Given an infinite dimensional Hilbert Space $\mathcal{H}$ what is the underlying locally compact Hausdorff etale groupoid $G$ such that $C_r^{\ast}(G)$ is $\ast$-isomorphic to $\mathcal{B}(\mathcal{H})$ as a $C^{\ast}$-algebra?
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3$\begingroup$ I believe it is not known what such a $G$ exists. See mathoverflow.net/questions/466749/… $\endgroup$– David GaoCommented Oct 16 at 18:55
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2$\begingroup$ Apologies (to @YemonChoi and @PKOA) for "group" in the title—I tried to edit to a more informative title, and didn't notice I'd changed "groupoid" to "group". $\endgroup$– LSpiceCommented Oct 17 at 2:50
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3$\begingroup$ @PKOA The issue is simply that the matrix units generate finite-dimensional matrix algebras as $C^\ast$-algebras, but only generate infinite-dimensional $B(H)$ as von Neumann algebras, and the latter is not enough to get a groupoid $C^\ast$-algebra structure. There’s also no other obvious choice, which is probably why this question is supposedly still open. $\endgroup$– David GaoCommented Oct 17 at 7:56
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3$\begingroup$ A groupoid C*-algebra $\mathrm{C}^\ast_{\mathrm{r}}G$ has a faithful conditional expectation $E$ onto the Cartan subalgebra (in the $\mathrm{C}^*$-sense) $C_0(G^{(0)})$. I guess faithfulness of $E$ (restricted to $K(H)$) implies $C_0(G^{(0)})$ atomic and $C_0(G^{(0)})$ cannot be a Cartan subalgebra. I haven't checked the detail. (The norm-closure of those $T$ on $\ell_2\mathbb{N}$ such that $\sup_i |\{ j : |T_{i,j}|+|T_{j,i}|>0 \}|<\infty$ is an etale groupoid $\mathrm{C}^\ast$-algebra, but it is strictly smaller than $B(\ell_2\mathbb{N})$.) $\endgroup$– Narutaka OZAWACommented Oct 18 at 0:56
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3$\begingroup$ @David Gao: If a $\mathrm{C}^\ast$-algebra $A\subset B(H)$ is the range of a faithful conditional expectation $E$, then it is a von Neumann algebra. Indeed, if $a$ is the SOT-limit of a bounded increasing net $(a_i)$ in $A_+$, then $E(a) - a \geq 0$, because $E(a)\geq E(a_i)=a_i$ for all $i$, which implies $E(a) = a$ by faithfulness of $E$. $\endgroup$– Narutaka OZAWACommented Oct 23 at 2:36
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