Proposition 3

Tag 0AE4, permanent. Cite it as [Bourbaki, Tag 0AE4].

Lie Groups and Lie Algebras, chapter III, § 5. Formal calculations in Lie groups

(27)

L = \sum_{p=1}^{\infty} \frac{(-1)^{p-1}}{p} \psi_p

(28)

E = \sum_{p=1}^{\infty} \frac{1}{p!} \psi_{p,p}

(Recall that

\psi_{p,p}

is the homogeneous component of

\psi_p

of degree

p

or, by (24),

E(tx) = (E(x))^{[t]}

The two sides are formal power series in $t$ and $x$ . Equating the terms of first degree in $t$ , we obtain

x = \sum_{m \geq 0} \phi_{1,m}(E(x)).

Now the inversion of a system of formal power series, when it is possible, is possible in only one way (Algebra, Chapter IV, § 6, Corollary to Proposition 8). Then

L(x) = \sum_{m \geq 0} \phi_{1,m}(x) \quad \text{by (29)}

= \sum_{p,m} \frac{(-1)^p}{p} \psi_{p,m}(x) \quad \text{by (25)}

= \sum_p \frac{(-1)^p}{p} \psi_p(x),

whence (i). Similarly, for

t \neq 0

L(tx) = tL((tx)^{[t^{-1}]})

= tL\left( \sum_{m \geq 0} \sum_{1 \leq r \leq m} t^{m-r} \phi_{r,m}(x) \right) \quad \text{by (24)}.

Equating the terms of first degree in t, we obtain

x = L \left( \sum_{m \geq 0} \phi_{m,m}(x) \right)

whence

E(x) = \sum_{m \geq 0} \phi_{m,m}(x)

= \sum_{m \geq 0} \frac{1}{m!} \psi_{m,m}(x) \quad \text{by (26).}

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

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