## Matrix multiplication as a convolution

1. The product of two power series/polynomials is

$\displaystyle \begin{array}{rl} & (a_0 + a_1 x + a_2 x^2 + \cdots) (b_0 + b_1 x + b_2 x^2 + \cdots) \\ = & c_0 + c_1 x + c_2 x^2 + \cdots \end{array}$

The coefficients given by

$\displaystyle c_n = \sum_{k=0}^n a_k b_{n-k}$

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

$\displaystyle f = \sum_x f(x) x, g = \sum_y g(y) y$

be two elements in ${C[G]}$. The product is

$\displaystyle f g = \sum_z ( \sum_{x y = z} f(x) g(y) ) z.$

When ${f}$ and ${g}$ are treated as functions ${f, g : G \rightarrow C}$,

$\displaystyle (f g) (z) = \sum_{x y = z} f(x) g(y)$

is a convolution. In fact, ${C[G]}$ is an example of convolution algebra ${L^1 (G)}$.

3. Let ${J = \{1, ... ,n \}}$ and ${G' = J \times J = \{ ( i, j ) : 1\le i, j \le n \}}$. ${G'}$ is a groupoid: not every pair of elements in ${G'}$ 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 ${C[G']}$ as above. The convolution in this case is then

$\displaystyle (f g) (i,k) = \sum_{ (i, j) (j, k) = (i, k) } f(i, j) g(j, k).$

If we rewrite this in another format, it should look more familiar:

$\displaystyle (AB)_{ik} = \sum_{i=1}^n A_{ij} B_{jk}.$

This is matrix multiplication.