Local Superderivations on Solvable Lie and Leibniz Superalgebras

Abstract

Throughout this paper, we show on one hand, that there are nilpotent and solvable Lie superalgebras with infinitely many local superderivations which are not standard superderivations. On the other hand, we show that every local superderivation is a superderivation on the maximal-dimensional solvable Lie superalgebras with model filiform or model nilpotent nilradical. Moreover, we extend the latter result for Leibniz superalgebras by showing that every local superderivation is a superderivation on the maximal-dimensional solvable Leibniz superalgebras with model filiform or model nilpotent non-Lie nilradical.

1 Introduction

Local derivations were considered for the first time in 1990 by Kadison [22] and also by Larson and Sourour [24]. In particular, Kadison showed that each continuous local derivation from a von Neumann algebra into the dual bimodule is a derivation. Let us note then, that the main problem studied in relation to this research topic is to determine when a local derivation is a derivation, see for instance [9, 20]. Additionally, other problem that has been largely studied is to find types of algebras containing local derivations which are not derivations [1]. More recently, in [4, 5, 13] the authors studied the aforementioned problems for Lie algebras, proving in particular that every local derivation of a semi-simple Lie algebra is a derivation and giving examples of solvable Lie algebras with local derivations which are not derivations. Likewise, an analogous study has been developed for Leibniz algebras, see for instance [6] and references therein.

Recently, studying local superderivations on semi-simple Lie superalgebras have drawn a lot of attention [14,15,16, 27], however none of them tackle local superderivations on solvable Lie superalgebras nor Leibniz superalgebras. Thus, this is the context of our work, studying local superderivations on solvable Lie and Leibniz superalgebras. Note that studying solvable Lie superalgebras presents more difficulties than studying solvable Lie algebras [26]. In particular, Lie’s theorem is not verified in general and neither its corollaries. Therefore, for a solvable Lie superalgebra L, $L^2:=[L,L]$ can not be nilpotent, see [25]. Nevertheless, in [11] the authors proved that under the condition of being $L^2$ nilpotent, any solvable Lie and Leibniz superalgebra over the real or complex field can be obtained by means of outer non-nilpotent superderivations of the nilradical in the same way as occurs for Lie and Leibniz algebras.

In this frame, we investigate local superderivations of solvable Lie and Leibniz superalgebras. First, we prove that there are nilpotent and solvable Lie superalgebras with infinitely many local superderivations which are not ordinary superderivations (see Sects. 3 and 5). Second, we prove on the maximal-dimensional solvable Lie superalgebras with model filiform or model nilpotent nilradical that every local superderivation is a superderivation (see Sect. 4). Finally, we extend this last result for the maximal-dimensional solvable Leibniz superalgebras with model filiform and model nilpotent non-Lie nilradical (see Sect. 6).

2 Preliminary Results

2.1 Preliminary for Lie Superalgebras

A vector space V is said to be $\mathbb {Z}_2$-graded if it admits a decomposition into a direct sum, $V=V_{\bar{0}} \oplus V_{\bar{1}}$, where ${\bar{0}}, {\bar{1}} \in \mathbb {Z}_2$. An element $X \in V$ is called homogeneous of degree |x| if it is an element of $V_{|x|}, {|x|} \in \mathbb {Z}_2$.

In particular, the elements of $V_{\bar{0}}$ (resp. $V_{\bar{1}}$) are also called even (resp. odd). A Lie superalgebra (see [21]) is a $\mathbb {Z}_2$-graded vector space ${\mathfrak g}={\mathfrak g}_{\bar{0}} \oplus {\mathfrak g}_{\bar{1}}$, with an even bilinear commutation operation (or “supercommutation”) $[\cdot ,\cdot ]$, which satisfies the conditions

• 1. $[x,y]=-(-1)^{|x| |y|}[y,x],$

• 2. $(-1)^{|x| |z|}[x,[y,z]] + (-1)^{|x| |y|}[y,[z,x]] + (-1)^{|y| |z|}[z,[x,y]]=0$ (super Jacobi identity)

for all homogeneous elements $x, y, z\in {\mathfrak g}$.

Thus, ${\mathfrak g}_{\bar{0}}$ is an ordinary Lie algebra, and ${\mathfrak g}_{\bar{1}}$ is a module over ${\mathfrak g}_{\bar{0}}$; the Lie superalgebra structure also contains the symmetric pairing $S^2 {\mathfrak g}_{\bar{1}} \longrightarrow {\mathfrak g}_{\bar{0}}$.

Let us note that both the descending central sequence and the derived sequence of a Lie superalgebra ${\mathfrak g}={\mathfrak g}_{\bar{0}} \oplus {\mathfrak g}_{\bar{1}}$ are defined in the same way as for Lie algebras: ${\mathcal C}^0({\mathfrak g}): ={\mathfrak g}$, ${\mathcal C}^{k+1}({\mathfrak g}):=[{\mathcal C}^k({\mathfrak g}),{\mathfrak g}]$ and ${\mathcal D}^0({\mathfrak g}): ={\mathfrak g}$, ${\mathcal D}^{k+1}({\mathfrak g}):=[{\mathcal D}^k({\mathfrak g}),{\mathcal D}^k({\mathfrak g})]$ respectively, for all $k\ge 0$. Thus, if ${\mathcal C}^k({\mathfrak g})=\{0\}$ (resp. ${\mathcal D}^k({\mathfrak g})=\{0\}$) for some k, then the Lie superalgebra is called nilpotent (resp. solvable). Note that nilpotent Lie superalgebras are in particular solvable. Remark also, that Engel’s theorem and its corollaries are still valid for Lie superalgebras. Then, a Lie superalgebra L is nilpotent if and only if $ad_{L} x$ is nilpotent for every homogeneous element x of L. Additionally, a Lie superalgebra L is solvable if and only if its even part $L_{\overline{0}}$ (a Lie algebra) is solvable. However, Lie’s Theorem does not hold for solvable Lie superalgebras.

At the same time, there are also defined two other crucial sequences denoted by ${\mathcal C}^{k}({\mathfrak g}_{\bar{0}})$ and ${\mathcal C}^{k}({\mathfrak g}_{\bar{1}})$ which will play an important role in our study. They are defined as follows:

$$\begin{aligned} {\mathcal C}^0({\mathfrak g}_{\bar{i}}):={\mathfrak g}_{\bar{i}}, {\mathcal C}^{k+1}({\mathfrak g}_{\bar{i}}) := [{\mathfrak g}_{\bar{0}}, {\mathcal C}^k({\mathfrak g}_{\bar{i}})], \ k\ge 0, {\bar{i}} \in \mathbb {Z}_2. \end{aligned}$$

Let us recall now, the definition of superderivations of superalgebras [21]. A superderivation of degree s of a superalgebra L, $s\in \mathbb {Z}_2$, is an endomorphism $D \in End_s L$ with the property

$$\begin{aligned} D(a b)=D(a)b + (-1)^{s \cdot deg a} a D(b) \end{aligned}$$

denote $Der_{s}(L) \subset End_s L$ the space of all superderivations of degree s. Then $Der( L)=Der _{\overline{0}}(L)\oplus Der _{\overline{1}}(L)$ is the Lie superalgebra of superderivations of L, with $Der _{\overline{0}}(L)$ composed by even superderivations and $Der _{\overline{1}}(L)$ by odd ones.

On the other hand, recall also that a homogeneous linear mapping $\Delta : L \longrightarrow L$ of degree s is called a local homogeneous superderivation of degree s if for any element $x \in L$, there exists a superderivation $D_x: L \longrightarrow L$ (depending on x) such that $\Delta (x)=D_x(x)$. Then, the set of all local superderivations can be expressed

$$\begin{aligned} LocDer(L)=LocDer _{\overline{0}}(L)\oplus LocDer _{\overline{1}}(L) \end{aligned}$$

with $LocDer _{\overline{0}}(L)$ (resp. $LocDer _{\overline{1}}(L)$) composed by even (resp. odd) local superderivations. For more details it can be consulted [14].

2.2 Preliminaries for Leibniz Superalgebras

Let us note that many results and definitions of the above sub-section can be extended for Leibniz superalgebras.

Definition 2.1

[2]. A $\mathbb {Z}_2$-graded vector space $L=L_{\bar{0}} \oplus L_{\bar{1}}$ is called a Leibniz superalgebra if it is equipped with a product $[\cdot ,\cdot ]$ which for an arbitrary element x and homogeneous elements y, z satisfies the condition

$$\begin{aligned}{}[x,[y,z]]=[[x,y],z]-(-1)^{|y| |z|}[[x,z],y] \hbox {{(super Leibniz identity)}}. \end{aligned}$$

Note that if a Leibniz superalgebra L satisfies the identity $[x,y]=-(-1)^{|x| |y|}[y,x]$ for any homogeneous elements $x, y \in L$, then the super Leibniz identity becomes the super Jacobi identity. Consequently, Leibniz superalgebras are a generalization of Lie superalgebras. Also and in the same way as for Lie superalgebras, isomorphisms are assumed to be consistent with the $\mathbb {Z}_2$-graduation.

Let us now denote by $R_x$ the right multiplication operator, i.e., $R_x : L \rightarrow L$ given as $R_x(y) := [y,x]$ for $y \in L$, then the super Leibniz identity can be expressed as $R_{[x,y]} = R_yR_x-(-1)^{|x| |y|}R_xR_y.$

If we denote by R(L) the set of all right multiplication operators, then R(L) with respect to the following multiplication

$$\begin{aligned} \langle R_a,R_b\rangle :=R_aR_b-(-1)^{\bar{i}\bar{j}}R_bR_a \end{aligned}$$

(2.1)

for $R_a\in {R(L)_{\bar{i}}}$, $R_b\in {R(L)_{\bar{j}}}$, forms a Lie superalgebra. Note that $R_a$ is a derivation. In fact, the condition for being a derivation of a Leibniz superalgebra (for more details see [23]) is $d([x,y])=(-1)^{\mid d \mid \mid y \mid } [d(x),y]+[x,d(y)]$. Since the degree of $R_z$ as homomorphism between $\mathbb {Z}_2$-graded vector spaces is the same as the degree of the homogeneous element z, that is $\vert R_z\vert =\vert z \vert$, then the condition for $R_z$ to be a derivation is exactly $R_z([x,y])=(-1)^{\mid z \mid \mid y \mid } [R_z(x),y]+[x,R_z(y)]$. This last condition can be rewritten $[[x,y],z]=(-1)^{\mid z \mid \mid y \mid } [[x,z],y]+[x,[y,z]]$ which is nothing but the super (graded) Leibniz identity. Let us remark that the definition of local superderivation is a natural extension from Lie theory.

Let us note also that the concepts of descending central sequence, nilindex, the variety of Leibniz superalgebras and Engel’s theorem are natural extensions from Lie theory.

Let $V = V_{\bar{0}} \oplus V_{\bar{1}}$ be the underlying vector space of L, $L = L_{\bar{0}} \oplus L_{\bar{1}} \in Leib^{n,m}$, being $Leib^{n,m}$ the variety of Leibniz superalgebras, and let G(V) be the group of the invertible linear mappings of the form $f=f_{\bar{0}}+f_{\bar{1}}$, such that $f_{\bar{0}} \in GL(n,\mathbb {C})$ and $f_{\bar{1}} \in GL(m,\mathbb {C})$ (then $G(V) = GL(n,\mathbb {C})\oplus GL(m,\mathbb {C}))$. The action of G(V) on $Leib^{n,m}$ induces an action on the Leibniz superalgebras variety: two laws $\lambda _1, \lambda _2$ are isomorphic if there exists a linear mapping $f = f_{\bar{0}}+f_{\bar{1}} \in G(V)$, such that

$$\begin{aligned} \lambda _2(x,y) = f_{\bar{i}+\bar{j}}^{-1}(\lambda _1(f_{\bar{i}}(x), f_{\bar{j}}(y))), \hbox {for any} \ x \in V_{\bar{i}}, y \in V_{\bar{j}}. \end{aligned}$$

Furthermore, the description of the variety of any class of algebras or superalgebras is a difficult problem. Different works (for example, [3, 7, 10, 18, 19]) are regarding the applications of algebraic groups theory to the description of the variety of Lie and Leibniz algebras.

Definition 2.2

For a Leibniz superalgebra $L=L_{\bar{0}} \oplus L_{\bar{1}}$ we define the right annihilator of L as the set $Ann(L):=\{x \in L : [L,x]=0\}$.

It is easy to see that Ann(L) is a two-sided ideal of L and $[x,x] \in Ann(L)$ for any $x \in L_{\bar{0}}$. This notion is compatible with the right annihilator in Leibniz algebras. If we consider the ideal $I:=ideal\langle [x,y]+(-1)^{ |x| |y|}[y,x]\rangle$, then $I \subset Ann(L).$

Let $L=L_{\bar{0}} \oplus L_{\bar{1}}$ be a nilpotent Leibniz superalgebra with $\dim L_{\bar{0}}=n$ and $\dim L_{\bar{1}}=m.$ From Equation (2.1) we have that R(L) is a Lie superalgebra, and in particular $R(L_{\bar{0}})$ is a Lie algebra. As $L_{\bar{1}}$ has $L_{\bar{0}}$-module structure we can consider $R(L_{\bar{0}})$ as a subset of $GL(V_{\bar{1}})$ , where $V_{\bar{1}}$ is the underlying vector space of $L_{\bar{1}}$. So, we have a Lie algebra formed by nilpotent endomorphisms of $V_{\bar{1}}$. Applying Engel’s theorem we have the existence of a sequence of subspaces of $V_{\bar{1}}$, $V_0 \subset V_1 \subset V_2 \subset \dots \subset V_m = V_{\bar{1}},$ with $R(L_{\bar{0}})(V_{{i+1}})\subset V_{ i}.$ Then, it can be defined the descending sequences $C^k(L_{\bar{0}})$ and $C^k(L_{\bar{1}})$ and the super-nilindex in the same way as for Lie superalgebras. That is, ${\mathcal C}^0(L_{\bar{i}}):=L_{\bar{i}}, {\mathcal C}^{k+1}(L_{\bar{i}}) := [ {\mathcal C}^k(L_{\bar{i}}),L_{\bar{0}}], \quad k\ge 0, {\bar{i}} \in \mathbb {Z}_2.$ If $L=L_{\bar{0}} \oplus L_{\bar{1}}$ is a nilpotent Leibniz superalgebra, then L has super-nilindex or s-nilindex (p, q) if satisfies

$$\begin{aligned} {\mathcal C}^{p-1}(L_{\bar{0}}) \ne 0, \qquad {\mathcal C}^{q-1}(L_{\bar{1}})\ne 0, \qquad {\mathcal C}^{p}(L_{\bar{0}})={\mathcal C}^{q}(L_{\bar{1}})=0. \end{aligned}$$

3 Local Superderivations of the Model Filiform Lie Superalgebra

We start our study with one case of nilpotent Lie superalgebra. Among all of them one that has been proved to be very relevant due to its properties is the model filiform Lie superalgebra since all the other filiform Lie superalgebras can be obtained from it by means of infinitesimal deformations [8]. These infinitesimal deformations are given by the even 2-cocycles $Z^2_0(L^{n,m},L^{n,m})$. We consider then, the model filiform Lie superalgebra $L^{n,m},$ that is, the simplest filiform Lie superalgebra which is defined by the only non-zero bracket products that follow

with a basis $\{ x_1, \dots , x_n \}$ of $(L^{n,m})_{\bar{0}}$ and a basis $\{ y_1, \dots , y_m \}$ of $(L^{n,m})_{\bar{1}}.$

For an even superderivation D of $L^{n,m}$ we have $D(L^{n,m}_{\overline{0}})\subset L^{n,m}_{\overline{0}}$ and $D(L^{n,m}_{\overline{1}})\subset L^{n,m}_{\overline{1}}$. Then we set

$$\begin{aligned} D(x_1)=\displaystyle \sum _{k=1}^n a_k x_k,\quad D(x_2)=\displaystyle \sum _{k=1}^n b_k x_k,\quad D(y_1)=\displaystyle \sum _{t=1}^m c_t y_t. \end{aligned}$$

Applying induction and the even superderivation condition for the products $[x_1,x_i]$ we derive

$$\begin{aligned} D(x_i)=((i-2)a_1+b_2)x_i+\displaystyle \sum _{k=i+1}^n b_{k-i+2}x_k,\quad 3\le i\le n. \end{aligned}$$

Similarly, from the products $[x_1,y_j]$ we get

$$\begin{aligned} D(y_j)=((j-1)a_1+c_1)y_j+\displaystyle \sum _{t=j+1}^m c_{t-j+1}y_t,\quad 2\le j\le m. \end{aligned}$$

Finally, from the product $[x_2,y_1]$ we obtain $b_1=0.$ Thus, we conclude

$$\begin{aligned}{} & {} Der _{\overline{0}}(L^{n,m})\\{} & {} =\left( \begin{array}{ccccccccc} a_1&{}\quad a_2&{}\quad a_3&{}\quad \dots &{}\quad a_n&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0 \\ 0&{}\quad b_2&{}\quad b_3&{}\quad \dots &{}\quad b_n&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0 \\ 0&{}\quad 0&{}\quad a_1+b_2&{}\quad \dots &{}\quad b_{n-1}&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad (n-2)a_1+b_2&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad c_1&{}\quad c_2&{}\quad \dots &{}\quad c_m\\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad 0&{}\quad a_1+c_1&{}\quad \dots &{}\quad c_{m-1}\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad (m-1)a_1+c_1 \end{array}\right) . \end{aligned}$$

Let now D be an odd superderivation of $L^{n,m}.$ Then we have

$$\begin{aligned} D(x_1)=\displaystyle \sum _{k=1}^m a_k y_k,\quad D(x_2)=\displaystyle \sum _{k=1}^m b_k y_k,\quad D(y_1)=\displaystyle \sum _{t=1}^n c_t x_t. \end{aligned}$$

According to the odd superderivation condition on the products of $L^{n,m}$ and induction, similar to even superderivation case, we obtain

$$\begin{aligned} D(x_i)= & {} \displaystyle \sum _{k=i-1}^m b_{k-i+2} y_k,\quad 3\le i\le n,\\ D(y_j)= & {} \displaystyle \sum _{t=j+1}^n c_{t-j+1} x_t,\quad 2\le j\le m. \end{aligned}$$

Considering superderivation property for the products $[x_2,y_1]$ and $[x_1, x_n]$, we get $c_1$ and $b_i=0,$ $1\le i\le m-n+1$ with $m\ge n.$ Thus, we conclude

$$\begin{aligned} Der _{\overline{1}}(L^{n,m})=\left( \begin{array}{ccccccccc} 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad a_1&{}\quad a_2&{}\quad \dots &{}\quad a_m \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad b_1&{}\quad b_2&{}\quad \dots &{}\quad b_m \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad 0&{}\quad b_1&{}\quad \dots &{}\quad b_{m-1}\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad b_1\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0\\ 0&{}\quad c_2&{}\quad c_3&{}\quad \dots &{}\quad c_n&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0\\ 0&{}\quad 0&{}\quad c_2&{}\quad \dots &{}\quad c_{n-1}&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad c_2&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0 \end{array}\right) , \end{aligned}$$

where $b_i=0,$ $1\le i\le m-n+1$ with $m\ge n.$

Theorem 3.1

$Der(L^{n,m})\subsetneqq LocDer(L^{n,m}).$

Proof

Consider the homogeneous linear mappings $\Delta _t : L^{n,m} \longrightarrow L^{n,m}, \ t\ne 2$ of degree 0 defined on the basis vectors of $L^{n,m}$ by

$$\begin{aligned} \Delta _t(x_1)= & {} x_1, \ \Delta _t(x_2)=x_2, \ \Delta _t(x_3)=tx_3, \ \Delta _t(x_i)=\Delta _t(y_j)\\= & {} 0, \ 4 \le i \le n, \ 1 \le j \le m. \end{aligned}$$

Clearly, $\Delta _t$ is not an even superderivation (because its matrix does not fit with the general matrix of even superderivations).

Consider the following superderivations:

• $d_1$ is the resultant even superderivation after replacing $a_1$ by 1 and all the rest of parameters by 0 on the general matrix of $Der _{\overline{0}}(L^{n,m}),$

• $d_2$ is the resultant even superderivation after replacing $b_2$ by 1 and all the rest of parameters by 0 on the general matrix of $Der _{\overline{0}}(L^{n,m}),$

• $d_t$ is the even superderivations defined by $d_t:=d_1+(t-1)d_2$.

Clearly,

$$\begin{aligned}{} & {} \Delta _t(x_1)=d_1(x_1), \ \Delta _t(x_2)=d_2(x_2), \ \Delta _t(x_3)=d_t(x_3), \\{} & {} \Delta _t(x_i)=d_0(x_i)=\Delta _t(y_j)=d_0(y_j), \ 4 \le i \le n, \ 1 \le j \le m \end{aligned}$$

being $d_0$ the null superderivation.

For an arbitrary element $e=\alpha _1 x_1+\dots +\alpha _n x_n+\beta _1 y_1+\dots +\beta _m y_m$ of $L^{n,m}$ we have

$$\begin{aligned} \Delta _t(e)=\alpha _1 x_1+\alpha _2x_2+\alpha _3tx_3. \end{aligned}$$

Define superderivation $d_e$ as follows $d_e(x_i)=d_e(y_j)=0$ with $4\le i\le n$ and $1\le j\le m$ and

$$\begin{aligned} d_e(x_1)= & {} \beta _1x_1+\beta _2x_2+\beta _3x_3,\quad d_e(x_2)=\gamma _2x_2+\gamma _3x_3,\\ d_e(x_3)= & {} (\beta _1+\gamma _2)x_3, \end{aligned}$$

where $\beta _1,\beta _2,\beta _3, \gamma _2, \gamma _3$ are some unknowns parameters. From $\Delta _t(e)=d_e(e)$ we derive $\beta _1=1$ and the following linear system of equations

$$\begin{aligned} \left( \begin{array}{llll} \alpha _1 &{}\quad 0&{}\quad \alpha _2&{}\quad 0\\ 0&{}\quad \alpha _1&{}\quad \alpha _3&{}\quad \alpha _2 \end{array} \right) \left( \begin{array}{l} \beta _2 \\ \beta _3\\ \gamma _2\\ \gamma _3 \end{array} \right) =\left( \begin{array}{c} \alpha _2\\ \alpha _3(t-1) \end{array} \right) , \end{aligned}$$

which always has a solution with respect to unknowns $\beta _2,\beta _3,\gamma _2,\gamma _3.$ Thus, we obtain the existence a superderivation $d_e$ such that $d_e(e)=\Delta _t(e)$. The proof is complete. $\square$

Remark 3.1

In fact, in the proof of Theorem 3.1 we show the existence of infinitely many local superderivations on the model filiform Lie superalgebra $L^{n,m}$ ($n\ge 3$) which are not superderivations. Note also, that analogously it can be found infinitely many odd local superderivations which are not odd superderivations.

Along the next sections, we consider non-nilpotent solvable Lie and Leibniz superalgebras with different types of nilradical, starting with abelian nilradical.

4 Local Superderivations of Maximal-Dimensional Solvable Lie Superalgebras with Model Filiform and Model Nilpotent Nilradical

In this section first, we consider the maximal-dimensional solvable Lie superalgebra with model filiform nilradical [12]. This superalgebra is unique for each pair of dimensions (n, m) and can be expressed by the only non-null bracket products that follow:

with $\{ x_1, \dots , x_n, t_1,t_2,t_3\}$ a basis of $(SL^{n,m})_{\bar{0}}$ and $\{ y_1, \dots , y_m\}$ a basis of $(SL^{n,m})_{\bar{1}}$.

Its superalgebra of superderivations was obtained in [12]. Next, we prove the following result.

Theorem 4.1

On the maximal-dimensional solvable Lie superalgebra with model filiform nilradical, every local superderivation is a superderivation.

Proof

First, we are going to express in a more suitable way for our purpose the solvable Lie superalgebra $SL^{n,m}$. Thus, after applying an elementary basis transformation one can express the table of multiplications $SL^{n,m}$ in a new basis $\{e_1,\dots ,e_n,e_{n+1},\dots ,e_{n+m}\}$ as follows:

In [12] the authors proved every superderivation is exactly the adjoint operator of an element of $SL^{n,m}$. Let us fix an arbitrary element $z=\gamma _1 t_1+\gamma _2 t_2+\gamma _3 t_3+\sum _{p=1}^{n+m} \beta _p e_p$ of the superalgebra $SL^{n,m},$ then for its adjoint operator $ad_z$ we obtain

$$\begin{aligned} ad_z(t_1)= & {} -\displaystyle \sum _{p=1}^{n+m} p\beta _p e_p, ad_z(t_2)=-\displaystyle \sum _{p=2}^{n} \beta _p e_p, ad_z(t_3)=-\displaystyle \sum _{p=n+1}^{n+m} \beta _p e_p,\\ ad_z(e_1)= & {} \gamma _1e_1-\displaystyle \sum _{p=2}^{n-1} \beta _p e_{p+1}-\displaystyle \sum _{p=n+1}^{n+m-1} \beta _p e_{p+1}, \\ \\ ad_z(e_i)= & {} (i \gamma _1+\gamma _2)e_i+\beta _1e_{i+1}, \ 2 \le i \le n, \\ \\ ad_z(e_{n+j})= & {} ((n+j) \gamma _1+\gamma _3)e_{n+j}+\beta _1 e_{n+j+1}, \quad 1 \le j \le m. \end{aligned}$$

Later on, when needed, from this general expression we will distinguish between even and odd superderivations. Let us consider now an arbitrary local superderivation $\Delta :SL^{n,m} \longrightarrow SL^{n,m}$. Since the value of a local superderivation on any vector coincides with the value on this vector of a superderivation, in particular on the basis vectors we have the following expression:

The goal now is to show that the expressions for $ad_z$ and $\Delta$ coincide. Firstly, we will show this coincidence on the generators of the basis vectors, i.e. $t_1,t_2,t_3, e_1,e_2$ and $e_{n+1}$. Let us consider $\Delta (st_2-t_1)$ with a fixed s verifying $2 \le s \le n$. Thus as $\Delta$ is linear we obtain

$$\begin{aligned} \Delta (st_2-t_1)= & {} s\Delta (t_2)-\Delta (t_1)=\displaystyle \sum _{p=1}^{n+m} p\beta _{1,p} e_p-s\displaystyle \sum _{p=2}^{n} \beta _{2,p} e_p\\= & {} \beta _{1,1}e_1+\displaystyle \sum _{p=2}^{n}( p\beta _{1,p}-s \beta _{2,p})e_p+\displaystyle \sum _{p=n+1}^{n+m} p\beta _{1,p} e_p, \end{aligned}$$

on the other hand and by definition the above coincides with the value of a superderivation, named $d^{e}$ on the vector $st_2-t_1$, thus

$$\begin{aligned} \Delta (st_2-t_1)=d^{e}(st_2-t_1)=\beta _{1}^{e}e_1+\displaystyle \sum _{p=2}^{n}( p\beta _{p}^{e}-s \beta _{p}^{e})e_p+\displaystyle \sum _{p=n+1}^{n+m} p\beta _{p}^{e} e_p, \end{aligned}$$

comparing the coefficients of $e_s$ on both expressions it follows that $s(\beta _{1,s}- \beta _{2,s})=0$ and $\beta _{1,s}= \beta _{2,s}$. Repeating this process for all the possible s with $2\le s \le n$, leads to $\beta _{1,s}=\beta _{2,s}$, $2 \le s \le n$.

Let us consider now $\Delta (st_3-t_1)$ with a fixed s verifying $n+1 \le s \le n+m$. Since $\Delta$ is linear map, we obtain

$$\begin{aligned} s\Delta (t_3)-\Delta (t_1)= & {} \displaystyle \sum _{p=1}^{n+m} p\beta _{1,p} e_p-s\displaystyle \sum _{p=n+1}^{n+m} \beta _{3,p} e_p=\beta _{1,1}e_1+\displaystyle \sum _{p=2}^{n} p\beta _{1,p}e_p\\{} & {} +\displaystyle \sum _{p=n+1}^{n+m} (p\beta _{1,p} e_p-s \beta _{3,p})e_p, \end{aligned}$$

on the other hand and by definition the above coincides with the value of a superderivation, named $d^{e}$ on the vector $st_3-t_1$, thus

$$\begin{aligned} \Delta (st_3-t_1)=d^{e}(st_3-t_1)=\beta _{1}^{e}e_1+\displaystyle \sum _{p=2}^{n} p\beta _{p}^{e}e_p+\displaystyle \sum _{p=n+1}^{n+m} (p\beta _{p}^{e}-s \beta _{p}^{e}) e_p, \end{aligned}$$

as above, comparing the coefficients of $e_s$ on both expressions it follows that $s(\beta _{1,s}- \beta _{3,s})=0$ and $\beta _{1,s}= \beta _{3,s}$. Repeating this process for all the possible s with $n+1 \le s \le n+m$, leads to $\beta _{1,s}=\beta _{3,s}$, $n+1 \le s \le n+m$. Renaming $\beta _{i,p}$ we have

$$\begin{aligned} \Delta (t_1)=-\displaystyle \sum _{p=1}^{n+m} p\beta _p e_p, \quad \Delta (t_2)=-\displaystyle \sum _{p=2}^{n} \beta _p e_p, \quad \Delta (t_3)=-\displaystyle \sum _{p=n+1}^{n+m} \beta _p e_p. \end{aligned}$$

Let us now consider $\Delta (t_1-3t_2-e_1)$, on one hand we have

$$\begin{aligned} \Delta (t_1)-3\Delta (t_2)-\Delta (e_1)= & {} -\displaystyle \sum _{p=1}^{n+m} p\beta _p e_p+3\displaystyle \sum _{p=2}^{n} \beta _p e_p-\gamma _{1,1}e_1+\displaystyle \sum _{p=2}^{n-1} \delta _{1,p} e_{p+1}\\{} & {} +\displaystyle \sum _{p=n+1}^{n+m-1} \delta _{1,p} e_{p+1}= (-\beta _1-\gamma _{1,1})e_1\\{} & {} +\beta _2e_2+\displaystyle \sum _{p=3}^{n}(3 \beta _p-p\beta _p+\delta _{1,p-1}) e_p\\{} & {} -(n+1)\beta _{n+1}e_{n+1} +\displaystyle \sum _{p=n+2}^{n+m}( \delta _{1,p-1}-p\beta _p) e_{p}, \end{aligned}$$

on the other hand and by definition the above coincides with the value of a superderivation, named $d^{e}$ on the vector $t_1-3t_2-e_1$, thus

$$\begin{aligned}{} & {} \Delta (t_1-3t_2-e_1)\\= & {} d^{e}(t_1-3t_2-e_1)=-\displaystyle \sum _{p=1}^{n+m} p\beta _p^{e} e_p+3\displaystyle \sum _{p=2}^{n} \beta _p^{e} e_p\\{} & {} -\gamma _1^{e}e_1+\displaystyle \sum _{p=2}^{n-1} \beta _p^{e} e_{p+1}+\displaystyle \sum _{p=n+1}^{n+m-1} \beta _p^{e} e_{p+1}\\= & {} (-\beta _1^{e}-\gamma _1^{e})e_1+\beta _2^{e}e_2+\displaystyle \sum _{p=3}^{n}(3 \beta _p^{e}-p\beta _p^{e}+\beta _{p-1}^{e}) \\{} & {} \times e_p-(n+1)\beta _{n+1}^{e}e_{n+1}+\displaystyle \sum _{p=n+2}^{n+m}( \beta _{p-1}^{e}-p\beta _p^{e}) e_{p}, \end{aligned}$$

considering the coefficients of $e_2$ and $e_3$ we obtain $\delta _{1,2}=\beta _2$. On account of $\Delta (t_1-(i+1)t_2-e_1)$ inductively we get $\delta _{1,i}=\beta _i$ for i verifying $2 \le i \le n-1$.

In a similar way from $\Delta (t_1-(n+2)t_3-e_1)$ and considering the coefficients of $e_{n+1}$ and $e_{n+2}$ we obtain $\delta _{1,n+1}=\beta _{n+1}$, after and considering the coefficients of $e_{n+1}, e_{n+2}$ and $e_{n+3}$ in $\Delta (t_1-(n+3)t_3-e_1)$ we obtain $\delta _{1,n+2}=\beta _{n+2}$. Therefore, by considering $\Delta (t_1-(n+j+1)t_3-e_1)$ inductively we get $\delta _{1,n+j}=\beta _{n+j}$ with $1 \le j \le m-1$. In summary, we have then

$$\begin{aligned} \Delta (e_1)=\gamma _1e_1-\displaystyle \sum _{p=2}^{n-1} \beta _p e_{p+1}-\displaystyle \sum _{p=n+1}^{n+m-1} \beta _p e_{p+1}. \end{aligned}$$

Let us now consider $\Delta (t_1-(i+1)t_2+e_i)$ with a fixed i verifying $2\le i \le n-1$. Then

$$\begin{aligned} \Delta (t_1)-(i+1)\Delta (t_2)+\Delta (e_i)= & {} -\displaystyle \sum _{p=1}^{n+m} p\beta _p e_p+(i+1)\displaystyle \sum _{p=2}^{n} \beta _p e_p\\{} & {} +(i \gamma _{i,1}+\gamma _{i,2})e_i+\delta _{i,1}e_{i+1}\\= & {} -\beta _1 e_1 +\dots +\delta _{i,1}e_{i+1}. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \Delta (t_1-(i+1)t_2+e_i)= & {} d^{e}(t_1-(i+1)t_2+e_i)\\= & {} -\displaystyle \sum _{p=1}^{n+m} p\beta _p^{e} e_p+(i+1)\displaystyle \sum _{p=2}^{n} \beta _p^{e} e_p\\{} & {} +(i \gamma _1^{e}+\gamma _2^{e})e_i+\beta _1^{e}e_{i+1}= -\beta _1^{e} e_1 +\dots +\beta _1^{e}e_{i+1} \end{aligned}$$

which leads to $\delta _{i,1}=\beta _1$. By repeating this process for all i with $2\le i \le n-1$, we conclude that $\delta _{i,1}=\beta _1$ for all i, $2\le i \le n-1$. Summing up

$$\begin{aligned} \Delta (e_i)=(i \gamma _{i,1}+\gamma _{i,2})e_i+\beta _1e_{i+1}, \ 2 \le i \le n. \end{aligned}$$

From $\Delta (e_2+e_{n+j})$ we have

$$\begin{aligned} \Delta (e_2)+\Delta (e_{n+j})= & {} (2\gamma _{2,1}+\gamma _{2,2})e_2+\beta _1e_{3}+((n+j) \gamma _{n+j,1}\\{} & {} +\gamma _{n+j,3})e_{n+j}+\delta _{n+j,1} e_{n+j+1}, \end{aligned}$$

and on the other hand

$$\begin{aligned} \Delta (e_2+e_{n+j})=d^{e}(e_2+e_{n+j})= & {} (2 \gamma _1^{e}+\gamma _2^{e})e_2+\beta _1^{e}e_{3}+((n+j) \gamma _1^{e}\\{} & {} +\gamma _3^{e})e_{n+j}+\beta _1^{e} e_{n+j+1}, \end{aligned}$$

which leads to $\delta _{n+j,1}=\beta _1$, $1 \le j \le m-1$.

Consequently, there is no loss of generality in supposing

$$\begin{aligned} \begin{array}{l} \Delta (t_1)=-\displaystyle \sum _{p=1}^{n+m} p\beta _p e_p, \quad \Delta (t_2)=-\displaystyle \sum _{p=2}^{n} \beta _p e_p, \quad \Delta (t_3)=-\displaystyle \sum _{p=n+1}^{n+m} \beta _p e_p, \\ \Delta (e_1)=\gamma _1e_1-\displaystyle \sum _{p=2}^{n-1} \beta _p e_{p+1}-\displaystyle \sum _{p=n+1}^{n+m-1} \beta _p e_{p+1}, \quad \Delta (e_2)=(2\gamma _1+\gamma _2)e_2+\beta _1e_{3},\\ \\ \Delta (e_i)=(i \gamma _{i,1}+\gamma _{i,2})e_i+\beta _1e_{i+1}, \ 3 \le i \le n, \\ \Delta (e_{n+1}) =((n+1) \gamma _1+\gamma _3)e_{n+1}+\beta _1 e_{n+2},\\ \\ \Delta (e_{n+j})=((n+j) \gamma _{n+j,1}+\gamma _{n+j,3})e_{n+j}+\beta _1 e_{n+j+1}, \ 2 \le j \le m-1,\\ \Delta (e_{n+m})=((n+m) \gamma _{n+m,1}+\gamma _{n+m,3})e_{n+m}. \end{array} \end{aligned}$$

At this point we are going to distinguish between even and odd local superderivations. Recall that $\{ t_1, t_2, t_3, e_1, \dots , e_n\}$ are even basis vectors of $SL^{n,m}$ and $\{e_{n+1}, \dots , e_{n+m}\}$ odd ones. Thus, if $\Delta$ is an odd local superderivation in particular $\Delta$ is a homogeneous linear mapping of degree 1,

$$\begin{aligned} \Delta : (SL^{n,m})_{\bar{0}} \longrightarrow (SL^{n,m})_{\bar{1}}\ \text{ and } \ \Delta : (SL^{n,m})_{\bar{1}} \longrightarrow (SL^{n,m})_{\bar{0}}. \end{aligned}$$

Therefore, the only non-null values on the basis vectors for an odd local superderivation are exactly:

$$\begin{aligned} \Delta (t_1)=-\displaystyle \sum _{p=n+1}^{n+m} p\beta _p e_p, \quad \Delta (t_3)=-\displaystyle \sum _{p=n+1}^{n+m} \beta _p e_p, \quad \Delta (e_1)=-\displaystyle \sum _{p=n+1}^{n+m-1} \beta _p e_{p+1}. \end{aligned}$$

Then every odd local superderivation is a standard odd superderivation. Regarding even local superderivations $\Delta : (SL^{n,m})_{\bar{0}} \longrightarrow (SL^{n,m})_{\bar{0}}\ \text{ and } \ \Delta : (SL^{n,m})_{\bar{1}} \longrightarrow (SL^{n,m})_{\bar{1}}$ we have

$$\begin{aligned} \Delta (t_1)= & {} -\displaystyle \sum _{p=1}^{n} p\beta _p e_p, \quad \Delta (t_2)=-\displaystyle \sum _{p=2}^{n} \beta _p e_p, \quad \Delta (t_3)=0, \quad \Delta (e_1)\\= & {} \gamma _1e_1-\displaystyle \sum _{p=2}^{n-1} \beta _p e_{p+1}, \\ \Delta (e_2)= & {} (2\gamma _1+\gamma _2)e_2+\beta _1e_{3},\quad \Delta (e_i)=(i \gamma _{i,1}+\gamma _{i,2})e_i+\beta _1e_{i+1}, \ 3 \le i \le n, \\ \Delta (e_{n+1})= & {} ((n+1) \gamma _1+\gamma _3)e_{n+1}+\beta _1 e_{n+2},\\ \Delta (e_{n+j})= & {} ((n+j) \gamma _{n+j,1}+\gamma _{n+j,3})e_{n+j}+\beta _1 e_{n+j+1}, \ 2 \le j \le m. \end{aligned}$$

Only rest to prove that $\gamma _{i,1}=\gamma _{n+j,1}=\gamma _1$, $\gamma _{i,2}=\gamma _2$ and $\gamma _{n+j,3}=\gamma _3$ in order to have a standard even superderivation. Let us consider $\Delta (t_1-(j+1)t_2+e_1-(j-1)e_2+\frac{1}{(j-2)!}e_{j+1})$ with a fixed j verifying $2 \le j \le n-1$, on one hand we have

$$\begin{aligned}{} & {} \Delta (t_1)-(j+1)\Delta ( t_2)+\Delta (e_1)-(j-1)\Delta (e_2)+\displaystyle \frac{1}{(j-2)!}\Delta (e_{j+1})\\{} & {} \quad =-\displaystyle \sum _{p=1}^{n} p\beta _p e_p +(j+1)\displaystyle \sum _{p=2}^{n} \beta _p e_p + \gamma _1e_1\\{} & {} \qquad -\displaystyle \sum _{p=2}^{n-1} \beta _p e_{p+1}-(j-1)(2\gamma _1+\gamma _2)e_2-(j-1)\beta _1e_{3}+\\{} & {} \qquad +\displaystyle \frac{1}{(j-2)!}((j+1) \gamma _{j+1,1}+\gamma _{j+1,2})e_{j+1}+\displaystyle \frac{1}{(j-2)!} \beta _1e_{j+2}\\{} & {} \quad = (\gamma _1-\beta _1)e_1-(j-1)(2\gamma _1+\gamma _2-\beta _2)e_2-[(j-1)\beta _1\\{} & {} \qquad +\beta _2-(j-2)\beta _3]e_3-\displaystyle \sum _{k=4}^{j}[ \beta _{k-1}-(j+1-k)\beta _k]e_k-[\beta _j\\{} & {} \qquad -\displaystyle \frac{1}{(j-2)!}((j+1) \gamma _{j+1,1}+\gamma _{j+1,2})]e_{j+1}+\dots \end{aligned}$$

on the other hand

$$\begin{aligned}{} & {} d^{e}(t_1-(j+1)t_2+e_1-(j-1)e_2+\frac{1}{(j-2)!}e_{j+1})\\{} & {} \quad = (\gamma _1^{e}-\beta _1^{e})e_1-(j-1)(2\gamma _1^{e}+\gamma _2^{e}-\beta _2^{e})e_2-[(j-1)\beta _1^{e}\\{} & {} \qquad +\beta _2^{e}-(j-2)\beta _3^{e}]e_3 -\displaystyle \sum _{k=4}^{j}[ \beta _{k-1}^{e}-(j+1-k)\beta _k^{e}]e_k-[\beta _j^{e}\\{} & {} \qquad -\displaystyle \frac{1}{(j-2)!}((j+1) \gamma _{1}^{e}+\gamma _{2}^{e})]e_{j+1}+\dots \end{aligned}$$

On account of the coefficients of $e_1,\dots ,e_{j+1}$ we have

The following linear combination of the above equations

$$\begin{aligned} (j-1)(1)+(2)+(3)+\displaystyle \sum _{k=4}^{j+1}\displaystyle \frac{(j-2)!}{(j+1-k)!}(k) \end{aligned}$$

leads to $(j+1) \gamma _{j+1,1}+\gamma _{j+1,2}=(j+1) \gamma _{1}+\gamma _{2}$. Repeating this process for all j with $2\le j \le n-1$ allow us to assume

$$\begin{aligned} \Delta (e_i)=(i \gamma _{1}+\gamma _{2})e_i+\beta _1e_{i+1}, \quad 3 \le i \le n. \end{aligned}$$

Finally, let us consider $\Delta (t_1+e_1-e_{n+1}+e_{n+j})$ for a fixed j verifying $2 \le j \le m$. Then on one hand, we have

$$\begin{aligned}{} & {} \Delta (t_1)+\Delta (e_1)+\Delta (e_{n+1})+\Delta (e_{n+j})\\{} & {} \quad = -\displaystyle \sum _{p=1}^{n} p\beta _p e_p + \gamma _1e_1-\displaystyle \sum _{p=2}^{n-1} \beta _p e_{p+1}+((n+1)\gamma _1+\gamma _3)e_{n+1}+\beta _1e_{n+2} \\{} & {} \qquad +((n+j) \gamma _{n+j,1}+\gamma _{n+j,3})e_{n+j}+ \beta _1e_{n+j+1}\\{} & {} \quad =(\gamma _1-\beta _1)e_1+\dots +((n+1)\gamma _1+\gamma _3)e_{n+1}+\beta _1e_{n+2}+((n+j) \gamma _{n+j,1}\\{} & {} \qquad +\gamma _{n+j,3})e_{n+j}+ \beta _1e_{n+j+1}. \end{aligned}$$

On the other hand, we get

$$\begin{aligned}{} & {} d^{e}(t_1)+d^{e} (e_1)+d^{e} (e_{n+1})+d^{e} (e_{n+j})\\{} & {} \quad = (\gamma _1^{e}-\beta _1^{e})e_1+\dots +((n+1)\gamma _1^{e}+\gamma _3^{e})e_{n+1}+\beta _1^{e}e_{n+2}+((n+j) \gamma _{1}^{e}\\{} & {} \qquad +\gamma _{3}^{e})e_{n+j}+ \beta _1^{e}e_{n+j+1} \end{aligned}$$

which leads to $((n+j) \gamma _{n+j,1}+\gamma _{n+j,3})=((n+j) \gamma _{1}+\gamma _{3})$. Hence, we obtain

$$\begin{aligned} \Delta (e_{n+j})=((n+j) \gamma _{1}+\gamma _{3})e_{n+j}+\beta _1 e_{n+j+1}, \ 2 \le j \le m, \end{aligned}$$

which completes the proof of the theorem. $\square$

Let us consider now the maximal-dimensional solvable Lie superalgebra with model nilpotent nilradical [12]. We denote this superalgebra by $SN(n_1,\ldots ,n_k,1 | m_1, \ldots , m_p)$ and it can expressed by the following products:

with $\{ x_1, \dots , x_{n_1+\cdots n_k +1},$ $t_1,\dots ,t_{k+1},$ $t'_1,\dots ,t'_p\}$ even basis vectors and $\{ y_1, \dots ,$ $y_{m_1+\cdots +m_p}\}$ odd basis vectors.

In [12] it is proved that all the superderivations are inner, then and following the spirit of the proof of the theorem for model filiform nilradical we have the next result. We omit the computations because they are rather cumbersome and do not contain any new idea.

Theorem 4.2

On the maximal-dimensional solvable Lie superalgebra with model nilpotent nilradical, $SN(n_1, \ldots ,n_k,1 |$ $m_1, \ldots , m_p)$ every local superderivation is a superderivation.

5 Local Superderivations of Solvable Lie Superalgebras with Non-model Nilradical

Along this section, we use an example of solvable Lie superalgebra whose nilradical is a non-model one, in particular the nilradical is the only one Lie superalgebra of maximal nilindex $K^{2,m}$ (for more details regarding $K^{2,m}$ see Theorem 4.17 of [17]). We build over this solvable Lie superalgebra infinitely many local superderivations which are not superderivations.

Thus, consider for any m odd positive integer $m \ge 3$, the $(m+3)$-dimensional solvable Lie superalgebra $L^{m+3}$ (named $L^{m+3}_{1, \frac{2-m}{2}, 1,0,\dots ,0}$ in [11]). For that Lie superalgebra there exists a basis, namely $\{z,x_1,x_2,y_1,\dots , y_m\}$ with $\{ z,x_1,x_2\}$ even basis vectors and $\{y_1,\dots , y_m\}$ odd basis vectors, in which $L^{m+3}$ can be expressed by the only non-null bracket products that follow:

being its nilradical:

Note that $[\cdot , \cdot ]$ is the standard skew-symmetric bracket product whereas $(\cdot , \cdot )$ denotes the symmetrical ones, recall that a Lie superalgebra structure ${\mathfrak g}={\mathfrak g}_{\bar{0}} \oplus {\mathfrak g}_{\bar{1}}$ contains in particular the symmetric pairing $S^2 {\mathfrak g}_{\bar{1}} \longrightarrow {\mathfrak g}_{\bar{0}}$. Moreover, because of these symmetric products $K^{2,m}$ is not a model nilpotent Lie superalgebra.

Consider now the even superderivations on $L^{m+3}$, that is $d\in Der _{\overline{0}} (L^{m+3})$ with

$$\begin{aligned} \begin{array}{l} d(z)=\alpha _0 z+\alpha _1 x_1+\alpha _2 x_2, \\ d(x_i)=\alpha _{i0} z+\alpha _{i1} x_1+\alpha _{i2} x_2, \quad 1 \le i \le 2, \\ d(y_j)=\beta _{j1}y_1+\beta _{j2} y_2+\cdots +\beta _{jm}y_m, \quad 1 \le j \le m. \end{array} \end{aligned}$$

A straightforward computation leads to the following general matrix of any even superderivation on $L^{m+3}:$

$$\begin{aligned} Der _{\overline{0}}(L^{m+3})=\left( \begin{array}{cccccccc} 0&{}\quad \alpha _1&{}\quad \alpha _2&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0 \\ 0&{}\quad \alpha _{11}&{}\quad \alpha _{12}&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0 \\ 0&{}\quad 0&{}\quad \alpha _{11}&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad 0 \\ 0&{}\quad 0&{}\quad 0&{}\quad \beta _{11}&{}\quad -\alpha _1&{}\quad 0&{}\quad \dots &{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \beta _{22}&{}\quad -\alpha _1&{}\quad \dots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad \vdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad -\alpha _1 \\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \dots &{}\quad \beta _{mm} \end{array}\right) \end{aligned}$$

with $\beta _{ii}=((i-1)+\frac{2-m}{2}) \alpha _{11}$ for i, $1 \le i \le m$. Note also, from the matrix that $\beta _{i,i+1}=-\alpha _1$ and $\beta _{ij}=0$ for the remaining possibilities.

Theorem 5.1

$Der(L^{m+3})\subsetneqq LocDer(L^{m+3}).$

Proof

Consider the homogeneous linear mappings $\Delta _t : L^{m+3} \longrightarrow L^{m+3}$ ($t\ne 1$) of degree 0 defined on the basis vectors of $L^{m+3}$ by

$$\begin{aligned} \Delta (z)=0, \ \Delta (x_1)=x_1, \ \Delta (x_2)=tx_2, \ \Delta (y_j)=0, \ 1 \le j \le m. \end{aligned}$$

It can be easily checked that $\Delta$ is not an even superderivation because its matrix does not fit with the general matrix of even superderivations. Nevertheless, for any t the map $\Delta _t$ is an even local superderivation. Indeed, we have

$$\begin{aligned} \Delta _t(z)= & {} d_0(z), \ \Delta _t(x_1)=d_1(x_1), \ \Delta _t(x_2)=td_1(x_2), \\ \Delta _t(y_j)= & {} d_0(y_j), \ 1 \le j \le m, \end{aligned}$$

where $d_1$ is the resultant even superderivation after replacing $\alpha _{11}$ by 1 and all the rest of parameters by 0 on the general matrix of $Der _{\overline{0}}(L^{m+3})$ and $d_0$ is the null superderivation.

Analogously to proof of Theorem 3.1 for a fixed element $e=\alpha _0 z+\alpha _1 x_1+\alpha _2 x_2+\beta _1 y_1+\dots +\beta _m y_m$ we have $\Delta _t(e)=\alpha _1 x_1+\alpha _2 t x_2.$

Now our goal is to prove the existence of a derivation $d_e$ such that $\Delta _t(e)=d_e(e).$

Let $d_e(z)=d_e(y_j)=0$ be with $1\le i\le m$ and

$$\begin{aligned} d_e(x_1)=a_1x_1+a_2x_2,\qquad d_e(x_2)=a_3x_2 \end{aligned}$$

where $a_1,a_2,a_3$ are unknowns. From the constraint $\Delta _t(e)=d_e(e)$ we have that $a_1=1$ and the following equation $\alpha _1 a_2+\alpha _2 (a_3-t)=0.$ This equation always has solution with respect to unknowns $a_2,a_3.$ Replacing one of theses solutions of $d_e$ we get that $\Delta _t$ is a local superderivation.

$\square$

Remark 5.1

In fact, in the proof of Theorem 5.1 we show the existence of infinitely many local superderivations on the $(m+3)$-dimensional solvable Lie superalgebra $L^{m+3}$ which are not superderivations. Note also, that analogously it can be found infinitely many odd local superderivations which are not odd superderivations.

6 Local Superderivations of Maximal-Dimensional Solvable Leibniz Superalgebras with Model Filiform and Model Nilpotent Non-Lie Nilradical

We consider the maximal-dimensional solvable Leibniz superalgebra with filiform nilradical [12]. This superalgebra is unique for each pair of dimensions (n, m) and can be expressed by:

$$\begin{aligned} SLP^{n,m}:\left\{ \begin{array}{ll} [x_i,x_1]=x_{i+1},&{}\quad 2\le i\le n-1;\\ {} [y_j,x_1]=y_{j+1},&{}\quad 1\le j\le m-1;\\ {} [t_1,x_1]=-x_1,&{}\quad \\ {} [x_1,t_1]=x_1,&{}\quad \\ {} [x_i,t_1]=(i-2)x_i,&{}\quad 3\le i\le n;\\ {} [y_j,t_1]=(j-1)y_j,&{}\quad 2\le j\le m;\\ {} [x_i,t_2]=x_i,&{}\quad 2\le i\le n;\\ {} [y_j,t_3]=y_j,&{}\quad 1\le j\le m; \end{array}\right. \end{aligned}$$

with $\{x_1,x_2,\dots ,x_n,t_1,t_2,t_3\}$ a basis of $(SLP^{n,m})_{\overline{0}}$ and $\{y_1,y_2,\dots ,y_m\}$ a basis of $(SLP^{n,m})_{\overline{1}}.$ Its superalgebra of superderivations was obtained in [12]. Next, we prove the following result.

Theorem 6.1

On the maximal-dimensional solvable Leibniz superalgebra with model filiform nilradical, every local superderivation is a superderivation.

Proof

We rewrite the table of multiplications of the superalgebra $SLP^{n,m}$ as follows:

$$\begin{aligned} SLP^{n,m}:\left\{ \begin{array}{ll} [e_i,e_1]=e_{i+1},&{}\quad 2\le i\le n-1;\\ {} [e_{n+j},e_1]=e_{n+j+1},&{}\quad 1\le j\le m-1;\\ {} [t_1,e_1]=-e_1,&{}\quad \\ {} [e_1,t_1]=e_1,&{}\quad \\ {} [e_i,t_1]=(i-2)e_i,&{}\quad 3\le i\le n;\\ {} [e_{n+j},t_1]=(j-1)e_{n+j},&{}\quad 2\le j\le m;\\ {} [e_i,t_2]=e_i,&{}\quad 2\le i\le n;\\ {} [e_{n+j},t_3]=e_{n+j},&{}\quad 1\le j\le m; \end{array}\right. \end{aligned}$$

The superderivations of $SLP^{n,m}$ are inner [12]. Let us fix an arbitrary element

$$\begin{aligned} z=\gamma _1 t_1+\gamma _2 t_2+\gamma _3t_3+\sum _{p=1}^{n+m} \beta _p e_p \end{aligned}$$

of $SLP^{n,m}$. Then for its right operator $R_z$ we obtain

$$\begin{aligned} R_z(t_1)= & {} -\beta _1 e_1\quad R_z(t_2)=R_z(t_3)=0,\\ R_z(e_1)= & {} \gamma _1 e_1\\ R_z(e_i)= & {} ((i-2)\gamma _1+\gamma _2)e_i+\beta _1 e_{i+1},\ \ 2\le i\le n-1,\\ R_z(e_{n})= & {} ((n-2)\gamma _1+\gamma _2)e_n,\\ R_z(e_{n+j})= & {} ((j-1)\gamma _1+\gamma _3)e_{n+j}+\beta _1 e_{n+j+1},\ 1\le j\le m-1,\\ R_z(e_{n+m})= & {} ((m-1)\gamma _1+\gamma _3)e_{n+m}. \end{aligned}$$

In [12], we prove that all superderivations are even. Let us consider an arbitrary local superderivation $\Delta : SLP^{n,m}\longrightarrow SLP^{n,m}.$ Similar to Lie superalgebra we have that:

$$\begin{aligned} \begin{array}{lll} \Delta (t_1)=-\beta _{1,1} e_1&{}\Delta (t_2)=\Delta (t_3)=0,&{} \\ \Delta (e_1)=\gamma _{1,1} e_1&{}&{}\\ \Delta (e_i)=((i-2)\gamma _{i,1}+\gamma _{i,2})e_i+\beta _{i,1} e_{i+1},&{} 2\le i\le n-1,&{}\\ \Delta (e_{n})=((n-2)\gamma _{n,1}+\gamma _{n,2})e_n,&{}&{}\\ \Delta (e_{n+j})=((j-1)\gamma _{n+j,1}+\gamma _{n+j,3})e_{n+j}+\beta _{n+j,1} e_{n+j+1},&{}1\le j\le m-1,&{}\\ \Delta (e_{n+m})=((m-1)\gamma _{n+m,1}+\gamma _{n+m,3})e_{n+m}.&{}&{} \end{array} \end{aligned}$$

The goal now is to show that the expressions for $R_z$ and $\Delta$ coincide. Firstly, we will show this coincidence on the generators of the basis vectors, i.e. $t_1,t_2,t_3, e_1,e_2$ and $e_{n+1}$. Let us consider $\Delta (t_1+e_2)$. Thus as $\Delta$ is linear we obtain

$$\begin{aligned} \Delta (t_1+e_2)=\Delta (t_1)+\Delta (e_2)=-\beta _{1,1} e_1+\gamma _{2,1}e_2+\beta _{2,1}e_3, \end{aligned}$$

on the other hand and by definition the above coincides with the value of a superderivation, named $R^{e}$ on the vector $t_1+e_2$, thus

$$\begin{aligned} \Delta (t_1+e_2)=R^{e}(t_1+e_2)=-\beta _{1}^{e}e_1+\gamma _2^e e_2+\beta _1^e e_3, \end{aligned}$$

comparing the coefficients of $e_1$ and $e_3$ leads to $\beta _{2,1}=\beta _{1,1}$. Let us consider $\Delta (t_1+e_j)=-\beta _{1,1} e_1+((j-2)\gamma _{j,1}+\gamma _{j,2})e_j+\beta _{j,1}e_{j+1}$ for $3\le j\le n-1$ and on the other hand and by the definition of local derivation we get that $R^{e}(t_1+e_j)=-\beta _1^e e_1 +((j-2)\gamma _1^e+\gamma _2^e)e_j+\beta _1^e e_{j+1}.$ Comparing the coefficients of $e_1$ and $e_{j+1}$ we obtain $\beta _{j,1}=\beta _{1,1}$ for $3\le j\le n-1.$

We consider now $\Delta (t_1+e_{n+1})$ and similar as the above we have that $\beta _{n+1,1}=\beta _{1,1}$. Analogously, if we take $\Delta (t_1+e_{n+j})$ with $2\le j\le m-1$ we have that $\beta _{n+j,1}=\beta _{1,1}$ with $2\le j\le m-1.$

Consequently, there is no loss of generality in supposing

$$\begin{aligned} \begin{array}{ll} \Delta (t_1)=-\beta _1 e_1, &{} \Delta (t_2)=\Delta (t_3)=0, \\ \Delta (e_1)=\gamma _1e_1&{} \Delta (e_2)=\gamma _2e_2+\beta _1e_{3}, \\ \Delta (e_i)=((i-2) \gamma _{i,1}+\gamma _{i,2})e_i+\beta _1e_{i+1}, \ 3 \le i \le n-1, &{}\Delta (e_n)=((n-2) \gamma _{n,1}+\gamma _{n,2})e_n,\\ \Delta (e_{n+1})=\gamma _3 e_{n+1}+\beta _1 e_{n+2},&{} \\ \Delta (e_{n+j})=((j-1) \gamma _{n+j,1}+\gamma _{n+j,3})e_{n+j}+\beta _1 e_{n+j+1}, &{} 2 \le j \le m-1,\\ \Delta (e_{n+m})=((m-1) \gamma _{n+m,1}+\gamma _{n+m,3})e_{n+m}.&{} \end{array} \end{aligned}$$

Let us consider $\Delta (t_1+e_1+e_2+e_3)=(\gamma _1-\beta _1)e_1+\gamma _2 e_2+(\beta _1+\gamma _{3,1}+\gamma _{3,2})e_3+\beta _1 e_4$ on the other hand and by definition the above coincides with the value of a superderivation, named $R^{e}$ on the vector $t_1+e_1+e_2+e_3$, thus

$$\begin{aligned} \Delta (t_1+e_1+e_2+e_3)= & {} R^{e}(t_1+e_1+e_2+e_3)=(\gamma _1^e-\beta _1^e)e_1+\gamma _2^e e_2\\{} & {} +(\beta _1^e +\gamma _1^e+\gamma _2^e)e_3+\beta _1^e e_4. \end{aligned}$$

On account of the coefficients of $e_1,\dots ,e_{4}$ we have

$$\begin{aligned} \begin{array}{l} (1)\ \gamma _1-\beta _1=\gamma _1^e-\beta _1^e,\\ (2)\ \gamma _2=\gamma _2^e,\\ (3)\ \gamma _{3,1}+\gamma _{3,2}+\beta _1=\gamma _1^e+\gamma _2^e+\beta _1^e,\\ (4)\ \beta _1=\beta _1^e. \end{array} \end{aligned}$$

From $(1)+(2)+(4)$ we have that $\gamma _1+\gamma _2=\gamma _1^e+\gamma _2^e$ and from $(3)-(4)$ we get $\gamma _{3,1}+\gamma _{3,2}=\gamma _1^e+\gamma _2^e$. Then, $\gamma _{3,1}+\gamma _{3,2}=\gamma _1+\gamma _2.$

Now, we study $\Delta (t_1+e_1+e_2+e_j)$ for $4\le j\le n$ and we obtain the following equations:

$$\begin{aligned} \begin{array}{ll} (1)\ \gamma _1-\beta _1=\gamma _1^e-\beta _1^e,\\ (2)\ \gamma _2=\gamma _2^e,\\ (3)\ \beta _1=\beta _1^e,\\ (4)\ (j-2)\gamma _{j,1}+\gamma _{j,2}=(j-2)\gamma _1^e+\gamma _2^e, \end{array} \end{aligned}$$

From $(j-2)\times (1)+(2)+(j-2)\times (3)$ we get $(j-2)\gamma _1^e+\gamma _2^e=(j-2)\gamma _1+\gamma _2$ and from (4) we leads to $(j-2)\gamma _{j,1}+\gamma _{j,2}=(j-2)\gamma _1+\gamma _2.$

We now consider the vectors $t_1+e_1+e_{n+1}+e_{n+j}$ for a fixed j verifying $2\le j\le m.$ For $j=2,$ we have that

$$\begin{aligned}{} & {} \Delta (t_1)+\Delta (e_1)+\Delta (e_{n+1})+\Delta (e_{n+2})= ( \gamma _1-\beta _1)e_1+\gamma _3 e_{n+1}+\\{} & {} \quad (\gamma _{n+2,1}+\gamma _{n+2,3}+\beta _1)e_{n+2}+\beta _1e_{n+3} \end{aligned}$$

on the other hand, we get

$$\begin{aligned}{} & {} R^{e}(t_1)+R^{e} (e_1)+R^{e} (e_{n+1})+R^{e} (e_{n+2})= ( \gamma _1^e-\beta _1^e)e_1+\gamma _3^e e_{n+1}\\{} & {} \quad + (\gamma _{1}^e+\gamma _{3}^e+\beta _1^e)e_{n+2}+\beta _1^e e_{n+3}, \end{aligned}$$

which leads to

$$\begin{aligned} \begin{array}{l} (1)\ \gamma _1-\beta _1=\gamma _1^e-\beta _1^e,\\ (2)\ \gamma _3=\gamma _3^e,\\ (3)\ \gamma _{n+2,1}+\gamma _{n+2,2}+\beta _1=\gamma _1^e+\gamma _2^e+\beta _1^e,\\ (4)\ \beta _1=\beta _1^e. \end{array} \end{aligned}$$

From $(1)+(2)+(3)$ we obtain $\gamma _1+\gamma _3=\gamma _1^e+\gamma _3^e$ and from $(3)-(4)$ we get $\gamma _{n+2,1}+\gamma _{n+2,3}=\gamma _1+\gamma _3.$

We repeat the calculations for the vectors $t_1+e_1+e_{n+1}+e_{n+j}$ with $3\le j\le m$ and we derive the following equations:

$$\begin{aligned} \begin{array}{ll} (1)\ \gamma _1-\beta _1=\gamma _1^e-\beta _1^e,\\ (2)\ \gamma _3=\gamma _3^e,\\ (3)\ \beta _1=\beta _1^e,\\ (4)\ (j-1)\gamma _{n+j,1}+\gamma _{n+j,3}=(j-1)\gamma _1^e+\gamma _3^e, \end{array} \end{aligned}$$

From $(j-1)\times (1)+(2)+(j-1)\times (3)$ we get $(j-1)\gamma _1^e+\gamma _3^e=(j-1)\gamma _1+\gamma _3$ and from (4) we derive $(j-1)\gamma _{n+j,1}+\gamma _{n+j,3}=(j-1)\gamma _1+\gamma _3$ with $3\le j\le m$ which completes the proof of the theorem. $\square$

Let us consider now the complex maximal-dimensional solvable Leibniz superalgebra with model nilpotent non-Lie nilradical [12]. We denote this superalgebra by $SNP(n_1, \cdots ,n_k,1 | m_1, \ldots , m_p)$ given by:

with $\{ x_1, \dots ,$ $x_{n_1+\cdots n_k +1},$ $t_1,\dots ,t_{k+1},$ $t'_1,\dots ,t'_p\}$ even basis vectors and $\{ y_1, \dots ,$ $y_{m_1+\cdots +m_p}\}$ odd basis vectors.

In [12] it is proved that all superderivation are inner. Analogously to Theorem 6.1, we have the following result.

Theorem 6.2

On the maximal-dimensional solvable Leibniz superalgebra with model nilpotent non-Lie nilradical, $SNP(n_1, \ldots ,n_k,1 | m_1, \ldots , m_p)$ every local superderivation is a superderivation.

Proof

The proof is carrying out by arguments that used in the previous theorem. We omit the proof of this theorem because the computations are rather cumbersome and do not contain any new idea. $\square$