Bourbaki

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. Then the mapping π:g↦AdUs(G)(g)\pi : g \mapsto \mathrm{Ad}_{U_s(G)}(g) is an analytic linear representation of G on Us(G)U_s(G) and L(π)a=adUs(G)aL(\pi)a = \mathrm{ad}_{U_s(G)}a for all a∈L(G)a \in L(G).

The linear representation π\pi is a quotient of ⨁r=0sTr(Ad)\bigoplus_{r=0}^s \mathrm{Tr}(\mathrm{Ad}) and is hence analytic. For a∈L(G)a \in L(G) and x1,x2,…,xsx_1, x_2, \ldots, x_s in L(G)L(G),

(L(π)a)(x1x2…xs)=∑i=1sx1…(L(Ad)a.xi)…xs(Proposition 41)=∑i=1sx1…([a,xi])…xs(Proposition 44)=(AdUs(G)a)(x1x2…xs). \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*}

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

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