There are two $C^\ast$-algebras associated to the $\ast$-algebra (under a convolution and the usual involution) $$C_c(G) := \{ f:G\longrightarrow \mathbb C :\:f \text{ has compact support}\}$$ of compactly supported continuous functions on a locally compact groupoid $G$: its completions in the norms $$ \|f\|_{\max} := \sup \{\|\pi(f)\|:\:\pi \text{ is a bounded involutive representation of C_c(G)}\}, $$ $$ \|f\|_r := \sup \{\|\pi_x(f)\|:\: x\in G^0 \}, $$ where $\pi_x$ is a regular representation in the fiber of $x$ under the target map.

Many references on groupoid $C^\ast$-algebras give the following example. Let $G$ be the groupoid of the equivalence relation on the set $$X:= [0,1]\times\{0,1\}$$ given by $(x,0)\sim (x,1)$ for $x \in (0,1).$ Marcolli writes in her ``Lectures on Arithmetic Noncommutative Geometry'' that the corresponding quotient space has no interesting functions, while the convolution algebra on $G$ is $$C_c(G) = \{f:G\longrightarrow C([0,1])\otimes M_{2}(\mathbb C):\: f(0),\,f(1) \text{ are diagonal} \}.$$ None of the sources I've found explain this description of $C_c(G)$ in detail sufficient for my understanding. So I ask $$\fbox{Question: Why does $C_c(G)$ have the above form for this groupoid?}$$

  • $\begingroup$ Did you mean $(x,0)\sim (x,1)$ iff $x\not\in\{0,1\}$? $\endgroup$ – Rasmus Apr 10 '12 at 17:12
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    $\begingroup$ Hint: Do you understand the case where $X=\{a,b\}$ and $a\sim b$? And the case where $X=\{a,b\}$ with the trivial equivalence relation? $\endgroup$ – Rasmus Apr 10 '12 at 17:14
  • $\begingroup$ @Rasmus yes, I do mean that. Corrected in the question. Thanks. I will think about those cases. $\endgroup$ – user3120 Apr 10 '12 at 18:51

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