Suppose that $(E,H)$ is a rigged (infinite dimensional, separable) Hilbert space, i.e. $H$ is a Hilbert space, and $E$ is a Fréchet space, equipped with a continuous linear injection $E \rightarrow H$ with dense image.

Define now the group $U(E,H)$ to consist of those invertible, continuous, linear transformations on $E$, which extend to unitary transformations of $H$.

What is known about the group $U(E,H)$?

For example:

  • Does the map $U(E,H) \rightarrow U(H)$ have dense image?
  • Is it a Fréchet Lie group? If so:
    • What is its Lie algebra? Is it just the continuous linear skew-adjoint operators on $E$?
    • Is the map $U(E,H) \times E \rightarrow E$ smooth?
    • Do the smooth vectors of the representation $U(E,H) \times H \rightarrow H$ consist exactly of $E$?
  • Is $U(E,H)$ contractible?

I am most interested in the case that $E$ is a nuclear Fréchet space.

Perhaps a good first step towards a Fréchet Lie group structure on $U(E,H)$ would be to equip the space of skew operators on $E$ with the structure of Fréchet (Lie) algebra. It's not so obvious to me how this should go. Is there anything known in this direction?

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    $\begingroup$ Presumably you mean for H to be infinite-dimensional... An interesting question! $\endgroup$
    – David Roberts
    Jul 22, 2020 at 13:15
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    $\begingroup$ I am not very familiar with definition of rigged Hilbert space, so does it allow $E$ to be a Banach space, such as $\ell_1$ sitting inside $\ell_2$? $\endgroup$
    – Yemon Choi
    Jul 22, 2020 at 16:02
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    $\begingroup$ Perhaps @Peter Michor can help? $\endgroup$ Jul 22, 2020 at 18:53
  • $\begingroup$ @YemonChoi Yes, in general this would be allowed. However, the only Banach spaces that are nuclear are finite dimensional. So in this case that I'm mostly interested in, $E$ will not be Banach. A good example of what I'm thinking about is $C^{\infty}(S^{1})$ sitting inside $L^{2}(S^{1})$. $\endgroup$
    – Peter
    Jul 22, 2020 at 21:12
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    $\begingroup$ The simplest interesting example is $E=s(\mathbb{N})$ the space of rapidly decaying sequences sitting inside $\ell^2(\mathbb{N})$. I would have to double check but linear continuous maps $E\rightarrow E$ are represented by infinite matrices $m_{i,j}$ with $|m_{i,j}|\le a_i b_j$ where $a_i$ is of rapid decay and $b_j$ is of at most polynomial growth. This is a very concrete starting point from which I think a complete understanding of $U(E,H)$ can be derived. $\endgroup$ Jul 24, 2020 at 13:01

2 Answers 2


These are just some thoughts on you question, so not an answer but too long for a comment. I waited a few days before posting this since, rather than addressing your post directly, I am going to look at a variant. This is usually not the done thing but my excuse is that I hope that it will contain material which could be of interest to you.

First of all, as has already been pointed out, your questions only make sense if you specify the relevant topologies. Experience suggests that the norm is not suitable for such questions. However, there are available a range of well-behaved and well-studied weaker ones which might do very well.

My main thesis is that most of the rigged Hilbert spaces which arise in mathematical physics have a special form and it is worth while starting there. The setting is an unbounded self adjoint operator $T$ on a Hilbert space. $E$ is then the intersection of the domains of definition of its powers. This is a Fréchet space with its natural structure, even Fréchet nuclear (Pietsch) under conditions on the spectral behaviour of $T$ which are satisfied by many of the classical operators (Laplace, Laplace-Beltrami, Schrödinger).

My main suggestion is that one study the space of unitary operators on the Hilbert space which commute with $T$. Such operators then map $E$ continuously into itself.

This space can be explicitly calculated for most of the situations mentioned above and this may provide some insight into your original question.

As an example, for the case of the standard one-dimensional Schrödinger operator the Fréchet space is that of the rapidly decreasing smooth functions and the corresponding operator space is the countable product of circle groups, regarded as multipliers on the coefficients of the Hermite expansion. As a compact group it won’t be dense anywhere in a non-trivial way but it will, presumably, be an infinite dimensional Lie group.

Similar remarks apply to the other classical cases where the spectral behaviour of the operator is known.

  • $\begingroup$ This sounds like an interesting approach. Do you have any references for this? Maybe in particular explaining the details of your example of the one-dimensional Schrödinger operator? Do you think you could relax the condition that the operators need to commute with T? Perhaps we could consider the unitary operators $U$ such that $[T,U]$ is a compact operator for example? $\endgroup$
    – Peter
    Jul 29, 2020 at 8:38

I've found a construction of a Fréchet Lie group structure on a subgroup of $\mathrm{GL}(E,H)$ in the literature that works under the additional assumption that $E$ is countably normed. Here is an overview:

First, let me recall the definition of a Sobolev Chain from page 1 of Omori's book.

Definition: A Sobolev chain is a system $\{ E, E_{k}\}_{k \in \mathbb{N}}$ where each $E_{k}$ is a Banach space, and for each $k \in \mathbb{N}$, there is a linear and dense embedding $E_{k+1} \rightarrow E_{k}$. And $E$ is the intersection of $E_{k}$ equipped with the inverse limit topology.

Let $L(E)$ then be the collection of continuous linear operators on $E$. Following Chapter IX in Omori we let $L_{0}(E)$ consist of those $T \in L(E)$ that extend to bounded linear operators $E_{k} \rightarrow E_{k}$ for all $k$. We let $L_{\infty}(E)$ consist of those $T \in L_{0}(E)$ such that there exist positive constants $C$ and $D_{k}$ such that \begin{equation} \| Tu \|_{k} \leqslant C\|u\|_{k} + D_{k}\|u\|_{k-1}. \end{equation} We then define $|T|$ to be the infimum over all such $C$. The strong PLN-topology is then defined by the semi-norm $| \cdot |$ and the projective limit topology of the Banach spaces of bounded operators on $E_{k}$ equipped with the norm topologies. We then define $\mathrm{GL}_{\infty}(E)$ to consist of those $T \in L_{\infty}(E)$ such that $T^{-1}$ exists and is an element of $L_{\infty}(E)$ itself. Omori then proves that $\mathrm{GL}_{\infty}(E)$ is open in $L_{\infty}(E)$, and is, moreover, a strong ILB-Lie group. In section 6 of this paper, it is proved that each strong ILB-Lie group is a regular Fréchet Lie group.

Now, to go back to the setting I was originally interested in, if $(E,H)$ is a rigged Hilbert space, then it admits a fundamental system of seminorms $\| \cdot \|_{1} \leqslant \| \cdot \|_{2} \leqslant \| \cdot \|_{3} \leqslant ...$

For each $k$ we let $E_{k}$ be the completion of $E$ with respect to $\| \cdot \|_{k}$. The assumption that $E$ is countably normed then means that all the linking maps $E_{k+1} \rightarrow E_{k}$ are linear and dense embeddings, and thus that $\{E, E_{k}\}_{k \in \mathbb{N}}$ is a Sobolev chain. We thus obtain a regular Fréchet Lie group $\mathrm{GL}_{\infty}(E)$.

This is (to some extent) an affirmative answer to the second question. What remains are its subquestions. Furthermore, we have not used nuclearity of $E$, I would be interested to see if this allows us to simplify anything or say anything more.

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    $\begingroup$ Passing to the completion may destroy injectivity. Frechet spaces weich are inverse limits of Banach spaces with injective connecting maps are called countably normed. $\endgroup$ Jul 28, 2020 at 6:12
  • $\begingroup$ @JochenWengenroth Thanks for pointing that out! I changed my answer accordingly. $\endgroup$
    – Peter
    Jul 29, 2020 at 8:32

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