Why does $S_n$ act on $\mathbb{R}[x_1,\dots,x_n]$ but not on $\mathbb{R}^n$?

Question

I'm working on a pair of unassessed course problems,

1. Show that $S_n$ acts on polynomials in $n$ variables by permuting variables.

2. For $\sigma\in S_n$ and $\underline{v}=(c_1,\dots,c_n)\in\mathbb{R}^n$ define $$\sigma\star(c_1,\dots,c_n)=(c_{\sigma(1)},\dots,c_{\sigma(n)}).$$ Decide whether $\star$ defines an action of $S_n$ on $\mathbb{R}^n$.

Solutions are given which treat the second group action axiom as follows.

\begin{align} \sigma\star(\sigma'\star f(x_1,\dots,x_n)) & = \sigma\star f(x_{\sigma'(1)},\dots,x_{\sigma'(n)}) \tag{1.1} \\ & = f(x_{\sigma(\sigma'(1))},\dots,x_{\sigma(\sigma'(n))}) \tag{1.2} \\ & = f(x_{(\sigma\sigma')(1)},\dots,x_{(\sigma\sigma')(n)}) \tag{1.3} \\ & = (\sigma\sigma')\star f(x_1,\dots,x_n), \tag{1.4} \end{align} so GA2 is satisfied.

\begin{align} \sigma\star(\sigma'\star(c_1,\dots,c_n) & = \sigma\star(c_{\sigma'(1)},\dots,c_{\sigma'(n)}) \tag{2.1} \\ & = \sigma\star(d_1,\dots,d_n) \text{ where } d_i=c_{\sigma'(i)} \tag{2.2} \\ & = (d_{\sigma(1)},\dots,d_{\sigma(n)}) \tag{2.3} \\ & = (c_{\sigma'(\sigma(1))},\dots,c_{\sigma'(\sigma(n))}) \tag{2.4} \\ & = (c_{(\sigma'\sigma)(1)},\dots,c_{(\sigma'\sigma)(n)}) \tag{2.5} \\ & = (\sigma'\sigma)\star(c_1,\dots,c_n) \tag{2.6} \end{align} and $S_n$ is not commutative, so in general $\sigma\sigma'\neq\sigma'\sigma$, so GA2 is not satisfied.

Why do the two different arguments apply in the two different cases? Why couldn't I swap the arguments around to reach the opposite conclusions?

Answer 1

The discussion that @ArturoMagidin links in the comments is very useful, but I'd like to present a different way of looking at the $\mathbb R^n$ case, because the notation and presentation used in your courses solution make it incredibly difficult for me to think about clearly.

Instead of thinking of $c \in \mathbb R^n$ as an ordered tuple, where you list its components in order, we'll convert that to thinking of it as a function from $\{1, 2, ..., n\}$ to $\mathbb R$, and so instead of referring to $(c_1, c_2, ....)$ we'll say that "the first component of $c$ is $c(1)$, ...".

Now lets take a super-simple $\sigma$, $\sigma =(1,4, 5)$. According to your definition

$$\sigma \star (c_1, c_2, ....) = (c_{\sigma(1)}, c_{\sigma(2), ...)} = (c_4, c_2, ...)$$

So, translating to the alternate notation, if we ask "what's the first component of $\sigma \star c$, i.e. $(\sigma \star c)(1)$, the answer is $c(4)$, i.e. $$(\sigma \star c)(i) = c(\sigma(i))$$ And now, if you remember that elements of $S_n$ can also be thought of as functions from $\{1,2, ...n\}$ to $\{1,2, ...n\}$, we can see that the line above is equivalent to $$ \sigma \star c = c \circ \sigma$$ where the right-hand-side is the compostion of two functions, first $\sigma$ from $\{1,2, ...n\}$ to $\{1,2, ...n\}$, then $c$ from $\{1,2, ...n\}$ to $\mathbb R$.

But notice that $\sigma$ has "flipped sides"! In the linked discussion they play with how this works out depending on whether you're applying things on the left or on the right. But with the conventions you're using, this leads us to

$$ \begin{align} \tau \star (\sigma \star c) &= \tau \star (c \circ \sigma)\\ &= (c \circ \sigma) \circ \tau\\ &= c \circ (\sigma \circ \tau) &\text{(basic rule of function composition)}\\ &= c \circ( \sigma\tau) &(\text{multiplication in } S_n)\\ &= (\sigma \tau) \star c\\ \end{align} $$

which is the same conclusion derived in your answer, but I find it easier to understand when presented this way. Maybe you do too?

(There's still your question of "why doesn't the same problem happen for $\mathbb R[x_1, x_2, ...., x_n]$?". And to be honest, I find this stuff hard to think about, and I don't have an answer for that. I'd like to be able to come up with a similar understanding of the action in the first case that clarifies things, but have yet to find one.)

Answer 2

Let’s begin with a little abstraction. Suppose some group $G$ acts on a set $X$ from the left (that is, $g_1(g_2x) = (g_1g_2)x$). Then, for any set $Y$, this left action on $X$ induces a right action on the set $X\to Y$ of functions from $X$ to $Y$ by $(fg)(x) = f(gx)$. Repeating once more for a set $Z$, this right action on $X\to Y$ induces a left action on $(X\to Y)\to Z$ by $(g f_1)(f_2)=f_1(f_2g)$.

In your case $G=S_n$, $X=\{1, \ldots, n\}$, $\mathbb R^n = X \to \mathbb R$ and $\mathbb R[x_1, \ldots, x_n] \subset (X \to \mathbb N) \to \mathbb R$ (those functions that are non-zero only in a finite number of arguments). From the general considerations above: $S_n$ is a left action on $X$, a right action on $\mathbb R^n$, and a left action on $\mathbb R[x_1, \ldots, x_n]$.

Answer 3

OP answering own question. I think the function case is well-answered by the linked answer of Arturo Magidin. I now address the vector case.

Consider the case $n=3$. Let $$\underline{c}=(c_1,c_2,c_3), \hspace{1em} \sigma=(12), \hspace{1em} \sigma'=(13).$$ Then (applying composed permutations from right to left) $$\sigma\sigma'=(12)(13)=(132), \hspace{1em} \sigma'\sigma=(13)(12)=(123).$$

Now, \begin{align} \sigma\star(\sigma'\star\underline{c}) & = (12)\star((13)\star(c_1,c_2,c_3)) \\ & = (12)\star(c_3,c_2,c_1) \\ & = \begin{cases} \begin{aligned} & (c_3,c_1,c_2)=(123)\underline{c}=(c_{(\sigma'\sigma)(1)},c_{(\sigma'\sigma)(2)},c_{(\sigma'\sigma)(3)}) \\ & \hspace{2em} \text{if we permute by } \textit{starting} \text{ position,} \end{aligned} \\ \begin{aligned} & (c_2,c_3,c_1)=(132)\underline{c}=(c_{(\sigma\sigma')(1)},c_{(\sigma\sigma')(2)},c_{(\sigma\sigma')(3)}) \\ & \hspace{2em} \text{if we permute by } \textit{current} \text{ position.} \end{aligned} \end{cases} \end{align} What confused me was that if $c_3$ (i.e. the element starting in position $3$) is moved to position $1$ then, for the purposes of further permutation, it is now element $1$, as if it were labeled $c_1$.

Thanks to all who commented and answered.