1
$\begingroup$

I am reading representations of Lie Algebra in Humphreys.He is defining representation as $L$-modules. In case of group representation we have the correspondence between representation and modules over the group ring. But in both the places we have different definition for modules like in group case a representation is a module over group ring which is actually the usual definition of module over ring. But in $L$-module case, we have extra condition $[xy].v =xy.v-yx.v.$ Why do we need this condition and one more question is given any algebraic structure and its representation is there any general definition of module over that structure? How to see modules say "Categorically"?

$\endgroup$
5
$\begingroup$

This definition of an $L$-module is the analogue of $\mathbb{C}G$-modules. A difference comes from the fact that $L$ is "only" a Lie algebra - it is not an associative algebra.

Consider the following. A representation of a finite group $G$ is essentially a homomorphism of groups $G\to GL(V)$ for some vector space $V$, where we use the natural group structure of composing linear isomorphisms in $GL(V)$. A representation of a Lie algebra $L$ is a homomorphism of Lie algebras $L\to {\frak gl}(V)$, where we use the natural Lie algebra structure of commutators in ${\frak gl}(V)$. That is precisely what the condition $$[xy]\cdot v=x\cdot(y\cdot v)-y\cdot(x\cdot v)\qquad(*) $$ conveys. In other words, it is not an "extra" requirement any more than the group action requirement $(gh)\cdot v=g\cdot(h\cdot v)$ is for $G$-modules.

This may appear to be a bit confusing at first, because Lie algebras are often the first (and in most cases the only) non-associative structure we encounter. Do remember the trick of turning an associative algebra into a Lie algebra by "forgetting" the associative product, and replacing it with the commutator. Again you see that condition $(*)$ is then a natural consequence.

After you have made quite a bit more headway into Humphreys, you will learn to "reverse" this process. Associated with every Lie algebra $L$ is an associative algebra $U(L)$, called the universal enveloping algebra. It is in a sense the smallest associative algebra with the property that every $L$-module is also a $U(L)$-module (in the sense of modules of algebras). It has the disadvantage of being infinite dimensional.

Historically the Lie algebras arose (IIRC) as tangent spaces of Lie groups, groups that are also differentiable manifolds, such as $SL_n(\mathbb{C})$. A Lie group $G$ has a Lie algebra associated with it. Not surprisingly the Lie algebra of $SL_n(\mathbb{C})$ turns out to be what you know as ${\frak sl}_n(\mathbb{C}).$ If a Lie group $G$ has a representation (as a group) with the extra requirement that the group action is also differentiable in a natural sense, then such a $G$-module becomes also a module of the Lie algebra in a relatively natural way. For example, the condition $(*)$ arises (loosely speaking) by differentiating the $G$-module condition along the commutator of two paths going via the identity element of $G$.

Humphreys is not interested in the Lie group side. He has done a lot of research on the algebraic group side, which is a purely algebraic way of getting a theory similar to that of Lie group representations, but with the manifolds and derivatives replaced with structures from algebraic geometry, and with extra difficulties in positive characteristics. So his goal is to get the students quickly to the rep theory and skip the analytic machinery required by Lie groups, as those are neither needed nor available in the algebraic group side anyway.

Finally, you do get a category of $L$-modules in the usual way. The objects of the category are $L$-modules, and morphisms are homomorphisms of $L$-modules. Nothing new or unusual there.

$\endgroup$
2
  • $\begingroup$ thanks. I am convinced with your answer for all my questions except the last one. My question is not about the category of L-modules. My question is, take any "algebraic" category wherever we can talk about module over that algebraic structure. is there any categorical general definition of module in such category? your answer's second paragraph justifying something about this. but I cant understand clearly. please explain. thanks. $\endgroup$ – GA316 Jul 23 '13 at 13:20
  • $\begingroup$ @GA316: The upshot is that the endomorphism ring $L(V)$ of a vector space $V$ can be given a structure of an associative ring, and hence also a Lie algebra. It also has a large group of units, $GL(V)$. Therefore it can be a codomain of homomorphisms of groups, rings, associative algebras and Lie algebras. In some wew only need $V$ to be an abelian group, and use its ring of endomorphisms as a codomain. $\endgroup$ – Jyrki Lahtonen Jul 23 '13 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.