# Does this groupoid have a quasi-diagonal reduced $C^*$-algebra?

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 $$a, b \in A$$.

• What topology are you giving? Are you giving the non-Hausdorff topology that any neighborhood of $(g,\infty)$ contains all but finitely many $k\in \mathbb N$? Jul 26, 2021 at 16:50
• 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. Jul 26, 2021 at 16:52
• @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 $k$. Jul 26, 2021 at 17:02
• @DiegoMartínez Can you describe the unitary representation of $H$ that generates $C^*_r(G)$? Jul 27, 2021 at 14:01
• @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. Jul 28, 2021 at 17:32