I would like to ask a number of questions about the theory of analytic vectors and the integrability of Lie-algebra representations, but before I do so, let me fix the terminology to be used in this OP.
Terminology
For each $ n \in \mathbb{N} $, let $ [n] \stackrel{\text{df}}{=} \mathbb{N}_{\leq n} $.
All Lie algebras mentioned here are assumed to have $ \mathbb{R} $ as their base field.
Let $ \mathcal{H} $ be a Hilbert space.
- A skew-symmetric operator on $ \mathcal{H} $ is a densely defined linear operator $ S $ on $ \mathcal{H} $ satisfying $$ \forall x,y \in \text{Dom}(S): \quad \langle S x,y \rangle = - \langle x,S y \rangle, $$ whence $ \text{Dom}(S) \subseteq \text{Dom}(S^{*}) $.
- A skew-adjoint operator on $ \mathcal{H} $ is a skew-symmetric operator $ A $ on $ \mathcal{H} $ satisfying $$ \text{Dom}(A) = \text{Dom}(A^{*}). $$
Here are some definitions to get started.
Definition. Let $ \mathcal{H} $ be a Hilbert space and $ T $ a densely defined linear operator on $ \mathcal{H} $. Then a vector $ v \in \mathcal{H} $ is said to be analytic for $ T $ if $ {T^{k}}(v) $ is defined for all $ k \in \mathbb{N} $ and there exists an $ s > 0 $ such that $$ \sum_{k = 0}^{\infty} \frac{s^{k}}{k!} \left\| {T^{k}}(v) \right\| < \infty. $$
Definition. Let $ \mathcal{H} $ be a Hilbert space and $ (T_{i})_{i \in [n]} $ a finite sequence of densely defined linear operators on $ \mathcal{H} $. Then a vector $ v \in \mathcal{H} $ is said to be jointly analytic for $ (T_{i})_{i \in [n]} $ if $$ \left[ T_{\alpha(1)} \circ \cdots \circ T_{\alpha(k)} \right] \! (v) $$ is defined for all $ k $-tuples $ \alpha: [k] \to [n] $, for all $ k \in \mathbb{N} $, and there exists an $ s > 0 $ such that $$ \sum_{k = 0}^{\infty} \left[ \sum_{\alpha: [k] \to [n]} \frac{s^{k}}{k!} \left\| \left[ T_{\alpha(1)} \circ \cdots \circ T_{\alpha(k)} \right] \! (v) \right\| \right] < \infty. $$
From Lemma 9.1 of Edward Nelson’s 1959 paper Analytic Vectors, we get the following:
Theorem 1 (Nelson). Suppose that:
- $ {\frak{g}} $ is an $ n $-dimensional Lie algebra and $ (X_{i})_{i \in [n]} $ an ordered basis for $ {\frak{g}} $.
- $ G $ denotes the unique simply connected Lie group whose Lie algebra is $ {\frak{g}} $.
- $ \mathcal{H} $ is a Hilbert space.
- $ T $ is a representation of $ {\frak{g}} $ on $ \mathcal{H} $ by skew-adjoint operators on $ \mathcal{H} $ having a dense common invariant domain.
- $ D $ denotes the largest common invariant domain of $ T[{\frak{g}}] $.
If there exists a dense linear subspace $ D' \subseteq D $ of vectors that are jointly analytic for $ (T(X_{i}))_{i \in [n]} $, then $ T $ is integrable to a strongly continuous unitary representation of $ G $ on $ \mathcal{H} $.
Note: $ D' $ is not required to be $ T[{\frak{g}}] $-invariant.
In their 1972 paper Simple Facts About Analytic Vectors and Integrability, Moshé Flato and his colleagues showed that one can relax the ‘jointly analytic’ requirement, but this at the expense of requiring a common dense invariant domain of analytic vectors.
Theorem 2 (Flato, Simon, Snellman & Sternheimer (a.k.a. FS$ ^{3} $)). Suppose that:
- $ {\frak{g}} $ is an $ n $-dimensional Lie algebra and $ (X_{i})_{i \in [n]} $ an ordered basis for $ {\frak{g}} $.
- $ G $ denotes the unique simply connected Lie group whose Lie algebra is $ {\frak{g}} $.
- $ \mathcal{H} $ is a Hilbert space.
- $ T $ is a representation of $ {\frak{g}} $ on $ \mathcal{H} $ by skew-adjoint operators on $ \mathcal{H} $ having a dense common invariant domain.
- $ D $ denotes the largest common invariant domain of $ T[{\frak{g}}] $.
If there exists a dense $ T[{\frak{g}}] $-invariant linear subspace $ D' \subseteq D $ of vectors that are analytic for $ T(X_{i}) $ for each $ i \in [n] $, then $ T $ is integrable to a strongly continuous unitary representation of $ G $ on $ \mathcal{H} $.
Note: $ D' $ is required this time to be $ T[{\frak{g}}] $-invariant.
My first question is:
Question 1. Is there an intuitive explanation for why the $ T[{\frak{g}}] $-invariance of $ D' $ is crucial for Theorem 2 whereas it does not play a role in Theorem 1?
Now, Lemma 7.1 and Theorem 3 of Nelson’s paper yield an almost complete converse to Theorem 1:
Theorem 3. Suppose that
- $ {\frak{g}} $ is an $ n $-dimensional Lie algebra and $ \beta = (X_{i})_{i \in [n]} $ an ordered basis for $ {\frak{g}} $.
- $ G $ denotes the unique simply connected Lie group whose Lie algebra is $ {\frak{g}} $.
- $ \mathcal{H} $ is a Hilbert space.
- $ T $ is a representation of $ {\frak{g}} $ on $ \mathcal{H} $ by skew-adjoint operators on $ \mathcal{H} $ having a dense common invariant domain.
- $ D $ denotes the largest common invariant domain of $ T[{\frak{g}}] $.
If $ T $ is integrable to a strongly continuous unitary representation $ \pi $ of $ G $ on $ \mathcal{H} $, so that for all $ X \in {\frak{g}} $, $$ T(X) = {\partial \pi}(X) \stackrel{\text{df}}{=} \text{Infinitesimal skew-adjoint generator of $ t \mapsto \pi(\exp(t X)) $}, $$ then the linear subspace $ D_{T,\beta} $ of all vectors that are jointly analytic for $ (T(X_{i}))_{i \in [n]} $ is dense in $ \mathcal{H} $.
The reason why I say ‘almost complete’ is that it is not obvious at first sight that $ D_{T,\beta} \subseteq D $. Hence, here is my second question:
Question 2: Is $ D_{T,\beta} \subseteq D $?
I am guessing that the answer is ‘yes’, and my reasoning is as follows:
If $ T $ is integrable to a strongly continuous unitary representation $ \pi $ of $ G $ on $ \mathcal{H} $, then the Gårding space $ \mathcal{G}(\pi) $ of $ \pi $ is contained in $ D $ because $ \mathcal{G}(\pi) $ is invariant under $ T[{\frak{g}}] $. By the Dixmier-Malliavin Theorem, $ \mathcal{G}(\pi) $ is the set of all vectors $ v \in \mathcal{H} $ such that the function $ g \mapsto [\pi(g)](v) $ from $ G $ to $ \mathcal{H} $ is smooth, so $ \mathcal{G}(\pi) $ contains $ \mathcal{A}(\pi) $, which denotes the set of all vectors $ v \in \mathcal{H} $ such that the function $ g \mapsto [\pi(g)](v) $ from $ G $ to $ \mathcal{H} $ is analytic, i.e., has a power series expansion in some local analytic chart around each point of $ G $. Lemma 7.1 of Nelson’s paper then says that $ \mathcal{A}(\pi) = D_{T,\beta} $, so we obtain $ D_{T,\beta} \subseteq \mathcal{G}(\pi) \subseteq D $.
This argument also shows that $ D_{T,\beta} $ does not depend on the ordered basis $ \beta $ chosen.
I might have made a mistake in my reasoning, so I would appreciate any critique. Even if my argument is correct, any suggestion toward a simpler proof is warmly welcome.
I was also wondering: Is there a converse to Theorem 2?
Conjecture. Suppose that
- $ {\frak{g}} $ is an $ n $-dimensional Lie algebra and $ \beta = (X_{i})_{i \in [n]} $ an ordered basis for $ {\frak{g}} $.
- $ G $ denotes the unique simply connected Lie group whose Lie algebra is $ {\frak{g}} $.
- $ \mathcal{H} $ is a Hilbert space.
- $ T $ is a representation of $ {\frak{g}} $ on $ \mathcal{H} $ by skew-adjoint operators on $ \mathcal{H} $ having a dense common invariant domain.
- $ D $ denotes the largest common invariant domain of $ T[{\frak{g}}] $.
If $ T $ is integrable to a strongly continuous unitary representation of $ G $ on $ \mathcal{H} $, then there exists a dense $ T[{\frak{g}}] $-invariant linear subspace $ D' \subseteq D $ of vectors that are analytic for $ T(X_{i}) $ for each $ i \in [n] $.
Note: I do not expect that the linear subspace $ \widetilde{D}_{T,\beta} $ of all vectors that are analytic for $ T(X_{i}) $ for each $ i \in [n] $ be $ T[{\frak{g}}] $-invariant or even be contained in $ D $, so we have to cut down to a smaller linear subspace.