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.)

notnoncommutative spaces, but (can be regarded as)coordinate ringsof noncommutative spaces. I notice this is what you said in your earlier question about group rings. – Yemon Choi Oct 6 '10 at 7:17