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Proceedings of the American Mathematical Society

ISSN 1088-6826(online) ISSN 0002-9939(print)

 

 

On the strong unbounded commutant of an $ \mathcal{O}^*$-algebra


Authors: A. Van Daele and A. Kasparek
Journal: Proc. Amer. Math. Soc. 105 (1989), 111-116
MSC: Primary 47D40; Secondary 46K10
DOI: https://doi.org/10.1090/S0002-9939-1989-0973842-7
MathSciNet review: 973842
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Abstract: Let $ \mathcal{D}$ be a dense subspace of a Hilbert space $ \mathcal{H}$. An $ {\mathcal{O}^*}$-algebra $ \mathcal{A}$ on $ \mathcal{D}$ is a $ *$-algebra of linear operators defined on $ \mathcal{D}$ and leaving $ \mathcal{D}$ invariant which contains the identity map $ I$ of $ \mathcal{D}$. The involution in $ \mathcal{A}$ is the map $ A \to {A^ + }: = {A^*} \upharpoonright \mathcal{D}$ (see [1]). It is possible to define different types of unbounded commutants of an $ {\mathcal{O}^*}$-algebra $ \mathcal{A}$. We follow the definitions given in [2]. Let $ {L_\mathcal{A}}(\mathcal{D},\mathcal{H})$ be the vector space of all continuous linear mappings of $ \mathcal{D}[{t_\mathcal{A}}]$ into $ \mathcal{H}$ with respect to the graph topology $ {t_\mathcal{A}}$ on $ \mathcal{D}$ introduced by the operators from $ \mathcal{A}$. Then the strong unbounded commutant is defined as $ \mathcal{A}_s^c: = \{ T \in {L_\mathcal{A}}(\mathcal{D},\mathcal{H}):T\mathcal{D} \subset \mathcal{D},TAx = ATx$ for all $ x \in \mathcal{D}{\text{and}}A \in \mathcal{A}\} $. $ \mathcal{A}_s^c$ is an algebra, but in general however it will not be $ *$-invariant (see [2]). In this paper we show that even worse can happen. We give an example of such an $ {\mathcal{O}^*}$-algebra $ \mathcal{A}$ and an operator $ T \in \mathcal{A}_s^c$ such that $ \mathcal{D}({T^*}) = \{ 0\} $.

In particular this shows that the strong unbounded commutant of an $ {\mathcal{O}^*}$-algebra may contain operators which are not closable. Furthermore the constructed $ {\mathcal{O}^*}$-algebra $ \mathcal{A}$ gives another example for an $ {\mathcal{O}^*}$-algebra whose socalled form commutant $ \mathcal{A}_f^c$ contains a sesquilinear form which is not an operator (see [2]).


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DOI: https://doi.org/10.1090/S0002-9939-1989-0973842-7
Article copyright: © Copyright 1989 American Mathematical Society