TITLE: Geometric interpretation of Universal enveloping algebras
QUESTION [13 upvotes]: Given a complex Lie algebra $\mathfrak g$, we can form its universal enveloping algebra and interpret it as a noncommutative space. 
Is this perspective useful? What does this space "look like"? 
How is it related to the space $\mathbb C[G]$ where $G$ is an algebraic group with Lie algebra $\mathfrak g$?
How is it related to the flag variety (in the case of a semisimple Lie algebra)? 
This question has also an incarnation which asks about 
group algebras.

REPLY [7 votes]: This remark is just a caution for when one tries to think about what the Spec of $U\mathfrak g$ might mean.
Suppose that $\mathfrak g$ is semisimple, and let $I$ be the kernel of the natural augmentation $U\mathfrak g \to \mathbb C$.  (Alternatively, $I$ is the kernel of the action of $U\mathfrak g$ on the trivial representation of $\mathfrak g$.)  From the semisimplicity one finds that $I^2 = I$, and hence that $I^n = I$ for every $n \geq 1$.  
One also checks (e.g. by a grading argument) that $U\mathfrak g$ is a non-commutative domain, in the sense that it has no zero divisors, and that $U\mathfrak g$ is Noetherian.
Now one easily checks that if $A$ is a commutative Noetherian domain, and if $I$ is 
a non-unit ideal such that $I^2 = I$, then necessarily $I = 0$.  The point is that if
$I = I^2$, the zero-locus of $I$ in Spec $A$ is "formally" isolated from its complement,
and so morally provides a disconnection of Spec $A$ (and this moral argument is made rigorous via an application of the Krull intersection theorem).  If $A$ is a domain, on the other hand, then Spec $A$ is irreducible, and so in particular connected, and so the only possibility
is that $I = 0$.
So $U\mathfrak g$ has various properties that would be mutually incompatible in a commutative ring, and hence one has to be careful in importing geometric intuition naively
in thinking about some kind of non-commutative Spec $U\mathfrak g$.
(Of course, the other answers here give a very non-naive discussion of various geometric points of view on $U\mathfrak g$, but I hope that the above remark will be useful to some people.  Let me note that it is also somewhat related to this answer.)

REPLY [4 votes]: Orbit method
As Mariano and Ben have already mentioned, $U(\mathfrak{g})$ quantizes coadjoint orbits of $\mathfrak{g}.$ 
In general, for a real algebraic group $G$ with Lie algebra $\mathfrak{g},$ the following three spaces are closely related:

The space of primitive ideals of $U(\mathfrak{g}).$ 
The set of isomorphism classes of irreducible unitary representations of $G.$
The set of coadjoint orbits of $G.$

The original result is due to Kirillov and states that if $G$ is nilpotent and simply-connected then spaces 2 and 3 are in a natural bijection. Later this was extended in various directions by Auslander and Kostant, Gabriel, Borho and Rentschler, Duflo, and many others. For the case $\mathfrak{g}=\mathfrak{gl}_n,$ the Jacobson topology on the space of primitive ideals was determined by Oshima in the Advances Math paper. It is related to the topology on the space consisting of Zariski closures of the conjugacy classes of $n\times n$ matrices, but there are some subtle differences.

Relation to flag variety In the case when $G$ is a compact semisimple Lie group, the coadjoint orbit $\mathcal{O}_\lambda=G\cdot\lambda$ is equivariantly isomorphic to generalized flag variety $G_{\mathbb{C}}/P_\lambda;$ moreover, this isomorphism is compatible with symplectic structures and line bundles. Here $\lambda$ is an integral weight (after natural identification between $\mathfrak{g}$ and its dual) and when $\lambda$ is a regular integral dominant weight, $P_\lambda=B,$ so that the coadjoint orbit $\mathcal{O}_\lambda$ may be identified with the full flag variety $G_{\mathbb{C}}/B$ polarized by the line bundle $\mathcal{L}_\lambda.$