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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 $\Phi(\cdot)$?

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    $\begingroup$ If $x \mapsto gx$ is a left action, what would $xg^{-1}$ even mean? $\endgroup$
    – Randall
    Dec 3 '21 at 17:46
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    $\begingroup$ I think what you mean: if $x \mapsto g \cdot x$ is a left action, then we can create an "equivalent" right action by $x \mapsto x \odot g$ where $x \odot g = g^{-1} \cdot x$. If you check this, you will see why applying the inverse is necessary. $\endgroup$
    – Randall
    Dec 3 '21 at 18:04
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    $\begingroup$ Ok, yes that's probably what I meant.. I'm simply trying to better understand this topic. So in my question case, I'm free to define the action however I want, whether it may be using $g$ or $g^{-1}$ because there's no need to specify the other case. When one might have both left and right actions, must be more careful. What about the action on the output? $\endgroup$ Dec 3 '21 at 18:06
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    $\begingroup$ I think so. It's good to check that these things are equivalent, so the choice doesn't matter that much (though a choice must be made). Sometimes the choice matters depending on if you write your functions on arguments as $f(x)$ or $xf$. $\endgroup$
    – Randall
    Dec 3 '21 at 18:07
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    $\begingroup$ It's not about input or output, it's about how you like to write functions, or if you prefer to compose left-to-right or vice versa. $\endgroup$
    – Randall
    Dec 3 '21 at 18:11
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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.

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  • $\begingroup$ Many thanks, but exactly what would be wrong in writing: $\Psi(X):=\frac{1}{|G|}\sum_{g \in G}g \cdot \Phi(g\cdot X)$ $\endgroup$ Dec 3 '21 at 18:31
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    $\begingroup$ If you look at the paper, you will see that the aim is that the map $\Psi$ should be equivariant; i.e. $g.\Psi(X) = \Psi(g.X)$ for all $g \in G$. With the given definition the map is indeed equivariant. With your proposed alternative, it would not be. $\endgroup$
    – Derek Holt
    Dec 3 '21 at 21:25
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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. :)

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