Generalized derivations of current Lie algebras

Abstract

Let L be a Lie algebra and let A be an associative commutative algebra with unity, both over the same field F. We consider the following question. Is every generalized derivation (resp. quasiderivation) of $L\otimes A$ the sum of a derivation and a map from the centroid of $L\otimes A$, if the same holds true for L?

Keywords:

• Current Lie algebra
• derivation
• generalized derivation
• Lie algebra
• quasiderivation
• tensor product of algebras

2020 Mathematics Subject Classification:

• 17B40

1 Introduction

Let L be a Lie algebra over a field F. A linear map $d:L\to L$ is called a derivation if $d(\left[x,y\right])=\left[d(x),y\right]+\left[x,d(y)\right]$ for all $x,y\in L$. As usual, we denote the set of all derivations of L by $\text{Der}(L)$. Obviously, $\text{Der}(L)$ is a Lie subalgebra of the general linear algebra $\mathfrak{g}\mathfrak{l}(L)$. There are several generalizations of the notion of a derivation. In this paper we consider generalized derivations and quasiderivations of Lie algebras as defined by Leger and Luks in [6]. Let $f:L\to L$ be a linear map. If there exist linear maps $g,h:L\to L$ such that $$\left[f(x),y\right]+\left[x,g(y)\right]=h\left(\left[x,y\right]\right)$$ for all $x,y\in L$, then f is called a generalized derivation. In case there exists a linear map $h:L\to L$ such that $$\left[f(x),y\right]+\left[x,f(y)\right]=h\left(\left[x,y\right]\right)$$ for all $x,y\in L$, then f is said to be a quasiderivation. By $\text{GDer}(L)$ we shall denote the set of all generalized derivations of L and by $\text{QDer}(L)$ the set of all quasiderivations of L. Obviously, $\text{QDer}(L)$ and $\text{GDer}(L)$ are Lie subalgebras of $\mathfrak{g}\mathfrak{l}(L)$ such that $$\text{Der}(L)\subseteq \text{QDer}(L)\subseteq \text{GDer}(L)\subseteq \mathfrak{g}\mathfrak{l}(L).$$

Yet another Lie subalgebra of $\mathfrak{g}\mathfrak{l}(L)$ is the centroid of L, which is defined as $$\text{Cent}(L)=\left\{\gamma \in \mathfrak{g}\mathfrak{l}(L)|\gamma ([x,y])=[x,\gamma (y)]\text{ for all }x,y\in L\right\}.$$

For each map $\gamma \in \text{Cent}(L)$ we have $$[\gamma (x),y]+[x,\gamma (y)]=2\gamma ([x,y])$$ for all $x,y\in L$. Thus, $\text{Cent}(L)\subseteq \text{QDer}(L)$ and so $$\text{Der}(L)+\text{Cent}(L)\subseteq \text{QDer}(L).$$

In several cases this is a strict inclusion. However, for some Lie algebras we have (1.1) $$\text{Der}(L)+\text{Cent}(L)=\text{QDer}(L)$$(1.1) or even (1.2) $$\text{Der}(L)+\text{Cent}(L)=\text{GDer}(L).$$(1.2)

Let us mention that Leger and Luks [6, Corollary 4.16] proved that (1.1) holds true for each centerless Lie algebra L generated by special weight spaces. Examples of Lie algebras satisfying (1.2) can be found in Brešar’s paper [1], where the structure of near-derivations was described for certain Lie algebras arising from associative ones. Note that the notion of a near-derivation, which was introduced in [1], is even more general than the notion of a generalized derivation.

Suppose that L and A are algebras over a field F, where L is a Lie algebra and A is an associative commutative algebra with unity. The tensor product algebra $L{\otimes }_{F}A$ (or shortly $L\otimes A$) is also a Lie algebra over F, which is called a current Lie algebra. Recall that the Lie product on $L\otimes A$ is defined as a bilinear map such that $$[x\otimes a,y\otimes b]=[x,y]\otimes \mathit{\text{ab}}$$ for any simple tensors $x\otimes a,y\otimes b\in L\otimes A$.

The aim of this paper is to consider the following two questions.

1. Does $L\otimes A$ satisfy (1.1), if L satisfies (1.1)?

2. Does $L\otimes A$ satisfy (1.2), if L satisfies (1.2)?

Our research was motivated by [6] and by Brešar’s papers [2, 3], where the study of functional identities on tensor products of algebras was initiated.

2 The results

Let L be a Lie algebra over a field F. Recall that the center $$Z(L):=\left\{x\in L\left|\right[x,y]=0\text{ for all }y\in L\right\}$$ and the derived algebra $$\left[L,L\right]:=\text{Span}(\left\{[x,y]|x,y\in L\right\})$$ are ideals of L. If $Z(L)=\left\{0\right\}$, we say that L is centerless. For any subset S of L the set $$Z_L(S):=\left\{x\in L\left|\right[x,s]=0\text{ for all }s\in S\right\}$$ is called the annihilator of S in L. If I is an ideal of L then $Z_L(I)$ is also an ideal of L. Thus, $Z_L([L,L])$ is an ideal of L and $$Z(L)=Z_L(L)\subseteq Z_L([L,L]).$$

Note that for any centerless Lie algebra L the sum $\text{Der}(L)+\text{Cent}(L)=\text{Der}(L)\oplus \text{Cent}(L)$ is a direct sum of vector spaces. Recall that a Lie algebra L is prime, if L has no nonzero ideals I, J such that $[I,J]=0$. Clearly, all prime Lie algebras are centerless. A Lie algebra is said to be perfect if $\left[L,L\right]=L$. Let us state our main result on the form of quasiderivations of a current Lie algebra $L\otimes A$.

Theorem 2.1.

Let $L\otimes A$ be a current Lie algebra over a field F, where L is centerless and $\text{char}(F)\ne 2$. Suppose that L is either perfect or prime. If $\text{QDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$, then $\text{QDer}(L\otimes A)=\text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

Another aim of this paper is to obtain a similar result for generalized derivations. Recall that the notion of a quasicentroid $\text{QCent}(L)$ of a Lie algebra L was defined in [6] as $$\text{QCent}(L)=\left\{f\in \mathfrak{g}\mathfrak{l}(L)\left|\right[f(x),y]=[x,f(y)]\text{ for all }x,y\in L\right\}.$$

Obviously, $\text{Cent}(L)\subseteq \text{QCent}(L)$ and (2.1) $$\text{GDer}(L)=\text{QDer}(L)+\text{QCent}(L)$$(2.1) (see [6, Proposition 3.3]). Note that each commuting linear map $f:L\to L$ (i.e. $\left[f(x),x\right]=0$ for all $x\in L$) belongs to $\text{QCent}(L)$. Moreover, if $\text{char}(F)\ne 2$ then $\text{QCent}(L)$ coincides with the set of all commuting linear maps of L.

Let L be a centerless Lie algebra over a field F with $\text{char}(F)\ne 2$. Suppose that L is either perfect or prime. Then $Z_L(\left[L,L\right])=0$ and hence the result of Brešar and Zhao [4, Corollary 3.3] implies that the set of all commuting linear maps of L coincides with $\text{Cent}(L)$. Thus, $\text{QCent}(L)=\text{Cent}(L)\subseteq \text{QDer}(L)$ and hence (2.1) implies $\text{GDer}(L)=\text{QDer}(L)$. Since $Z_L(\left[L,L\right])=0$ it follows that $Z_{L\otimes A}(\left[L\otimes A,L\otimes A\right])=0$ and so $\text{GDer}(L\otimes A)=\text{QDer}(L\otimes A)$ as well. Hence, Theorem 2.1 implies the following corollary.

Corollary 2.2.

Let $L\otimes A$ be a current Lie algebra over a field F, where L is centerless and $\text{char}(F)\ne 2$. Suppose that L is either perfect or prime. Then $\text{GDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$ implies $\text{GDer}(L\otimes A)=\text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

If $L\otimes A$ is a current Lie algebra, where A is finite dimensional, then we obtain the same conclusion assuming only that L is centerless.

Theorem 2.3.

Let $L\otimes A$ be a current Lie algebra over a field F with $\text{char}(F)\ne 2$. Suppose that L is centerless and A is finite dimensional.

1. If $\text{GenDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$, then $\text{GenDer}(L\otimes A)=\text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

2. If $\text{QDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$, then $\text{QDer}(L\otimes A)=\text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

The proofs of Theorems 2.1 and 2.3 are given in the next section.

3 The proofs

Let L and A be algebras over a field F, where L is a Lie algebra and A is an associative commutative algebra with unity. Pick a basis $\mathcal{B}=\left\{b_i|i\in I\right\}$ of A. Hence every element in $L\otimes A$ can be written uniquely in the form $x_{i_1}\otimes b_{i_1}+x_{i_2}\otimes b_{i_2}+\cdots +x_{i_n}\otimes b_{i_n}$ where $n\ge 1$ and $x_i\in L$.

Let $f:L\otimes A\to L\otimes A$ be a linear map. For any element $x\in L$ there exist unique elements $f_i(x)\in L,i\in I$, such that (3.1) $$f(x\otimes 1)=\sum _{i\in I}{f}_i(x)\otimes b_i,$$(3.1) where $f_i(x)=0$ for all but finitely many $i\in I$. For each $i\in I$ the map $f_i:L\to L$ defined by $f_i:x\mapsto f_i(x)$ is obviously linear. Let $f_{\mathcal{B}}:L\otimes A\to L\otimes A$ be a linear map such that (3.2) $$f_{\mathcal{B}}(x\otimes a)=\sum _{i\in I}{f}_i(x)\otimes a{b}_i$$(3.2) for each simple tensor $x\otimes a\in L\otimes A$. Obviously, $f_{\mathcal{B}}$ is well-defined since for each $x\in L$ we have $f_i(x)\ne 0$ for only finitely many elements $i\in I$. Note that $f(x\otimes 1)=f_{\mathcal{B}}(x\otimes 1)$ for all $x\in L$.

The following proposition shows that certain properties of a linear map f are inherited to the map $f_{\mathcal{B}}$.

Proposition 3.1.

Let $L\otimes A$ be a current Lie algebra over a field F. For any basis $\mathcal{B}$ of A the following assertions hold true:

1. If $f\in \text{GenDer}(L\otimes A)$, then $f_{\mathcal{B}}\in \text{GenDer}(L\otimes A)$.

2. If $f\in \text{QDer}(L\otimes A)$, then $f_{\mathcal{B}}\in \text{QDer}(L\otimes A)$.

Proof.

First, suppose that $f\in \text{GenDer}(L\otimes A)$. Then there exist linear maps $g,h:L\otimes A\to L\otimes A$ such that (3.3) $$\left[f(x),y\right]+\left[x,g(y)\right]=h\left(\left[x,y\right]\right)$$(3.3) for all $x,y\in L\otimes A$. Pick any basis $\mathcal{B}=\left\{b_i|i\in I\right\}$ of A. Then according to (3.2) the linear maps $g_{\mathcal{B}},h_{\mathcal{B}}:L\otimes A\to L\otimes A$ are given by (3.4) $$g_{\mathcal{B}}(x\otimes a)=\sum _{i\in I}{g}_i(x)\otimes a{b}_i\hspace{1em}\text{and}\hspace{1em}{h}_{\mathcal{B}}(x\otimes a)=\sum _{i\in I}{h}_i(x)\otimes a{b}_i$$(3.4) for any simple tensor $x\otimes a\in L\otimes A$. In order to prove that $f_{\mathcal{B}}\in \text{GenDer}(L\otimes A)$, let us show that (3.5) $$\left[f_{\mathcal{B}}(x),y\right]+\left[x,g_{\mathcal{B}}(y)\right]=h_{\mathcal{B}}\left(\left[x,y\right]\right)$$(3.5) for all $x,y\in L\otimes A$. Setting simple tensors $x\otimes 1,y\otimes 1$ in (3.3) and using $\left[x\otimes 1,y\otimes 1\right]=\left[x,y\right]\otimes 1$, we get $$h\left(\left[x,y\right]\otimes 1\right)=\left[f(x\otimes 1),y\otimes 1\right]+\left[x\otimes 1,g(y\otimes 1)\right].$$

According to (3.2) this identity can be rewritten as $$\begin{array}{c}\sum _{i\in I}{h}_i(\left[x,y\right])\otimes b_i=\left[\sum _{i\in I}{f}_i(x)\otimes b_i,y\otimes 1\right]+\left[x\otimes 1,\sum _{i\in I}{g}_i(y)\otimes b_i\right]\\ =\sum _{i\in I}\left[f_i(x),y\right]\otimes b_i+\sum _{i\in I}\left[x,g_i(y)\right]\otimes b_i\end{array}$$ and consequently $$\sum _{i\in I}(\left[f_i(x),y\right]+\left[x,g_i(y)\right]-h_i(\left[x,y\right]))\otimes b_i=0.$$ for all $x,y\in L$. Thus, for any $i\in I$ we have (3.6) $$\left[f_i(x),y\right]+\left[x,g_i(y)\right]=h_i(\left[x,y\right])$$(3.6) for all $x,y\in L$. Using (3.6) we obtain $$\begin{array}{c}{h}_{\mathcal{B}}\left(\left[x\otimes a,y\otimes b\right]\right)=h_{\mathcal{B}}\left(\left[x,y\right]\otimes \mathit{\text{ab}}\right)=\sum _{i\in I}{h}_i(\left[x,y\right])\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}(\left[f_i(x),y\right]+\left[x,g_i(y)\right])\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}\left[f_i(x),y\right]\otimes \mathit{\text{ab}}{b}_i+\sum _{i\in I}\left[x,g_i(y)\right]\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}\left[f_i(x)\otimes a{b}_i,y\otimes b\right]+\sum _{i\in I}\left[x\otimes a,g_i(y)\otimes b{b}_i\right]\\ =\left[\sum _{i\in I}{f}_i(x)\otimes a{b}_i,y\otimes b\right]+\left[x\otimes a,\sum _{i\in I}{g}_i(y)\otimes b{b}_i\right]\\ =\left[f_{\mathcal{B}}(x\otimes a),y\otimes b\right]+\left[x\otimes a,g_{\mathcal{B}}(y\otimes b)\right].\end{array}$$ for all simple tensors $x\otimes a,y\otimes b\in L\otimes A$. Since $f_{\mathcal{B}},g_{\mathcal{B}}$, and $h_{\mathcal{B}}$ are linear maps it follows that (3.5) holds true. Thus, $f_{\mathcal{B}}\in \text{GenDer}(L\otimes A)$ and so the proof of (i) is complete.

Note that (ii) can be proved analogously by setting f = g in the arguments above. □

Lemma 3.2.

Let $L\otimes A$ be a current Lie algebra over a field F and let $\mathcal{B}=\left\{b_i|i\in I\right\}$ be a basis of A. Suppose that $\left\{f_i:L\to L|i\in I\right\}$ is a family of linear maps such that for any $x\in L$ we have $f_i(x)\ne 0$ for only finitely many elements $i\in I$. Let a linear map $f_{\mathcal{B}}:L\otimes A\to L\otimes A$ be defined as in (3.2).

1. If $f_i\in \text{Der}(L)$ for all $i\in I$, then $f_{\mathcal{B}}\in \text{Der}(L\otimes A)$.

2. If $f_i\in \text{Cent}(L)$ for all $i\in I$, then $f_{\mathcal{B}}\in \text{Cent}(L\otimes A)$.

Proof.

(i) Suppose that $f_i\in \text{Der}(L)$ for all $i\in I$. Thus, for each $i\in I$ $$\left[f_i(x),y\right]+\left[x,f_i(y)\right]=f_i(\left[x,y\right])$$ for all $x,y\in L$. For any simple tensors $x\otimes a,y\otimes b\in L\otimes A$ we have $$\begin{array}{c}{f}_{\mathcal{B}}\left(\left[x\otimes a,y\otimes b\right]\right)=f_{\mathcal{B}}\left(\left[x,y\right]\otimes \mathit{\text{ab}}\right)=\sum _{i\in I}{f}_i(\left[x,y\right])\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}(\left[f_i(x),y\right]+\left[x,f_i(y)\right])\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}\left[f_i(x),y\right]\otimes \mathit{\text{ab}}{b}_i+\sum _{i\in I}\left[x,f_i(y)\right]\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}\left[f_i(x)\otimes a{b}_i,y\otimes b\right]+\sum _{i\in I}\left[x\otimes a,f_i(y)\otimes b{b}_i\right]\\ =\left[\sum _{i\in I}{f}_i(x)\otimes a{b}_i,y\otimes b\right]+\left[x\otimes a,\sum _{i\in I}{f}_i(y)\otimes b{b}_i\right]\\ =\left[f_{\mathcal{B}}(x\otimes a),y\otimes b\right]+\left[x\otimes a,f_{\mathcal{B}}(y\otimes b)\right].\end{array}$$

Since $f_{\mathcal{B}}$ is linear it follows that $f_{\mathcal{B}}\in \text{Der}(L\otimes A)$.

(ii) Let’s assume that $f_i\in \text{Cent}(L)$ for all $i\in I$. Thus, for each $i\in I$ $$\left[f_i(x),y\right]=f_i(\left[x,y\right])$$

for all $x,y\in L$. For any simple tensors $x\otimes a,y\otimes b\in L\otimes A$ we have $$\begin{array}{c}{f}_{\mathcal{B}}\left(\left[x\otimes a,y\otimes b\right]\right)=f_{\mathcal{B}}\left(\left[x,y\right]\otimes \mathit{\text{ab}}\right)=\sum _{i\in I}{f}_i(\left[x,y\right])\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}\left[f_i(x),y\right]\otimes \mathit{\text{ab}}{b}_i\\ =\sum _{i\in I}\left[f_i(x)\otimes a{b}_i,y\otimes b\right]\\ =\left[\sum _{i\in I}{f}_i(x)\otimes a{b}_i,y\otimes b\right]\\ =\left[f_{\mathcal{B}}(x\otimes a),y\otimes b\right].\end{array}$$

Since $f_{\mathcal{B}}$ is linear it follows that $f_{\mathcal{B}}\in \text{Cent}(L\otimes A)$. □

Lemma 3.3.

Let $L\otimes A$ be a current Lie algebra over a field F and let $\mathcal{B}=\left\{b_i|i\in I\right\}$ be a basis of A. Suppose that L is perfect or L is prime. Furthermore, assume that $\text{QDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$. If $f\in \text{QDer}(L\otimes A)$, then $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

Proof.

Let us pick an arbitrary quasi-derivation $f\in \text{QDer}(L\otimes A)$. Then there exists a linear map $h:L\otimes A\to L\otimes A$ such that (3.7) $$\left[f(x),y\right]+\left[x,f(y)\right]=h\left(\left[x,y\right]\right)$$(3.7) for all $x,y\in L\otimes A$. Recall that there exist families of linear maps $\left\{f_i:L\to L|i\in I\right\}$ and $\{h_i:L\to$ $L|i\in I\}$ such that $$f(x\otimes 1)=\sum _{i\in I}{f}_i(x)\otimes b_i\text{ and }h(x\otimes 1)=\sum _{i\in I}{h}_i(x)\otimes b_i$$ for all $x\in L$, where for each $x\in L$ we have $f_i(x)\ne 0$ for only finitely many elements $i\in I$ and $h_i(x)\ne 0$ for only finitely many elements $i\in I$ (see (3.1)). Similarly as in the proof of Proposition 3.1 we see that for any $i\in I$ we get (3.8) $$\left[f_i(x),y\right]+\left[x,f_i(y)\right]=h_i(\left[x,y\right])$$(3.8) for all $x,y\in L$. Hence, each f_i is a quasi-derivation of L. Consequently, our assumption implies $f_i\in \text{Der}(L)\oplus \text{Cent}(L)$ for all $i\in I$. Thus, for each $i\in I$ there exist maps $d_i\in \text{Der}(L)$ and ${\gamma }_i\in \text{Cent}(L)$ such that $f_i=d_i+{\gamma }_i$. Hence, (3.8) can be rewritten as (3.9) $$\begin{array}{c}{h}_i(\left[x,y\right])=\left[d_i\left(x\right)+{\gamma }_i\left(x\right),y\right]+\left[x,d_i\left(y\right)+{\gamma }_i\left(y\right)\right]\\ =\left[d_i\left(x\right),y\right]+\left[x,d_i\left(y\right)\right]+\left[{\gamma }_i\left(x\right),y\right]+\left[x,{\gamma }_i\left(y\right)\right]\\ =d_i\left(\left[x,y\right]\right)+{\gamma }_i\left(\left[x,y\right]\right)+{\gamma }_i\left(\left[x,y\right]\right)\\ =f_i\left(\left[x,y\right]\right)+{\gamma }_i\left(\left[x,y\right]\right)\end{array}$$(3.9) for all $x,y\in L$ and $i\in I$.

First, suppose that L is perfect. Since $\left[L,L\right]=L$ it follows from (3.9) that ${\gamma }_i(x)=h_i(x)-f_i(x)$ for all $x\in L$ and $i\in I$. Hence, for each $x\in L$ we have ${\gamma }_i(x)=0$ for all but finitely many elements $i\in I$ and consequently $d_i(x)=f_i(x)-{\gamma }_i(x)=0$ for all but finitely many elements $i\in I$.

Next, suppose that L is prime. According to (3.9) we see that for each pair $x,y\in L$ we have ${\gamma }_i(\left[x,y\right])=0$ for all but finitely many elements $i\in I$. Without loss of generality, we may assume that L is nonzero. Since L is prime it follows that $\left[L,L\right]\ne \left\{0\right\}$. Hence, there exist elements $x_0,y_0\in L$ such that $\left[x_0,y_0\right]\ne 0$. Since L is torsion free $\text{Cent}(L)$-module (see [5, Theorem 1.1]) and since ${\gamma }_i(\left[x_0,y_0\right])=0$ for all but finitely many elements $i\in I$ it follows that ${\gamma }_i=0$ for all but finitely many elements $i\in I$. Consequently, for each $x\in L$ also $d_i(x)=f_i(x)-{\gamma }_i(x)=0$ for all but finitely many elements $i\in I$.

Let $d_{\mathcal{B}},{\gamma }_{\mathcal{B}}:L\otimes A\to L\otimes A$ be linear maps such that (3.10) $$d_{\mathcal{B}}(x\otimes a)=\sum _{i\in I}{d}_i(x)\otimes a{b}_i\hspace{1em}\text{and}\hspace{1em}{\gamma }_{\mathcal{B}}(x\otimes a)=\sum _{i\in I}{\gamma }_i(x)\otimes a{b}_i$$(3.10) for each simple tensor $x\otimes a\in L\otimes A$. Obviously, $d_{\mathcal{B}}$ and ${\gamma }_{\mathcal{B}}$ are well-defined, since in case L is perfect or prime, both sums in (3.10) are finite. Namely, for any $x\in L$ in both cases $d_i(x)=0$ for all but finitely many elements $i\in I$ and ${\gamma }_i(x)=0$ for all but finitely many elements $i\in I$. Now, Lemma 3.2 implies that $d_{\mathcal{B}}\in \text{Der}(L\otimes A)$ and ${\gamma }_{\mathcal{B}}\in \text{Cent}(L\otimes A)$. We can now conclude that $$\begin{array}{c}{f}_{\mathcal{B}}(x\otimes a)=\sum _{i\in I}{f}_i(x)\otimes a{b}_i=\sum _{i\in I}(d_i\left(x\right)+{\gamma }_i\left(x\right))\otimes a{b}_i\\ =\sum _{i\in I}{d}_i\left(x\right)\otimes a{b}_i+\sum _{i\in I}{\gamma }_i\left(x\right)\otimes a{b}_i\\ =d_{\mathcal{B}}(x\otimes a)+{\gamma }_{\mathcal{B}}(x\otimes a)\end{array}$$ for each simple tensor $x\otimes a\in L\otimes A$. Since $f_{\mathcal{B}},d_{\mathcal{B}}$, and ${\gamma }_{\mathcal{B}}$ are linear maps it follows $f_{\mathcal{B}}=d_{\mathcal{B}}+{\gamma }_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$. □

If we assume that A is a finite dimensional algebra in Lemma 3.3, then we can drop the assumption of L being perfect or prime. Namely, in this case both sums in (3.10) are finite and so the maps $d_{\mathcal{B}}$ and ${\gamma }_{\mathcal{B}}$ are well-defined. Thus, using similar arguments as in the proof of Lemma 3.3 we obtain the following proposition.

Proposition 3.4.

Let $L\otimes A$ be a current Lie algebra over a field F and let $\mathcal{B}$ be a basis of A. Suppose that ${\mathrm{\text{dim}}}_{\mathbb{F}}A<\infty$.

1. If $f\in \text{GenDer}(L\otimes A)$ and $\text{GenDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$, then $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

2. If $f\in \text{QDer}(L\otimes A)$ and $\text{QDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$, then $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$.

Recall that a map $f:L\to L$ is commuting if $\left[f(x),x\right]=0$ for all $x\in L$. The following lemma, which will be used in the proof of Theorem 2.3, follows directly from [6, Proposition 5.26].

Lemma 3.5.

Let L be a centerless Lie algebra over a field F with $\text{char}(F)\ne 2$. If a quasi-derivation $f\in \text{QDer}(L)$ is commuting, then $f\in \text{Cent}(L)$.

Now, we can prove our main results, Theorems 2.1 and 2.3.

Proof of Theorem 2.1.

Suppose that $\text{QDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$. Pick any basis $\mathcal{B}=\left\{b_i|i\in I\right\}$ of A. Let $f\in \text{QDer}(L\otimes A)$ be an arbitrary quasi-derivation. According to Proposition 3.1 the map $f_{\mathcal{B}}$ is a quasi-derivation of $L\otimes A$. Moreover, Lemma 3.3 implies that $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$. Let $F=f-f_{\mathcal{B}}$. Obviously, $F\in \text{QDer}(L\otimes A)$ and $F(x\otimes 1)=f(x\otimes 1)-f_{\mathcal{B}}(x\otimes 1)=0$ for all $x\in L$. Since F is a quasi-derivation there exists a linear map $H:L\otimes A\to L\otimes A$ such that (3.11) $$\left[F(x),y\right]+\left[x,F(y)\right]=H\left(\left[x,y\right]\right)$$(3.11) for all $x,y\in L\otimes A$. For each simple tensor $x\otimes a\in L\otimes A$ there exist unique elements $F_i(x\otimes a)\in L,i\in I$, such that $$F(x\otimes a)=\sum _{i\in I}{F}_i(x\otimes a)\otimes b_i,$$ where $F_i(x\otimes a)\ne 0$ for only finitely many $i\in I$. For each $a\in A$ and each $i\in I$ we define a map $F_{a,i}:L\to L$ by $F_{a,i}:x\mapsto F_i(x\otimes a)$, which is obviously linear. First, we shall show that $F_{a,i}\in \text{Cent}(L)$ for any $a\in A$ and any $i\in I$. Let us fix an arbitrary $a\in A$. Setting simple tensors $x\otimes a$ and $x\otimes 1$ in (3.11) and using $F(x\otimes 1)=0$ we obtain $$\left[F(x\otimes a),x\otimes 1\right]=H\left(\left[x\otimes a,x\otimes 1\right]\right)=H\left(\left[x,x\right]\otimes a\right)=0$$ for all $x\in L$. Hence, $$\begin{array}{c}0=\left[F(x\otimes a),x\otimes 1\right]=\left[\sum _{i\in I}{F}_i(x\otimes a)\otimes b_i,x\otimes 1\right]\\ =\sum _{i\in I}\left[F_{a,i}(x),x\right]\otimes b_i\end{array}$$ for all $x\in L$. Consequently, $\left[F_{a,i}(x),x\right]=0$ for all $x\in L$ and all $i\in I$. Thus, for each $a\in A$ and each $i\in I$ the map $F_{a,i}$ is commuting. According to our assumption L is perfect and centerless or L is prime. In both cases it follows that $Z_L(\left[L,L\right])=\left\{0\right\}$. Hence, [4, Corollary 3.3] yields that $F_{a,i}\in \text{Cent}(L)$ for all $i\in I,a\in A$. Next, we claim that $F\in \text{Der}(L\otimes A)$. Namely, setting simple tensors $x\otimes a$ in $y\otimes b$ in (3.11) we get (3.12) $$H\left(\left[x\otimes a,y\otimes b\right]\right)=\left[F(x\otimes a),y\otimes b\right]+\left[x\otimes a,F(y\otimes b)\right].$$(3.12)

On the other hand, since $\left[x\otimes a,y\otimes b\right]=\left[x\otimes \mathit{\text{ab}},y\otimes 1\right]$ and $F(y\otimes 1)=0$, we obtain $$H\left(\left[x\otimes a,y\otimes b\right]\right)=\left[F\left(x\otimes \mathit{\text{ab}}\right),y\otimes 1\right]=\left[\sum _{i\in I}{F}_{\mathit{\text{ab}},i}(x)\otimes b_i,y\otimes 1\right]$$ for all $x,y\in L$ and $a,b\in A$. Since $F_{\mathit{\text{ab}},i}\in \text{Cent}(L)$ it follows $$H\left(\left[x\otimes a,y\otimes b\right]\right)=\sum _{i\in I}\left[F_{\mathit{\text{ab}},i}(x),y\right]\otimes b_i=\sum _{i\in I}{F}_{\mathit{\text{ab}},i}(\left[x,y\right])\otimes b_i=F\left(\left[x,y\right]\otimes \mathit{\text{ab}}\right)$$ for all $x,y\in L$ and $a,b\in A$. Hence, (3.13) $$H\left(\left[x\otimes a,y\otimes b\right]\right)=F\left(\left[x\otimes a,y\otimes b\right]\right)$$(3.13) for all $x\otimes a,y\otimes b\in L\otimes A$. Consequently, (3.12) can be rewritten as $$F\left(\left[x\otimes a,y\otimes b\right]\right)=\left[F(x\otimes a),y\otimes b\right]+\left[x\otimes a,F(y\otimes b)\right]$$ for all $x\otimes a,y\otimes b\in L\otimes A$. Since F is linear it follows that $F\in \text{Der}(L\otimes A).$ We can now conclude that $$f=f_{\mathcal{B}}+F,$$ where $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$ and $F\in \text{Der}(L\otimes A)$. Thus, $f\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$ and so the proof is complete. □

Proof of Theorem 2.3.

In order to prove (i) let us assume that $\text{GenDer}(L)=\text{Der}(L)\oplus \text{Cent}(L)$. Pick any basis $\mathcal{B}=\left\{b_i\right|i\in I\}$ of A. Here, $I=\{1,2,\dots ,n\}$, since A is finite dimensional. Let $f\in \text{GenDer}(L\otimes A)$. Then there exist linear maps $g,h:L\otimes A\to L\otimes A$ such that (3.14) $$\left[f(x),y\right]+\left[x,g(y)\right]=h\left(\left[x,y\right]\right)$$(3.14) for all $x,y\in L\otimes A$. Proposition 3.1 implies $f_{\mathcal{B}}\in \text{GenDer}(L\otimes A)$. Moreover, (3.15) $$\left[f_{\mathcal{B}}(x),y\right]+\left[x,g_{\mathcal{B}}(y)\right]=h_{\mathcal{B}}\left(\left[x,y\right]\right)$$(3.15) for all $x,y\in L\otimes A$ (see the proof of Proposition 3.1). According to Proposition 3.4 we know that $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$. Let us define the following maps: $F=f-f_{\mathcal{B}},G=g-g_{\mathcal{B}}$, and $H=h-h_{\mathcal{B}}$. Obviously, F, G, and H are linear maps. Using (3.14) and (3.15) we get (3.16) $$\left[F(x),y\right]+\left[x,G(y)\right]=H\left(\left[x,y\right]\right)$$(3.16) for all $x,y\in L\otimes A.$ Moreover, $F(x\otimes 1)=0=G(x\otimes 1)$ for all $x\in L$. For each simple tensor $x\otimes a\in L\otimes A$ there exist unique elements $F_i(x\otimes a),G_i(x\otimes a),H_i(x\otimes a)\in L,i\in I$, such that (3.17) $$\begin{array}{c}F(x\otimes a)=\sum _{i\in I}{F}_i(x\otimes a)\otimes b_i,\\ G(x\otimes a)=\sum _{i\in I}{G}_i(x\otimes a)\otimes b_i,\\ H(x\otimes a)=\sum _{i\in I}{H}_i(x\otimes a)\otimes b_i.\end{array}$$(3.17)

For each $a\in A$ and each $i\in I$ we define maps $F_{a,i},G_{a,i},H_{a,i}:L\to L$ by $F_{a,i}:x\mapsto F_i(x\otimes a),G_{a,i}:x\mapsto G_i(x\otimes a)$, and $H_{a,i}:x\mapsto H_i(x\otimes a)$. It is easy to see that $F_{a,i},G_{a,i},H_{a,i}$ are linear maps. Our aim is to prove that F is a derivation. First, let us prove that $F_{a,i}\in \text{Cent}(L)$ for all $a\in A$ and $i\in I$. Since $G(y\otimes 1)=0$ it follows from (3.16) that (3.18) $$\left[F(x\otimes a),y\otimes 1\right]=H\left(\left[x\otimes a,y\otimes 1\right]\right)=H\left(\left[x,y\right]\otimes a\right)$$(3.18) for all $x,y\in L$ and $a\in A$. Fix an arbitrary element $a\in A$. Using (3.17) we can rewrite (3.18) as $$\begin{array}{c}0=\left[F(x\otimes a),y\otimes 1\right]-H\left(\left[x,y\right]\otimes a\right)\\ =\left[\sum _{i\in I}{F}_i(x\otimes a)\otimes b_i,y\otimes 1\right]-\sum _{i\in I}{H}_i([x,y]\otimes a)\otimes b_i\\ =\sum _{i\in I}(\left[F_{i,a}(x),y\right]-H_{i,a}(\left[x,y\right]))\otimes b_i\end{array}$$ for all $x,y\in L$. Hence, (3.19) $$\left[F_{a,i}(x),y\right]=H_{a,i}(\left[x,y\right])$$(3.19) for all $x,y\in L$. Consequently, $\left[F_{a,i}(x),x\right]=0$ for all $x\in L$. Thus, $F_{a,i}$ is a commuting linear map for each $a\in A$ and each $i\in I$. By interchanging the roles of x and y in (3.19) and using $\left[x,y\right]=-\left[y,x\right]$ we get $\left[x,F_{a,i}(y)\right]=H_{a,i}(\left[x,y\right])$ and so $$\left[F_{a,i}(x),y\right]+\left[x,F_{a,i}(y)\right]=2H_{a,i}(\left[x,y\right])$$ for all $x,y\in L$. This means that $F_{a,i}\in \text{QDer}(L)$ and hence Lemma 3.5 yields that $F_{i,a}\in \text{Cent}(L)$ for each $a\in A$ and each $i\in I$. Next, we shall prove that F = G. Namely, since $G(y\otimes 1)=0=F(x\otimes 1)$ we see that (3.16) yields $$\begin{array}{c}\left[F(x\otimes a),y\otimes 1\right]=H\left(\left[x\otimes a,y\otimes 1\right]\right)=H\left(\left[x\otimes 1,y\otimes a\right]\right)\\ =\left[x\otimes 1,G(y\otimes a)\right]\end{array}$$ for all $x,y\in L$ and $a\in A$. Fix an arbitrary $a\in A$. Using (3.17) we can rewrite the last identity as $$\begin{array}{c}0=\left[\sum _{i\in I}{F}_{a,i}(x)\otimes b_i,y\otimes 1\right]-\left[x\otimes 1,\sum _{i\in I}{G}_{a,i}(y)\otimes b_i\right]\\ =\sum _{i\in I}\left[F_{a,i}(x),y\right]\otimes b_i-\sum _{i\in I}\left[x,G_{a,i}(y)\right]\otimes b_i\\ =\sum _{i\in I}(\left[F_{a,i}(x),y\right]-\left[x,G_{a,i}(y)\right])\otimes b_i\end{array}$$ for all $x,y\in L$. Consequently, for each $i\in I$ we have $$\left[F_{a,i}(x),y\right]=\left[x,G_{a,i}(y)\right]$$ for all $x,y\in L$. Since $F_{a,i}\in \text{Cent}(L)$ it follows $$\left[x,G_{a,i}(y)\right]=\left[F_{a,i}(x),y\right]=F_{a,i}(\left[x,y\right])=\left[x,F_{a,i}(y)\right]$$ and hence $\left[x,G_{a,i}(y)-F_{a,i}(y)\right]=0$ for all $x,y\in L$ and $i\in I$. Thus, $G_{a,i}(y)-F_{a,i}(y)$ belongs to the center of L for all $y\in L$ and $i\in I$. Since L is centerless it now follows that $G_{a,i}=F_{a,i}$ for all $i\in I$ and all $a\in A$. Accordingly, (3.17) implies $F(x\otimes a)=G(x\otimes a)$ for each simple tensor $x\otimes a\in L\otimes A$. However, since F and G are linear it follows F = G. By using the same arguments as in the proof of Theorem 2.1 we can now show that F is a derivation. Thus, since $f=f_{\mathcal{B}}+F$, where $f_{\mathcal{B}}\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$ and $F\in \text{Der}(L\otimes A)$, it follows that $f\in \text{Der}(L\otimes A)\oplus \text{Cent}(L\otimes A)$. The proof of (i) is now complete.

Note that (ii) can be proved analogously by setting f = g in the arguments above. □

Remark.

According to Theorem 2.3 one might conjecture that Theorem 2.1 holds true even without the assumption that a Lie algebra L is either perfect or prime. Unfortunately, we were not able to prove nor disprove this conjecture.