Is the convolution operation some kind of group operation?

Question

I'm just curious but will the convolution operation be any sort of group operation? A motivating example would be to see that the natural exponential family of distribution functions are closed under convolution (although I'm not sure about the inverse). I know the irreducible unitary representation is a possible link of convolution to groups, but its just a hunch.

If not, could it relate to any weaker structure? like a monoid, semigroup?

Answer 1

Let $X$ be any set and $F$ a field. The space of functions $\hom(X,F)$ is a vector space over $F$; given any function $f:X\to F$ and scalar $a\in F$, we have $(af)(x):=af(x)$, and given any other function $g$ we have $(f+g)(x):=f(x)+f(x)$, for all $x\in X$. These are called pointwise operations, because you act on the function by acting on its value at each individual argument.

If $M$ is any monoid, we can equip $\hom(M,F)$ with a convolution operation

$$(f*g)(x):=\sum_{ab=x}f(a)g(b).$$

This operation is commutative if $M$ is commutative, but otherwise not necessarily. This is because the tuples $(a,b)$ for which $ab=x$ may not be the same tuples for which $ba=x$. But the operation is associative. And the function $f$ defined by $f(e_M)=1_F$ and $f(x)=0$ for $x\ne e_M$ is a two-sided identity element with respect to this operation. Moreover, convolution distributes over addition!

This means that $\hom(M,F)$ forms a unital, associative algebra over a field when equipped with pointwise addition and convolution. One can also restrict to certain subspaces of $\hom(M,F)$ (for instance, if $M$ and $F$ are topological, then we can take compactly supported continuous maps).

Example: if we take the subspace of functions with finite support, we get the monoid algebra denoted $F[M]$. This is usually presented as the space of formal $F$-linear combinations of elements of $M$ (with the obvious multiplication extended using the distributive property). Such a formal sum of the form $\sum_{x\in M} a_xx$ corresponds to the function $x\mapsto a_x$. Check for yourself that multiplication of formal sums corresponds to convolution of functions.

Example: if $M=(\Bbb N,+)$ is the naturals (containing $0$) under addition, then $F[M]\cong F[x]$ is the polynomial ring in one variable. (Can you figure out what the isomorphism is?)

Answer 2

The space of integrable functions under convolution is a commutative semicategory. That is, it is commutative and associative, but it is not defined for all integrable functions. If you only consider those functions for which their convolutions are defined (i.e. compactly supported continuous functions, then it is closed, but at best, you get a commutative semigroup.