Ties between Lie algebras and ring theory

Solution 1:

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.