Timeline for Casimir operators for Jacobi Lie algebra

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Apr 26, 2017 at 9:43	comment	added	Paul Levy		Well, by the Duflo isomorphism, $Z(U({\mathfrak g}))$ is isomorphic to $S({\mathfrak g})^{\mathfrak g}$ for any finite-dimensional Lie algebra. Now if ${\mathfrak g}=\mathfrak{sp}_{2(n+1)}$ and our Lie algebra is ${\mathfrak g}_0$, then obviously $S(\mathfrak{g})^{\mathfrak g}\subset S(\mathfrak{g})^{{\mathfrak g}_0}$, and further we have a ${\mathfrak g}_0$-stable decomposition ${\mathfrak g}={\mathfrak g}_0+I$, so I think we have a map from $S(\mathfrak{g})^{\mathfrak g}$ to $S(\mathfrak{g}_0)^{{\mathfrak g}_0}$. It's a start though it certainly isn't surjective.
Apr 26, 2017 at 7:37	comment	added	Arnold Neumaier		I edited my question to make it more precise, answering your initial questions. The problem is indeed that the Jacobi Lie algebra is not semisimple; so that the standard stuff does not apply. I don't see how the embedding into sp(2n+2) helps; it seems to me that its Casimirs are unlike to generate those of the Jacobi Lie algebra.
Apr 25, 2017 at 22:42	history	answered	Paul Levy	CC BY-SA 3.0