On the left ideal in the universal enveloping algebra of a Lie group generated by a complex Lie subalgebra

Abstract:

Let ${G_0}$ be a connected Lie group with Lie algebra ${g_0}$ and let h be a Lie subalgebra of the complexification g of ${g_0}$ . Let ${C^\infty }{({G_0})^h}$ be the annihilator of h in ${C^\infty }({G_0})$ and let $\mathcal {A} = \mathcal {A}({C^\infty }{({G_0})^h})$ be the annihilator of ${C^\infty }{({G_0})^h}$ in the universal enveloping algebra $\mathcal {U}(g)$ of g. If h is the complexification of the Lie algebra ${h_0}$ of a Lie subgroup ${H_0}$ of ${G_0}$ then $\mathcal {A} = \mathcal {U}(g)h$ whenever ${H_0}$ is closed, is a known result, and the point of this paper is to prove the converse assertion. The paper has two distinct parts, one for ${C^\infty }$, the other for holomorphic functions. In the first part the Lie algebra ${\bar h_0}$ of the closure of ${H_0}$ is characterized as the annihilator in ${g_0}$ of ${C^\infty }{({G_0})^h}$, and it is proved that ${h_0}$ is an ideal in ${\bar h_0}$ and that ${\bar h_0} = {h_0} \oplus v$ where v is an abelian subalgebra of ${\bar h_0}$. In the second part we consider a complexification G of ${G_0}$ and assume that h is the Lie algebra of a closed connected subgroup H of G. Then we establish that $\mathcal {A}(\mathcal {O}{(G)^h}) = \mathcal {U}(g)h$ if and only if $G/H$ has many holomorphic functions. This is the case if $G/H$ is a quasi-affine variety. From this we get that if H is a unipotent subgroup of G or if G and H are reductive groups then $\mathcal {A}({C^\infty }{({G_0})^h}) = \mathcal {U}(g)h$.