Definition
Smooth vector of a Lie-group representation
A Hilbert-space vector whose orbit map under a Lie-group representation is infinitely differentiable in norm.
Let be a finite-dimensional Lie group and let be a strongly continuous unitary representation. A vector is a smooth vector if its orbit map
is as a map from the smooth manifold to the Hilbert space , with differentiability taken in the norm topology. The space of all smooth vectors is denoted . It is a -invariant linear subspace and is dense in .
Finite differentiability and topology
A differentiable vector only requires a orbit map; a smooth vector requires all orders. The smooth-vector Fréchet topology records convergence of every representation derivative and is generally stronger than the inherited Hilbert norm. Equivalently, smooth vectors form the intersection of all ordered generator domains.
Some authors call a representation smooth when its smooth vectors are dense. In the finite-dimensional Lie-group setting, Gårding's theorem guarantees that property; it does not say that every Hilbert vector is smooth. For the Schrödinger representation the smooth vectors are exactly Schwartz functions.
Differentiation
For in the Lie algebra and , the norm derivative
exists. These operators share the invariant dense domain and assemble into the derived representation. They are generally unbounded as operators on , so their common domain is essential data.
Density and regularization
For , the integrated vector
is smooth. Approximate identities of such functions converge strongly to the identity, proving the density of ; this is commonly called Gårding's argument.
Relation to generators
For each , Stone's theorem gives a self-adjoint operator with . On smooth vectors,
Thus is the skew-symmetric infinitesimal action, while is the self-adjoint Stone generator. Neither operator should be described without its domain.
References
- G. Warner, Harmonic Analysis on Semi-Simple Lie Groups I, Springer, 1972. DOI record. Relevant: §4.4 on differentiable and smooth vectors.