Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. Then $\mathfrak{g}$ may be interpreted as the Lie algebra of right (equivalently left) invariant vector fields. Let $\mathcal{U}(\mathfrak{g})$ be its enveloping algebra.

Why $\mathcal{U}(\mathfrak{g})$ may be interpreted as the algebra of all right invariant differential operators?

In other words: as there are no algebra relations within $\mathcal{U}(\mathfrak{g})$ except $XY-YX=[X,Y]$ the same should be true for invariant differential operators: however, is it obvious?

  • 1
    $\begingroup$ Here is one way to see this: There is a map from the universal enveloping algebra to the algebra of differential operators landing in right-invariant operators. Your question seems to be why this map is an isomorphism; for this, it suffices to check this statement after taking associated gradeds w/r/t the PBW&order filtrations. The associated graded of $\mathcal{D}_G$ is isomorphic to $\mathcal{O}_G\otimes\operatorname{Sym}^{\bullet}\mathfrak{g},$ with $G$ acting purely on the first factor. $\endgroup$
    – dhy
    Apr 3, 2018 at 18:05
  • $\begingroup$ Thank you for your comment: could you please provide some more details (maybe as an answer)? $\endgroup$
    – truebaran
    Apr 4, 2018 at 13:53

1 Answer 1


From the Lie algebra inclusion $\mathfrak{g}\rightarrow \Gamma(T_G)$ of right invariant vector fields we get a map $U(\mathfrak{g})\rightarrow\Gamma(\mathcal{D}_G).$ This induces a map $U(\mathfrak{g})\otimes\mathcal{O}_G\rightarrow\mathcal{D}_G$ of $\mathcal{O}_G$-modules. We have a structure of $G$-equivariant sheaf on both sides, with respect to the right translation action of $G$ on itself. (The $G$-equivariant structure on the left hand side comes from tensoring up the $G$-equivariant structure on $\mathcal{O}_G,$ i.e., the $G$ action on the $U(\mathfrak{g})$ part is trivial.) Our map is $G$-equivariant, and if we know it is an isomorphism, we will get an isomorphism $\Gamma(U(\mathfrak{g})\otimes\mathcal{O}_G)^G\rightarrow\Gamma(\mathcal{D}_G)^G,$ or $U(g)\rightarrow\Gamma(\mathcal{D}_G)^G$, as desired.

Now let us show that $U(\mathfrak{g})\otimes\mathcal{O}_G\rightarrow\mathcal{D}_G$ is an isomorphism. We have a canonical filtration $F_i\mathcal{D}_G$ on $\mathcal{D}_G$ by order of differential operators. On the left hand side, the PBW filtration $F_iU(\mathfrak{g})$ induces a filtration $F_iU(\mathfrak{g})\otimes\mathcal{O}_G$.

As our map sends $F_iU(\mathfrak{g})\otimes\mathcal{O}_G$ to $F_i\mathcal{D}_G$ and both filtrations are exhaustive, it suffices to check that the map on associated gradeds is an isomorphism. The associated graded of the left hand side is $\operatorname{Sym}^{\bullet}\mathfrak{g}\otimes\mathcal{O}_G.$

For the right hand side, we know that $F_i\mathcal{D}_G/F_{i-1}\mathcal{D}_G$ can be canonically identified with $\operatorname{Sym}^iT_G.$ The inclusion $\mathfrak{g}\rightarrow \Gamma(T_G)$ gives rise to a map $\mathfrak{g}\otimes\mathcal{O}_G\rightarrow T_G$ which can be seen to be an isomorphism. This identifies the associated graded of the right hand side with $\operatorname{Sym}^{\bullet}\mathfrak{g}\otimes\mathcal{O}_G,$ and the map on associated gradeds becomes the identity.


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