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

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.

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