Proposition 51

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

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let $G$ be a Lie group, of finite dimension if $K$ is of characteristic $> 0$ , and let $a_1, \ldots, a_p$ be elements of $L(G)$ , $F$ a complete normable space and $\alpha$ a differential form of degree $p-1$ on $G$ with values in $F$ . If $\alpha$ is left invariant, then

(d\alpha)_e(a_1, \ldots, a_p) = \sum_{i<j} (-1)^{i+j} \alpha_e([a_i, a_j], a_1, \ldots, a_{i-1}, a_{i+1}, \ldots, a_{j-1}, a_{j+1}, \ldots, a_p).

If $\alpha$ is right invariant, then

(d\alpha)_e(a_1, \ldots, a_p) = - \sum_{i < j} (-1)^{i+j} \alpha_e([a_i, a_j], a_1, \ldots, a_{i-1}, a_{i+1}, \ldots, a_{j-1}, a_{j+1}, \ldots, a_p).

Suppose that

\alpha

is left invariant. By Differentiable and Analytic Manifolds, R, 8.5.7, then

(d\alpha)(L_{a_1}, \ldots, L_{a_p}) = \sum_i (-1)^{i-1} L_{a_i} \alpha(L_{a_1}, \ldots, L_{a_{i-1}}, L_{a_{i+1}}, \ldots, L_{a_p}) + \sum_{i < j} (-1)^{i+j} \alpha([L_{a_i}, L_{a_j}], L_{a_1}, \ldots, L_{a_{i-1}}, L_{a_{i+1}}, \ldots, L_{a_{j-1}}, L_{a_{j+1}}, \ldots, L_{a_p}).

But the functions

\alpha(L_{a_1}, \ldots, L_{a_{i-1}}, L_{a_{i+1}}, \ldots, L_{a_p})

on G are left invariant and therefore constant. Hence

L_{a_i} \alpha(L_{a_1}, \ldots, L_{a_{i-1}}, L_{a_{i+1}}, \ldots, L_{a_p}) = 0.

Moreover,

[L_{a_i}, L_{a_j}] = L_{[a_i, a_j]}

(Proposition 23), whence the first formula of Proposition 51. The second can be established analogously, this time using the relation

[R_{a_i}, R_{a_j}] = -R_{[a_i, a_j]}

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

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Cites

Proposition 23