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

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]).