A question about finite group acting on inputs and outputs of maps between vector spaces. I'm reading this paper and considering the following notations/definitions:
Let $G$ be a finite group, $V,W$ finite dimensional vector spaces and consider two maps $\phi:V \rightarrow \mathbb{R}$, $\Phi:V \rightarrow W$.
Now we define:
$$\begin{align}
\psi(X) &:= \frac{1}{|G|}\sum_{g \in G}\phi(g^{-1} \cdot X) \\ \Psi(X) &:= \frac{1}{|G|}\sum_{g \in G}g \cdot \Phi(g^{-1} \cdot X),
\end{align}$$
where $\cdot$ denotes a group action, $X \subset V$.
First question:
Why do we take the inverse of $g$ to act on the set $X$?
As far as I'm concerned (but not 100% sure), this should be arbitrary, because

*

*if $x \mapsto gx$ is a left action, then $x \mapsto xg^{-1}$ is a right
action.


*if $x \mapsto xg$ is a right action, then $x \mapsto g^{-1}x$ is a left
action.
So for example, I could modify the above definitions saying that $\phi(g \cdot X)$ is the left action, while $\phi(X\cdot g^{-1})$ would be the right one.
Second Question:
Given that we assume $g^{-1}$ for the left action, why do we need to take $g$ to act on the output $\Phi(\cdot)$?
 A: To answer your questions, for $\psi(X)$, it makes no difference whether you write $g \cdot X$ or $g^{-1} \cdot X$, because we are summing over all $g \in G$.
For the second question, you could replace $g \cdot \Phi(g^{-1} \cdot x)$ by $g^{-1} \cdot \Phi(g \cdot x)$ without affecting the result of the summation, but I am guessing that the author definitely wants to have  one $g$ and one $g^{-1}$, just as we do when we define conjugation in a group.
A: About left-versus-right actions: first, there are two distinct parts to this, one, about notation, the other about literal "physical" action.
The notational question is whether we want to write $g\cdot x$ or $x\cdot g$ for "the action of $g$ on $x$". Either way, we would surely want associativity, namely $g\cdot (h\cdot x)=(gh)\cdot x$, while $(x\cdot g)\cdot h)=x\cdot (gh)$.
The "physical" aspect is about which side of a group we multiply on, and whether it's by $g$ or $g^{-1}$, when we have a group acting on itself, for example. If the action is left multiplication, for associativity that left action should be just $g\cdot x=gx$. For the action of right multiplication, the action should be $g\cdot x=xg^{-1}$, for associativity:
$$
g\cdot (h\cdot x) \;=\; g\cdot (xh^{-1}) \;=\; (xh^{-1})g^{-1} \;=\; x(gh)^{-1}
$$
Yes, we can dodge the inverse in the latter case by the notational ploy of declaring that we want a right action (with the slightly different associativity). :)
For $G$ acting on functions $f$ on a set $X$ on which $G$ notationally-acts on the left, for a notationally-left action, associativity again compels an inverse : $(g\cdot f)(x)=f(g^{-1}x)$.
And so on. :)
