Remark

Let $\sigma$ be a permutation in $\mathfrak{S}_n$; as the mapping

of $M^n$ into $T^n(M)$ is A-multilinear, it may be written uniquely as

where $u_\sigma$ is an endomorphism of the A-module $T^n(M)$, also denoted by $z \mapsto \sigma . z$. Clearly, if $\sigma$ is the identity element of $\mathfrak{S}_n$, $u_\sigma$ is the identity; on the other hand, writing $y_i = x_{\sigma^{-1}(i)}$, we obtain, for every permutation $\tau \in \mathfrak{S}_n, y_{\tau^{-1}(i)} = x_{\sigma^{-1}(\tau^{-1}(i))}$ and hence $\tau . (\sigma . z) = (\tau \sigma) . z$; in other words, the A-module $T^n(M)$ is a left $\mathfrak{S}_n$-*set* under the operation $(\sigma, z) \mapsto \sigma . z$ (I, § 5, no. 1). The elements of $T^n(M)$ such that $\sigma . z = z$ for all $\sigma \in \mathfrak{S}_n$ are called (contravariant) symmetric tensors of order n; they form a sub-A-module $S'_n(M)$ of $T^n(M)$.

For all $z \in T^n(M)$, we write $s . z = \sum_{\sigma \in \mathfrak{S}_n} \sigma . z$ and call $s . z$ the symmetrization of the tensor $z$; clearly $s . z$ is a symmetric tensor and therefore $z \mapsto s . z$ is an endomorphism of $T^n(M)$ whose image $S''_n(M)$ is contained in $S'_n(M)$; in general, $S''_n(M) \neq S'_n(M)$ (Exercise 5). If $z$ is a symmetric tensor, then $s . z = n! z$; it follows that when n! is invertible in A, the endomorphism $z \mapsto (n!)^{-1} s . z$ is a projector of $T^n(M)$ (II, § 1, no. 8), whose image is $S'_n(M) = S''_n(M)$; moreover the kernel on this projector is just $\mathfrak{J}'_n$. For, obviously $\sigma(\mathfrak{J}'_n) \subset \mathfrak{J}'_n$ for all $\sigma \in \mathfrak{S}_n$ and $\mathfrak{J}'_n$ is by definition generated by the tensors $z - \rho . z$, where $\rho$ is a transposition exchanging two consecutive numbers in $\{1, n\}$; also, if $\sigma, \tau$ are two permutations in $\mathfrak{S}_n$, then $z - (\sigma \tau) . z = z - \sigma . z + \sigma . (z - \tau . z)$, whence it follows (since every permutation in $\mathfrak{S}_n$ is a product of transpositions exchanging two consecutive numbers) that $z - \sigma . z \in \mathfrak{J}'_n$ for all $z \in T^n(M)$ and $\sigma \in \mathfrak{S}_n$. Therefore (always supposing that n! is invertible in A), it is seen that

for all $z \in T^n(M)$, which proves our assertion.

When $n!$ is invertible in $A$, the submodules $S'_n(M)$ and $\mathfrak{g}'_n$ of $T^n(M)$ are therefore supplementary and the restriction to $S'_n(M)$ of the canonical homomorphism $T^n(M) \to S^n(M) = T^n(M)/\mathfrak{g}'_n$ is an $A$-module isomorphism, which allows us in the case envisaged to identify symmetric tensors of order $n$ with the elements of the $n$-th symmetric power of $M$. Note however that this identification is not compatible with multiplication, the product (in $T(M)$) of two symmetric tensors not being symmetric in general and not therefore having as image in $S(M)$ the product of the images of the symmetric tensors considered.