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 $SL(2, \mathbf{C})$?
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The difficulty with using "explicit monomials" is that this requires a choice of a basis |
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This may not answer directly the question about Lie groups and their Lie algebras, but the parallel treatment of an affine group scheme 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 opaque-looking 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. |
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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 :) |
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