Some questions on analytic vectors and the integrability of Lie-algebra representations 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.

 A: To demonstrate the delicate nature of questions concerning the integrability of Lie-algebra representations, I would like to highlight a major error that I committed while I was formulating the OP.
The last item in the hypothesis set for each theorem defines $ D $ to be the largest invariant domain of $ T[{\frak{g}}] $. This is absolute nonsense because it may no longer be true for $ X,Y \in {\frak{g}} $ and $ \alpha \in \mathbb{R} $ that
$$
T(\alpha X + Y)|_{D} = \alpha \cdot T(X)|_{D} + T(Y)|_{D} \quad \text{or} \quad
T([X,Y])|_{D} = \left[ T(X)|_{D},T(Y)|_{D} \right].
$$
These relations are only assumed to hold on the original invariant domain specified in the second-last item of the hypothesis set, so enlarging the domain while preserving the validity of the relations is not what we can get for free.
What I should have said was:

Let $ \displaystyle D \subseteq \bigcap_{X \in {\frak{g}}} \text{Dom}(T(X)) $ be the largest invariant domain on which $ T $ is still a representation of $ {\frak{g}} $, i.e., the relations given above still hold.

$ D $ exists and is dense because it is the linear span of all invariant domains on which $ T $ is a representation of $ {\frak{g}} $, and the original invariant domain, being one of them, is dense.

Changes to my answer to Question 2
With this clarification, one only has to add a small clause to my answer to Question 2:

$ T $ is a representation of $ {\frak{g}} $ on $ \mathcal{G}(\pi) $ (because the differential $ \partial \pi $ is a representation of $ {\frak{g}} $ on $ \mathcal{G}(\pi) $, which is a standard fact in the theory of smooth vectors of Lie-group representations).

Hence, by definition, we obtain $ \mathcal{G}(\pi) \subseteq D $. Then by Lemma 7.1 and Theorem 3 of Nelson’s paper, $ \mathcal{A}(\pi) $ is a dense linear subspace of $ \mathcal{G}(\pi) $ consisting of vectors that are jointly analytic for
$$
  \left( {\partial \pi}(X_{i})|_{\mathcal{G}(\pi)} \right)_{i \in [n]}
= \left( T(X_{i})|_{\mathcal{G}(\pi)} \right)_{i \in [n]}.
$$
Therefore, $ \mathcal{A}(\pi) \subseteq D $ is a dense linear subspace of vectors that are jointly analytic for $ (T(X_{i}))_{i \in [n]} $.
Also, one minor mistake that I made in the argument was to say that Lemma 7.1 yields $ \mathcal{A}(\pi) = D_{T,\beta} $. As far as I know, we can only say that $ \mathcal{A}(\pi) \subseteq D_{T,\beta} $. However, we do not require $ D_{T,\beta} $ because $ \mathcal{A}(\pi) $ is all we need.


Answer to Question 1: It is not necessary to impose the $ T[{\frak{g}}] $-invariance of $ D' $ for the conclusion of Theorem 2 to hold.

Proof Suppose that $ D' \subseteq D $ is a dense non-$ T[{\frak{g}}] $-invariant linear subspace of vectors that are analytic for $ T(X_{i}) $ for each $ i \in [n] $. We can then enlarge $ D' $ to a $ T[{\frak{g}}] $-invariant linear subspace $ D'' \subseteq D $ of vectors that are analytic for $ T(X_{i}) $ for each $ i \in [n] $.
Let $ D'' $ be the smallest $ T[{\frak{g}}] $-invariant linear subspace of $ \mathcal{H} $ containing $ D' $. This is simply the linear span of all elements of $ \mathcal{H} $ of the form
$$
[T(Y_{1}) \circ \cdots \circ T(Y_{m})](v),
$$
where (i) $ m \in \mathbb{N}_{0} $, (ii) $ Y_{1},\ldots,Y_{m} \in {\frak{g}} $ and (iii) $ v \in D' $.
Obviously, $ D'' \subseteq D $. By a proposition of Flato and Simon (Proposition 1 of their paper Separate and Joint Analyticity in Lie Groups Representations), every element of the form above is analytic for $ T(X_{i}) $ for each $ i \in [n] $. Therefore, every element of $ D'' $ is analytic for $ T(X_{i}) $ for each $ i \in [n] $.
Finally, as $ D' $ is dense, $ D'' $ is automatically dense, so the hypotheses of Theorem 2 are satisfied. $ \quad \blacksquare $


Answer to Question 3: Yes, there is a converse to Theorem 2.

Proof By Theorem 3, if $ T $ is integrable, then there exists a dense linear subspace $ D' \subseteq D $ of vectors that are jointly analytic for $ (T(X_{i}))_{i \in [n]} $. Joint analyticity is a much stronger condition than separate analyticity, so every element of $ D' $ is analytic for $ T(X_{i}) $ for each $ i \in [n] $. If $ D' $ is not already $ T[{\frak{g}}] $-invariant, then we can enlarge it to a $ T[{\frak{g}}] $-invariant linear subspace $ D'' $ as prescribed in the previous section. $ \quad \blacksquare $
Note: By the answer to Question 1, the requirement of $ T[{\frak{g}}] $-invariance in the statement of Theorem 2 and its converse can be dropped.
