Smooth vectors as iterated generator domains

Statement

Let $\pi$ be a strongly continuous unitary representation of a finite-dimensional Lie group $G$ on $H$ . For $X\in\mathfrak g$ , let $K_X=iA_X$ be the maximal skew-adjoint generator of $t\mapsto\pi(\exp(tX))$ . Then its smooth-vector space is exactly

H^\infty= \bigcap_{r\geq1}\ \bigcap_{X_1,\ldots,X_r\in\mathfrak g} \operatorname{Dom}(K_{X_1}\cdots K_{X_r}).

Products use the usual domains of unbounded operators, with the rightmost factor applied first.

What a product domain requires

For example,

\operatorname{Dom}(K_XK_Y) =\{v\in\operatorname{Dom}(K_Y):K_Yv\in\operatorname{Dom}(K_X)\}.

It is not enough that $v$ separately belong to both generator domains. Requiring the analogous condition for all finite words supplies one common domain closed under every derived action.

Finite differentiability

The $C^k$ -vector space is the intersection of these domains for $1\leq r\leq k$ . In particular, first derivatives in all directions give $H^1$ , not automatically $H^\infty$ . The restriction to a finite-dimensional Lie group is part of the statement; arbitrary infinite-dimensional groups require additional hypotheses.

Relation to a single regularity operator

The same smooth space is the intersection of the domains of all powers of the closed Nelson Laplacian. The word-domain formulation records each infinitesimal direction, while the Laplacian packages the regularity into one operator.

References

Karl-Hermann Neeb, On Differentiable Vectors for Representations of Infinite Dimensional Lie Groups. Definition 3.1(c)–(d), Theorem 9.4, specialized to finite-dimensional groups.