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Let $G$ be Lie group and $g$ be its Lie algebra. Is it true (and if not generally, then under which circumstances) that the the algebra of its differential operators is isomorphic to the universal envelopping algebra $U(g)$ of $g$? If yes, can you give me an isomorphism between these two algebras?

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The algebra of differential operators on what? – Mariano Suárez-Alvarez Dec 8 '11 at 1:21
$\mathfrak{g}$ can be identified with left-invariant vector fields on $G$, hence with a certain Lie algebra of derivations on $C^{\infty}(G)$. This gives a natural map from $U(\mathfrak{g})$ to the algebra generated by these derivations, but I don't know if this map is injective, and I doubt it is surjective unless you restrict yourself to left-invariant differential operators (I think this is meaningful anyway). – Qiaochu Yuan Dec 8 '11 at 1:23
up vote 8 down vote accepted

$U(g)$ is isomorphic to the algebra $D_{l}(G)$ of left invariant differential operators on the Lie group $G$. Its center $Z(U(g))$ is the algebra of bi-invariant differential operators.

To prove the first statement, one defines the map $U(g)\to D_l(G)$ by extending the inclusion $g\to D_l(G)$ (remember $g$ is the space of left invariant vector fields on $G$, which are precisely the left invariant differential operators of order one), after an easy check that this works. The surjectivity is more involved, of course. IIRC it is done in detail in Helgason's book on symmetric spaces.

The algebra of differential operators on $G$, on the other hand, is considerably larger. Even the algebra of differential operators on $G$ with polynomial coefficients (that is, regular functions---I am assuming $G$ is an algebraic group here) has Gelfand-Kirillov dimension twice that of $U(g)$.

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