# Definition of a module in Humphreys

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".

• Any $\mathfrak{g}$-module corresponds bijectively to a $U(\mathfrak{g})$-module. So we have our ring $R=U(\mathfrak{g})$, the universal enveloping algebra. – Dietrich Burde Nov 7 '19 at 20:04
• @DietrichBurde Just to clarify, by this you mean that a $\mathfrak g$-module as given in Humphreys is equivalent to a module in the usual sense over the universal enveloping algebra? – Bence Racskó Nov 7 '19 at 20:07
• Yes, exactly. Also a $G$-module for a group $G$ is just a $\Bbb Z[G]$-module for the group ring $R=\Bbb Z[G]$. – Dietrich Burde Nov 7 '19 at 20:08
• @DietrichBurde Ah, this makes sense, and I probably should have thought of it. Although I would have appreciated it if the author mentioned this when he defined it... I know that your comment is short, but it did answer my question so if you make it into an answer I'll accept. – Bence Racskó Nov 7 '19 at 20:09

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.

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.