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$\newcommand{\g}{\mathfrak g}$ Let $\g$ be a Lie algebra, and observe that since $U(\g)$ is a cocommutative Hopf algebra, it makes sense to look for (naturally arising and perhaps commutative?) algebras in the (symmetric monoidal) category of $\g$-modules (call these $\g$-algebras).

My question is quite simple: let $G$ be a Lie group with Lie algebra $\g$. Is there a natural way to construct commutative $\g$-algebras? For example, suppose that $M$ is a manifold with a left $G$-action. Can one produce on $C^\infty(M)$ a $\g$-action compatible with the product (in the sense that $g(xy) = (gx)y + x(gy)$) that encodes such $G$-action, in the way Lie-Rinehart pairs encode Lie algebroids?

Can you think of any one natural/geometric context where $\g$-algebras arise?

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  • $\begingroup$ Hi Pedro. Any context in mind? If $\mathfrak g=0$ then Tour question is "what is a natural comm algebra?". If the answer is "a space", then with $\mathfrak g$ then answer is "a space with some vector fields".. I know you know that answer, but what are you looking for? The symmetric algebra in some natural representation? $\endgroup$
    – Marco Farinati
    Commented Jun 28, 2020 at 4:11
  • $\begingroup$ Deriving the action of G in C infty you always get an action of the Lie algebra by derivations on functions $\endgroup$
    – Marco Farinati
    Commented Jun 28, 2020 at 4:14
  • $\begingroup$ Hola @MarcoFarinati ! Of course. I am looking for something similar to: a Lie algebroid on $M$ is the same as a Lie—Rinehart pair $(C^\infty(M),\mathfrak g)$. So I guess the equivalent I want here is just some vector fields that satisfy the defining relations on $\mathfrak g$, yeah. Perhaps there are more interesting examples... $\endgroup$
    – Pedro
    Commented Jun 28, 2020 at 12:30

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