G}}=\text {id}_\mathcal {G},$ and therefore $\theta \circ \varphi _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}=\text {id}_\mathcal {I}.$ Hence, $\varphi =\text {id}_\mathcal {L}.$

Case 2$\lambda =t=-1.$ In this case, $\omega =1,$ and we obtain an automorphism of form (20).

Lemma 33

Let $\nabla$ be a 2-local automorphism of the simple Leibniz algebra $\mathcal {L}=\mathfrak {sl_2}\oplus \mathcal {I},$ where $\mathcal {I}=\text {span}\{x_0, x_1, x_2\},$ such that $\nabla (h)=h$ and $\nabla (h+i_0)=h+i_0,$ where $i_0=x_0+x_1+x_2.$ Then, $\nabla =\text {id}_\mathcal {L}.$

Proof

Let $x\in \mathfrak {sl}_2.$ Take an automorphism $\varphi ^{x,h}=\varphi ^{x,h}_{\mathfrak {sl}_2,\mathfrak {sl}_2}+\varphi ^{x,h}_{\mathfrak {sl}_2,\mathcal {I}} +\varphi ^{x,h}_{\mathcal {I},\mathcal {I}}\in \text {Aut}(\mathcal {L})$ such that $\nabla (x)=\varphi ^{x,h}(x)$ and $\nabla (h)=\varphi ^{x,h}(h).$ Since

$$\begin{aligned} h=\nabla (h)=\varphi ^{x,h}(h)=\varphi ^{x,h}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(h)+\varphi ^{x,h}_{\mathfrak {sl}_2, \mathcal {I}}(h), \end{aligned}$$

it follows that $\varphi ^{x,h}_{\mathfrak {sl}_2,\mathcal {I}}\equiv 0.$ Then

$$\begin{aligned} \nabla (x)=\varphi ^{x,h}(x)=\varphi ^{x,h}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(x)\in \mathfrak {sl}_2, \ \forall \ x\in \mathfrak {sl}_2. \end{aligned}$$

Let now $x\in \mathfrak {sl}_2$ be a nonzero element. Take an automorphism $\varphi ^{x,h+i_0}=\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}+ \varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathcal {I}}+\varphi ^{x,h+i_0}_{\mathcal {I},\mathcal {I}}$ such that $\nabla (x)=\varphi ^{x,h+i_0}(x)$ and $\nabla (h+i_0)=\varphi ^{x,h+i_0}(h+i_0).$ Since $\nabla (x)\in \mathfrak {sl}_2$ and

$$\begin{aligned} \nabla (x)=\varphi ^{x,h+i_0}(x)=\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(x) +\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathcal {I}}(x), \end{aligned}$$

we have that $\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathcal {I}}\equiv 0.$ Then

$$\begin{aligned} h+i_0=\nabla (h+i_0)=\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(h)+ \varphi ^{x,h+i_0}_{\mathcal {I},\mathcal {I}}(i_0), \end{aligned}$$

implies that $\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(h)=h$ and $\varphi ^{x,h+i_0}_{\mathcal {I},\mathcal {I}}(i_0)=i_0.$ By Example 32, we have that $\varphi ^{x,h+i_0}\equiv \text {id}_{\mathcal {L}},$ and therefore

$$\begin{aligned} \nabla (x)=\varphi ^{x,h+i_0}(x)=x. \end{aligned}$$

Let $x\in \mathfrak {sl}_2$ be a nonzero element, and let $i\in \mathcal {I}.$ Take an automorphism $\varphi ^{x,x+i}=\varphi ^{x,x+i}_{\mathfrak {sl}_2,\mathfrak {sl}_2}+ \varphi ^{x,x+i}_{\mathfrak {sl}_2,\mathcal {I}}+\varphi ^{x,x+i}_{\mathcal {I},\mathcal {I}}$ such that $\nabla (x)=\varphi ^{x,x+i}(x)$ and $\nabla (x+i)=\varphi ^{x,x+i}(x+i).$ Since

$$\begin{aligned} x=\nabla (x)=\varphi ^{x,x+i}(x)=\varphi ^{x,x+i}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(x)+\varphi ^{x,x+i}_{\mathfrak {sl}_2,\mathcal {I}}(x), \end{aligned}$$

it follows that $\varphi ^{x,x+i}_{\mathfrak {sl}_2,\mathcal {I}}\equiv 0.$ Then

$$\begin{aligned} \nabla (x+i)=\varphi ^{x,x+i}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(x)+\varphi ^{x,x+i}_{\mathcal {I},\mathcal {I}}(i)=x+i', \end{aligned}$$

where $i'\in \mathcal {I}.$

Now we shall show that $\nabla (x+i)=x+i$ for all $x\in \mathfrak {sl}_2,$$i\in \mathcal {I}.$

Case 1 Let $x=c_1 h+c_0 e +c_2 f,$ where $|c_0|+|c_2|\ne 0,$ and let $i\in \mathcal {I}.$ Take an automorphism $\varphi ^{x,h+i_0}=\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}+ \varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathcal {I}}+\varphi ^{x,h+i_0}_{\mathcal {I},\mathcal {I}}$ such that $\nabla (x)=\varphi ^{x,h+i_0}(x)$ and $\nabla (h+i_0)=\varphi ^{x,h+i_0}(h+i_0).$ Since $\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(h)=h,$ by Lemma 29 there exists a nonzero number $t$ such that $\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(e)=t e$ and $\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(f)=t^{-1} f.$ Then,

$$\begin{aligned} x=\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(x)=c_1 h+t c_0 e+ t^{-1} c_2 f. \end{aligned}$$

Since $|c_0|+|c_2|\ne 0,$ it follows that $t=1.$ Thus, $\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}=\text {id}_{\mathfrak {sl}_2}$ and $\varphi ^{x,h+i_0}_{\mathcal {I},\mathcal {I}}=\lambda \,\text {id}_\mathcal {I}$ (see Lemma 27). Further,

$$\begin{aligned} i_0=\varphi ^{x,h+i_0}_{\mathfrak {sl}_2,\mathcal {I}}(h)+\varphi ^{x,h+i_0}_{\mathcal {I},\mathcal {I}}(i_0)=\omega \theta (h)+\lambda i_0=(2\omega +\lambda )x_1+\lambda (x_0+x_2). \end{aligned}$$

Thus, $\lambda =1$ and $\omega =0,$ and therefore $\varphi ^{x,h+i_0}=\text {id}_{\mathfrak {sl}_2}.$ So,

$$\begin{aligned} \nabla (x+i)=x+i. \end{aligned}$$

Case 2 Let $x=c_1 h+c_0 e +c_2 f,$ where $c_1\ne 0,$ and let $i\in \mathcal {I}.$ Take an automorphism $\varphi ^{x,e+i_0}=\varphi ^{x,e+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}+ \varphi ^{x,e+i_0}_{\mathfrak {sl}_2,\mathcal {I}}+\varphi ^{x,e+i_0}_{\mathcal {I},\mathcal {I}}$ such that $\nabla (x)=\varphi ^{x,e+i_0}(x)$ and $\nabla (e+i_0)=\varphi ^{x,e+i_0}(e+i_0).$ From Case 1, it follows that $\nabla (e+i_0)=e+i_0.$ Then,

$\varphi ^{x,e+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(h)=h$ and $\varphi ^{x,e+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(e)=e,$

because $c_1\ne 0.$ Thus, Lemma 29 implies that $\varphi ^{x,e+i_0}_{\mathfrak {sl}_2,\mathfrak {sl}_2}\equiv \text {id}_{\mathfrak {sl}_2}.$ By Lemma  27, we have $\varphi ^{x,e+i_0}_{\mathcal {I},\mathcal {I}}=\lambda \text {id}_\mathcal {I}.$ Further,

$$\begin{aligned} i_0=\varphi ^{x,e+i_0}_{\mathfrak {sl}_2,\mathcal {I}}(e)+\varphi ^{x,e+i_0}_{\mathcal {I},\mathcal {I}}(i_0)=\omega \theta (e)+\lambda i_0=(2\omega +\lambda )x_0+\lambda (x_1+x_2). \end{aligned}$$

Thus, $\lambda =1$ and $\omega =0,$ and therefore $\varphi ^{x,e+i_0}=\text {id}_{\mathfrak {sl}_2}.$ So,

$$\begin{aligned} \nabla (x+i)=x+i. \end{aligned}$$

Case 3 Let $i\in \mathcal {I}.$ Take an automorphism $\varphi ^{i,h+i}$ such that $\nabla (i)=\varphi ^{i,h+i}(i)$ and $\nabla (h+i)=\varphi ^{i,h+i}(h+i).$ From Case 2, it follows that

$$\begin{aligned} h+i=\nabla (h+i)=\varphi ^{i,h+i}_{\mathfrak {sl}_2,\mathfrak {sl}_2}(h)+ \varphi ^{i,h+i}_{\mathfrak {sl}_2,\mathcal {I}}(h)+ \varphi ^{i,h+i}_{\mathcal {I},\mathcal {I}}(i), \end{aligned}$$

and therefore

$$\begin{aligned} i=\varphi ^{i,h+i}_{\mathfrak {sl}_2,\mathcal {I}}(h)+ \varphi ^{i,h+i}_{\mathcal {I},\mathcal {I}}(i). \end{aligned}$$

Then,

$$\begin{aligned} i-\nabla (i)=(\varphi ^{i,h+i}_{\mathfrak {sl}_2,\mathcal {I}}(h)+ \varphi ^{i,h+i}_{\mathcal {I},\mathcal {I}}(i))-\varphi ^{i,h+i}_{\mathcal {I},\mathcal {I}}(i)= \varphi ^{i,h+i}_{\mathfrak {sl}_2,\mathcal {I}}(h)=*x_1. \end{aligned}$$

(21)

Now take an automorphism $\varphi ^{i,e+i}$ such that

$\nabla (i)=\varphi ^{i,e+i}(i)$ and $\nabla (e+i)=\varphi ^{i,e+i}(e+i).$

From Case 1, it follows that $\nabla (e+i)=e+i.$ Then,

$$\begin{aligned} i-\nabla (i)=(\varphi ^{i,e+i}_{\mathfrak {sl}_2,\mathcal {I}}(e)+ \varphi ^{i,e+i}_{\mathcal {I},\mathcal {I}}(i))-\varphi ^{i,e+i}_{\mathcal {I},\mathcal {I}}(i)= \varphi ^{i,e+i}_{\mathfrak {sl}_2,\mathcal {I}}(e)=*x_0. \end{aligned}$$

(22)

Combining (21) and (22), we obtain that $\nabla (i)=i.$$\square$

Theorem 34

Any 2-local automorphism of a simple Leibniz algebra $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}$ is an automorphism.

Proof

Case 1 Let $\dim \mathcal {G}\ne \dim \mathcal {I}$ or $\dim \mathcal {H}\ge 2.$ Let $\nabla$ be a 2-local automorphism and $\nabla (h_0+i_0)=\varphi _{h_0+i_0}(h_0+i_0)$ for some $\varphi _{h_0+i_0}\in Aut(\mathcal {L})$. Denote $\widetilde{\nabla }=\varphi ^{-1}_{h_0+i_0}\circ \nabla .$ Then, for a 2-local automorphism $\widetilde{\nabla }$, we have $\widetilde{\nabla }(h_0+i_0)=h_0+i_0$. For an element $x\in \mathcal {L}$, there exists $\widetilde{\varphi }_{x, h_0+i_0}\in Aut(\mathcal {L})$ such that

$$\begin{aligned} \widetilde{\varphi }_{x, h_0+i_0}(h_0+i_0)=\widetilde{\nabla }(h_0+i_0)=h_0+i_0\hbox { and }\widetilde{\nabla }(x)=\widetilde{\varphi }_{x, h_0+i_0}(x). \end{aligned}$$

Using Lemmas 30 and 31, we conclude that $\widetilde{\varphi }_{x, h_0+i_0}=\text {id}_\mathcal {L}$. Thus, $\widetilde{\nabla }(x)=\widetilde{\varphi }_{x, h_0+i_0}(x)=x$ for each $x\in \mathcal {L},$ and therefore $\varphi ^{-1}_{h_0+i_0}\circ \nabla =\text {id}_\mathcal {L}.$ Hence, $\nabla =\varphi _{h_0+i_0}$ is an automorphism.

Case 2 Let $\mathcal {L}$ be an algebra from Example 32, and let $\nabla$ be a 2-local automorphism on $\mathcal {L}.$ Take a 2-local automorphism $\varphi _{h, h+i_0}$ such that $\nabla (h)=\varphi _{h, h+i_0}(h)$ and $\nabla (h+i_0)=\varphi _{h, h+i_0}(h+i_0).$ Then, $h$ and $h+i_0$ both are fixed points of 2-local automorphism $\varphi ^{-1}_{h, h+i_0}\circ \nabla ,$ and therefore by Lemma  33, it is an identical automorphism. Thus, $\nabla =\varphi _{h, h+i_0}$ is an automorphism. $\square$

6.2 2-Local Automorphisms on Filiform Leibniz Algebras

The following theorems are similar to the corresponding theorems for the Lie algebras case and their proofs are obtained by replacing the words “Lie algebra” by “Leibniz algebra” (see [7]).

Theorem 35

Let $\mathcal {L}$ be an n-dimensional Leibniz algebra with $n\ge 2$. Suppose that

• (i) $\dim [\mathcal {L},\mathcal {L}] \le n-2;$

• (ii) $\text {Ann}(\mathcal {L})\cap [\mathcal {L},\mathcal {L}]\ne \{0\}.$

Then, $\mathcal {L}$ admits a 2-local automorphism which is not an automorphism.

Theorem 36

Let $\mathcal {L}$ be a finite-dimensional nonnull-filiform nilpotent Leibniz algebra with $\dim \mathcal {L}\ge 2.$ Then, $\mathcal {L}$ admits a 2-local automorphism which is not an automorphism.

Let $NF_n$ be the unique n-dimensional nilpotent Leibniz with condition $\dim [\mathcal {L},\mathcal {L}]=n-1$ (see the end of Sect. 3).

Let $\varphi \in Aut(NF_n)$ and $\varphi (e_1)=\sum \nolimits _{i=1}^{n}\alpha _i e_i$ for some $\alpha _i\in \mathbb {C}, \ \alpha _1\ne 0$, then it is easy to check that

$$\begin{aligned} \varphi (e_j)=\alpha _1^{j-1}\sum _{i=1}^{n+1-j}\alpha _ie_{i+j-1}, \ 2\le j \le n. \end{aligned}$$

Using this property, as in the case of derivations, we conclude that an automorphism of $NF_n$ is uniquely defined by its value on the element $e_1$ and any 2-local automorphism of this algebra is an automorphism.

6.3 Local Automorphisms on Simple Leibniz Algebras

The following result shows that the problem concerning local automorphism of simple Leibniz algebras is reduced to the similar problem for simple Lie algebras.

Theorem 37

Let $\nabla$ be a local automorphism of simple Leibniz algebra $\mathcal {L}=\mathcal {G}\dot{+}\mathcal {I}.$ Then, $\nabla$ is an automorphism if and only if its $\nabla _{\mathcal {G},\mathcal {G}}$ part is an automorphism of the Lie algebra $\mathcal {G}$.

Proof

The necessity part is evident and we shall consider the sufficient part.

Case 1$\dim \mathcal {G}=\dim \mathcal {I}.$ Take the bases $\{x_1, \ldots , x_m\}$ and $\{y_i: y_i=\theta (x_i), i\in \overline{1, m}\}$ on $\mathcal {G}$ and $\mathcal {I},$ respectively, as in the proof of Lemma 21.

Suppose that $\nabla$ is a local automorphism of $\mathcal {L}$ such that its $\nabla _{\mathcal {G},\mathcal {G}}$ part is an automorphism. Consider an automorphism $\psi =\nabla _{\mathcal {G},\mathcal {G}}+ \theta \circ \nabla _{\mathcal {G},\mathcal {G}}\circ \theta ^{-1}.$ Then, $\psi ^{-1} \circ \nabla$ is a local automorphism of $\mathcal {L}$ such that $(\psi ^{-1} \circ \nabla )_{\mathcal {G},\mathcal {G}}=\text {id}_\mathcal {G}.$ So, below it suffices to consider a local automorphism $\nabla$ such that $\nabla _{\mathcal {G},\mathcal {G}}=\text {id}_\mathcal {G}.$

Let $x_k\in \mathcal {G}.$ Then,

$$\begin{aligned} \nabla (x_k)= & {} \nabla _{\mathcal {G},\mathcal {G}}(x_k)+\nabla _{\mathcal {G},\mathcal {I}}(x_k) =x_k+\nabla _{\mathcal {G},\mathcal {I}}(x_k). \end{aligned}$$

Take an automorphism $\varphi ^{x_k}$ such that $\nabla (x_k)=\varphi ^{x_k}(x_k).$ Then,

$$\begin{aligned} \nabla (x_k)= & {} \varphi ^{x_k}(x_k)=\varphi ^{x_k}_{\mathcal {G},\mathcal {G}}(x_k)+\omega _{x_k}\theta (\varphi ^{x_k}_{\mathcal {G},\mathcal {G}}(x_k)). \end{aligned}$$

Comparing the last two equalities, we obtain that $\varphi ^{x_k}_{\mathcal {G},\mathcal {G}}(x_k)=x_k,$ and therefore

$$\begin{aligned} \nabla (x_k)= & {} \varphi ^{x_k}_{\mathcal {G},\mathcal {G}}(x_k)+\omega _{x_k}\theta (\varphi ^{x_k}_{\mathcal {G},\mathcal {G}}(x_k))\\= & {} x_k+\omega _{x_k}\theta (x_k)=x_k+\omega _{x_k}y_k. \end{aligned}$$

Likewise for an element $x=x_k+x_s\in \mathcal {G}$, we have that

$$\begin{aligned} \nabla (x)= & {} \varphi ^{x}(x)=\varphi ^{x}_{\mathcal {G},\mathcal {G}}(x_k+x_s)+ \omega _{x_k+x_s}\theta (\varphi ^{x}_{\mathcal {G},\mathcal {G}}(x_k+x_s))\\= & {} x_k+x_s+\omega _{x_k+x_s}\theta (x_k+x_s)=x_k+x_s+\omega _{x_k+x_s}(y_k+y_s). \end{aligned}$$

Since

$$\begin{aligned} \nabla (x)= & {} \nabla (x_k+x_s)=\nabla (x_k)+\nabla (x_s)=x_k+x_s+\omega _{x_k}y_k+\omega _{x_s}y_s, \end{aligned}$$

we have that $\omega _{x_k+x_s}=\omega _{x_k}=\omega _{x_s}.$ This means that there exists $\omega \in \mathbb {C}$ such that $\nabla (x)=x+\omega \theta (x),$ i.e., $\nabla _{\mathcal {G},\mathcal {I}}=\omega \theta .$

Now take an element $x=x_k+y_k\in \mathcal {G}+\mathcal {I}$ and an automorphism $\varphi ^{x}$ such that $\nabla (x)=\varphi ^{x}(x).$ Then,

$$\begin{aligned} \nabla (x)= & {} \varphi ^{x}(x)=\varphi ^{x}_{\mathcal {G},\mathcal {G}}(x_k)+ \omega _{x_k}\theta (\varphi ^{x}_{\mathcal {G},\mathcal {G}}(x_k))+\lambda _{x}\theta (\varphi ^{x}_{\mathcal {G},\mathcal {G}}(\theta ^{-1}(y_k)))\\= & {} x_k+\omega _{x}\theta (x_k)+\lambda _{x}y_k=x_k+(\omega _{x}+\lambda _x)y_k. \end{aligned}$$

Further, for an element $x=x_k+x_s+y_k+y_s\in \mathcal {G}+\mathcal {I}$ take an automorphism $\varphi ^{x}$ such that $\nabla (x)=\varphi ^{x}(x).$ Then,

$$\begin{aligned} \nabla (x)= & {} \varphi ^{x}(x)=\varphi ^{x}_{\mathcal {G},\mathcal {G}}(x_k+x_s)+ \omega _{x_k}\theta (\varphi ^{x}_{\mathcal {G},\mathcal {G}}(x_k+x_s))\\&+ \lambda _{x}\theta (\varphi ^{x}_{\mathcal {G},\mathcal {G}}(\theta ^{-1}(y_k+y_s)))\\= & {} x_k+x_s+\omega _{x}\theta (x_k+x_s)+\lambda _{x}(y_k+y_s)=x_k+x_s+(\omega _{x}+\lambda _x)(y_k+y_s). \end{aligned}$$

Combing the last two equalities, we have that there exists $t\in \mathbb {C}$ such that $\nabla (x_k+y_s)=x_k+ty_k.$ Then,

$$\begin{aligned} \nabla (y_k)= & {} \nabla (x_k+y_k)-\nabla (x_k)\\= & {} x_k+ty_k-(x_k+\omega y_k)=(t-\omega )y_k. \end{aligned}$$

This means that $\nabla _{\mathcal {I},\mathcal {I}}=(t-\omega )\text {id}_\mathcal {I}.$ Hence, $\nabla =\text {id}_\mathcal {G}+\omega \theta +(t-\omega )\text {id}_\mathcal {I}.$

Case 2$\dim \mathcal {G}\ne \dim \mathcal {I}.$ Take an element of the form (19), i.e.,

$$\begin{aligned} i_0=\sum _{\beta \in \varGamma }\sum _{k=1}^{s_\beta }i^{(k)}_{l_\beta }. \end{aligned}$$

Let $\varphi _{h_0+i_0}$ be an automorphism such that $\nabla (h_0+i_0)=\varphi _{h_0+i_0}(h_0+i_0).$ If necessary, we can replace $\nabla$ by $\varphi ^{-1}_{h_0+i_0}\circ \nabla ,$ and suppose that $\nabla (h_0+i_0)=h_0+i_0.$

Since

$$\begin{aligned} h_0+i_0=\nabla (h_0+i_0)=\nabla _{\mathcal {G},\mathcal {G}}(h_0)+\nabla _{\mathcal {I},\mathcal {I}}(i_0), \end{aligned}$$

it follows that $\nabla _{\mathcal {G},\mathcal {G}}(h_0)=h_0$ and $\nabla _{\mathcal {I},\mathcal {I}}(i_0)=i_0.$ Since $\nabla _{\mathcal {G},\mathcal {G}}$ is an automorphism, by Lemma 29, for every $\alpha \in \varPhi$ there exists a nonzero $t_\alpha \in \mathbb {C}$ such that $\nabla _{\mathcal {G},\mathcal {G}}(e_\alpha )=t_\alpha e_\alpha ,$$\nabla _{\mathcal {G},\mathcal {G}}(e_{-\alpha })=t^{-1}_\alpha e_{-\alpha }$ and $\nabla _{\mathcal {G},\mathcal {G}}(h)=h$ for all $h\in \mathcal {H}.$

Let $i_\beta \in \mathcal {I}_\beta .$ Then,

$$\begin{aligned} \nabla (h_0+i_\beta )= & {} \nabla (h_0)+\nabla (i_\beta ) =h_0+\nabla (i_\beta ). \end{aligned}$$

Take an automorphism $\varphi ^{h_0+i_\beta }$ such that $\nabla (h_0+i_\beta )=\varphi ^{h_0+i_\beta }(h_0+i_\beta ).$ Then,

$$\begin{aligned} \nabla (h_0+i_\beta )= & {} \varphi ^{h_0+i_\beta }(h_0+i_\beta )=\varphi ^{h_0}_{\mathcal {G},\mathcal {G}}(h_0)+ \varphi ^{h_0+i_\beta }_{\mathcal {I},\mathcal {I}}(i_\beta ). \end{aligned}$$

Comparing the last two equalities, we obtain that $\varphi ^{h_0+i_\beta }_{\mathcal {G},\mathcal {G}}(h_0)=h_0,$ and therefore $\varphi ^{h_0+i_\beta }$ acts as diagonal matrix on $\mathcal {L}.$ Thus,

$$\begin{aligned} \nabla (i_\beta )= & {} \varphi ^{h_0+i_\beta }_{\mathcal {I},\mathcal {I}}(i_\beta )=c_\beta i_\beta . \end{aligned}$$

Since $\nabla (i_0)=i_0,$ it follows that $\nabla (i_\beta )=i_\beta$ for all $\beta .$ So, $\nabla _{\mathcal {I},\mathcal {I}}=\text {id}_\mathcal {I}.$

Let $\alpha \in \varPhi .$ Considering $\mathcal {I}$ as $(\mathfrak {sl_2})_\alpha$-module, where $(\mathfrak {sl_2})_\alpha \equiv \text {span}\{e_\alpha , e_{-\alpha }, h_\alpha =[e_{\alpha }, e_{-\alpha }]\},$ we can find a nontrivial irreducible submodule $\mathcal {J}_\alpha$ of $\mathcal {I}.$ Then, $\mathcal {J}_\alpha$ admits a basis $\{x_0^{\alpha }, \ldots , x_n^{\alpha }\}$ such that [24]:

$$\begin{aligned} \begin{array}{lll} \, [x_k^{\alpha }, e_\alpha ]=x_{k+1}^{\alpha }, &{} k\in \{0, \ldots , n-1\}, &{} \\ \, [x_k^{\alpha }, e_{-\alpha }]=- k(n + 1 - k)x_{k-1}^{\alpha }, &{} k\in \{1,\ldots , n\},&{}\\ \, [x_k^{\alpha }, h_\alpha ]=(n-2k)x_k^{\alpha }, &{} k\in \{0,\ldots , n\}.&{}\\ \end{array} \end{aligned}$$

The matrix of the right multiplication operator $R_{h_\alpha +e_\alpha }$ on $\mathcal {J}_\alpha$ has the following form:

$$\begin{aligned} \left( \begin{array}{ccccccc} n &{} 0 &{} 0 &{} \cdots &{} 0 &{} 0 &{} 0\\ 1 &{} n-2 &{} 0 &{} \cdots &{} 0 &{} 0 &{} 0\\ 0 &{} 1 &{} n-4 &{} \cdots &{} 0 &{} 0 &{} 0\\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} 0 &{} 1 &{} -(n-4) &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} -(n-2) &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} -n \\ \end{array} \right) . \end{aligned}$$

Direct computations show that $n$ is a eigenvalue of this matrix, and we take a nonzero eigenvector $i=\sum \nolimits _{s=0}^n t_s x_s^{\alpha } \in \mathcal {J}_\alpha$ corresponding to this eigenvalue, i.e., $[i, h_\alpha +e_\alpha ]=n i,$ with $t_0\ne 0.$ For an element $x=h_\alpha +e_\alpha +i$, choose an automorphism $\varphi ^x$ such that $\nabla (x)=\varphi ^x(x).$ Then,

$$\begin{aligned}{}[\nabla (x), \nabla (x)]= & {} [h_\alpha +t_\alpha e_\alpha +i, h+t_\alpha e_\alpha +i]=[i, h_\alpha +t_\alpha e_\alpha ] \end{aligned}$$

and

$$\begin{aligned}{}[\nabla (x), \nabla (x)]= & {} [\varphi ^x(h_\alpha +e_\alpha +i), \varphi ^x(h_\alpha +e_\alpha +i)]\\= & {} \varphi ^x([h_\alpha +e_\alpha +i, h_\alpha +e_\alpha +i])\\= & {} \varphi ^x([i, h_\alpha +e_\alpha ])=n\varphi ^x(i)=n \nabla (i)=n i. \end{aligned}$$

The last two equalities imply that

$$\begin{aligned}{}[i, h_\alpha +t_\alpha e_\alpha ]=n i= [i, h_\alpha +e_\alpha ]. \end{aligned}$$

Comparing coefficients at the basis element $x^\alpha _1$ in the above equality, we conclude that

$$\begin{aligned} t_\alpha t_0+(n-2)t_1=t_0+(n-2)t_1. \end{aligned}$$

Thus $t_\alpha =1,$ and therefore $\nabla (e_\alpha )=e_{\alpha }.$ So, $\nabla =\text {id}_\mathcal {L}.$$\square$

Remark 38

We conjecture that every local automorphism of simple Lie algebra (and hence of simple Leibniz algebra) over an algebraically closed field of zero characteristic is an automorphism.