**Question**


I am working on $C^\*$-algebras and I've been given Alain Connes's book [Noncommutative Geometry](http://www.alainconnes.org/docs/book94bigpdf.pdf). I am having troubles with understanding the examples on pages 91-93 (86-88 in the printed book). Maybe the problem is, that I do not understand, how those algebras are constructed. Could somebody explain the second way of the construction (page 91 (85))? I am talking about the part starting:


2) The second is to consider the larger algebra $B \supset C(Y)$ of all $2 \times 2$ matrices:
$$
f =
\begin{pmatrix}
f\_{aa} & f\_{ab} \\
f\_{ba} & f\_{bb} \\
\end{pmatrix}.
$$
...


I do not really understand, how to construct this matrix in general. For example - what is $f\_{aa}, f\_{ab},\ldots$? How are those constructed/represented e.g. in the example $2.\beta$ **The dual of the infinite dihedral group** (pages 92-93 (87-88))?


Many thanks.


**[Update 2]**


I have already asked this question [here](https://math.stackexchange.com/q/1262207/112628) (about two days ago, without any answer yet), but maybe this site is more suitable for this type of question (although maybe only the same people will see it)? I am not sure, so I might delete one of them later (I really don't want to *spam* every question I don't get answered immediately.


**[Update]**


I would be glad to receive any tips or guesses. I am aware, that this is broad topic, I am not looking for precise and rigorous explanation.



Examples of $C^*$-algebras in Noncommutative Geometry from A. Connes
Maybe it will help to see how the algebra $B$ is a special case of the general construction in $2.\alpha$. The compact manifold is the space $Y = \{a, b\}$ consisting of two points. The open cover of $Y$ consists of the sets $U\_1 = \{a\}$ and $U\_2 = \{b\}$. The equivalence relation on $Y$ is just $a \sim a$, $b \sim b$, $a \sim b$, $b \sim a$, and thus its graph $\mathcal{R}$ in $Y \times Y$ is just $Y \times Y$ itself.


So let's build the C\*-algebra $C^\*(\mathcal{R})$ described in Proposition 1. The underlying vector space is $C\_0(\mathcal{R})$, and this is just:
$$C\_0(\mathcal{R}) = C(\{a,a\}) \oplus C(\{a,b\}) \oplus C(\{b,a\}) \oplus C(\{b,b\})$$
Of course each algebra in the direct sum is isomorphic to $\mathbb{C}$, so $C\_0(\mathcal{R}) \cong \mathbb{C}^4$. Still, let's label elements of $C(\{a,a\}) \cong \mathbb{C}$ as $f\_{aa}$ (and similarly with the other summands) to keep track. In other words, an element of $C\_0(\mathcal{R}) \cong \mathbb{C}^4$ has the form $f = (f\_{aa},f\_{ab},f\_{ba},f\_{bb})$. 


Now let's write out the product in $C^\*(\mathcal{R})$ using the formula in Proposition 1:


$$fg(a,a) = \sum\_{a \sim x \sim a} f(a,x)g(x,a) = f\_{aa}g\_{aa} + f\_{ab}g\_{ba}$$
$$fg(a,b) = \sum\_{a \sim x \sim b} f(a,x)g(x,b) = f\_{aa}g\_{ab} + f\_{ab}g\_{bb}$$


and so on. In other words, the product $fg$ is just the matrix product:


$$\left(\begin{array}{cc} f\_{aa} & f\_{ab} \\ f\_{ba} & f\_{bb} \end{array}\right)
\left(\begin{array}{cc} g\_{aa} & g\_{ab} \\ g\_{ba} & g\_{bb} \end{array}\right)$$


We find that $C^\*(\mathcal{R})$ is isomorphic to the matrix algebra $M\_2(\mathbb{C})$. But certainly the construction is more interesting if the manifold and equivalence relation are not so trivial.


