# 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

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 $$\Phi(\cdot)$$?

• If $x \mapsto gx$ is a left action, what would $xg^{-1}$ even mean? Dec 3 '21 at 17:46
• 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. Dec 3 '21 at 18:04
• 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? Dec 3 '21 at 18:06
• 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 $xf$. Dec 3 '21 at 18:07
• 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. Dec 3 '21 at 18:11

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.

• Many thanks, but exactly what would be wrong in writing: $\Psi(X):=\frac{1}{|G|}\sum_{g \in G}g \cdot \Phi(g\cdot X)$ Dec 3 '21 at 18:31
• 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. Dec 3 '21 at 21:25

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. :)