Proposition 2

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

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

— (i) The A-algebra $TS(M)$ is associative, commutative and unital.

(ii) Let $p_1, \ldots, p_n$ be integers $> 0$ , and let $S_{p_1| \ldots |p_n}$ be the set of all $\sigma \in S_{p_1+\cdots+p_n}$ leaving the following intervals of $\mathbf{N}$ stable:

(1, p_1), (p_1+1, p_1+p_2), \ldots, (p_1+\cdots+p_{n-1}+1, p_1+\cdots+p_n).

Let $z_1 \in \mathbf{TS}^{p_1}(M) , \ldots , z_n \in \mathbf{TS}^{p_n}(M)$ , then

z_1 z_2 \ldots z_n = \operatorname{Tr}_{\mathfrak{S}_{p_1 + \cdots + p_n}/\mathfrak{S}_{p_1}| \cdots | p_n} (z_1 \otimes z_2 \otimes \ldots \otimes z_n) .

In particular if $x_1, \ldots, x_r \in M$ , we have $x_1 \ldots x_r = s(x_1 \otimes \ldots \otimes x_r)$ .

The assertion (ii) is clear for $n = 1$ . Assume that the relation

z_2 \ldots z_n = \operatorname{Tr}_{\mathfrak{S}_{p_2 + \cdots + p_n}/\mathfrak{S}_{p_2}| \cdots | p_n} (z_2 \otimes \ldots \otimes z_n)

has been proved, and let us identify $\mathfrak{S}_{p_2 + \cdots + p_n}$ with the subgroup of $\mathfrak{S}_{p_1 + \cdots + p_n}$ consisting of all permutations whose restriction to $(1, p_1)$ is the identity. Then

\operatorname{Tr}_{\mathfrak{S}_{p_1}|p_2+\cdots+p_n/\mathfrak{S}_{p_1}|p_2| \cdots | p_n} (z_1 \otimes z_2 \otimes \ldots \otimes z_r) =

= z_1 \otimes \operatorname{Tr}_{\mathfrak{S}_{p_2+\cdots+p_n}/\mathfrak{S}_{p_2}| \cdots | p_n} (z_2 \otimes \ldots \otimes z_r) = z_1 \otimes (z_2 \ldots z_n) .

Hence we have

z_1 z_2 \ldots z_r = z_1 (z_2 \ldots z_r) =

= \operatorname{Tr}_{\mathfrak{S}_{p_1+p_2+\cdots+p_n}/\mathfrak{S}_{p_1}|p_2-\cdots+p_n} (z_1 \otimes (z_2 \ldots z_n))

= \operatorname{Tr}_{\mathfrak{S}_{p_1+\cdots+p_n}/\mathfrak{S}_{p_1}|p_2+\cdots+p_n} (\operatorname{Tr}_{\mathfrak{S}_{p_1}|p_2+\cdots+p_n/\mathfrak{S}_{p_1}|p_2| \cdots | p_n} (z_1 \otimes z_2 \otimes \ldots \otimes z_n))

= \operatorname{Tr}_{\mathfrak{S}_{p_1-\cdots+p_n}/\mathfrak{S}_{p_1}| \cdots | p_n} (z_1 \otimes z_2 \otimes \ldots \otimes z_n)

by Prop. 1, (ii) of IV, p. 42. Thus (ii) is established.

In particular,

z_1 (z_2 z_3) = \operatorname{Tr}_{\mathfrak{S}_{p_1+p_2+p_3}/\mathfrak{S}_{p_1}|p_2|p_3} (z_1 \otimes z_2 \otimes z_3) ,

and in the same way we show that

(z_1 z_2) z_3 = \operatorname{Tr}_{\mathfrak{S}_{p_1+p_2+p_3}/\mathfrak{S}_{p_1}|p_2|p_3} (z_1 \otimes z_2 \otimes z_3) .

Hence the algebra $\mathbf{TS}(M)$ is associative.

Let $o$ be the element of $\mathfrak{S}_{p_1+p_2}$ such that

\begin{align*} \sigma(1) &= p_2 + 1, \; \sigma(2) = p_2 + 2, \; \ldots, \; \sigma(p_1) = p_2 + p_1 , \\ \sigma(p_1 + 1) &= 1, \; \sigma(p_1 + 2) = 2, \; \ldots, \; \sigma(p_1 + p_2) = p_2 . \end{align*}

Then

z_2 z_1 = \operatorname{Tr}_{\mathfrak{S}_{p_1+p_2}/\mathfrak{S}_{p_2}|p_1} (z_2 \otimes z_1)

= \operatorname{Tr}_{\mathfrak{S}_{p_1+p_2}/\sigma \mathfrak{S}_{p_1}|p_2 \sigma^{-1}\sigma} (z_1 \otimes z_2)

= \operatorname{Tr}_{\mathfrak{S}_{p_1+p_2}/\mathfrak{S}_{p_1}|p_2} (z_1 \otimes z_2) \quad \text{by Prop 1, (i)}

= z_1 z_2 .

Hence the algebra $\mathbf{TS}(M)$ is commutative.

It is to be noted that the canonical injection of $\mathbf{TS}(M)$ in $\mathbf{T}(M)$ is not in general an algebra homomorphism. Worse still, $\mathbf{TS}(M)$ is not in general stable under the multiplication of $\mathbf{T}(M)$ .

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

Cited by

Cites

Prop. 1, (ii) of IV, p. 42