Exercise 1, § 6 Lie Groups and Lie Algebras, II, 201 Show that exp(U)⋅exp(V)⋅exp(−U)=exp(∑n=0∞1n!(ad U)n(V))\exp(U) \cdot \exp(V) \cdot \exp(-U) = \exp\left( \sum_{n=0}^\infty \frac{1}{n!} (\mathrm{ad}\, U)^n(V) \right)exp(U)⋅exp(V)⋅exp(−U)=exp(∑n=0∞n!1(adU)n(V)). Deduce that H(U,H(V,−U))=∑n=0∞1n!(ad U)n(V)=(exp(ad U))(V). H(U, H(V, -U)) = \sum_{n=0}^\infty \frac{1}{n!} (\mathrm{ad}\, U)^n(V) = (\exp(\mathrm{ad}\, U))(V). H(U,H(V,−U))=n=0∑∞n!1(adU)n(V)=(exp(adU))(V). Solution No solution has been written yet.