Given a group acting on a set $X$, there is a standard way to define an action of the group on the set of functions of $X$. This can be extended to the set of functions of functions of $X$ as I show below, and it can also be extended to functions of functions of functions (and of course to higher orders beyond that). However, for the latter (and higher) cases, I can't see a way to write out the action explicitly in terms of function compositions, rather than in terms of the action on "lower-order" functions. My question is about whether it's possible to do this.
The following paragraphs present the problem in greater detail. Consider a group $G$ acting upon a set $X$. If we consider the set $A$ of functions $a:X\to P$ for some set $P$, there is a natural action of $G$ upon $A$ given by $(g.a)(x) = a(g^{-1}.x)$ for all $g\in G$, $a\in A$, $x\in X$. Here the period '$.$' is used to represent both the action of $G$ upon $X$ and the action of $G$ upon $A$. As a concrete example, let $X$ be the set of faces of a cube and $G$ be its group of rotational symmetries. Then $A$ can be thought of as the set of colourings of the cube's faces, with $P$ being the set of colours.
If we write the action of $G$ upon $X$ as $g(x)$ instead of $g.x$ then we can write the action of $G$ upon $F$ as $g.a = a\circ g^{-1}$. This is useful because it allows us to eliminate $x$ from the notation, and allows us to think in terms of function composition rather than the more abstract notion of a group action.
Let us now consider the set $B$ of functions of functions of $X$, that is, the set of functions $b:A\to Q$ for some set $Q$. An example might be a functiom that counts the number of blue faces that are adjacent to red faces. We want to define a natural action of $G$ upon $B$.
Since we already have an action of $G$ on $A$ we can apply the same trick again and write $(g.b)(a) = b(g^{-1}.a)$, for all $a\in A$, $b\in B$, $g\in G$. In terms of funtion composition this becomes $(g.b)(a) = b(a\circ g)$, but I can't see an obvious way to eliminate $a$ from the notation as we were able to do with $x$ above.
Finally, let us consider the set $C$ of functions $c:B\to R$ for some set $R$. That is, functions of functions of functions of $X$. As before we can write $(g.c)(b) = c(g^{-1}.b)$. However, what I can't see is how to write out this action explicitly in terms of function composition, rather than in terms of the action on $B$. That is, I want to get rid of the '$.$' in the right-hand side of this equation, but I can't see a way to do it.
My question is whether it is possible to do this, and if so, how. If it can be done for functions of functions of functions, can it also be done for functions of functions of functions of functions, etc.?
Best Answer
you define $g.x=g(x)$, so you can also define that $g.a=g(a)$, then $$(g.b)(a)=b(g^{-1} .a) =(b\circ g^{-1})(a)$$ so you have $$g.b=b\circ g^{-1}.$$ Furthermore, $$(g.c)(b)=c(g^{-1}.b)$$ then we can as before define $g.b=g(b)$ and we have $(g.c)(b)=c(g^{-1}(b))$, then we get $g.c=c\circ g^{-1}$.