Remark

Tag 0A29, permanent. Cite it as [Bourbaki, Tag 0A29].

Lie Groups and Lie Algebras, chapter II, § 1. Enveloping bigebra of a Lie algebra

(1) $U_n = \sum_{i=0}^n U^i$ (Chapter I, § 2, no. 7, Corollary 4 to Theorem 1).

(2) The mapping $\eta$ is the unique morphism of graded cogebras of $S(g)$ into $U$ such that $\eta(1) = 1$ and $\eta(x) = \sigma(x)$ for $x \in g$ . For if $\eta'$ is a morphism satisfying these conditions, we prove by induction on $n$ that $\eta'(x^n) = \eta(x^n)$ for $x \in g$ and $n > 1$ . As $c_S^+(x^n) = \sum_{i=1}^{n-1} \binom{n}{i} x^i \otimes x^{n-i}$ by (3) and (11),

(\eta \otimes \eta)(c_S^+(x^n)) = (\eta' \otimes \eta')(c_S^+(x^n))

by the induction hypothesis. It follows that

c^+(\eta(x^n)) = c^+(\eta'(x^n))

; it follows that

\eta(x^n) - \eta'(x^n)

is a primitive element of degree

n

and hence is zero (Corollary to Proposition 9).

(3) Let $\psi$ be the canonical isomorphism of the bigebra $TS(g)$ onto the bigebra $S(g)$ (Algebra, Chapter IV, § 5, Corollary 1 to Proposition 12). The mapping

\eta \circ \psi : TS(g) \to U

is called canonical. It is the unique morphism

\eta'

of graded cogebras of

TS(g)

into

U

such that

\eta'(1) = 1

and

\eta'(x) = \sigma(x)

for all

x \in g

(4) Let $V$ be a vector space. The primitive elements of the bigebra $S(V)$ are the elements of degree 1. This follows from the Corollary to Proposition 9 applied to the commutative Lie algebra $V$ .

Let $(e_i)_{i \in I}$ be a basis of the vector $K$ -space $g$ , where the indexing set $I$ is totally ordered. For all $\alpha \in \mathbf{N}^{(I)}$ , we write

e_\alpha = \prod_{i \in I} \frac{\sigma(e_i)^{\alpha(i)}}{\alpha(i)!}.

The

e_\alpha

, for

|\alpha| \leq n

, form a basis of the vector

K

-space

U_n

(Chapter I, § 2, no. 7, Corollary 3 to Theorem 1). Then

e_0 = 1, \quad e_{e_i} = \sigma(e_i) \quad \text{for } i \in I.

As the graded algebra associated with the filtered algebra U is commutative (loc. cit., Theorem 1), for $\alpha, \beta$ in $\mathbf{N}^{(\Omega)}$ ,

e_{\alpha} \cdot e_{\beta} \equiv ((\alpha, \beta)) \cdot e_{\alpha + \beta} \mod U_{|\alpha| + |\beta| - 1}.

where

((\alpha, \beta)) = \prod_{i \in I} \frac{(\alpha(i) + \beta(i))!}{\alpha(i)! \beta(i)!}.

On the other hand, we have immediately

\varepsilon(e_0) = 1, \quad \varepsilon(e_{\alpha}) = 0 \quad \text{for } |\alpha| \geq 1.

Finally, formula (12) implies that, for $\alpha \in \mathbf{N}^{(\Omega)}$ ,

c(e_{\alpha}) = \sum_{\beta + \gamma = \alpha} e_{\beta} \otimes e_{\gamma}.

This formula allows us to determine the algebra $U' = \mathrm{Hom}(U, K)$ dual to the cogebra U (Algebra, Chapter III, § 11, no. 2). For let $K[[X_i]]_{i \in I}$ be the algebra of formal power series in indeterminates $(X_i)_{i \in I}$ (cf. Algebra, Chapter III, § 2, no. 11); if $\lambda \in U'$ , let $f_{\lambda}$ denote the formal power series

f_{\lambda} = \sum_{\alpha} \langle \lambda, e_{\alpha} \rangle X^{\alpha}, \quad \text{where } X^{\alpha} = \prod_{i \in I} X_i^{\alpha(i)}

and the summation index $\alpha$ runs through $\mathbf{N}^{(\Omega)}$ .

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

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