Library / Lie Groups and Lie Algebras / Chapter III / § 3
Proposition 45
Tag 0ABI, permanent. Cite it as [Bourbaki, Tag 0ABI].
Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra
Suppose that G is finite-dimensional and that K is of characteristic 0. Let s be an integer ⩾ 0 \geqslant 0 ⩾ 0 . Then the mapping π : g ↦ A d U s ( G ) ( g ) \pi : g \mapsto \mathrm{Ad}_{U_s(G)}(g) π : g ↦ Ad U s ( G ) ( g ) is an analytic linear representation of G on U s ( G ) U_s(G) U s ( G ) and L ( π ) a = a d U s ( G ) a L(\pi)a = \mathrm{ad}_{U_s(G)}a L ( π ) a = ad U s ( G ) a for all a ∈ L ( G ) a \in L(G) a ∈ L ( G ) .
The linear representation π \pi π is a quotient of ⨁ r = 0 s T r ( A d ) \bigoplus_{r=0}^s \mathrm{Tr}(\mathrm{Ad}) ⨁ r = 0 s Tr ( Ad ) and is hence analytic. For a ∈ L ( G ) a \in L(G) a ∈ L ( G ) and x 1 , x 2 , … , x s x_1, x_2, \ldots, x_s x 1 , x 2 , … , x s in L ( G ) L(G) L ( G ) ,
( L ( π ) a ) ( x 1 x 2 … x s ) = ∑ i = 1 s x 1 … ( L ( A d ) a . x i ) … x s (Proposition 41) = ∑ i = 1 s x 1 … ( [ a , x i ] ) … x s (Proposition 44) = ( A d U s ( G ) a ) ( x 1 x 2 … x s ) .
\begin{align*}
(L(\pi)a)(x_1x_2\ldots x_s) &= \sum_{i=1}^s x_1 \ldots (L(\mathrm{Ad})a.x_i) \ldots x_s & \text{(Proposition 41)} \\
&= \sum_{i=1}^s x_1 \ldots ([a, x_i]) \ldots x_s & \text{(Proposition 44)} \\
&= (\mathrm{Ad}_{U_s(G)}a)(x_1x_2\ldots x_s).
\end{align*}
( L ( π ) a ) ( x 1 x 2 … x s ) = i = 1 ∑ s x 1 … ( L ( Ad ) a . x i ) … x s = i = 1 ∑ s x 1 … ([ a , x i ]) … x s = ( Ad U s ( G ) a ) ( x 1 x 2 … x s ) . (Proposition 41) (Proposition 44)
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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