Natural permutation representation definition? I was wondering about a definition I encountered towards the beginning of Gruson-Serganova:
for G = $S_n$, and $k$ a field, and $s \in S_n$ and $(x_1, ..., x_n) \in k^n$, the natural permutation representation is defined as $\rho_s$ $(x_1, ..., x_n)$ = $(x_{s^{-1} (1)},...,x_{s^{-1}(n)})$.
I understand the permutation representation, but why do the inverses show up here? What makes this 'natural' if the permutation representation isn't? I asked my professor about why the inverses show up, and he said that otherwise it wouldn't be a homomorphism, but I see no reason why it wouldn't end up being a homomorphism.
I feel like I'm missing something pretty basic here.
 A: For $\rho$ to be a representation you want that $\rho_{gh}=\rho_g\rho_h$.
If $e_1,\ldots,e_n$ be the standard basis of $k^n$
The action of $g\in S_n$ is defined by $\rho_g(e_i)=e_{g(i)}$. You can check this gives a representation since $\rho_g\rho_h(e_i) = \rho_g(e_{h(i)}) = e_{gh(i)}=\rho_{gh}(e_i)$. (This does not work if you replace it with the inverse!)
Take $n=3$ and $g=(123)$ for example.
Then $g(e_1)=e_2$ means that $\rho_g(1,0,0)=(0,1,0)$.
In general, the action of $g$ in coordinates is
$\rho_g(x_1,x_2,x_3)= (x_3,x_1,x_2)=(x_{g^{-1}(1)},x_{g^{-1}(2)},x_{g^{-1}(3)})$. This is were the inverses come from
A: For a concrete example I refer to this answer I gave to a similar question. Here is an argument explaining why the definition with inverses is in fact the more natural one.
First of all, I would like to address of the relation  between permutations as values versus permutations as operations. Mathematically permutations of a set $X$ are maps $X\to X$, and hence logically compose as maps usually do, the one written on the right being applied first (unless one belongs to the school of people writing function applications with the functions to the right of their argument, which I will exclude here). While "permuting objects" has a clear informal sense as activity, this does not yet associate a particular permutation to each permutation action. It seems however natural that the action corresponding to a permutation of $X=\{1,\ldots,n\}$ that sends an element $i$ to $k$, takes the object at position $i$ and moves it to position $k$. I will therefore assume that "permuting by a permutation $\sigma$" means that for each $i$ the value initially at position $i$ ends up at position $\sigma(i)$. This rule has the natural order of composition: permuting by $\sigma\circ\tau$ can be achieved by first moving any value from position $i$ to position $\tau(i)$ (permuting by $\tau$) from where it then is moved to position $\sigma(\tau(i))$ (further permuting by $\sigma$).
If one writes that as a formula, the permutation is
$$
  (x_1,x_2,\ldots,x_n)\mapsto
  (x_{\sigma^{-1}(1)},x_{\sigma^{-1}(2)},\ldots,x_{\sigma^{-1}(n)}).
$$
That may seem surprising at first. But the right hand expression has to what value arrives at each position; for the first position, the answer is "whatever value was moved to the first position", which is the value initially at position $\sigma^{-1}(1)$, in other words $x_{\sigma^{-1}(1)}$; a similar description goes for other output positions.
The $K$-linear action of $\mathbf S_n$ is a special case of this permutation action, on $n$-tuples of scalars in $K^n$. It is a morphism of groups $\mathbf S_n\to\mathbf{GL}(n,k)$ by virtue of respecting the natural order of composition, as explained. One can also characterise it by the fact that it sends the canonical basis vector $e_i$ to the basis vector $e_{\sigma(i)}$, as one would expect. One would hope that this obviously natural way of making the correspondence would inspire those who define the "permutation matrix of $\sigma$" to be the matrix for this action of $\sigma$; that matrix is $(\delta_{i,\sigma(j)})_{i,j=1}^n$, with $e_{\sigma(j)}$ as its column$~j$ for each$~j$. But this is unfortunately not the case. The Wikipedia article on Permutation Matrix is an example of a wrong choice made, while presenting this as one of two equivalent options, which it is not.
