Matrix multiplication as a convolution

1. The product of two power series/polynomials is $$\begin{array}{rl}& (a_0+a_{1}x+a_2{x}^2+\cdots )(b_0+b_{1}x+b_2{x}^2+\cdots )\\ =& c_0+c_{1}x+c_2{x}^2+\cdots \end{array}$$ The coefficients given by $$c_n=\sum _{k=0}^n{a}_k{b}_{n-k}$$ is sometimes called the Cauchy product. This is a convolution.

2. Let $G$ be a finite group and $C[G]$ be the group algebra with complex coefficients. Let $$f=\sum _{x}f(x)x,g=\sum _{y}g(y)y$$ be two elements in $C[G]$. The product is $$fg=\sum _z(\sum _{xy=z}f(x)g(y))z.$$ When $f$ and $g$ are treated as functions $f,g:G\to C$, $$(fg)(z)=\sum _{xy=z}f(x)g(y)$$ is a convolution. In fact, $C[G]$ is an example of convolution algebra $L^1(G)$.

3. Let $J=\{1,...,n\}$ and $G^{\prime }=J\times J=\{(i,j):1\le i,j\le n\}$. $G^{\prime }$ is a groupoid: not every pair of elements in $G^{\prime }$ can be composed, $(a,b)$ can only be composed with $(c,d)$ when $b=c$. In such case, $(i,j)(j,k)=(i,k).$ 

Let us consider the groupoid convolution algebra $C[G^{\prime }]$ as above. The convolution in this case is then $$(fg)(i,k)=\sum _{(i,j)(j,k)=(i,k)}f(i,j)g(j,k).$$ If we rewrite this in another format, it should look more familiar: $$(AB{)}_{ik}=\sum _{i=1}^n{A}_{ij}{B}_{jk}.$$ This is matrix multiplication.