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.




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**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
$$
\text{Dom}(A) = \text{Dom}(A^{\*}).
$$




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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
$$
 \sum\_{k = 0}^{\infty}
 \left[
 \sum\_{\alpha: [k] \to [n]} \frac{s^{k}}{k!}
 \left\| \left[ T\_{\alpha(1)} \circ \cdots \circ T\_{\alpha(k)} \right] \! (v) \right\|
 \right]
< \infty.
$$




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From Lemma 9.1 of Edward Nelson’s 1959 paper [*Analytic Vectors*](http://www.jstor.org/discover/10.2307/1970331?uid=3739672&uid=2129&uid=2&uid=70&uid=4&uid=3739256&sid=21104400927691), 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*](https://eudml.org/doc/81902), 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.
> 
> 
> 



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




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**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.




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> 
> **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*](http://www.sciencedirect.com/science/article/pii/0022123673900359)), 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 $




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> 
> **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.


