I have a question about the difference in the definition of the action of the symmetric group on abstract tensor spaces and concrete tensor spaces consisting of multilinear maps.
Let $n$ be an integer greater than or equal to $1$, $V$ be a vector space over the real numbers $\mathbb{R}$, and $S_n = \{ \sigma : \{1,2,…,n\} \rightarrow \{1,2,…,n\} \mid \sigma \text{ is a bijection} \}$ (the symmetric group of order $n$). Consider $M$ as the tensor product of $n$ copies of $V^*$. The action of $S_n$ on $M$ is defined for $\sigma \in S_n$ and $u_1, \ldots, u_n \in V^*$ by
$$ \sigma(u_1 \otimes \cdots \otimes u_n) = u_{\sigma(1)} \otimes \cdots \otimes u_{\sigma(n)} $$
and $u \in M$ is defined as symmetric if $\sigma u = u$ for all $\sigma \in M$.
On the other hand, in texts like Tu's 'An Introduction to Manifolds', when considering $L = \{f: V^n \rightarrow \mathbb{R} \mid f \text{ is a } n\text{-linear map} \}$, the action of $S_n$ on $L$ is defined for $\sigma \in S_n$ and $f \in L$ as
$$(\sigma f)(x_1, \ldots, x_n) = f(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) \quad (x_i \in V),$$
and $f \in L$ is defined as symmetric if $\sigma f = f$ for all $\sigma \in S_n$.
Here, $M$ and $L$ are isomorphic via the homomorphism $\phi: M \rightarrow L$ defined by $$(\phi(f_1 \otimes \cdots \otimes f_n))(x_1, \ldots, x_n) = f_1(x_1) \cdots f_n(x_n) \quad (x_i \in V).$$ Therefore, I expected that if $\Phi(u) = f$, then $\phi(\sigma u)$ should equal $\sigma f$. However, in reality, $\phi(\sigma u) = \sigma^{-1}f$.
For example, consider $n=3$, $u = e_1 \otimes e_2 \otimes e_3$ (where $e_i$ are basis elements of $V^*$), and $\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}
$. Here, $f(x_1, x_2, x_3) := (\phi(u))(x_1, x_2, x_3) = e_1(x_1)e_2(x_2)e_3(x_3)$, and $(\sigma f)(x_1, x_2, x_3) = e_2(x_1)e_3(x_2)e_1(x_3)$, but $(\phi(\sigma u))(x_1, x_2, x_3) = e_3(x_1)e_1(x_2)e_2(x_3)$, so $\sigma f \neq \phi(\sigma u) = \sigma^{-1}f$.
Thus, it seems to me more appropriate to define the action on $L$ as $(\sigma f)(x_1,\ldots,x_n) = f(x_{\sigma^{-1}(1)},\ldots,x_{\sigma^{-1}(n)})$ to maintain the correspondence. Defining it this way still results in an equivalent definition of symmetry (since $\sigma f = f \text{ for any } \sigma \in S_n$ is equivalent to $\sigma^{-1} f = f \text{ for any } \sigma \in S_n$).
Why is it not defined this way? Are there any advantages to defining it as in Tu's 'An Introduction to Manifolds'?
Best Answer
This is a place that is particularly susceptible to typos, mistakes and misinterpretations. In what follows, the composition of permutations is defined as $(\sigma\tau)(i) = \sigma(\tau(i))$ (as it should be, but you can find the opposite order in some sources).
The correct way to define the left action of the symmetric group $S_n$ on tensors is then $$\sigma (v_1 \otimes v_2 \otimes \dots \otimes v_n) = v_{\sigma^{-1}(1)} \otimes v_{\sigma^{-1}(2)} \otimes \dots \otimes v_{\sigma^{-1}(n)}$$ It is a bit hard to see, but this definition coincides with the informal definition: move the $i$-th factor of the tensor to the $\sigma(i)$-th place. It is easier to understand in this form: $$\sigma (v_1 \otimes v_2 \otimes \dots \otimes v_n) = w_1 \otimes w_2 \otimes \dots \otimes w_n \Leftrightarrow v_i = w_{\sigma(i)} \Leftrightarrow w_i = v_{\sigma^{-1}(i)}$$ If we forget the inverses like in your question, we'll obtain a right action!
Now to get an action on multilinear functions $f \colon V^{\times n} \to k$, we need to take another inverse and get $$\sigma f(x_1, \dots, x_n) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)})$$ and accordingly for the corresponding linear map on tensors $F \colon V^{\otimes n} \to \mathbb{R}$ we define $$\sigma F(x_1 \otimes \dots \otimes x_n) = F(x_{\sigma(1)} \otimes \dots \otimes x_{\sigma(n)}) = F(\sigma^{-1} x),$$ so that $\sigma F(\sigma x) = F(x)$.