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It is pointed out in Geometry from the spectral point of view the following: If one considers a discrete space, say, the two-point space $\{1,2\}$, after identifying its points $X=\{1,2\}/\sim$, the algebra of functions $A=C(\{1,2\}/\sim,\mathbb{C})$ is the algebra of matrices $M_2(\mathbb{C})$ with usual matrix product. According to the author, the noncommutativity of this algebra is a result of the relation between the "points".

(If one takes the quotient one has $X=\{*\}$, whose algebra is $\mathbb{C}$. I still have no problem with this apparent ambiguity, i.e. $M_2(\mathbb{C})$ is Morita equivalent to $\mathbb{C}$, so it will have the same "noncommutative topology", so to say). However, concerning Connes' statement,

Question: where does the usual matrix product comes from?

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up vote 3 down vote accepted

It comes from composition of isomorphisms. One version of the "algebra of functions" on, say, a finite groupoid $G$ is its groupoid algebra $\mathbb{C}[G]$, which is a direct generalization of the group algebra: take the free vector space on the morphisms in $G$ with multiplication given by composition (or $0$ if there is no composition). If $G$ is an equivalence relation, then $\mathbb{C}[G]$ is a finite direct product of matrix algebras.

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Just to flesh out a tad, since this took at least one reader who's never used a groupoid algebra a minute to follow: $\{1,2\}/\sim$ is a groupoid on the four elements $1,2,R,L$ where $R:1\to 2, L:2\to 1$, $RL=1,LR=2$. Then one computes $(a1+bL+cR+d2)(e1+fL+gR+h2)=(ae+bg)0+(af+bh)L+(ce+dg)R+(cf+dh)1$, which is the same as the multiplication of the $2\times 2$ matrices with entries $a,b,c,d$ and $e,f,g,h$. – Kevin Carlson Nov 1 '12 at 4:56
@Kevin: you don't need to work quite this hard and you can be more general. Recall that as a vector space, the $n \times n$ matrix algebra is spanned by elements $e_{ij}$ satisfying $e_{ij} e_{jk} = e_{ik}$. Now think of $e_{ij}$ as a morphism $i \to j$ in a groupoid with objects $\{ 1, 2, ... n \}$ (with the appropriate convention for composition). – Qiaochu Yuan Nov 1 '12 at 5:17
Ah, I see, thanks, Qiaochu! – Kevin Carlson Nov 1 '12 at 5:55

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