Bourbaki

Exercise 2, § 6

Lie Groups and Lie Algebras, II, 201

Show that H(U,V)=∑n=0∞1n!(ad U)n(H(V,U))H(U, V) = \sum_{n=0}^\infty \frac{1}{n!} (\mathrm{ad}\, U)^n(H(V, U)). (Apply Exercise 1, noting that H(U,H(H(V,U),−U))=H(U,V)H(U, H(H(V, U), -U)) = H(U, V).)

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.