A question about finite group acting on inputs and outputs of maps between vector spaces.

Question

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

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

2. 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 $\Phi(\cdot)$?

Answer 1

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.

Answer 2

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. :)