Let $H$ and $K$ be infinite dimensional (separable) Hilbert spaces and $X=B(H,K)$ denote the space of bounded linear operators. For $T_1, T_2$ in $X$, we define $D_{T_1,T_2}:X \to X$ as $D_{T_1,T_2}(T)=TT_1^*T_2$. Finally define $V^0=\operatorname{span}\{D_{T_1,T_2}, T_1, T_2 \in X\}$. One can check that $V^0$ is a pre-$C^*$-algebra with involution $D_{T_1,T_2}^*=D_{T_2,T_1}$. Let $V$ denotes the closure of $V^0$ inside $B(X)$.

Is $V$ isomorphic to some well known $C^*$-algebra?

P.S: This question was first posted on Math Stackexchange here. Also, this question is particular case of a more general construction given at Trying to understand construction of $C^*$-algebra corresponding to a ternary $C^*$-ring from a paper.

  • 2
    $\begingroup$ TeX note: use $V^0 = \operatorname{span} \{…\}$ $V^0 = \operatorname{span} \{…\}$, not $V^0 =$span$\{…\}$ $V^0 =$span$\{…\}$. I have edited accordingly. $\endgroup$
    – LSpice
    Oct 15 '21 at 1:01
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    $\begingroup$ @LSpice: Much thanks! $\endgroup$
    – Math Lover
    Oct 15 '21 at 2:19

Assuming $H$ is separable? If $K$ is finite dimensional and $H$ is infinite dimensional then this gives you the compact operators on $H$. Otherwise you get $B(H)$.

If $H$ can have uncountable dimension then there are more possibilities because there are more closed ideals of $B(H)$.

Some details: $B(H)$ acts on $X = B(H,K)$ by multiplication from the right, and this isometrically embeds $B(H)$ in $B(X)$. If ${\rm dim}(H) \leq {\rm dim}(K)$ then the operators of the form $T_1^*T_2$ with $T_1, T_2 \in X$ comprise all of $B(H)$. If $H$ is infinite dimensional and $K$ is finite dimensional then all operators of the form $T_1^*T_2$ are compact, and as long as ${\rm dim}(K) \geq 1$ they include all rank 1 operators, so their closed span equals the compact operators.


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