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Let $\mathcal{H}$ be a complex Hilbert space, and $\mathcal{B}(\mathcal{H})$ be the $C^{\ast}$-algebra of bounded operators on $\mathcal{H}$. Is there an étale groupoid $\mathcal{G}$ such that its $C^{\ast}$-algebra (full or reduced) is isomorphic to $\mathcal{B}(\mathcal{H})$?

In this paper, Alcides Buss and Aidan Sims proved that not all $C^{\ast}$-algebras are groupoid algebras. Note that $\mathcal{B}(\mathcal{H})$ satisfies the criteria given in their paper.

A weaker form of this question is the following: Is every $C^{\ast}$-algebra a closed $\ast$-subalgebra of an étale groupoid $C^{\ast}$-algebra?

Edit: In discussion with Prof. Ruy Exel, he suggested some improvements to this question: Let $D \subseteq \mathcal{B}(\mathcal{H})$ be the $\ast$-subalgebra consisting of all diagonal operators relative to some orthonormal basis.

He advised me to specialize this question by demanding that the étale groupoid $\mathcal{G}$, with $C^{\ast}_{\mathrm{red}}(\mathcal{G}) \cong \mathcal{B}(\mathcal{H})$, satisfies that the isomorphism induces a correspondence between $D$ and $C(\mathcal{G}^{(0)})$.

His suggestion is based on the fact that most of the time, when someone is tasked with finding a groupoid such that its reduced algebra is a specific $C^{\ast}$-algebra $A$, one specifies the subalgebra that corresponds to $C(\mathcal{G}^{(0)})$.

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    $\begingroup$ When you speak of "the" Cstar algebra of a groupoid - do you mean full? reduced? $\endgroup$
    – Yemon Choi
    Commented Mar 11 at 7:55
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    $\begingroup$ Full or reduced. Edited. @YemonChoi $\endgroup$
    – Luiz Felipe Garcia
    Commented Mar 11 at 10:54
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    $\begingroup$ Are you sure $\mathcal{B}(\mathcal{H})$ is not isomorphic to its opposite algebra? Do you have a reference for that? $\endgroup$
    – Diego Martinez
    Commented Mar 16 at 13:16
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    $\begingroup$ It is isomorphic to its opposite algebra. This is why it might be a grupoid $C^*$-algebra. $\endgroup$
    – Luiz Felipe Garcia
    Commented Mar 16 at 13:55
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    $\begingroup$ Actually, let me rephrase my previous comment: Ah, so you mean B(H) does not satisfy the criteria in that paper. In any case, the answer to your question may be "yes". The easiest that comes to mind is taking N with the distance d(n,m)=1 when n≠m. It then follows that the uniform Roe algebra of that space is B(ℓ2(N). $\endgroup$
    – Diego Martinez
    Commented Mar 16 at 15:12

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