Proposition 5

Tag 0A6E, permanent. Cite it as [Bourbaki, Tag 0A6E].

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

Let $K_{r,s}$ be the component of $H(U + V, -U)$ of multidegree $(r, s)$ . Then

K_{n,1}(U, V) = \frac{1}{(n+1)!} (\operatorname{ad} U)^n(V) \quad \text{for } n \geq 0.

We write $K(U, V) = H(U + V, -U)$ , $K_1(U, V) = \sum_{n \geq 0} K_{n,1}(U, V)$ . We denote by L (resp. R) left (resp. right) multiplication by $U$ on $\hat{A}(\{U, V\})$ .

We can write

e^{U}V e^{-U} = \sum_{p, q} \frac{U^p}{p!} V \frac{(-U)^q}{q!} = \sum_{n \geq 0} \frac{1}{n!} \left( \sum_{p+q=n} \frac{n!}{p!q!} (L^p(-R)^q) \cdot V \right) = \sum_{n \geq 0} \frac{1}{n!} (L - R)^n V

and therefore

e^{U}V e^{-U} = \sum_{n \geq 0} \frac{1}{n!} (\operatorname{ad} U)^n V.

We now calculate modulo the ideal

\sum_{m \geq 0} \sum_{n \geq 2} A^{m,n}(\{U, V\})

A(\{U, V\})

. For all

n \geq 1

(U + V)^n \equiv U^n + \sum_{i=1}^{n-1} U^i V U^{n-1-i}

whence

(\operatorname{ad} U)(U + V)^n \equiv ((L - R) \sum_{i=1}^{n-1} L^i R^{n-i}) \cdot V \\ \equiv (L^n - R^n) \cdot V \\ \equiv U^n V - V U^n.

Therefore

(\operatorname{ad} U) \cdot e^{U+V} \equiv e^{U}V - V e^{U}

summing over

n

On the other hand, $K_1(U, V) \equiv K(U, V)$ and $e^{K_1(U, V)} \equiv 1 + K_1(U, V)$ and hence

K_1 \equiv e^K - 1 \equiv e^{U+V}e^{-U} - 1

by Proposition 3. We deduce that

\begin{align*} (\mathrm{ad}\ U)K_1 &\equiv Ue^{U+V}e^{-U} - e^{U+V}e^{-U}U \\ &\equiv (e^{UV} - Ve^U)e^{-U} \quad \text{(by (23))} \\ &\equiv e^{UV}e^{-U} - V \\ &\equiv \sum_{n \geq 1} \frac{1}{n!} (\mathrm{ad}\ U)^n V \quad \text{(by (22))} \\ &\equiv (\mathrm{ad}\ U)\left( \sum_{n \geq 0} \frac{(\mathrm{ad}\ U)^n}{(n+1)!} V \right). \end{align*}

It then suffices to apply the Remark of § 2, no. 11.

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

Proposition 5

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