Posted by Jason Polak on 06. May 2017 · 2 comments · Categories: group-theory

When I first saw convolution I happily used it without thinking about why we should define it in the first place. Here’s a post that might tell you why it makes sense from an algebraic viewpoint. Let’s recall the convolution product, as it is for functions $f:\R\to\R$. If $f,g:\R\to\R$ are two such functions, then the convolution $f*g$ is defined as
$$
(f*g)(y) := \int_\R f(y-x)g(x){\rm d}x$$
provided the integral exists. Why do we do this? For convolution, the explanation starts with polynomials. Multiplying polynomials is simplicity itself. For example, here is the product of two real-valued polynomials:
$$(1 + 2t + 5t^2)(1 + 3t) = 1 + 5t + 11t^2 + 15y^3$$
But a real-valued polynomial could also be thought of as a function $f:\Z\to \R$, where $f(x)$ is zero for all but finitely many $x$ and nonzero only if $x \geq 0$. In this setup, $f(x)$ represents the coefficient of $t^x$. Addition of polynomials becomes pointwise addition of functions. Now here’s the good part: polynomial multiplication becomes convolution: if $f,g:\Z\to\R$ are two polynomials then
$$(fg)(y) = \sum_{x\in \Z} f(y-x)g(x).$$
For a finite group $G$ then, complex representations of $G$ correspond to representations of the algebra of complex-valued functions on $G$ where the multiplication is the convolution product.

2 Comments

  1. I don’t think this explanation really make sense. I see your argument at Serge Lang’s book (where he use it as an example). now I would like to think it as a natural construction involving the algebraic structure of the domain.

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