Proposition 1

Tag 017V, permanent. Cite it as [Bourbaki, Tag 017V].

Lie Groups and Lie Algebras, chapter VIII, § 12. Chevalley orders

Let A be a bigebra, with coproduct $c$ , and $x$ a primitive element (Chap. II, §1, no. 2) of A. Then

c(x^{(n,d)}) =\sum x^{(p,d)}\otimes x^{(q,d)} \tag{4}

p\in \mathbf{N},q\in \mathbf{N},p+q=n

The proposition is trivial for $n\leq 0$ . We argue by induction on $n$ . If formula (4) is true for $n$ , then

(n+ 1)c(x^{(n+1,d)}) =c(x-dn)c(x^{(n,d)})

= (x\otimes 1 + 1\otimes x-dn1\otimes 1)c(x^{(n,d)})

=\sum_{p+q=n}[xx^{(p,d)}\otimes x^{(q,d)}+x^{(p,d)}\otimes xx^{(q,d)}-(p+q)dx^{(p,d)}\otimes x^{(q,d)}]

=\sum_{p+q=n}(x-pd)x^{(p,d)}\otimes x^{(q,d)}+\sum_{p+q=n}x^{(p,d)}\otimes (x-qd)x^{(q,d)}

=\sum_{p+q=n}(p+ 1)x^{(p+1,d)}\otimes x^{(q,d)}+\sum_{p+q=n}(q+ 1)x^{(p,d)}\otimes x^{(q+1,d)}

=\sum_{r+s=n+1}rx^{(r,d)}\otimes x^{(s,d)}+\sum_{r+s=n+1}sx^{(r,d)}\otimes x^{(s,d)}

= (n+ 1)\sum_{r+s=n+1}x^{(r,d)}\otimes x^{(s,d)}

hence formula (4) for $n+ 1$ .

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Cites

Chap. II, §1, no. 2