Bourbaki

Remark

Tag 0A69, permanent. Cite it as [Bourbaki, Tag 0A69].

Lie Groups and Lie Algebras, chapter II, § 6. The Hausdorff series

(1) We have defined (§ 3, no. 2, Remark) a projector QQ of A(X)A(X) onto L(X)L(X) such that Q(am)=0Q(a^m) = 0 for a∈L(X)a \in L(X) and m⩾2m \geqslant 2 and Q(1)=0Q(1) = 0. Then H=Q(exp⁡H)=Q(exp⁡U.exp⁡V)H = Q(\exp H) = Q(\exp U.\exp V), whence immediately

(14)Hr,s=Q(UrVsr!s!)for r+s⩾1. (14)\quad H_{r,s} = Q \left( \frac{U^r V^s}{r! s!} \right) \quad \text{for } r + s \geqslant 1.

(2) We have

(15)H(U,V)≡U+V+12[U,V]+112[U,[U,V]]+112[V,[V,U]]−124[U,[V,[U,V]]] (15)\quad H(U, V) \equiv U + V + \frac{1}{2}[U, V] + \frac{1}{12}[U, [U, V]] \\ + \frac{1}{12}[V, [V, U]] - \frac{1}{24}[U, [V, [U, V]]]
modulo ∑n⩾5Ln({U,V})\sum_{n \geqslant 5} L^n(\{U, V\}).

(3) H0,n=Hn,0=0H_{0,n} = H_{n,0} = 0 for every integer n≠1n \neq 1, whence

(16)H(U,0)=H(0,U)=U. (16)\quad H(U, 0) = H(0, U) = U.

On the other hand, as [U,−U]=0[U, -U] = 0,

(17)H(U,−U)=0. (17)\quad H(U, -U) = 0.

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

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Cites

Transcribed from the printed volume. Not translated.