Examples of $C^*$-algebras in Noncommutative Geometry from A. Connes Question
I am working on $C^*$-algebras and I've been given Alain Connes's  book Noncommutative Geometry. 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 (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.
 A: 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.
