Again, look at example 4.9 in the paper. Any étale groupoid is a groupoid of germs of an inverse semigroup action (some people may disagree what 'germs' means here, but what I said is true with the notion of germs in that paper). And, thus, some groupoids of germs are amenable, some are not. It depends on the groupoid itself.

What do you mean? Whether I can provide some examples of amenable groupoids? Examples 4.8 and 4.9 in the paper I referenced are somewhat enlightening.

You may find the definition of amenable groupoid (for non Hausdorff étale groupoids as well) in my paper with Buss arxiv.org/pdf/2302.14466 (see definition 4.23). Nevertheless, as Yemon Choi pointed out, non Hausdorffness has nothing to do with separability of the algebra.

@JamieGabe thanks! I think homogeneous algebras are as above with only 1 possible dimension for $H$ though. I agree that subhomogeneous do embed into $M_n(C_0(X))$, but I was hoping for a more explicit description. Thanks in any case!

In any case, the isomorphism you're looking for can be written down for the dense subalgebras. Any $f \in C_c(G^0) \rtimes_{alg} S$ can be written (by definition) as a finite linear combination $f = f_{s_1} + \dots + f_{s_k}$ where $f_{s_i}$ is a function on $G$ that is continuous and supported on the open bisection $s_i$. Then this map is continuous in the max norm, in the reduced one, and also in the essential seminorm (this is only a seminorm in general).

Yes, the idea is to construct a Fell bundle (but you don't really need to, since Fell bundles are much more general). In any case, elements $s \in S$ are open bisections of $G$, and $A_s = C_0(s)$, which will be embedded into $C^*(G)$ in the usual way (similarly for the reduced and essential). The formula that you want should then be that the product of $a_s$ and $a_t$ looks like $[gh \mapsto a_s(g) a_t(h)]$, where $g \in s, h \in t$ (and thus $gh \in st$). This should be the formula I think you meant to write: $ \mu_{s, t}(a_s \otimes a_t) = \beta_s(\beta_{s^*}(a_s)a_t)$

By the way, the isomorphism is just the same map, which is continuous both in the max norm, the reduced one, and the "essential" one (the latter only being different from the reduced for non-Hausdorff groupoids).

This is developed in several papers of Buss, Exel, Meyer and Kwaśniewski, but I think Sections 2 and 4 of my paper with Buss is a nice summary: sciencedirect.com/science/article/abs/pii/… The literature in the previous papers is somewhat scattered.

Well, by the very definition of what it means to be of finite propagation. $t \in \mathcal{B}(\ell^2(\mathbb{N}))$ has propagation at most $r \geq 0$ if $p_B t p_A = 0$ whenever $d(A, B) > r$. Since the space $(\mathbb{N}, d)$ is bounded, every operator has finite propagation (has propagation at most $1$). Hence $C_u^*(\mathbb{N}, d) = \mathcal{B}(\ell^2(\mathbb{N}))$.

While the above is true, it is not true that the above model yields a Cartan subalgebra, since $\ell^\infty(\mathbb{N})$ is not a (C*-)Cartan subalgebra of $\mathcal{B}(\mathcal{\ell^2(\mathbb{N}}))$, as it lacks enough normalizers. So I do not know the answer to your question. You may check Section 6.4 in my arxiv.org/pdf/2312.08907.pdf, where we do actually need local finiteness of the space to get a groupoid model.

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).

Are you sure $\mathcal{B}(\mathcal{H})$ is not isomorphic to its opposite algebra? Do you have a reference for that?

Yes, take $\ell^\infty(\mathbb{N})$. It is not separable, and then the state $\text{ev}_n$ yields a 1-dimensional GNS. The answer to your question is yes even in the case for some state of full support as well.

You may want to work with a explicit description of $H$, such as $\ell^2(\mathbb{N})$. In such a case you may want to construct $n$ (explicit) copies of $H$ within itself, take unitaries between $H$ and the copies (so isometries from $H$ onto a proper closed subspace) and then conjugate by these. (Another hint: think of the phenomena behind Hilbert's hotel, or Dedekind's definition of an infinite set).

@AntonioLorenzin yes, but you can always substitute $A$ by the C*-subalgebra generated by $a$ and $1$, which is separable, and then run the same argument.

I was going to do something more complicated, but @CalebEckhardt's approach is much easier.