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 $B(X)$.



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> Is $V$ isomorphic to some well known $C^\*$-algebra?
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P.S: This question was first posted on Math Stackexchange [here](https://math.stackexchange.com/questions/4274449/trying-to-recognise-a-c-algebra). 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](https://mathoverflow.net/questions/402950/trying-to-understand-construction-of-c-algebra-corresponding-to-a-ternary-c).



Trying to recognise a $C^*$-algebra
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.


