Let $H$ be a discrete group, and let $X$ be the one-point compactification of $\mathbb{N}$. Consider the étale groupoid $G = H \times \{\infty\} \sqcup \mathbb{N}$, whose unit space is $X$, and with operations:

  1. $G$ is a group bundle: $s(g) = r(g)$ for every $g \in G$
  2. The isotropy at every $k \in \mathbb{N}$ is trivial: $G_k = \{g \in G \mid s(g) = k\} = \{k\}$
  3. The isotropy at $\infty$ is $H$: $G_\infty = \{g \in G \mid s(g) = \infty\} = H$

Question: Can $C_r^*(G)$ be quasi-diagonal, and $H$ be non-amenable?

Note that (modulo the Tikuisis-White-Winter theorem) it is not hard to prove that if $H$ is amenable then $C_r^*(G)$ is quasi-diagonal. Moreover, a non-amenable fiber can only happen in a non-isolated point of $X$.

Recall: We say that a separable $C^*$-algebra $A$ is quasi-diagonal if there are contractive and completely positive maps $\varphi_n \colon A \rightarrow M_{k(n)}$ such that $\|\varphi_n(a)\| \rightarrow \|a\|$ and $\|\varphi_n(ab) - \varphi_n(a)\varphi_n(b)\| \rightarrow 0$ for every $a, b \in A$.

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    $\begingroup$ What topology are you giving? Are you giving the non-Hausdorff topology that any neighborhood of $(g,\infty)$ contains all but finitely many $k\in \mathbb N$? $\endgroup$ Jul 26, 2021 at 16:50
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    $\begingroup$ This is essentially an inverse semigroup $C^*$-algebra but I don't know if it helps. You can take the inverse monoid $M$ with group of units $H$, a zero and a countably infinite set of orthogonal idempotents which are stabilized by the elements of $H$ on either side and this will be Paterson's universal groupoid. $\endgroup$ Jul 26, 2021 at 16:52
  • $\begingroup$ @BenjaminSteinberg thanks, I knew this already. This is actually the point of view I come from, though I thought people who knew about quasi-diagonality would know more about groupoids than inverse semigroups. And yes, the topology I'm equipping it with is non Hausdorff, and every neighborhood of $(g, \infty)$ contains cofinitely many $k$. $\endgroup$
    – user147609
    Jul 26, 2021 at 17:02
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    $\begingroup$ @DiegoMartínez Can you describe the unitary representation of $H$ that generates $C^*_r(G)$? $\endgroup$ Jul 27, 2021 at 14:01
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    $\begingroup$ @DiegoMartínez In that case it will only be QD when $H$ is amenable. Suppose $H$ is not amenable. Since $\pi$ is the direct sum of the left regular and infinitely many trivial representations and the trivial representation does not weakly contain the left regular it follows that $C^*_\pi(H)\cong C^*_r(H)\oplus \mathbb{C}.$ If $C^*_r(H)\oplus \mathbb{C}$ were QD then $C^*_r(H)$ would also be since it is a subalgebra. But $C^*_r(H)$ isn't QD since $H$ is not amenable. $\endgroup$ Jul 28, 2021 at 17:32


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