Monday, February 3, 2014

Matrix multiplication as a convolution

1. The product of two power series/polynomials is $$ \begin{align*} & (a_0 + a_1 x + a_2 x^2 + \cdots) (b_0 + b_1 x + b_2 x^2 + \cdots)  \\ = & c_0 + c_1 x + c_2 x^2 + \cdots  \end{align*}$$ 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 $$ f g = \sum_z ( \sum_{x y = z} f(x) g(y) ) z. $$ When $f$ and $g$ are treated as functions $f, g : G \to C$, $$ (f g) (z) = \sum_{x y = z} f(x) g(y)$$ is a convolution. In fact, $C[G]$ is an example of convolution algebra $L^1 (G)$.

3. Let $J = \{1, ... ,n \}$ and $G' = J \times J = \{ ( i, j ) : 1≤ i, j ≤n \}$. $G'$ 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 $$(f g) (i,k) = \sum_{ (i, j) (j, k) = (i, k) } f(i, j) g(j, k).$$ If we rewrite this in another format, it should look more familiar: $$(AB)_{ik} = \sum_{i=1}^n A_{ij} B_{jk}. $$ This is matrix multiplication.

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