Higher order Approximation of Lie groups - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-26T02:50:29Z http://mathoverflow.net/feeds/question/73333 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73333/higher-order-approximation-of-lie-groups Higher order Approximation of Lie groups Mirco 2011-08-21T15:00:17Z 2011-08-21T19:37:30Z <p>Maybe the following is trivial or folklore, but I can't find any concrete proof of the theorem, that higher order derivatives of Lie groups don't give any new information above what is coded in its Lie algebra.</p> <p>Can someone explain why this is true?</p> http://mathoverflow.net/questions/73333/higher-order-approximation-of-lie-groups/73353#73353 Answer by Ben McKay for Higher order Approximation of Lie groups Ben McKay 2011-08-21T19:37:30Z 2011-08-21T19:37:30Z <p>The reason is the Campbell-Baker-Hausdorff equation, which proves that all higher derivatives of the multiplication map are expressed in exponential coordinates explicitly in terms of iterated Lie brackets. Once you know the Lie bracket operation, you can calculate the Taylor series expansion of the multiplication operation explicitly, order by order. The Taylor series has positive radius of convergence, so the multiplication is given, near the identity element, by this series expansion. (I shouldn't be giving this answer, as it is in all basic introductions to Lie groups, not a research level question.) The explicit formula is in Serre's book Lie Algebras and Lie Groups, chapter 4, p. 28.</p>