Definition of a module in Humphreys

Question

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, $[x,y]v=xyv-yxv$.

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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.

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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".

Answer 1

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.

Answer 2

You are right to say that those two definitions do not match: a $\mathfrak{g}$-module wher $\mathfrak{g}$ is seen as a Lie algebra is not a module where $\mathfrak{g}$ is seen as a nonassociative algebra.

There is a possible reconciliation in the comments, with the fact that $\mathfrak{g}$-modules are (usual) modules over the enveloping algebra.

Here is another take: for any $F$-vector space, $\operatorname{End}_F(V)$ is an associative $F$-algebra. Then for any other associative algebra $A$, a structure of $A$-module on $V$ is an algebra morphism $A\to \operatorname{End}_F(V)$. It also makes sense if $A$ is nonassociative (we can just forget that $\operatorname{End}_F(V)$ is associative too).

Now $\operatorname{End}_F(V)$ is also a Lie algebra, for the usual Lie bracket $[x,y]=xy-yx$. Usually, to avoid confusion with the associative algebra structure, we write $\mathfrak{gl}(V)$ for this Lie algebra. Then for any Lie algebra $\mathfrak{g}$, a structure of $\mathfrak{g}$-module on $V$ is a Lie algebra morphism $\mathfrak{g}\to \mathfrak{gl}(V)$.

The point of this answer is to notice that this definition is in essence the same, except that we shift the focus on the structure of $\operatorname{End}_F(V)$: instead of looking at it as a usual algebra, we can look at it as a Lie algebra (with a different product).

There are ways to unify all this properly, for instance using the notion of operads, but I hope this little argument is at least convincing for you.