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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.?

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  • $\begingroup$ Previous related question by me: math.stackexchange.com/questions/785930/… $\endgroup$
    – N. Virgo
    Commented May 19, 2014 at 2:24
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    $\begingroup$ I don't really understand what kind of answer you're expecting here. All you know about a function $(X \to P) \to Q$ is that it takes as input a function $X \to P$ and returns an element of $Q$. Anything you want to say about a group action on such functions has to make use of this fact, so you need to refer to the fact that you know how to act on functions $X \to P$ somehow, either by explicitly mentioning it in your notation or by naming that group action. What else could you possibly do? $\endgroup$ Commented May 19, 2014 at 3:10
  • $\begingroup$ @mweiss thanks, fixed. $\endgroup$
    – N. Virgo
    Commented May 19, 2014 at 3:22
  • $\begingroup$ @BrunoJoyal I've fixed both errors. ($Q$ in the second paragraph was supposed to be $P$.) $\endgroup$
    – N. Virgo
    Commented May 19, 2014 at 3:23
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    $\begingroup$ I'm not sure why you can't just do the same thing at a higher level that you've already done: Once you have defined how $G$ acts on $A$ you can write $g(a)$ for $g.a$. Then for any $b \in B$ you have $(g.b)(a)=(b \circ g^{-1})(a)$, so you can write the action of $G$ on $B$ as $g.b = b \circ g^{-1}$. Right? $\endgroup$
    – mweiss
    Commented May 19, 2014 at 3:40

2 Answers 2

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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 $g.c=c\circ g^{-1}$.

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  • $\begingroup$ before this you should b $\endgroup$
    – tangyao
    Commented May 28, 2017 at 10:37
  • $\begingroup$ Without proper formatting, this is difficult to read. $\endgroup$ Commented May 28, 2017 at 11:05
  • $\begingroup$ tangyao, I edited your post to use LaTeX markup to make it more readable. Can you check it's all correct and approve if so? It's a while since I thought about this sort of thing, but I think this approach works and is the right one, though I'd like to mull it over a bit before accepting - thank you for the answer! $\endgroup$
    – N. Virgo
    Commented May 28, 2017 at 13:49
  • $\begingroup$ I think you should prove that the actions you defined are really actions!,I have tried to prove it with the definition of action,then I found they are really actions,then the answer is clear;and your edit is right $\endgroup$
    – tangyao
    Commented May 29, 2017 at 9:25
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I don't know if this is interesting for you, but it might give you some inspiration. Let $G$ be a (finite) group acting faithfully on a (finite) set $X$ (written on the left as g.x). Define the wreath product of $G$ by $G$, as the group $G\wr G = G^X \rtimes G$, with the second $G$ acting as you mentioned above (on the argument of functions). Then there is a natural action of $G\wr G$ on the function set $X^X$: for $f\in X^X$, $x\in X$ $\alpha \in G^X$ and $\beta\in G$ define $$ (\alpha\beta).f(x) = \alpha(\beta.x). (f(\beta.x)). $$

The "bottom" $G^X$ permutes the values of functions and the "top" $G$ permutes the arguments.

This action is "usually" primitive. The nice thing about it is that it makes sense to iterate it! For $n\in \mathbb{N}$, define $W_1 = G$ and $W_{n+1} = G \wr W_n$. Then $W_n$ comes with an action on $X^{(X^{\ldots^X})}$.

This is not exactly what you were asking, but I hope it helps. If I may ask: where did your interest in this come from?

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  • $\begingroup$ +1, this seems interesting and I will work my way through it. It's an old question and I can't clearly remember the thought process that led to it, but it had something to do with wanting to characterise the symmetry group of a cellular automaton's update rule. These act on configurations of the lattice, which can be seen as functions of space, and one can also consider quantities that are functions of the lattice state (hence functions of functions). Some of these may be invariant under the dynamics. I'm not sure why I needed functions of functions of functions, though. $\endgroup$
    – N. Virgo
    Commented Feb 21, 2020 at 7:28

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