I am having some troubles reconciling the concept of a module as I know it vs. what is written in Introduction to Lie algebras and representation theory by Humphreys.
Suppose that $\mathfrak g$ is a Lie algebra over the field $\mathbb F$ and $V$ is an $\mathbb F$-vector space.
Humphreys defines a $\mathfrak g$-module to be such a vector space together with a map $\mathfrak g\times V\rightarrow V$, $(x,v)\mapsto xv$ such that this map is bilinear over $\mathbb F$ and in addition, $[x,y]v=xyv-yxv$.
Now, what I know is that a module is a "vector space" over a ring. Now, a ring is usually defined to be unital and associative, which a Lie algebra isn't, so $\mathfrak g$ is not a ring.
But I know that authors sometimes relax these requirements, so let us assume that rings don't have to possess multiplicative identities and don't need to be associative, in which sense $\mathfrak g$ is a ring.
With this concept of a ring, employing the usual definition of a module, $M$ is a (left) $\mathfrak g$-module, if there is a commutative and associative addition within $M$ with a unique null element and for each element a unique additive inverse, and there is a "scalar multiplication" map $\mathfrak g\times M\rightarrow M$, $(x,m)\mapsto xm$ such that
- $x(m+m^\prime)=xm+xm^\prime$;
- $(x+y)m=xm+ym$
- $[x,y]m=x(ym)$.
The third property here is in direct conflict with Humphreys' definition which would be $[x,y]m=x(ym)-y(xm)$. (In addition, I get the impression that the third property might be unpleasant if the ring isn't associative, truthfully I have never seen a module defined over a nonassociative ring before)
Moreover, I do not see why $M$ should be an $\mathbb F$-vector space. Based on the (usual) definition of a module, only $\mathfrak g$ acts via scalar multiplication on $M$ and $\mathbb F$ does not embed naturally into $\mathfrak g$.
Question: How is the definition of a $\mathfrak g$-module given by Humphreys related to the usual definition of a module over a ring? Because it seems to me that they are not only not equivalent, but the definition Humphreys gives is not even a special case, due to the conflicting "third property".
Best Answer
An $L$-module $V$ is just a Lie algebra representation $\phi\colon L\rightarrow \mathfrak{gl}(V)$, where $x.v=\phi(x)(v)$. By the universal property of $U(L)$, the universal enveloping algebra of $L$, $\phi$ extends to a representation of $U(L)$ on $V$. Conversely, every representation of $U(L)$ on $V$ restricts to a representation of $L$ on $V$. In this sense, $L$-modules correspond to $U(L)$-modules.