Certain groupoid and its $C^{*}$ algebra

Question

Let $X$ be a finite subset of real numbers. Let $G$ be the collection of all non empty subsets of $X$ and $G_{0}$ be the collection of all singleton subsets of $X$.

We define two maps $r,s:G \to G_{0}$ with $$r(A)=\text{The singletone consisting of Maximum} \;\; of\;\; A$$ and $$s(A)=\text{The singleton consisting of minimum}\;\; of \;\; A$$

This provide a semi groupoid structure on the pair $(G,G_{0})$ whose composition law is the union operation $A\circ B= A \cup B$.

What is the precise description of the groupoid associated with this semigroupoid? The resulting groupoid gives us a groupoid $C^{*}$ algebra. Is this (finite dimensional) $C^{*}$ algebra studied, already?What is its precise description?

What about if we replace the pair $(G,G_{0})$ with the collection of compact subsets of the interval and its singleton subsets, respectively?

Answer 1

I will explain how I ended up with the description given in my comment above.

Let $\{1, \dots ,n \}$ be your finite subset of $\mathbb{R}$ (only their order is important, we do not really care about the precise values).

A first remark is that your semi-groupoid is in fact a category: $\{i\}$ is an identity from $i$ to $i$. As identity are idempotent, they always become identity in the groupoid completion so one can just consider this semi-groupoid as a category and take the associated groupoid.

A second remarks is that as a category it can be described in the following way:

It is freely generated by $n$ objects $1, \dots ,n$, with for each $i <j$ a (freely added) morphism from $i$ to $j$.

Indeed, a morphism in this free category will just be a sequence of composable arrow of this form. Such a morphism necessarily goes from $i $ to $j$ with $i \leqslant j$ because we did not add any morphism in the other direction. And an arrow in the free category is just describe by the list of $v$ between $i$ and $j$ into which it factors, hence a subset pf $\{1,\dots,n\}$ whose minimum is $i$ and maximum is $j$, composition being exactly the union of such subsets.

This implies that the associated groupoid is the groupoid freely generated by the same generators. If we only add the generators of the form $n<n+1$ one gets the pair groupoid on $n$-objects, it is hence equivalent to the trivial groupoid. For each other generator tat we need to freely add one can add instead an automorphism of the object $1$ in order to get an equivalent groupoids. Hence your groupoids is equivalent to the groups with one generator for each pair $i<j$ of $\{1 \dots,n\}$ whcih are at distance at least $2$ and there is exactly $k=(n-2)(n-3)/2$ such pairs.

So your groupoid can be describe as the groupoid on $n$ objects such that for all $x,y$, $Hom(x,y)=F_k$ the free group on $k=(n-3)(n-2)/2$ generators and the corresponding $C^*$-algebra will be the algebra of $n$ by $n$ -matrix with coefficient in the group $C^*$ -algebra of $F_k$ (reduced or maximal depending on if you consider the reduced or maximal $C^*$-algebra of the groupoid) and it will not be finite dimensional.

If you want to consider compact subset instead of finite subset, then if you don't topologize I think you will get a free group one an uncountable number of generators which does not seem to be a very interesting things to study. If you topologize (for example with the Hausdorff topology) that produces a pretty interesting groupoid but it does not seem locally compact to me and does not come with a Haar system so I don't see how you can attach a $C$-algebra to it...