Library / Lie Groups and Lie Algebras / Chapter II / § 7
Lemma 1
Tag 0A6L, permanent. Cite it as [Bourbaki, Tag 0A6L].
Lie Groups and Lie Algebras, chapter II, § 7. Convergence of the Hausdorff series (real or complex case)
∥ H ~ r , s ∥ ≤ M r + s − 1 η r , s .
\| \tilde{H}_{r,s} \| \leq M^{r+s-1} \eta_{r,s}.
∥ H ~ r , s ∥ ≤ M r + s − 1 η r , s . Let
r i , s i r_i, s_i r i , s i be in
N \mathbf{N} N for
1 ≤ i ≤ m 1 \leq i \leq m 1 ≤ i ≤ m , with
s m = 1 s_m = 1 s m = 1 ; we write
r = ∑ i r i , s = ∑ i s i r = \sum_i r_i, s = \sum_i s_i r = ∑ i r i , s = ∑ i s i and consider the following element of
L ( { U , V } ) L(\{U, V\}) L ({ U , V }) :
Z = ( ( ∑ i = 1 m − 1 ( a d U ) r i ( a d V ) s i ) ( a d U ) r m ) ( V ) .
Z = \left( \left( \sum_{i=1}^{m-1} (\mathrm{ad}\ U)^{r_i} (\mathrm{ad}\ V)^{s_i} \right) (\mathrm{ad}\ U)^{r_m} \right)(V).
Z = ( ( i = 1 ∑ m − 1 ( ad U ) r i ( ad V ) s i ) ( ad U ) r m ) ( V ) . Then
Z ~ = f ∘ p \tilde{Z} = f \circ p Z ~ = f ∘ p , where
f f f is the following
( r + s ) (r+s) ( r + s ) -linear mapping of
g r + s g^{r+s} g r + s into
g g g :
( x 1 , … , x r , y 1 , … , y s ) ↦ ( a d ( x 1 ) ∘ ⋯ ∘ a d ( x r ) ∘ a d ( y 1 ) ∘ ⋯ ∘ a d ( y s ) ) ( y s )
(x_1, \ldots, x_r, y_1, \ldots, y_s) \mapsto (\mathrm{ad}(x_1) \circ \cdots \circ \mathrm{ad}(x_r) \circ \mathrm{ad}(y_1) \circ \cdots \circ \mathrm{ad}(y_s))(y_s)
( x 1 , … , x r , y 1 , … , y s ) ↦ ( ad ( x 1 ) ∘ ⋯ ∘ ad ( x r ) ∘ ad ( y 1 ) ∘ ⋯ ∘ ad ( y s )) ( y s ) and where
p p p is the following mapping of
g 2 g^2 g 2 into
g r + s g^{r+s} g r + s :
( x , y ) ↦ ( x , … , x ⏟ r , y , … , y ⏟ s ) ;
(x, y) \mapsto (\underbrace{x, \ldots, x}_{r}, \underbrace{y, \ldots, y}_{s});
( x , y ) ↦ ( r x , … , x , s y , … , y ) ; hence
∥ Z ~ ∥ ≤ ∥ f ∥ ≤ M r + s − 1 \| \tilde{Z} \| \leq \| f \| \leq M^{r+s-1} ∥ Z ~ ∥ ≤ ∥ f ∥ ≤ M r + s − 1 (Differentiable and Analytic Manifolds, R, Appendix). Applying these inequalities to the various terms on the right hand side of formula (9) of § 6, no. 4, we obtain:
∥ ( H r , s ′ ) ∼ ∥ ≤ M r + s − 1 r + s ∑ m ≥ 1 1 m ∑ r 1 + ⋯ + r m = r s 1 + ⋯ + s m = s r 1 + s 1 ≥ 1 , … , r m − 1 + s m − 1 ≥ 1 1 r 1 ! … r m ! s 1 ! … s m ! .
\|(H'_{r,s})\sim\|
\leq \frac{M^{r+s-1}}{r+s} \sum_{m \geq 1} \frac{1}{m} \sum_{\substack{r_1 + \cdots + r_m = r \\ s_1 + \cdots + s_m = s \\ r_1 + s_1 \geq 1, \ldots, r_{m-1} + s_{m-1} \geq 1}} \frac{1}{r_1! \ldots r_m! s_1! \ldots s_m!}.
∥ ( H r , s ′ ) ∼ ∥ ≤ r + s M r + s − 1 m ≥ 1 ∑ m 1 r 1 + ⋯ + r m = r s 1 + ⋯ + s m = s r 1 + s 1 ≥ 1 , … , r m − 1 + s m − 1 ≥ 1 ∑ r 1 ! … r m ! s 1 ! … s m ! 1 . A similar argument gives
∥ ( H r , s ′ ′ ) ∼ ∥ ≤ M r + s − 1 r + s ∑ m ≥ 1 1 m ∑ r 1 + ⋯ + r m − 1 = r − 1 s 1 + ⋯ + s m − 1 = s r 1 + s 1 ≥ 1 , … , r m − 1 + s m − 1 ≥ 1 1 r 1 ! … r m − 1 ! s 1 ! … s m − 1 !
\|(H''_{r,s})\sim\|
\leq \frac{M^{r+s-1}}{r+s} \sum_{m \geq 1} \frac{1}{m} \sum_{\substack{r_1 + \cdots + r_{m-1} = r-1 \\ s_1 + \cdots + s_{m-1} = s \\ r_1 + s_1 \geq 1, \ldots, r_{m-1} + s_{m-1} \geq 1}} \frac{1}{r_1! \ldots r_{m-1}! s_1! \ldots s_{m-1}!}
∥ ( H r , s ′′ ) ∼ ∥ ≤ r + s M r + s − 1 m ≥ 1 ∑ m 1 r 1 + ⋯ + r m − 1 = r − 1 s 1 + ⋯ + s m − 1 = s r 1 + s 1 ≥ 1 , … , r m − 1 + s m − 1 ≥ 1 ∑ r 1 ! … r m − 1 ! s 1 ! … s m − 1 ! 1 whence, by (8)
∥ H ~ r , s ∥ < η r , s M r + s − 1 r + s ≤ η r , s M r + s − 1 ,
\|\tilde{H}_{r,s}\| < \eta_{r,s} \frac{M^{r+s-1}}{r+s} \leq \eta_{r,s} M^{r+s-1},
∥ H ~ r , s ∥ < η r , s r + s M r + s − 1 ≤ η r , s M r + s − 1 , which proves the lemma.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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