Corollary 1

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

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

Let E be a complete normable space over $\mathbf{C}$ and F the space of continuous n-linear mappings of $E^n$ into E. For all $v \in \mathscr{L}(E)$ , let $\sigma(v)$ be the element of $\mathscr{L}(F)$ defined by

(\sigma(v)f)(x_1, \ldots, x_n) = v(f(x_1, \ldots, x_n)) - \sum_{i=1}^n f(x_1, \ldots, vx_i, \ldots, x_n).

For all $g \in \mathbf{GL}(E)$ , let $\rho(g)$ be the element of $\mathbf{GL}(F)$ defined by

(\rho(g)f)(x_1, \ldots, x_n) = g(f(g^{-1}x_1, \ldots, g^{-1}x_n)).

Let $u \in \mathscr{L}(E)$ be such that every $z \in \mathrm{Sp}\,u$ satisfies $|\mathscr{J}(z)| < \frac{2\pi}{n+1}$ . Then, for all $f \in F$ , the conditions $\sigma(u)f = 0$ and $\rho(\exp u)f = f$ are equivalent.

$\mathbf{L}(\rho) = \sigma$ (§ 3, no. 11, Corollary 1 to Proposition 41) and hence

\rho(\exp u) = \exp \sigma(u)

(no. 4, Corollary 3 to Proposition 10). By Proposition 18 it then suffices to prove that $\mathrm{Sp}\,\sigma(u)$ does not meet $2i\pi(\mathbf{Z} - \{0\})$ . But this follows from the following lemma:

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

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Corollary 2, § 6

Corollary 1

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