# Functions space of discrete space: how does taking quotients lead to noncommutativity?

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?

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.
• 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 $e,f,g,h$. Nov 1, 2012 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). Nov 1, 2012 at 5:17