Equivariant map between two vector spaces definition and formulation. Reading the following paper (Proposition 2.2)
https://arxiv.org/pdf/1804.10306.pdf
I'm stuck trying to understand the following:
We have a compact group $\Gamma$, $V,U$ two vector spaces carrying a linear representation of $\Gamma$.
Then a map $f : V \rightarrow U$ is called $\Gamma$-equivariant if $f(R_\gamma \textbf{x}) = R_\gamma f(\textbf{x})$.
I don't understand why both in the propostion and in the explicit form of equivariant map $f$ suddenly a $R_\gamma^{-1}$ appears. Where is this term coming from? I thought initially that we take the inverse because of the natural induced action on functions which typically requires the inverse to be taken in order to satisfy the associativity property. But in the above definition of equivariance, which I've reported here exactly as stated in the paper, I only see $R_\gamma$.
EDIT: some comments given @PaulGarrett answer:
So let's say we consider two linear automorphisms on the two vector spaces (fixing $\gamma$):
\begin{equation}
R_\gamma : V \rightarrow V \, \, \, , \, \, \, R_\gamma : U \rightarrow U
\end{equation}
Then consider a function $f : V \rightarrow U$. Again, according to what I had understood about the extension of actions to functions, my definition of equivariance in this case would be
$$f(R^{-1}_\gamma \textbf{x}) = R^{-1}_\gamma f(\textbf{x})$$
but the main point to be taken into account, as mentioned in the answer is that $R^{-1}_\gamma = R_{\gamma^{-1}}$ cause a representation is a group homomorphism and therefore, saying for example $\theta = \gamma^{-1}$ we could redefine the above thing to be $f(R_\theta\textbf{x}) = R_\theta f(\textbf{x})$ as equivariance definition.
This is pretty clear to me now, anyway, by looking at again at Proposition 2.2 in the paper, he uses both $R^{-1}_\gamma$ and $R_\gamma$ to define his $\hat{f}: V \rightarrow U$ equivariant map:
$$\hat{f}(\textbf{x}) = \int_\Gamma \sum_{n=1}^N R^{-1}_\gamma \textbf{y}_n \sigma(l_n(R_\gamma(\textbf{x})+h_n) \, d\gamma$$
Our $f$ in this case is the non-linear pointwise activation function $\sigma: \mathbb{R} \rightarrow \mathbb{R}$.
So why does he still mix things up here? ^^"
Apologize whether this might be a trivial observation.
P.S. also, probably you meant $R_\gamma \circ f = f \circ R_\gamma$ in your answer.
 A: Ah, yes, an iconic issue! :) I assure you that, yes, while there is some mathematical content here, the extent of it is indicated by your correct remark about wanting/needing associativity.
Yes, a map $f:V\to U$ could be acted upon by $(R_\gamma f)(x)=f(R_\gamma^{-1}x)$, and this would be more compatible with other actions of a group on a set, and thereby on functions on the set. And, yes, consistent with that, perhaps $R_\gamma$ should act on the output $f(x)$ by $R_\gamma^{-1}(f(x))$.
But/and the assertion that $f$ is equivariant becomes $R_\gamma^{-1}f(x)=(R_\gamma^{-1}f)(x)$, or $R_\gamma^{-1}\circ f = f\circ R_\gamma^{-1}$. Whether or not everyone has thought through the inverse-for-associativity thing, it is clear that the collection of all these conditions is the same as $R_\gamma\circ f=f\circ \gamma$, since $R_{\gamma^{-1}}=R_\gamma^{-1}$.
As a notational variant: yes, we could also require an associative action of $\Gamma$ on linear maps $f:U\to V$, so that $f$ is $\Gamma$-equivariant if and only if it is invariant under this action (thus allowing the possibility of $f$ transforming by some other representation of $\Gamma$...) This might be more of a reason to be more scrupulous about inverse-or-not. :)
EDIT: yes, first, oops, meant $R_\gamma\circ f=f\circ R_\gamma$.
Filling out my last remark: again, this is substantially about notation, yes, but misjudging non-commutativity in a non-commutative setting can lead to ridiculous (false) conclusions.
First, the iconic pseudo-problem that with a group $G$ acting on a set $S$, "the" action of $G$ on functions_on $S$ needs an inverse, however we denote the action on the set. (The requirement of associativity is definitive...) For example, it is reasonable to write $g\times s\to g\cdot s$ for the action of $g\in G$ on $s\in S$. The "definition/requirement" of a group action on a set is the associativity $g\cdot (h\cdot s)=(gh)\cdot s$. So far, this is abstract, and is not really about left-or-right at all.
To have $G$ act on functions on $S$, with associativity, as usual, requires $(g\cdot f)(s)=f(g^{-1}s)$. As sometimes happens, we can create an illusion that the inverse is not necessary, by "declaring" that the original group action on the set is "on the right", or ... something. That does not change the mathematics. :)  (More can be said about circumstances in which left actions are equivalent to right actions of related objects, and vice-versa...)
Continuing as in the revised question, for example, if we have an action of a group $G$ on $V$ and $H$ on $W$, in fact there is an action of $G\times H$ on $\varphi\in {\mathrm {Hom}}(V,W)$ by
$$(g,h)\cdot \varphi \;=\; h\circ \varphi\circ g^{-1} $$
Again, the $g^{-1}$ is for associativity. :)
When both vector spaces are $G$-spaces/representations, then things simplify, and $g\circ \varphi g^{-1}$ appears.
Of course, as usual, the family of conditions $g\circ \varphi \circ g^{-1}$ for all $g$ is equivalent to the family of conditions with $g$ replaced by $g^{-1}$.
(There are analogous, more complicated, issues when it is a ring of some sort acting...)
"That's all I know". :)
