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Let $\mathfrak{h}$ be a Cartan subalgebra of a Lie group $G$. It is said that $U(\mathfrak{h})$ is isomorphic to $\mathcal{O}(\mathfrak{h}^*)$. Here $\mathcal{O}(\mathfrak{h}^*)$ is the ring of polynomial functions on $\mathfrak{h}^*$. Thank you very much.

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  • $\begingroup$ Observe that since $\mathfrak{h}$ is abelian, $U(\mathfrak{h})$ is isomorphic to the symmetric algebra $\operatorname{S}(\mathfrak{h})$ of $\mathfrak{h}$, which in turn is isomorphic to the algebra of polynomial functions on $\mathfrak{h}^*$. $\endgroup$ – ivanpenev Jun 11 '14 at 18:50
  • $\begingroup$ I implicitly assumed that the Lie algebra of $G$ is reductive (as this is the case most familiar to me), which implies that $\mathfrak{h}$ is abelian. A Cartan subalgebra of an arbitrary (finite-dimensional) Lie algebra $\mathfrak{g}$ is a self-normalising nilpotent subalgebra of $\mathfrak{g}$. Thus, if $\mathfrak{g}$ is e.g. the algebra of all strictly upper-triangular matrices of degree $n\geq3$, then $\mathfrak{g}$ is its own Cartan subalgebra, and is non-commutative. $\endgroup$ – ivanpenev Jun 11 '14 at 19:23
  • $\begingroup$ On the other hand, in order for $U(\mathfrak{h})$ to be isomorphic to $\mathcal{O}(\mathfrak{h}^*)$, the universal enveloping algebra $U(\mathfrak{h})$ must be commutative, as so is $\mathcal{O}(\mathfrak{h}^*)$. If $U(\mathfrak{h})$ is commutative, then so is $\mathfrak{h}$. Thus, no matter what $G$ is, the commutativity of $\mathfrak{h}$ is necessary and sufficient for the isomorphism $U(\mathfrak{h})\simeq \mathcal{O}(\mathfrak{h}^*)$ to hold. $\endgroup$ – ivanpenev Jun 11 '14 at 19:23

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