I would like to get a general understanding of the relationship between (noncommutative) ring theory and Lie algebra theory. All Lie algebras are finite dimensional and over a field $k$ of characteristic $0$ (algebraically closed if necessary.) All rings are associative and are unital. Lie algebras will be called $L$ and rings $R$. What I know of both I learnt in Humphrey's Introduction to Lie algebras and representation theory and Lam's A first course in noncommutative rings. I apologize in advance for the mess, and the vagueness of the question. I self study these topics and realize there are connections but it's a mess in my head...

Lam tells me Wedderburn tried to recreate Cartan's theory and was thus led to define radicals for rings, so the Lie algebra radical predates the Wedderburn and Jacobson radical in ring theory.

Every Lie algebra has its universal enveloping algebra $U(L)$, in such a way that representations of $L$ are the same thing as $U(L)$-modules. Also every ring has a Lie algebra structure. Edit Thus, via the enveloping algebra Lie algebras and their representations can be studied from a ring theoretic point of view. Is the converse true in some sense?

Both theories have radicals, concepts of simplicity and semisimplicity. Both have structure theorems for semisimple objects and their representations : there is the Artin-Wedderburn theory for semisimple rings and simple artinian rings, and complete reducibility of any modules on one hand, and there is the root space decomposition for semisimple Lie algebras and the not so trivial Weyl theorem on complete reducibility of finite dimensional representations of semisimple Lie algebras.

What I would like to know is how these notions interact, for instance, what relationship is there between the Lie algebra radical and the Jacobson radical of its enveloping algebra? What kind of ring is $U(L)$? It can't be semisimple I think because it is infinite dimensional, but nonetheless it almost verifies the criteria of complete reducibility of its representations via Weyl's theorem... What is the ring theory of the universal enveloping algebra? A reference that explains the ties would be welcome too!

Thank you for your time!

  • $\begingroup$ I am far from an expert on Lie algebras, but: "Is the converse true?" Here I think you are asking whether one can recover representation theory / module theory of a ring $R$ via modules over its associated Lie ring $\operatorname{Lie}(R)$. This can't be true in general: if $R$ is a commutative $k$-algebra, then $\operatorname{Lie}(R)$ depends on nothing but the $k$-vector space structure of $R$. $\endgroup$ – Pete L. Clark May 6 '12 at 20:51
  • $\begingroup$ @Pete L. Clark absolutely. I nearly wrote something to that effect but prefered to stay vague. Sometimes you start with something, do stuff to it, seemingly lose all structure or most of it only to find that the structure somehow resurfaces somewhere else. This is only a weak argument for what I just described, but in the commutative case $A$ is not completely lost as $A^{Lie}$ is a (commutative) Lie algebra and its envelopping algebra is $A[X]$. $\endgroup$ – Olivier Bégassat May 6 '12 at 21:05
  • $\begingroup$ I should say : if you work in the category of Lie algebras over the commutative ring $A$. $\endgroup$ – Olivier Bégassat May 6 '12 at 21:15
  • $\begingroup$ I do understand your question and I'm currently looking for some references that presents Lie theory as much as possible as a corollary of de theory developed for associative algebras (where obviously the associative algebra would be the enveloping algebra). Did you find any reference in the meanwhile? Please put the result of your findings here math.stackexchange.com/questions/2024458/… I will upvote $\endgroup$ – Dac0 Jan 27 '17 at 17:07

What kind of ring is $U(L)$?

Since representations of Lie algebras behave like representations of groups (the category has tensor products and duals, for example), you should expect that the universal enveloping algebra $U(\mathfrak{g})$ has some extra structure which causes this, and it does: namely, it is a Hopf algebra (a structure shared by group algebras). The comultiplication is defined on basis elements $x \in \mathfrak{g}$ by $$x \mapsto 1 \otimes x + x \otimes 1$$

(this is necessary for it to exponentiate to the usual comultiplication $g \mapsto g \otimes g$ on group algebras) and the antipode is defined by $$x \mapsto -x$$

(again necessary to exponentiate to the usual antipode $g \mapsto g^{-1}$ on group algebras).

This is an important observation in the theory of quantum groups, among other things.

Thus, via the envelopping algebra Lie algebras and their represnetations cn be studied from a ring theoretic point of view. Is the cconverse true in some sense?

Not in the naive sense, the basic problem being that if $A$ is an algebra and $L(A)$ that same algebra regarded as a Lie algebra under the bracket $[a, b] = ab - ba$, then a representation of $L(A)$ does not in general extend to a representation of $A$, but to a representation of $U(L(A))$, which may be a very different algebra (take for example $A = \text{End}(\mathbb{C}^2)$).

Of course there are other relationships between ring theory and Lie theory. For example, if $A$ is a $k$-algebra then $\text{Der}_k(A)$, the space of $k$-linear derivations of $A$, naturally forms a Lie algebra under the commutator bracket. Roughly speaking this is the "Lie algebra of $\text{Aut}(A)$" in a way that is made precise for example in this blog post.

  • $\begingroup$ Thank you for your answer. I was aware of the coalgebra structure but didn't realize the implications and that something alike should be expected. Do you know about the radical maybe? I also wonder about the center. You get central elements in the semisimple case as Casimir elements of representations, I wonder if these generate the center along with the field. What would be great would be to have something like "the radical of $L$ generates the radical of $U(L)$". $\endgroup$ – Olivier Bégassat May 5 '12 at 21:07
  • $\begingroup$ @Olivier: in the semisimple case the Harish-Chandra isomorphism (en.wikipedia.org/wiki/Harish-Chandra_isomorphism) gives an explicit description of the center. Your conjecture about the radical is false (take $\mathfrak{g}$ to be finite-dimensional abelian). $\endgroup$ – Qiaochu Yuan May 5 '12 at 21:34
  • $\begingroup$ I should add the requirement of semisimplicity, which was implicit in my mind as Humphreys defines Casimir elements of representations in the semisimple case. $\endgroup$ – Olivier Bégassat May 5 '12 at 21:51
  • $\begingroup$ @Olivier: if I am reading this paper right, the Jacobson radical of $U(\mathfrak{g})$ actually has nothing to do with the Lie algebra radical of $\mathfrak{g}$: ams.org/journals/proc/1986-098-04/S0002-9939-1986-0861746-7/… $\endgroup$ – Qiaochu Yuan May 5 '12 at 22:05
  • $\begingroup$ That short article was veey helpful. Seems like the universal envelopping algebra is always Jacobson semisimple. And the proof is so simple I feel silly for raising the question ^^ $\endgroup$ – Olivier Bégassat May 6 '12 at 20:41

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