Representation of $S_n$ by $V^{\otimes n}$, Let $V$ be a real and finite dimensional vectorspace. Then
$$
\sigma.(v_1 \otimes \cdots \otimes v_n) := (-1)^{\sigma} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(n)}.
$$
My question: Why is this action on $V^{\otimes n}$ a representation of $S_n$ ? How do I check this ?
 A: This action on $V^{\otimes n}$ gives a homomorphism from $S_n$ to the automorphism group of $V^{\otimes n}$.
Define $$\phi : S_{n} \to \text{Aut}(V^{\otimes n})$$ $$\phi(\sigma)\biggl(v_1 \otimes \cdots \otimes v_n\biggr) := (-1)^{\sigma} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(n)}$$
and then extend $\phi(\sigma)$ by linearity.
Then you have to check that $\phi(\lambda \sigma) = \phi(\lambda) \cdot \phi(\sigma)$ and that each $\phi(\sigma)$ is bijective.
A: *

*Let $\sigma$ be an element of $S_n$. There is a function $$m_\sigma:\underbrace{V\times\cdots\times V}_{\text{$n$ factors}}\to V^{\otimes n}$$ such that for all $v_1$, $\dots$, $v_n\in V$, $$m_\sigma(v_1,\dots,v_n)=(-1)^{\operatorname{sgn}\sigma} v_{\sigma^{-1}(1)}\otimes\cdots\otimes v_{\sigma^{-1}(n)}.$$ You can check that $m_\sigma$ is $n$-multilinear, so it induces a linear map $\bar m_\sigma:V^{\otimes n}\to V^{\otimes n}$.

*Next, you can check that if $\sigma$ and $\tau$ are elements of $S_n$, then $\bar m_\sigma\circ\bar m_\tau=\bar m_{\sigma\tau}$, and that if $e$ is the identity element in $S_n$, then $\bar m_{e}$ is the identity map of $V^{\otimes n}$.

*Using that, you can check that the linear maps $\bar m_{\sigma}$ and $\bar m_{\sigma^{-1}}$, both $V^{\otimes n}\to V^{\otimes n}$, are mutually inverse.

*It follows from this that we have a function $\rho:\sigma\in S_n\mapsto\bar m_{\sigma}\in\operatorname{Aut}(V^{\otimes n})$. Now it is easy to show that this function is even a group homomorphism.
A: Once you see that this determines a linear map $U(\sigma)$ from $V^{\otimes n}$ to itself, all you have to verify is that $\sigma \to U(\sigma)$ is a homomorphism, i.e. that $U(\sigma \tau) = U(\sigma) U(\tau)$ for permutations $\sigma$ and $\tau$.
A: I think that an explicit example might be helpful.  Let's look at $n = 3$.  I will write the elements of $S_3$ in cycle notation:
$$
S_3 = \{ 1, \underbrace{(12), (13), (23)}_{\text{transpositions}}, \underbrace{(123), (132)}_{3-\text{cycles}} \}.
$$
Let's fix the dimension of the underlying vector space as well.  Say, $\dim V = 3$ with basis $\{ v_1, v_2, v_3 \}$.  Now, $S_3$ will act on the $3$-fold tensor product:  $V^{\otimes 3} = V \otimes V \otimes V$, which has dimension $3^3 = 27$ and basis:
$$
\left\{\begin{align}
v_1 \otimes &v_1 \otimes v_1 \\
v_1 \otimes &v_1 \otimes v_2 \\
v_1 \otimes &v_1 \otimes v_3 \\
v_1 \otimes &v_2 \otimes v_1 \\
v_1 \otimes &v_2 \otimes v_2 \\
&\vdots \\
v_3 \otimes &v_3 \otimes v_2 \\
v_3 \otimes &v_3 \otimes v_3
\end{align}\right\}.
$$
The representation can be thought of as a group homomorphism $S_3 \to GL(V^{\otimes 3})$, so we can associate a $27 \times 27$ matrix to each permutation, by using the given ordered basis.
As mentioned in the comments, in order to act on the left (write your functions on the left), you have to use the inverse of each permutation.  Study the following calculation to see why.
$$
v_1 \otimes v_2 \otimes v_3 \mathrel{\mathop{\mapsto}^{(12)}} - v_2 \otimes v_1 \otimes v_3 \mathrel{\mathop{\mapsto}^{(23)}} v_2 \otimes v_3 \otimes v_1
$$
Now, $(23)(12) = (132)$ in $S_3$, and
$$
v_1 \otimes v_2 \otimes v_3 \mathrel{\mathop{\mapsto}^{(132)}} v_2 \otimes v_3 \otimes v_1
$$
only if the permutation acts by its inverse.  Here's the general calculation which uses the fact that the sign $\sigma \mapsto (-1)^{\sigma}$ is a group homomorphism.
$$
\begin{align}
(\sigma \tau).(v_{i_1} \otimes v_{i_2} \otimes v_{i_3}) &= (-1)^{\sigma \tau} v_{(\sigma \tau)^{-1}(i_1)} \otimes v_{(\sigma \tau)^{-1}(i_2)} \otimes v_{(\sigma \tau)^{-1}(i_3)} \\
&= (-1)^{\sigma} (-1)^{\tau} v_{\tau^{-1} (\sigma^{-1}(i_1))} \otimes v_{\tau^{-1} (\sigma^{-1}(i_2))} \otimes v_{\tau^{-1} (\sigma^{-1}(i_3))} \\
&= \tau.((-1)^{\sigma} v_{\sigma^{-1}(i_1)} \otimes v_{\sigma^{-1}(i_2)} \otimes v_{\sigma^{-1}(i_3)}) \\
&= \sigma.(\tau.(v_{i_1} \otimes v_{i_2} \otimes v_{i_3}))
\end{align}
$$
I hope that this helps.
