I'm trying to get my head around the concept of Fourier Transform on a compact group. The standard definition is
$$\widehat{f}(\pi)=\int_Gdg\,f(g)\pi(g)$$
where $\pi\in\widehat G$, the Pontryagin dual of $G$. In my case, I have an operator in the form
$$\mathcal{T}[A]=\int_Gdg\,f(g)\pi_q(g)\,A\,\pi_{q'}^\dagger(g)$$ where $\pi_q$ and $\pi_{q'}$ are two inequivalent irreps of $G$, $A$ is a linear bounded map from the carrier space of $\pi_{q'}$ to the carrier space of $\pi_{q}$ and $f\in L^1(G)$.
I would like to cast the operator $\mathcal T$ in terms of the Fourier Transform of $f$. I think that I need to use tensor product representations (TPR) of $\pi_q$ and $\pi_{q'}$ to obtain a form that can be connected to the standard definition, because with a TPR one has $\pi_q(g)\,A\,\pi_{q'}^\dagger(g)=(\pi_q\otimes\pi_{q'})(g)A$.
My first question is: what is the explicit form of $(\pi_q\otimes\pi_{q'})(g)$ and how does it work on $A$? My confusion comes from the fact that if $\pi_q$ is $n\times n$ and $\pi_{q'}$ is $m\times m$, then $A$ is $n\times m$, so how can a $mn\times mn$ linear operator work on an $n\times m$ one?
First question solved: the operator $A$ has to be writen in Liouville form, basically by stacking all the columns in an $mn\times 1$ vector.
Then, a TPR is generally reducible, so one can write $$(\pi_q\otimes\pi_{q'})(g) = \bigoplus_\alpha\tau^\alpha(g)$$ with $\tau^\alpha$ irreps of $G$.
My second question is: does the following hold? $$\widehat f(\tau^\alpha)=\int_Gdg\,f(g)\tau^\alpha(g)$$ $$\mathcal T=\bigoplus_\alpha \widehat f(\tau^\alpha)$$
My third question is: what could I learn from some simple cases? For instance, what if $f$ were constant, or what if $G$ were abelian, or what if $q=q'$?