Proposition 12

Tag 06D9, permanent. Cite it as [Bourbaki, Tag 06D9].

Algebra, chapter IV, § 5. Symmetric tensors and polynomial mappings

— (i) If $u \in S^n(M)$ , then $\psi_M(\varphi_M(u)) = n!u$ .

(ii) If $v \in TS^n(M)$ then $\varphi_M(\psi_M(v)) = n!v$ .

Let $x_1, \ldots, x_n \in M$ and let $u$ be the product $x_1 \ldots x_n$ calculated in $S(M)$ . Then $\varphi_M(u)$ is the product $x_1 \ldots x_n$ calculated in $TS(M)$ , that is

\sum_{\sigma \in \mathfrak{S}_n} x_{\sigma(1)} \otimes \ldots \otimes x_{\sigma(n)}.

Hence $\psi_M(\varphi_M(u))$ equals $\sum_{\sigma \in \mathfrak{S}_n} x_{\sigma(1)} \ldots x_{\sigma(n)}$ calculated in $S(M)$ , that is

n! x_1 \ldots x_n = n!u.

Let $v = \sum_{i=1}^p x_1^i \otimes x_2^i \otimes \ldots \otimes x_n^i$ be an element of $TS^n(M)$ , where the $x_j^i$ belong to $M$ ; then $\psi_M(v)$ is equal to $\sum_{i=1}^p x_1^i x_2^i \ldots x_n^i$ calculated in $S(M)$ , whence

\varphi_M(\psi_M(v)) = \sum_{i=1}^p s(x_1^i \otimes x_2^i \otimes \ldots \otimes x_n^i) = s(v) = n!v.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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