Nuclear smooth vectors and trace-class representations

Statement

Let $\pi$ be a strongly continuous unitary representation of a finite-dimensional Lie group $G$ . With the smooth-vector Fréchet topology,

H^\infty\text{ is nuclear} \quad\Longleftrightarrow\quad \pi(f)\text{ is trace class for every }f\in C_c^\infty(G).

In other words, nuclearity of $H^\infty$ is equivalent to $\pi$ being a trace-class representation.

Consequence for Hilbert-space riggings

The continuous dense inclusions

H^\infty\hookrightarrow H\hookrightarrow H^{-\infty}

exist for every such representation, with distribution vectors on the right. Under the nuclear convention for a rigged Hilbert space, this becomes a nuclear Gelfand triple precisely in the trace-class case.

Type I is a different condition

Being a type I group controls decomposition into irreducible representations. It does not guarantee nuclear smooth vectors, even for every irreducible representation. For $n\geq2$ ,

G=\operatorname{SL}_n(\mathbb R)\ltimes\mathbb R^n

is type I but has an irreducible unitary representation that is not trace class. Its smooth-vector space is therefore not nuclear. Here the semidirect product uses the usual linear action on $\mathbb R^n$ .

Regularity versus summability

The Nelson Laplacian controls smooth-vector regularity. The trace-class condition additionally requires a sufficiently high negative power of one plus its nonnegative self-adjoint closure to be trace class. Smoothness alone is not this summability requirement.

References

Gerrit van Dijk, Karl-Hermann Neeb, Hadi Salmasian, and Christoph Zellner, On the characterization of trace class representations and Schwartz operators. Proposition 1.11; Theorem 1.3 for the equivalent Laplacian criterion.
Anton Deitmar and Gerrit van Dijk, Trace class groups. Proposition 1.9(b) for the type I counterexample.