The universal enveloping algebra of a Lie algebra $\mathfrak{g}$ is isomorphic to the algebra of distributions on the Lie group $G$ with support at the identity. The proof I have of this fact uses the universal property of the universal enveloping algebra and hence it is not constructive. I was wondering if there is an explicit map? For example what would happen for $\mathfrak{g} = \mathfrak{sl}(2, \mathbf{C})$ and $G = SL(2, \mathbf{C})$? For an explicit monomial in $\mathscr{U} \ \left ( \mathfrak{sl}(2, \mathbf{C}) \right )$ can we write the corresponding distribution and how it acts on functions on $SL(2, \mathbf{C})$?
3 Answers
The difficulty with using "explicit monomials" is that this requires a choice of a basis $\{X_1,\ldots, X_{\text{dim}\mathfrak{g}}\}$. Having said that, the PBW theorem states that each element of $U(\mathfrak{g})$ is uniquely represented as a linear combination of monomials in generators of the form $X_{i_1}\ldots X_{i_n}$ with $i_1\leq\ldots\leq i_n$ and $n\geq 0.$ You may identify the elements of this basis of $\mathfrak{g}$ with left invariant vector fields on $G,$ which form a Lie subalgebra of the associative algebra $D(G)$ of differential operators on $G.$ Each monomial thus becomes an element of $D(G)$ and for $D\in D(G),$ the corresponding distribution is $D\cdot\delta_e,$ where $\delta_e$ is the delta function at the identity of $G.$ The effect of a monomial of degree $n$ on a function can be found by the standard calculation involving "integration by parts" $n$ times.

$\begingroup$ A question of notation: How does $D\cdot \delta_e$ act on a function? I am assuming that $\delta_e (f) = f(e)$ which is a constant. $\endgroup$– NajdorfOct 8, 2010 at 10:42

$\begingroup$ @Najdorf It seems that $D.\delta_e$ should act as $D$ according to $$ <(D.\delta_e)\varphi,\psi>=<\delta_e\varphi,D\psi>$$ $\endgroup$– HanyOct 8, 2010 at 11:25

$\begingroup$ @Hany: i don't understand your comment: a distribution $T$ acts on a single test function $\varphi$ as $\varphi\mapsto\langle T,\varphi\rangle$. So, supposing to give some meaning to $T=D.\delta_e$ as adistribution, what does the notation $\langle T\vaprhi,\psi\rangle$ mean? $\endgroup$– QfwfqOct 8, 2010 at 11:48


$\begingroup$ @Hany: so, it's not clear to me how the question asked by Najdorf should be answered. $\endgroup$– QfwfqOct 8, 2010 at 11:51
This may not answer directly the question about Lie groups and their Lie algebras, but the parallel treatment of an affine group scheme $G$ over a rather general base field $K$ (or even commutative ring) offers an interesting perspective. The algebra of distributions on $K[G]$ and its relationship with the universal enveloping algebra of the Lie algebra of $G$ are discussed carefully in Chapter I.7 (especially I.7.10) of Representations of Algebraic Groups by J.C. Jantzen (2nd ed., AMS, 2003). He refers to the foundational treatment in the older book Groupes algebriques by Demazure and Gabriel for some details of the isomorphism in characteristic 0. which does use the defining universal property of the enveloping algebra. In fact, I find it hard to see how any proof not relying on a fixed choice of basis could avoid this property. Without a basis how would one provide any "constructive" description of the enveloping algebra or of the desired isomorphism with the distribution algebra?
The main advantage of this broader algebraic framework is that you also see where the isomorphism between the distribution algebra and the enveloping algebra fails in prime characteristic. (There the distribution algebra or hyperalgebra also has a concrete realization using a Chevalley basis of the Lie algebra, but involves divided powers starting over the ring of integers, etc.)
In any approach, there is likely to be a certain amount of opaquelooking formalism involved when working abstractly with the Lie algebra of a group. For this reason, the PBW basis does have a lot of appeal in practical situations such as those coming up in parts of physics.

$\begingroup$ I looked at Jantzen but nothing there is explicit. He does not give you a map and he does not even do it for an example. $\endgroup$– NajdorfOct 14, 2010 at 1:43
It seems to me the question is not fully correct because there is NO canonical identification of U(g) and distribitions on G (sup. at e).
In Victor's answer you can take action LEFT invariant vec. fields and you can take RIGHT  the answer would be different.
It seems if you take "left" than U(g) will act as left invariant differential operator, if you take "right" it will act as "right" invariant differential operator.
It seems to me that it is general fact which should be checked on R^n  take any differential operator D, consider distribution d= D\delta, then action of this distribution by convolution is exactly the same action of D. The proof seems obvious  just integration by parts
I have not checked all this, hopefully it is nevertheless true :)