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 analyticfor $ (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.
- $ 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} $ ofallvectors 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.