How does the symmetric group act on tuples?

Question

Given a set $X$, I've seen people write the action of a permutation $ \sigma \in S_{n}$ on an $n$-tuple of elements of $X$ as $$ \;\; \sigma (x_{1},...,x_{n})=(x_{\sigma^{-1}(1)},...,x_{\sigma^{-1}(n)}). \;\;$$ But this does not seem to me to give a left action of $S_{n}$ on $n$-tuples, since $$\sigma(\tau(x_{1},...,x_{n}))=\sigma(x_{\tau^{-1}(1)},...,x_{\tau^{-1}(n)})=(x_{\sigma^{-1}\tau^{-1}(1)},...,x_{\sigma^{-1}\tau^{-1}(n)})$$ is not in general equal to $$(\sigma \tau)(x_{1},...,x_{n})=(x_{(\sigma \tau)^{-1}(1)},...,x_{(\sigma \tau)^{-1}(n)})=(x_{\tau^{-1}\sigma^{-1}(1)},...,x_{\tau^{-1}\sigma^{-1}(n)}).$$

On the other hand, if we don't write out symbols, but think of a tuple as a map from the $n$-element set $\{1,...,n\}$ to $X$, then the first formula seems to make sense: the $\sigma^{-1}$ comes from precomposing with the right action of $S_{n}$ on the set $\{1,...,n\}$ determined by the left action of $S_{n}$ on $\{1,...,n\}$, where we use the convention that functions act on the left of elements.

So the question is whether the first formula defines a left or a right action on tuples. The symbols seem to suggest a right action while interpreting tuples as maps seems to suggest a left action. I think that in my formulas above I must be getting confused.

Answer 1

The first formula defines a left action on tuples. Note that $\sigma \in S_n$ acts on the positions, not on the indices. You did wrong in computing $\sigma\bigl(\tau(x_1, \ldots, x_n)\bigr)$.

We have $\tau(x_1, \ldots, x_n) = (x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)})$ by definition. Now $\sigma$ acts from the left on this tuple. Acting on a tuple, $\sigma$ "moves" the entry in position $\sigma^{-1}(1)$ to the front, right? But which $x_j$ is that? If we write $y_i = x_{\tau^{-1}(i)}$ we have $$ \sigma(x_{\tau^{-1}(1)}, \ldots, x_{\tau^{-1}(n)}) = \sigma(y_1, \ldots, y_n) = (y_{\sigma^{-1}(1)}, \ldots, y_{\sigma^{-1}(n)}) $$ So the new first entry is $y_{\sigma^{-1}(1)} = x_{\tau^{-1}\sigma^{-1}(1)}$, giving $$ \sigma\bigl(\tau(x_1, \ldots, x_n)\bigr) = (x_{\tau^{-1}\sigma^{-1}(1)}, \ldots, x_{\tau^{-1}\sigma^{-1}(n)}) = (x_{(\sigma\tau)^{-1}(1)}, \ldots, x_{(\sigma\tau)^{-1}(n)}). $$

Answer 2

Your computation is wrong. You read $\sigma (x_1,...,x_n)=(x_{\sigma^{-1}(1)},...,x_{\sigma^{-1}(n)})$ as if the index $i$ is an atttribute of the entry $x_i$, and as if $\sigma$ acts on individual entries (leaving them in place!) by applying $\sigma^{-1}$ to the attribute. The way it is written, that interpretation is tempting, but if you look at what you really want to describe it is clear that this cannot be right.

The index is a selection key that tells you which one of the values is chosen; the values themselves are not affected at all. So you should interpret the rule as a follows: the first component of the result is the one that was previously at position $\sigma^{-1}(1)$, and more generally the component at position$~i$ of the result is the one that was previously at position $\sigma^{-1}(i)$. Therefore one gets $$ \sigma(x_{\tau^{-1}(1)},...,x_{\tau^{-1}(n)})=(x_{\tau^{-1}(\sigma^{-1}(1))},...,x_{\tau^{-1}(\sigma^{-1}(n))}), $$ because the component that is in position $\sigma^{-1}(1)$ in the argument tuple on the left is $x_{\tau^{-1}(\sigma^{-1}(1))}$: more generally the component that is in position $k$ on the left is $x_{\tau^{-1}(k)}$, and with $k=\sigma^{-1}(i)$ this gives that the component that is in position $\sigma^{-1}(i)$ on the left is $x_{\tau^{-1}(\sigma^{-1}(i))}$, which is the value that goes to position$~i$ on the right.

See also this answer.

Answer 3

Given $ \sigma \in S_n $ and $n $-tuple $ (x_1, \ldots, x_n) $, it is natural to define

$$\sigma * (x_1, \ldots, x_n) := [ (x_1, \ldots, x_n) \text{ after sending each } x_i \text{ into slot } \sigma (i) ] $$

That is, $ (x_1 , \ldots, x_n) \rightsquigarrow \sigma * (x_1, \ldots, x_n) $ amounts to sending whatever is in slot $ i $ to slot $ \sigma (i) $.

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We see $ \sigma * (\tau * (x_1, \ldots, x_n)) = (\sigma \tau)*(x_1, \ldots, x_n) $

[ Because in $ (x_1, \ldots, x_n) \rightsquigarrow \sigma * (\tau * (x_1, \ldots, x_n)) $, $ x_i $ is first sent to slot $ \tau(i) $ and then to slot $ \sigma (\tau(i)) $. This is also what $ (x_1, \ldots, x_n) \rightsquigarrow (\sigma \tau)*(x_1, \ldots, x_n) $ does. ]

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Also $ \sigma * (x_1, \ldots, x_n) = (x_{\sigma^{-1} (1)}, \ldots, x_{\sigma^{-1} (n)} ) $

[ Because the $ x_t $ which gets sent to slot $ 1 $ satisfies $ \sigma (t) = 1 $ ie $ t = \sigma^{-1} (1) $. And so on. ]