What is the meaning of Lie algebra action?

Question

Let $U(L)$ be the universal enveloping algebra of a Lie algebra $L$. Let $A$ be an associative algebra such that $L$ acts on $A$ as derivation. I do not understand what is the meaning of L acts on A as derivation? In the following the $L$-action is extended to the $U(L)$-action as $x_1 \dots x_n \cdot a = x_1\cdot(\dots(x_n \cdot a) \dots )$ for all $x_i \in L$ and $a \in A$. $A$ is called algebra with derivation.

Answer 1

First, I think you want to say that the Lie algebra $L$ acts on the (associative) algebra $A$ by derivations, plural. That is, for every $x\in L$ and for every $a,b\in A$, $x\cdot (ab)=(x\cdot a)b+a(x\cdot b)$ (and $a\to x\cdot a$ is linear). Different $x\in L$ may act differently. This "derivation" property is an abstraction of Leibniz' rule for derivatives of products: $(ab)'=a'b+ab'$.

It seems to me likely that you do also want the action of $L$ on $A$ to respect the Lie bracket, in the sense that $$ [x,y]\cdot a \;=\; x\cdot(y\cdot a) - y\cdot (x\cdot a) $$

The extension of the action of a Lie algebra to an action of the universal enveloping algebra (with the condition satisfied by the Lie bracket... otherwise it doesn't quite work) is just by induction, that for all $x\in L$ and $\alpha\in U(L)$ and $a\in A$, $(x\alpha)\cdot a=x\cdot (\alpha\cdot a)$.

Answer 2

The following definition is frequently used.

Definition: Let $L$ be a Lie algebra over a field $K$, and $A$ be an associative $K$-algebra. Then by an action of $L$ on $A$, we mean a Lie algebra homomorphism
$$ \phi\colon L\rightarrow \operatorname{Der}(A). $$ So the action is given by $x.v=\phi(x)(v)$ for $x\in L$.