Let us consider the enveloping algebra $\mathcal{U}(\mathfrak{g})$ of some Lie algebra $\mathfrak{g}$.
Under what assumptions about $\mathfrak{g}$, does the enveloping algebra generate a locally compact quantum group in the sense of Kustermans and Vaes?
By that I mean, we generate the von Neumann algebra from the elements of $\mathcal{U}(\mathfrak{g})$ understood as unbounded densely defined (differential) operators on $L^2(G)$.
Is it true that the generated von Neumann algebra is the group von Neumann algebra of the associated simply connected Lie group? Or does the algebra need to be (for example) semi-simple?
Edit: by 'the von Neumann algebra generated by an unbounded operator $T$', I mean: $$ W^*(T) := \{ x \in B(H) \; | \; xT \subset Tx \text{ and } xT^* \subset T^*x \}' $$ so the usual generation ($T''$) but the 'first commutant' is the appropriate one for unbounded operators.
Thank you!
