When is convolution not commutative?

Let $$G$$ be a locally compact Hausdorff group with a left Haar measure $$\lambda$$. Define the convolution of two functions $$f,g \in L^1(G)$$ by

$$(f \ast g)(x) = \int f(y) g(y^{-1}x) d\lambda (y), ~~~ \forall x \in G$$

If the group $$G$$ is abelian the convolution is commutative: $$f \ast g = g \ast f$$.

In general, for any $$x \in G$$ we have (written multiplicatively)

$$(f \ast g)(x) = \int f(y) g(y^{-1}x) d\lambda(y) = \int f(xy) g((xy)^{-1}x) d\lambda(y) = \int f(xy) g(y^{-1}) d\lambda(y)$$

In the second equality, we apply a left shift by $$x^{-1}$$ which does not change the integral since $$\lambda$$ is left invariant.

Precomposing with inversion yields

$$\int f(xy^{-1}) g(y) d\rho(y)$$

where $$\rho$$ is the associated right Haar measure defined by $$\rho(B) = \lambda(B^{-1})$$ for any Borel set $$B \subseteq G$$.

Finally, commuting $$x$$ and $$y^{-1}$$ gives

$$\int g(y) f(y^{-1}x) d\rho(y)$$

Now, if $$G$$ is unimodular, $$\rho$$ and $$\lambda$$ coincide, so the last expression is the convolution $$g \ast f$$. Also, since both $$y^{-1} \in G$$ and $$x \in G$$ are arbitrary, the step requires $$G$$ to be abelian (which then also makes it unimodular).

I am looking for an explicit counterexample to the claim that $$f \ast g = g \ast f$$ in general, and conditions under which the formula is true (which are hopefully weaker than $$G$$ being abelian).

Thank you very much in advance!

• Typically $G$ does have to be abelian for this to work. I don't know off the top of my head if this is literally true in general, but any non-abelian groups with commutative convolution would need to have some very strange properties. I'm pretty sure it couldn't be metrizable, for example. Aug 3 '20 at 16:53
• As for an explicit counterexample, try any non-abelian finite or countable group, where $f,g$ take the value 1 at a single element and 0 elsewhere. Aug 3 '20 at 16:54

Convolution of two $$C_c$$ functions commute $$\iff$$ $$G$$ is abelian

As you noted if $$G$$ is abelian then it is trivial that convolutions commute.

For the converse, let convolution of any two $$C_c$$ functions commute. Let $$f,g \in C_c(G)$$

Then $$\forall x \in G \text{ we have }$$ $$0= f*g(x)-g*f(x)=\int_G f(xy)g(y^{-1}) d\lambda(y) - \int_{G} g(y)f(y^{-1}x)d\lambda(y)$$ $$=\int_G f(xy^{-1})g(y)\Delta(y^{-1}) d\lambda(y) - \int_{G} g(y)f(y^{-1}x)d\lambda(y)$$ $$\implies \int_G g(y)(\Delta(y^{-1})f(xy^{-1})-f(y^{-1}x))d\lambda(y)=0$$

Since, $$g \in C_c(G)$$ was arbitrarily chosen, it follows that $$\Delta(y^{-1})f(xy^{-1})=f(y^{-1}x), \forall x,y \in G$$ So put $$x=1$$ above and note that $$\Delta(y^{-1})f(y^{-1})=f(y^{-1})$$ . Again $$f \in C_c(G)$$ was arbitrarily chosen thus $$f$$ can very well be non-zero at $$y^{-1}$$. So we get, $$\Delta(y^{-1})=1, \forall y \in G$$

Hence, $$f(xy^{-1})=f(y^{-1}x) \forall x,y \in G$$ . Then just replace $$y$$ by $$y^{-1}$$ and we get $$f(xy)=f(yx) \forall f \in C_c(G) \implies xy=yx, \forall x,y \in G$$

Since, you have the result for $$C_c(G)$$, it follows for $$L^1(G)$$

For a locally compact group $$G$$, one has that $$L^1(G)$$ is commutative if and only if $$G$$ is commutative. See, for example, Theorem 1.6.4 in Principles of Harmonic Analysis by Deitmar and Echterhoff. As for your desire to see an example of noncommutativity in $$L^1(G)$$, the aforementioned fact says that any choice of nonabelian $$G$$ must lead to one. A simple way to proceed is to take $$G$$ to be a noncommutative discrete group (or even a finite group) as, in this case, one has an inclusion $$G \subset L^1(G)$$. This is because each $$g \in G$$ may be identified with the function $$\delta_g \in L^1(G)$$ defined by $$\delta_g(g)=1$$ and $$\delta_g(h)=0$$ if $$h \neq g$$. One can check that $$\delta_g * \delta _h = \delta_{gh}$$. This example is not unrelated to the methods by which one would prove the equivalence between abelianness of $$L^1(G)$$ and $$G$$. The idea is to use approximations to such delta functions, or to construct a larger algebra than $$L^1(G)$$ (a sort of multiplier algebra) which contains them.

This is a rather long comment.

An explicit example is $$G=SL(2,\mathbb{R})$$, (the group of all real 2x2 matrices of determinant 1).

Looking at this non-abelian group it is interesting to consider the subgroup $$K$$ of all rotations, which is abelian.

We may then consider the double coset space $$G\backslash\!\backslash K$$. This is a space of equivalent classes $$\bar{g}$$ where we identify all elements $$h, g\in G$$ provided there are $$k_1,k_2\in K$$ such that $$h= k_1gk_2$$ Now $$G\backslash\!\backslash K$$ is not a group, but the Haar measure on $$G$$ induce a measure on $$G\backslash\!\backslash K$$ and it is remarkable that $$f * g = g* f$$ on $$L^1(G\backslash\!\backslash K)$$. (See e.g Sugiura ”Unitary representations and Harmonic Analysis”). Around 1960-ish Naimark worked on translation operators on $$L^1$$ -algebras in an attempt to understand Harmonic analysis in a wider sense, however, I cannot recall the sources for this.