$K-$structure of ${\cal U}(\mathfrak{g})$ for $\mathfrak{s}\mathfrak{u}(n,1)$ and $\mathfrak{s}\mathfrak{o}(n,1)$

Research paper by Hrvoje Kraljević

Indexed on: 23 Nov '16Published on: 23 Nov '16Published in: arXiv - Mathematics - Representation Theory


Let $G$ be the adjoint group of a real simple Lie algebra $\mathfrak{g}_0$ equal either $\mathfrak{s}\mathfrak{u}(n,1)$ or $\mathfrak{s}\mathfrak{o}(n,1),$ $K$ its maximal compact subgroup, ${\cal U}(\mathfrak{g})$ the universal enveloping algebra of the complexification $\mathfrak{g}$ of $\mathfrak{g}_0$ and ${\cal U}(\mathfrak{g})^K$ its subalgebra of $K-$invariant elements. By a result of F. Knopp [3] ${\cal U}(\mathfrak{g})$ is free as a ${\cal U}(\mathfrak{g})^K-$module, so there exists a $K-$submodule $E$ of ${\cal U}(\mathfrak{g})$ such that the multiplication defines an isomorphism of $K-$modules ${\cal U}(\mathfrak{g})^K\otimes E\longrightarrow{\cal U}(\mathfrak{g}).$ We prove that $E$ is equivalent to the regular representation of $K,$ i.e. that the multiplicity of every $\delta\in\hat{K}$ in $E$ equals its dimension. As a consequence we get that for any finitedimensional complex $K-$module $V$ the space $({\cal U}(\mathfrak{g})\otimes V)^K$ of $K-$invariants is free ${\cal U}(\mathfrak{g})^K-$module of rank $\dim\,V.$